From e95bf23159aef7622acd76336a7d315f38b9c115 Mon Sep 17 00:00:00 2001 From: Morten Hjorth-Jensen Date: Wed, 19 Oct 2022 07:28:21 +0200 Subject: [PATCH] update with refs --- doc/pub/week42/html/._week42-bs000.html | 187 +++---- doc/pub/week42/html/._week42-bs001.html | 187 +++---- doc/pub/week42/html/._week42-bs002.html | 187 +++---- doc/pub/week42/html/._week42-bs003.html | 224 ++++---- doc/pub/week42/html/._week42-bs004.html | 233 ++++---- doc/pub/week42/html/._week42-bs005.html | 217 ++++---- doc/pub/week42/html/._week42-bs006.html | 224 ++++---- doc/pub/week42/html/._week42-bs007.html | 215 ++++---- doc/pub/week42/html/._week42-bs008.html | 214 ++++---- doc/pub/week42/html/._week42-bs009.html | 208 +++---- doc/pub/week42/html/._week42-bs010.html | 200 +++---- doc/pub/week42/html/._week42-bs011.html | 207 ++++--- doc/pub/week42/html/._week42-bs012.html | 212 ++++---- doc/pub/week42/html/._week42-bs013.html | 212 ++++---- doc/pub/week42/html/._week42-bs014.html | 227 ++++---- 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b/doc/pub/week42/html/._week42-bs000.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -442,7 +447,7 @@ MathJax.Hub.Config({
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    • diff --git a/doc/pub/week42/html/._week42-bs001.html b/doc/pub/week42/html/._week42-bs001.html index cb2813b44..573c075c2 100644 --- a/doc/pub/week42/html/._week42-bs001.html +++ b/doc/pub/week42/html/._week42-bs001.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -459,7 +464,7 @@ MathJax.Hub.Config({
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    • diff --git a/doc/pub/week42/html/._week42-bs002.html b/doc/pub/week42/html/._week42-bs002.html index 6b0deae8b..e8f31ec12 100644 --- a/doc/pub/week42/html/._week42-bs002.html +++ b/doc/pub/week42/html/._week42-bs002.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -422,7 +427,7 @@ we will also study the usage of 11
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    • diff --git a/doc/pub/week42/html/._week42-bs003.html b/doc/pub/week42/html/._week42-bs003.html index 88668420e..c4b7d9426 100644 --- a/doc/pub/week42/html/._week42-bs003.html +++ b/doc/pub/week42/html/._week42-bs003.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,37 +404,14 @@ MathJax.Hub.Config({

     

     

     

    -

    Solving ODEs with Deep Learning

    - -
    -
    - -

    The Universal Approximation Theorem states that a neural network can -approximate any function at a single hidden layer along with one input -and output layer to any given precision. -

    -
    -
    - - -
    -
    - -

    An Introduction to Neural Network Methods for Differential Equations, by Yadav and Kumar.

    -
    -
    - - -
    -
    - -

    The lectures on differential equations were developed by Kristine Baluka Hein, now PhD student at IFI. -A great thanks to Kristine. -

    -
    -
    - +

    Back propagation and automatic differentiation

    +

    For more details on the back propagation algorithm and automatic differentiation see

    +
      +
    1. https://www.jmlr.org/papers/volume18/17-468/17-468.pdf
      2. +
    2. https://deepimaging.github.io/lectures/lecture_11_Backpropagation.pdf
      3. +
    3. Slides 12-44 at URL":http://cs231n.stanford.edu/slides/2017/cs231n_2017_lecture4.pdf"
      4. +

    diff --git a/doc/pub/week42/html/._week42-bs004.html b/doc/pub/week42/html/._week42-bs004.html index 1cd65ec3c..82ca6e260 100644 --- a/doc/pub/week42/html/._week42-bs004.html +++ b/doc/pub/week42/html/._week42-bs004.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,26 +404,36 @@ MathJax.Hub.Config({

     

     

     

    -

    Ordinary Differential Equations

    +

    Solving ODEs with Deep Learning

    -

    An ordinary differential equation (ODE) is an equation involving functions having one variable.

    - -

    In general, an ordinary differential equation looks like

    - -$$ -\begin{equation} \tag{1} -f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0 -\end{equation} -$$ - -

    where \( g(x) \) is the function to find, and \( g^{(n)}(x) \) is the \( n \)-th derivative of \( g(x) \).

    - -

    The \( f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right) \) is just a way to write that there is an expression involving \( x \) and \( g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x) \) on the left side of the equality sign in (1). -The highest order of derivative, that is the value of \( n \), determines to the order of the equation. -The equation is referred to as a \( n \)-th order ODE. -Along with (1), some additional conditions of the function \( g(x) \) are typically given -for the solution to be unique. +

    +
    + +

    The Universal Approximation Theorem states that a neural network can +approximate any function at a single hidden layer along with one input +and output layer to any given precision.

    +
    +
    + + +
    +
    + +

    An Introduction to Neural Network Methods for Differential Equations, by Yadav and Kumar.

    +
    +
    + + +
    +
    + +

    The lectures on differential equations were developed by Kristine Baluka Hein, now PhD student at IFI. +A great thanks to Kristine. +

    +
    +
    +

    @@ -439,7 +454,7 @@ for the solution to be unique.

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    • diff --git a/doc/pub/week42/html/._week42-bs005.html b/doc/pub/week42/html/._week42-bs005.html index f7bebd540..195e9d0c5 100644 --- a/doc/pub/week42/html/._week42-bs005.html +++ b/doc/pub/week42/html/._week42-bs005.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,31 +404,27 @@ MathJax.Hub.Config({

     

     

     

    -

    The trial solution

    +

    Ordinary Differential Equations

    -

    Let the trial solution \( g_t(x) \) be

    +

    An ordinary differential equation (ODE) is an equation involving functions having one variable.

    + +

    In general, an ordinary differential equation looks like

    $$ -\begin{equation} - g_t(x) = h_1(x) + h_2(x,N(x,P)) -\tag{2} +\begin{equation} \tag{1} +f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0 \end{equation} $$ -

    where \( h_1(x) \) is a function that makes \( g_t(x) \) satisfy a given set -of conditions, \( N(x,P) \) a neural network with weights and biases -described by \( P \) and \( h_2(x, N(x,P)) \) some expression involving the -neural network. The role of the function \( h_2(x, N(x,P)) \), is to -ensure that the output from \( N(x,P) \) is zero when \( g_t(x) \) is -evaluated at the values of \( x \) where the given conditions must be -satisfied. The function \( h_1(x) \) should alone make \( g_t(x) \) satisfy -the conditions. +

    where \( g(x) \) is the function to find, and \( g^{(n)}(x) \) is the \( n \)-th derivative of \( g(x) \).

    + +

    The \( f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right) \) is just a way to write that there is an expression involving \( x \) and \( g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x) \) on the left side of the equality sign in (1). +The highest order of derivative, that is the value of \( n \), determines to the order of the equation. +The equation is referred to as a \( n \)-th order ODE. +Along with (1), some additional conditions of the function \( g(x) \) are typically given +for the solution to be unique.

    -

    But what about the network \( N(x,P) \)?

    - -

    As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation.

    -

    diff --git a/doc/pub/week42/html/._week42-bs006.html b/doc/pub/week42/html/._week42-bs006.html index 3a7f239a3..66ac9f0c9 100644 --- a/doc/pub/week42/html/._week42-bs006.html +++ b/doc/pub/week42/html/._week42-bs006.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,34 +404,31 @@ MathJax.Hub.Config({

     

     

     

    -

    Minimization process

    +

    The trial solution

    -

    For the minimization to be defined, we need to have a cost function at hand to minimize.

    - -

    It is given that \( f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) \) should be equal to zero in (1). -We can choose to consider the mean squared error as the cost function for an input \( x \). -Since we are looking at one input, the cost function is just \( f \) squared. -The cost function \( c\left(x, P \right) \) can therefore be expressed as -

    +

    Let the trial solution \( g_t(x) \) be

    $$ -C\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2 -$$ - -

    If \( N \) inputs are given as a vector \( \boldsymbol{x} \) with elements \( x_i \) for \( i = 1,\dots,N \), -the cost function becomes -

    - -$$ -\begin{equation} \tag{3} - C\left(\boldsymbol{x}, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2 +\begin{equation} + g_t(x) = h_1(x) + h_2(x,N(x,P)) +\tag{2} \end{equation} $$ -

    The neural net should then find the parameters \( P \) that minimizes the cost function in -(3) for a set of \( N \) training samples \( x_i \). +

    where \( h_1(x) \) is a function that makes \( g_t(x) \) satisfy a given set +of conditions, \( N(x,P) \) a neural network with weights and biases +described by \( P \) and \( h_2(x, N(x,P)) \) some expression involving the +neural network. The role of the function \( h_2(x, N(x,P)) \), is to +ensure that the output from \( N(x,P) \) is zero when \( g_t(x) \) is +evaluated at the values of \( x \) where the given conditions must be +satisfied. The function \( h_1(x) \) should alone make \( g_t(x) \) satisfy +the conditions.

    +

    But what about the network \( N(x,P) \)?

    + +

    As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation.

    +

    diff --git a/doc/pub/week42/html/._week42-bs007.html b/doc/pub/week42/html/._week42-bs007.html index fd71d6bb6..ae4837d21 100644 --- a/doc/pub/week42/html/._week42-bs007.html +++ b/doc/pub/week42/html/._week42-bs007.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,14 +404,32 @@ MathJax.Hub.Config({

     

     

     

    -

    Minimizing the cost function using gradient descent and automatic differentiation

    +

    Minimization process

    -

    To perform the minimization using gradient descent, the gradient of \( C\left(\boldsymbol{x}, P\right) \) is needed. -It might happen so that finding an analytical expression of the gradient of \( C(\boldsymbol{x}, P) \) from (3) gets too messy, depending on which cost function one desires to use. +

    For the minimization to be defined, we need to have a cost function at hand to minimize.

    + +

    It is given that \( f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) \) should be equal to zero in (1). +We can choose to consider the mean squared error as the cost function for an input \( x \). +Since we are looking at one input, the cost function is just \( f \) squared. +The cost function \( c\left(x, P \right) \) can therefore be expressed as

    -

    Luckily, there exists libraries that makes the job for us through automatic differentiation. -Automatic differentiation is a method of finding the derivatives numerically with very high precision. +$$ +C\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2 +$$ + +

    If \( N \) inputs are given as a vector \( \boldsymbol{x} \) with elements \( x_i \) for \( i = 1,\dots,N \), +the cost function becomes +

    + +$$ +\begin{equation} \tag{3} + C\left(\boldsymbol{x}, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2 +\end{equation} +$$ + +

    The neural net should then find the parameters \( P \) that minimizes the cost function in +(3) for a set of \( N \) training samples \( x_i \).

    @@ -431,7 +454,7 @@ Automatic differentiation is a method of finding the derivatives numerically wit

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    -

    Example: Exponential decay

    +

    Minimizing the cost function using gradient descent and automatic differentiation

    -

    An exponential decay of a quantity \( g(x) \) is described by the equation

    +

    To perform the minimization using gradient descent, the gradient of \( C\left(\boldsymbol{x}, P\right) \) is needed. +It might happen so that finding an analytical expression of the gradient of \( C(\boldsymbol{x}, P) \) from (3) gets too messy, depending on which cost function one desires to use. +

    -$$ -\begin{equation} \tag{4} - g'(x) = -\gamma g(x) -\end{equation} -$$ - -

    with \( g(0) = g_0 \) for some chosen initial value \( g_0 \).

    - -

    The analytical solution of (4) is

    - -$$ -\begin{equation} - g(x) = g_0 \exp\left(-\gamma x\right) -\tag{5} -\end{equation} -$$ - -

    Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of (4).

    +

    Luckily, there exists libraries that makes the job for us through automatic differentiation. +Automatic differentiation is a method of finding the derivatives numerically with very high precision. +

    @@ -445,7 +437,7 @@ $$

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    -

    The function to solve for

    +

    Example: Exponential decay

    -

    The program will use a neural network to solve

    +

    An exponential decay of a quantity \( g(x) \) is described by the equation

    $$ -\begin{equation} \tag{6} -g'(x) = -\gamma g(x) +\begin{equation} \tag{4} + g'(x) = -\gamma g(x) \end{equation} $$ -

    where \( g(0) = g_0 \) with \( \gamma \) and \( g_0 \) being some chosen values.

    +

    with \( g(0) = g_0 \) for some chosen initial value \( g_0 \).

    -

    In this example, \( \gamma = 2 \) and \( g_0 = 10 \).

    +

    The analytical solution of (4) is

    + +$$ +\begin{equation} + g(x) = g_0 \exp\left(-\gamma x\right) +\tag{5} +\end{equation} +$$ + +

    Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of (4).

    @@ -437,7 +451,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs010.html b/doc/pub/week42/html/._week42-bs010.html index 9111ccde3..594c125e8 100644 --- a/doc/pub/week42/html/._week42-bs010.html +++ b/doc/pub/week42/html/._week42-bs010.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,14 +404,19 @@ MathJax.Hub.Config({

     

     

     

    -

    The trial solution

    -

    To begin with, a trial solution \( g_t(t) \) must be chosen. A general trial solution for ordinary differential equations could be

    +

    The function to solve for

    + +

    The program will use a neural network to solve

    $$ -g_t(x, P) = h_1(x) + h_2(x, N(x, P)) +\begin{equation} \tag{6} +g'(x) = -\gamma g(x) +\end{equation} $$ -

    with \( h_1(x) \) ensuring that \( g_t(x) \) satisfies some conditions and \( h_2(x,N(x, P)) \) an expression involving \( x \) and the output from the neural network \( N(x,P) \) with \( P \) being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer.

    +

    where \( g(0) = g_0 \) with \( \gamma \) and \( g_0 \) being some chosen values.

    + +

    In this example, \( \gamma = 2 \) and \( g_0 = 10 \).

    @@ -433,7 +443,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs011.html b/doc/pub/week42/html/._week42-bs011.html index 9c19c1892..8fb4a6ef4 100644 --- a/doc/pub/week42/html/._week42-bs011.html +++ b/doc/pub/week42/html/._week42-bs011.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,26 +404,14 @@ MathJax.Hub.Config({

     

     

     

    -

    Setup of Network

    - -

    In this network, there are no weights and bias at the input layer, so \( P = \{ P_{\text{hidden}}, P_{\text{output}} \} \). -If there are \( N_{\text{hidden} } \) neurons in the hidden layer, then \( P_{\text{hidden}} \) is a \( N_{\text{hidden} } \times (1 + N_{\text{input}}) \) matrix, given that there are \( N_{\text{input}} \) neurons in the input layer. -

    - -

    The first column in \( P_{\text{hidden} } \) represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer. -If there are \( N_{\text{output} } \) neurons in the output layer, then \( P_{\text{output}} \) is a \( N_{\text{output} } \times (1 + N_{\text{hidden} }) \) matrix. -

    - -

    Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron.

    - -

    It is given that \( g(0) = g_0 \). The trial solution must fulfill this condition to be a proper solution of (6). A possible way to ensure that \( g_t(0, P) = g_0 \), is to let \( F(N(x,P)) = x \cdot N(x,P) \) and \( A(x) = g_0 \). This gives the following trial solution:

    +

    The trial solution

    +

    To begin with, a trial solution \( g_t(t) \) must be chosen. A general trial solution for ordinary differential equations could be

    $$ -\begin{equation} \tag{7} -g_t(x, P) = g_0 + x \cdot N(x, P) -\end{equation} +g_t(x, P) = h_1(x) + h_2(x, N(x, P)) $$ +

    with \( h_1(x) \) ensuring that \( g_t(x) \) satisfies some conditions and \( h_2(x,N(x, P)) \) an expression involving \( x \) and the output from the neural network \( N(x,P) \) with \( P \) being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer.

    @@ -445,7 +438,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs012.html b/doc/pub/week42/html/._week42-bs012.html index e34835ed8..15881d0f5 100644 --- a/doc/pub/week42/html/._week42-bs012.html +++ b/doc/pub/week42/html/._week42-bs012.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,31 +404,26 @@ MathJax.Hub.Config({

     

     

     

    -

    Reformulating the problem

    +

    Setup of Network

    -

    We wish that our neural network manages to minimize a given cost function.

    - -

    A reformulation of out equation, (6), must therefore be done, -such that it describes the problem a neural network can solve for. +

    In this network, there are no weights and bias at the input layer, so \( P = \{ P_{\text{hidden}}, P_{\text{output}} \} \). +If there are \( N_{\text{hidden} } \) neurons in the hidden layer, then \( P_{\text{hidden}} \) is a \( N_{\text{hidden} } \times (1 + N_{\text{input}}) \) matrix, given that there are \( N_{\text{input}} \) neurons in the input layer.

    -

    The neural network must find the set of weights and biases \( P \) such that the trial solution in (7) satisfies (6).

    +

    The first column in \( P_{\text{hidden} } \) represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer. +If there are \( N_{\text{output} } \) neurons in the output layer, then \( P_{\text{output}} \) is a \( N_{\text{output} } \times (1 + N_{\text{hidden} }) \) matrix. +

    -

    The trial solution

    +

    Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron.

    + +

    It is given that \( g(0) = g_0 \). The trial solution must fulfill this condition to be a proper solution of (6). A possible way to ensure that \( g_t(0, P) = g_0 \), is to let \( F(N(x,P)) = x \cdot N(x,P) \) and \( A(x) = g_0 \). This gives the following trial solution:

    $$ +\begin{equation} \tag{7} g_t(x, P) = g_0 + x \cdot N(x, P) -$$ - -

    has been chosen such that it already solves the condition \( g(0) = g_0 \). What remains, is to find \( P \) such that

    - -$$ -\begin{equation} \tag{8} -g_t'(x, P) = - \gamma g_t(x, P) \end{equation} $$ -

    is fulfilled as best as possible.

    @@ -450,7 +450,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs013.html b/doc/pub/week42/html/._week42-bs013.html index 09b384011..4069f29c7 100644 --- a/doc/pub/week42/html/._week42-bs013.html +++ b/doc/pub/week42/html/._week42-bs013.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,28 +404,31 @@ MathJax.Hub.Config({

     

     

     

    -

    More technicalities

    +

    Reformulating the problem

    -

    The left hand side and right hand side of (8) must be computed separately, and then the neural network must choose weights and biases, contained in \( P \), such that the sides are equal as best as possible. -This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. -In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to \( P \) of the neural network. +

    We wish that our neural network manages to minimize a given cost function.

    + +

    A reformulation of out equation, (6), must therefore be done, +such that it describes the problem a neural network can solve for.

    -

    This gives the following cost function our neural network must solve for:

    +

    The neural network must find the set of weights and biases \( P \) such that the trial solution in (7) satisfies (6).

    + +

    The trial solution

    $$ -\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\} +g_t(x, P) = g_0 + x \cdot N(x, P) $$ -

    (the notation \( \min_{P}\{ f(x, P) \} \) means that we desire to find \( P \) that yields the minimum of \( f(x, P) \))

    - -

    or, in terms of weights and biases for the hidden and output layer in our network:

    +

    has been chosen such that it already solves the condition \( g(0) = g_0 \). What remains, is to find \( P \) such that

    $$ -\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\} +\begin{equation} \tag{8} +g_t'(x, P) = - \gamma g_t(x, P) +\end{equation} $$ -

    for an input value \( x \).

    +

    is fulfilled as best as possible.

    @@ -447,7 +455,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs014.html b/doc/pub/week42/html/._week42-bs014.html index 5f7594b54..998c79c68 100644 --- a/doc/pub/week42/html/._week42-bs014.html +++ b/doc/pub/week42/html/._week42-bs014.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,29 +404,29 @@ MathJax.Hub.Config({

     

     

     

    -

    More details

    +

    More technicalities

    -

    If the neural network evaluates \( g_t(x, P) \) at more values for \( x \), say \( N \) values \( x_i \) for \( i = 1, \dots, N \), then the total error to minimize becomes

    - -$$ -\begin{equation} \tag{9} -\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\} -\end{equation} -$$ - -

    Letting \( \boldsymbol{x} \) be a vector with elements \( x_i \) and \( C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \) denote the cost function, the minimization problem that our network must solve, becomes

    - -$$ -\min_{P} C(\boldsymbol{x}, P) -$$ - -

    In terms of \( P_{\text{hidden} } \) and \( P_{\text{output} } \), this could also be expressed as

    - -

    $$ -\min_{P_{\text{hidden} }, \ P_{\text{output} }} C(\boldsymbol{x}, \{P_{\text{hidden} }, P_{\text{output} }\}) -$$ +

    The left hand side and right hand side of (8) must be computed separately, and then the neural network must choose weights and biases, contained in \( P \), such that the sides are equal as best as possible. +This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. +In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to \( P \) of the neural network.

    +

    This gives the following cost function our neural network must solve for:

    + +$$ +\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\} +$$ + +

    (the notation \( \min_{P}\{ f(x, P) \} \) means that we desire to find \( P \) that yields the minimum of \( f(x, P) \))

    + +

    or, in terms of weights and biases for the hidden and output layer in our network:

    + +$$ +\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\} +$$ + +

    for an input value \( x \).

    +

    diff --git a/doc/pub/week42/html/._week42-bs015.html b/doc/pub/week42/html/._week42-bs015.html index 3fe901c33..3e1f1984e 100644 --- a/doc/pub/week42/html/._week42-bs015.html +++ b/doc/pub/week42/html/._week42-bs015.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,13 +404,27 @@ MathJax.Hub.Config({

     

     

     

    -

    A possible implementation of a neural network

    +

    More details

    -

    For simplicity, it is assumed that the input is an array \( \boldsymbol{x} = (x_1, \dots, x_N) \) with \( N \) elements. It is at these points the neural network should find \( P \) such that it fulfills (9).

    +

    If the neural network evaluates \( g_t(x, P) \) at more values for \( x \), say \( N \) values \( x_i \) for \( i = 1, \dots, N \), then the total error to minimize becomes

    -

    First, the neural network must feed forward the inputs. -This means that \( \boldsymbol{x}s \) must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further. -The input layer will consist of \( N_{\text{input} } \) neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be \( N_{\text{hidden} } \). +$$ +\begin{equation} \tag{9} +\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\} +\end{equation} +$$ + +

    Letting \( \boldsymbol{x} \) be a vector with elements \( x_i \) and \( C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \) denote the cost function, the minimization problem that our network must solve, becomes

    + +$$ +\min_{P} C(\boldsymbol{x}, P) +$$ + +

    In terms of \( P_{\text{hidden} } \) and \( P_{\text{output} } \), this could also be expressed as

    + +

    $$ +\min_{P_{\text{hidden} }, \ P_{\text{output} }} C(\boldsymbol{x}, \{P_{\text{hidden} }, P_{\text{output} }\}) +$$

    @@ -433,7 +452,7 @@ The input layer will consist of \( N_{\text{input} } \) neurons, passing its ele

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    • diff --git a/doc/pub/week42/html/._week42-bs016.html b/doc/pub/week42/html/._week42-bs016.html index bc6977901..c31edcb10 100644 --- a/doc/pub/week42/html/._week42-bs016.html +++ b/doc/pub/week42/html/._week42-bs016.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,24 +404,14 @@ MathJax.Hub.Config({

     

     

     

    -

    Technicalities

    +

    A possible implementation of a neural network

    -

    For the \( i \)-th in the hidden layer with weight \( w_i^{\text{hidden} } \) and bias \( b_i^{\text{hidden} } \), the weighting from the \( j \)-th neuron at the input layer is:

    - -$$ -\begin{aligned} -z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\ -&= -\begin{pmatrix} -b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -x_j -\end{pmatrix} -\end{aligned} -$$ +

    For simplicity, it is assumed that the input is an array \( \boldsymbol{x} = (x_1, \dots, x_N) \) with \( N \) elements. It is at these points the neural network should find \( P \) such that it fulfills (9).

    +

    First, the neural network must feed forward the inputs. +This means that \( \boldsymbol{x}s \) must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further. +The input layer will consist of \( N_{\text{input} } \) neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be \( N_{\text{hidden} } \). +

    @@ -443,7 +438,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs017.html b/doc/pub/week42/html/._week42-bs017.html index 466cfc964..5505259fc 100644 --- a/doc/pub/week42/html/._week42-bs017.html +++ b/doc/pub/week42/html/._week42-bs017.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,22 +404,21 @@ MathJax.Hub.Config({

     

     

     

    -

    Final technicalities I

    +

    Technicalities

    -

    The result after weighting the inputs at the \( i \)-th hidden neuron can be written as a vector:

    +

    For the \( i \)-th in the hidden layer with weight \( w_i^{\text{hidden} } \) and bias \( b_i^{\text{hidden} } \), the weighting from the \( j \)-th neuron at the input layer is:

    $$ \begin{aligned} -\boldsymbol{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\ +z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\ &= \begin{pmatrix} - b_i^{\text{hidden}} & w_i^{\text{hidden}} +b_i^{\text{hidden}} & w_i^{\text{hidden}} \end{pmatrix} \begin{pmatrix} -1 & 1 & \dots & 1 \\ -x_1 & x_2 & \dots & x_N -\end{pmatrix} \\ -&= \boldsymbol{p}_{i, \text{hidden}}^T X +1 \\ +x_j +\end{pmatrix} \end{aligned} $$ @@ -444,7 +448,7 @@ $$
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    • diff --git a/doc/pub/week42/html/._week42-bs018.html b/doc/pub/week42/html/._week42-bs018.html index 34be191e7..07d55b9ce 100644 --- a/doc/pub/week42/html/._week42-bs018.html +++ b/doc/pub/week42/html/._week42-bs018.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,35 +404,25 @@ MathJax.Hub.Config({

     

     

     

    -

    Final technicalities II

    +

    Final technicalities I

    -

    The vector \( \boldsymbol{p}_{i, \text{hidden}}^T \) constitutes each row in \( P_{\text{hidden} } \), which contains the weights for the neural network to minimize according to (9).

    - -

    After having found \( \boldsymbol{z}_{i}^{\text{hidden}} \) for every \( i \)-th neuron within the hidden layer, the vector will be sent to an activation function \( a_i(\boldsymbol{z}) \).

    - -

    In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron:

    +

    The result after weighting the inputs at the \( i \)-th hidden neuron can be written as a vector:

    $$ -f(z) = \frac{1}{1 + \exp{(-z)}} +\begin{aligned} +\boldsymbol{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\ +&= +\begin{pmatrix} + b_i^{\text{hidden}} & w_i^{\text{hidden}} +\end{pmatrix} +\begin{pmatrix} +1 & 1 & \dots & 1 \\ +x_1 & x_2 & \dots & x_N +\end{pmatrix} \\ +&= \boldsymbol{p}_{i, \text{hidden}}^T X +\end{aligned} $$ -

    It is possible to use other activations functions for the hidden layer also.

    - -

    The output \( \boldsymbol{x}_i^{\text{hidden}} \) from each \( i \)-th hidden neuron is:

    - -

    $$ -\boldsymbol{x}_i^{\text{hidden} } = f\big( \boldsymbol{z}_{i}^{\text{hidden}} \big) -$$ -

    - -

    The outputs \( \boldsymbol{x}_i^{\text{hidden} } \) are then sent to the output layer.

    - -

    The output layer consists of one neuron in this case, and combines the -output from each of the neurons in the hidden layers. The output layer -combines the results from the hidden layer using some weights \( w_i^{\text{output}} \) -and biases \( b_i^{\text{output}} \). In this case, -it is assumes that the number of neurons in the output layer is one. -

    @@ -454,7 +449,7 @@ it is assumes that the number of neurons in the output layer is one.

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    • diff --git a/doc/pub/week42/html/._week42-bs019.html b/doc/pub/week42/html/._week42-bs019.html index a5d874fd0..bef66727e 100644 --- a/doc/pub/week42/html/._week42-bs019.html +++ b/doc/pub/week42/html/._week42-bs019.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,23 +404,35 @@ MathJax.Hub.Config({

     

     

     

    -

    Final technicalities III

    +

    Final technicalities II

    -

    The procedure of weighting the output neuron \( j \) in the hidden layer to the \( i \)-th neuron in the output layer is similar as for the hidden layer described previously.

    +

    The vector \( \boldsymbol{p}_{i, \text{hidden}}^T \) constitutes each row in \( P_{\text{hidden} } \), which contains the weights for the neural network to minimize according to (9).

    + +

    After having found \( \boldsymbol{z}_{i}^{\text{hidden}} \) for every \( i \)-th neuron within the hidden layer, the vector will be sent to an activation function \( a_i(\boldsymbol{z}) \).

    + +

    In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron:

    $$ -\begin{aligned} -z_{1,j}^{\text{output}} & = -\begin{pmatrix} -b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -\boldsymbol{x}_j^{\text{hidden}} -\end{pmatrix} -\end{aligned} +f(z) = \frac{1}{1 + \exp{(-z)}} $$ +

    It is possible to use other activations functions for the hidden layer also.

    + +

    The output \( \boldsymbol{x}_i^{\text{hidden}} \) from each \( i \)-th hidden neuron is:

    + +

    $$ +\boldsymbol{x}_i^{\text{hidden} } = f\big( \boldsymbol{z}_{i}^{\text{hidden}} \big) +$$ +

    + +

    The outputs \( \boldsymbol{x}_i^{\text{hidden} } \) are then sent to the output layer.

    + +

    The output layer consists of one neuron in this case, and combines the +output from each of the neurons in the hidden layers. The output layer +combines the results from the hidden layer using some weights \( w_i^{\text{output}} \) +and biases \( b_i^{\text{output}} \). In this case, +it is assumes that the number of neurons in the output layer is one. +

    @@ -442,7 +459,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs020.html b/doc/pub/week42/html/._week42-bs020.html index c0c441f31..4515dd14a 100644 --- a/doc/pub/week42/html/._week42-bs020.html +++ b/doc/pub/week42/html/._week42-bs020.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,22 +404,23 @@ MathJax.Hub.Config({

     

     

     

    -

    Final technicalities IV

    +

    Final technicalities III

    -

    Expressing \( z_{1,j}^{\text{output}} \) as a vector gives the following way of weighting the inputs from the hidden layer:

    +

    The procedure of weighting the output neuron \( j \) in the hidden layer to the \( i \)-th neuron in the output layer is similar as for the hidden layer described previously.

    $$ -\boldsymbol{z}_{1}^{\text{output}} = +\begin{aligned} +z_{1,j}^{\text{output}} & = \begin{pmatrix} b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}} \end{pmatrix} \begin{pmatrix} -1 & 1 & \dots & 1 \\ -\boldsymbol{x}_1^{\text{hidden}} & \boldsymbol{x}_2^{\text{hidden}} & \dots & \boldsymbol{x}_N^{\text{hidden}} +1 \\ +\boldsymbol{x}_j^{\text{hidden}} \end{pmatrix} +\end{aligned} $$ -

    In this case we seek a continuous range of values since we are approximating a function. This means that after computing \( \boldsymbol{z}_{1}^{\text{output}} \) the neural network has finished its feed forward step, and \( \boldsymbol{z}_{1}^{\text{output}} \) is the final output of the network.

    @@ -441,7 +447,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs021.html b/doc/pub/week42/html/._week42-bs021.html index e0180ad63..2390b5250 100644 --- a/doc/pub/week42/html/._week42-bs021.html +++ b/doc/pub/week42/html/._week42-bs021.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,19 +404,22 @@ MathJax.Hub.Config({

     

     

     

    -

    Back propagation

    +

    Final technicalities IV

    -

    The next step is to decide how the parameters should be changed such that they minimize the cost function.

    - -

    The chosen cost function for this problem is

    +

    Expressing \( z_{1,j}^{\text{output}} \) as a vector gives the following way of weighting the inputs from the hidden layer:

    $$ -C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 +\boldsymbol{z}_{1}^{\text{output}} = +\begin{pmatrix} +b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}} +\end{pmatrix} +\begin{pmatrix} +1 & 1 & \dots & 1 \\ +\boldsymbol{x}_1^{\text{hidden}} & \boldsymbol{x}_2^{\text{hidden}} & \dots & \boldsymbol{x}_N^{\text{hidden}} +\end{pmatrix} $$ -

    In order to minimize the cost function, an optimization method must be chosen.

    - -

    Here, gradient descent with a constant step size has been chosen.

    +

    In this case we seek a continuous range of values since we are approximating a function. This means that after computing \( \boldsymbol{z}_{1}^{\text{output}} \) the neural network has finished its feed forward step, and \( \boldsymbol{z}_{1}^{\text{output}} \) is the final output of the network.

    @@ -438,7 +446,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs022.html b/doc/pub/week42/html/._week42-bs022.html index 7af301066..3ea351b2f 100644 --- a/doc/pub/week42/html/._week42-bs022.html +++ b/doc/pub/week42/html/._week42-bs022.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,44 +404,19 @@ MathJax.Hub.Config({

     

     

     

    -

    Gradient descent

    +

    Back propagation

    -

    The idea of the gradient descent algorithm is to update parameters in -a direction where the cost function decreases goes to a minimum. -

    +

    The next step is to decide how the parameters should be changed such that they minimize the cost function.

    -

    In general, the update of some parameters \( \boldsymbol{\omega} \) given a cost -function defined by some weights \( \boldsymbol{\omega} \), \( C(\boldsymbol{x}, -\boldsymbol{\omega}) \), goes as follows: -

    +

    The chosen cost function for this problem is

    $$ -\boldsymbol{\omega}_{\text{new} } = \boldsymbol{\omega} - \lambda \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega}) +C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 $$ -

    for a number of iterations or until $ \big|\big| \boldsymbol{\omega}_{\text{new} } - \boldsymbol{\omega} \big|\big|$ becomes smaller than some given tolerance.

    - -

    The value of \( \lambda \) decides how large steps the algorithm must take -in the direction of $ \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega})$. -The notation \( \nabla_{\boldsymbol{\omega}} \) express the gradient with respect -to the elements in \( \boldsymbol{\omega} \). -

    - -

    In our case, we have to minimize the cost function \( C(\boldsymbol{x}, P) \) with -respect to the two sets of weights and biases, that is for the hidden -layer \( P_{\text{hidden} } \) and for the output layer \( P_{\text{output} -} \) . -

    - -

    This means that \( P_{\text{hidden} } \) and \( P_{\text{output} } \) is updated by

    - -$$ -\begin{aligned} -P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} C(\boldsymbol{x}, P) \\ -P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} C(\boldsymbol{x}, P) -\end{aligned} -$$ +

    In order to minimize the cost function, an optimization method must be chosen.

    +

    Here, gradient descent with a constant step size has been chosen.

    @@ -463,7 +443,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs023.html b/doc/pub/week42/html/._week42-bs023.html index de5d9ebc8..ffe5c1256 100644 --- a/doc/pub/week42/html/._week42-bs023.html +++ b/doc/pub/week42/html/._week42-bs023.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,173 +404,43 @@ MathJax.Hub.Config({

     

     

     

    -

    The code for solving the ODE

    +

    Gradient descent

    +

    The idea of the gradient descent algorithm is to update parameters in +a direction where the cost function decreases goes to a minimum. +

    - -
    -
    -
    -
    -
    - 
    import autograd.numpy as np
-from autograd import grad, elementwise_grad
-import autograd.numpy.random as npr
-from matplotlib import pyplot as plt
+
    In general, the update of some parameters \( \boldsymbol{\omega} \) given a cost
+function defined by some weights \( \boldsymbol{\omega} \), \( C(\boldsymbol{x},
+\boldsymbol{\omega}) \), goes as follows:
+
    
 
-def sigmoid(z):
-    return 1/(1 + np.exp(-z))
+$$
+\boldsymbol{\omega}_{\text{new} } = \boldsymbol{\omega} - \lambda \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega})
+$$
 
-# Assuming one input, hidden, and output layer
-def neural_network(params, x):
+
    for a number of iterations or until $ \big|\big| \boldsymbol{\omega}_{\text{new} } - \boldsymbol{\omega} \big|\big|$ becomes smaller than some given tolerance.
    
