From 59d0e80451e04cb09b443189f8dd29231c5ba0eb Mon Sep 17 00:00:00 2001 From: Morten Hjorth-Jensen Date: Tue, 19 Oct 2021 08:25:18 +0200 Subject: [PATCH] updating files --- doc/BookChapters/chapter10.ipynb | 2047 ++++++++++++++++++++ doc/BookChapters/chapter11.ipynb | 3023 ++++++++++++++++++++++++++++++ doc/LectureNotes/chapter11.ipynb | 3023 ++++++++++++++++++++++++++++++ doc/LectureNotes/chapter12.ipynb | 718 +++++++ doc/LectureNotes/schedule.md | 2 +- 5 files changed, 8812 insertions(+), 1 deletion(-) create mode 100644 doc/BookChapters/chapter10.ipynb create mode 100644 doc/BookChapters/chapter11.ipynb create mode 100644 doc/LectureNotes/chapter11.ipynb create mode 100644 doc/LectureNotes/chapter12.ipynb diff --git a/doc/BookChapters/chapter10.ipynb b/doc/BookChapters/chapter10.ipynb new file mode 100644 index 000000000..cc65cfe33 --- /dev/null +++ b/doc/BookChapters/chapter10.ipynb @@ -0,0 +1,2047 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Building a Feed Forward Neural Network\n", + "\n", + "We are now gong to develop an example based on the MNIST data\n", + "base. This is a classification problem and we need to use our\n", + "cross-entropy function we discussed in connection with logistic\n", + "regression. The cross-entropy defines our cost function for the\n", + "classificaton problems with neural networks.\n", + "\n", + "In binary classification with two classes $(0, 1)$ we define the\n", + "logistic/sigmoid function as the probability that a particular input\n", + "is in class $0$ or $1$. This is possible because the logistic\n", + "function takes any input from the real numbers and inputs a number\n", + "between 0 and 1, and can therefore be interpreted as a probability. It\n", + "also has other nice properties, such as a derivative that is simple to\n", + "calculate.\n", + "\n", + "For an input $\\boldsymbol{a}$ from the hidden layer, the probability that the input $\\boldsymbol{x}$\n", + "is in class 0 or 1 is just. We let $\\theta$ represent the unknown weights and biases to be adjusted by our equations). The variable $x$\n", + "represents our activation values $z$. We have" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "P(y = 0 \\mid \\hat{x}, \\hat{\\theta}) = \\frac{1}{1 + \\exp{(- \\hat{x}})} ,\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "and" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "P(y = 1 \\mid \\hat{x}, \\hat{\\theta}) = 1 - P(y = 0 \\mid \\hat{x}, \\hat{\\theta}) ,\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $y \\in \\{0, 1\\}$ and $\\hat{\\theta}$ represents the weights and biases\n", + "of our network.\n", + "\n", + "\n", + "\n", + "## Defining the cost function\n", + "\n", + "Our cost function is given as (see the Logistic regression lectures)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\mathcal{C}(\\hat{\\theta}) = - \\ln P(\\mathcal{D} \\mid \\hat{\\theta}) = - \\sum_{i=1}^n\n", + "y_i \\ln[P(y_i = 0)] + (1 - y_i) \\ln [1 - P(y_i = 0)] = \\sum_{i=1}^n \\mathcal{L}_i(\\hat{\\theta}) .\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "This last equality means that we can interpret our *cost* function as a sum over the *loss* function\n", + "for each point in the dataset $\\mathcal{L}_i(\\hat{\\theta})$. \n", + "The negative sign is just so that we can think about our algorithm as minimizing a positive number, rather\n", + "than maximizing a negative number. \n", + "\n", + "In *multiclass* classification it is common to treat each integer label as a so called *one-hot* vector: \n", + "\n", + "$y = 5 \\quad \\rightarrow \\quad \\hat{y} = (0, 0, 0, 0, 0, 1, 0, 0, 0, 0) ,$ and\n", + "\n", + "\n", + "$y = 1 \\quad \\rightarrow \\quad \\hat{y} = (0, 1, 0, 0, 0, 0, 0, 0, 0, 0) ,$ \n", + "\n", + "\n", + "i.e. a binary bit string of length $C$, where $C = 10$ is the number of classes in the MNIST dataset (numbers from $0$ to $9$).. \n", + "\n", + "If $\\hat{x}_i$ is the $i$-th input (image), $y_{ic}$ refers to the $c$-th component of the $i$-th\n", + "output vector $\\hat{y}_i$. \n", + "The probability of $\\hat{x}_i$ being in class $c$ will be given by the softmax function:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "P(y_{ic} = 1 \\mid \\hat{x}_i, \\hat{\\theta}) = \\frac{\\exp{((\\hat{a}_i^{hidden})^T \\hat{w}_c)}}\n", + "{\\sum_{c'=0}^{C-1} \\exp{((\\hat{a}_i^{hidden})^T \\hat{w}_{c'})}} ,\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "which reduces to the logistic function in the binary case. \n", + "The likelihood of this $C$-class classifier\n", + "is now given as:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "P(\\mathcal{D} \\mid \\hat{\\theta}) = \\prod_{i=1}^n \\prod_{c=0}^{C-1} [P(y_{ic} = 1)]^{y_{ic}} .\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Again we take the negative log-likelihood to define our cost function:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\mathcal{C}(\\hat{\\theta}) = - \\log{P(\\mathcal{D} \\mid \\hat{\\theta})}.\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "See the logistic regression lectures for a full definition of the cost function.\n", + "\n", + "The back propagation equations need now only a small change, namely the definition of a new cost function. We are thus ready to use the same equations as before!\n", + "\n", + "\n", + "### Example: binary classification problem\n", + "\n", + "As an example of the above, relevant for project 2 as well, let us consider a binary class. As discussed in our logistic regression lectures, we defined a cost function in terms of the parameters $\\beta$ as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\mathcal{C}(\\hat{\\beta}) = - \\sum_{i=1}^n \\left(y_i\\log{p(y_i \\vert x_i,\\hat{\\beta})}+(1-y_i)\\log{1-p(y_i \\vert x_i,\\hat{\\beta})}\\right),\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where we had defined the logistic (sigmoid) function" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "p(y_i =1\\vert x_i,\\hat{\\beta})=\\frac{\\exp{(\\beta_0+\\beta_1 x_i)}}{1+\\exp{(\\beta_0+\\beta_1 x_i)}},\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "and" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "p(y_i =0\\vert x_i,\\hat{\\beta})=1-p(y_i =1\\vert x_i,\\hat{\\beta}).\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The parameters $\\hat{\\beta}$ were defined using a minimization method like gradient descent or Newton-Raphson's method. \n", + "\n", + "Now we replace $x_i$ with the activation $z_i^l$ for a given layer $l$ and the outputs as $y_i=a_i^l=f(z_i^l)$, with $z_i^l$ now being a function of the weights $w_{ij}^l$ and biases $b_i^l$. \n", + "We have then" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "a_i^l = y_i = \\frac{\\exp{(z_i^l)}}{1+\\exp{(z_i^l)}},\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "z_i^l = \\sum_{j}w_{ij}^l a_j^{l-1}+b_i^l,\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where the superscript $l-1$ indicates that these are the outputs from layer $l-1$.\n", + "Our cost function at the final layer $l=L$ is now" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\mathcal{C}(\\hat{W}) = - \\sum_{i=1}^n \\left(t_i\\log{a_i^L}+(1-t_i)\\log{(1-a_i^L)}\\right),\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where we have defined the targets $t_i$. The derivatives of the cost function with respect to the output $a_i^L$ are then easily calculated and we get" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\frac{\\partial \\mathcal{C}(\\hat{W})}{\\partial a_i^L} = \\frac{a_i^L-t_i}{a_i^L(1-a_i^L)}.\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In case we use another activation function than the logistic one, we need to evaluate other derivatives. \n", + "\n", + "\n", + "\n", + "### The Softmax function\n", + "\n", + "In case we employ the more general case given by the Softmax equation, we need to evaluate the derivative of the activation function with respect to the activation $z_i^l$, that is we need" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\frac{\\partial f(z_i^l)}{\\partial w_{jk}^l} =\n", + "\\frac{\\partial f(z_i^l)}{\\partial z_j^l} \\frac{\\partial z_j^l}{\\partial w_{jk}^l}= \\frac{\\partial f(z_i^l)}{\\partial z_j^l}a_k^{l-1}.\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "For the Softmax function we have" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "f(z_i^l) = \\frac{\\exp{(z_i^l)}}{\\sum_{m=1}^K\\exp{(z_m^l)}}.