# 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()
-
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
- 39
- 40
- ... -
- 93 +
- 94
- »
-
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 @@ $$
-
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()
-
-@@ -599,7 +460,7 @@ 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 @@ $$
-
Setting up the code
+Comparing with a numerical scheme
-We can then compare the result from this numerical scheme with the output from our network using Autograd:
+The Poisson equation is possible to solve using Taylor series to approximate the second derivative.
+Using Taylor series, the second derivative can be expressed as
- -import autograd.numpy as np
-from autograd import grad, elementwise_grad
-import autograd.numpy.random as npr
-from matplotlib import pyplot as plt
+$$
+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()
-
-which makes it possible to solve for the vector \( \boldsymbol{g} \).
@@ -641,7 +516,7 @@ 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 @@ $$
-
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
-
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 @@ $$
-
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 @@ $$
-
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 @@ $$
-
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
-
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
- 50
- 51
- ... -
- 93 +
- 94
- »
-
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 \( u(x) = \sin(\pi x) \).
-
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
-
-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
-
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 @@ $$
-
The problem to solve for
- -The wave equation to solve for, is
+Example: Solving the wave equation with Neural Networks
+The wave equation is
$$ -\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 @@ $$
-
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 @@ $$
-
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({
-
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()
+
+-
-
- Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al -
- Neural networks for solving differential equations by A. Honchar -
- Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener -
- Introduction to Partial Differential Equations by A. Tveito, R. Winther -
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
+-
+
- Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al +
- Neural networks for solving differential equations by A. Honchar +
- Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener +
- Introduction to Partial Differential Equations by A. Tveito, R. Winther +
-
@@ -445,7 +437,7 @@ Networks still apply (back propagation, gradient descent etc etc).
- 60
- 61
- ... -
- 93 +
- 94
- »
-
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.
+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).
-
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.
-
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.
- -
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. -
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.
- 67
- 68
- ... -
- 93 +
- 94
- »
-
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
-
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
-
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 @@ $$
-
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 @@ $$
-
Efficient Polynomial Multiplication
+Convolution Examples: Polynomial multiplication
-Computing polynomial products can be implemented efficiently if we rewrite the more brute force multiplications using convolution. -We note first that the new coefficients are given as +
We have already met such an example in project 1 when we tried to set +up the design matrix for a two-dimensional function. This was an +example of polynomial multiplication. Let us recast such a problem in terms of the convolution operation. +Let us look a the following polynomials to second and third order, respectively:
- $$ -\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 @@ $$
-
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
-
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 @@ $$
-
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.
-
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 @@ $$
-
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 @@ $$
-
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 @@ $$
-
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()
-
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 @@ $$
-
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 @@ $$
-
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 @@ $$
-
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.
+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. +
-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).
- 85
- 86
- ... -
- 93 +
- 94
- »
-
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 @@ $$
-
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.
-
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
- -
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.
- -
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.
-
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()
-
Importing Keras and Tensorflow
+Prerequisites: Collect and pre-process data
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()
- -
Running with Keras
- + +Importing Keras and Tensorflow
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)
- -
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)
-
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()
-
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()
-
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
-
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()
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()
-
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()
-
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()
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
For more details on the back propagation algorithm and automatic differentiation seeFinally, 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))
+
+
+
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
+-
+
- https://www.jmlr.org/papers/volume18/17-468/17-468.pdf +
- https://deepimaging.github.io/lectures/lecture_11_Backpropagation.pdf +
- Slides 12-44 at URL":http://cs231n.stanford.edu/slides/2017/cs231n_2017_lecture4.pdf" +
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
+-
+
- https://www.jmlr.org/papers/volume18/17-468/17-468.pdf +
- https://deepimaging.github.io/lectures/lecture_11_Backpropagation.pdf +
- Slides 12-44 at URL":http://cs231n.stanford.edu/slides/2017/cs231n_2017_lecture4.pdf" +