 
-    # Find the weights (including and biases) for the hidden and output layer.
-    # Assume that params is a list of parameters for each layer.
-    # The biases are the first element for each array in params,
-    # and the weights are the remaning elements in each array in params.
+
    The value of \( \lambda \) decides how large steps the algorithm must take
+in the direction of $ \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega})$.
+The notation \( \nabla_{\boldsymbol{\omega}} \) express the gradient with respect
+to the elements in \( \boldsymbol{\omega} \).
+
    
 
-    w_hidden = params[0]
-    w_output = params[1]
+
    In our case, we have to minimize the cost function \( C(\boldsymbol{x}, P) \) with
+respect to the two sets of weights and biases, that is for the hidden
+layer \( P_{\text{hidden} } \) and for the output layer \( P_{\text{output}
+} \) .
+
    
 
-    # Assumes input x being an one-dimensional array
-    num_values = np.size(x)
-    x = x.reshape(-1, num_values)
+
    This means that \( P_{\text{hidden} } \) and \( P_{\text{output} } \) is updated by
    
 
-    # Assume that the input layer does nothing to the input x
-    x_input = x
-
-    ## Hidden layer:
-
-    # Add a row of ones to include bias
-    x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0)
-
-    z_hidden = np.matmul(w_hidden, x_input)
-    x_hidden = sigmoid(z_hidden)
-
-    ## Output layer:
-
-    # Include bias:
-    x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0)
-
-    z_output = np.matmul(w_output, x_hidden)
-    x_output = z_output
-
-    return x_output
-
-# The trial solution using the deep neural network:
-def g_trial(x,params, g0 = 10):
-    return g0 + x*neural_network(params,x)
-
-# The right side of the ODE:
-def g(x, g_trial, gamma = 2):
-    return -gamma*g_trial
-
-# The cost function:
-def cost_function(P, x):
-
-    # Evaluate the trial function with the current parameters P
-    g_t = g_trial(x,P)
-
-    # Find the derivative w.r.t x of the neural network
-    d_net_out = elementwise_grad(neural_network,1)(P,x)
-
-    # Find the derivative w.r.t x of the trial function
-    d_g_t = elementwise_grad(g_trial,0)(x,P)
-
-    # The right side of the ODE
-    func = g(x, g_t)
-
-    err_sqr = (d_g_t - func)**2
-    cost_sum = np.sum(err_sqr)
-
-    return cost_sum / np.size(err_sqr)
-
-# Solve the exponential decay ODE using neural network with one input, hidden, and output layer
-def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb):
-    ## Set up initial weights and biases
-
-    # For the hidden layer
-    p0 = npr.randn(num_neurons_hidden, 2 )
-
-    # For the output layer
-    p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included
-
-    P = [p0, p1]
-
-    print('Initial cost: %g'%cost_function(P, x))
-
-    ## Start finding the optimal weights using gradient descent
-
-    # Find the Python function that represents the gradient of the cost function
-    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
-    cost_function_grad = grad(cost_function,0)
-
-    # Let the update be done num_iter times
-    for i in range(num_iter):
-        # Evaluate the gradient at the current weights and biases in P.
-        # The cost_grad consist now of two arrays;
-        # one for the gradient w.r.t P_hidden and
-        # one for the gradient w.r.t P_output
-        cost_grad =  cost_function_grad(P, x)
-
-        P[0] = P[0] - lmb * cost_grad[0]
-        P[1] = P[1] - lmb * cost_grad[1]
-
-    print('Final cost: %g'%cost_function(P, x))
-
-    return P
-
-def g_analytic(x, gamma = 2, g0 = 10):
-    return g0*np.exp(-gamma*x)
-
-# Solve the given problem
-if __name__ == '__main__':
-    # Set seed such that the weight are initialized
-    # with same weights and biases for every run.
-    npr.seed(15)
-
-    ## Decide the vales of arguments to the function to solve
-    N = 10
-    x = np.linspace(0, 1, N)
-
-    ## Set up the initial parameters
-    num_hidden_neurons = 10
-    num_iter = 10000
-    lmb = 0.001
-
-    # Use the network
-    P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb)
-
-    # Print the deviation from the trial solution and true solution
-    res = g_trial(x,P)
-    res_analytical = g_analytic(x)
-
-    print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical)))
-
-    # Plot the results
-    plt.figure(figsize=(10,10))
-
-    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
-    plt.plot(x, res_analytical)
-    plt.plot(x, res[0,:])
-    plt.legend(['analytical','nn'])
-    plt.xlabel('x')
-    plt.ylabel('g(x)')
-    plt.show()
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    +$$ +\begin{aligned} +P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} C(\boldsymbol{x}, P) \\ +P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} C(\boldsymbol{x}, P) +\end{aligned} +$$

    @@ -593,7 +468,7 @@ MathJax.Hub.Config({

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    -

    The network with one input layer, specified number of hidden layers, and one output layer

    - -

    It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers.

    - -

    The number of neurons within each hidden layer are given as a list of integers in the program below.

    +

    The code for solving the ODE

    @@ -420,14 +421,16 @@ MathJax.Hub.Config({ def sigmoid(z): return 1/(1 + np.exp(-z)) -# The neural network with one input layer and one output layer, -# but with number of hidden layers specified by the user. -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers +# Assuming one input, hidden, and output layer +def neural_network(params, x): - N_hidden = np.size(deep_params) - 1 # -1 since params consists of - # parameters to all the hidden - # layers AND the output layer. + # Find the weights (including and biases) for the hidden and output layer. + # Assume that params is a list of parameters for each layer. + # The biases are the first element for each array in params, + # and the weights are the remaning elements in each array in params. + + w_hidden = params[0] + w_output = params[1] # Assumes input x being an one-dimensional array num_values = np.size(x) @@ -436,57 +439,43 @@ MathJax.Hub.Config({ # Assume that the input layer does nothing to the input x x_input = x - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input + ## Hidden layer: - ## Hidden layers: + # Add a row of ones to include bias + x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0) - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden + z_hidden = np.matmul(w_hidden, x_input) + x_hidden = sigmoid(z_hidden) ## Output layer: - # Get the weights and bias for this layer - w_output = deep_params[-1] - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) + x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0) - z_output = np.matmul(w_output, x_prev) + z_output = np.matmul(w_output, x_hidden) x_output = z_output return x_output # The trial solution using the deep neural network: -def g_trial_deep(x,params, g0 = 10): - return g0 + x*deep_neural_network(params, x) +def g_trial(x,params, g0 = 10): + return g0 + x*neural_network(params,x) # The right side of the ODE: def g(x, g_trial, gamma = 2): return -gamma*g_trial -# The same cost function as before, but calls deep_neural_network instead. -def cost_function_deep(P, x): +# The cost function: +def cost_function(P, x): # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) + g_t = g_trial(x,P) # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(deep_neural_network,1)(P,x) + d_net_out = elementwise_grad(neural_network,1)(P,x) # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) + d_g_t = elementwise_grad(g_trial,0)(x,P) # The right side of the ODE func = g(x, g_t) @@ -496,48 +485,38 @@ MathJax.Hub.Config({ return cost_sum / np.size(err_sqr) -# Solve the exponential decay ODE using neural network with one input and one output layer, -# but with specified number of hidden layers from the user. -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # The number of elements in the list num_hidden_neurons thus represents - # the number of hidden layers. - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - +# Solve the exponential decay ODE using neural network with one input, hidden, and output layer +def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb): ## Set up initial weights and biases - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + # For the hidden layer + p0 = npr.randn(num_neurons_hidden, 2 ) # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included - print('Initial cost: %g'%cost_function_deep(P, x)) + P = [p0, p1] + + print('Initial cost: %g'%cost_function(P, x)) ## Start finding the optimal weights using gradient descent # Find the Python function that represents the gradient of the cost function # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) + cost_function_grad = grad(cost_function,0) # Let the update be done num_iter times for i in range(num_iter): # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) + # The cost_grad consist now of two arrays; + # one for the gradient w.r.t P_hidden and + # one for the gradient w.r.t P_output + cost_grad = cost_function_grad(P, x) - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] + P[0] = P[0] - lmb * cost_grad[0] + P[1] = P[1] - lmb * cost_grad[1] - print('Final cost: %g'%cost_function_deep(P, x)) + print('Final cost: %g'%cost_function(P, x)) return P @@ -546,6 +525,8 @@ MathJax.Hub.Config({ # Solve the given problem if __name__ == '__main__': + # Set seed such that the weight are initialized + # with same weights and biases for every run. npr.seed(15) ## Decide the vales of arguments to the function to solve @@ -553,21 +534,27 @@ MathJax.Hub.Config({ x = np.linspace(0, 1, N) ## Set up the initial parameters - num_hidden_neurons = np.array([10,10]) + num_hidden_neurons = 10 num_iter = 10000 lmb = 0.001 - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) + # Use the network + P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb) - res = g_trial_deep(x,P) + # Print the deviation from the trial solution and true solution + res = g_trial(x,P) res_analytical = g_analytic(x) + print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical))) + + # Plot the results plt.figure(figsize=(10,10)) - plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution') + plt.title('Performance of neural network solving an ODE compared to the analytical solution') plt.plot(x, res_analytical) plt.plot(x, res[0,:]) - plt.legend(['analytical','dnn']) + plt.legend(['analytical','nn']) + plt.xlabel('x') plt.ylabel('g(x)') plt.show() @@ -611,7 +598,7 @@ MathJax.Hub.Config({
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    -

    Example: Population growth

    +

    The network with one input layer, specified number of hidden layers, and one output layer

    -

    A logistic model of population growth assumes that a population converges toward an equilibrium. -The population growth can be modeled by -

    +

    It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers.

    -$$ -\begin{equation} \tag{10} - g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -$$ +

    The number of neurons within each hidden layer are given as a list of integers in the program below.

    -

    where \( g(t) \) is the population density at time \( t \), \( \alpha > 0 \) the growth rate and \( A > 0 \) is the maximum population number in the environment. -Also, at \( t = 0 \) the population has the size \( g(0) = g_0 \), where \( g_0 \) is some chosen constant. -

    -

    In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability -and high execution time (this might be more apparent in the examples solving PDEs), -using a library like TensorFlow is recommended. -Here, we stay with a more simple approach and implement for comparison, the simple forward Euler method. -

    + +
    +
    +
    +
    +
    + 
    import autograd.numpy as np
+from autograd import grad, elementwise_grad
+import autograd.numpy.random as npr
+from matplotlib import pyplot as plt
+
+def sigmoid(z):
+    return 1/(1 + np.exp(-z))
+
+# The neural network with one input layer and one output layer,
+# but with number of hidden layers specified by the user.
+def deep_neural_network(deep_params, x):
+    # N_hidden is the number of hidden layers
+
+    N_hidden = np.size(deep_params) - 1 # -1 since params consists of
+                                        # parameters to all the hidden
+                                        # layers AND the output layer.
+
+    # Assumes input x being an one-dimensional array
+    num_values = np.size(x)
+    x = x.reshape(-1, num_values)
+
+    # Assume that the input layer does nothing to the input x
+    x_input = x
+
+    # Due to multiple hidden layers, define a variable referencing to the
+    # output of the previous layer:
+    x_prev = x_input
+
+    ## Hidden layers:
+
+    for l in range(N_hidden):
+        # From the list of parameters P; find the correct weigths and bias for this layer
+        w_hidden = deep_params[l]
+
+        # Add a row of ones to include bias
+        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
+
+        z_hidden = np.matmul(w_hidden, x_prev)
+        x_hidden = sigmoid(z_hidden)
+
+        # Update x_prev such that next layer can use the output from this layer
+        x_prev = x_hidden
+
+    ## Output layer:
+
+    # Get the weights and bias for this layer
+    w_output = deep_params[-1]
+
+    # Include bias:
+    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
+
+    z_output = np.matmul(w_output, x_prev)
+    x_output = z_output
+
+    return x_output
+
+# The trial solution using the deep neural network:
+def g_trial_deep(x,params, g0 = 10):
+    return g0 + x*deep_neural_network(params, x)
+
+# The right side of the ODE:
+def g(x, g_trial, gamma = 2):
+    return -gamma*g_trial
+
+# The same cost function as before, but calls deep_neural_network instead.
+def cost_function_deep(P, x):
+
+    # Evaluate the trial function with the current parameters P
+    g_t = g_trial_deep(x,P)
+
+    # Find the derivative w.r.t x of the neural network
+    d_net_out = elementwise_grad(deep_neural_network,1)(P,x)
+
+    # Find the derivative w.r.t x of the trial function
+    d_g_t = elementwise_grad(g_trial_deep,0)(x,P)
+
+    # The right side of the ODE
+    func = g(x, g_t)
+
+    err_sqr = (d_g_t - func)**2
+    cost_sum = np.sum(err_sqr)
+
+    return cost_sum / np.size(err_sqr)
+
+# Solve the exponential decay ODE using neural network with one input and one output layer,
+# but with specified number of hidden layers from the user.
+def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
+    # num_hidden_neurons is now a list of number of neurons within each hidden layer
+
+    # The number of elements in the list num_hidden_neurons thus represents
+    # the number of hidden layers.
+
+    # Find the number of hidden layers:
+    N_hidden = np.size(num_neurons)
+
+    ## Set up initial weights and biases
+
+    # Initialize the list of parameters:
+    P = [None]*(N_hidden + 1) # + 1 to include the output layer
+
+    P[0] = npr.randn(num_neurons[0], 2 )
+    for l in range(1,N_hidden):
+        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
+
+    # For the output layer
+    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
+
+    print('Initial cost: %g'%cost_function_deep(P, x))
+
+    ## Start finding the optimal weights using gradient descent
+
+    # Find the Python function that represents the gradient of the cost function
+    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
+    cost_function_deep_grad = grad(cost_function_deep,0)
+
+    # Let the update be done num_iter times
+    for i in range(num_iter):
+        # Evaluate the gradient at the current weights and biases in P.
+        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
+        # in the hidden layers and output layers evaluated at x.
+        cost_deep_grad =  cost_function_deep_grad(P, x)
+
+        for l in range(N_hidden+1):
+            P[l] = P[l] - lmb * cost_deep_grad[l]
+
+    print('Final cost: %g'%cost_function_deep(P, x))
+
+    return P
+
+def g_analytic(x, gamma = 2, g0 = 10):
+    return g0*np.exp(-gamma*x)
+
+# Solve the given problem
+if __name__ == '__main__':
+    npr.seed(15)
+
+    ## Decide the vales of arguments to the function to solve
+    N = 10
+    x = np.linspace(0, 1, N)
+
+    ## Set up the initial parameters
+    num_hidden_neurons = np.array([10,10])
+    num_iter = 10000
+    lmb = 0.001
+
+    P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)
+
+    res = g_trial_deep(x,P)
+    res_analytical = g_analytic(x)
+
+    plt.figure(figsize=(10,10))
+
+    plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution')
+    plt.plot(x, res_analytical)
+    plt.plot(x, res[0,:])
+    plt.legend(['analytical','dnn'])
+    plt.ylabel('g(x)')
+    plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +

    @@ -446,7 +616,7 @@ Here, we stay with a more simple approach and implement for comparison, the simp

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    • diff --git a/doc/pub/week42/html/._week42-bs026.html b/doc/pub/week42/html/._week42-bs026.html index 0f01ece80..b35e8c7af 100644 --- a/doc/pub/week42/html/._week42-bs026.html +++ b/doc/pub/week42/html/._week42-bs026.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,21 +404,27 @@ MathJax.Hub.Config({

     

     

     

    -

    Setting up the problem

    +

    Example: Population growth

    -

    Here, we will model a population \( g(t) \) in an environment having carrying capacity \( A \). -The population follows the model +

    A logistic model of population growth assumes that a population converges toward an equilibrium. +The population growth can be modeled by

    $$ -\begin{equation} \tag{11} -g'(t) = \alpha g(t)(A - g(t)) +\begin{equation} \tag{10} + g'(t) = \alpha g(t)(A - g(t)) \end{equation} $$ -

    where \( g(0) = g_0 \).

    +

    where \( g(t) \) is the population density at time \( t \), \( \alpha > 0 \) the growth rate and \( A > 0 \) is the maximum population number in the environment. +Also, at \( t = 0 \) the population has the size \( g(0) = g_0 \), where \( g_0 \) is some chosen constant. +

    -

    In this example, we let \( \alpha = 2 \), \( A = 1 \), and \( g_0 = 1.2 \).

    +

    In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability +and high execution time (this might be more apparent in the examples solving PDEs), +using a library like TensorFlow is recommended. +Here, we stay with a more simple approach and implement for comparison, the simple forward Euler method. +

    @@ -440,7 +451,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs027.html b/doc/pub/week42/html/._week42-bs027.html index 603402aff..f4d747248 100644 --- a/doc/pub/week42/html/._week42-bs027.html +++ b/doc/pub/week42/html/._week42-bs027.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,27 +404,21 @@ MathJax.Hub.Config({

     

     

     

    -

    The trial solution

    +

    Setting up the problem

    -

    We will get a slightly different trial solution, as the boundary conditions are different -compared to the case for exponential decay. +

    Here, we will model a population \( g(t) \) in an environment having carrying capacity \( A \). +The population follows the model

    -

    A possible trial solution satisfying the condition \( g(0) = g_0 \) could be

    - -

    $$ -h_1(t) = g_0 + t \cdot N(t,P) $$ -

    - -

    with \( N(t,P) \) being the output from the neural network with weights and biases for each layer collected in the set \( P \).

    - -

    The analytical solution is

    - -

    $$ -g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)} +\begin{equation} \tag{11} +g'(t) = \alpha g(t)(A - g(t)) +\end{equation} $$ -

    + +

    where \( g(0) = g_0 \).

    + +

    In this example, we let \( \alpha = 2 \), \( A = 1 \), and \( g_0 = 1.2 \).

    @@ -446,7 +445,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs028.html b/doc/pub/week42/html/._week42-bs028.html index 2d99b7fe5..e816f1b55 100644 --- a/doc/pub/week42/html/._week42-bs028.html +++ b/doc/pub/week42/html/._week42-bs028.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,195 +404,27 @@ MathJax.Hub.Config({

     

     

     

    -

    The program using Autograd

    +

    The trial solution

    -

    The network will be the similar as for the exponential decay example, but with some small modifications for our problem.

    +

    We will get a slightly different trial solution, as the boundary conditions are different +compared to the case for exponential decay. +

    +

    A possible trial solution satisfying the condition \( g(0) = g_0 \) could be

    - -
    -
    -
    -
    -
    - 
    import autograd.numpy as np
-from autograd import grad, elementwise_grad
-import autograd.numpy.random as npr
-from matplotlib import pyplot as plt
+
    $$
+h_1(t) = g_0 + t \cdot N(t,P)
+$$
+
    
 
-def sigmoid(z):
-    return 1/(1 + np.exp(-z))
+
    with \( N(t,P) \) being the output from the neural network with weights and biases for each layer collected in the set \( P \).
    
 
-# Function to get the parameters.
-# Done such that one can easily change the paramaters after one's liking.
-def get_parameters():
-    alpha = 2
-    A = 1
-    g0 = 1.2
-    return alpha, A, g0
-
-def deep_neural_network(P, x):
-    # N_hidden is the number of hidden layers
-    N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
-
-    # Assumes input x being an one-dimensional array
-    num_values = np.size(x)
-    x = x.reshape(-1, num_values)
-
-    # Assume that the input layer does nothing to the input x
-    x_input = x
-
-    # Due to multiple hidden layers, define a variable referencing to the
-    # output of the previous layer:
-    x_prev = x_input
-
-    ## Hidden layers:
-
-    for l in range(N_hidden):
-        # From the list of parameters P; find the correct weigths and bias for this layer
-        w_hidden = P[l]
-
-        # Add a row of ones to include bias
-        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
-
-        z_hidden = np.matmul(w_hidden, x_prev)
-        x_hidden = sigmoid(z_hidden)
-
-        # Update x_prev such that next layer can use the output from this layer
-        x_prev = x_hidden
-
-    ## Output layer:
-
-    # Get the weights and bias for this layer
-    w_output = P[-1]
-
-    # Include bias:
-    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
-
-    z_output = np.matmul(w_output, x_prev)
-    x_output = z_output
-
-    return x_output
-
-
-def cost_function_deep(P, x):
-
-    # Evaluate the trial function with the current parameters P
-    g_t = g_trial_deep(x,P)
-
-    # Find the derivative w.r.t x of the trial function
-    d_g_t = elementwise_grad(g_trial_deep,0)(x,P)
-
-    # The right side of the ODE
-    func = f(x, g_t)
-
-    err_sqr = (d_g_t - func)**2
-    cost_sum = np.sum(err_sqr)
-
-    return cost_sum / np.size(err_sqr)
-
-# The right side of the ODE:
-def f(x, g_trial):
-    alpha,A, g0 = get_parameters()
-    return alpha*g_trial*(A - g_trial)
-
-# The trial solution using the deep neural network:
-def g_trial_deep(x, params):
-    alpha,A, g0 = get_parameters()
-    return g0 + x*deep_neural_network(params,x)
-
-# The analytical solution:
-def g_analytic(t):
-    alpha,A, g0 = get_parameters()
-    return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t))
-
-def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
-    # num_hidden_neurons is now a list of number of neurons within each hidden layer
-
-    # Find the number of hidden layers:
-    N_hidden = np.size(num_neurons)
-
-    ## Set up initial weigths and biases
-
-    # Initialize the list of parameters:
-    P = [None]*(N_hidden + 1) # + 1 to include the output layer
-
-    P[0] = npr.randn(num_neurons[0], 2 )
-    for l in range(1,N_hidden):
-        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
-
-    # For the output layer
-    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
-
-    print('Initial cost: %g'%cost_function_deep(P, x))
-
-    ## Start finding the optimal weigths using gradient descent
-
-    # Find the Python function that represents the gradient of the cost function
-    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
-    cost_function_deep_grad = grad(cost_function_deep,0)
-
-    # Let the update be done num_iter times
-    for i in range(num_iter):
-        # Evaluate the gradient at the current weights and biases in P.
-        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
-        # in the hidden layers and output layers evaluated at x.
-        cost_deep_grad =  cost_function_deep_grad(P, x)
-
-        for l in range(N_hidden+1):
-            P[l] = P[l] - lmb * cost_deep_grad[l]
-
-    print('Final cost: %g'%cost_function_deep(P, x))
-
-    return P
-
-if __name__ == '__main__':
-    npr.seed(4155)
-
-    ## Decide the vales of arguments to the function to solve
-    Nt = 10
-    T = 1
-    t = np.linspace(0,T, Nt)
-
-    ## Set up the initial parameters
-    num_hidden_neurons = [100, 50, 25]
-    num_iter = 1000
-    lmb = 1e-3
-
-    P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)
-
-    g_dnn_ag = g_trial_deep(t,P)
-    g_analytical = g_analytic(t)
-
-    # Find the maximum absolute difference between the solutons:
-    diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))
-    print("The max absolute difference between the solutions is: %g"%diff_ag)
-
-    plt.figure(figsize=(10,10))
-
-    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
-    plt.plot(t, g_analytical)
-    plt.plot(t, g_dnn_ag[0,:])
-    plt.legend(['analytical','nn'])
-    plt.xlabel('t')
-    plt.ylabel('g(t)')
-
-    plt.show()
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    +

    The analytical solution is

    +

    $$ +g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)} +$$ +

    @@ -614,7 +451,7 @@ MathJax.Hub.Config({

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    • diff --git a/doc/pub/week42/html/._week42-bs029.html b/doc/pub/week42/html/._week42-bs029.html index ac929111f..e62070267 100644 --- a/doc/pub/week42/html/._week42-bs029.html +++ b/doc/pub/week42/html/._week42-bs029.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,57 +404,9 @@ MathJax.Hub.Config({

     

     

     

    -

    Using forward Euler to solve the ODE

    +

    The program using Autograd

    -

    A straightforward way of solving an ODE numerically, is to use Euler's method.

    - -

    Euler's method uses Taylor series to approximate the value at a function \( f \) at a step \( \Delta x \) from \( x \):

    - -

    $$ -f(x + \Delta x) \approx f(x) + \Delta x f'(x) -$$ -

    - -

    In our case, using Euler's method to approximate the value of \( g \) at a step \( \Delta t \) from \( t \) yields

    - -$$ -\begin{aligned} - g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\ - &= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big) -\end{aligned} -$$ - -

    along with the condition that \( g(0) = g_0 \).

    - -

    Let \( t_i = i \cdot \Delta t \) where \( \Delta t = \frac{T}{N_t-1} \) where \( T \) is the final time our solver must solve for and \( N_t \) the number of values for \( t \in [0, T] \) for \( i = 0, \dots, N_t-1 \).

    - -

    For \( i \geq 1 \), we have that

    -$$ -\begin{aligned} -t_i &= i\Delta t \\ -&= (i - 1)\Delta t + \Delta t \\ -&= t_{i-1} + \Delta t -\end{aligned} -$$ - -

    Now, if \( g_i = g(t_i) \) then

    - -$$ -\begin{equation} - \begin{aligned} - g_i &= g(t_i) \\ - &= g(t_{i-1} + \Delta t) \\ - &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\ - &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big) - \end{aligned} -\end{equation} \tag{12} -$$ - -

    for \( i \geq 1 \) and \( g_0 = g(t_0) = g(0) = g_0 \).

    - -

    Equation (12) could be implemented in the following way, -extending the program that uses the network using Autograd: -

    +

    The network will be the similar as for the exponential decay example, but with some small modifications for our problem.

    @@ -458,8 +415,136 @@ extending the program that uses the network using Autograd:
    - 
    # Assume that all function definitions from the example program using Autograd
-# are located here.
+  
    import autograd.numpy as np
+from autograd import grad, elementwise_grad
+import autograd.numpy.random as npr
+from matplotlib import pyplot as plt
+
+def sigmoid(z):
+    return 1/(1 + np.exp(-z))
+
+# Function to get the parameters.
+# Done such that one can easily change the paramaters after one's liking.
+def get_parameters():
+    alpha = 2
+    A = 1
+    g0 = 1.2
+    return alpha, A, g0
+
+def deep_neural_network(P, x):
+    # N_hidden is the number of hidden layers
+    N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
+
+    # Assumes input x being an one-dimensional array
+    num_values = np.size(x)
+    x = x.reshape(-1, num_values)
+
+    # Assume that the input layer does nothing to the input x
+    x_input = x
+
+    # Due to multiple hidden layers, define a variable referencing to the
+    # output of the previous layer:
+    x_prev = x_input
+
+    ## Hidden layers:
+
+    for l in range(N_hidden):
+        # From the list of parameters P; find the correct weigths and bias for this layer
+        w_hidden = P[l]
+
+        # Add a row of ones to include bias
+        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
+
+        z_hidden = np.matmul(w_hidden, x_prev)
+        x_hidden = sigmoid(z_hidden)
+
+        # Update x_prev such that next layer can use the output from this layer
+        x_prev = x_hidden
+
+    ## Output layer:
+
+    # Get the weights and bias for this layer
+    w_output = P[-1]
+
+    # Include bias:
+    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
+
+    z_output = np.matmul(w_output, x_prev)
+    x_output = z_output
+
+    return x_output
+
+
+def cost_function_deep(P, x):
+
+    # Evaluate the trial function with the current parameters P
+    g_t = g_trial_deep(x,P)
+
+    # Find the derivative w.r.t x of the trial function
+    d_g_t = elementwise_grad(g_trial_deep,0)(x,P)
+
+    # The right side of the ODE
+    func = f(x, g_t)
+
+    err_sqr = (d_g_t - func)**2
+    cost_sum = np.sum(err_sqr)
+
+    return cost_sum / np.size(err_sqr)
+
+# The right side of the ODE:
+def f(x, g_trial):
+    alpha,A, g0 = get_parameters()
+    return alpha*g_trial*(A - g_trial)
+
+# The trial solution using the deep neural network:
+def g_trial_deep(x, params):
+    alpha,A, g0 = get_parameters()
+    return g0 + x*deep_neural_network(params,x)
+
+# The analytical solution:
+def g_analytic(t):
+    alpha,A, g0 = get_parameters()
+    return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t))
+
+def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
+    # num_hidden_neurons is now a list of number of neurons within each hidden layer
+
+    # Find the number of hidden layers:
+    N_hidden = np.size(num_neurons)
+
+    ## Set up initial weigths and biases
+
+    # Initialize the list of parameters:
+    P = [None]*(N_hidden + 1) # + 1 to include the output layer
+
+    P[0] = npr.randn(num_neurons[0], 2 )
+    for l in range(1,N_hidden):
+        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
+
+    # For the output layer
+    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
+
+    print('Initial cost: %g'%cost_function_deep(P, x))
+
+    ## Start finding the optimal weigths using gradient descent
+
+    # Find the Python function that represents the gradient of the cost function
+    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
+    cost_function_deep_grad = grad(cost_function_deep,0)
+
+    # Let the update be done num_iter times
+    for i in range(num_iter):
+        # Evaluate the gradient at the current weights and biases in P.
+        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
+        # in the hidden layers and output layers evaluated at x.
+        cost_deep_grad =  cost_function_deep_grad(P, x)
+
+        for l in range(N_hidden+1):
+            P[l] = P[l] - lmb * cost_deep_grad[l]
+
+    print('Final cost: %g'%cost_function_deep(P, x))
+
+    return P
 
 if __name__ == '__main__':
     npr.seed(4155)
@@ -470,7 +555,7 @@ extending the program that uses the network using Autograd:
     t = np.linspace(0,T, Nt)
 
     ## Set up the initial parameters
-    num_hidden_neurons = [100,50,25]
+    num_hidden_neurons = [100, 50, 25]
     num_iter = 1000
     lmb = 1e-3
 
@@ -492,36 +577,6 @@ extending the program that uses the network using Autograd:
     plt.xlabel('t')
     plt.ylabel('g(t)')
 
-    ## Find an approximation to the funtion using forward Euler
-
-    alpha, A, g0 = get_parameters()
-    dt = T/(Nt - 1)
-
-    # Perform forward Euler to solve the ODE
-    g_euler = np.zeros(Nt)
-    g_euler[0] = g0
-
-    for i in range(1,Nt):
-        g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1]))
-
-    # Print the errors done by each method
-    diff1 = np.max(np.abs(g_euler - g_analytical))
-    diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical))
-
-    print('Max absolute difference between Euler method and analytical: %g'%diff1)
-    print('Max absolute difference between deep neural network and analytical: %g'%diff2)
-
-    # Plot results
-    plt.figure(figsize=(10,10))
-
-    plt.plot(t,g_euler)
-    plt.plot(t,g_analytical)
-    plt.plot(t,g_dnn_ag[0,:])
-
-    plt.legend(['euler','analytical','dnn'])
-    plt.xlabel('Time t')
-    plt.ylabel('g(t)')
-
     plt.show()
    @@ -564,7 +619,7 @@ extending the program that uses the network using Autograd:
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    • diff --git a/doc/pub/week42/html/._week42-bs030.html b/doc/pub/week42/html/._week42-bs030.html index 5c1f64add..5300f08d4 100644 --- a/doc/pub/week42/html/._week42-bs030.html +++ b/doc/pub/week42/html/._week42-bs030.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,31 +404,146 @@ MathJax.Hub.Config({

     

     

     

    -

    Example: Solving the one dimensional Poisson equation

    +

    Using forward Euler to solve the ODE

    -

    The Poisson equation for \( g(x) \) in one dimension is

    +

    A straightforward way of solving an ODE numerically, is to use Euler's method.

    +

    Euler's method uses Taylor series to approximate the value at a function \( f \) at a step \( \Delta x \) from \( x \):

    + +

    $$ +f(x + \Delta x) \approx f(x) + \Delta x f'(x) $$ -\begin{equation} \tag{13} - -g''(x) = f(x) -\end{equation} -$$ - -

    where \( f(x) \) is a given function for \( x \in (0,1) \).

    - -

    The conditions that \( g(x) \) is chosen to fulfill, are

    -$$ -\begin{align*} - g(0) &= 0 \\ - g(1) &= 0 -\end{align*} -$$ - -

    This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used. -The results from the networks can then be compared to the analytical solution. -In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks.

    +

    In our case, using Euler's method to approximate the value of \( g \) at a step \( \Delta t \) from \( t \) yields

    + +$$ +\begin{aligned} + g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\ + &= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big) +\end{aligned} +$$ + +

    along with the condition that \( g(0) = g_0 \).

    + +

    Let \( t_i = i \cdot \Delta t \) where \( \Delta t = \frac{T}{N_t-1} \) where \( T \) is the final time our solver must solve for and \( N_t \) the number of values for \( t \in [0, T] \) for \( i = 0, \dots, N_t-1 \).

    + +

    For \( i \geq 1 \), we have that

    +$$ +\begin{aligned} +t_i &= i\Delta t \\ +&= (i - 1)\Delta t + \Delta t \\ +&= t_{i-1} + \Delta t +\end{aligned} +$$ + +

    Now, if \( g_i = g(t_i) \) then

    + +$$ +\begin{equation} + \begin{aligned} + g_i &= g(t_i) \\ + &= g(t_{i-1} + \Delta t) \\ + &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\ + &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big) + \end{aligned} +\end{equation} \tag{12} +$$ + +

    for \( i \geq 1 \) and \( g_0 = g(t_0) = g(0) = g_0 \).

    + +

    Equation (12) could be implemented in the following way, +extending the program that uses the network using Autograd: +

    + + + +
    +
    +
    +
    +
    + 
    # Assume that all function definitions from the example program using Autograd
+# are located here.
+
+if __name__ == '__main__':
+    npr.seed(4155)
+
+    ## Decide the vales of arguments to the function to solve
+    Nt = 10
+    T = 1
+    t = np.linspace(0,T, Nt)
+
+    ## Set up the initial parameters
+    num_hidden_neurons = [100,50,25]
+    num_iter = 1000
+    lmb = 1e-3
+
+    P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)
+
+    g_dnn_ag = g_trial_deep(t,P)
+    g_analytical = g_analytic(t)
+
+    # Find the maximum absolute difference between the solutons:
+    diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))
+    print("The max absolute difference between the solutions is: %g"%diff_ag)
+
+    plt.figure(figsize=(10,10))
+
+    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
+    plt.plot(t, g_analytical)
+    plt.plot(t, g_dnn_ag[0,:])
+    plt.legend(['analytical','nn'])
+    plt.xlabel('t')
+    plt.ylabel('g(t)')
+
+    ## Find an approximation to the funtion using forward Euler
+
+    alpha, A, g0 = get_parameters()
+    dt = T/(Nt - 1)
+
+    # Perform forward Euler to solve the ODE
+    g_euler = np.zeros(Nt)
+    g_euler[0] = g0
+
+    for i in range(1,Nt):
+        g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1]))
+
+    # Print the errors done by each method
+    diff1 = np.max(np.abs(g_euler - g_analytical))
+    diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical))
+
+    print('Max absolute difference between Euler method and analytical: %g'%diff1)
+    print('Max absolute difference between deep neural network and analytical: %g'%diff2)
+
+    # Plot results
+    plt.figure(figsize=(10,10))
+
+    plt.plot(t,g_euler)
+    plt.plot(t,g_analytical)
+    plt.plot(t,g_dnn_ag[0,:])
+
+    plt.legend(['euler','analytical','dnn'])
+    plt.xlabel('Time t')
+    plt.ylabel('g(t)')
+
+    plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +

      @@ -449,7 +569,7 @@ In addition, it could be interesting to see how a typical method for numerically
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    diff --git a/doc/pub/week42/html/._week42-bs031.html b/doc/pub/week42/html/._week42-bs031.html index f141f8a6f..0ae5c7e99 100644 --- a/doc/pub/week42/html/._week42-bs031.html +++ b/doc/pub/week42/html/._week42-bs031.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,36 +404,30 @@ MathJax.Hub.Config({

     

     

     

    -

    The specific equation to solve for

    +

    Example: Solving the one dimensional Poisson equation

    -

    Here, the function \( g(x) \) to solve for follows the equation

    +

    The Poisson equation for \( g(x) \) in one dimension is

    $$ --g''(x) = f(x),\qquad x \in (0,1) +\begin{equation} \tag{13} + -g''(x) = f(x) +\end{equation} $$ -

    where \( f(x) \) is a given function, along with the chosen conditions

    +

    where \( f(x) \) is a given function for \( x \in (0,1) \).

    +

    The conditions that \( g(x) \) is chosen to fulfill, are

    $$ -\begin{aligned} -g(0) = g(1) = 0 -\end{aligned}\tag{14} -$$ - -

    In this example, we consider the case when \( f(x) = (3x + x^2)\exp(x) \).