\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Its derivative with respect to $z_j^l$ gives" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\frac{\\partial f(z_i^l)}{\\partial z_j^l}= f(z_i^l)\\left(\\delta_{ij}-f(z_j^l)\\right),\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "which in case of the simply binary model reduces to having $i=j$. \n", + "\n", + "\n", + "## Developing a code for doing neural networks with back propagation\n", + "\n", + "\n", + "One can identify a set of key steps when using neural networks to solve supervised learning problems: \n", + "\n", + "1. Collect and pre-process data \n", + "\n", + "2. Define model and architecture \n", + "\n", + "3. Choose cost function and optimizer \n", + "\n", + "4. Train the model \n", + "\n", + "5. Evaluate model performance on test data \n", + "\n", + "6. Adjust hyperparameters (if necessary, network architecture)\n", + "\n", + "### Collect and pre-process data\n", + "\n", + "Here we will be using the MNIST dataset, which is readily available through the **scikit-learn**\n", + "package. You may also find it for example [here](http://yann.lecun.com/exdb/mnist/). \n", + "The *MNIST* (Modified National Institute of Standards and Technology) database is a large database\n", + "of handwritten digits that is commonly used for training various image processing systems. \n", + "The MNIST dataset consists of 70 000 images of size $28\\times 28$ pixels, each labeled from 0 to 9. \n", + "The scikit-learn dataset we will use consists of a selection of 1797 images of size $8\\times 8$ collected and processed from this database. \n", + "\n", + "To feed data into a feed-forward neural network we need to represent\n", + "the inputs as a design/feature matrix $X = (n_{inputs}, n_{features})$. Each\n", + "row represents an *input*, in this case a handwritten digit, and\n", + "each column represents a *feature*, in this case a pixel. The\n", + "correct answers, also known as *labels* or *targets* are\n", + "represented as a 1D array of integers \n", + "$Y = (n_{inputs}) = (5, 3, 1, 8,...)$.\n", + "\n", + "As an example, say we want to build a neural network using supervised learning to predict Body-Mass Index (BMI) from\n", + "measurements of height (in m) \n", + "and weight (in kg). If we have measurements of 5 people the design/feature matrix could be for example: \n", + "\n", + "$$ X = \\begin{bmatrix}\n", + "1.85 & 81\\\\\n", + "1.71 & 65\\\\\n", + "1.95 & 103\\\\\n", + "1.55 & 42\\\\\n", + "1.63 & 56\n", + "\\end{bmatrix} ,$$ \n", + "\n", + "and the targets would be: \n", + "\n", + "$$ Y = (23.7, 22.2, 27.1, 17.5, 21.1) $$ \n", + "\n", + "Since each input image is a 2D matrix, we need to flatten the image\n", + "(i.e. \"unravel\" the 2D matrix into a 1D array) to turn the data into a\n", + "design/feature matrix. This means we lose all spatial information in the\n", + "image, such as locality and translational invariance. More complicated\n", + "architectures such as Convolutional Neural Networks can take advantage\n", + "of such information, and are most commonly applied when analyzing\n", + "images." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "%matplotlib inline\n", + "\n", + "# import necessary packages\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "from sklearn import datasets\n", + "\n", + "\n", + "# ensure the same random numbers appear every time\n", + "np.random.seed(0)\n", + "\n", + "# display images in notebook\n", + "%matplotlib inline\n", + "plt.rcParams['figure.figsize'] = (12,12)\n", + "\n", + "\n", + "# download MNIST dataset\n", + "digits = datasets.load_digits()\n", + "\n", + "# define inputs and labels\n", + "inputs = digits.images\n", + "labels = digits.target\n", + "\n", + "print(\"inputs = (n_inputs, pixel_width, pixel_height) = \" + str(inputs.shape))\n", + "print(\"labels = (n_inputs) = \" + str(labels.shape))\n", + "\n", + "\n", + "# flatten the image\n", + "# the value -1 means dimension is inferred from the remaining dimensions: 8x8 = 64\n", + "n_inputs = len(inputs)\n", + "inputs = inputs.reshape(n_inputs, -1)\n", + "print(\"X = (n_inputs, n_features) = \" + str(inputs.shape))\n", + "\n", + "\n", + "# choose some random images to display\n", + "indices = np.arange(n_inputs)\n", + "random_indices = np.random.choice(indices, size=5)\n", + "\n", + "for i, image in enumerate(digits.images[random_indices]):\n", + " plt.subplot(1, 5, i+1)\n", + " plt.axis('off')\n", + " plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest')\n", + " plt.title(\"Label: %d\" % digits.target[random_indices[i]])\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Train and test datasets\n", + "\n", + "Performing analysis before partitioning the dataset is a major error, that can lead to incorrect conclusions. \n", + "\n", + "We will reserve $80 \\%$ of our dataset for training and $20 \\%$ for testing. \n", + "\n", + "It is important that the train and test datasets are drawn randomly from our dataset, to ensure\n", + "no bias in the sampling. \n", + "Say you are taking measurements of weather data to predict the weather in the coming 5 days.\n", + "You don't want to train your model on measurements taken from the hours 00.00 to 12.00, and then test it on data\n", + "collected from 12.00 to 24.00." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "from sklearn.model_selection import train_test_split\n", + "\n", + "# one-liner from scikit-learn library\n", + "train_size = 0.8\n", + "test_size = 1 - train_size\n", + "X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size,\n", + " test_size=test_size)\n", + "\n", + "# equivalently in numpy\n", + "def train_test_split_numpy(inputs, labels, train_size, test_size):\n", + " n_inputs = len(inputs)\n", + " inputs_shuffled = inputs.copy()\n", + " labels_shuffled = labels.copy()\n", + " \n", + " np.random.shuffle(inputs_shuffled)\n", + " np.random.shuffle(labels_shuffled)\n", + " \n", + " train_end = int(n_inputs*train_size)\n", + " X_train, X_test = inputs_shuffled[:train_end], inputs_shuffled[train_end:]\n", + " Y_train, Y_test = labels_shuffled[:train_end], labels_shuffled[train_end:]\n", + " \n", + " return X_train, X_test, Y_train, Y_test\n", + "\n", + "#X_train, X_test, Y_train, Y_test = train_test_split_numpy(inputs, labels, train_size, test_size)\n", + "\n", + "print(\"Number of training images: \" + str(len(X_train)))\n", + "print(\"Number of test images: \" + str(len(X_test)))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Define model and architecture\n", + "\n", + "Our simple feed-forward neural network will consist of an *input* layer, a single *hidden* layer and an *output* layer. The activation $y$ of each neuron is a weighted sum of inputs, passed through an activation function. In case of the simple perceptron model we have \n", + "\n", + "$$ z = \\sum_{i=1}^n w_i a_i ,$$\n", + "\n", + "$$ y = f(z) ,$$\n", + "\n", + "where $f$ is the activation function, $a_i$ represents input from neuron $i$ in the preceding layer\n", + "and $w_i$ is the weight to input $i$. \n", + "The activation of the neurons in the input layer is just the features (e.g. a pixel value). \n", + "\n", + "The simplest activation function for a neuron is the *Heaviside* function:\n", + "\n", + "$$ f(z) = \n", + "\\begin{cases}\n", + "1, & z > 0\\\\\n", + "0, & \\text{otherwise}\n", + "\\end{cases}\n", + "$$\n", + "\n", + "A feed-forward neural network with this activation is known as a *perceptron*. \n", + "For a binary classifier (i.e. two classes, 0 or 1, dog or not-dog) we can also use this in our output layer. \n", + "This activation can be generalized to $k$ classes (using e.g. the *one-against-all* strategy), \n", + "and we call these architectures *multiclass perceptrons*. \n", + "\n", + "However, it is now common to use the terms Single Layer Perceptron (SLP) (1 hidden layer) and \n", + "Multilayer Perceptron (MLP) (2 or more hidden layers) to refer to feed-forward neural networks with any activation function. \n", + "\n", + "Typical choices for activation functions include the sigmoid function, hyperbolic tangent, and Rectified Linear Unit (ReLU). \n", + "We will be using the sigmoid function $\\sigma(x)$: \n", + "\n", + "$$ f(x) = \\sigma(x) = \\frac{1}{1 + e^{-x}} ,$$\n", + "\n", + "which is inspired by probability theory (see logistic regression) and was most commonly used until about 2011. See the discussion below concerning other activation functions.\n", + "\n", + "### Layers\n", + "\n", + "* Input \n", + "\n", + "Since each input image has 8x8 = 64 pixels or features, we have an input layer of 64 neurons. \n", + "\n", + "* Hidden layer\n", + "\n", + "We will use 50 neurons in the hidden layer receiving input from the neurons in the input layer. \n", + "Since each neuron in the hidden layer is connected to the 64 inputs we have 64x50 = 3200 weights to the hidden layer. \n", + "\n", + "* Output\n", + "\n", + "If we were building a binary classifier, it would be sufficient with a single neuron in the output layer,\n", + "which could output 0 or 1 according to the Heaviside function. This would be an example of a *hard* classifier, meaning it outputs the class of the input directly. However, if we are dealing with noisy data it is often beneficial to use a *soft* classifier, which outputs the probability of being in class 0 or 1. \n", + "\n", + "For a soft binary classifier, we could use a single neuron and interpret the output as either being the probability of being in class 0 or the probability of being in class 1. Alternatively we could use 2 neurons, and interpret each neuron as the probability of being in each class. \n", + "\n", + "Since we are doing multiclass classification, with 10 categories, it is natural to use 10 neurons in the output layer. We number the neurons $j = 0,1,...,9$. The activation of each output neuron $j$ will be according to the *softmax* function: \n", + "\n", + "$$ P(\\text{class $j$} \\mid \\text{input $\\hat{a}$}) = \\frac{\\exp{(\\hat{a}^T \\hat{w}_j)}}\n", + "{\\sum_{c=0}^{9} \\exp{(\\hat{a}^T \\hat{w}_c)}} ,$$ \n", + "\n", + "i.e. each neuron $j$ outputs the probability of being in class $j$ given an input from the hidden layer $\\hat{a}$, with $\\hat{w}_j$ the weights of neuron $j$ to the inputs. \n", + "The denominator is a normalization factor to ensure the outputs (probabilities) sum up to 1. \n", + "The exponent is just the weighted sum of inputs as before: \n", + "\n", + "$$ z_j = \\sum_{i=1}^n w_ {ij} a_i+b_j.$$ \n", + "\n", + "Since each neuron in the output layer is connected to the 50 inputs from the hidden layer we have 50x10 = 500\n", + "weights to the output layer.\n", + "\n", + "\n", + "Typically weights are initialized with small values distributed around zero, drawn from a uniform\n", + "or normal distribution. Setting all weights to zero means all neurons give the same output, making the network useless. \n", + "\n", + "Adding a bias value to the weighted sum of inputs allows the neural network to represent a greater range\n", + "of values. Without it, any input with the value 0 will be mapped to zero (before being passed through the activation). The bias unit has an output of 1, and a weight to each neuron $j$, $b_j$: \n", + "\n", + "$$ z_j = \\sum_{i=1}^n w_ {ij} a_i + b_j.$$ \n", + "\n", + "The bias weights $\\hat{b}$ are often initialized to zero, but a small value like $0.01$ ensures all neurons have some output which can be backpropagated in the first training cycle." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# building our neural network\n", + "\n", + "n_inputs, n_features = X_train.shape\n", + "n_hidden_neurons = 50\n", + "n_categories = 10\n", + "\n", + "# we make the weights normally distributed using numpy.random.randn\n", + "\n", + "# weights and bias in the hidden layer\n", + "hidden_weights = np.random.randn(n_features, n_hidden_neurons)\n", + "hidden_bias = np.zeros(n_hidden_neurons) + 0.01\n", + "\n", + "# weights and bias in the output layer\n", + "output_weights = np.random.randn(n_hidden_neurons, n_categories)\n", + "output_bias = np.zeros(n_categories) + 0.01" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Feed-forward pass\n", + "\n", + "Denote $F$ the number of features, $H$ the number of hidden neurons and $C$ the number of categories. \n", + "For each input image we calculate a weighted sum of input features (pixel values) to each neuron $j$ in the hidden layer $l$: \n", + "\n", + "$$ z_{j}^{l} = \\sum_{i=1}^{F} w_{ij}^{l} x_i + b_{j}^{l},$$\n", + "\n", + "this is then passed through our activation function \n", + "\n", + "$$ a_{j}^{l} = f(z_{j}^{l}) .$$ \n", + "\n", + "We calculate a weighted sum of inputs (activations in the hidden layer) to each neuron $j$ in the output layer: \n", + "\n", + "$$ z_{j}^{L} = \\sum_{i=1}^{H} w_{ij}^{L} a_{i}^{l} + b_{j}^{L}.$$ \n", + "\n", + "Finally we calculate the output of neuron $j$ in the output layer using the softmax function: \n", + "\n", + "$$ a_{j}^{L} = \\frac{\\exp{(z_j^{L})}}\n", + "{\\sum_{c=0}^{C-1} \\exp{(z_c^{L})}} .$$ \n", + "\n", + "\n", + "Since our data has the dimensions $X = (n_{inputs}, n_{features})$ and our weights to the hidden\n", + "layer have the dimensions \n", + "$W_{hidden} = (n_{features}, n_{hidden})$,\n", + "we can easily feed the network all our training data in one go by taking the matrix product \n", + "\n", + "$$ X W^{h} = (n_{inputs}, n_{hidden}),$$ \n", + "\n", + "and obtain a matrix that holds the weighted sum of inputs to the hidden layer\n", + "for each input image and each hidden neuron. \n", + "We also add the bias to obtain a matrix of weighted sums to the hidden layer $Z^{h}$: \n", + "\n", + "$$ \\hat{z}^{l} = \\hat{X} \\hat{W}^{l} + \\hat{b}^{l} ,$$\n", + "\n", + "meaning the same bias (1D array with size equal number of hidden neurons) is added to each input image. \n", + "This is then passed through the activation: \n", + "\n", + "$$ \\hat{a}^{l} = f(\\hat{z}^l) .$$ \n", + "\n", + "This is fed to the output layer: \n", + "\n", + "$$ \\hat{z}^{L} = \\hat{a}^{L} \\hat{W}^{L} + \\hat{b}^{L} .$$\n", + "\n", + "Finally we receive our output values for each image and each category by passing it through the softmax function: \n", + "\n", + "$$ output = softmax (\\hat{z}^{L}) = (n_{inputs}, n_{categories}) .$$" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# setup the feed-forward pass, subscript h = hidden layer\n", + "\n", + "def sigmoid(x):\n", + " return 1/(1 + np.exp(-x))\n", + "\n", + "def feed_forward(X):\n", + " # weighted sum of inputs to the hidden layer\n", + " z_h = np.matmul(X, hidden_weights) + hidden_bias\n", + " # activation in the hidden layer\n", + " a_h = sigmoid(z_h)\n", + " \n", + " # weighted sum of inputs to the output layer\n", + " z_o = np.matmul(a_h, output_weights) + output_bias\n", + " # softmax output\n", + " # axis 0 holds each input and axis 1 the probabilities of each category\n", + " exp_term = np.exp(z_o)\n", + " probabilities = exp_term / np.sum(exp_term, axis=1, keepdims=True)\n", + " \n", + " return probabilities\n", + "\n", + "probabilities = feed_forward(X_train)\n", + "print(\"probabilities = (n_inputs, n_categories) = \" + str(probabilities.shape))\n", + "print(\"probability that image 0 is in category 0,1,2,...,9 = \\n\" + str(probabilities[0]))\n", + "print(\"probabilities sum up to: \" + str(probabilities[0].sum()))\n", + "print()\n", + "\n", + "# we obtain a prediction by taking the class with the highest likelihood\n", + "def predict(X):\n", + " probabilities = feed_forward(X)\n", + " return np.argmax(probabilities, axis=1)\n", + "\n", + "predictions = predict(X_train)\n", + "print(\"predictions = (n_inputs) = \" + str(predictions.shape))\n", + "print(\"prediction for image 0: \" + str(predictions[0]))\n", + "print(\"correct label for image 0: \" + str(Y_train[0]))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Choose cost function and optimizer\n", + "\n", + "To measure how well our neural network is doing we need to introduce a cost function. \n", + "We will call the function that gives the error of a single sample output the *loss* function, and the function\n", + "that gives the total error of our network across all samples the *cost* function.\n", + "A typical choice for multiclass classification is the *cross-entropy* loss, also known as the negative log likelihood. \n", + "\n", + "In *multiclass* classification it is common to treat each integer label as a so called *one-hot* vector: \n", + "\n", + "$$ y = 5 \\quad \\rightarrow \\quad \\hat{y} = (0, 0, 0, 0, 0, 1, 0, 0, 0, 0) ,$$ \n", + "\n", + "\n", + "$$ y = 1 \\quad \\rightarrow \\quad \\hat{y} = (0, 1, 0, 0, 0, 0, 0, 0, 0, 0) ,$$ \n", + "\n", + "\n", + "i.e. a binary bit string of length $C$, where $C = 10$ is the number of classes in the MNIST dataset. \n", + "\n", + "Let $y_{ic}$ denote the $c$-th component of the $i$-th one-hot vector. \n", + "We define the cost function $\\mathcal{C}$ as a sum over the cross-entropy loss for each point $\\hat{x}_i$ in the dataset.