    - -

    For this case, a possible trial solution satisfying the conditions could be

    - -$$ -g_t(x) = x \cdot (1-x) \cdot N(P,x) -$$ - -

    The analytical solution for this problem is

    - -$$ -g(x) = x(1 - x)\exp(x) +\begin{align*} + g(0) &= 0 \\ + g(1) &= 0 +\end{align*} $$ +

    This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used. +The results from the networks can then be compared to the analytical solution. +In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks. +

    @@ -455,7 +454,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs032.html b/doc/pub/week42/html/._week42-bs032.html index 5bb659abc..09a60e17d 100644 --- a/doc/pub/week42/html/._week42-bs032.html +++ b/doc/pub/week42/html/._week42-bs032.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,179 +404,35 @@ MathJax.Hub.Config({

     

     

     

    -

    Solving the equation using Autograd

    +

    The specific equation to solve for

    +

    Here, the function \( g(x) \) to solve for follows the equation

    - -
    -
    -
    -
    -
    - 
    import autograd.numpy as np
-from autograd import grad, elementwise_grad
-import autograd.numpy.random as npr
-from matplotlib import pyplot as plt
+$$
+-g''(x) = f(x),\qquad x \in (0,1)
+$$
 
-def sigmoid(z):
-    return 1/(1 + np.exp(-z))
+
    where \( f(x) \) is a given function, along with the chosen conditions
    
 
-def deep_neural_network(deep_params, x):
-    # N_hidden is the number of hidden layers
-    N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
+$$
+\begin{aligned}
+g(0) = g(1) = 0
+\end{aligned}\tag{14}
+$$
 
-    # Assumes input x being an one-dimensional array
-    num_values = np.size(x)
-    x = x.reshape(-1, num_values)
+
    In this example, we consider the case when \( f(x) = (3x + x^2)\exp(x) \).
    
 
-    # Assume that the input layer does nothing to the input x
-    x_input = x
+
    For this case, a possible trial solution satisfying the conditions could be
    
 
-    # Due to multiple hidden layers, define a variable referencing to the
-    # output of the previous layer:
-    x_prev = x_input
+$$
+g_t(x) = x \cdot (1-x) \cdot N(P,x)
+$$
 
-    ## Hidden layers:
+
    The analytical solution for this problem is
    
 
-    for l in range(N_hidden):
-        # From the list of parameters P; find the correct weigths and bias for this layer
-        w_hidden = deep_params[l]
-
-        # Add a row of ones to include bias
-        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
-
-        z_hidden = np.matmul(w_hidden, x_prev)
-        x_hidden = sigmoid(z_hidden)
-
-        # Update x_prev such that next layer can use the output from this layer
-        x_prev = x_hidden
-
-    ## Output layer:
-
-    # Get the weights and bias for this layer
-    w_output = deep_params[-1]
-
-    # Include bias:
-    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
-
-    z_output = np.matmul(w_output, x_prev)
-    x_output = z_output
-
-    return x_output
-
-def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
-    # num_hidden_neurons is now a list of number of neurons within each hidden layer
-
-    # Find the number of hidden layers:
-    N_hidden = np.size(num_neurons)
-
-    ## Set up initial weigths and biases
-
-    # Initialize the list of parameters:
-    P = [None]*(N_hidden + 1) # + 1 to include the output layer
-
-    P[0] = npr.randn(num_neurons[0], 2 )
-    for l in range(1,N_hidden):
-        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
-
-    # For the output layer
-    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
-
-    print('Initial cost: %g'%cost_function_deep(P, x))
-
-    ## Start finding the optimal weigths using gradient descent
-
-    # Find the Python function that represents the gradient of the cost function
-    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
-    cost_function_deep_grad = grad(cost_function_deep,0)
-
-    # Let the update be done num_iter times
-    for i in range(num_iter):
-        # Evaluate the gradient at the current weights and biases in P.
-        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
-        # in the hidden layers and output layers evaluated at x.
-        cost_deep_grad =  cost_function_deep_grad(P, x)
-
-        for l in range(N_hidden+1):
-            P[l] = P[l] - lmb * cost_deep_grad[l]
-
-    print('Final cost: %g'%cost_function_deep(P, x))
-
-    return P
-
-## Set up the cost function specified for this Poisson equation:
-
-# The right side of the ODE
-def f(x):
-    return (3*x + x**2)*np.exp(x)
-
-def cost_function_deep(P, x):
-
-    # Evaluate the trial function with the current parameters P
-    g_t = g_trial_deep(x,P)
-
-    # Find the derivative w.r.t x of the trial function
-    d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)
-
-    right_side = f(x)
-
-    err_sqr = (-d2_g_t - right_side)**2
-    cost_sum = np.sum(err_sqr)
-
-    return cost_sum/np.size(err_sqr)
-
-# The trial solution:
-def g_trial_deep(x,P):
-    return x*(1-x)*deep_neural_network(P,x)
-
-# The analytic solution;
-def g_analytic(x):
-    return x*(1-x)*np.exp(x)
-
-if __name__ == '__main__':
-    npr.seed(4155)
-
-    ## Decide the vales of arguments to the function to solve
-    Nx = 10
-    x = np.linspace(0,1, Nx)
-
-    ## Set up the initial parameters
-    num_hidden_neurons = [200,100]
-    num_iter = 1000
-    lmb = 1e-3
-
-    P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)
-
-    g_dnn_ag = g_trial_deep(x,P)
-    g_analytical = g_analytic(x)
-
-    # Find the maximum absolute difference between the solutons:
-    max_diff = np.max(np.abs(g_dnn_ag - g_analytical))
-    print("The max absolute difference between the solutions is: %g"%max_diff)
-
-    plt.figure(figsize=(10,10))
-
-    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
-    plt.plot(x, g_analytical)
-    plt.plot(x, g_dnn_ag[0,:])
-    plt.legend(['analytical','nn'])
-    plt.xlabel('x')
-    plt.ylabel('g(x)')
-    plt.show()
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    +$$ +g(x) = x(1 - x)\exp(x) +$$

    @@ -599,7 +460,7 @@ MathJax.Hub.Config({

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    • diff --git a/doc/pub/week42/html/._week42-bs033.html b/doc/pub/week42/html/._week42-bs033.html index 42410a314..ae3c20e6d 100644 --- a/doc/pub/week42/html/._week42-bs033.html +++ b/doc/pub/week42/html/._week42-bs033.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,92 +404,180 @@ MathJax.Hub.Config({

     

     

     

    -

    Comparing with a numerical scheme

    +

    Solving the equation using Autograd

    -

    The Poisson equation is possible to solve using Taylor series to approximate the second derivative.

    -

    Using Taylor series, the second derivative can be expressed as

    + +
    +
    +
    +
    +
    + 
    import autograd.numpy as np
+from autograd import grad, elementwise_grad
+import autograd.numpy.random as npr
+from matplotlib import pyplot as plt
 
-
    $$
-g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x)
-$$
-
    
+def sigmoid(z):
+    return 1/(1 + np.exp(-z))
 
-
    where \( \Delta x \) is a small step size and \( E_{\Delta x}(x) \) being the error term.
    
+def deep_neural_network(deep_params, x):
+    # N_hidden is the number of hidden layers
+    N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
 
-
    Looking away from the error terms gives an approximation to the second derivative:
    
+    # Assumes input x being an one-dimensional array
+    num_values = np.size(x)
+    x = x.reshape(-1, num_values)
 
-$$
-\begin{equation} \tag{15}
-g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2}
-\end{equation}
-$$
+    # Assume that the input layer does nothing to the input x
+    x_input = x
 
-
    If \( x_i = i \Delta x = x_{i-1} + \Delta x \) and \( g_i = g(x_i) \) for \( i = 1,\dots N_x - 2 \) with \( N_x \) being the number of values for \( x \), (15) becomes
    
+    # Due to multiple hidden layers, define a variable referencing to the
+    # output of the previous layer:
+    x_prev = x_input
 
-$$
-\begin{aligned}
-g''(x_i) &\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\
-&= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2}
-\end{aligned}
-$$
+    ## Hidden layers:
 
-
    Since we know from our problem that
    
+    for l in range(N_hidden):
+        # From the list of parameters P; find the correct weigths and bias for this layer
+        w_hidden = deep_params[l]
 
-$$
-\begin{aligned}
--g''(x) &= f(x) \\
-&= (3x + x^2)\exp(x)
-\end{aligned}
-$$
+        # Add a row of ones to include bias
+        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
 
-
    along with the conditions \( g(0) = g(1) = 0 \),
-the following scheme can be used to find an approximate solution for \( g(x) \) numerically:
-
    
+        z_hidden = np.matmul(w_hidden, x_prev)
+        x_hidden = sigmoid(z_hidden)
 
-$$
-\begin{equation}
-  \begin{aligned}
-  -\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &= f(x_i) \\
-  -g_{i+1} + 2g_i - g_{i-1} &= \Delta x^2 f(x_i)
-  \end{aligned}
-\end{equation} \tag{16}
-$$
+        # Update x_prev such that next layer can use the output from this layer
+        x_prev = x_hidden
 
-
    for \( i = 1, \dots, N_x - 2 \) where \( g_0 = g_{N_x - 1} = 0 \) and \( f(x_i) = (3x_i + x_i^2)\exp(x_i) \), which is given for our specific problem.
    
+    ## Output layer:
 
-
    The equation can be rewritten into a matrix equation:
    
+    # Get the weights and bias for this layer
+    w_output = deep_params[-1]
 
-$$
-\begin{aligned}
-\begin{pmatrix}
-2 & -1 & 0 & \dots & 0 \\
--1 & 2 & -1 & \dots & 0 \\
-\vdots & & \ddots & & \vdots \\
-0 & \dots & -1 & 2 & -1  \\
-0 & \dots & 0 & -1 & 2\\
-\end{pmatrix}
-\begin{pmatrix}
-g_1 \\
-g_2 \\
-\vdots \\
-g_{N_x - 3} \\
-g_{N_x - 2}
-\end{pmatrix}
-&=
-\Delta x^2
-\begin{pmatrix}
-f(x_1) \\
-f(x_2) \\
-\vdots \\
-f(x_{N_x - 3}) \\
-f(x_{N_x - 2})
-\end{pmatrix} \\
-\boldsymbol{A}\boldsymbol{g} &= \boldsymbol{f},
-\end{aligned}
-$$
+    # Include bias:
+    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
+
+    z_output = np.matmul(w_output, x_prev)
+    x_output = z_output
+
+    return x_output
+
+def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
+    # num_hidden_neurons is now a list of number of neurons within each hidden layer
+
+    # Find the number of hidden layers:
+    N_hidden = np.size(num_neurons)
+
+    ## Set up initial weigths and biases
+
+    # Initialize the list of parameters:
+    P = [None]*(N_hidden + 1) # + 1 to include the output layer
+
+    P[0] = npr.randn(num_neurons[0], 2 )
+    for l in range(1,N_hidden):
+        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
+
+    # For the output layer
+    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
+
+    print('Initial cost: %g'%cost_function_deep(P, x))
+
+    ## Start finding the optimal weigths using gradient descent
+
+    # Find the Python function that represents the gradient of the cost function
+    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
+    cost_function_deep_grad = grad(cost_function_deep,0)
+
+    # Let the update be done num_iter times
+    for i in range(num_iter):
+        # Evaluate the gradient at the current weights and biases in P.
+        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
+        # in the hidden layers and output layers evaluated at x.
+        cost_deep_grad =  cost_function_deep_grad(P, x)
+
+        for l in range(N_hidden+1):
+            P[l] = P[l] - lmb * cost_deep_grad[l]
+
+    print('Final cost: %g'%cost_function_deep(P, x))
+
+    return P
+
+## Set up the cost function specified for this Poisson equation:
+
+# The right side of the ODE
+def f(x):
+    return (3*x + x**2)*np.exp(x)
+
+def cost_function_deep(P, x):
+
+    # Evaluate the trial function with the current parameters P
+    g_t = g_trial_deep(x,P)
+
+    # Find the derivative w.r.t x of the trial function
+    d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)
+
+    right_side = f(x)
+
+    err_sqr = (-d2_g_t - right_side)**2
+    cost_sum = np.sum(err_sqr)
+
+    return cost_sum/np.size(err_sqr)
+
+# The trial solution:
+def g_trial_deep(x,P):
+    return x*(1-x)*deep_neural_network(P,x)
+
+# The analytic solution;
+def g_analytic(x):
+    return x*(1-x)*np.exp(x)
+
+if __name__ == '__main__':
+    npr.seed(4155)
+
+    ## Decide the vales of arguments to the function to solve
+    Nx = 10
+    x = np.linspace(0,1, Nx)
+
+    ## Set up the initial parameters
+    num_hidden_neurons = [200,100]
+    num_iter = 1000
+    lmb = 1e-3
+
+    P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)
+
+    g_dnn_ag = g_trial_deep(x,P)
+    g_analytical = g_analytic(x)
+
+    # Find the maximum absolute difference between the solutons:
+    max_diff = np.max(np.abs(g_dnn_ag - g_analytical))
+    print("The max absolute difference between the solutions is: %g"%max_diff)
+
+    plt.figure(figsize=(10,10))
+
+    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
+    plt.plot(x, g_analytical)
+    plt.plot(x, g_dnn_ag[0,:])
+    plt.legend(['analytical','nn'])
+    plt.xlabel('x')
+    plt.ylabel('g(x)')
+    plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    -

    which makes it possible to solve for the vector \( \boldsymbol{g} \).

    @@ -511,7 +604,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs034.html b/doc/pub/week42/html/._week42-bs034.html index 85ef83710..0b10a9bb5 100644 --- a/doc/pub/week42/html/._week42-bs034.html +++ b/doc/pub/week42/html/._week42-bs034.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,222 +404,92 @@ MathJax.Hub.Config({

     

     

     

    -

    Setting up the code

    +

    Comparing with a numerical scheme

    -

    We can then compare the result from this numerical scheme with the output from our network using Autograd:

    +

    The Poisson equation is possible to solve using Taylor series to approximate the second derivative.

    +

    Using Taylor series, the second derivative can be expressed as

    - -
    -
    -
    -
    -
    - 
    import autograd.numpy as np
-from autograd import grad, elementwise_grad
-import autograd.numpy.random as npr
-from matplotlib import pyplot as plt
+
    $$
+g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x)
+$$
+
    
 
-def sigmoid(z):
-    return 1/(1 + np.exp(-z))
+
    where \( \Delta x \) is a small step size and \( E_{\Delta x}(x) \) being the error term.
    
 
-def deep_neural_network(deep_params, x):
-    # N_hidden is the number of hidden layers
-    N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
+
    Looking away from the error terms gives an approximation to the second derivative:
    
 
-    # Assumes input x being an one-dimensional array
-    num_values = np.size(x)
-    x = x.reshape(-1, num_values)
+$$
+\begin{equation} \tag{15}
+g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2}
+\end{equation}
+$$
 
-    # Assume that the input layer does nothing to the input x
-    x_input = x
+
    If \( x_i = i \Delta x = x_{i-1} + \Delta x \) and \( g_i = g(x_i) \) for \( i = 1,\dots N_x - 2 \) with \( N_x \) being the number of values for \( x \), (15) becomes
    
 
-    # Due to multiple hidden layers, define a variable referencing to the
-    # output of the previous layer:
-    x_prev = x_input
+$$
+\begin{aligned}
+g''(x_i) &\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\
+&= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2}
+\end{aligned}
+$$
 
-    ## Hidden layers:
+
    Since we know from our problem that
    
 
-    for l in range(N_hidden):
-        # From the list of parameters P; find the correct weigths and bias for this layer
-        w_hidden = deep_params[l]
+$$
+\begin{aligned}
+-g''(x) &= f(x) \\
+&= (3x + x^2)\exp(x)
+\end{aligned}
+$$
 
-        # Add a row of ones to include bias
-        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
+
    along with the conditions \( g(0) = g(1) = 0 \),
+the following scheme can be used to find an approximate solution for \( g(x) \) numerically:
+
    
 
-        z_hidden = np.matmul(w_hidden, x_prev)
-        x_hidden = sigmoid(z_hidden)
+$$
+\begin{equation}
+  \begin{aligned}
+  -\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &= f(x_i) \\
+  -g_{i+1} + 2g_i - g_{i-1} &= \Delta x^2 f(x_i)
+  \end{aligned}
+\end{equation} \tag{16}
+$$
 
-        # Update x_prev such that next layer can use the output from this layer
-        x_prev = x_hidden
+
    for \( i = 1, \dots, N_x - 2 \) where \( g_0 = g_{N_x - 1} = 0 \) and \( f(x_i) = (3x_i + x_i^2)\exp(x_i) \), which is given for our specific problem.
    
 
-    ## Output layer:
+
    The equation can be rewritten into a matrix equation:
    
 
-    # Get the weights and bias for this layer
-    w_output = deep_params[-1]
-
-    # Include bias:
-    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
-
-    z_output = np.matmul(w_output, x_prev)
-    x_output = z_output
-
-    return x_output
-
-def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
-    # num_hidden_neurons is now a list of number of neurons within each hidden layer
-
-    # Find the number of hidden layers:
-    N_hidden = np.size(num_neurons)
-
-    ## Set up initial weigths and biases
-
-    # Initialize the list of parameters:
-    P = [None]*(N_hidden + 1) # + 1 to include the output layer
-
-    P[0] = npr.randn(num_neurons[0], 2 )
-    for l in range(1,N_hidden):
-        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
-
-    # For the output layer
-    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
-
-    print('Initial cost: %g'%cost_function_deep(P, x))
-
-    ## Start finding the optimal weigths using gradient descent
-
-    # Find the Python function that represents the gradient of the cost function
-    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
-    cost_function_deep_grad = grad(cost_function_deep,0)
-
-    # Let the update be done num_iter times
-    for i in range(num_iter):
-        # Evaluate the gradient at the current weights and biases in P.
-        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
-        # in the hidden layers and output layers evaluated at x.
-        cost_deep_grad =  cost_function_deep_grad(P, x)
-
-        for l in range(N_hidden+1):
-            P[l] = P[l] - lmb * cost_deep_grad[l]
-
-    print('Final cost: %g'%cost_function_deep(P, x))
-
-    return P
-
-## Set up the cost function specified for this Poisson equation:
-
-# The right side of the ODE
-def f(x):
-    return (3*x + x**2)*np.exp(x)
-
-def cost_function_deep(P, x):
-
-    # Evaluate the trial function with the current parameters P
-    g_t = g_trial_deep(x,P)
-
-    # Find the derivative w.r.t x of the trial function
-    d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)
-
-    right_side = f(x)
-
-    err_sqr = (-d2_g_t - right_side)**2
-    cost_sum = np.sum(err_sqr)
-
-    return cost_sum/np.size(err_sqr)
-
-# The trial solution:
-def g_trial_deep(x,P):
-    return x*(1-x)*deep_neural_network(P,x)
-
-# The analytic solution;
-def g_analytic(x):
-    return x*(1-x)*np.exp(x)
-
-if __name__ == '__main__':
-    npr.seed(4155)
-
-    ## Decide the vales of arguments to the function to solve
-    Nx = 10
-    x = np.linspace(0,1, Nx)
-
-    ## Set up the initial parameters
-    num_hidden_neurons = [200,100]
-    num_iter = 1000
-    lmb = 1e-3
-
-    P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)
-
-    g_dnn_ag = g_trial_deep(x,P)
-    g_analytical = g_analytic(x)
-
-    # Find the maximum absolute difference between the solutons:
-
-    plt.figure(figsize=(10,10))
-
-    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
-    plt.plot(x, g_analytical)
-    plt.plot(x, g_dnn_ag[0,:])
-    plt.legend(['analytical','nn'])
-    plt.xlabel('x')
-    plt.ylabel('g(x)')
-
-    ## Perform the computation using the numerical scheme
-
-    dx = 1/(Nx - 1)
-
-    # Set up the matrix A
-    A = np.zeros((Nx-2,Nx-2))
-
-    A[0,0] = 2
-    A[0,1] = -1
-
-    for i in range(1,Nx-3):
-        A[i,i-1] = -1
-        A[i,i] = 2
-        A[i,i+1] = -1
-
-    A[Nx - 3, Nx - 4] = -1
-    A[Nx - 3, Nx - 3] = 2
-
-    # Set up the vector f
-    f_vec = dx**2 * f(x[1:-1])
-
-    # Solve the equation
-    g_res = np.linalg.solve(A,f_vec)
-
-    g_vec = np.zeros(Nx)
-    g_vec[1:-1] = g_res
-
-    # Print the differences between each method
-    max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical))
-    max_diff2 = np.max(np.abs(g_vec - g_analytical))
-    print("The max absolute difference between the analytical solution and DNN Autograd: %g"%max_diff1)
-    print("The max absolute difference between the analytical solution and numerical scheme: %g"%max_diff2)
-
-    # Plot the results
-    plt.figure(figsize=(10,10))
-
-    plt.plot(x,g_vec)
-    plt.plot(x,g_analytical)
-    plt.plot(x,g_dnn_ag[0,:])
-
-    plt.legend(['numerical scheme','analytical','dnn'])
-    plt.show()
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    +$$ +\begin{aligned} +\begin{pmatrix} +2 & -1 & 0 & \dots & 0 \\ +-1 & 2 & -1 & \dots & 0 \\ +\vdots & & \ddots & & \vdots \\ +0 & \dots & -1 & 2 & -1 \\ +0 & \dots & 0 & -1 & 2\\ +\end{pmatrix} +\begin{pmatrix} +g_1 \\ +g_2 \\ +\vdots \\ +g_{N_x - 3} \\ +g_{N_x - 2} +\end{pmatrix} +&= +\Delta x^2 +\begin{pmatrix} +f(x_1) \\ +f(x_2) \\ +\vdots \\ +f(x_{N_x - 3}) \\ +f(x_{N_x - 2}) +\end{pmatrix} \\ +\boldsymbol{A}\boldsymbol{g} &= \boldsymbol{f}, +\end{aligned} +$$ +

    which makes it possible to solve for the vector \( \boldsymbol{g} \).

    @@ -641,7 +516,7 @@ MathJax.Hub.Config({

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    • diff --git a/doc/pub/week42/html/._week42-bs035.html b/doc/pub/week42/html/._week42-bs035.html index a4e122316..948118068 100644 --- a/doc/pub/week42/html/._week42-bs035.html +++ b/doc/pub/week42/html/._week42-bs035.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,23 +404,222 @@ MathJax.Hub.Config({

     

     

     

    -

    Partial Differential Equations

    +

    Setting up the code

    -

    A partial differential equation (PDE) has a solution here the function -is defined by multiple variables. The equation may involve all kinds -of combinations of which variables the function is differentiated with -respect to. -

    +

    We can then compare the result from this numerical scheme with the output from our network using Autograd:

    -

    In general, a partial differential equation for a function \( g(x_1,\dots,x_N) \) with \( N \) variables may be expressed as

    -$$ -\begin{equation} \tag{17} - f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0 -\end{equation} -$$ + +
    +
    +
    +
    +
    + 
    import autograd.numpy as np
+from autograd import grad, elementwise_grad
+import autograd.numpy.random as npr
+from matplotlib import pyplot as plt
+
+def sigmoid(z):
+    return 1/(1 + np.exp(-z))
+
+def deep_neural_network(deep_params, x):
+    # N_hidden is the number of hidden layers
+    N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
+
+    # Assumes input x being an one-dimensional array
+    num_values = np.size(x)
+    x = x.reshape(-1, num_values)
+
+    # Assume that the input layer does nothing to the input x
+    x_input = x
+
+    # Due to multiple hidden layers, define a variable referencing to the
+    # output of the previous layer:
+    x_prev = x_input
+
+    ## Hidden layers:
+
+    for l in range(N_hidden):
+        # From the list of parameters P; find the correct weigths and bias for this layer
+        w_hidden = deep_params[l]
+
+        # Add a row of ones to include bias
+        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
+
+        z_hidden = np.matmul(w_hidden, x_prev)
+        x_hidden = sigmoid(z_hidden)
+
+        # Update x_prev such that next layer can use the output from this layer
+        x_prev = x_hidden
+
+    ## Output layer:
+
+    # Get the weights and bias for this layer
+    w_output = deep_params[-1]
+
+    # Include bias:
+    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
+
+    z_output = np.matmul(w_output, x_prev)
+    x_output = z_output
+
+    return x_output
+
+def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
+    # num_hidden_neurons is now a list of number of neurons within each hidden layer
+
+    # Find the number of hidden layers:
+    N_hidden = np.size(num_neurons)
+
+    ## Set up initial weigths and biases
+
+    # Initialize the list of parameters:
+    P = [None]*(N_hidden + 1) # + 1 to include the output layer
+
+    P[0] = npr.randn(num_neurons[0], 2 )
+    for l in range(1,N_hidden):
+        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
+
+    # For the output layer
+    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
+
+    print('Initial cost: %g'%cost_function_deep(P, x))
+
+    ## Start finding the optimal weigths using gradient descent
+
+    # Find the Python function that represents the gradient of the cost function
+    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
+    cost_function_deep_grad = grad(cost_function_deep,0)
+
+    # Let the update be done num_iter times
+    for i in range(num_iter):
+        # Evaluate the gradient at the current weights and biases in P.
+        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
+        # in the hidden layers and output layers evaluated at x.
+        cost_deep_grad =  cost_function_deep_grad(P, x)
+
+        for l in range(N_hidden+1):
+            P[l] = P[l] - lmb * cost_deep_grad[l]
+
+    print('Final cost: %g'%cost_function_deep(P, x))
+
+    return P
+
+## Set up the cost function specified for this Poisson equation:
+
+# The right side of the ODE
+def f(x):
+    return (3*x + x**2)*np.exp(x)
+
+def cost_function_deep(P, x):
+
+    # Evaluate the trial function with the current parameters P
+    g_t = g_trial_deep(x,P)
+
+    # Find the derivative w.r.t x of the trial function
+    d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)
+
+    right_side = f(x)
+
+    err_sqr = (-d2_g_t - right_side)**2
+    cost_sum = np.sum(err_sqr)
+
+    return cost_sum/np.size(err_sqr)
+
+# The trial solution:
+def g_trial_deep(x,P):
+    return x*(1-x)*deep_neural_network(P,x)
+
+# The analytic solution;
+def g_analytic(x):
+    return x*(1-x)*np.exp(x)
+
+if __name__ == '__main__':
+    npr.seed(4155)
+
+    ## Decide the vales of arguments to the function to solve
+    Nx = 10
+    x = np.linspace(0,1, Nx)
+
+    ## Set up the initial parameters
+    num_hidden_neurons = [200,100]
+    num_iter = 1000
+    lmb = 1e-3
+
+    P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)
+
+    g_dnn_ag = g_trial_deep(x,P)
+    g_analytical = g_analytic(x)
+
+    # Find the maximum absolute difference between the solutons:
+
+    plt.figure(figsize=(10,10))
+
+    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
+    plt.plot(x, g_analytical)
+    plt.plot(x, g_dnn_ag[0,:])
+    plt.legend(['analytical','nn'])
+    plt.xlabel('x')
+    plt.ylabel('g(x)')
+
+    ## Perform the computation using the numerical scheme
+
+    dx = 1/(Nx - 1)
+
+    # Set up the matrix A
+    A = np.zeros((Nx-2,Nx-2))
+
+    A[0,0] = 2
+    A[0,1] = -1
+
+    for i in range(1,Nx-3):
+        A[i,i-1] = -1
+        A[i,i] = 2
+        A[i,i+1] = -1
+
+    A[Nx - 3, Nx - 4] = -1
+    A[Nx - 3, Nx - 3] = 2
+
+    # Set up the vector f
+    f_vec = dx**2 * f(x[1:-1])
+
+    # Solve the equation
+    g_res = np.linalg.solve(A,f_vec)
+
+    g_vec = np.zeros(Nx)
+    g_vec[1:-1] = g_res
+
+    # Print the differences between each method
+    max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical))
+    max_diff2 = np.max(np.abs(g_vec - g_analytical))
+    print("The max absolute difference between the analytical solution and DNN Autograd: %g"%max_diff1)
+    print("The max absolute difference between the analytical solution and numerical scheme: %g"%max_diff2)
+
+    # Plot the results
+    plt.figure(figsize=(10,10))
+
+    plt.plot(x,g_vec)
+    plt.plot(x,g_analytical)
+    plt.plot(x,g_dnn_ag[0,:])
+
+    plt.legend(['numerical scheme','analytical','dnn'])
+    plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    -

    where \( f \) is an expression involving all kinds of possible mixed derivatives of \( g(x_1,\dots,x_N) \) up to an order \( n \). In order for the solution to be unique, some additional conditions must also be given.

    @@ -442,7 +646,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs036.html b/doc/pub/week42/html/._week42-bs036.html index aa7233e7c..15ef7d918 100644 --- a/doc/pub/week42/html/._week42-bs036.html +++ b/doc/pub/week42/html/._week42-bs036.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,24 +404,23 @@ MathJax.Hub.Config({

     

     

     

    -

    Type of problem

    +

    Partial Differential Equations

    -

    The problem our network must solve for, is similar to the ODE case. -We must have a trial solution \( g_t \) at hand. +

    A partial differential equation (PDE) has a solution here the function +is defined by multiple variables. The equation may involve all kinds +of combinations of which variables the function is differentiated with +respect to.

    -

    For instance, the trial solution could be expressed as

    +

    In general, a partial differential equation for a function \( g(x_1,\dots,x_N) \) with \( N \) variables may be expressed as

    + $$ -\begin{align*} - g_t(x_1,\dots,x_N) = h_1(x_1,\dots,x_N) + h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) -\end{align*} +\begin{equation} \tag{17} + f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0 +\end{equation} $$ -

    where \( h_1(x_1,\dots,x_N) \) is a function that ensures \( g_t(x_1,\dots,x_N) \) satisfies some given conditions. -The neural network \( N(x_1,\dots,x_N,P) \) has weights and biases described by \( P \) and \( h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) \) is an expression using the output from the neural network in some way. -

    - -

    The role of the function \( h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) \), is to ensure that the output of \( N(x_1,\dots,x_N,P) \) is zero when \( g_t(x_1,\dots,x_N) \) is evaluated at the values of \( x_1,\dots,x_N \) where the given conditions must be satisfied. The function \( h_1(x_1,\dots,x_N) \) should alone make \( g_t(x_1,\dots,x_N) \) satisfy the conditions.

    +

    where \( f \) is an expression involving all kinds of possible mixed derivatives of \( g(x_1,\dots,x_N) \) up to an order \( n \). In order for the solution to be unique, some additional conditions must also be given.

    @@ -443,7 +447,7 @@ The neural network \( N(x_1,\dots,x_N,P) \) has weights and biases described by

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    • diff --git a/doc/pub/week42/html/._week42-bs037.html b/doc/pub/week42/html/._week42-bs037.html index 6a37e51d0..94b373791 100644 --- a/doc/pub/week42/html/._week42-bs037.html +++ b/doc/pub/week42/html/._week42-bs037.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,26 +404,24 @@ MathJax.Hub.Config({

     

     

     

    -

    Network requirements

    +

    Type of problem

    -

    The network tries then the minimize the cost function following the -same ideas as described for the ODE case, but now with more than one -variables to consider. The concept still remains the same; find a set -of parameters \( P \) such that the expression \( f \) in (17) is as -close to zero as possible. +

    The problem our network must solve for, is similar to the ODE case. +We must have a trial solution \( g_t \) at hand.

    -

    As for the ODE case, the cost function is the mean squared error that -the network must try to minimize. The cost function for the network to -minimize is +

    For instance, the trial solution could be expressed as

    +$$ +\begin{align*} + g_t(x_1,\dots,x_N) = h_1(x_1,\dots,x_N) + h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) +\end{align*} +$$ + +

    where \( h_1(x_1,\dots,x_N) \) is a function that ensures \( g_t(x_1,\dots,x_N) \) satisfies some given conditions. +The neural network \( N(x_1,\dots,x_N,P) \) has weights and biases described by \( P \) and \( h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) \) is an expression using the output from the neural network in some way.

    -$$ -\begin{equation*} -C\left(x_1, \dots, x_N, P\right) = \left( f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -$$ - +

    The role of the function \( h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) \), is to ensure that the output of \( N(x_1,\dots,x_N,P) \) is zero when \( g_t(x_1,\dots,x_N) \) is evaluated at the values of \( x_1,\dots,x_N \) where the given conditions must be satisfied. The function \( h_1(x_1,\dots,x_N) \) should alone make \( g_t(x_1,\dots,x_N) \) satisfy the conditions.

    @@ -445,7 +448,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs038.html b/doc/pub/week42/html/._week42-bs038.html index 1bb646ac8..082e6308e 100644 --- a/doc/pub/week42/html/._week42-bs038.html +++ b/doc/pub/week42/html/._week42-bs038.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,17 +404,23 @@ MathJax.Hub.Config({

     

     

     

    -

    More details

    +

    Network requirements

    -

    If we let \( \boldsymbol{x} = \big( x_1, \dots, x_N \big) \) be an array containing the values for \( x_1, \dots, x_N \) respectively, the cost function can be reformulated into the following:

    -$$ - C\left(\boldsymbol{x}, P\right) = f\left( \left( \boldsymbol{x}, \frac{\partial g(\boldsymbol{x}) }{\partial x_1}, \dots , \frac{\partial g(\boldsymbol{x}) }{\partial x_N}, \frac{\partial g(\boldsymbol{x}) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\boldsymbol{x}) }{\partial x_N^n} \right) \right)^2 -$$ +

    The network tries then the minimize the cost function following the +same ideas as described for the ODE case, but now with more than one +variables to consider. The concept still remains the same; find a set +of parameters \( P \) such that the expression \( f \) in (17) is as +close to zero as possible. +

    + +

    As for the ODE case, the cost function is the mean squared error that +the network must try to minimize. The cost function for the network to +minimize is +

    -

    If we also have \( M \) different sets of values for \( x_1, \dots, x_N \), that is \( \boldsymbol{x}_i = \big(x_1^{(i)}, \dots, x_N^{(i)}\big) \) for \( i = 1,\dots,M \) being the rows in matrix \( X \), the cost function can be generalized into

    $$ \begin{equation*} -C\left(X, P \right) = \sum_{i=1}^M f\left( \left( \boldsymbol{x}_i, \frac{\partial g(\boldsymbol{x}_i) }{\partial x_1}, \dots , \frac{\partial g(\boldsymbol{x}_i) }{\partial x_N}, \frac{\partial g(\boldsymbol{x}_i) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\boldsymbol{x}_i) }{\partial x_N^n} \right) \right)^2. +C\left(x_1, \dots, x_N, P\right) = \left( f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) \right)^2 \end{equation*} $$ @@ -439,7 +450,7 @@ $$
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    • diff --git a/doc/pub/week42/html/._week42-bs039.html b/doc/pub/week42/html/._week42-bs039.html index 575daed6c..99135c155 100644 --- a/doc/pub/week42/html/._week42-bs039.html +++ b/doc/pub/week42/html/._week42-bs039.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,25 +404,20 @@ MathJax.Hub.Config({

     

     

     

    -

    Example: The diffusion equation

    +

    More details

    -

    In one spatial dimension, the equation reads

    +

    If we let \( \boldsymbol{x} = \big( x_1, \dots, x_N \big) \) be an array containing the values for \( x_1, \dots, x_N \) respectively, the cost function can be reformulated into the following:

    +$$ + C\left(\boldsymbol{x}, P\right) = f\left( \left( \boldsymbol{x}, \frac{\partial g(\boldsymbol{x}) }{\partial x_1}, \dots , \frac{\partial g(\boldsymbol{x}) }{\partial x_N}, \frac{\partial g(\boldsymbol{x}) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\boldsymbol{x}) }{\partial x_N^n} \right) \right)^2 +$$ + +

    If we also have \( M \) different sets of values for \( x_1, \dots, x_N \), that is \( \boldsymbol{x}_i = \big(x_1^{(i)}, \dots, x_N^{(i)}\big) \) for \( i = 1,\dots,M \) being the rows in matrix \( X \), the cost function can be generalized into

    $$ \begin{equation*} - \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} +C\left(X, P \right) = \sum_{i=1}^M f\left( \left( \boldsymbol{x}_i, \frac{\partial g(\boldsymbol{x}_i) }{\partial x_1}, \dots , \frac{\partial g(\boldsymbol{x}_i) }{\partial x_N}, \frac{\partial g(\boldsymbol{x}_i) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\boldsymbol{x}_i) }{\partial x_N^n} \right) \right)^2. \end{equation*} $$ -

    where a possible choice of conditions are

    -$$ -\begin{align*} -g(0,t) &= 0 ,\qquad t \geq 0 \\ -g(1,t) &= 0, \qquad t \geq 0 \\ -g(x,0) &= u(x),\qquad x\in [0,1] -\end{align*} -$$ - -

    with \( u(x) \) being some given function.

    @@ -444,7 +444,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs040.html b/doc/pub/week42/html/._week42-bs040.html index 6642fad61..21cb2bc7e 100644 --- a/doc/pub/week42/html/._week42-bs040.html +++ b/doc/pub/week42/html/._week42-bs040.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,18 +404,16 @@ MathJax.Hub.Config({

     

     

     

    -

    Defining the problem

    - -

    For this case, we want to find \( g(x,t) \) such that

    +

    Example: The diffusion equation

    +

    In one spatial dimension, the equation reads

    $$ -\begin{equation} +\begin{equation*} \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation} \tag{18} +\end{equation*} $$ -

    and

    - +

    where a possible choice of conditions are

    $$ \begin{align*} g(0,t) &= 0 ,\qquad t \geq 0 \\ @@ -419,12 +422,7 @@ g(x,0) &= u(x),\qquad x\in [0,1] \end{align*} $$ -

    with \( u(x) = \sin(\pi x) \).

    - -

    First, let us set up the deep neural network. -The deep neural network will follow the same structure as discussed in the examples solving the ODEs. -First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions. -

    +

    with \( u(x) \) being some given function.

    @@ -451,7 +449,7 @@ First, we will look into how Autograd could be used in a network tailored to sol

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    • diff --git a/doc/pub/week42/html/._week42-bs041.html b/doc/pub/week42/html/._week42-bs041.html index e3ec9e8a1..823888eaa 100644 --- a/doc/pub/week42/html/._week42-bs041.html +++ b/doc/pub/week42/html/._week42-bs041.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,84 +404,33 @@ MathJax.Hub.Config({

     

     

     

    -

    Setting up the network using Autograd

    +

    Defining the problem

    -

    The only change to do here, is to extend our network such that -functions of multiple parameters are correctly handled. In this case -we have two variables in our function to solve for, that is time \( t \) -and position \( x \). The variables will be represented by a -one-dimensional array in the program. The program will evaluate the -network at each possible pair \( (x,t) \), given an array for the desired -\( x \)-values and \( t \)-values to approximate the solution at. +

    For this case, we want to find \( g(x,t) \) such that

    + +$$ +\begin{equation} + \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} +\end{equation} \tag{18} +$$ + +

    and

    + +$$ +\begin{align*} +g(0,t) &= 0 ,\qquad t \geq 0 \\ +g(1,t) &= 0, \qquad t \geq 0 \\ +g(x,0) &= u(x),\qquad x\in [0,1] +\end{align*} +$$ + +

    with \( u(x) = \sin(\pi x) \).

    + +

    First, let us set up the deep neural network. +The deep neural network will follow the same structure as discussed in the examples solving the ODEs. +First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions.

    - - -
    -
    -
    -
    -
    - 
    def sigmoid(z):
-    return 1/(1 + np.exp(-z))
-
-def deep_neural_network(deep_params, x):
-    # x is now a point and a 1D numpy array; make it a column vector
-    num_coordinates = np.size(x,0)
-    x = x.reshape(num_coordinates,-1)
-
-    num_points = np.size(x,1)
-
-    # N_hidden is the number of hidden layers
-    N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
-
-    # Assume that the input layer does nothing to the input x
-    x_input = x
-    x_prev = x_input
-
-    ## Hidden layers:
-
-    for l in range(N_hidden):
-        # From the list of parameters P; find the correct weigths and bias for this layer
-        w_hidden = deep_params[l]
-
-        # Add a row of ones to include bias
-        x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)
-
-        z_hidden = np.matmul(w_hidden, x_prev)
-        x_hidden = sigmoid(z_hidden)
-
-        # Update x_prev such that next layer can use the output from this layer
-        x_prev = x_hidden
-
-    ## Output layer:
-
-    # Get the weights and bias for this layer
-    w_output = deep_params[-1]
-
-    # Include bias:
-    x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)
-
-    z_output = np.matmul(w_output, x_prev)
-    x_output = z_output
-
-    return x_output[0][0]
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - -

      @@ -502,7 +456,7 @@ network at each possible pair \( (x,t) \), given an array for the desired
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    diff --git a/doc/pub/week42/html/._week42-bs042.html b/doc/pub/week42/html/._week42-bs042.html index 73f075ab6..181ad62a4 100644 --- a/doc/pub/week42/html/._week42-bs042.html +++ b/doc/pub/week42/html/._week42-bs042.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,30 +404,83 @@ MathJax.Hub.Config({

     

     

     

    -

    Setting up the network using Autograd; The trial solution

    +

    Setting up the network using Autograd

    -

    The cost function must then iterate through the given arrays -containing values for \( x \) and \( t \), defines a point \( (x,t) \) the deep -neural network and the trial solution is evaluated at, and then finds -the Jacobian of the trial solution. +

    The only change to do here, is to extend our network such that +functions of multiple parameters are correctly handled. In this case +we have two variables in our function to solve for, that is time \( t \) +and position \( x \). The variables will be represented by a +one-dimensional array in the program. The program will evaluate the +network at each possible pair \( (x,t) \), given an array for the desired +\( x \)-values and \( t \)-values to approximate the solution at.

    -

    A possible trial solution for this PDE is

    -

    $$ -g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P) -$$ -

    + +
    +
    +
    +
    +
    + 
    def sigmoid(z):
+    return 1/(1 + np.exp(-z))
 
-
    with \( A(x,t) \) being a function ensuring that \( g_t(x,t) \) satisfies our given conditions, and \( N(x,t,P) \) being the output from the deep neural network using weights and biases for each layer from \( P \).
    
+def deep_neural_network(deep_params, x):
+    # x is now a point and a 1D numpy array; make it a column vector
+    num_coordinates = np.size(x,0)
+    x = x.reshape(num_coordinates,-1)
 
-
    To fulfill the conditions, \( A(x,t) \) could be:
    
+    num_points = np.size(x,1)
+
+    # N_hidden is the number of hidden layers
+    N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
+
+    # Assume that the input layer does nothing to the input x
+    x_input = x
+    x_prev = x_input
+
+    ## Hidden layers:
+
+    for l in range(N_hidden):
+        # From the list of parameters P; find the correct weigths and bias for this layer
+        w_hidden = deep_params[l]
+
+        # Add a row of ones to include bias
+        x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)
+
+        z_hidden = np.matmul(w_hidden, x_prev)
+        x_hidden = sigmoid(z_hidden)
+
+        # Update x_prev such that next layer can use the output from this layer
+        x_prev = x_hidden
+
+    ## Output layer:
+
+    # Get the weights and bias for this layer
+    w_output = deep_params[-1]
+
+    # Include bias:
+    x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)
+
+    z_output = np.matmul(w_output, x_prev)
+    x_output = z_output
+
+    return x_output[0][0]
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    -

    $$ -h_1(x,t) = (1-t)\Big(u(x) - \big((1-x)u(0) + x u(1)\big)\Big) = (1-t)u(x) = (1-t)\sin(\pi x) -$$ -since \( (0) = u(1) = 0 \) and \( u(x) = \sin(\pi x) \). -

    @@ -449,7 +507,7 @@ since \( (0) = u(1) = 0 \) and \( u(x) = \sin(\pi x) \).

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    -

    Why the jacobian?

    +

    Setting up the network using Autograd; The trial solution

    -

    The Jacobian is used because the program must find the derivative of -the trial solution with respect to \( x \) and \( t \). +

    The cost function must then iterate through the given arrays +containing values for \( x \) and \( t \), defines a point \( (x,t) \) the deep +neural network and the trial solution is evaluated at, and then finds +the Jacobian of the trial solution.

    -

    This gives the necessity of computing the Jacobian matrix, as we want -to evaluate the gradient with respect to \( x \) and \( t \) (note that the -Jacobian of a scalar-valued multivariate function is simply its -gradient). +

    A possible trial solution for this PDE is

    + +

    $$ +g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P) +$$

    -

    In Autograd, the differentiation is by default done with respect to -the first input argument of your Python function. Since the points is -an array representing \( x \) and \( t \), the Jacobian is calculated using -the values of \( x \) and \( t \). +

    with \( A(x,t) \) being a function ensuring that \( g_t(x,t) \) satisfies our given conditions, and \( N(x,t,P) \) being the output from the deep neural network using weights and biases for each layer from \( P \).

    + +

    To fulfill the conditions, \( A(x,t) \) could be:

    + +

    $$ +h_1(x,t) = (1-t)\Big(u(x) - \big((1-x)u(0) + x u(1)\big)\Big) = (1-t)u(x) = (1-t)\sin(\pi x) +$$ +since \( (0) = u(1) = 0 \) and \( u(x) = \sin(\pi x) \).

    -

    To find the second derivative with respect to \( x \) and \( t \), the -Jacobian can be found for the second time. The result is a Hessian -matrix, which is the matrix containing all the possible second order -mixed derivatives of \( g(x,t) \). -

    - - - -
    -
    -
    -
    -
    - 
    # Set up the trial function:
-def u(x):
-    return np.sin(np.pi*x)
-
-def g_trial(point,P):
-    x,t = point
-    return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)
-
-# The right side of the ODE:
-def f(point):
-    return 0.
-
-# The cost function:
-def cost_function(P, x, t):
-    cost_sum = 0
-
-    g_t_jacobian_func = jacobian(g_trial)
-    g_t_hessian_func = hessian(g_trial)
-
-    for x_ in x:
-        for t_ in t:
-            point = np.array([x_,t_])
-
-            g_t = g_trial(point,P)
-            g_t_jacobian = g_t_jacobian_func(point,P)
-            g_t_hessian = g_t_hessian_func(point,P)
-
-            g_t_dt = g_t_jacobian[1]
-            g_t_d2x = g_t_hessian[0][0]
-
-            func = f(point)
-
-            err_sqr = ( (g_t_dt - g_t_d2x) - func)**2
-            cost_sum += err_sqr
-
-    return cost_sum
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - -

      @@ -507,7 +454,7 @@ mixed derivatives of \( g(x,t) \).
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    diff --git a/doc/pub/week42/html/._week42-bs044.html b/doc/pub/week42/html/._week42-bs044.html index 70b032d86..82a006009 100644 --- a/doc/pub/week42/html/._week42-bs044.html +++ b/doc/pub/week42/html/._week42-bs044.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,24 +404,28 @@ MathJax.Hub.Config({

     

     

     

    -

    Setting up the network using Autograd; The full program

    +

    Why the jacobian?

    -

    Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution.

    - -

    The analytical solution of our problem is

    - -

    $$ -g(x,t) = \exp(-\pi^2 t)\sin(\pi x) -$$ +

    The Jacobian is used because the program must find the derivative of +the trial solution with respect to \( x \) and \( t \).

    -

    A possible way to implement a neural network solving the PDE, is given below. -Be aware, though, that it is fairly slow for the parameters used. -A better result is possible, but requires more iterations, and thus longer time to complete. +

    This gives the necessity of computing the Jacobian matrix, as we want +to evaluate the gradient with respect to \( x \) and \( t \) (note that the +Jacobian of a scalar-valued multivariate function is simply its +gradient).

    -

    Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE. -Using TensorFlow results in a much better execution time. Try it! +

    In Autograd, the differentiation is by default done with respect to +the first input argument of your Python function. Since the points is +an array representing \( x \) and \( t \), the Jacobian is calculated using +the values of \( x \) and \( t \). +

    + +

    To find the second derivative with respect to \( x \) and \( t \), the +Jacobian can be found for the second time. The result is a Hessian +matrix, which is the matrix containing all the possible second order +mixed derivatives of \( g(x,t) \).

    @@ -426,61 +435,7 @@ Using TensorFlow results in a much better execution time. Try it!
    - 
    import autograd.numpy as np
-from autograd import jacobian,hessian,grad
-import autograd.numpy.random as npr
-from matplotlib import cm
-from matplotlib import pyplot as plt
-from mpl_toolkits.mplot3d import axes3d
-
-## Set up the network
-
-def sigmoid(z):
-    return 1/(1 + np.exp(-z))
-
-def deep_neural_network(deep_params, x):
-    # x is now a point and a 1D numpy array; make it a column vector
-    num_coordinates = np.size(x,0)
-    x = x.reshape(num_coordinates,-1)
-
-    num_points = np.size(x,1)
-
-    # N_hidden is the number of hidden layers
-    N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
-
-    # Assume that the input layer does nothing to the input x
-    x_input = x
-    x_prev = x_input
-
-    ## Hidden layers:
-
-    for l in range(N_hidden):
-        # From the list of parameters P; find the correct weigths and bias for this layer
-        w_hidden = deep_params[l]
-
-        # Add a row of ones to include bias
-        x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)
-
-        z_hidden = np.matmul(w_hidden, x_prev)
-        x_hidden = sigmoid(z_hidden)
-
-        # Update x_prev such that next layer can use the output from this layer
-        x_prev = x_hidden
-
-    ## Output layer:
-
-    # Get the weights and bias for this layer
-    w_output = deep_params[-1]
-
-    # Include bias:
-    x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)
-
-    z_output = np.matmul(w_output, x_prev)
-    x_output = z_output
-
-    return x_output[0][0]
-
-## Define the trial solution and cost function
+  
    # Set up the trial function:
 def u(x):
     return np.sin(np.pi*x)
 
@@ -515,140 +470,7 @@ Using TensorFlow results in a much better execution time. Try it!
             err_sqr = ( (g_t_dt - g_t_d2x) - func)**2
             cost_sum += err_sqr
 
-    return cost_sum /( np.size(x)*np.size(t) )
-
-## For comparison, define the analytical solution
-def g_analytic(point):
-    x,t = point
-    return np.exp(-np.pi**2*t)*np.sin(np.pi*x)
-
-## Set up a function for training the network to solve for the equation
-def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):
-    ## Set up initial weigths and biases
-    N_hidden = np.size(num_neurons)
-
-    ## Set up initial weigths and biases
-
-    # Initialize the list of parameters:
-    P = [None]*(N_hidden + 1) # + 1 to include the output layer
-
-    P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias
-    for l in range(1,N_hidden):
-        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
-
-    # For the output layer
-    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
-
-    print('Initial cost: ',cost_function(P, x, t))
-
-    cost_function_grad = grad(cost_function,0)
-
-    # Let the update be done num_iter times
-    for i in range(num_iter):
-        cost_grad =  cost_function_grad(P, x , t)
-
-        for l in range(N_hidden+1):
-            P[l] = P[l] - lmb * cost_grad[l]
-
-    print('Final cost: ',cost_function(P, x, t))
-
-    return P
-
-if __name__ == '__main__':
-    ### Use the neural network:
-    npr.seed(15)
-
-    ## Decide the vales of arguments to the function to solve
-    Nx = 10; Nt = 10
-    x = np.linspace(0, 1, Nx)
-    t = np.linspace(0,1,Nt)
-
-    ## Set up the parameters for the network
-    num_hidden_neurons = [100, 25]
-    num_iter = 250
-    lmb = 0.01
-
-    P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)
-
-    ## Store the results
-    g_dnn_ag = np.zeros((Nx, Nt))
-    G_analytical = np.zeros((Nx, Nt))
-    for i,x_ in enumerate(x):
-        for j, t_ in enumerate(t):
-            point = np.array([x_, t_])
-            g_dnn_ag[i,j] = g_trial(point,P)
-
-            G_analytical[i,j] = g_analytic(point)
-
-    # Find the map difference between the analytical and the computed solution
-    diff_ag = np.abs(g_dnn_ag - G_analytical)
-    print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag))
-
-    ## Plot the solutions in two dimensions, that being in position and time
-
-    T,X = np.meshgrid(t,x)
-
-    fig = plt.figure(figsize=(10,10))
-    ax = fig.gca(projection='3d')
-    ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))
-    s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis)
-    ax.set_xlabel('Time $t$')
-    ax.set_ylabel('Position $x$');
-
-
-    fig = plt.figure(figsize=(10,10))
-    ax = fig.gca(projection='3d')
-    ax.set_title('Analytical solution')
-    s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)
-    ax.set_xlabel('Time $t$')
-    ax.set_ylabel('Position $x$');
-
-    fig = plt.figure(figsize=(10,10))
-    ax = fig.gca(projection='3d')
-    ax.set_title('Difference')
-    s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis)
-    ax.set_xlabel('Time $t$')
-    ax.set_ylabel('Position $x$');
-
-    ## Take some slices of the 3D plots just to see the solutions at particular times
-    indx1 = 0
-    indx2 = int(Nt/2)
-    indx3 = Nt-1
-
-    t1 = t[indx1]
-    t2 = t[indx2]
-    t3 = t[indx3]
-
-    # Slice the results from the DNN
-    res1 = g_dnn_ag[:,indx1]
-    res2 = g_dnn_ag[:,indx2]
-    res3 = g_dnn_ag[:,indx3]
-
-    # Slice the analytical results
-    res_analytical1 = G_analytical[:,indx1]
-    res_analytical2 = G_analytical[:,indx2]
-    res_analytical3 = G_analytical[:,indx3]
-
-    # Plot the slices
-    plt.figure(figsize=(10,10))
-    plt.title("Computed solutions at time = %g"%t1)
-    plt.plot(x, res1)
-    plt.plot(x,res_analytical1)
-    plt.legend(['dnn','analytical'])
-
-    plt.figure(figsize=(10,10))
-    plt.title("Computed solutions at time = %g"%t2)
-    plt.plot(x, res2)
-    plt.plot(x,res_analytical2)
-    plt.legend(['dnn','analytical'])
-
-    plt.figure(figsize=(10,10))
-    plt.title("Computed solutions at time = %g"%t3)
-    plt.plot(x, res3)
-    plt.plot(x,res_analytical3)
-    plt.legend(['dnn','analytical'])
-
-    plt.show()
+    return cost_sum
    @@ -690,7 +512,7 @@ Using TensorFlow results in a much better execution time. Try it!
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    • diff --git a/doc/pub/week42/html/._week42-bs045.html b/doc/pub/week42/html/._week42-bs045.html index eccd75c2f..69324d757 100644 --- a/doc/pub/week42/html/._week42-bs045.html +++ b/doc/pub/week42/html/._week42-bs045.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,28 +404,271 @@ MathJax.Hub.Config({

     

     

     

    -

    Example: Solving the wave equation with Neural Networks

    +

    Setting up the network using Autograd; The full program

    -

    The wave equation is

    -$$ -\begin{equation*} - \frac{\partial^2 g(x,t)}{\partial t^2} = c^2\frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation*} -$$ +

    Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution.

    -

    with \( c \) being the specified wave speed.

    +

    The analytical solution of our problem is

    -

    Here, the chosen conditions are

    -$$ -\begin{align*} - g(0,t) &= 0 \\ - g(1,t) &= 0 \\ - g(x,0) &= u(x) \\ - \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} &= v(x) -\end{align*} +

    $$ +g(x,t) = \exp(-\pi^2 t)\sin(\pi x) $$ +

    + +

    A possible way to implement a neural network solving the PDE, is given below. +Be aware, though, that it is fairly slow for the parameters used. +A better result is possible, but requires more iterations, and thus longer time to complete. +

    + +

    Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE. +Using TensorFlow results in a much better execution time. Try it! +

    + + + +
    +
    +
    +
    +
    + 
    import autograd.numpy as np
+from autograd import jacobian,hessian,grad
+import autograd.numpy.random as npr
+from matplotlib import cm
+from matplotlib import pyplot as plt
+from mpl_toolkits.mplot3d import axes3d
+
+## Set up the network
+
+def sigmoid(z):
+    return 1/(1 + np.exp(-z))
+
+def deep_neural_network(deep_params, x):
+    # x is now a point and a 1D numpy array; make it a column vector
+    num_coordinates = np.size(x,0)
+    x = x.reshape(num_coordinates,-1)
+
+    num_points = np.size(x,1)
+
+    # N_hidden is the number of hidden layers
+    N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
+
+    # Assume that the input layer does nothing to the input x
+    x_input = x
+    x_prev = x_input
+
+    ## Hidden layers:
+
+    for l in range(N_hidden):
+        # From the list of parameters P; find the correct weigths and bias for this layer
+        w_hidden = deep_params[l]
+
+        # Add a row of ones to include bias
+        x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)
+
+        z_hidden = np.matmul(w_hidden, x_prev)
+        x_hidden = sigmoid(z_hidden)
+
+        # Update x_prev such that next layer can use the output from this layer
+        x_prev = x_hidden
+
+    ## Output layer:
+
+    # Get the weights and bias for this layer
+    w_output = deep_params[-1]
+
+    # Include bias:
+    x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)
+
+    z_output = np.matmul(w_output, x_prev)
+    x_output = z_output
+
+    return x_output[0][0]
+
+## Define the trial solution and cost function
+def u(x):
+    return np.sin(np.pi*x)
+
+def g_trial(point,P):
+    x,t = point
+    return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)
+
+# The right side of the ODE:
+def f(point):
+    return 0.
+
+# The cost function:
+def cost_function(P, x, t):
+    cost_sum = 0
+
+    g_t_jacobian_func = jacobian(g_trial)
+    g_t_hessian_func = hessian(g_trial)
+
+    for x_ in x:
+        for t_ in t:
+            point = np.array([x_,t_])
+
+            g_t = g_trial(point,P)
+            g_t_jacobian = g_t_jacobian_func(point,P)
+            g_t_hessian = g_t_hessian_func(point,P)
+
+            g_t_dt = g_t_jacobian[1]
+            g_t_d2x = g_t_hessian[0][0]
+
+            func = f(point)
+
+            err_sqr = ( (g_t_dt - g_t_d2x) - func)**2
+            cost_sum += err_sqr
+
+    return cost_sum /( np.size(x)*np.size(t) )
+
+## For comparison, define the analytical solution
+def g_analytic(point):
+    x,t = point
+    return np.exp(-np.pi**2*t)*np.sin(np.pi*x)
+
+## Set up a function for training the network to solve for the equation
+def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):
+    ## Set up initial weigths and biases
+    N_hidden = np.size(num_neurons)
+
+    ## Set up initial weigths and biases
+
+    # Initialize the list of parameters:
+    P = [None]*(N_hidden + 1) # + 1 to include the output layer
+
+    P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias
+    for l in range(1,N_hidden):
+        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
+
+    # For the output layer
+    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
+
+    print('Initial cost: ',cost_function(P, x, t))
+
+    cost_function_grad = grad(cost_function,0)
+
+    # Let the update be done num_iter times
+    for i in range(num_iter):
+        cost_grad =  cost_function_grad(P, x , t)
+
+        for l in range(N_hidden+1):
+            P[l] = P[l] - lmb * cost_grad[l]
+
+    print('Final cost: ',cost_function(P, x, t))
+
+    return P
+
+if __name__ == '__main__':
+    ### Use the neural network:
+    npr.seed(15)
+
+    ## Decide the vales of arguments to the function to solve
+    Nx = 10; Nt = 10
+    x = np.linspace(0, 1, Nx)
+    t = np.linspace(0,1,Nt)
+
+    ## Set up the parameters for the network
+    num_hidden_neurons = [100, 25]
+    num_iter = 250
+    lmb = 0.01
+
+    P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)
+
+    ## Store the results
+    g_dnn_ag = np.zeros((Nx, Nt))
+    G_analytical = np.zeros((Nx, Nt))
+    for i,x_ in enumerate(x):
+        for j, t_ in enumerate(t):
+            point = np.array([x_, t_])
+            g_dnn_ag[i,j] = g_trial(point,P)
+
+            G_analytical[i,j] = g_analytic(point)
+
+    # Find the map difference between the analytical and the computed solution
+    diff_ag = np.abs(g_dnn_ag - G_analytical)
+    print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag))
+
+    ## Plot the solutions in two dimensions, that being in position and time
+
+    T,X = np.meshgrid(t,x)
+
+    fig = plt.figure(figsize=(10,10))
+    ax = fig.gca(projection='3d')
+    ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))
+    s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis)
+    ax.set_xlabel('Time $t$')
+    ax.set_ylabel('Position $x$');
+
+
+    fig = plt.figure(figsize=(10,10))
+    ax = fig.gca(projection='3d')
+    ax.set_title('Analytical solution')
+    s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)
+    ax.set_xlabel('Time $t$')
+    ax.set_ylabel('Position $x$');
+
+    fig = plt.figure(figsize=(10,10))
+    ax = fig.gca(projection='3d')
+    ax.set_title('Difference')
+    s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis)
+    ax.set_xlabel('Time $t$')
+    ax.set_ylabel('Position $x$');
+
+    ## Take some slices of the 3D plots just to see the solutions at particular times
+    indx1 = 0
+    indx2 = int(Nt/2)
+    indx3 = Nt-1
+
+    t1 = t[indx1]
+    t2 = t[indx2]
+    t3 = t[indx3]
+
+    # Slice the results from the DNN
+    res1 = g_dnn_ag[:,indx1]
+    res2 = g_dnn_ag[:,indx2]
+    res3 = g_dnn_ag[:,indx3]
+
+    # Slice the analytical results
+    res_analytical1 = G_analytical[:,indx1]
+    res_analytical2 = G_analytical[:,indx2]
+    res_analytical3 = G_analytical[:,indx3]
+
+    # Plot the slices
+    plt.figure(figsize=(10,10))
+    plt.title("Computed solutions at time = %g"%t1)
+    plt.plot(x, res1)
+    plt.plot(x,res_analytical1)
+    plt.legend(['dnn','analytical'])
+
+    plt.figure(figsize=(10,10))
+    plt.title("Computed solutions at time = %g"%t2)
+    plt.plot(x, res2)
+    plt.plot(x,res_analytical2)
+    plt.legend(['dnn','analytical'])
+
+    plt.figure(figsize=(10,10))
+    plt.title("Computed solutions at time = %g"%t3)
+    plt.plot(x, res3)
+    plt.plot(x,res_analytical3)
+    plt.legend(['dnn','analytical'])
+
+    plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    -

    where \( \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} \) means the derivative of \( g(x,t) \) with respect to \( t \) is evaluated at \( t = 0 \), and \( u(x) \) and \( v(x) \) being given functions.

    @@ -447,7 +695,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs046.html b/doc/pub/week42/html/._week42-bs046.html index 8f02c7dbb..ee4636ac3 100644 --- a/doc/pub/week42/html/._week42-bs046.html +++ b/doc/pub/week42/html/._week42-bs046.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,30 +404,28 @@ MathJax.Hub.Config({

     

     

     

    -

    The problem to solve for

    - -

    The wave equation to solve for, is

    +

    Example: Solving the wave equation with Neural Networks

    +

    The wave equation is

    $$ -\begin{equation} \tag{19} -\frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation} +\begin{equation*} + \frac{\partial^2 g(x,t)}{\partial t^2} = c^2\frac{\partial^2 g(x,t)}{\partial x^2} +\end{equation*} $$ -

    where \( c \) is the given wave speed. -The chosen conditions for this equation are -

    +

    with \( c \) being the specified wave speed.

    +

    Here, the chosen conditions are

    $$ -\begin{aligned} -g(0,t) &= 0, &t \geq 0 \\ -g(1,t) &= 0, &t \geq 0 \\ -g(x,0) &= u(x), &x\in[0,1] \\ -\frac{\partial g(x,t)}{\partial t}\Big |_{t = 0} &= v(x), &x \in [0,1] -\end{aligned} \tag{20} +\begin{align*} + g(0,t) &= 0 \\ + g(1,t) &= 0 \\ + g(x,0) &= u(x) \\ + \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} &= v(x) +\end{align*} $$ -

    In this example, let \( c = 1 \) and \( u(x) = \sin(\pi x) \) and \( v(x) = -\pi\sin(\pi x) \).

    +

    where \( \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} \) means the derivative of \( g(x,t) \) with respect to \( t \) is evaluated at \( t = 0 \), and \( u(x) \) and \( v(x) \) being given functions.

    @@ -449,7 +452,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs047.html b/doc/pub/week42/html/._week42-bs047.html index be4fbe9c1..e8b611fdd 100644 --- a/doc/pub/week42/html/._week42-bs047.html +++ b/doc/pub/week42/html/._week42-bs047.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,26 +404,30 @@ MathJax.Hub.Config({

     

     

     

    -

    The trial solution

    -

    Setting up the network is done in similar matter as for the example of solving the diffusion equation. -The only things we have to change, is the trial solution such that it satisfies the conditions from (20) and the cost function. -

    +

    The problem to solve for

    -

    The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution \( g_t(x,t) \) is

    +

    The wave equation to solve for, is

    -

    $$ -g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P) $$ -

    - -

    where

    - -

    $$ -h_1(x,t) = (1-t^2)u(x) + tv(x) +\begin{equation} \tag{19} +\frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2} +\end{equation} $$ + +

    where \( c \) is the given wave speed. +The chosen conditions for this equation are

    -

    Note that this trial solution satisfies the conditions only if \( u(0) = v(0) = u(1) = v(1) = 0 \), which is the case in this example.

    +$$ +\begin{aligned} +g(0,t) &= 0, &t \geq 0 \\ +g(1,t) &= 0, &t \geq 0 \\ +g(x,0) &= u(x), &x\in[0,1] \\ +\frac{\partial g(x,t)}{\partial t}\Big |_{t = 0} &= v(x), &x \in [0,1] +\end{aligned} \tag{20} +$$ + +

    In this example, let \( c = 1 \) and \( u(x) = \sin(\pi x) \) and \( v(x) = -\pi\sin(\pi x) \).

    @@ -445,7 +454,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs048.html b/doc/pub/week42/html/._week42-bs048.html index 3778d6ecf..3719c430f 100644 --- a/doc/pub/week42/html/._week42-bs048.html +++ b/doc/pub/week42/html/._week42-bs048.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,15 +404,27 @@ MathJax.Hub.Config({

     

     

     

    -

    The analytical solution

    +

    The trial solution

    +

    Setting up the network is done in similar matter as for the example of solving the diffusion equation. +The only things we have to change, is the trial solution such that it satisfies the conditions from (20) and the cost function. +

    -

    The analytical solution for our specific problem, is

    +

    The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution \( g_t(x,t) \) is

    $$ -g(x,t) = \sin(\pi x)\cos(\pi t) - \sin(\pi x)\sin(\pi t) +g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P) $$

    +

    where

    + +

    $$ +h_1(x,t) = (1-t^2)u(x) + tv(x) +$$ +

    + +

    Note that this trial solution satisfies the conditions only if \( u(0) = v(0) = u(1) = v(1) = 0 \), which is the case in this example.

    +

    diff --git a/doc/pub/week42/html/._week42-bs049.html b/doc/pub/week42/html/._week42-bs049.html index 8b8f22241..973c0495e 100644 --- a/doc/pub/week42/html/._week42-bs049.html +++ b/doc/pub/week42/html/._week42-bs049.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,250 +404,14 @@ MathJax.Hub.Config({

     

     

     

    -

    Solving the wave equation - the full program using Autograd

    +

    The analytical solution

    +

    The analytical solution for our specific problem, is

    - -
    -
    -
    -
    -
    - 
    import autograd.numpy as np
-from autograd import hessian,grad
-import autograd.numpy.random as npr
-from matplotlib import cm
-from matplotlib import pyplot as plt
-from mpl_toolkits.mplot3d import axes3d
-
-## Set up the trial function:
-def u(x):
-    return np.sin(np.pi*x)
-
-def v(x):
-    return -np.pi*np.sin(np.pi*x)
-
-def h1(point):
-    x,t = point
-    return (1 - t**2)*u(x) + t*v(x)
-
-def g_trial(point,P):
-    x,t = point
-    return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point)
-
-## Define the cost function
-def cost_function(P, x, t):
-    cost_sum = 0
-
-    g_t_hessian_func = hessian(g_trial)
-
-    for x_ in x:
-        for t_ in t:
-            point = np.array([x_,t_])
-
-            g_t_hessian = g_t_hessian_func(point,P)
-
-            g_t_d2x = g_t_hessian[0][0]
-            g_t_d2t = g_t_hessian[1][1]
-
-            err_sqr = ( (g_t_d2t - g_t_d2x) )**2
-            cost_sum += err_sqr
-
-    return cost_sum / (np.size(t) * np.size(x))
-
-## The neural network
-def sigmoid(z):
-    return 1/(1 + np.exp(-z))
-
-def deep_neural_network(deep_params, x):
-    # x is now a point and a 1D numpy array; make it a column vector
-    num_coordinates = np.size(x,0)
-    x = x.reshape(num_coordinates,-1)
-
-    num_points = np.size(x,1)
-
-    # N_hidden is the number of hidden layers
-    N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
-
-    # Assume that the input layer does nothing to the input x
-    x_input = x
-    x_prev = x_input
-
-    ## Hidden layers:
-
-    for l in range(N_hidden):
-        # From the list of parameters P; find the correct weigths and bias for this layer
-        w_hidden = deep_params[l]
-
-        # Add a row of ones to include bias
-        x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)
-
-        z_hidden = np.matmul(w_hidden, x_prev)
-        x_hidden = sigmoid(z_hidden)
-
-        # Update x_prev such that next layer can use the output from this layer
-        x_prev = x_hidden
-
-    ## Output layer:
-
-    # Get the weights and bias for this layer
-    w_output = deep_params[-1]
-
-    # Include bias:
-    x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)
-
-    z_output = np.matmul(w_output, x_prev)
-    x_output = z_output
-
-    return x_output[0][0]
-
-## The analytical solution
-def g_analytic(point):
-    x,t = point
-    return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t)
-
-def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):
-    ## Set up initial weigths and biases
-    N_hidden = np.size(num_neurons)
-
-    ## Set up initial weigths and biases
-
-    # Initialize the list of parameters:
-    P = [None]*(N_hidden + 1) # + 1 to include the output layer
-
-    P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias
-    for l in range(1,N_hidden):
-        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
-
-    # For the output layer
-    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
-
-    print('Initial cost: ',cost_function(P, x, t))
-
-    cost_function_grad = grad(cost_function,0)
-
-    # Let the update be done num_iter times
-    for i in range(num_iter):
-        cost_grad =  cost_function_grad(P, x , t)
-
-        for l in range(N_hidden+1):
-            P[l] = P[l] - lmb * cost_grad[l]
-
-
-    print('Final cost: ',cost_function(P, x, t))
-
-    return P
-
-if __name__ == '__main__':
-    ### Use the neural network:
-    npr.seed(15)
-
-    ## Decide the vales of arguments to the function to solve
-    Nx = 10; Nt = 10
-    x = np.linspace(0, 1, Nx)
-    t = np.linspace(0,1,Nt)
-
-    ## Set up the parameters for the network
-    num_hidden_neurons = [50,20]
-    num_iter = 1000
-    lmb = 0.01
-
-    P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)
-
-    ## Store the results
-    res = np.zeros((Nx, Nt))
-    res_analytical = np.zeros((Nx, Nt))
-    for i,x_ in enumerate(x):
-        for j, t_ in enumerate(t):
-            point = np.array([x_, t_])
-            res[i,j] = g_trial(point,P)
-
-            res_analytical[i,j] = g_analytic(point)
-
-    diff = np.abs(res - res_analytical)
-    print("Max difference between analytical and solution from nn: %g"%np.max(diff))
-
-    ## Plot the solutions in two dimensions, that being in position and time
-
-    T,X = np.meshgrid(t,x)
-
-    fig = plt.figure(figsize=(10,10))
-    ax = fig.gca(projection='3d')
-    ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))
-    s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis)
-    ax.set_xlabel('Time $t$')
-    ax.set_ylabel('Position $x$');
-
-
-    fig = plt.figure(figsize=(10,10))
-    ax = fig.gca(projection='3d')
-    ax.set_title('Analytical solution')
-    s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)
-    ax.set_xlabel('Time $t$')
-    ax.set_ylabel('Position $x$');
-
-
-    fig = plt.figure(figsize=(10,10))
-    ax = fig.gca(projection='3d')
-    ax.set_title('Difference')
-    s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis)
-    ax.set_xlabel('Time $t$')
-    ax.set_ylabel('Position $x$');
-
-    ## Take some slices of the 3D plots just to see the solutions at particular times
-    indx1 = 0
-    indx2 = int(Nt/2)
-    indx3 = Nt-1
-
-    t1 = t[indx1]
-    t2 = t[indx2]
-    t3 = t[indx3]
-
-    # Slice the results from the DNN
-    res1 = res[:,indx1]
-    res2 = res[:,indx2]
-    res3 = res[:,indx3]
-
-    # Slice the analytical results
-    res_analytical1 = res_analytical[:,indx1]
-    res_analytical2 = res_analytical[:,indx2]
-    res_analytical3 = res_analytical[:,indx3]
-
-    # Plot the slices
-    plt.figure(figsize=(10,10))
-    plt.title("Computed solutions at time = %g"%t1)
-    plt.plot(x, res1)
-    plt.plot(x,res_analytical1)
-    plt.legend(['dnn','analytical'])
-
-    plt.figure(figsize=(10,10))
-    plt.title("Computed solutions at time = %g"%t2)
-    plt.plot(x, res2)
-    plt.plot(x,res_analytical2)
-    plt.legend(['dnn','analytical'])
-
-    plt.figure(figsize=(10,10))
-    plt.title("Computed solutions at time = %g"%t3)
-    plt.plot(x, res3)
-    plt.plot(x,res_analytical3)
-    plt.legend(['dnn','analytical'])
-
-    plt.show()
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - +

    $$ +g(x,t) = \sin(\pi x)\cos(\pi t) - \sin(\pi x)\sin(\pi t) +$$ +

    @@ -669,7 +438,7 @@ MathJax.Hub.Config({

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    • diff --git a/doc/pub/week42/html/._week42-bs050.html b/doc/pub/week42/html/._week42-bs050.html index 0e6847cd6..9d959a937 100644 --- a/doc/pub/week42/html/._week42-bs050.html +++ b/doc/pub/week42/html/._week42-bs050.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,14 +404,251 @@ MathJax.Hub.Config({