\n", + "\n", + "In the one-hot representation only one of the terms in the loss function is non-zero, namely the\n", + "probability of the correct category $c'$ \n", + "(i.e. the category $c'$ such that $y_{ic'} = 1$). This means that the cross entropy loss only punishes you for how wrong\n", + "you got the correct label. The probability of category $c$ is given by the softmax function. The vector $\\hat{\\theta}$ represents the parameters of our network, i.e. all the weights and biases. \n", + "\n", + "\n", + "\n", + "### Optimizing the cost function\n", + "\n", + "The network is trained by finding the weights and biases that minimize the cost function. One of the most widely used classes of methods is *gradient descent* and its generalizations. The idea behind gradient descent\n", + "is simply to adjust the weights in the direction where the gradient of the cost function is large and negative. This ensures we flow toward a *local* minimum of the cost function. \n", + "Each parameter $\\theta$ is iteratively adjusted according to the rule \n", + "\n", + "$$ \\theta_{i+1} = \\theta_i - \\eta \\nabla \\mathcal{C}(\\theta_i) ,$$\n", + "\n", + "where $\\eta$ is known as the *learning rate*, which controls how big a step we take towards the minimum. \n", + "This update can be repeated for any number of iterations, or until we are satisfied with the result. \n", + "\n", + "A simple and effective improvement is a variant called *Batch Gradient Descent*. \n", + "Instead of calculating the gradient on the whole dataset, we calculate an approximation of the gradient\n", + "on a subset of the data called a *minibatch*. \n", + "If there are $N$ data points and we have a minibatch size of $M$, the total number of batches\n", + "is $N/M$. \n", + "We denote each minibatch $B_k$, with $k = 1, 2,...,N/M$. The gradient then becomes: \n", + "\n", + "$$ \\nabla \\mathcal{C}(\\theta) = \\frac{1}{N} \\sum_{i=1}^N \\nabla \\mathcal{L}_i(\\theta) \\quad \\rightarrow \\quad\n", + "\\frac{1}{M} \\sum_{i \\in B_k} \\nabla \\mathcal{L}_i(\\theta) ,$$\n", + "\n", + "i.e. instead of averaging the loss over the entire dataset, we average over a minibatch. \n", + "\n", + "This has two important benefits: \n", + "1. Introducing stochasticity decreases the chance that the algorithm becomes stuck in a local minima. \n", + "\n", + "2. It significantly speeds up the calculation, since we do not have to use the entire dataset to calculate the gradient. \n", + "\n", + "The various optmization methods, with codes and algorithms, are discussed in our lectures on [Gradient descent approaches](https://compphysics.github.io/MachineLearning/doc/pub/Splines/html/Splines-bs.html).\n", + "\n", + "\n", + "### Regularization\n", + "\n", + "It is common to add an extra term to the cost function, proportional\n", + "to the size of the weights. This is equivalent to constraining the\n", + "size of the weights, so that they do not grow out of control.\n", + "Constraining the size of the weights means that the weights cannot\n", + "grow arbitrarily large to fit the training data, and in this way\n", + "reduces *overfitting*.\n", + "\n", + "We will measure the size of the weights using the so called *L2-norm*, meaning our cost function becomes: \n", + "\n", + "$$ \\mathcal{C}(\\theta) = \\frac{1}{N} \\sum_{i=1}^N \\mathcal{L}_i(\\theta) \\quad \\rightarrow \\quad\n", + "\\frac{1}{N} \\sum_{i=1}^N \\mathcal{L}_i(\\theta) + \\lambda \\lvert \\lvert \\hat{w} \\rvert \\rvert_2^2 \n", + "= \\frac{1}{N} \\sum_{i=1}^N \\mathcal{L}(\\theta) + \\lambda \\sum_{ij} w_{ij}^2,$$ \n", + "\n", + "i.e. we sum up all the weights squared. The factor $\\lambda$ is known as a regularization parameter.\n", + "\n", + "\n", + "In order to train the model, we need to calculate the derivative of\n", + "the cost function with respect to every bias and weight in the\n", + "network. In total our network has $(64 + 1)\\times 50=3250$ weights in\n", + "the hidden layer and $(50 + 1)\\times 10=510$ weights to the output\n", + "layer ($+1$ for the bias), and the gradient must be calculated for\n", + "every parameter. We use the *backpropagation* algorithm discussed\n", + "above. This is a clever use of the chain rule that allows us to\n", + "calculate the gradient efficently. \n", + "\n", + "\n", + "### Matrix multiplication\n", + "\n", + "To more efficently train our network these equations are implemented using matrix operations. \n", + "The error in the output layer is calculated simply as, with $\\hat{t}$ being our targets, \n", + "\n", + "$$ \\delta_L = \\hat{t} - \\hat{y} = (n_{inputs}, n_{categories}) .$$ \n", + "\n", + "The gradient for the output weights is calculated as \n", + "\n", + "$$ \\nabla W_{L} = \\hat{a}^T \\delta_L = (n_{hidden}, n_{categories}) ,$$\n", + "\n", + "where $\\hat{a} = (n_{inputs}, n_{hidden})$. This simply means that we are summing up the gradients for each input. \n", + "Since we are going backwards we have to transpose the activation matrix. \n", + "\n", + "The gradient with respect to the output bias is then \n", + "\n", + "$$ \\nabla \\hat{b}_{L} = \\sum_{i=1}^{n_{inputs}} \\delta_L = (n_{categories}) .$$ \n", + "\n", + "The error in the hidden layer is \n", + "\n", + "$$ \\Delta_h = \\delta_L W_{L}^T \\circ f'(z_{h}) = \\delta_L W_{L}^T \\circ a_{h} \\circ (1 - a_{h}) = (n_{inputs}, n_{hidden}) ,$$ \n", + "\n", + "where $f'(a_{h})$ is the derivative of the activation in the hidden layer. The matrix products mean\n", + "that we are summing up the products for each neuron in the output layer. The symbol $\\circ$ denotes\n", + "the *Hadamard product*, meaning element-wise multiplication. \n", + "\n", + "This again gives us the gradients in the hidden layer: \n", + "\n", + "$$ \\nabla W_{h} = X^T \\delta_h = (n_{features}, n_{hidden}) ,$$ \n", + "\n", + "$$ \\nabla b_{h} = \\sum_{i=1}^{n_{inputs}} \\delta_h = (n_{hidden}) .$$" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# to categorical turns our integer vector into a onehot representation\n", + "from sklearn.metrics import accuracy_score\n", + "\n", + "# one-hot in numpy\n", + "def to_categorical_numpy(integer_vector):\n", + " n_inputs = len(integer_vector)\n", + " n_categories = np.max(integer_vector) + 1\n", + " onehot_vector = np.zeros((n_inputs, n_categories))\n", + " onehot_vector[range(n_inputs), integer_vector] = 1\n", + " \n", + " return onehot_vector\n", + "\n", + "#Y_train_onehot, Y_test_onehot = to_categorical(Y_train), to_categorical(Y_test)\n", + "Y_train_onehot, Y_test_onehot = to_categorical_numpy(Y_train), to_categorical_numpy(Y_test)\n", + "\n", + "def feed_forward_train(X):\n", + " # weighted sum of inputs to the hidden layer\n", + " z_h = np.matmul(X, hidden_weights) + hidden_bias\n", + " # activation in the hidden layer\n", + " a_h = sigmoid(z_h)\n", + " \n", + " # weighted sum of inputs to the output layer\n", + " z_o = np.matmul(a_h, output_weights) + output_bias\n", + " # softmax output\n", + " # axis 0 holds each input and axis 1 the probabilities of each category\n", + " exp_term = np.exp(z_o)\n", + " probabilities = exp_term / np.sum(exp_term, axis=1, keepdims=True)\n", + " \n", + " # for backpropagation need activations in hidden and output layers\n", + " return a_h, probabilities\n", + "\n", + "def backpropagation(X, Y):\n", + " a_h, probabilities = feed_forward_train(X)\n", + " \n", + " # error in the output layer\n", + " error_output = probabilities - Y\n", + " # error in the hidden layer\n", + " error_hidden = np.matmul(error_output, output_weights.T) * a_h * (1 - a_h)\n", + " \n", + " # gradients for the output layer\n", + " output_weights_gradient = np.matmul(a_h.T, error_output)\n", + " output_bias_gradient = np.sum(error_output, axis=0)\n", + " \n", + " # gradient for the hidden layer\n", + " hidden_weights_gradient = np.matmul(X.T, error_hidden)\n", + " hidden_bias_gradient = np.sum(error_hidden, axis=0)\n", + "\n", + " return output_weights_gradient, output_bias_gradient, hidden_weights_gradient, hidden_bias_gradient\n", + "\n", + "print(\"Old accuracy on training data: \" + str(accuracy_score(predict(X_train), Y_train)))\n", + "\n", + "eta = 0.01\n", + "lmbd = 0.01\n", + "for i in range(1000):\n", + " # calculate gradients\n", + " dWo, dBo, dWh, dBh = backpropagation(X_train, Y_train_onehot)\n", + " \n", + " # regularization term gradients\n", + " dWo += lmbd * output_weights\n", + " dWh += lmbd * hidden_weights\n", + " \n", + " # update weights and biases\n", + " output_weights -= eta * dWo\n", + " output_bias -= eta * dBo\n", + " hidden_weights -= eta * dWh\n", + " hidden_bias -= eta * dBh\n", + "\n", + "print(\"New accuracy on training data: \" + str(accuracy_score(predict(X_train), Y_train)))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Improving performance\n", + "\n", + "As we can see the network does not seem to be learning at all. It seems to be just guessing the label for each image. \n", + "In order to obtain a network that does something useful, we will have to do a bit more work. \n", + "\n", + "The choice of *hyperparameters* such as learning rate and regularization parameter is hugely influential for the performance of the network. Typically a *grid-search* is performed, wherein we test different hyperparameters separated by orders of magnitude. For example we could test the learning rates $\\eta = 10^{-6}, 10^{-5},...,10^{-1}$ with different regularization parameters $\\lambda = 10^{-6},...,10^{-0}$. \n", + "\n", + "Next, we haven't implemented minibatching yet, which introduces stochasticity and is though to act as an important regularizer on the weights. We call a feed-forward + backward pass with a minibatch an *iteration*, and a full training period\n", + "going through the entire dataset ($n/M$ batches) an *epoch*.