     

     

     

    -

    Resources on differential equations and deep learning

    +

    Solving the wave equation - the full program using Autograd

    + + + +
    +
    +
    +
    +
    + 
    import autograd.numpy as np
+from autograd import hessian,grad
+import autograd.numpy.random as npr
+from matplotlib import cm
+from matplotlib import pyplot as plt
+from mpl_toolkits.mplot3d import axes3d
+
+## Set up the trial function:
+def u(x):
+    return np.sin(np.pi*x)
+
+def v(x):
+    return -np.pi*np.sin(np.pi*x)
+
+def h1(point):
+    x,t = point
+    return (1 - t**2)*u(x) + t*v(x)
+
+def g_trial(point,P):
+    x,t = point
+    return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point)
+
+## Define the cost function
+def cost_function(P, x, t):
+    cost_sum = 0
+
+    g_t_hessian_func = hessian(g_trial)
+
+    for x_ in x:
+        for t_ in t:
+            point = np.array([x_,t_])
+
+            g_t_hessian = g_t_hessian_func(point,P)
+
+            g_t_d2x = g_t_hessian[0][0]
+            g_t_d2t = g_t_hessian[1][1]
+
+            err_sqr = ( (g_t_d2t - g_t_d2x) )**2
+            cost_sum += err_sqr
+
+    return cost_sum / (np.size(t) * np.size(x))
+
+## The neural network
+def sigmoid(z):
+    return 1/(1 + np.exp(-z))
+
+def deep_neural_network(deep_params, x):
+    # x is now a point and a 1D numpy array; make it a column vector
+    num_coordinates = np.size(x,0)
+    x = x.reshape(num_coordinates,-1)
+
+    num_points = np.size(x,1)
+
+    # N_hidden is the number of hidden layers
+    N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
+
+    # Assume that the input layer does nothing to the input x
+    x_input = x
+    x_prev = x_input
+
+    ## Hidden layers:
+
+    for l in range(N_hidden):
+        # From the list of parameters P; find the correct weigths and bias for this layer
+        w_hidden = deep_params[l]
+
+        # Add a row of ones to include bias
+        x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)
+
+        z_hidden = np.matmul(w_hidden, x_prev)
+        x_hidden = sigmoid(z_hidden)
+
+        # Update x_prev such that next layer can use the output from this layer
+        x_prev = x_hidden
+
+    ## Output layer:
+
+    # Get the weights and bias for this layer
+    w_output = deep_params[-1]
+
+    # Include bias:
+    x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)
+
+    z_output = np.matmul(w_output, x_prev)
+    x_output = z_output
+
+    return x_output[0][0]
+
+## The analytical solution
+def g_analytic(point):
+    x,t = point
+    return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t)
+
+def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):
+    ## Set up initial weigths and biases
+    N_hidden = np.size(num_neurons)
+
+    ## Set up initial weigths and biases
+
+    # Initialize the list of parameters:
+    P = [None]*(N_hidden + 1) # + 1 to include the output layer
+
+    P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias
+    for l in range(1,N_hidden):
+        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
+
+    # For the output layer
+    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
+
+    print('Initial cost: ',cost_function(P, x, t))
+
+    cost_function_grad = grad(cost_function,0)
+
+    # Let the update be done num_iter times
+    for i in range(num_iter):
+        cost_grad =  cost_function_grad(P, x , t)
+
+        for l in range(N_hidden+1):
+            P[l] = P[l] - lmb * cost_grad[l]
+
+
+    print('Final cost: ',cost_function(P, x, t))
+
+    return P
+
+if __name__ == '__main__':
+    ### Use the neural network:
+    npr.seed(15)
+
+    ## Decide the vales of arguments to the function to solve
+    Nx = 10; Nt = 10
+    x = np.linspace(0, 1, Nx)
+    t = np.linspace(0,1,Nt)
+
+    ## Set up the parameters for the network
+    num_hidden_neurons = [50,20]
+    num_iter = 1000
+    lmb = 0.01
+
+    P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)
+
+    ## Store the results
+    res = np.zeros((Nx, Nt))
+    res_analytical = np.zeros((Nx, Nt))
+    for i,x_ in enumerate(x):
+        for j, t_ in enumerate(t):
+            point = np.array([x_, t_])
+            res[i,j] = g_trial(point,P)
+
+            res_analytical[i,j] = g_analytic(point)
+
+    diff = np.abs(res - res_analytical)
+    print("Max difference between analytical and solution from nn: %g"%np.max(diff))
+
+    ## Plot the solutions in two dimensions, that being in position and time
+
+    T,X = np.meshgrid(t,x)
+
+    fig = plt.figure(figsize=(10,10))
+    ax = fig.gca(projection='3d')
+    ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))
+    s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis)
+    ax.set_xlabel('Time $t$')
+    ax.set_ylabel('Position $x$');
+
+
+    fig = plt.figure(figsize=(10,10))
+    ax = fig.gca(projection='3d')
+    ax.set_title('Analytical solution')
+    s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)
+    ax.set_xlabel('Time $t$')
+    ax.set_ylabel('Position $x$');
+
+
+    fig = plt.figure(figsize=(10,10))
+    ax = fig.gca(projection='3d')
+    ax.set_title('Difference')
+    s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis)
+    ax.set_xlabel('Time $t$')
+    ax.set_ylabel('Position $x$');
+
+    ## Take some slices of the 3D plots just to see the solutions at particular times
+    indx1 = 0
+    indx2 = int(Nt/2)
+    indx3 = Nt-1
+
+    t1 = t[indx1]
+    t2 = t[indx2]
+    t3 = t[indx3]
+
+    # Slice the results from the DNN
+    res1 = res[:,indx1]
+    res2 = res[:,indx2]
+    res3 = res[:,indx3]
+
+    # Slice the analytical results
+    res_analytical1 = res_analytical[:,indx1]
+    res_analytical2 = res_analytical[:,indx2]
+    res_analytical3 = res_analytical[:,indx3]
+
+    # Plot the slices
+    plt.figure(figsize=(10,10))
+    plt.title("Computed solutions at time = %g"%t1)
+    plt.plot(x, res1)
+    plt.plot(x,res_analytical1)
+    plt.legend(['dnn','analytical'])
+
+    plt.figure(figsize=(10,10))
+    plt.title("Computed solutions at time = %g"%t2)
+    plt.plot(x, res2)
+    plt.plot(x,res_analytical2)
+    plt.legend(['dnn','analytical'])
+
+    plt.figure(figsize=(10,10))
+    plt.title("Computed solutions at time = %g"%t3)
+    plt.plot(x, res3)
+    plt.plot(x,res_analytical3)
+    plt.legend(['dnn','analytical'])
+
+    plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + -
      -
    1. Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al
      2. -
    2. Neural networks for solving differential equations by A. Honchar
      3. -
    3. Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener
      4. -
    4. Introduction to Partial Differential Equations by A. Tveito, R. Winther
      5. -

      @@ -432,7 +674,7 @@ MathJax.Hub.Config({
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    diff --git a/doc/pub/week42/html/._week42-bs051.html b/doc/pub/week42/html/._week42-bs051.html index 8b9fcb702..d0832d57e 100644 --- a/doc/pub/week42/html/._week42-bs051.html +++ b/doc/pub/week42/html/._week42-bs051.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,27 +404,14 @@ MathJax.Hub.Config({

     

     

     

    -

    Convolutional Neural Networks (recognizing images)

    - -

    Convolutional neural networks (CNNs) were developed during the last -decade of the previous century, with a focus on character recognition -tasks. Nowadays, CNNs are a central element in the spectacular success -of deep learning methods. The success in for example image -classifications have made them a central tool for most machine -learning practitioners. -

    - -

    CNNs are very similar to ordinary Neural Networks. -They are made up of neurons that have learnable weights and -biases. Each neuron receives some inputs, performs a dot product and -optionally follows it with a non-linearity. The whole network still -expresses a single differentiable score function: from the raw image -pixels on one end to class scores at the other. And they still have a -loss function (for example Softmax) on the last (fully-connected) layer -and all the tips/tricks we developed for learning regular Neural -Networks still apply (back propagation, gradient descent etc etc). -

    +

    Resources on differential equations and deep learning

    +
      +
    1. Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al
      2. +
    2. Neural networks for solving differential equations by A. Honchar
      3. +
    3. Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener
      4. +
    4. Introduction to Partial Differential Equations by A. Tveito, R. Winther
      5. +

      @@ -445,7 +437,7 @@ Networks still apply (back propagation, gradient descent etc etc).
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    • »
    diff --git a/doc/pub/week42/html/._week42-bs052.html b/doc/pub/week42/html/._week42-bs052.html index 37a8f5744..2fdd92b7d 100644 --- a/doc/pub/week42/html/._week42-bs052.html +++ b/doc/pub/week42/html/._week42-bs052.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,22 +404,27 @@ MathJax.Hub.Config({

     

     

     

    -

    What is the Difference

    +

    Convolutional Neural Networks (recognizing images)

    -

    CNN architectures make the explicit assumption that -the inputs are images, which allows us to encode certain properties -into the architecture. These then make the forward function more -efficient to implement and vastly reduce the amount of parameters in -the network. +

    Convolutional neural networks (CNNs) were developed during the last +decade of the previous century, with a focus on character recognition +tasks. Nowadays, CNNs are a central element in the spectacular success +of deep learning methods. The success in for example image +classifications have made them a central tool for most machine +learning practitioners.

    -

    Here we provide only a superficial overview, for the more interested, we recommend highly the course -IN5400 – Machine Learning for Image Analysis -and the slides of CS231. +

    CNNs are very similar to ordinary Neural Networks. +They are made up of neurons that have learnable weights and +biases. Each neuron receives some inputs, performs a dot product and +optionally follows it with a non-linearity. The whole network still +expresses a single differentiable score function: from the raw image +pixels on one end to class scores at the other. And they still have a +loss function (for example Softmax) on the last (fully-connected) layer +and all the tips/tricks we developed for learning regular Neural +Networks still apply (back propagation, gradient descent etc etc).

    -

    Another good read is the article here https://arxiv.org/pdf/1603.07285.pdf.

    -

    diff --git a/doc/pub/week42/html/._week42-bs053.html b/doc/pub/week42/html/._week42-bs053.html index b09c73f14..55108a3d9 100644 --- a/doc/pub/week42/html/._week42-bs053.html +++ b/doc/pub/week42/html/._week42-bs053.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,17 +404,22 @@ MathJax.Hub.Config({

     

     

     

    -

    Neural Networks vs CNNs

    +

    What is the Difference

    -

    Neural networks are defined as affine transformations, that is -a vector is received as input and is multiplied with a matrix of so-called weights (our unknown paramters) to produce an -output (to which a bias vector is usually added before passing the result -through a nonlinear activation function). This is applicable to any type of input, be it an -image, a sound clip or an unordered collection of features: whatever their -dimensionality, their representation can always be flattened into a vector -before the transformation. +

    CNN architectures make the explicit assumption that +the inputs are images, which allows us to encode certain properties +into the architecture. These then make the forward function more +efficient to implement and vastly reduce the amount of parameters in +the network.

    +

    Here we provide only a superficial overview, for the more interested, we recommend highly the course +IN5400 – Machine Learning for Image Analysis +and the slides of CS231. +

    + +

    Another good read is the article here https://arxiv.org/pdf/1603.07285.pdf.

    +

      @@ -435,7 +445,7 @@ before the transformation.
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    • »
    diff --git a/doc/pub/week42/html/._week42-bs054.html b/doc/pub/week42/html/._week42-bs054.html index b3221653a..9c1526265 100644 --- a/doc/pub/week42/html/._week42-bs054.html +++ b/doc/pub/week42/html/._week42-bs054.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,28 +404,15 @@ MathJax.Hub.Config({

     

     

     

    -

    Why CNNS for images, sound files, medical images from CT scans etc?

    +

    Neural Networks vs CNNs

    -

    However, when we consider images, sound clips and many other similar kinds of data, these data have an intrinsic -structure. More formally, they share these important properties: -

    -
      -
    • They are stored as multi-dimensional arrays (think of the pixels of a figure) .
      • -
    • They feature one or more axes for which ordering matters (e.g., width and height axes for an image, time axis for a sound clip).
      • -
    • One axis, called the channel axis, is used to access different views of the data (e.g., the red, green and blue channels of a color image, or the left and right channels of a stereo audio track).
      • -
    -

    These properties are not exploited when an affine transformation is applied; in -fact, all the axes are treated in the same way and the topological information -is not taken into account. Still, taking advantage of the implicit structure of -the data may prove very handy in solving some tasks, like computer vision and -speech recognition, and in these cases it would be best to preserve it. This is -where discrete convolutions come into play. -

    - -

    A discrete convolution is a linear transformation that preserves this notion of -ordering. It is sparse (only a few input units contribute to a given output -unit) and reuses parameters (the same weights are applied to multiple locations -in the input). +

    Neural networks are defined as affine transformations, that is +a vector is received as input and is multiplied with a matrix of so-called weights (our unknown paramters) to produce an +output (to which a bias vector is usually added before passing the result +through a nonlinear activation function). This is applicable to any type of input, be it an +image, a sound clip or an unordered collection of features: whatever their +dimensionality, their representation can always be flattened into a vector +before the transformation.

    @@ -448,7 +440,7 @@ in the input).

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    • diff --git a/doc/pub/week42/html/._week42-bs055.html b/doc/pub/week42/html/._week42-bs055.html index 067be749f..3e09653fa 100644 --- a/doc/pub/week42/html/._week42-bs055.html +++ b/doc/pub/week42/html/._week42-bs055.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,32 +404,30 @@ MathJax.Hub.Config({

     

     

     

    -

    Regular NNs don’t scale well to full images

    +

    Why CNNS for images, sound files, medical images from CT scans etc?

    -

    As an example, consider -an image of size \( 32\times 32\times 3 \) (32 wide, 32 high, 3 color channels), so a -single fully-connected neuron in a first hidden layer of a regular -Neural Network would have \( 32\times 32\times 3 = 3072 \) weights. This amount still -seems manageable, but clearly this fully-connected structure does not -scale to larger images. For example, an image of more respectable -size, say \( 200\times 200\times 3 \), would lead to neurons that have -\( 200\times 200\times 3 = 120,000 \) weights. +

    However, when we consider images, sound clips and many other similar kinds of data, these data have an intrinsic +structure. More formally, they share these important properties: +

    +
      +
    • They are stored as multi-dimensional arrays (think of the pixels of a figure) .
      • +
    • They feature one or more axes for which ordering matters (e.g., width and height axes for an image, time axis for a sound clip).
      • +
    • One axis, called the channel axis, is used to access different views of the data (e.g., the red, green and blue channels of a color image, or the left and right channels of a stereo audio track).
      • +
    +

    These properties are not exploited when an affine transformation is applied; in +fact, all the axes are treated in the same way and the topological information +is not taken into account. Still, taking advantage of the implicit structure of +the data may prove very handy in solving some tasks, like computer vision and +speech recognition, and in these cases it would be best to preserve it. This is +where discrete convolutions come into play.

    -

    We could have -several such neurons, and the parameters would add up quickly! Clearly, -this full connectivity is wasteful and the huge number of parameters -would quickly lead to possible overfitting. +

    A discrete convolution is a linear transformation that preserves this notion of +ordering. It is sparse (only a few input units contribute to a given output +unit) and reuses parameters (the same weights are applied to multiple locations +in the input).

    -
    -
    -
    -

    Figure 1: A regular 3-layer Neural Network.

    -
    -

    -
    -

      @@ -450,7 +453,7 @@ would quickly lead to possible overfitting.
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    diff --git a/doc/pub/week42/html/._week42-bs056.html b/doc/pub/week42/html/._week42-bs056.html index 40c377744..991fc7de6 100644 --- a/doc/pub/week42/html/._week42-bs056.html +++ b/doc/pub/week42/html/._week42-bs056.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,42 +404,30 @@ MathJax.Hub.Config({

     

     

     

    -

    3D volumes of neurons

    +

    Regular NNs don’t scale well to full images

    -

    Convolutional Neural Networks take advantage of the fact that the -input consists of images and they constrain the architecture in a more -sensible way. +

    As an example, consider +an image of size \( 32\times 32\times 3 \) (32 wide, 32 high, 3 color channels), so a +single fully-connected neuron in a first hidden layer of a regular +Neural Network would have \( 32\times 32\times 3 = 3072 \) weights. This amount still +seems manageable, but clearly this fully-connected structure does not +scale to larger images. For example, an image of more respectable +size, say \( 200\times 200\times 3 \), would lead to neurons that have +\( 200\times 200\times 3 = 120,000 \) weights.

    -

    In particular, unlike a regular Neural Network, the -layers of a CNN have neurons arranged in 3 dimensions: width, -height, depth. (Note that the word depth here refers to the third -dimension of an activation volume, not to the depth of a full Neural -Network, which can refer to the total number of layers in a network.) -

    - -

    To understand it better, the above example of an image -with an input volume of -activations has dimensions \( 32\times 32\times 3 \) (width, height, -depth respectively). -

    - -

    The neurons in a layer will -only be connected to a small region of the layer before it, instead of -all of the neurons in a fully-connected manner. Moreover, the final -output layer could for this specific image have dimensions \( 1\times 1 \times 10 \), -because by the -end of the CNN architecture we will reduce the full image into a -single vector of class scores, arranged along the depth -dimension. +

    We could have +several such neurons, and the parameters would add up quickly! Clearly, +this full connectivity is wasteful and the huge number of parameters +would quickly lead to possible overfitting.

    -

    Figure 2: A CNN arranges its neurons in three dimensions (width, height, depth), as visualized in one of the layers. Every layer of a CNN transforms the 3D input volume to a 3D output volume of neuron activations. In this example, the red input layer holds the image, so its width and height would be the dimensions of the image, and the depth would be 3 (Red, Green, Blue channels).

    +

    Figure 1: A regular 3-layer Neural Network.

    -

    +

    @@ -462,7 +455,7 @@ dimension.

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    • diff --git a/doc/pub/week42/html/._week42-bs057.html b/doc/pub/week42/html/._week42-bs057.html index 1d010781d..4ad410c45 100644 --- a/doc/pub/week42/html/._week42-bs057.html +++ b/doc/pub/week42/html/._week42-bs057.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -398,26 +403,45 @@ MathJax.Hub.Config({

     

     

     

    - -

    Layers used to build CNNs

    + +

    3D volumes of neurons

    -

    A simple CNN is a sequence of layers, and every layer of a CNN -transforms one volume of activations to another through a -differentiable function. We use three main types of layers to build -CNN architectures: Convolutional Layer, Pooling Layer, and -Fully-Connected Layer (exactly as seen in regular Neural Networks). We -will stack these layers to form a full CNN architecture. +

    Convolutional Neural Networks take advantage of the fact that the +input consists of images and they constrain the architecture in a more +sensible way.

    -

    A simple CNN for image classification could have the architecture:

    +

    In particular, unlike a regular Neural Network, the +layers of a CNN have neurons arranged in 3 dimensions: width, +height, depth. (Note that the word depth here refers to the third +dimension of an activation volume, not to the depth of a full Neural +Network, which can refer to the total number of layers in a network.) +

    + +

    To understand it better, the above example of an image +with an input volume of +activations has dimensions \( 32\times 32\times 3 \) (width, height, +depth respectively). +

    + +

    The neurons in a layer will +only be connected to a small region of the layer before it, instead of +all of the neurons in a fully-connected manner. Moreover, the final +output layer could for this specific image have dimensions \( 1\times 1 \times 10 \), +because by the +end of the CNN architecture we will reduce the full image into a +single vector of class scores, arranged along the depth +dimension. +

    + +
    +
    +
    +

    Figure 2: A CNN arranges its neurons in three dimensions (width, height, depth), as visualized in one of the layers. Every layer of a CNN transforms the 3D input volume to a 3D output volume of neuron activations. In this example, the red input layer holds the image, so its width and height would be the dimensions of the image, and the depth would be 3 (Red, Green, Blue channels).

    +
    +

    +
    -
      -
    • INPUT (\( 32\times 32 \times 3 \)) will hold the raw pixel values of the image, in this case an image of width 32, height 32, and with three color channels R,G,B.
      • -
    • CONV (convolutional )layer will compute the output of neurons that are connected to local regions in the input, each computing a dot product between their weights and a small region they are connected to in the input volume. This may result in volume such as \( [32\times 32\times 12] \) if we decided to use 12 filters.
      • -
    • RELU layer will apply an elementwise activation function, such as the \( max(0,x) \) thresholding at zero. This leaves the size of the volume unchanged (\( [32\times 32\times 12] \)).
      • -
    • POOL (pooling) layer will perform a downsampling operation along the spatial dimensions (width, height), resulting in volume such as \( [16\times 16\times 12] \).
      • -
    • FC (i.e. fully-connected) layer will compute the class scores, resulting in volume of size \( [1\times 1\times 10] \), where each of the 10 numbers correspond to a class score, such as among the 10 categories of the MNIST images we considered above . As with ordinary Neural Networks and as the name implies, each neuron in this layer will be connected to all the numbers in the previous volume.
      • -

      @@ -443,7 +467,7 @@ will stack these layers to form a full CNN architecture.
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    diff --git a/doc/pub/week42/html/._week42-bs058.html b/doc/pub/week42/html/._week42-bs058.html index 37c84855f..962241d1d 100644 --- a/doc/pub/week42/html/._week42-bs058.html +++ b/doc/pub/week42/html/._week42-bs058.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -398,23 +403,26 @@ MathJax.Hub.Config({

     

     

     

    - -

    Transforming images

    + +

    Layers used to build CNNs

    -

    CNNs transform the original image layer by layer from the original -pixel values to the final class scores. +

    A simple CNN is a sequence of layers, and every layer of a CNN +transforms one volume of activations to another through a +differentiable function. We use three main types of layers to build +CNN architectures: Convolutional Layer, Pooling Layer, and +Fully-Connected Layer (exactly as seen in regular Neural Networks). We +will stack these layers to form a full CNN architecture.

    -

    Observe that some layers contain -parameters and other don’t. In particular, the CNN layers perform -transformations that are a function of not only the activations in the -input volume, but also of the parameters (the weights and biases of -the neurons). On the other hand, the RELU/POOL layers will implement a -fixed function. The parameters in the CONV/FC layers will be trained -with gradient descent so that the class scores that the CNN computes -are consistent with the labels in the training set for each image. -

    +

    A simple CNN for image classification could have the architecture:

    +
      +
    • INPUT (\( 32\times 32 \times 3 \)) will hold the raw pixel values of the image, in this case an image of width 32, height 32, and with three color channels R,G,B.
      • +
    • CONV (convolutional )layer will compute the output of neurons that are connected to local regions in the input, each computing a dot product between their weights and a small region they are connected to in the input volume. This may result in volume such as \( [32\times 32\times 12] \) if we decided to use 12 filters.
      • +
    • RELU layer will apply an elementwise activation function, such as the \( max(0,x) \) thresholding at zero. This leaves the size of the volume unchanged (\( [32\times 32\times 12] \)).
      • +
    • POOL (pooling) layer will perform a downsampling operation along the spatial dimensions (width, height), resulting in volume such as \( [16\times 16\times 12] \).
      • +
    • FC (i.e. fully-connected) layer will compute the class scores, resulting in volume of size \( [1\times 1\times 10] \), where each of the 10 numbers correspond to a class score, such as among the 10 categories of the MNIST images we considered above . As with ordinary Neural Networks and as the name implies, each neuron in this layer will be connected to all the numbers in the previous volume.
      • +

      @@ -440,7 +448,7 @@ are consistent with the labels in the training set for each image.
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    diff --git a/doc/pub/week42/html/._week42-bs059.html b/doc/pub/week42/html/._week42-bs059.html index 4b45decbd..89062cd5d 100644 --- a/doc/pub/week42/html/._week42-bs059.html +++ b/doc/pub/week42/html/._week42-bs059.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,24 +404,21 @@ MathJax.Hub.Config({

     

     

     

    -

    CNNs in brief

    +

    Transforming images

    -

    In summary:

    - -
      -
    • A CNN architecture is in the simplest case a list of Layers that transform the image volume into an output volume (e.g. holding the class scores)
      • -
    • There are a few distinct types of Layers (e.g. CONV/FC/RELU/POOL are by far the most popular)
      • -
    • Each Layer accepts an input 3D volume and transforms it to an output 3D volume through a differentiable function
      • -
    • Each Layer may or may not have parameters (e.g. CONV/FC do, RELU/POOL don’t)
      • -
    • Each Layer may or may not have additional hyperparameters (e.g. CONV/FC/POOL do, RELU doesn’t)
      • -
    -

    For more material on convolutional networks, we strongly recommend -the course -IN5400 – Machine Learning for Image Analysis -and the slides of CS231 which is taught at Stanford University (consistently ranked as one of the top computer science programs in the world). Michael Nielsen's book is a must read, in particular chapter 6 which deals with CNNs. +

    CNNs transform the original image layer by layer from the original +pixel values to the final class scores.

    -

    The textbook by Goodfellow et al, see chapter 9 contains an in depth discussion as well.

    +

    Observe that some layers contain +parameters and other don’t. In particular, the CNN layers perform +transformations that are a function of not only the activations in the +input volume, but also of the parameters (the weights and biases of +the neurons). On the other hand, the RELU/POOL layers will implement a +fixed function. The parameters in the CONV/FC layers will be trained +with gradient descent so that the class scores that the CNN computes +are consistent with the labels in the training set for each image. +

    @@ -443,7 +445,7 @@ and the slides of 68

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    • diff --git a/doc/pub/week42/html/._week42-bs060.html b/doc/pub/week42/html/._week42-bs060.html index dbdcdd9d5..9c92d10c3 100644 --- a/doc/pub/week42/html/._week42-bs060.html +++ b/doc/pub/week42/html/._week42-bs060.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,15 +404,24 @@ MathJax.Hub.Config({

     

     

     

    -

    Key Idea

    +

    CNNs in brief

    -

    A dense neural network is representd by an affine operation (like matrix-matrix multiplication) where all parameters are included.

    +

    In summary:

    -

    The key idea in CNNs for say imaging is that in images neighbor pixels tend to be related! So we connect -only neighboring neurons in the input instead of connecting all with the first hidden layer. +

      +
    • A CNN architecture is in the simplest case a list of Layers that transform the image volume into an output volume (e.g. holding the class scores)
      • +
    • There are a few distinct types of Layers (e.g. CONV/FC/RELU/POOL are by far the most popular)
      • +
    • Each Layer accepts an input 3D volume and transforms it to an output 3D volume through a differentiable function
      • +
    • Each Layer may or may not have parameters (e.g. CONV/FC do, RELU/POOL don’t)
      • +
    • Each Layer may or may not have additional hyperparameters (e.g. CONV/FC/POOL do, RELU doesn’t)
      • +
    +

    For more material on convolutional networks, we strongly recommend +the course +IN5400 – Machine Learning for Image Analysis +and the slides of CS231 which is taught at Stanford University (consistently ranked as one of the top computer science programs in the world). Michael Nielsen's book is a must read, in particular chapter 6 which deals with CNNs.

    -

    We say we perform a filtering (convolution is the mathematical operation).

    +

    The textbook by Goodfellow et al, see chapter 9 contains an in depth discussion as well.

    @@ -434,7 +448,7 @@ only neighboring neurons in the input instead of connecting all with the first h

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    • diff --git a/doc/pub/week42/html/._week42-bs061.html b/doc/pub/week42/html/._week42-bs061.html index 6fdc145c2..ad067e592 100644 --- a/doc/pub/week42/html/._week42-bs061.html +++ b/doc/pub/week42/html/._week42-bs061.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,38 +404,15 @@ MathJax.Hub.Config({

     

     

     

    -

    Mathematics of CNNs

    +

    Key Idea

    -

    The mathematics of CNNs is based on the mathematical operation of -convolution. In mathematics (in particular in functional analysis), -convolution is represented by matheematical operation (integration, -summation etc) on two function in order to produce a third function -that expresses how the shape of one gets modified by the other. -Convolution has a plethora of applications in a variety of disciplines, spanning from statistics to signal processing, computer vision, solutions of differential equations,linear algebra, engineering, and yes, machine learning. +

    A dense neural network is representd by an affine operation (like matrix-matrix multiplication) where all parameters are included.

    + +

    The key idea in CNNs for say imaging is that in images neighbor pixels tend to be related! So we connect +only neighboring neurons in the input instead of connecting all with the first hidden layer.

    -

    Mathematically, convolution is defined as follows (one-dimensional example): -Let us define a continuous function \( y(t) \) given by -

    -$$ -y(t) = \int x(a) w(t-a) da, -$$ - -

    where \( x(a) \) represents a so-called input and \( w(t-a) \) is normally called the weight function or kernel.

    - -

    The above integral is written in a more compact form as

    -$$ -y(t) = \left(x * w\right)(t). -$$ - -

    The discretized version reads

    -$$ -y(t) = \sum_{a=-\infty}^{a=\infty}x(a)w(t-a). -$$ - -

    Computing the inverse of the above convolution operations is known as deconvolution.

    - -

    How can we use this? And what does it mean? Let us study some familiar examples first.

    +

    We say we perform a filtering (convolution is the mathematical operation).

    @@ -457,7 +439,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs062.html b/doc/pub/week42/html/._week42-bs062.html index 78d61a175..d4c64c777 100644 --- a/doc/pub/week42/html/._week42-bs062.html +++ b/doc/pub/week42/html/._week42-bs062.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,27 +404,38 @@ MathJax.Hub.Config({

     

     

     

    -

    Convolution Examples: Polynomial multiplication

    +

    Mathematics of CNNs

    -

    We have already met such an example in project 1 when we tried to set -up the design matrix for a two-dimensional function. This was an -example of polynomial multiplication. Let us recast such a problem in terms of the convolution operation. -Let us look a the following polynomials to second and third order, respectively: +

    The mathematics of CNNs is based on the mathematical operation of +convolution. In mathematics (in particular in functional analysis), +convolution is represented by matheematical operation (integration, +summation etc) on two function in order to produce a third function +that expresses how the shape of one gets modified by the other. +Convolution has a plethora of applications in a variety of disciplines, spanning from statistics to signal processing, computer vision, solutions of differential equations,linear algebra, engineering, and yes, machine learning. +

    + +

    Mathematically, convolution is defined as follows (one-dimensional example): +Let us define a continuous function \( y(t) \) given by

    $$ -p(t) = \alpha_0+\alpha_1 t+\alpha_2 t^2, +y(t) = \int x(a) w(t-a) da, $$ -

    and

    +

    where \( x(a) \) represents a so-called input and \( w(t-a) \) is normally called the weight function or kernel.

    + +

    The above integral is written in a more compact form as

    $$ -s(t) = \beta_0+\beta_1 t+\beta_2 t^2+\beta_3 t^3. +y(t) = \left(x * w\right)(t). $$ -

    The polynomial multiplication gives us a new polynomial of degree \( 5 \)

    +

    The discretized version reads

    $$ -z(t) = \delta_0+\delta_1 t+\delta_2 t^2+\delta_3 t^3+\delta_4 t^4+\delta_5 t^5. +y(t) = \sum_{a=-\infty}^{a=\infty}x(a)w(t-a). $$ +

    Computing the inverse of the above convolution operations is known as deconvolution.

    + +

    How can we use this? And what does it mean? Let us study some familiar examples first.

    @@ -446,7 +462,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs063.html b/doc/pub/week42/html/._week42-bs063.html index 96baf8a35..71ab93e87 100644 --- a/doc/pub/week42/html/._week42-bs063.html +++ b/doc/pub/week42/html/._week42-bs063.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,36 +404,27 @@ MathJax.Hub.Config({

     

     

     

    -

    Efficient Polynomial Multiplication

    +

    Convolution Examples: Polynomial multiplication

    -

    Computing polynomial products can be implemented efficiently if we rewrite the more brute force multiplications using convolution. -We note first that the new coefficients are given as +

    We have already met such an example in project 1 when we tried to set +up the design matrix for a two-dimensional function. This was an +example of polynomial multiplication. Let us recast such a problem in terms of the convolution operation. +Let us look a the following polynomials to second and third order, respectively:

    - $$ -\begin{split} -\delta_0=&\alpha_0\beta_0\\ -\delta_1=&\alpha_1\beta_0+\alpha_1\beta_0\\ -\delta_2=&\alpha_0\beta_2+\alpha_1\beta_1+\alpha_2\beta_0\\ -\delta_3=&\alpha_1\beta_2+\alpha_2\beta_1+\alpha_0\beta_3\\ -\delta_4=&\alpha_2\beta_2+\alpha_1\beta_3\\ -\delta_5=&\alpha_2\beta_3.\\ -\end{split} +p(t) = \alpha_0+\alpha_1 t+\alpha_2 t^2, $$ -

    We note that \( \alpha_i=0 \) except for \( i\in \left\{0,1,2\right\} \) and \( \beta_i=0 \) except for \( i\in\left\{0,1,2,3\right\} \).

    - -

    We can then rewrite the coefficients \( \delta_j \) using a discrete convolution as

    +

    and

    $$ -\delta_j = \sum_{i=-\infty}^{i=\infty}\alpha_i\beta_{j-i}=(\alpha * \beta)_j, +s(t) = \beta_0+\beta_1 t+\beta_2 t^2+\beta_3 t^3. $$ -

    or as a double sum with restriction \( l=i+j \)

    +

    The polynomial multiplication gives us a new polynomial of degree \( 5 \)

    $$ -\delta_l = \sum_{ij}\alpha_i\beta_{j}. +z(t) = \delta_0+\delta_1 t+\delta_2 t^2+\delta_3 t^3+\delta_4 t^4+\delta_5 t^5. $$ -

    Do you see a potential drawback with these equations?

    @@ -455,7 +451,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs064.html b/doc/pub/week42/html/._week42-bs064.html index 82661151a..71c797f90 100644 --- a/doc/pub/week42/html/._week42-bs064.html +++ b/doc/pub/week42/html/._week42-bs064.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,42 +404,36 @@ MathJax.Hub.Config({

     

     

     

    -

    A more efficient way of coding the above Convolution

    +

    Efficient Polynomial Multiplication

    -

    Since we only have a finite number of \( \alpha \) and \( \beta \) values -which are non-zero, we can rewrite the above convolution expressions -as a matrix-vector multiplication +

    Computing polynomial products can be implemented efficiently if we rewrite the more brute force multiplications using convolution. +We note first that the new coefficients are given as

    $$ -\boldsymbol{\delta}=\begin{bmatrix}\alpha_0 & 0 & 0 & 0 \\ - \alpha_1 & \alpha_0 & 0 & 0 \\ - \alpha_2 & \alpha_1 & \alpha_0 & 0 \\ - 0 & \alpha_2 & \alpha_1 & \alpha_0 \\ - 0 & 0 & \alpha_2 & \alpha_1 \\ - 0 & 0 & 0 & \alpha_2 - \end{bmatrix}\begin{bmatrix} \beta_0 \\ \beta_1 \\ \beta_2 \\ \beta_3\end{bmatrix}. +\begin{split} +\delta_0=&\alpha_0\beta_0\\ +\delta_1=&\alpha_1\beta_0+\alpha_1\beta_0\\ +\delta_2=&\alpha_0\beta_2+\alpha_1\beta_1+\alpha_2\beta_0\\ +\delta_3=&\alpha_1\beta_2+\alpha_2\beta_1+\alpha_0\beta_3\\ +\delta_4=&\alpha_2\beta_2+\alpha_1\beta_3\\ +\delta_5=&\alpha_2\beta_3.\\ +\end{split} $$ -

    The process is commutative and we can easily see that we can rewrite the multiplication in terms of a matrix holding \( \beta \) and a vector holding \( \alpha \). -In this case we have -

    +

    We note that \( \alpha_i=0 \) except for \( i\in \left\{0,1,2\right\} \) and \( \beta_i=0 \) except for \( i\in\left\{0,1,2,3\right\} \).

    + +

    We can then rewrite the coefficients \( \delta_j \) using a discrete convolution as

    $$ -\boldsymbol{\delta}=\begin{bmatrix}\beta_0 & 0 & 0 \\ - \beta_1 & \beta_0 & 0 \\ - \beta_2 & \beta_1 & \beta_0 \\ - \beta_3 & \beta_2 & \beta_1 \\ - 0 & \beta_3 & \beta_2 \\ - 0 & 0 & \beta_3 - \end{bmatrix}\begin{bmatrix} \alpha_0 \\ \alpha_1 \\ \alpha_2\end{bmatrix}. +\delta_j = \sum_{i=-\infty}^{i=\infty}\alpha_i\beta_{j-i}=(\alpha * \beta)_j, $$ -

    Note that the use of these matrices is for mathematical purposes only and not implementation purposes. -When implementing the above equation we do not encode (and allocate memory) the matrices explicitely. -We rather code the convolutions in the minimal memory footprint that they require. -

    +

    or as a double sum with restriction \( l=i+j \)

    +$$ +\delta_l = \sum_{ij}\alpha_i\beta_{j}. +$$ -

    Does the number of floating point operations change here when we use the commutative property?

    +

    Do you see a potential drawback with these equations?

    @@ -461,7 +460,7 @@ We rather code the convolutions in the minimal memory footprint that they requir

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    • diff --git a/doc/pub/week42/html/._week42-bs065.html b/doc/pub/week42/html/._week42-bs065.html index bc813339c..1542e6b77 100644 --- a/doc/pub/week42/html/._week42-bs065.html +++ b/doc/pub/week42/html/._week42-bs065.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,27 +404,42 @@ MathJax.Hub.Config({

     

     

     

    -

    Convolution Examples: Principle of Superposition and Periodic Forces (Fourier Transforms)

    +

    A more efficient way of coding the above Convolution

    -

    For problems with so-called harmonic oscillations, given by for example the following differential equation

    -$$ -m\frac{d^2x}{dt^2}+\eta\frac{dx}{dt}+x(t)=F(t), -$$ - -

    where \( F(t) \) is an applied external force acting on the system (often called a driving force), one can use the theory of Fourier transformations to find the solutions of this type of equations.

    - -

    If one has several driving forces, \( F(t)=\sum_n F_n(t) \), one can find -the particular solution to each \( F_n \), \( x_{pn}(t) \), and the particular -solution for the entire driving force is then given by a series like +

    Since we only have a finite number of \( \alpha \) and \( \beta \) values +which are non-zero, we can rewrite the above convolution expressions +as a matrix-vector multiplication

    $$ -\begin{equation} -x_p(t)=\sum_nx_{pn}(t). -\tag{21} -\end{equation} +\boldsymbol{\delta}=\begin{bmatrix}\alpha_0 & 0 & 0 & 0 \\ + \alpha_1 & \alpha_0 & 0 & 0 \\ + \alpha_2 & \alpha_1 & \alpha_0 & 0 \\ + 0 & \alpha_2 & \alpha_1 & \alpha_0 \\ + 0 & 0 & \alpha_2 & \alpha_1 \\ + 0 & 0 & 0 & \alpha_2 + \end{bmatrix}\begin{bmatrix} \beta_0 \\ \beta_1 \\ \beta_2 \\ \beta_3\end{bmatrix}. $$ +

    The process is commutative and we can easily see that we can rewrite the multiplication in terms of a matrix holding \( \beta \) and a vector holding \( \alpha \). +In this case we have +

    +$$ +\boldsymbol{\delta}=\begin{bmatrix}\beta_0 & 0 & 0 \\ + \beta_1 & \beta_0 & 0 \\ + \beta_2 & \beta_1 & \beta_0 \\ + \beta_3 & \beta_2 & \beta_1 \\ + 0 & \beta_3 & \beta_2 \\ + 0 & 0 & \beta_3 + \end{bmatrix}\begin{bmatrix} \alpha_0 \\ \alpha_1 \\ \alpha_2\end{bmatrix}. +$$ + +

    Note that the use of these matrices is for mathematical purposes only and not implementation purposes. +When implementing the above equation we do not encode (and allocate memory) the matrices explicitely. +We rather code the convolutions in the minimal memory footprint that they require. +

    + +

    Does the number of floating point operations change here when we use the commutative property?

    @@ -446,7 +466,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs066.html b/doc/pub/week42/html/._week42-bs066.html index ce4f32769..62b3f9f94 100644 --- a/doc/pub/week42/html/._week42-bs066.html +++ b/doc/pub/week42/html/._week42-bs066.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,32 +404,27 @@ MathJax.Hub.Config({

     

     

     

    -

    Principle of Superposition

    +

    Convolution Examples: Principle of Superposition and Periodic Forces (Fourier Transforms)

    -

    This is known as the principle of superposition. It only applies when -the homogenous equation is linear. If there were an anharmonic term -such as \( x^3 \) in the homogenous equation, then when one summed various -solutions, \( x=(\sum_n x_n)^2 \), one would get cross -terms. Superposition is especially useful when \( F(t) \) can be written -as a sum of sinusoidal terms, because the solutions for each -sinusoidal (sine or cosine) term is analytic. -

    +

    For problems with so-called harmonic oscillations, given by for example the following differential equation

    +$$ +m\frac{d^2x}{dt^2}+\eta\frac{dx}{dt}+x(t)=F(t), +$$ -

    Driving forces are often periodic, even when they are not -sinusoidal. Periodicity implies that for some time \( \tau \) +

    where \( F(t) \) is an applied external force acting on the system (often called a driving force), one can use the theory of Fourier transformations to find the solutions of this type of equations.

    + +

    If one has several driving forces, \( F(t)=\sum_n F_n(t) \), one can find +the particular solution to each \( F_n \), \( x_{pn}(t) \), and the particular +solution for the entire driving force is then given by a series like

    $$ -\begin{eqnarray} -F(t+\tau)=F(t). -\end{eqnarray} +\begin{equation} +x_p(t)=\sum_nx_{pn}(t). +\tag{21} +\end{equation} $$ -

    One example of a non-sinusoidal periodic force is a square wave. Many -components in electric circuits are non-linear, e.g. diodes, which -makes many wave forms non-sinusoidal even when the circuits are being -driven by purely sinusoidal sources. -

    @@ -451,7 +451,7 @@ driven by purely sinusoidal sources.

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    • diff --git a/doc/pub/week42/html/._week42-bs067.html b/doc/pub/week42/html/._week42-bs067.html index 35efebba2..055bc9178 100644 --- a/doc/pub/week42/html/._week42-bs067.html +++ b/doc/pub/week42/html/._week42-bs067.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,63 +404,32 @@ MathJax.Hub.Config({

     

     

     

    -

    Simple Code Example

    +

    Principle of Superposition

    -

    The code here shows a typical example of such a square wave generated using the functionality included in the scipy Python package. We have used a period of \( \tau=0.2 \).

    +

    This is known as the principle of superposition. It only applies when +the homogenous equation is linear. If there were an anharmonic term +such as \( x^3 \) in the homogenous equation, then when one summed various +solutions, \( x=(\sum_n x_n)^2 \), one would get cross +terms. Superposition is especially useful when \( F(t) \) can be written +as a sum of sinusoidal terms, because the solutions for each +sinusoidal (sine or cosine) term is analytic. +

    - - -
    -
    -
    -
    -
    - 
    import numpy as np
-import math
-from scipy import signal
-import matplotlib.pyplot as plt
-
-# number of points                                                                                       
-n = 500
-# start and final times                                                                                  
-t0 = 0.0
-tn = 1.0
-# Period                                                                                                 
-t = np.linspace(t0, tn, n, endpoint=False)
-SqrSignal = np.zeros(n)
-SqrSignal = 1.0+signal.square(2*np.pi*5*t)
-plt.plot(t, SqrSignal)
-plt.ylim(-0.5, 2.5)
-plt.show()
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - -

    For the sinusoidal example the -period is \( \tau=2\pi/\omega \). However, higher harmonics can also -satisfy the periodicity requirement. In general, any force that -satisfies the periodicity requirement can be expressed as a sum over -harmonics, +

    Driving forces are often periodic, even when they are not +sinusoidal. Periodicity implies that for some time \( \tau \)

    $$ -\begin{equation} -F(t)=\frac{f_0}{2}+\sum_{n>0} f_n\cos(2n\pi t/\tau)+g_n\sin(2n\pi t/\tau). -\tag{22} -\end{equation} +\begin{eqnarray} +F(t+\tau)=F(t). +\end{eqnarray} $$ +

    One example of a non-sinusoidal periodic force is a square wave. Many +components in electric circuits are non-linear, e.g. diodes, which +makes many wave forms non-sinusoidal even when the circuits are being +driven by purely sinusoidal sources. +

    @@ -482,7 +456,7 @@ $$

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    -

    Wrapping up Fourier transforms

    +

    Simple Code Example

    -

    We can write down the answer for -\( x_{pn}(t) \), by substituting \( f_n/m \) or \( g_n/m \) for \( F_0/m \). By -writing each factor \( 2n\pi t/\tau \) as \( n\omega t \), with \( \omega\equiv -2\pi/\tau \), +

    The code here shows a typical example of such a square wave generated using the functionality included in the scipy Python package. We have used a period of \( \tau=0.2 \).

    + + + +
    +
    +
    +
    +
    + 
    import numpy as np
+import math
+from scipy import signal
+import matplotlib.pyplot as plt
+
+# number of points                                                                                       
+n = 500
+# start and final times                                                                                  
+t0 = 0.0
+tn = 1.0
+# Period                                                                                                 
+t = np.linspace(t0, tn, n, endpoint=False)
+SqrSignal = np.zeros(n)
+SqrSignal = 1.0+signal.square(2*np.pi*5*t)
+plt.plot(t, SqrSignal)
+plt.ylim(-0.5, 2.5)
+plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +

    For the sinusoidal example the +period is \( \tau=2\pi/\omega \). However, higher harmonics can also +satisfy the periodicity requirement. In general, any force that +satisfies the periodicity requirement can be expressed as a sum over +harmonics,

    $$ \begin{equation} -\tag{23} -F(t)=\frac{f_0}{2}+\sum_{n>0}f_n\cos(n\omega t)+g_n\sin(n\omega t). +F(t)=\frac{f_0}{2}+\sum_{n>0} f_n\cos(2n\pi t/\tau)+g_n\sin(2n\pi t/\tau). +\tag{22} \end{equation} $$ -

    The solutions for \( x(t) \) then come from replacing \( \omega \) with -\( n\omega \) for each term in the particular solution, -

    - -$$ -\begin{eqnarray} -x_p(t)&=&\frac{f_0}{2k}+\sum_{n>0} \alpha_n\cos(n\omega t-\delta_n)+\beta_n\sin(n\omega t-\delta_n),\\ -\nonumber -\alpha_n&=&\frac{f_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\ -\nonumber -\beta_n&=&\frac{g_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\ -\nonumber -\delta_n&=&\tan^{-1}\left(\frac{2\beta n\omega}{\omega_0^2-n^2\omega^2}\right). -\end{eqnarray} -$$ -

    @@ -456,7 +487,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs069.html b/doc/pub/week42/html/._week42-bs069.html index aa44c09b0..26ee969a4 100644 --- a/doc/pub/week42/html/._week42-bs069.html +++ b/doc/pub/week42/html/._week42-bs069.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,70 +404,37 @@ MathJax.Hub.Config({

     

     

     

    -

    Finding the Coefficients

    +

    Wrapping up Fourier transforms

    -

    Because the forces have been applied for a long time, any non-zero -damping eliminates the homogenous parts of the solution, so one need -only consider the particular solution for each \( n \). -

    - -

    The problem is considered solved if one can find expressions for the -coefficients \( f_n \) and \( g_n \), even though the solutions are expressed -as an infinite sum. The coefficients can be extracted from the -function \( F(t) \) by -

    - -$$ -\begin{eqnarray} -\tag{24} -f_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\cos(2n\pi t/\tau),\\ -\nonumber -g_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\sin(2n\pi t/\tau). -\end{eqnarray} -$$ - -

    To check the consistency of these expressions and to verify -Eq. (24), one can insert the expansion of \( F(t) \) in -Eq. (23) into the expression for the coefficients in -Eq. (24) and see whether -

    - -$$ -\begin{eqnarray} -f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~\left\{ -\frac{f_0}{2}+\sum_{m>0}f_m\cos(m\omega t)+g_m\sin(m\omega t) -\right\}\cos(n\omega t). -\end{eqnarray} -$$ - -

    Immediately, one can throw away all the terms with \( g_m \) because they -convolute an even and an odd function. The term with \( f_0/2 \) -disappears because \( \cos(n\omega t) \) is equally positive and negative -over the interval and will integrate to zero. For all the terms -\( f_m\cos(m\omega t) \) appearing in the sum, one can use angle addition -formulas to see that \( \cos(m\omega t)\cos(n\omega -t)=(1/2)(\cos[(m+n)\omega t]+\cos[(m-n)\omega t] \). This will integrate -to zero unless \( m=n \). In that case the \( m=n \) term gives +

    We can write down the answer for +\( x_{pn}(t) \), by substituting \( f_n/m \) or \( g_n/m \) for \( F_0/m \). By +writing each factor \( 2n\pi t/\tau \) as \( n\omega t \), with \( \omega\equiv +2\pi/\tau \),

    $$ \begin{equation} -\int_{-\tau/2}^{\tau/2}dt~\cos^2(m\omega t)=\frac{\tau}{2}, -\tag{25} +\tag{23} +F(t)=\frac{f_0}{2}+\sum_{n>0}f_n\cos(n\omega t)+g_n\sin(n\omega t). \end{equation} $$ -

    and

    +

    The solutions for \( x(t) \) then come from replacing \( \omega \) with +\( n\omega \) for each term in the particular solution, +

    $$ \begin{eqnarray} -f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~f_n/2\\ +x_p(t)&=&\frac{f_0}{2k}+\sum_{n>0} \alpha_n\cos(n\omega t-\delta_n)+\beta_n\sin(n\omega t-\delta_n),\\ \nonumber -&=&f_n~\checkmark. +\alpha_n&=&\frac{f_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\ +\nonumber +\beta_n&=&\frac{g_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\ +\nonumber +\delta_n&=&\tan^{-1}\left(\frac{2\beta n\omega}{\omega_0^2-n^2\omega^2}\right). \end{eqnarray} $$ -

    The same method can be used to check for the consistency of \( g_n \).

    @@ -489,7 +461,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs070.html b/doc/pub/week42/html/._week42-bs070.html index 6c999bd66..524b5a298 100644 --- a/doc/pub/week42/html/._week42-bs070.html +++ b/doc/pub/week42/html/._week42-bs070.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,68 +404,70 @@ MathJax.Hub.Config({

     

     

     

    -

    Final words on Fourier Transforms

    +

    Finding the Coefficients

    -

    The code here uses the Fourier series applied to a -square wave signal. The code here -visualizes the various approximations given by Fourier series compared -with a square wave with period \( T=0.2 \) (dimensionless time), width \( 0.1 \) and max value of the force \( F=2 \). We -see that when we increase the number of components in the Fourier -series, the Fourier series approximation gets closer and closer to the -square wave signal. +

    Because the forces have been applied for a long time, any non-zero +damping eliminates the homogenous parts of the solution, so one need +only consider the particular solution for each \( n \).

    +

    The problem is considered solved if one can find expressions for the +coefficients \( f_n \) and \( g_n \), even though the solutions are expressed +as an infinite sum. The coefficients can be extracted from the +function \( F(t) \) by +

    - -
    -
    -
    -
    -
    - 
    import numpy as np
-import math
-from scipy import signal
-import matplotlib.pyplot as plt
+$$
+\begin{eqnarray}
+\tag{24}
+f_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\cos(2n\pi t/\tau),\\
+\nonumber
+g_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\sin(2n\pi t/\tau).
+\end{eqnarray}
+$$
 
-# number of points                                                                                       
-n = 500
-# start and final times                                                                                  
-t0 = 0.0
-tn = 1.0
-# Period                                                                                                 
-T =0.2
-# Max value of square signal                                                                             
-Fmax= 2.0
-# Width of signal   
-Width = 0.1
-t = np.linspace(t0, tn, n, endpoint=False)
-SqrSignal = np.zeros(n)
-FourierSeriesSignal = np.zeros(n)
-SqrSignal = 1.0+signal.square(2*np.pi*5*t+np.pi*Width/T)
-a0 = Fmax*Width/T
-FourierSeriesSignal = a0
-Factor = 2.0*Fmax/np.pi
-for i in range(1,500):
-    FourierSeriesSignal += Factor/(i)*np.sin(np.pi*i*Width/T)*np.cos(i*t*2*np.pi/T)
-plt.plot(t, SqrSignal)
-plt.plot(t, FourierSeriesSignal)
-plt.ylim(-0.5, 2.5)
-plt.show()
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    +

    To check the consistency of these expressions and to verify +Eq. (24), one can insert the expansion of \( F(t) \) in +Eq. (23) into the expression for the coefficients in +Eq. (24) and see whether +

    +$$ +\begin{eqnarray} +f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~\left\{ +\frac{f_0}{2}+\sum_{m>0}f_m\cos(m\omega t)+g_m\sin(m\omega t) +\right\}\cos(n\omega t). +\end{eqnarray} +$$ + +

    Immediately, one can throw away all the terms with \( g_m \) because they +convolute an even and an odd function. The term with \( f_0/2 \) +disappears because \( \cos(n\omega t) \) is equally positive and negative +over the interval and will integrate to zero. For all the terms +\( f_m\cos(m\omega t) \) appearing in the sum, one can use angle addition +formulas to see that \( \cos(m\omega t)\cos(n\omega +t)=(1/2)(\cos[(m+n)\omega t]+\cos[(m-n)\omega t] \). This will integrate +to zero unless \( m=n \). In that case the \( m=n \) term gives +

    + +$$ +\begin{equation} +\int_{-\tau/2}^{\tau/2}dt~\cos^2(m\omega t)=\frac{\tau}{2}, +\tag{25} +\end{equation} +$$ + +

    and

    + +$$ +\begin{eqnarray} +f_n&=?&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~f_n/2\\ +\nonumber +&=&f_n~\checkmark. +\end{eqnarray} +$$ + +

    The same method can be used to check for the consistency of \( g_n \).

    @@ -487,7 +494,7 @@ plt.show()

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    • diff --git a/doc/pub/week42/html/._week42-bs071.html b/doc/pub/week42/html/._week42-bs071.html index 7c3d041c7..dbc056f34 100644 --- a/doc/pub/week42/html/._week42-bs071.html +++ b/doc/pub/week42/html/._week42-bs071.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,23 +404,68 @@ MathJax.Hub.Config({

     

     

     

    -

    Two-dimensional Objects

    +

    Final words on Fourier Transforms

    -

    We often use convolutions over more than one dimension at a time. If -we have a two-dimensional image \( I \) as input, we can have a filter -defined by a two-dimensional kernel \( K \). This leads to an output \( S \) +

    The code here uses the Fourier series applied to a +square wave signal. The code here +visualizes the various approximations given by Fourier series compared +with a square wave with period \( T=0.2 \) (dimensionless time), width \( 0.1 \) and max value of the force \( F=2 \). We +see that when we increase the number of components in the Fourier +series, the Fourier series approximation gets closer and closer to the +square wave signal.

    -$$ -S_(i,j)=(I * K)(i,j) = \sum_m\sum_n I(m,n)K(i-m,j-n). -$$ -

    Convolution is a commutatitave process, which means we can rewrite this equation as

    -$$ -S_(i,j)=(I * K)(i,j) = \sum_m\sum_n I(i-m,j-n)K(m,n). -$$ + +
    +
    +
    +
    +
    + 
    import numpy as np
+import math
+from scipy import signal
+import matplotlib.pyplot as plt
+
+# number of points                                                                                       
+n = 500
+# start and final times                                                                                  
+t0 = 0.0
+tn = 1.0
+# Period                                                                                                 
+T =0.2
+# Max value of square signal                                                                             
+Fmax= 2.0
+# Width of signal   
+Width = 0.1
+t = np.linspace(t0, tn, n, endpoint=False)
+SqrSignal = np.zeros(n)
+FourierSeriesSignal = np.zeros(n)
+SqrSignal = 1.0+signal.square(2*np.pi*5*t+np.pi*Width/T)
+a0 = Fmax*Width/T
+FourierSeriesSignal = a0
+Factor = 2.0*Fmax/np.pi
+for i in range(1,500):
+    FourierSeriesSignal += Factor/(i)*np.sin(np.pi*i*Width/T)*np.cos(i*t*2*np.pi/T)
+plt.plot(t, SqrSignal)
+plt.plot(t, FourierSeriesSignal)
+plt.ylim(-0.5, 2.5)
+plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    -

    Normally the latter is more straightforward to implement in a machine elarning library since there is less variation in the range of values of \( m \) and \( n \).

    @@ -442,7 +492,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs072.html b/doc/pub/week42/html/._week42-bs072.html index fd753b163..7e45b2d9f 100644 --- a/doc/pub/week42/html/._week42-bs072.html +++ b/doc/pub/week42/html/._week42-bs072.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,13 +404,23 @@ MathJax.Hub.Config({

     

     

     

    -

    Cross-Correlation

    +

    Two-dimensional Objects

    + +

    We often use convolutions over more than one dimension at a time. If +we have a two-dimensional image \( I \) as input, we can have a filter +defined by a two-dimensional kernel \( K \). This leads to an output \( S \) +

    -

    Many deep learning libraries implement cross-correlation instead of convolution

    $$ -S_(i,j)=(I * K)(i,j) = \sum_m\sum_n I(i+m,j-+)K(m,n). +S_(i,j)=(I * K)(i,j) = \sum_m\sum_n I(m,n)K(i-m,j-n). $$ +

    Convolution is a commutatitave process, which means we can rewrite this equation as

    +$$ +S_(i,j)=(I * K)(i,j) = \sum_m\sum_n I(i-m,j-n)K(m,n). +$$ + +

    Normally the latter is more straightforward to implement in a machine elarning library since there is less variation in the range of values of \( m \) and \( n \).

    @@ -432,7 +447,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs073.html b/doc/pub/week42/html/._week42-bs073.html index 11c996cb7..fbfeff6dd 100644 --- a/doc/pub/week42/html/._week42-bs073.html +++ b/doc/pub/week42/html/._week42-bs073.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,33 +404,13 @@ MathJax.Hub.Config({

     

     

     

    -

    More on Dimensionalities

    - -

    In feilds like signal processing (and imaging as well), one designs -so-called filters. These filters are defined by the convolutions and -are often hand-crafted. One may specify filters for smoothing, edge -detection, frequency reshaping, and similar operations. However with -neural networks the idea is to automatically learn the filters and use -many of them in conjunction with non-linear operations (activation -functions). -

    - -

    As an example consider a neural network operating on sound sequence -data. Assume that we an input vector \( \boldsymbol{x} \) of length \( d=10^6 \). We -construct then a neural network with onle hidden layer only with -\( 10^4 \) nodes. This means that we will have a weight matrix with -\( 10^4\times 10^6=10^{10} \) weights to be determined, together with \( 10^4 \) biases. -

    - -

    Assume furthermore that we have an output layer which is meant to train whether the sound sequence represents a human voice (true) or something else (false). -It means that we have only one output node. But since this output node connects to \( 10^4 \) nodes in the hidden layer, there are in total \( 10^4 \) weights to be determined for the output layer, plus one bias. In total we have -

    +

    Cross-Correlation

    +

    Many deep learning libraries implement cross-correlation instead of convolution

    $$ -\mathrm{NumberParameters}=10^{10}+10^4+10^4+1 \approx 10^{10}, +S_(i,j)=(I * K)(i,j) = \sum_m\sum_n I(i+m,j-+)K(m,n). $$ -

    that is ten billion parameters to determine.

    @@ -452,7 +437,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs074.html b/doc/pub/week42/html/._week42-bs074.html index 385a1acbb..548b39ee3 100644 --- a/doc/pub/week42/html/._week42-bs074.html +++ b/doc/pub/week42/html/._week42-bs074.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,26 +404,34 @@ MathJax.Hub.Config({

     

     

     

    -

    Further Dimensionality Remarks

    +

    More on Dimensionalities

    -

    In today’s architecture one can train such neural networks, however -this is a huge number of parameters for the task at hand. In general, -it is a very wasteful and inefficient use of dense matrices as -parameters. Just as importantly, such trained network parameters are -very specific for the type of input data on which they were trained -and the network is not likely to generalize easily to variations in -the input. +

    In feilds like signal processing (and imaging as well), one designs +so-called filters. These filters are defined by the convolutions and +are often hand-crafted. One may specify filters for smoothing, edge +detection, frequency reshaping, and similar operations. However with +neural networks the idea is to automatically learn the filters and use +many of them in conjunction with non-linear operations (activation +functions).

    -

    The main principles that justify convolutions is locality of -information and repetion of patterns within the signal. Sound samples -of the input in adjacent spots are much more likely to affect each -other than those that are very far away. Similarly, sounds are -repeated in multiple times in the signal. While slightly simplistic, -reasoning about such a sound example demonstrates this. The same -principles then apply to images and other similar data. +

    As an example consider a neural network operating on sound sequence +data. Assume that we an input vector \( \boldsymbol{x} \) of length \( d=10^6 \). We +construct then a neural network with onle hidden layer only with +\( 10^4 \) nodes. This means that we will have a weight matrix with +\( 10^4\times 10^6=10^{10} \) weights to be determined, together with \( 10^4 \) biases.

    +

    Assume furthermore that we have an output layer which is meant to train whether the sound sequence represents a human voice (true) or something else (false). +It means that we have only one output node. But since this output node connects to \( 10^4 \) nodes in the hidden layer, there are in total \( 10^4 \) weights to be determined for the output layer, plus one bias. In total we have +

    + +$$ +\mathrm{NumberParameters}=10^{10}+10^4+10^4+1 \approx 10^{10}, +$$ + +

    that is ten billion parameters to determine.

    +

      @@ -444,7 +457,7 @@ principles then apply to images and other similar data.
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    diff --git a/doc/pub/week42/html/._week42-bs075.html b/doc/pub/week42/html/._week42-bs075.html index bee4f3d5c..07d8a7c2f 100644 --- a/doc/pub/week42/html/._week42-bs075.html +++ b/doc/pub/week42/html/._week42-bs075.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,11 +404,26 @@ MathJax.Hub.Config({

     

     

     

    -

    CNNs in more detail, Lecture from IN5400

    +

    Further Dimensionality Remarks

    + +

    In today’s architecture one can train such neural networks, however +this is a huge number of parameters for the task at hand. In general, +it is a very wasteful and inefficient use of dense matrices as +parameters. Just as importantly, such trained network parameters are +very specific for the type of input data on which they were trained +and the network is not likely to generalize easily to variations in +the input. +

    + +

    The main principles that justify convolutions is locality of +information and repetion of patterns within the signal. Sound samples +of the input in adjacent spots are much more likely to affect each +other than those that are very far away. Similarly, sounds are +repeated in multiple times in the signal. While slightly simplistic, +reasoning about such a sound example demonstrates this. The same +principles then apply to images and other similar data. +

    -

      @@ -429,7 +449,7 @@ MathJax.Hub.Config({
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    diff --git a/doc/pub/week42/html/._week42-bs076.html b/doc/pub/week42/html/._week42-bs076.html index 366a126ba..393e78c76 100644 --- a/doc/pub/week42/html/._week42-bs076.html +++ b/doc/pub/week42/html/._week42-bs076.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,19 +404,11 @@ MathJax.Hub.Config({

     

     

     

    -

    CNNs in more detail, building convolutional neural networks in Tensorflow and Keras

    - -

    As discussed above, CNNs are neural networks built from the assumption that the inputs -to the network are 2D images. This is important because the number of features or pixels in images -grows very fast with the image size, and an enormous number of weights and biases are needed in order to build an accurate network. -

    - -

    As before, we still have our input, a hidden layer and an output. What's novel about convolutional networks -are the convolutional and pooling layers stacked in pairs between the input and the hidden layer. -In addition, the data is no longer represented as a 2D feature matrix, instead each input is a number of 2D -matrices, typically 1 for each color dimension (Red, Green, Blue). -

    +

    CNNs in more detail, Lecture from IN5400

    +

      @@ -437,7 +434,7 @@ matrices, typically 1 for each color dimension (Red, Green, Blue).
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    diff --git a/doc/pub/week42/html/._week42-bs077.html b/doc/pub/week42/html/._week42-bs077.html index de0832176..e24efd447 100644 --- a/doc/pub/week42/html/._week42-bs077.html +++ b/doc/pub/week42/html/._week42-bs077.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,15 +404,18 @@ MathJax.Hub.Config({

     

     

     

    -

    Setting it up

    +

    CNNs in more detail, building convolutional neural networks in Tensorflow and Keras

    -

    It means that to represent the entire -dataset of images, we require a 4D matrix or tensor. This tensor has the dimensions: +

    As discussed above, CNNs are neural networks built from the assumption that the inputs +to the network are 2D images. This is important because the number of features or pixels in images +grows very fast with the image size, and an enormous number of weights and biases are needed in order to build an accurate network.

    -$$ -(n_{inputs},\, n_{pixels, width},\, n_{pixels, height},\, depth) . -$$ +

    As before, we still have our input, a hidden layer and an output. What's novel about convolutional networks +are the convolutional and pooling layers stacked in pairs between the input and the hidden layer. +In addition, the data is no longer represented as a 2D feature matrix, instead each input is a number of 2D +matrices, typically 1 for each color dimension (Red, Green, Blue). +

    @@ -434,7 +442,7 @@ $$

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    • diff --git a/doc/pub/week42/html/._week42-bs078.html b/doc/pub/week42/html/._week42-bs078.html index bb7ec3823..1ffd7879a 100644 --- a/doc/pub/week42/html/._week42-bs078.html +++ b/doc/pub/week42/html/._week42-bs078.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,20 +404,14 @@ MathJax.Hub.Config({

     

     

     

    -

    The MNIST dataset again

    +

    Setting it up

    -

    The MNIST dataset consists of grayscale images with a pixel size of -\( 28\times 28 \), meaning we require \( 28 \times 28 = 724 \) weights to each -neuron in the first hidden layer. -

    - -

    If we were to analyze images of size \( 128\times 128 \) we would require -\( 128 \times 128 = 16384 \) weights to each neuron. Even worse if we were -dealing with color images, as most images are, we have an image matrix -of size \( 128\times 128 \) for each color dimension (Red, Green, Blue), -meaning 3 times the number of weights \( = 49152 \) are required for every -single neuron in the first hidden layer. +

    It means that to represent the entire +dataset of images, we require a 4D matrix or tensor. This tensor has the dimensions:

    +$$ +(n_{inputs},\, n_{pixels, width},\, n_{pixels, height},\, depth) . +$$

    @@ -440,7 +439,7 @@ single neuron in the first hidden layer.

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    • diff --git a/doc/pub/week42/html/._week42-bs079.html b/doc/pub/week42/html/._week42-bs079.html index 03ccf2f06..f0179d41d 100644 --- a/doc/pub/week42/html/._week42-bs079.html +++ b/doc/pub/week42/html/._week42-bs079.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,19 +404,19 @@ MathJax.Hub.Config({

     

     

     

    -

    Strong correlations

    +

    The MNIST dataset again

    -

    Images typically have strong local correlations, meaning that a small -part of the image varies little from its neighboring regions. If for -example we have an image of a blue car, we can roughly assume that a -small blue part of the image is surrounded by other blue regions. +

    The MNIST dataset consists of grayscale images with a pixel size of +\( 28\times 28 \), meaning we require \( 28 \times 28 = 724 \) weights to each +neuron in the first hidden layer.

    -

    Therefore, instead of connecting every single pixel to a neuron in the -first hidden layer, as we have previously done with deep neural -networks, we can instead connect each neuron to a small part of the -image (in all 3 RGB depth dimensions). The size of each small area is -fixed, and known as a receptive. +

    If we were to analyze images of size \( 128\times 128 \) we would require +\( 128 \times 128 = 16384 \) weights to each neuron. Even worse if we were +dealing with color images, as most images are, we have an image matrix +of size \( 128\times 128 \) for each color dimension (Red, Green, Blue), +meaning 3 times the number of weights \( = 49152 \) are required for every +single neuron in the first hidden layer.

    @@ -440,7 +445,7 @@ fixed, and known as a 88
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    • diff --git a/doc/pub/week42/html/._week42-bs080.html b/doc/pub/week42/html/._week42-bs080.html index 26e275173..12d3d82fa 100644 --- a/doc/pub/week42/html/._week42-bs080.html +++ b/doc/pub/week42/html/._week42-bs080.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -398,26 +403,22 @@ MathJax.Hub.Config({

     

     

     

    - -

    Layers of a CNN

    -

    The layers of a convolutional neural network arrange neurons in 3D: width, height and depth. -The input image is typically a square matrix of depth 3. + +

    Strong correlations

    + +

    Images typically have strong local correlations, meaning that a small +part of the image varies little from its neighboring regions. If for +example we have an image of a blue car, we can roughly assume that a +small blue part of the image is surrounded by other blue regions.

    -

    A convolution is performed on the image which outputs -a 3D volume of neurons. The weights to the input are arranged in a number of 2D matrices, known as filters. +

    Therefore, instead of connecting every single pixel to a neuron in the +first hidden layer, as we have previously done with deep neural +networks, we can instead connect each neuron to a small part of the +image (in all 3 RGB depth dimensions). The size of each small area is +fixed, and known as a receptive.

    -

    Each filter slides along the input image, taking the dot product -between each small part of the image and the filter, in all depth -dimensions. This is then passed through a non-linear function, -typically the Rectified Linear (ReLu) function, which serves as the -activation of the neurons in the first convolutional layer. This is -further passed through a pooling layer, which reduces the size of the -convolutional layer, e.g. by taking the maximum or average across some -small regions, and this serves as input to the next convolutional -layer. -

    @@ -444,7 +445,7 @@ layer.

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    • diff --git a/doc/pub/week42/html/._week42-bs081.html b/doc/pub/week42/html/._week42-bs081.html index 4f0c66daa..a2bfcc3f0 100644 --- a/doc/pub/week42/html/._week42-bs081.html +++ b/doc/pub/week42/html/._week42-bs081.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -398,19 +403,26 @@ MathJax.Hub.Config({

     

     

     

    - -

    Systematic reduction

    - -

    By systematically reducing the size of the input volume, through -convolution and pooling, the network should create representations of -small parts of the input, and then from them assemble representations -of larger areas. The final pooling layer is flattened to serve as -input to a hidden layer, such that each neuron in the final pooling -layer is connected to every single neuron in the hidden layer. This -then serves as input to the output layer, e.g. a softmax output for -classification. + +

    Layers of a CNN

    +

    The layers of a convolutional neural network arrange neurons in 3D: width, height and depth. +The input image is typically a square matrix of depth 3.

    +

    A convolution is performed on the image which outputs +a 3D volume of neurons. The weights to the input are arranged in a number of 2D matrices, known as filters. +

    + +

    Each filter slides along the input image, taking the dot product +between each small part of the image and the filter, in all depth +dimensions. This is then passed through a non-linear function, +typically the Rectified Linear (ReLu) function, which serves as the +activation of the neurons in the first convolutional layer. This is +further passed through a pooling layer, which reduces the size of the +convolutional layer, e.g. by taking the maximum or average across some +small regions, and this serves as input to the next convolutional +layer. +

    @@ -437,7 +449,7 @@ classification.