\n", + "\n", + "If this does not improve network performance, you may want to consider altering the network architecture, adding more neurons or hidden layers. \n", + "Andrew Ng goes through some of these considerations in this [video](https://youtu.be/F1ka6a13S9I). You can find a summary of the video [here](https://kevinzakka.github.io/2016/09/26/applying-deep-learning/). \n", + "\n", + "\n", + "It is very natural to think of the network as an object, with specific instances of the network\n", + "being realizations of this object with different hyperparameters. An implementation using Python classes provides a clean structure and interface, and the full implementation of our neural network is given below." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "class NeuralNetwork:\n", + " def __init__(\n", + " self,\n", + " X_data,\n", + " Y_data,\n", + " n_hidden_neurons=50,\n", + " n_categories=10,\n", + " epochs=10,\n", + " batch_size=100,\n", + " eta=0.1,\n", + " lmbd=0.0):\n", + "\n", + " self.X_data_full = X_data\n", + " self.Y_data_full = Y_data\n", + "\n", + " self.n_inputs = X_data.shape[0]\n", + " self.n_features = X_data.shape[1]\n", + " self.n_hidden_neurons = n_hidden_neurons\n", + " self.n_categories = n_categories\n", + "\n", + " self.epochs = epochs\n", + " self.batch_size = batch_size\n", + " self.iterations = self.n_inputs // self.batch_size\n", + " self.eta = eta\n", + " self.lmbd = lmbd\n", + "\n", + " self.create_biases_and_weights()\n", + "\n", + " def create_biases_and_weights(self):\n", + " self.hidden_weights = np.random.randn(self.n_features, self.n_hidden_neurons)\n", + " self.hidden_bias = np.zeros(self.n_hidden_neurons) + 0.01\n", + "\n", + " self.output_weights = np.random.randn(self.n_hidden_neurons, self.n_categories)\n", + " self.output_bias = np.zeros(self.n_categories) + 0.01\n", + "\n", + " def feed_forward(self):\n", + " # feed-forward for training\n", + " self.z_h = np.matmul(self.X_data, self.hidden_weights) + self.hidden_bias\n", + " self.a_h = sigmoid(self.z_h)\n", + "\n", + " self.z_o = np.matmul(self.a_h, self.output_weights) + self.output_bias\n", + "\n", + " exp_term = np.exp(self.z_o)\n", + " self.probabilities = exp_term / np.sum(exp_term, axis=1, keepdims=True)\n", + "\n", + " def feed_forward_out(self, X):\n", + " # feed-forward for output\n", + " z_h = np.matmul(X, self.hidden_weights) + self.hidden_bias\n", + " a_h = sigmoid(z_h)\n", + "\n", + " z_o = np.matmul(a_h, self.output_weights) + self.output_bias\n", + " \n", + " exp_term = np.exp(z_o)\n", + " probabilities = exp_term / np.sum(exp_term, axis=1, keepdims=True)\n", + " return probabilities\n", + "\n", + " def backpropagation(self):\n", + " error_output = self.probabilities - self.Y_data\n", + " error_hidden = np.matmul(error_output, self.output_weights.T) * self.a_h * (1 - self.a_h)\n", + "\n", + " self.output_weights_gradient = np.matmul(self.a_h.T, error_output)\n", + " self.output_bias_gradient = np.sum(error_output, axis=0)\n", + "\n", + " self.hidden_weights_gradient = np.matmul(self.X_data.T, error_hidden)\n", + " self.hidden_bias_gradient = np.sum(error_hidden, axis=0)\n", + "\n", + " if self.lmbd > 0.0:\n", + " self.output_weights_gradient += self.lmbd * self.output_weights\n", + " self.hidden_weights_gradient += self.lmbd * self.hidden_weights\n", + "\n", + " self.output_weights -= self.eta * self.output_weights_gradient\n", + " self.output_bias -= self.eta * self.output_bias_gradient\n", + " self.hidden_weights -= self.eta * self.hidden_weights_gradient\n", + " self.hidden_bias -= self.eta * self.hidden_bias_gradient\n", + "\n", + " def predict(self, X):\n", + " probabilities = self.feed_forward_out(X)\n", + " return np.argmax(probabilities, axis=1)\n", + "\n", + " def predict_probabilities(self, X):\n", + " probabilities = self.feed_forward_out(X)\n", + " return probabilities\n", + "\n", + " def train(self):\n", + " data_indices = np.arange(self.n_inputs)\n", + "\n", + " for i in range(self.epochs):\n", + " for j in range(self.iterations):\n", + " # pick datapoints with replacement\n", + " chosen_datapoints = np.random.choice(\n", + " data_indices, size=self.batch_size, replace=False\n", + " )\n", + "\n", + " # minibatch training data\n", + " self.X_data = self.X_data_full[chosen_datapoints]\n", + " self.Y_data = self.Y_data_full[chosen_datapoints]\n", + "\n", + " self.feed_forward()\n", + " self.backpropagation()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Evaluate model performance on test data\n", + "\n", + "To measure the performance of our network we evaluate how well it does it data it has never seen before, i.e. the test data. \n", + "We measure the performance of the network using the *accuracy* score. \n", + "The accuracy is as you would expect just the number of images correctly labeled divided by the total number of images. A perfect classifier will have an accuracy score of $1$. \n", + "\n", + "$$ \\text{Accuracy} = \\frac{\\sum_{i=1}^n I(\\hat{y}_i = y_i)}{n} ,$$ \n", + "\n", + "where $I$ is the indicator function, $1$ if $\\hat{y}_i = y_i$ and $0$ otherwise." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "epochs = 100\n", + "batch_size = 100\n", + "\n", + "dnn = NeuralNetwork(X_train, Y_train_onehot, eta=eta, lmbd=lmbd, epochs=epochs, batch_size=batch_size,\n", + " n_hidden_neurons=n_hidden_neurons, n_categories=n_categories)\n", + "dnn.train()\n", + "test_predict = dnn.predict(X_test)\n", + "\n", + "# accuracy score from scikit library\n", + "print(\"Accuracy score on test set: \", accuracy_score(Y_test, test_predict))\n", + "\n", + "# equivalent in numpy\n", + "def accuracy_score_numpy(Y_test, Y_pred):\n", + " return np.sum(Y_test == Y_pred) / len(Y_test)\n", + "\n", + "#print(\"Accuracy score on test set: \", accuracy_score_numpy(Y_test, test_predict))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Adjust hyperparameters\n", + "\n", + "We now perform a grid search to find the optimal hyperparameters for the network. \n", + "Note that we are only using 1 layer with 50 neurons, and human performance is estimated to be around $98\\%$ ($2\\%$ error rate)." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "eta_vals = np.logspace(-5, 1, 7)\n", + "lmbd_vals = np.logspace(-5, 1, 7)\n", + "# store the models for later use\n", + "DNN_numpy = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)\n", + "\n", + "# grid search\n", + "for i, eta in enumerate(eta_vals):\n", + " for j, lmbd in enumerate(lmbd_vals):\n", + " dnn = NeuralNetwork(X_train, Y_train_onehot, eta=eta, lmbd=lmbd, epochs=epochs, batch_size=batch_size,\n", + " n_hidden_neurons=n_hidden_neurons, n_categories=n_categories)\n", + " dnn.train()\n", + " \n", + " DNN_numpy[i][j] = dnn\n", + " \n", + " test_predict = dnn.predict(X_test)\n", + " \n", + " print(\"Learning rate = \", eta)\n", + " print(\"Lambda = \", lmbd)\n", + " print(\"Accuracy score on test set: \", accuracy_score(Y_test, test_predict))\n", + " print()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Visualization" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# visual representation of grid search\n", + "# uses seaborn heatmap, you can also do this with matplotlib imshow\n", + "import seaborn as sns\n", + "\n", + "sns.set()\n", + "\n", + "train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n", + "test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n", + "\n", + "for i in range(len(eta_vals)):\n", + " for j in range(len(lmbd_vals)):\n", + " dnn = DNN_numpy[i][j]\n", + " \n", + " train_pred = dnn.predict(X_train) \n", + " test_pred = dnn.predict(X_test)\n", + "\n", + " train_accuracy[i][j] = accuracy_score(Y_train, train_pred)\n", + " test_accuracy[i][j] = accuracy_score(Y_test, test_pred)\n", + "\n", + " \n", + "fig, ax = plt.subplots(figsize = (10, 10))\n", + "sns.heatmap(train_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n", + "ax.set_title(\"Training Accuracy\")\n", + "ax.set_ylabel(\"$\\eta$\")\n", + "ax.set_xlabel(\"$\\lambda$\")\n", + "plt.show()\n", + "\n", + "fig, ax = plt.subplots(figsize = (10, 10))\n", + "sns.heatmap(test_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n", + "ax.set_title(\"Test Accuracy\")\n", + "ax.set_ylabel(\"$\\eta$\")\n", + "ax.set_xlabel(\"$\\lambda$\")\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## scikit-learn implementation\n", + "\n", + "**scikit-learn** focuses more\n", + "on traditional machine learning methods, such as regression,\n", + "clustering, decision trees, etc. As such, it has only two types of\n", + "neural networks: Multi Layer Perceptron outputting continuous values,\n", + "*MPLRegressor*, and Multi Layer Perceptron outputting labels,\n", + "*MLPClassifier*. We will see how simple it is to use these classes.