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    • diff --git a/doc/pub/week42/html/._week42-bs082.html b/doc/pub/week42/html/._week42-bs082.html index a01b973f2..7ccc8b4dc 100644 --- a/doc/pub/week42/html/._week42-bs082.html +++ b/doc/pub/week42/html/._week42-bs082.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,68 +404,17 @@ MathJax.Hub.Config({

     

     

     

    -

    Prerequisites: Collect and pre-process data

    +

    Systematic reduction

    - -
    -
    -
    -
    -
    - 
    # import necessary packages
-import numpy as np
-import matplotlib.pyplot as plt
-from sklearn import datasets
-
-
-# ensure the same random numbers appear every time
-np.random.seed(0)
-
-# display images in notebook
-%matplotlib inline
-plt.rcParams['figure.figsize'] = (12,12)
-
-
-# download MNIST dataset
-digits = datasets.load_digits()
-
-# define inputs and labels
-inputs = digits.images
-labels = digits.target
-
-# RGB images have a depth of 3
-# our images are grayscale so they should have a depth of 1
-inputs = inputs[:,:,:,np.newaxis]
-
-print("inputs = (n_inputs, pixel_width, pixel_height, depth) = " + str(inputs.shape))
-print("labels = (n_inputs) = " + str(labels.shape))
-
-
-# choose some random images to display
-n_inputs = len(inputs)
-indices = np.arange(n_inputs)
-random_indices = np.random.choice(indices, size=5)
-
-for i, image in enumerate(digits.images[random_indices]):
-    plt.subplot(1, 5, i+1)
-    plt.axis('off')
-    plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest')
-    plt.title("Label: %d" % digits.target[random_indices[i]])
-plt.show()
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    +

    By systematically reducing the size of the input volume, through +convolution and pooling, the network should create representations of +small parts of the input, and then from them assemble representations +of larger areas. The final pooling layer is flattened to serve as +input to a hidden layer, such that each neuron in the final pooling +layer is connected to every single neuron in the hidden layer. This +then serves as input to the output layer, e.g. a softmax output for +classification. +

    @@ -488,7 +442,7 @@ plt.show()

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    • diff --git a/doc/pub/week42/html/._week42-bs083.html b/doc/pub/week42/html/._week42-bs083.html index c8ff39bc2..0af4301f2 100644 --- a/doc/pub/week42/html/._week42-bs083.html +++ b/doc/pub/week42/html/._week42-bs083.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,7 +404,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Importing Keras and Tensorflow

    +

    Prerequisites: Collect and pre-process data

    @@ -407,28 +412,46 @@ MathJax.Hub.Config({
    - 
    from tensorflow.keras import datasets, layers, models
-from tensorflow.keras.layers import Input
-from tensorflow.keras.models import Sequential      #This allows appending layers to existing models
-from tensorflow.keras.layers import Dense           #This allows defining the characteristics of a particular layer
-from tensorflow.keras import optimizers             #This allows using whichever optimiser we want (sgd,adam,RMSprop)
-from tensorflow.keras import regularizers           #This allows using whichever regularizer we want (l1,l2,l1_l2)
-from tensorflow.keras.utils import to_categorical   #This allows using categorical cross entropy as the cost function
-#from tensorflow.keras import Conv2D
-#from tensorflow.keras import MaxPooling2D
-#from tensorflow.keras import Flatten
+  
    # import necessary packages
+import numpy as np
+import matplotlib.pyplot as plt
+from sklearn import datasets
 
-from sklearn.model_selection import train_test_split
 
-# representation of labels
-labels = to_categorical(labels)
+# ensure the same random numbers appear every time
+np.random.seed(0)
 
-# split into train and test data
-# one-liner from scikit-learn library
-train_size = 0.8
-test_size = 1 - train_size
-X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size,
-                                                    test_size=test_size)
+# display images in notebook
+%matplotlib inline
+plt.rcParams['figure.figsize'] = (12,12)
+
+
+# download MNIST dataset
+digits = datasets.load_digits()
+
+# define inputs and labels
+inputs = digits.images
+labels = digits.target
+
+# RGB images have a depth of 3
+# our images are grayscale so they should have a depth of 1
+inputs = inputs[:,:,:,np.newaxis]
+
+print("inputs = (n_inputs, pixel_width, pixel_height, depth) = " + str(inputs.shape))
+print("labels = (n_inputs) = " + str(labels.shape))
+
+
+# choose some random images to display
+n_inputs = len(inputs)
+indices = np.arange(n_inputs)
+random_indices = np.random.choice(indices, size=5)
+
+for i, image in enumerate(digits.images[random_indices]):
+    plt.subplot(1, 5, i+1)
+    plt.axis('off')
+    plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest')
+    plt.title("Label: %d" % digits.target[random_indices[i]])
+plt.show()
    @@ -469,6 +492,8 @@ X_train, X_test, Y_train, Y_test = train_tes
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    • diff --git a/doc/pub/week42/html/._week42-bs084.html b/doc/pub/week42/html/._week42-bs084.html index e8aa662ca..5168c554e 100644 --- a/doc/pub/week42/html/._week42-bs084.html +++ b/doc/pub/week42/html/._week42-bs084.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -398,9 +403,8 @@ MathJax.Hub.Config({

     

     

     

    - -

    Running with Keras

    - + +

    Importing Keras and Tensorflow

    @@ -408,32 +412,28 @@ MathJax.Hub.Config({
    - 
    def create_convolutional_neural_network_keras(input_shape, receptive_field,
-                                              n_filters, n_neurons_connected, n_categories,
-                                              eta, lmbd):
-    model = Sequential()
-    model.add(layers.Conv2D(n_filters, (receptive_field, receptive_field), input_shape=input_shape, padding='same',
-              activation='relu', kernel_regularizer=regularizers.l2(lmbd)))
-    model.add(layers.MaxPooling2D(pool_size=(2, 2)))
-    model.add(layers.Flatten())
-    model.add(layers.Dense(n_neurons_connected, activation='relu', kernel_regularizer=regularizers.l2(lmbd)))
-    model.add(layers.Dense(n_categories, activation='softmax', kernel_regularizer=regularizers.l2(lmbd)))
-    
-    sgd = optimizers.SGD(lr=eta)
-    model.compile(loss='categorical_crossentropy', optimizer=sgd, metrics=['accuracy'])
-    
-    return model
+  
    from tensorflow.keras import datasets, layers, models
+from tensorflow.keras.layers import Input
+from tensorflow.keras.models import Sequential      #This allows appending layers to existing models
+from tensorflow.keras.layers import Dense           #This allows defining the characteristics of a particular layer
+from tensorflow.keras import optimizers             #This allows using whichever optimiser we want (sgd,adam,RMSprop)
+from tensorflow.keras import regularizers           #This allows using whichever regularizer we want (l1,l2,l1_l2)
+from tensorflow.keras.utils import to_categorical   #This allows using categorical cross entropy as the cost function
+#from tensorflow.keras import Conv2D
+#from tensorflow.keras import MaxPooling2D
+#from tensorflow.keras import Flatten
 
-epochs = 100
-batch_size = 100
-input_shape = X_train.shape[1:4]
-receptive_field = 3
-n_filters = 10
-n_neurons_connected = 50
-n_categories = 10
+from sklearn.model_selection import train_test_split
 
-eta_vals = np.logspace(-5, 1, 7)
-lmbd_vals = np.logspace(-5, 1, 7)
+# representation of labels
+labels = to_categorical(labels)
+
+# split into train and test data
+# one-liner from scikit-learn library
+train_size = 0.8
+test_size = 1 - train_size
+X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size,
+                                                    test_size=test_size)
    @@ -473,6 +473,7 @@ lmbd_vals = np.
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    • diff --git a/doc/pub/week42/html/._week42-bs085.html b/doc/pub/week42/html/._week42-bs085.html index 0b103c795..b14a8fe0c 100644 --- a/doc/pub/week42/html/._week42-bs085.html +++ b/doc/pub/week42/html/._week42-bs085.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -398,8 +403,8 @@ MathJax.Hub.Config({

     

     

     

    - -

    Final part

    + +

    Running with Keras

    @@ -408,22 +413,32 @@ MathJax.Hub.Config({
    - 
    CNN_keras = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
-        
-for i, eta in enumerate(eta_vals):
-    for j, lmbd in enumerate(lmbd_vals):
-        CNN = create_convolutional_neural_network_keras(input_shape, receptive_field,
+  
    def create_convolutional_neural_network_keras(input_shape, receptive_field,
                                               n_filters, n_neurons_connected, n_categories,
-                                              eta, lmbd)
-        CNN.fit(X_train, Y_train, epochs=epochs, batch_size=batch_size, verbose=0)
-        scores = CNN.evaluate(X_test, Y_test)
-        
-        CNN_keras[i][j] = CNN
-        
-        print("Learning rate = ", eta)
-        print("Lambda = ", lmbd)
-        print("Test accuracy: %.3f" % scores[1])
-        print()
+                                              eta, lmbd):
+    model = Sequential()
+    model.add(layers.Conv2D(n_filters, (receptive_field, receptive_field), input_shape=input_shape, padding='same',
+              activation='relu', kernel_regularizer=regularizers.l2(lmbd)))
+    model.add(layers.MaxPooling2D(pool_size=(2, 2)))
+    model.add(layers.Flatten())
+    model.add(layers.Dense(n_neurons_connected, activation='relu', kernel_regularizer=regularizers.l2(lmbd)))
+    model.add(layers.Dense(n_categories, activation='softmax', kernel_regularizer=regularizers.l2(lmbd)))
+    
+    sgd = optimizers.SGD(lr=eta)
+    model.compile(loss='categorical_crossentropy', optimizer=sgd, metrics=['accuracy'])
+    
+    return model
+
+epochs = 100
+batch_size = 100
+input_shape = X_train.shape[1:4]
+receptive_field = 3
+n_filters = 10
+n_neurons_connected = 50
+n_categories = 10
+
+eta_vals = np.logspace(-5, 1, 7)
+lmbd_vals = np.logspace(-5, 1, 7)
    @@ -462,6 +477,7 @@ MathJax.Hub.Config({
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    • diff --git a/doc/pub/week42/html/._week42-bs086.html b/doc/pub/week42/html/._week42-bs086.html index 405baa9c5..956aa4410 100644 --- a/doc/pub/week42/html/._week42-bs086.html +++ b/doc/pub/week42/html/._week42-bs086.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,7 +404,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Final visualization

    +

    Final part

    @@ -408,36 +413,22 @@ MathJax.Hub.Config({
    - 
    # visual representation of grid search
-# uses seaborn heatmap, could probably do this in matplotlib
-import seaborn as sns
-
-sns.set()
-
-train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
-test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
-
-for i in range(len(eta_vals)):
-    for j in range(len(lmbd_vals)):
-        CNN = CNN_keras[i][j]
-
-        train_accuracy[i][j] = CNN.evaluate(X_train, Y_train)[1]
-        test_accuracy[i][j] = CNN.evaluate(X_test, Y_test)[1]
-
+  
    CNN_keras = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
         
-fig, ax = plt.subplots(figsize = (10, 10))
-sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis")
-ax.set_title("Training Accuracy")
-ax.set_ylabel("$\eta$")
-ax.set_xlabel("$\lambda$")
-plt.show()
-
-fig, ax = plt.subplots(figsize = (10, 10))
-sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
-ax.set_title("Test Accuracy")
-ax.set_ylabel("$\eta$")
-ax.set_xlabel("$\lambda$")
-plt.show()
+for i, eta in enumerate(eta_vals):
+    for j, lmbd in enumerate(lmbd_vals):
+        CNN = create_convolutional_neural_network_keras(input_shape, receptive_field,
+                                              n_filters, n_neurons_connected, n_categories,
+                                              eta, lmbd)
+        CNN.fit(X_train, Y_train, epochs=epochs, batch_size=batch_size, verbose=0)
+        scores = CNN.evaluate(X_test, Y_test)
+        
+        CNN_keras[i][j] = CNN
+        
+        print("Learning rate = ", eta)
+        print("Lambda = ", lmbd)
+        print("Test accuracy: %.3f" % scores[1])
+        print()
    @@ -475,6 +466,7 @@ plt.show()
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    • diff --git a/doc/pub/week42/html/._week42-bs087.html b/doc/pub/week42/html/._week42-bs087.html index 046d6ef56..c1ca5e5ff 100644 --- a/doc/pub/week42/html/._week42-bs087.html +++ b/doc/pub/week42/html/._week42-bs087.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,13 +404,7 @@ MathJax.Hub.Config({

     

     

     

    -

    The CIFAR01 data set

    - -

    The CIFAR10 dataset contains 60,000 color images in 10 classes, with -6,000 images in each class. The dataset is divided into 50,000 -training images and 10,000 testing images. The classes are mutually -exclusive and there is no overlap between them. -

    +

    Final visualization

    @@ -414,16 +413,36 @@ exclusive and there is no overlap between them.
    - 
    import tensorflow as tf
+  
    # visual representation of grid search
+# uses seaborn heatmap, could probably do this in matplotlib
+import seaborn as sns
 
-from tensorflow.keras import datasets, layers, models
-import matplotlib.pyplot as plt
+sns.set()
 
-# We import the data set
-(train_images, train_labels), (test_images, test_labels) = datasets.cifar10.load_data()
+train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
+test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
 
-# Normalize pixel values to be between 0 and 1 by dividing by 255. 
-train_images, test_images = train_images / 255.0, test_images / 255.0
+for i in range(len(eta_vals)):
+    for j in range(len(lmbd_vals)):
+        CNN = CNN_keras[i][j]
+
+        train_accuracy[i][j] = CNN.evaluate(X_train, Y_train)[1]
+        test_accuracy[i][j] = CNN.evaluate(X_test, Y_test)[1]
+
+        
+fig, ax = plt.subplots(figsize = (10, 10))
+sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis")
+ax.set_title("Training Accuracy")
+ax.set_ylabel("$\eta$")
+ax.set_xlabel("$\lambda$")
+plt.show()
+
+fig, ax = plt.subplots(figsize = (10, 10))
+sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
+ax.set_title("Test Accuracy")
+ax.set_ylabel("$\eta$")
+ax.set_xlabel("$\lambda$")
+plt.show()
    @@ -460,6 +479,7 @@ train_images, test_images = train_images 91
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    • diff --git a/doc/pub/week42/html/._week42-bs088.html b/doc/pub/week42/html/._week42-bs088.html index e606f5550..c9f718b9b 100644 --- a/doc/pub/week42/html/._week42-bs088.html +++ b/doc/pub/week42/html/._week42-bs088.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,9 +404,13 @@ MathJax.Hub.Config({

     

     

     

    -

    Verifying the data set

    +

    The CIFAR01 data set

    -

    To verify that the dataset looks correct, let's plot the first 25 images from the training set and display the class name below each image.

    +

    The CIFAR10 dataset contains 60,000 color images in 10 classes, with +6,000 images in each class. The dataset is divided into 50,000 +training images and 10,000 testing images. The classes are mutually +exclusive and there is no overlap between them. +

    @@ -410,20 +419,16 @@ MathJax.Hub.Config({
    - 
    class_names = ['airplane', 'automobile', 'bird', 'cat', 'deer',
-               'dog', 'frog', 'horse', 'ship', 'truck']
-​
-plt.figure(figsize=(10,10))
-for i in range(25):
-    plt.subplot(5,5,i+1)
-    plt.xticks([])
-    plt.yticks([])
-    plt.grid(False)
-    plt.imshow(train_images[i], cmap=plt.cm.binary)
-    # The CIFAR labels happen to be arrays, 
-    # which is why you need the extra index
-    plt.xlabel(class_names[train_labels[i][0]])
-plt.show()
+  
    import tensorflow as tf
+
+from tensorflow.keras import datasets, layers, models
+import matplotlib.pyplot as plt
+
+# We import the data set
+(train_images, train_labels), (test_images, test_labels) = datasets.cifar10.load_data()
+
+# Normalize pixel values to be between 0 and 1 by dividing by 255. 
+train_images, test_images = train_images / 255.0, test_images / 255.0
    @@ -459,6 +464,7 @@ plt.show()
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    • diff --git a/doc/pub/week42/html/._week42-bs089.html b/doc/pub/week42/html/._week42-bs089.html index 251d49082..10d733977 100644 --- a/doc/pub/week42/html/._week42-bs089.html +++ b/doc/pub/week42/html/._week42-bs089.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,11 +404,9 @@ MathJax.Hub.Config({

     

     

     

    -

    Set up the model

    +

    Verifying the data set

    -

    The 6 lines of code below define the convolutional base using a common pattern: a stack of Conv2D and MaxPooling2D layers.

    - -

    As input, a CNN takes tensors of shape (image_height, image_width, color_channels), ignoring the batch size. If you are new to these dimensions, color_channels refers to (R,G,B). In this example, you will configure our CNN to process inputs of shape (32, 32, 3), which is the format of CIFAR images. You can do this by passing the argument input_shape to our first layer.

    +

    To verify that the dataset looks correct, let's plot the first 25 images from the training set and display the class name below each image.

    @@ -412,16 +415,20 @@ MathJax.Hub.Config({
    - 
    model = models.Sequential()
-model.add(layers.Conv2D(32, (3, 3), activation='relu', input_shape=(32, 32, 3)))
-model.add(layers.MaxPooling2D((2, 2)))
-model.add(layers.Conv2D(64, (3, 3), activation='relu'))
-model.add(layers.MaxPooling2D((2, 2)))
-model.add(layers.Conv2D(64, (3, 3), activation='relu'))
-
-# Let's display the architecture of our model so far.
-
-model.summary()
+  
    class_names = ['airplane', 'automobile', 'bird', 'cat', 'deer',
+               'dog', 'frog', 'horse', 'ship', 'truck']
+​
+plt.figure(figsize=(10,10))
+for i in range(25):
+    plt.subplot(5,5,i+1)
+    plt.xticks([])
+    plt.yticks([])
+    plt.grid(False)
+    plt.imshow(train_images[i], cmap=plt.cm.binary)
+    # The CIFAR labels happen to be arrays, 
+    # which is why you need the extra index
+    plt.xlabel(class_names[train_labels[i][0]])
+plt.show()
    @@ -437,7 +444,6 @@ model.summary()
    -

    You can see that the output of every Conv2D and MaxPooling2D layer is a 3D tensor of shape (height, width, channels). The width and height dimensions tend to shrink as you go deeper in the network. The number of output channels for each Conv2D layer is controlled by the first argument (e.g., 32 or 64). Typically, as the width and height shrink, you can afford (computationally) to add more output channels in each Conv2D layer.

    @@ -457,6 +463,7 @@ model.summary()

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    • diff --git a/doc/pub/week42/html/._week42-bs090.html b/doc/pub/week42/html/._week42-bs090.html index 5ccd0423b..1da34d82c 100644 --- a/doc/pub/week42/html/._week42-bs090.html +++ b/doc/pub/week42/html/._week42-bs090.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,16 +404,11 @@ MathJax.Hub.Config({

     

     

     

    -

    Add Dense layers on top

    +

    Set up the model

    -

    To complete our model, you will feed the last output tensor from the -convolutional base (of shape (4, 4, 64)) into one or more Dense layers -to perform classification. Dense layers take vectors as input (which -are 1D), while the current output is a 3D tensor. First, you will -flatten (or unroll) the 3D output to 1D, then add one or more Dense -layers on top. CIFAR has 10 output classes, so you use a final Dense -layer with 10 outputs and a softmax activation. -

    +

    The 6 lines of code below define the convolutional base using a common pattern: a stack of Conv2D and MaxPooling2D layers.

    + +

    As input, a CNN takes tensors of shape (image_height, image_width, color_channels), ignoring the batch size. If you are new to these dimensions, color_channels refers to (R,G,B). In this example, you will configure our CNN to process inputs of shape (32, 32, 3), which is the format of CIFAR images. You can do this by passing the argument input_shape to our first layer.

    @@ -417,10 +417,14 @@ layer with 10 outputs and a softmax activation.
    - 
    model.add(layers.Flatten())
-model.add(layers.Dense(64, activation='relu'))
-model.add(layers.Dense(10))
-Here's the complete architecture of our model.
+  
    model = models.Sequential()
+model.add(layers.Conv2D(32, (3, 3), activation='relu', input_shape=(32, 32, 3)))
+model.add(layers.MaxPooling2D((2, 2)))
+model.add(layers.Conv2D(64, (3, 3), activation='relu'))
+model.add(layers.MaxPooling2D((2, 2)))
+model.add(layers.Conv2D(64, (3, 3), activation='relu'))
+
+# Let's display the architecture of our model so far.
 
 model.summary()
 
    
@@ -438,7 +442,7 @@ model.summary()
    -

    As you can see, our (4, 4, 64) outputs were flattened into vectors of shape (1024) before going through two Dense layers.

    +

    You can see that the output of every Conv2D and MaxPooling2D layer is a 3D tensor of shape (height, width, channels). The width and height dimensions tend to shrink as you go deeper in the network. The number of output channels for each Conv2D layer is controlled by the first argument (e.g., 32 or 64). Typically, as the width and height shrink, you can afford (computationally) to add more output channels in each Conv2D layer.

    @@ -457,6 +461,7 @@ model.summary()

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    • diff --git a/doc/pub/week42/html/._week42-bs091.html b/doc/pub/week42/html/._week42-bs091.html index 85313a0b5..e3f2ac47a 100644 --- a/doc/pub/week42/html/._week42-bs091.html +++ b/doc/pub/week42/html/._week42-bs091.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,7 +404,16 @@ MathJax.Hub.Config({

     

     

     

    -

    Compile and train the model

    +

    Add Dense layers on top

    + +

    To complete our model, you will feed the last output tensor from the +convolutional base (of shape (4, 4, 64)) into one or more Dense layers +to perform classification. Dense layers take vectors as input (which +are 1D), while the current output is a 3D tensor. First, you will +flatten (or unroll) the 3D output to 1D, then add one or more Dense +layers on top. CIFAR has 10 output classes, so you use a final Dense +layer with 10 outputs and a softmax activation. +

    @@ -408,12 +422,12 @@ MathJax.Hub.Config({
    - 
    model.compile(optimizer='adam',
-              loss=tf.keras.losses.SparseCategoricalCrossentropy(from_logits=True),
-              metrics=['accuracy'])
-​
-history = model.fit(train_images, train_labels, epochs=10, 
-                    validation_data=(test_images, test_labels))
+  
    model.add(layers.Flatten())
+model.add(layers.Dense(64, activation='relu'))
+model.add(layers.Dense(10))
+Here's the complete architecture of our model.
+
+model.summary()
    @@ -429,6 +443,7 @@ history = model
    +

    As you can see, our (4, 4, 64) outputs were flattened into vectors of shape (1024) before going through two Dense layers.

    @@ -446,6 +461,7 @@ history = model

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    • diff --git a/doc/pub/week42/html/._week42-bs092.html b/doc/pub/week42/html/._week42-bs092.html index f58d73b35..5469749d0 100644 --- a/doc/pub/week42/html/._week42-bs092.html +++ b/doc/pub/week42/html/._week42-bs092.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -399,7 +404,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Finally, evaluate the model

    +

    Compile and train the model

    @@ -408,16 +413,12 @@ MathJax.Hub.Config({
    - 
    plt.plot(history.history['accuracy'], label='accuracy')
-plt.plot(history.history['val_accuracy'], label = 'val_accuracy')
-plt.xlabel('Epoch')
-plt.ylabel('Accuracy')
-plt.ylim([0.5, 1])
-plt.legend(loc='lower right')
-
-test_loss, test_acc = model.evaluate(test_images,  test_labels, verbose=2)
-
-print(test_acc)
+  
    model.compile(optimizer='adam',
+              loss=tf.keras.losses.SparseCategoricalCrossentropy(from_logits=True),
+              metrics=['accuracy'])
+​
+history = model.fit(train_images, train_labels, epochs=10, 
+                    validation_data=(test_images, test_labels))
    @@ -449,6 +450,8 @@ test_loss, test_acc = model91
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  • »
    • diff --git a/doc/pub/week42/html/week42-bs.html b/doc/pub/week42/html/week42-bs.html index b35848fde..f0afc8252 100644 --- a/doc/pub/week42/html/week42-bs.html +++ b/doc/pub/week42/html/week42-bs.html @@ -41,6 +41,10 @@ doconce format html week42.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -298,96 +302,97 @@ MathJax.Hub.Config({ @@ -442,7 +447,7 @@ MathJax.Hub.Config({
  • 9
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  • ...
    • -
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  • »
    • diff --git a/doc/pub/week42/html/week42-reveal.html b/doc/pub/week42/html/week42-reveal.html index 74e0e03b9..b3656a2f0 100644 --- a/doc/pub/week42/html/week42-reveal.html +++ b/doc/pub/week42/html/week42-reveal.html @@ -245,6 +245,17 @@ we will also study the usage of Back propagation and automatic differentiation + +

    For more details on the back propagation algorithm and automatic differentiation see

    +
      +

    1. https://www.jmlr.org/papers/volume18/17-468/17-468.pdf
      2. +

    2. https://deepimaging.github.io/lectures/lecture_11_Backpropagation.pdf
      3. +

    3. Slides 12-44 at URL":http://cs231n.stanford.edu/slides/2017/cs231n_2017_lecture4.pdf"
      4. +
    + +

    Solving ODEs with Deep Learning

    diff --git a/doc/pub/week42/html/week42-solarized.html b/doc/pub/week42/html/week42-solarized.html index ce467b326..89b12a444 100644 --- a/doc/pub/week42/html/week42-solarized.html +++ b/doc/pub/week42/html/week42-solarized.html @@ -68,6 +68,10 @@ div.toc p,a { 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -379,6 +383,15 @@ MathJax.Hub.Config({ we will also study the usage of Autograd in computing gradients for deep learning. For the documentation of Autograd and examples see the lectures slides from week 39 and the Autograd documentation.

    +









    +

    Back propagation and automatic differentiation

    + +

    For more details on the back propagation algorithm and automatic differentiation see

    +
      +
    1. https://www.jmlr.org/papers/volume18/17-468/17-468.pdf
      2. +
    2. https://deepimaging.github.io/lectures/lecture_11_Backpropagation.pdf
      3. +
    3. Slides 12-44 at URL":http://cs231n.stanford.edu/slides/2017/cs231n_2017_lecture4.pdf"
      4. +










    Solving ODEs with Deep Learning

    diff --git a/doc/pub/week42/html/week42.html b/doc/pub/week42/html/week42.html index 2060ea45a..ada8ea3d8 100644 --- a/doc/pub/week42/html/week42.html +++ b/doc/pub/week42/html/week42.html @@ -145,6 +145,10 @@ div.toc p,a { 2, None, 'using-automatic-differentiation'), + ('Back propagation and automatic differentiation', + 2, + None, + 'back-propagation-and-automatic-differentiation'), ('Solving ODEs with Deep Learning', 2, None, @@ -456,6 +460,15 @@ MathJax.Hub.Config({ we will also study the usage of Autograd in computing gradients for deep learning. For the documentation of Autograd and examples see the lectures slides from week 39 and the Autograd documentation.

    +









    +

    Back propagation and automatic differentiation

    + +

    For more details on the back propagation algorithm and automatic differentiation see

    +
      +
    1. https://www.jmlr.org/papers/volume18/17-468/17-468.pdf
      2. +
    2. https://deepimaging.github.io/lectures/lecture_11_Backpropagation.pdf
      3. +
    3. Slides 12-44 at URL":http://cs231n.stanford.edu/slides/2017/cs231n_2017_lecture4.pdf"
      4. +