\n", + "\n", + "**scikit-learn** implements a few improvements from our neural network,\n", + "such as early stopping, a varying learning rate, different\n", + "optimization methods, etc. We would therefore expect a better\n", + "performance overall." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "from sklearn.neural_network import MLPClassifier\n", + "# store models for later use\n", + "DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)\n", + "\n", + "for i, eta in enumerate(eta_vals):\n", + " for j, lmbd in enumerate(lmbd_vals):\n", + " dnn = MLPClassifier(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',\n", + " alpha=lmbd, learning_rate_init=eta, max_iter=epochs)\n", + " dnn.fit(X_train, Y_train)\n", + " \n", + " DNN_scikit[i][j] = dnn\n", + " \n", + " print(\"Learning rate = \", eta)\n", + " print(\"Lambda = \", lmbd)\n", + " print(\"Accuracy score on test set: \", dnn.score(X_test, Y_test))\n", + " print()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Visualization" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# optional\n", + "# visual representation of grid search\n", + "# uses seaborn heatmap, could probably do this in matplotlib\n", + "import seaborn as sns\n", + "\n", + "sns.set()\n", + "\n", + "train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n", + "test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n", + "\n", + "for i in range(len(eta_vals)):\n", + " for j in range(len(lmbd_vals)):\n", + " dnn = DNN_scikit[i][j]\n", + " \n", + " train_pred = dnn.predict(X_train) \n", + " test_pred = dnn.predict(X_test)\n", + "\n", + " train_accuracy[i][j] = accuracy_score(Y_train, train_pred)\n", + " test_accuracy[i][j] = accuracy_score(Y_test, test_pred)\n", + "\n", + " \n", + "fig, ax = plt.subplots(figsize = (10, 10))\n", + "sns.heatmap(train_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n", + "ax.set_title(\"Training Accuracy\")\n", + "ax.set_ylabel(\"$\\eta$\")\n", + "ax.set_xlabel(\"$\\lambda$\")\n", + "plt.show()\n", + "\n", + "fig, ax = plt.subplots(figsize = (10, 10))\n", + "sns.heatmap(test_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n", + "ax.set_title(\"Test Accuracy\")\n", + "ax.set_ylabel(\"$\\eta$\")\n", + "ax.set_xlabel(\"$\\lambda$\")\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Building neural networks in Tensorflow and Keras\n", + "\n", + "Now we want to build on the experience gained from our neural network implementation in NumPy and scikit-learn\n", + "and use it to construct a neural network in Tensorflow. Once we have constructed a neural network in NumPy\n", + "and Tensorflow, building one in Keras is really quite trivial, though the performance may suffer. \n", + "\n", + "In our previous example we used only one hidden layer, and in this we will use two. From this it should be quite\n", + "clear how to build one using an arbitrary number of hidden layers, using data structures such as Python lists or\n", + "NumPy arrays.\n", + "\n", + "\n", + "Tensorflow is an open source library machine learning library\n", + "developed by the Google Brain team for internal use. It was released\n", + "under the Apache 2.0 open source license in November 9, 2015.\n", + "\n", + "Tensorflow is a computational framework that allows you to construct\n", + "machine learning models at different levels of abstraction, from\n", + "high-level, object-oriented APIs like Keras, down to the C++ kernels\n", + "that Tensorflow is built upon. The higher levels of abstraction are\n", + "simpler to use, but less flexible, and our choice of implementation\n", + "should reflect the problems we are trying to solve.\n", + "\n", + "[Tensorflow uses](https://www.tensorflow.org/guide/graphs) so-called graphs to represent your computation\n", + "in terms of the dependencies between individual operations, such that you first build a Tensorflow *graph*\n", + "to represent your model, and then create a Tensorflow *session* to run the graph.\n", + "\n", + "In this guide we will analyze the same data as we did in our NumPy and\n", + "scikit-learn tutorial, gathered from the MNIST database of images. We\n", + "will give an introduction to the lower level Python Application\n", + "Program Interfaces (APIs), and see how we use them to build our graph.\n", + "Then we will build (effectively) the same graph in Keras, to see just\n", + "how simple solving a machine learning problem can be.\n", + "\n", + "To install tensorflow on Unix/Linux systems, use pip as" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "pip3 install tensorflow" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "and/or if you use **anaconda**, just write (or install from the graphical user interface)\n", + "(current release of CPU-only TensorFlow)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "conda create -n tf tensorflow\n", + "conda activate tf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "To install the current release of GPU TensorFlow" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "conda create -n tf-gpu tensorflow-gpu\n", + "conda activate tf-gpu" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Keras is a high level [neural network](https://en.wikipedia.org/wiki/Application_programming_interface)\n", + "that supports Tensorflow, CTNK and Theano as backends. \n", + "If you have Anaconda installed you may run the following command" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "conda install keras" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "You can look up the [instructions here](https://keras.io/) for more information.\n", + "\n", + "We will to a large extent use **keras** in this course. \n", + "\n", + "\n", + "Let us look again at the MINST data set." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# import necessary packages\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "import tensorflow as tf\n", + "from sklearn import datasets\n", + "\n", + "\n", + "# ensure the same random numbers appear every time\n", + "np.random.seed(0)\n", + "\n", + "# display images in notebook\n", + "%matplotlib inline\n", + "plt.rcParams['figure.figsize'] = (12,12)\n", + "\n", + "\n", + "# download MNIST dataset\n", + "digits = datasets.load_digits()\n", + "\n", + "# define inputs and labels\n", + "inputs = digits.images\n", + "labels = digits.target\n", + "\n", + "print(\"inputs = (n_inputs, pixel_width, pixel_height) = \" + str(inputs.shape))\n", + "print(\"labels = (n_inputs) = \" + str(labels.shape))\n", + "\n", + "\n", + "# flatten the image\n", + "# the value -1 means dimension is inferred from the remaining dimensions: 8x8 = 64\n", + "n_inputs = len(inputs)\n", + "inputs = inputs.reshape(n_inputs, -1)\n", + "print(\"X = (n_inputs, n_features) = \" + str(inputs.shape))\n", + "\n", + "\n", + "# choose some random images to display\n", + "indices = np.arange(n_inputs)\n", + "random_indices = np.random.choice(indices, size=5)\n", + "\n", + "for i, image in enumerate(digits.images[random_indices]):\n", + " plt.subplot(1, 5, i+1)\n", + " plt.axis('off')\n", + " plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest')\n", + " plt.title(\"Label: %d\" % digits.target[random_indices[i]])\n", + "plt.show()" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "from tensorflow.keras.layers import Input\n", + "from tensorflow.keras.models import Sequential #This allows appending layers to existing models\n", + "from tensorflow.keras.layers import Dense #This allows defining the characteristics of a particular layer\n", + "from tensorflow.keras import optimizers #This allows using whichever optimiser we want (sgd,adam,RMSprop)\n", + "from tensorflow.keras import regularizers #This allows using whichever regularizer we want (l1,l2,l1_l2)\n", + "from tensorflow.keras.utils import to_categorical #This allows using categorical cross entropy as the cost function\n", + "\n", + "from sklearn.model_selection import train_test_split\n", + "\n", + "# one-hot representation of labels\n", + "labels = to_categorical(labels)\n", + "\n", + "# split into train and test data\n", + "train_size = 0.8\n", + "test_size = 1 - train_size\n", + "X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size,\n", + " test_size=test_size)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "\n", + "epochs = 100\n", + "batch_size = 100\n", + "n_neurons_layer1 = 100\n", + "n_neurons_layer2 = 50\n", + "n_categories = 10\n", + "eta_vals = np.logspace(-5, 1, 7)\n", + "lmbd_vals = np.logspace(-5, 1, 7)\n", + "def create_neural_network_keras(n_neurons_layer1, n_neurons_layer2, n_categories, eta, lmbd):\n", + " model = Sequential()\n", + " model.add(Dense(n_neurons_layer1, activation='sigmoid', kernel_regularizer=regularizers.l2(lmbd)))\n", + " model.add(Dense(n_neurons_layer2, activation='sigmoid', kernel_regularizer=regularizers.l2(lmbd)))\n", + " model.add(Dense(n_categories, activation='softmax'))\n", + " \n", + " sgd = optimizers.SGD(lr=eta)\n", + " model.compile(loss='categorical_crossentropy', optimizer=sgd, metrics=['accuracy'])\n", + " \n", + " return model" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "DNN_keras = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)\n", + " \n", + "for i, eta in enumerate(eta_vals):\n", + " for j, lmbd in enumerate(lmbd_vals):\n", + " DNN = create_neural_network_keras(n_neurons_layer1, n_neurons_layer2, n_categories,\n", + " eta=eta, lmbd=lmbd)\n", + " DNN.fit(X_train, Y_train, epochs=epochs, batch_size=batch_size, verbose=0)\n", + " scores = DNN.evaluate(X_test, Y_test)\n", + " \n", + " DNN_keras[i][j] = DNN\n", + " \n", + " print(\"Learning rate = \", eta)\n", + " print(\"Lambda = \", lmbd)\n", + " print(\"Test accuracy: %.3f\" % scores[1])\n", + " print()" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# optional\n", + "# visual representation of grid search\n", + "# uses seaborn heatmap, could probably do this in matplotlib\n", + "import seaborn as sns\n", + "\n", + "sns.set()\n", + "\n", + "train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n", + "test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n", + "\n", + "for i in range(len(eta_vals)):\n", + " for j in range(len(lmbd_vals)):\n", + " DNN = DNN_keras[i][j]\n", + "\n", + " train_accuracy[i][j] = DNN.evaluate(X_train, Y_train)[1]\n", + " test_accuracy[i][j] = DNN.evaluate(X_test, Y_test)[1]\n", + "\n", + " \n", + "fig, ax = plt.subplots(figsize = (10, 10))\n", + "sns.heatmap(train_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n", + "ax.set_title(\"Training Accuracy\")\n", + "ax.set_ylabel(\"$\\eta$\")\n", + "ax.set_xlabel(\"$\\lambda$\")\n", + "plt.show()\n", + "\n", + "fig, ax = plt.subplots(figsize = (10, 10))\n", + "sns.heatmap(test_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n", + "ax.set_title(\"Test Accuracy\")\n", + "ax.set_ylabel(\"$\\eta$\")\n", + "ax.set_xlabel(\"$\\lambda$\")\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## The Breast Cancer Data, now with Keras" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "\n", + "import tensorflow as tf\n", + "from tensorflow.keras.layers import Input\n", + "from tensorflow.keras.models import Sequential #This allows appending layers to existing models\n", + "from tensorflow.keras.layers import Dense #This allows defining the characteristics of a particular layer\n", + "from tensorflow.keras import optimizers #This allows using whichever optimiser we want (sgd,adam,RMSprop)\n", + "from tensorflow.keras import regularizers #This allows using whichever regularizer we want (l1,l2,l1_l2)\n", + "from tensorflow.keras.utils import to_categorical #This allows using categorical cross entropy as the cost function\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "import seaborn as sns\n", + "from sklearn.model_selection import train_test_split as splitter\n", + "from sklearn.datasets import load_breast_cancer\n", + "import pickle\n", + "import os \n", + "\n", + "\n", + "\"\"\"Load breast cancer dataset\"\"\"\n", + "\n", + "np.random.seed(0) #create same seed for random number every time\n", + "\n", + "cancer=load_breast_cancer() #Download breast cancer dataset\n", + "\n", + "inputs=cancer.data #Feature matrix of 569 rows (samples) and 30 columns (parameters)\n", + "outputs=cancer.target #Label array of 569 rows (0 for benign and 1 for malignant)\n", + "labels=cancer.feature_names[0:30]\n", + "\n", + "print('The content of the breast cancer dataset is:') #Print information about the datasets\n", + "print(labels)\n", + "print('-------------------------')\n", + "print(\"inputs = \" + str(inputs.shape))\n", + "print(\"outputs = \" + str(outputs.shape))\n", + "print(\"labels = \"+ str(labels.shape))\n", + "\n", + "x=inputs #Reassign the Feature and Label matrices to other variables\n", + "y=outputs\n", + "\n", + "#%% \n", + "\n", + "# Visualisation of dataset (for correlation analysis)\n", + "\n", + "plt.figure()\n", + "plt.scatter(x[:,0],x[:,2],s=40,c=y,cmap=plt.cm.Spectral)\n", + "plt.xlabel('Mean radius',fontweight='bold')\n", + "plt.ylabel('Mean perimeter',fontweight='bold')\n", + "plt.show()\n", + "\n", + "plt.figure()\n", + "plt.scatter(x[:,5],x[:,6],s=40,c=y, cmap=plt.cm.Spectral)\n", + "plt.xlabel('Mean compactness',fontweight='bold')\n", + "plt.ylabel('Mean concavity',fontweight='bold')\n", + "plt.show()\n", + "\n", + "\n", + "plt.figure()\n", + "plt.scatter(x[:,0],x[:,1],s=40,c=y,cmap=plt.cm.Spectral)\n", + "plt.xlabel('Mean radius',fontweight='bold')\n", + "plt.ylabel('Mean texture',fontweight='bold')\n", + "plt.show()\n", + "\n", + "plt.figure()\n", + "plt.scatter(x[:,2],x[:,1],s=40,c=y,cmap=plt.cm.Spectral)\n", + "plt.xlabel('Mean perimeter',fontweight='bold')\n", + "plt.ylabel('Mean compactness',fontweight='bold')\n", + "plt.show()\n", + "\n", + "\n", + "# Generate training and testing datasets\n", + "\n", + "#Select features relevant to classification (texture,perimeter,compactness and symmetery) \n", + "#and add to input matrix\n", + "\n", + "temp1=np.reshape(x[:,1],(len(x[:,1]),1))\n", + "temp2=np.reshape(x[:,2],(len(x[:,2]),1))\n", + "X=np.hstack((temp1,temp2)) \n", + "temp=np.reshape(x[:,5],(len(x[:,5]),1))\n", + "X=np.hstack((X,temp)) \n", + "temp=np.reshape(x[:,8],(len(x[:,8]),1))\n", + "X=np.hstack((X,temp)) \n", + "\n", + "X_train,X_test,y_train,y_test=splitter(X,y,test_size=0.1) #Split datasets into training and testing\n", + "\n", + "y_train=to_categorical(y_train) #Convert labels to categorical when using categorical cross entropy\n", + "y_test=to_categorical(y_test)\n", + "\n", + "del temp1,temp2,temp\n", + "\n", + "# %%\n", + "\n", + "# Define tunable parameters\"\n", + "\n", + "eta=np.logspace(-3,-1,3) #Define vector of learning rates (parameter to SGD optimiser)\n", + "lamda=0.01 #Define hyperparameter\n", + "n_layers=2 #Define number of hidden layers in the model\n", + "n_neuron=np.logspace(0,3,4,dtype=int) #Define number of neurons per layer\n", + "epochs=100 #Number of reiterations over the input data\n", + "batch_size=100 #Number of samples per gradient update\n", + "\n", + "# %%\n", + "\n", + "\"\"\"Define function to return Deep Neural Network model\"\"\"\n", + "\n", + "def NN_model(inputsize,n_layers,n_neuron,eta,lamda):\n", + " model=Sequential() \n", + " for i in range(n_layers): #Run loop to add hidden layers to the model\n", + " if (i==0): #First layer requires input dimensions\n", + " model.add(Dense(n_neuron,activation='relu',kernel_regularizer=regularizers.l2(lamda),input_dim=inputsize))\n", + " else: #Subsequent layers are capable of automatic shape inferencing\n", + " model.add(Dense(n_neuron,activation='relu',kernel_regularizer=regularizers.l2(lamda)))\n", + " model.add(Dense(2,activation='softmax')) #2 outputs - ordered and disordered (softmax for prob)\n", + " sgd=optimizers.SGD(lr=eta)\n", + " model.compile(loss='categorical_crossentropy',optimizer=sgd,metrics=['accuracy'])\n", + " return model\n", + "\n", + " \n", + "Train_accuracy=np.zeros((len(n_neuron),len(eta))) #Define matrices to store accuracy scores as a function\n", + "Test_accuracy=np.zeros((len(n_neuron),len(eta))) #of learning rate and number of hidden neurons for \n", + "\n", + "for i in range(len(n_neuron)): #run loops over hidden neurons and learning rates to calculate \n", + " for j in range(len(eta)): #accuracy scores \n", + " DNN_model=NN_model(X_train.shape[1],n_layers,n_neuron[i],eta[j],lamda)\n", + " DNN_model.fit(X_train,y_train,epochs=epochs,batch_size=batch_size,verbose=1)\n", + " Train_accuracy[i,j]=DNN_model.evaluate(X_train,y_train)[1]\n", + " Test_accuracy[i,j]=DNN_model.evaluate(X_test,y_test)[1]\n", + " \n", + "\n", + "def plot_data(x,y,data,title=None):\n", + "\n", + " # plot