    Solving ODEs with Deep Learning

    diff --git a/doc/pub/week42/ipynb/ipynb-week42-src.tar.gz b/doc/pub/week42/ipynb/ipynb-week42-src.tar.gz index 4d3f4e1e31a5c0e21b70a2058b97b288f9022fca..1f93a18ca6ecd195d0d82701e5e7d8dff0336926 100644 GIT binary patch delta 20 ccmcbyp7q9hRyO%=4u(5jjci-l7%znb09E$~qW}N^ delta 20 bcmcbyp7q9hRyO%=4u;N#Mz*bNjF-XzQoIL7 diff --git a/doc/pub/week42/ipynb/week42.ipynb b/doc/pub/week42/ipynb/week42.ipynb index b3e9846d5..b6be36723 100644 --- a/doc/pub/week42/ipynb/week42.ipynb +++ b/doc/pub/week42/ipynb/week42.ipynb @@ -2,7 +2,7 @@ "cells": [ { "cell_type": "markdown", - "id": "71969448", + "id": "abd8d829", "metadata": { "editable": true }, @@ -14,7 +14,7 @@ }, { "cell_type": "markdown", - "id": "5927a879", + "id": "094896b1", "metadata": { "editable": true }, @@ -29,7 +29,7 @@ }, { "cell_type": "markdown", - "id": "ab457c67", + "id": "154f7248", "metadata": { "editable": true }, @@ -69,7 +69,7 @@ }, { "cell_type": "markdown", - "id": "cb68f381", + "id": "b54219f8", "metadata": { "editable": true }, @@ -82,7 +82,24 @@ }, { "cell_type": "markdown", - "id": "e9757bb0", + "id": "e305d5ab", + "metadata": { + "editable": true + }, + "source": [ + "## Back propagation and automatic differentiation\n", + "\n", + "For more details on the back propagation algorithm and automatic differentiation see\n", + "1. \n", + "\n", + "2. \n", + "\n", + "3. Slides 12-44 at URL\":http://cs231n.stanford.edu/slides/2017/cs231n_2017_lecture4.pdf\"" + ] + }, + { + "cell_type": "markdown", + "id": "9e2ee8a5", "metadata": { "editable": true }, @@ -105,7 +122,7 @@ }, { "cell_type": "markdown", - "id": "c01e34d4", + "id": "7be55a45", "metadata": { "editable": true }, @@ -119,7 +136,7 @@ }, { "cell_type": "markdown", - "id": "73d51e66", + "id": "ae3ed46d", "metadata": { "editable": true }, @@ -136,7 +153,7 @@ }, { "cell_type": "markdown", - "id": "59bec857", + "id": "545ce0d6", "metadata": { "editable": true }, @@ -152,7 +169,7 @@ }, { "cell_type": "markdown", - "id": "2be8c0d1", + "id": "3886d27e", "metadata": { "editable": true }, @@ -164,7 +181,7 @@ }, { "cell_type": "markdown", - "id": "ce8cc983", + "id": "0615c51a", "metadata": { "editable": true }, @@ -182,7 +199,7 @@ }, { "cell_type": "markdown", - "id": "60181011", + "id": "760fd5b3", "metadata": { "editable": true }, @@ -203,7 +220,7 @@ }, { "cell_type": "markdown", - "id": "7bbdb7b0", + "id": "dd82a779", "metadata": { "editable": true }, @@ -220,7 +237,7 @@ }, { "cell_type": "markdown", - "id": "043b1c3d", + "id": "5a885957", "metadata": { "editable": true }, @@ -232,7 +249,7 @@ }, { "cell_type": "markdown", - "id": "ca974c55", + "id": "e0f8495d", "metadata": { "editable": true }, @@ -243,7 +260,7 @@ }, { "cell_type": "markdown", - "id": "e4e96ffc", + "id": "4656e257", "metadata": { "editable": true }, @@ -260,7 +277,7 @@ }, { "cell_type": "markdown", - "id": "d8794c9f", + "id": "58c60357", "metadata": { "editable": true }, @@ -271,7 +288,7 @@ }, { "cell_type": "markdown", - "id": "4bffa650", + "id": "a0e58166", "metadata": { "editable": true }, @@ -287,7 +304,7 @@ }, { "cell_type": "markdown", - "id": "0508c8f3", + "id": "9eb6e1a7", "metadata": { "editable": true }, @@ -299,7 +316,7 @@ }, { "cell_type": "markdown", - "id": "6f834b8e", + "id": "83e80946", "metadata": { "editable": true }, @@ -316,7 +333,7 @@ }, { "cell_type": "markdown", - "id": "dd0973d2", + "id": "8c004b4f", "metadata": { "editable": true }, @@ -328,7 +345,7 @@ }, { "cell_type": "markdown", - "id": "dcaa7eef", + "id": "d54fb1ec", "metadata": { "editable": true }, @@ -346,7 +363,7 @@ }, { "cell_type": "markdown", - "id": "b8c28d82", + "id": "18b9d06f", "metadata": { "editable": true }, @@ -356,7 +373,7 @@ }, { "cell_type": "markdown", - "id": "0f7655fc", + "id": "97866a0a", "metadata": { "editable": true }, @@ -368,7 +385,7 @@ }, { "cell_type": "markdown", - "id": "76091988", + "id": "69c6fcd3", "metadata": { "editable": true }, @@ -385,7 +402,7 @@ }, { "cell_type": "markdown", - "id": "6c67a7bf", + "id": "d499d3e5", "metadata": { "editable": true }, @@ -397,7 +414,7 @@ }, { "cell_type": "markdown", - "id": "cb573831", + "id": "c90ba23f", "metadata": { "editable": true }, @@ -408,7 +425,7 @@ }, { "cell_type": "markdown", - "id": "3eb0cc23", + "id": "f58b1555", "metadata": { "editable": true }, @@ -420,7 +437,7 @@ }, { "cell_type": "markdown", - "id": "31908004", + "id": "26c33544", "metadata": { "editable": true }, @@ -430,7 +447,7 @@ }, { "cell_type": "markdown", - "id": "f866092e", + "id": "dbad6700", "metadata": { "editable": true }, @@ -450,7 +467,7 @@ }, { "cell_type": "markdown", - "id": "2bec25cc", + "id": "d3f1064b", "metadata": { "editable": true }, @@ -467,7 +484,7 @@ }, { "cell_type": "markdown", - "id": "38507480", + "id": "eb16a920", "metadata": { "editable": true }, @@ -486,7 +503,7 @@ }, { "cell_type": "markdown", - "id": "7764da0d", + "id": "d168e8b4", "metadata": { "editable": true }, @@ -498,7 +515,7 @@ }, { "cell_type": "markdown", - "id": "f5de95d3", + "id": "d0d39e8c", "metadata": { "editable": true }, @@ -508,7 +525,7 @@ }, { "cell_type": "markdown", - "id": "7ce98a78", + "id": "32b55a44", "metadata": { "editable": true }, @@ -525,7 +542,7 @@ }, { "cell_type": "markdown", - "id": "04d06358", + "id": "01755fdb", "metadata": { "editable": true }, @@ -535,7 +552,7 @@ }, { "cell_type": "markdown", - "id": "2bf90658", + "id": "4f44cb94", "metadata": { "editable": true }, @@ -551,7 +568,7 @@ }, { "cell_type": "markdown", - "id": "045fd7f4", + "id": "a31c38af", "metadata": { "editable": true }, @@ -563,7 +580,7 @@ }, { "cell_type": "markdown", - "id": "016e04ba", + "id": "99ce7f31", "metadata": { "editable": true }, @@ -575,7 +592,7 @@ }, { "cell_type": "markdown", - "id": "8f78c934", + "id": "c483383a", "metadata": { "editable": true }, @@ -587,7 +604,7 @@ }, { "cell_type": "markdown", - "id": "99d301b8", + "id": "3c8ded88", "metadata": { "editable": true }, @@ -597,7 +614,7 @@ }, { "cell_type": "markdown", - "id": "ba694165", + "id": "649c7111", "metadata": { "editable": true }, @@ -609,7 +626,7 @@ }, { "cell_type": "markdown", - "id": "ba758927", + "id": "fb9dffec", "metadata": { "editable": true }, @@ -626,7 +643,7 @@ }, { "cell_type": "markdown", - "id": "997b874e", + "id": "52b7cf5b", "metadata": { "editable": true }, @@ -636,7 +653,7 @@ }, { "cell_type": "markdown", - "id": "9d9d67e1", + "id": "4466939f", "metadata": { "editable": true }, @@ -648,7 +665,7 @@ }, { "cell_type": "markdown", - "id": "8e4ecc83", + "id": "272354c9", "metadata": { "editable": true }, @@ -662,7 +679,7 @@ }, { "cell_type": "markdown", - "id": "f6f8703b", + "id": "74b1e07e", "metadata": { "editable": true }, @@ -678,7 +695,7 @@ }, { "cell_type": "markdown", - "id": "67c4966d", + "id": "c7253f4a", "metadata": { "editable": true }, @@ -690,7 +707,7 @@ }, { "cell_type": "markdown", - "id": "d1be2efb", + "id": "407ba791", "metadata": { "editable": true }, @@ -712,7 +729,7 @@ }, { "cell_type": "markdown", - "id": "29a8a509", + "id": "1c227fc8", "metadata": { "editable": true }, @@ -724,7 +741,7 @@ }, { "cell_type": "markdown", - "id": "f1687c57", + "id": "c2f07984", "metadata": { "editable": true }, @@ -747,7 +764,7 @@ }, { "cell_type": "markdown", - "id": "61a01126", + "id": "c2da4d4c", "metadata": { "editable": true }, @@ -763,7 +780,7 @@ }, { "cell_type": "markdown", - "id": "172c4d3e", + "id": "22eba71d", "metadata": { "editable": true }, @@ -775,7 +792,7 @@ }, { "cell_type": "markdown", - "id": "4244d014", + "id": "8418f5be", "metadata": { "editable": true }, @@ -799,7 +816,7 @@ }, { "cell_type": "markdown", - "id": "22bed77b", + "id": "c4fd8194", "metadata": { "editable": true }, @@ -811,7 +828,7 @@ }, { "cell_type": "markdown", - "id": "5eccb43c", + "id": "0fe93c43", "metadata": { "editable": true }, @@ -832,7 +849,7 @@ }, { "cell_type": "markdown", - "id": "03db9e20", + "id": "1a3fc54b", "metadata": { "editable": true }, @@ -844,7 +861,7 @@ }, { "cell_type": "markdown", - "id": "3f31c3f4", + "id": "df15e0a7", "metadata": { "editable": true }, @@ -863,7 +880,7 @@ }, { "cell_type": "markdown", - "id": "a21dbd3c", + "id": "b15b66c2", "metadata": { "editable": true }, @@ -873,7 +890,7 @@ }, { "cell_type": "markdown", - "id": "75e61111", + "id": "d81e1727", "metadata": { "editable": true }, @@ -887,7 +904,7 @@ }, { "cell_type": "markdown", - "id": "3a5ce8ec", + "id": "6db9a985", "metadata": { "editable": true }, @@ -899,7 +916,7 @@ }, { "cell_type": "markdown", - "id": "d50c7bd3", + "id": "f7f7ec9b", "metadata": { "editable": true }, @@ -911,7 +928,7 @@ }, { "cell_type": "markdown", - "id": "78f4451d", + "id": "913add99", "metadata": { "editable": true }, @@ -928,7 +945,7 @@ }, { "cell_type": "markdown", - "id": "1bf52e70", + "id": "ef0ca791", "metadata": { "editable": true }, @@ -940,7 +957,7 @@ }, { "cell_type": "markdown", - "id": "5e7a3453", + "id": "e054149f", "metadata": { "editable": true }, @@ -962,7 +979,7 @@ }, { "cell_type": "markdown", - "id": "1c3bcea4", + "id": "93cd08f7", "metadata": { "editable": true }, @@ -977,7 +994,7 @@ }, { "cell_type": "markdown", - "id": "fc0b0fe0", + "id": "bd12f38d", "metadata": { "editable": true }, @@ -988,7 +1005,7 @@ { "cell_type": "code", "execution_count": 1, - "id": "252b1de9", + "id": "a65da691", "metadata": { "collapsed": false, "editable": true @@ -1145,7 +1162,7 @@ }, { "cell_type": "markdown", - "id": "a7553a7c", + "id": "3fd575a2", "metadata": { "editable": true }, @@ -1160,7 +1177,7 @@ { "cell_type": "code", "execution_count": 2, - "id": "0896717e", + "id": "b02e1252", "metadata": { "collapsed": false, "editable": true @@ -1329,7 +1346,7 @@ }, { "cell_type": "markdown", - "id": "170ccb4a", + "id": "c1e9c039", "metadata": { "editable": true }, @@ -1342,7 +1359,7 @@ }, { "cell_type": "markdown", - "id": "37e97405", + "id": "c6a1e130", "metadata": { "editable": true }, @@ -1359,7 +1376,7 @@ }, { "cell_type": "markdown", - "id": "aaa64fe3", + "id": "9945d24c", "metadata": { "editable": true }, @@ -1375,7 +1392,7 @@ }, { "cell_type": "markdown", - "id": "0170d5e2", + "id": "f264d698", "metadata": { "editable": true }, @@ -1388,7 +1405,7 @@ }, { "cell_type": "markdown", - "id": "a89977c2", + "id": "4981f1e1", "metadata": { "editable": true }, @@ -1405,7 +1422,7 @@ }, { "cell_type": "markdown", - "id": "8a8f2604", + "id": "8fd7ea35", "metadata": { "editable": true }, @@ -1417,7 +1434,7 @@ }, { "cell_type": "markdown", - "id": "71220bc4", + "id": "98e96613", "metadata": { "editable": true }, @@ -1444,7 +1461,7 @@ }, { "cell_type": "markdown", - "id": "d154d14b", + "id": "64c1380d", "metadata": { "editable": true }, @@ -1457,7 +1474,7 @@ { "cell_type": "code", "execution_count": 3, - "id": "47695183", + "id": "a6e57516", "metadata": { "collapsed": false, "editable": true @@ -1631,7 +1648,7 @@ }, { "cell_type": "markdown", - "id": "0b4badf2", + "id": "1e447a21", "metadata": { "editable": true }, @@ -1651,7 +1668,7 @@ }, { "cell_type": "markdown", - "id": "14c42ad5", + "id": "5c370209", "metadata": { "editable": true }, @@ -1666,7 +1683,7 @@ }, { "cell_type": "markdown", - "id": "3e0342f3", + "id": "6af02d57", "metadata": { "editable": true }, @@ -1680,7 +1697,7 @@ }, { "cell_type": "markdown", - "id": "99084f97", + "id": "95ddd05b", "metadata": { "editable": true }, @@ -1696,7 +1713,7 @@ }, { "cell_type": "markdown", - "id": "509a25f3", + "id": "47e35a4b", "metadata": { "editable": true }, @@ -1706,7 +1723,7 @@ }, { "cell_type": "markdown", - "id": "460d8a50", + "id": "40698593", "metadata": { "editable": true }, @@ -1728,7 +1745,7 @@ }, { "cell_type": "markdown", - "id": "be0837e3", + "id": "ad3cd476", "metadata": { "editable": true }, @@ -1742,7 +1759,7 @@ { "cell_type": "code", "execution_count": 4, - "id": "71b7f89b", + "id": "dc1e94ba", "metadata": { "collapsed": false, "editable": true @@ -1818,7 +1835,7 @@ }, { "cell_type": "markdown", - "id": "7fa188b9", + "id": "548bed23", "metadata": { "editable": true }, @@ -1830,7 +1847,7 @@ }, { "cell_type": "markdown", - "id": "58338ea3", + "id": "2719f69d", "metadata": { "editable": true }, @@ -1847,7 +1864,7 @@ }, { "cell_type": "markdown", - "id": "e5816a56", + "id": "e22196f1", "metadata": { "editable": true }, @@ -1859,7 +1876,7 @@ }, { "cell_type": "markdown", - "id": "d987225a", + "id": "7d008f7f", "metadata": { "editable": true }, @@ -1874,7 +1891,7 @@ }, { "cell_type": "markdown", - "id": "0ecee4fa", + "id": "52d2549c", "metadata": { "editable": true }, @@ -1886,7 +1903,7 @@ }, { "cell_type": "markdown", - "id": "09d8fe88", + "id": "489dfc20", "metadata": { "editable": true }, @@ -1898,7 +1915,7 @@ }, { "cell_type": "markdown", - "id": "010c8d0e", + "id": "89cd1faa", "metadata": { "editable": true }, @@ -1910,7 +1927,7 @@ }, { "cell_type": "markdown", - "id": "8ff35418", + "id": "07cdc931", "metadata": { "editable": true }, @@ -1920,7 +1937,7 @@ }, { "cell_type": "markdown", - "id": "0400df1b", + "id": "9358036f", "metadata": { "editable": true }, @@ -1937,7 +1954,7 @@ }, { "cell_type": "markdown", - "id": "f57ef714", + "id": "bfc66633", "metadata": { "editable": true }, @@ -1949,7 +1966,7 @@ }, { "cell_type": "markdown", - "id": "0b530b7e", + "id": "9036d243", "metadata": { "editable": true }, @@ -1961,7 +1978,7 @@ }, { "cell_type": "markdown", - "id": "32074979", + "id": "cbe88af0", "metadata": { "editable": true }, @@ -1971,7 +1988,7 @@ }, { "cell_type": "markdown", - "id": "89ada272", + "id": "a347107b", "metadata": { "editable": true }, @@ -1983,7 +2000,7 @@ }, { "cell_type": "markdown", - "id": "c18937e6", + "id": "7433b8f3", "metadata": { "editable": true }, @@ -1994,7 +2011,7 @@ { "cell_type": "code", "execution_count": 5, - "id": "d4092b8d", + "id": "d0813f2d", "metadata": { "collapsed": false, "editable": true @@ -2155,7 +2172,7 @@ }, { "cell_type": "markdown", - "id": "7994cf0f", + "id": "f0abbbd3", "metadata": { "editable": true }, @@ -2177,7 +2194,7 @@ }, { "cell_type": "markdown", - "id": "9182ed7b", + "id": "5972f4a4", "metadata": { "editable": true }, @@ -2194,7 +2211,7 @@ }, { "cell_type": "markdown", - "id": "8d185853", + "id": "66486037", "metadata": { "editable": true }, @@ -2204,7 +2221,7 @@ }, { "cell_type": "markdown", - "id": "9208126e", + "id": "b95db726", "metadata": { "editable": true }, @@ -2219,7 +2236,7 @@ }, { "cell_type": "markdown", - "id": "8445769d", + "id": "56dfacfe", "metadata": { "editable": true }, @@ -2229,7 +2246,7 @@ }, { "cell_type": "markdown", - "id": "6dc8a1a6", + "id": "d6ea9a88", "metadata": { "editable": true }, @@ -2244,7 +2261,7 @@ }, { "cell_type": "markdown", - "id": "4f14fcad", + "id": "0cff1a42", "metadata": { "editable": true }, @@ -2255,7 +2272,7 @@ }, { "cell_type": "markdown", - "id": "2f125772", + "id": "44cd6a7b", "metadata": { "editable": true }, @@ -2275,7 +2292,7 @@ }, { "cell_type": "markdown", - "id": "5c2ac0df", + "id": "c85a0279", "metadata": { "editable": true }, @@ -2287,7 +2304,7 @@ }, { "cell_type": "markdown", - "id": "f317b533", + "id": "27958d4b", "metadata": { "editable": true }, @@ -2324,7 +2341,7 @@ }, { "cell_type": "markdown", - "id": "0d91012b", + "id": "ad6836c2", "metadata": { "editable": true }, @@ -2334,7 +2351,7 @@ }, { "cell_type": "markdown", - "id": "61d96b00", + "id": "0260be46", "metadata": { "editable": true }, @@ -2347,7 +2364,7 @@ { "cell_type": "code", "execution_count": 6, - "id": "7396cf44", + "id": "01fe4413", "metadata": { "collapsed": false, "editable": true @@ -2548,7 +2565,7 @@ }, { "cell_type": "markdown", - "id": "f4a5899e", + "id": "87ecdb6d", "metadata": { "editable": true }, @@ -2565,7 +2582,7 @@ }, { "cell_type": "markdown", - "id": "36d79444", + "id": "6420f00b", "metadata": { "editable": true }, @@ -2582,7 +2599,7 @@ }, { "cell_type": "markdown", - "id": "30e200d2", + "id": "ea80f33b", "metadata": { "editable": true }, @@ -2592,7 +2609,7 @@ }, { "cell_type": "markdown", - "id": "c6f5588d", + "id": "2701592c", "metadata": { "editable": true }, @@ -2607,7 +2624,7 @@ }, { "cell_type": "markdown", - "id": "33f64737", + "id": "d5e81572", "metadata": { "editable": true }, @@ -2621,7 +2638,7 @@ }, { "cell_type": "markdown", - "id": "f430c9fc", + "id": "5fad32fb", "metadata": { "editable": true }, @@ -2634,7 +2651,7 @@ }, { "cell_type": "markdown", - "id": "66874a09", + "id": "88dfb36f", "metadata": { "editable": true }, @@ -2654,7 +2671,7 @@ }, { "cell_type": "markdown", - "id": "a6c9f930", + "id": "ebaba66a", "metadata": { "editable": true }, @@ -2666,7 +2683,7 @@ }, { "cell_type": "markdown", - "id": "224769a1", + "id": "49e28269", "metadata": { "editable": true }, @@ -2678,7 +2695,7 @@ }, { "cell_type": "markdown", - "id": "a8a61cbd", + "id": "e76c1ce2", "metadata": { "editable": true }, @@ -2690,7 +2707,7 @@ }, { "cell_type": "markdown", - "id": "9354e37b", + "id": "1549d32f", "metadata": { "editable": true }, @@ -2700,7 +2717,7 @@ }, { "cell_type": "markdown", - "id": "cad003ef", + "id": "da6fb7be", "metadata": { "editable": true }, @@ -2712,7 +2729,7 @@ }, { "cell_type": "markdown", - "id": "b32db6fa", + "id": "b1e74a19", "metadata": { "editable": true }, @@ -2724,7 +2741,7 @@ }, { "cell_type": "markdown", - "id": "1c0a1d29", + "id": "48df8601", "metadata": { "editable": true }, @@ -2736,7 +2753,7 @@ }, { "cell_type": "markdown", - "id": "9702e3ce", + "id": "e23e991d", "metadata": { "editable": true }, @@ -2746,7 +2763,7 @@ }, { "cell_type": "markdown", - "id": "931352d7", + "id": "eef4c245", "metadata": { "editable": true }, @@ -2762,7 +2779,7 @@ }, { "cell_type": "markdown", - "id": "248122df", + "id": "00e49269", "metadata": { "editable": true }, @@ -2772,7 +2789,7 @@ }, { "cell_type": "markdown", - "id": "bbf13143", + "id": "4c8c0864", "metadata": { "editable": true }, @@ -2784,7 +2801,7 @@ }, { "cell_type": "markdown", - "id": "9e97a876", + "id": "cfd29570", "metadata": { "editable": true }, @@ -2801,7 +2818,7 @@ }, { "cell_type": "markdown", - "id": "1ab0fd56", + "id": "85cc21dd", "metadata": { "editable": true }, @@ -2811,7 +2828,7 @@ }, { "cell_type": "markdown", - "id": "a1974f42", + "id": "52446c6c", "metadata": { "editable": true }, @@ -2827,7 +2844,7 @@ }, { "cell_type": "markdown", - "id": "cceb6c23", + "id": "7a6a3a79", "metadata": { "editable": true }, @@ -2841,7 +2858,7 @@ }, { "cell_type": "markdown", - "id": "2f1b9b90", + "id": "51920eea", "metadata": { "editable": true }, @@ -2860,7 +2877,7 @@ { "cell_type": "code", "execution_count": 7, - "id": "6a5c64f3", + "id": "e79edf06", "metadata": { "collapsed": false, "editable": true @@ -2915,7 +2932,7 @@ }, { "cell_type": "markdown", - "id": "30af5bfc", + "id": "92235aff", "metadata": { "editable": true }, @@ -2945,7 +2962,7 @@ }, { "cell_type": "markdown", - "id": "4e462387", + "id": "ed32d0c3", "metadata": { "editable": true }, @@ -2974,7 +2991,7 @@ { "cell_type": "code", "execution_count": 8, - "id": "d77459b9", + "id": "df3d43b2", "metadata": { "collapsed": false, "editable": true @@ -3021,7 +3038,7 @@ }, { "cell_type": "markdown", - "id": "5a9b0863", + "id": "dfd5e194", "metadata": { "editable": true }, @@ -3047,7 +3064,7 @@ { "cell_type": "code", "execution_count": 9, - "id": "21da2908", + "id": "53d722cd", "metadata": { "collapsed": false, "editable": true @@ -3281,7 +3298,7 @@ }, { "cell_type": "markdown", - "id": "fa170659", + "id": "6e6643b1", "metadata": { "editable": true }, @@ -3293,7 +3310,7 @@ }, { "cell_type": "markdown", - "id": "b1c06ed3", + "id": "16cd3520", "metadata": { "editable": true }, @@ -3305,7 +3322,7 @@ }, { "cell_type": "markdown", - "id": "4f6fa124", + "id": "754d5669", "metadata": { "editable": true }, @@ -3317,7 +3334,7 @@ }, { "cell_type": "markdown", - "id": "aa1e0a64", + "id": "85b2c747", "metadata": { "editable": true }, @@ -3334,7 +3351,7 @@ }, { "cell_type": "markdown", - "id": "5dd980b7", + "id": "bcbf83a9", "metadata": { "editable": true }, @@ -3344,7 +3361,7 @@ }, { "cell_type": "markdown", - "id": "8297b7cb", + "id": "1659ec1a", "metadata": { "editable": true }, @@ -3356,7 +3373,7 @@ }, { "cell_type": "markdown", - "id": "374e70bc", + "id": "7c0fca91", "metadata": { "editable": true }, @@ -3373,7 +3390,7 @@ }, { "cell_type": "markdown", - "id": "3106b98d", + "id": "6e6aba88", "metadata": { "editable": true }, @@ -3384,7 +3401,7 @@ }, { "cell_type": "markdown", - "id": "a6fab25e", + "id": "4e881998", "metadata": { "editable": true }, @@ -3404,7 +3421,7 @@ }, { "cell_type": "markdown", - "id": "4323f63a", + "id": "1cafacfe", "metadata": { "editable": true }, @@ -3414,7 +3431,7 @@ }, { "cell_type": "markdown", - "id": "30cd4903", + "id": "cb906aba", "metadata": { "editable": true }, @@ -3440,7 +3457,7 @@ }, { "cell_type": "markdown", - "id": "05de0c2c", + "id": "32ccd5c7", "metadata": { "editable": true }, @@ -3456,7 +3473,7 @@ }, { "cell_type": "markdown", - "id": "be066349", + "id": "eae08701", "metadata": { "editable": true }, @@ -3467,7 +3484,7 @@ { "cell_type": "code", "execution_count": 10, - "id": "1b7f0fdc", + "id": "4c31a212", "metadata": { "collapsed": false, "editable": true @@ -3698,7 +3715,7 @@ }, { "cell_type": "markdown", - "id": "713c08cd", + "id": "01330e68", "metadata": { "editable": true }, @@ -3716,7 +3733,7 @@ }, { "cell_type": "markdown", - "id": "38146b2a", + "id": "4f7a9803", "metadata": { "editable": true }, @@ -3743,7 +3760,7 @@ }, { "cell_type": "markdown", - "id": "c0f0be03", + "id": "828e6019", "metadata": { "editable": true }, @@ -3765,7 +3782,7 @@ }, { "cell_type": "markdown", - "id": "b87e651f", + "id": "92fd4d62", "metadata": { "editable": true }, @@ -3783,7 +3800,7 @@ }, { "cell_type": "markdown", - "id": "75bf765b", + "id": "ac6ce257", "metadata": { "editable": true }, @@ -3813,7 +3830,7 @@ }, { "cell_type": "markdown", - "id": "c2fa3952", + "id": "ff9ebb73", "metadata": { "editable": true }, @@ -3843,7 +3860,7 @@ }, { "cell_type": "markdown", - "id": "2b6c7bd5", + "id": "db515315", "metadata": { "editable": true }, @@ -3883,7 +3900,7 @@ }, { "cell_type": "markdown", - "id": "699f8f55", + "id": "cdd402e7", "metadata": { "editable": true }, @@ -3912,7 +3929,7 @@ }, { "cell_type": "markdown", - "id": "a0e0f2f7", + "id": "49acc82f", "metadata": { "editable": true }, @@ -3934,7 +3951,7 @@ }, { "cell_type": "markdown", - "id": "fceed21f", + "id": "56d06b86", "metadata": { "editable": true }, @@ -3963,7 +3980,7 @@ }, { "cell_type": "markdown", - "id": "2177bf3e", + "id": "94c5721c", "metadata": { "editable": true }, @@ -3980,7 +3997,7 @@ }, { "cell_type": "markdown", - "id": "9279b9f6", + "id": "44c8b5a8", "metadata": { "editable": true }, @@ -4000,7 +4017,7 @@ }, { "cell_type": "markdown", - "id": "0d8b61f4", + "id": "3b32c3b2", "metadata": { "editable": true }, @@ -4012,7 +4029,7 @@ }, { "cell_type": "markdown", - "id": "c2cc9e41", + "id": "f9534bc9", "metadata": { "editable": true }, @@ -4024,7 +4041,7 @@ }, { "cell_type": "markdown", - "id": "6526c039", + "id": "0165e5ff", "metadata": { "editable": true }, @@ -4036,7 +4053,7 @@ }, { "cell_type": "markdown", - "id": "d525dcfd", + "id": "81d5bee2", "metadata": { "editable": true }, @@ -4046,7 +4063,7 @@ }, { "cell_type": "markdown", - "id": "45b90dfa", + "id": "596229fb", "metadata": { "editable": true }, @@ -4058,7 +4075,7 @@ }, { "cell_type": "markdown", - "id": "e1dadc3d", + "id": "82a4b22e", "metadata": { "editable": true }, @@ -4070,7 +4087,7 @@ }, { "cell_type": "markdown", - "id": "bf47cca8", + "id": "b3429713", "metadata": { "editable": true }, @@ -4085,7 +4102,7 @@ }, { "cell_type": "markdown", - "id": "762a9811", + "id": "d5817d7d", "metadata": { "editable": true }, @@ -4097,7 +4114,7 @@ }, { "cell_type": "markdown", - "id": "2b38b95f", + "id": "49b85586", "metadata": { "editable": true }, @@ -4107,7 +4124,7 @@ }, { "cell_type": "markdown", - "id": "8b22ca55", + "id": "97f1c627", "metadata": { "editable": true }, @@ -4119,7 +4136,7 @@ }, { "cell_type": "markdown", - "id": "85d3e8cb", + "id": "47abe7ce", "metadata": { "editable": true }, @@ -4129,7 +4146,7 @@ }, { "cell_type": "markdown", - "id": "adb52f17", + "id": "f4baf465", "metadata": { "editable": true }, @@ -4141,7 +4158,7 @@ }, { "cell_type": "markdown", - "id": "8612082c", + "id": "bf512f61", "metadata": { "editable": true }, @@ -4154,7 +4171,7 @@ }, { "cell_type": "markdown", - "id": "7087b85c", + "id": "8ee1612b", "metadata": { "editable": true }, @@ -4173,7 +4190,7 @@ }, { "cell_type": "markdown", - "id": "32fcb02b", + "id": "527003b3", "metadata": { "editable": true }, @@ -4185,7 +4202,7 @@ }, { "cell_type": "markdown", - "id": "0edb71dc", + "id": "ee0d7901", "metadata": { "editable": true }, @@ -4197,7 +4214,7 @@ }, { "cell_type": "markdown", - "id": "f7e5a188", + "id": "8eb6fdfb", "metadata": { "editable": true }, @@ -4207,7 +4224,7 @@ }, { "cell_type": "markdown", - "id": "cf163b16", + "id": "96c164e0", "metadata": { "editable": true }, @@ -4219,7 +4236,7 @@ }, { "cell_type": "markdown", - "id": "129bbc75", + "id": "412fb91f", "metadata": { "editable": true }, @@ -4229,7 +4246,7 @@ }, { "cell_type": "markdown", - "id": "226195f6", + "id": "70a30687", "metadata": { "editable": true }, @@ -4243,7 +4260,7 @@ }, { "cell_type": "markdown", - "id": "12d9f037", + "id": "0fced6a8", "metadata": { "editable": true }, @@ -4261,7 +4278,7 @@ }, { "cell_type": "markdown", - "id": "e2f90255", + "id": "7ce7e477", "metadata": { "editable": true }, @@ -4272,7 +4289,7 @@ }, { "cell_type": "markdown", - "id": "2081816f", + "id": "4c6024fb", "metadata": { "editable": true }, @@ -4290,7 +4307,7 @@ }, { "cell_type": "markdown", - "id": "cdc9d3ed", + "id": "25715e96", "metadata": { "editable": true }, @@ -4304,7 +4321,7 @@ }, { "cell_type": "markdown", - "id": "aaa777dd", + "id": "48e86964", "metadata": { "editable": true }, @@ -4316,7 +4333,7 @@ }, { "cell_type": "markdown", - "id": "11c5b5a8", + "id": "dd8b183e", "metadata": { "editable": true }, @@ -4328,7 +4345,7 @@ }, { "cell_type": "markdown", - "id": "3aa250ef", + "id": "acd194a6", "metadata": { "editable": true }, @@ -4342,7 +4359,7 @@ }, { "cell_type": "markdown", - "id": "961ede7c", + "id": "9cdc871b", "metadata": { "editable": true }, @@ -4360,7 +4377,7 @@ }, { "cell_type": "markdown", - "id": "576d6fdf", + "id": "963c1109", "metadata": { "editable": true }, @@ -4381,7 +4398,7 @@ }, { "cell_type": "markdown", - "id": "6cdc2b5e", + "id": "a21aea89", "metadata": { "editable": true }, @@ -4395,7 +4412,7 @@ }, { "cell_type": "markdown", - "id": "8c9c6cba", + "id": "2da231a6", "metadata": { "editable": true }, @@ -4408,7 +4425,7 @@ }, { "cell_type": "markdown", - "id": "e202ed85", + "id": "401164a3", "metadata": { "editable": true }, @@ -4421,7 +4438,7 @@ { "cell_type": "code", "execution_count": 11, - "id": "4f96f54f", + "id": "e3701802", "metadata": { "collapsed": false, "editable": true @@ -4449,7 +4466,7 @@ }, { "cell_type": "markdown", - "id": "3ca973bb", + "id": "586d9ba6", "metadata": { "editable": true }, @@ -4463,7 +4480,7 @@ }, { "cell_type": "markdown", - "id": "5093b52f", + "id": "6f38255c", "metadata": { "editable": true }, @@ -4481,7 +4498,7 @@ }, { "cell_type": "markdown", - "id": "c8221263", + "id": "ced5a425", "metadata": { "editable": true }, @@ -4496,7 +4513,7 @@ }, { "cell_type": "markdown", - "id": "b752f365", + "id": "d65e6b1a", "metadata": { "editable": true }, @@ -4514,7 +4531,7 @@ }, { "cell_type": "markdown", - "id": "33b723cf", + "id": "4780a0b5", "metadata": { "editable": true }, @@ -4525,7 +4542,7 @@ }, { "cell_type": "markdown", - "id": "7a44c734", + "id": "fdcb608d", "metadata": { "editable": true }, @@ -4545,7 +4562,7 @@ }, { "cell_type": "markdown", - "id": "d0471242", + "id": "5edd527a", "metadata": { "editable": true }, @@ -4564,7 +4581,7 @@ }, { "cell_type": "markdown", - "id": "1d098db1", + "id": "2a07ac4a", "metadata": { "editable": true }, @@ -4584,7 +4601,7 @@ }, { "cell_type": "markdown", - "id": "47dafb78", + "id": "08ddec1e", "metadata": { "editable": true }, @@ -4597,7 +4614,7 @@ }, { "cell_type": "markdown", - "id": "2a630583", + "id": "ebb221b1", "metadata": { "editable": true }, @@ -4613,7 +4630,7 @@ }, { "cell_type": "markdown", - "id": "e8398de8", + "id": "fb8cf771", "metadata": { "editable": true }, @@ -4630,7 +4647,7 @@ }, { "cell_type": "markdown", - "id": "f0942137", + "id": "2379d8ce", "metadata": { "editable": true }, @@ -4648,7 +4665,7 @@ }, { "cell_type": "markdown", - "id": "a12b2fb1", + "id": "36efca82", "metadata": { "editable": true }, @@ -4658,7 +4675,7 @@ }, { "cell_type": "markdown", - "id": "22383a13", + "id": "4b7c953d", "metadata": { "editable": true }, @@ -4674,7 +4691,7 @@ }, { "cell_type": "markdown", - "id": "681aa58e", + "id": "882e7174", "metadata": { "editable": true }, @@ -4684,7 +4701,7 @@ }, { "cell_type": "markdown", - "id": "afef3e8a", + "id": "9a1bfdd0", "metadata": { "editable": true }, @@ -4703,7 +4720,7 @@ { "cell_type": "code", "execution_count": 12, - "id": "f772f96e", + "id": "37f3dc9f", "metadata": { "collapsed": false, "editable": true @@ -4743,7 +4760,7 @@ }, { "cell_type": "markdown", - "id": "d5eb6970", + "id": "d6d7dbaf", "metadata": { "editable": true }, @@ -4757,7 +4774,7 @@ }, { "cell_type": "markdown", - "id": "7c489201", + "id": "b75ff7c6", "metadata": { "editable": true }, @@ -4769,7 +4786,7 @@ }, { "cell_type": "markdown", - "id": "73329326", + "id": "ab3c1523", "metadata": { "editable": true }, @@ -4779,7 +4796,7 @@ }, { "cell_type": "markdown", - "id": "5ba9ea72", + "id": "29af51e8", "metadata": { "editable": true }, @@ -4791,7 +4808,7 @@ }, { "cell_type": "markdown", - "id": "6f9772f7", + "id": "87064aee", "metadata": { "editable": true }, @@ -4801,7 +4818,7 @@ }, { "cell_type": "markdown", - "id": "7e26e8dd", + "id": "794e45d6", "metadata": { "editable": true }, @@ -4813,7 +4830,7 @@ }, { "cell_type": "markdown", - "id": "e648519f", + "id": "fa43e53c", "metadata": { "editable": true }, @@ -4825,7 +4842,7 @@ }, { "cell_type": "markdown", - "id": "e1f46d8c", + "id": "5c05185c", "metadata": { "editable": true }, @@ -4852,7 +4869,7 @@ }, { "cell_type": "markdown", - "id": "24a4f93b", + "id": "c4abb269", "metadata": { "editable": true }, @@ -4864,7 +4881,7 @@ }, { "cell_type": "markdown", - "id": "67fcc050", + "id": "89ce0cd0", "metadata": { "editable": true }, @@ -4874,7 +4891,7 @@ }, { "cell_type": "markdown", - "id": "faded8fc", + "id": "b6093d7d", "metadata": { "editable": true }, @@ -4900,7 +4917,7 @@ }, { "cell_type": "markdown", - "id": "43291108", + "id": "1c2f9d07", "metadata": { "editable": true }, @@ -4912,7 +4929,7 @@ }, { "cell_type": "markdown", - "id": "e2948c4e", + "id": "d25141a7", "metadata": { "editable": true }, @@ -4931,7 +4948,7 @@ }, { "cell_type": "markdown", - "id": "608385aa", + "id": "97f37684", "metadata": { "editable": true }, @@ -4944,7 +4961,7 @@ }, { "cell_type": "markdown", - "id": "a7dba021", + "id": "06ca30fb", "metadata": { "editable": true }, @@ -4956,7 +4973,7 @@ }, { "cell_type": "markdown", - "id": "e36164b3", + "id": "945d9820", "metadata": { "editable": true }, @@ -4977,7 +4994,7 @@ }, { "cell_type": "markdown", - "id": "5c330563", + "id": "78260f1c", "metadata": { "editable": true }, @@ -4998,7 +5015,7 @@ }, { "cell_type": "markdown", - "id": "e5200851", + "id": "aed50975", "metadata": { "editable": true }, @@ -5023,7 +5040,7 @@ }, { "cell_type": "markdown", - "id": "2b563a44", + "id": "dbb1f3d9", "metadata": { "editable": true }, @@ -5042,7 +5059,7 @@ }, { "cell_type": "markdown", - "id": "ac3282e0", + "id": "21f7a203", "metadata": { "editable": true }, @@ -5053,7 +5070,7 @@ { "cell_type": "code", "execution_count": 13, - "id": "b450060e", + "id": "599d4c52", "metadata": { "collapsed": false, "editable": true @@ -5104,7 +5121,7 @@ }, { "cell_type": "markdown", - "id": "600a3b38", + "id": "a762279b", "metadata": { "editable": true }, @@ -5115,7 +5132,7 @@ { "cell_type": "code", "execution_count": 14, - "id": "94601750", + "id": "3d283ac2", "metadata": { "collapsed": false, "editable": true @@ -5148,7 +5165,7 @@ }, { "cell_type": "markdown", - "id": "0794854d", + "id": "f554cf0e", "metadata": { "editable": true }, @@ -5159,7 +5176,7 @@ { "cell_type": "code", "execution_count": 15, - "id": "73d2352c", + "id": "395b2e91", "metadata": { "collapsed": false, "editable": true @@ -5196,7 +5213,7 @@ }, { "cell_type": "markdown", - "id": "bf108fb1", + "id": "8688bef2", "metadata": { "editable": true }, @@ -5207,7 +5224,7 @@ { "cell_type": "code", "execution_count": 16, - "id": "63b0726e", + "id": "ece60043", "metadata": { "collapsed": false, "editable": true @@ -5234,7 +5251,7 @@ }, { "cell_type": "markdown", - "id": "6419d516", + "id": "61a2ff97", "metadata": { "editable": true }, @@ -5245,7 +5262,7 @@ { "cell_type": "code", "execution_count": 17, - "id": "d4334277", + "id": "56482d32", "metadata": { "collapsed": false, "editable": true @@ -5286,7 +5303,7 @@ }, { "cell_type": "markdown", - "id": "8a85adb8", + "id": "6bf45693", "metadata": { "editable": true }, @@ -5302,7 +5319,7 @@ { "cell_type": "code", "execution_count": 18, - "id": "8b63d391", + "id": "cd9c8313", "metadata": { "collapsed": false, "editable": true @@ -5323,7 +5340,7 @@ }, { "cell_type": "markdown", - "id": "3f67780b", + "id": "ae2cd771", "metadata": { "editable": true }, @@ -5336,7 +5353,7 @@ { "cell_type": "code", "execution_count": 19, - "id": "bde09e2a", + "id": "fd0eb790", "metadata": { "collapsed": false, "editable": true @@ -5361,7 +5378,7 @@ }, { "cell_type": "markdown", - "id": "ba3900fd", + "id": "622519b8", "metadata": { "editable": true }, @@ -5376,7 +5393,7 @@ { "cell_type": "code", "execution_count": 20, - "id": "51ee395c", + "id": "6d573492", "metadata": { "collapsed": false, "editable": true @@ -5397,7 +5414,7 @@ }, { "cell_type": "markdown", - "id": "dc4d629b", + "id": "7d8938d9", "metadata": { "editable": true }, @@ -5407,7 +5424,7 @@ }, { "cell_type": "markdown", - "id": "cac88afb", + "id": "7e377497", "metadata": { "editable": true }, @@ -5426,7 +5443,7 @@ { "cell_type": "code", "execution_count": 21, - "id": "43e6acbf", + "id": "addb2ceb", "metadata": { "collapsed": false, "editable": true @@ -5443,7 +5460,7 @@ }, { "cell_type": "markdown", - "id": "20ec8e8f", + "id": "938b7da0", "metadata": { "editable": true }, @@ -5453,7 +5470,7 @@ }, { "cell_type": "markdown", - "id": "bdfe97c5", + "id": "2c3de061", "metadata": { "editable": true }, @@ -5464,7 +5481,7 @@ { "cell_type": "code", "execution_count": 22, - "id": "b6950248", + "id": "f01dce7e", "metadata": { "collapsed": false, "editable": true @@ -5481,7 +5498,7 @@ }, { "cell_type": "markdown", - "id": "2b499270", + "id": "55c40e2c", "metadata": { "editable": true }, @@ -5492,7 +5509,7 @@ { "cell_type": "code", "execution_count": 23, - "id": "4fe029be", + "id": "418f2f9d", "metadata": { "collapsed": false, "editable": true diff --git a/doc/src/week42/week42.do.txt b/doc/src/week42/week42.do.txt index 2e69ca0a2..62d7feb1c 100644 --- a/doc/src/week42/week42.do.txt +++ b/doc/src/week42/week42.do.txt @@ -39,6 +39,15 @@ In our discussions of ordinary differential equations we will also study the usage of "Autograd":"https://www.youtube.com/watch?v=fRf4l5qaX1M&ab_channel=AlexSmola" in computing gradients for deep learning. For the documentation of Autograd and examples see the lectures slides from "week 39":"https://compphysics.github.io/MachineLearning/doc/pub/week39/html/week39.html" and the "Autograd documentation":"https://github.com/HIPS/autograd". +!split +===== Back propagation and automatic differentiation ===== + +For more details on the back propagation algorithm and automatic differentiation see +o URL:"https://www.jmlr.org/papers/volume18/17-468/17-468.pdf" +o URL:"https://deepimaging.github.io/lectures/lecture_11_Backpropagation.pdf" +o Slides 12-44 at URL":http://cs231n.stanford.edu/slides/2017/cs231n_2017_lecture4.pdf" + + !split ===== Solving ODEs with Deep Learning =====