results\n", + " fontsize=16\n", + "\n", + "\n", + " fig = plt.figure()\n", + " ax = fig.add_subplot(111)\n", + " cax = ax.matshow(data, interpolation='nearest', vmin=0, vmax=1)\n", + " \n", + " cbar=fig.colorbar(cax)\n", + " cbar.ax.set_ylabel('accuracy (%)',rotation=90,fontsize=fontsize)\n", + " cbar.set_ticks([0,.2,.4,0.6,0.8,1.0])\n", + " cbar.set_ticklabels(['0%','20%','40%','60%','80%','100%'])\n", + "\n", + " # put text on matrix elements\n", + " for i, x_val in enumerate(np.arange(len(x))):\n", + " for j, y_val in enumerate(np.arange(len(y))):\n", + " c = \"${0:.1f}\\\\%$\".format( 100*data[j,i]) \n", + " ax.text(x_val, y_val, c, va='center', ha='center')\n", + "\n", + " # convert axis vaues to to string labels\n", + " x=[str(i) for i in x]\n", + " y=[str(i) for i in y]\n", + "\n", + "\n", + " ax.set_xticklabels(['']+x)\n", + " ax.set_yticklabels(['']+y)\n", + "\n", + " ax.set_xlabel('$\\\\mathrm{learning\\\\ rate}$',fontsize=fontsize)\n", + " ax.set_ylabel('$\\\\mathrm{hidden\\\\ neurons}$',fontsize=fontsize)\n", + " if title is not None:\n", + " ax.set_title(title)\n", + "\n", + " plt.tight_layout()\n", + "\n", + " plt.show()\n", + " \n", + "plot_data(eta,n_neuron,Train_accuracy, 'training')\n", + "plot_data(eta,n_neuron,Test_accuracy, 'testing')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Fine-tuning neural network hyperparameters\n", + "\n", + "The flexibility of neural networks is also one of their main\n", + "drawbacks: there are many hyperparameters to tweak. Not only can you\n", + "use any imaginable network topology (how neurons/nodes are interconnected),\n", + "but even in a simple FFNN you can change the number of layers, the\n", + "number of neurons per layer, the type of activation function to use in\n", + "each layer, the weight initialization logic, the stochastic gradient optmized and much more. How do you\n", + "know what combination of hyperparameters is the best for your task?\n", + "\n", + "* You can use grid search with cross-validation to find the right hyperparameters.\n", + "\n", + "However,since there are many hyperparameters to tune, and since\n", + "training a neural network on a large dataset takes a lot of time, you\n", + "will only be able to explore a tiny part of the hyperparameter space.\n", + "\n", + "\n", + "* You can use randomized search.\n", + "\n", + "* Or use tools like [Oscar](http://oscar.calldesk.ai/), which implements more complex algorithms to help you find a good set of hyperparameters quickly. \n", + "\n", + "For many problems you can start with just one or two hidden layers and it will work just fine.\n", + "For the MNIST data set you ca easily get a high accuracy using just one hidden layer with a\n", + "few hundred neurons.\n", + "You can reach for this data set above 98% accuracy using two hidden layers with the same total amount of\n", + "neurons, in roughly the same amount of training time. \n", + "\n", + "For more complex problems, you can gradually\n", + "ramp up the number of hidden layers, until you start overfitting the training set. Very complex tasks, such\n", + "as large image classification or speech recognition, typically require networks with dozens of layers\n", + "and they need a huge amount\n", + "of training data. However, you will rarely have to train such networks from scratch: it is much more\n", + "common to reuse parts of a pretrained state-of-the-art network that performs a similar task.\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "## Which activation function should I use?\n", + "\n", + "The Back propagation algorithm we derived above works by going from\n", + "the output layer to the input layer, propagating the error gradient on\n", + "the way. Once the algorithm has computed the gradient of the cost\n", + "function with regards to each parameter in the network, it uses these\n", + "gradients to update each parameter with a Gradient Descent (GD) step.\n", + "\n", + "\n", + "Unfortunately for us, the gradients often get smaller and smaller as the\n", + "algorithm progresses down to the first hidden layers. As a result, the\n", + "GD update leaves the lower layer connection weights\n", + "virtually unchanged, and training never converges to a good\n", + "solution. This is known in the literature as \n", + "**the vanishing gradients problem**. \n", + "\n", + "In other cases, the opposite can happen, namely the the gradients can grow bigger and\n", + "bigger. The result is that many of the layers get large updates of the \n", + "weights the\n", + "algorithm diverges. This is the **exploding gradients problem**, which is\n", + "mostly encountered in recurrent neural networks. More generally, deep\n", + "neural networks suffer from unstable gradients, different layers may\n", + "learn at widely different speeds\n", + "\n", + "\n", + "\n", + "\n", + "Although this unfortunate behavior has been empirically observed for\n", + "quite a while (it was one of the reasons why deep neural networks were\n", + "mostly abandoned for a long time), it is only around 2010 that\n", + "significant progress was made in understanding it.\n", + "\n", + "A paper titled [Understanding the Difficulty of Training Deep\n", + "Feedforward Neural Networks by Xavier Glorot and Yoshua Bengio](http://proceedings.mlr.press/v9/glorot10a.html) found that\n", + "the problems with the popular logistic\n", + "sigmoid activation function and the weight initialization technique\n", + "that was most popular at the time, namely random initialization using\n", + "a normal distribution with a mean of 0 and a standard deviation of\n", + "1. \n", + "\n", + "They showed that with this activation function and this\n", + "initialization scheme, the variance of the outputs of each layer is\n", + "much greater than the variance of its inputs. Going forward in the\n", + "network, the variance keeps increasing after each layer until the\n", + "activation function saturates at the top layers. This is actually made\n", + "worse by the fact that the logistic function has a mean of 0.5, not 0\n", + "(the hyperbolic tangent function has a mean of 0 and behaves slightly\n", + "better than the logistic function in deep networks).\n", + "\n", + "\n", + "\n", + "Looking at the logistic activation function, when inputs become large\n", + "(negative or positive), the function saturates at 0 or 1, with a\n", + "derivative extremely close to 0. Thus when backpropagation kicks in,\n", + "it has virtually no gradient to propagate back through the network,\n", + "and what little gradient exists keeps getting diluted as\n", + "backpropagation progresses down through the top layers, so there is\n", + "really nothing left for the lower layers.\n", + "\n", + "In their paper, Glorot and Bengio propose a way to significantly\n", + "alleviate this problem. We need the signal to flow properly in both\n", + "directions: in the forward direction when making predictions, and in\n", + "the reverse direction when backpropagating gradients. We don’t want\n", + "the signal to die out, nor do we want it to explode and saturate. For\n", + "the signal to flow properly, the authors argue that we need the\n", + "variance of the outputs of each layer to be equal to the variance of\n", + "its inputs, and we also need the gradients to have equal variance\n", + "before and after flowing through a layer in the reverse direction.\n", + "\n", + "\n", + "\n", + "One of the insights in the 2010 paper by Glorot and Bengio was that\n", + "the vanishing/exploding gradients problems were in part due to a poor\n", + "choice of activation function. Until then most people had assumed that\n", + "if Nature had chosen to use roughly sigmoid activation functions in\n", + "biological neurons, they must be an excellent choice. But it turns out\n", + "that other activation functions behave much better in deep neural\n", + "networks, in particular the ReLU activation function, mostly because\n", + "it does not saturate for positive values (and also because it is quite\n", + "fast to compute).\n", + "\n", + "\n", + "## The RELU function family\n", + "\n", + "The ReLU activation function suffers from a problem known as the dying\n", + "ReLUs: during training, some neurons effectively die, meaning they\n", + "stop outputting anything other than 0.\n", + "\n", + "In some cases, you may find that half of your network’s neurons are\n", + "dead, especially if you used a large learning rate. During training,\n", + "if a neuron’s weights get updated such that the weighted sum of the\n", + "neuron’s inputs is negative, it will start outputting 0. When this\n", + "happen, the neuron is unlikely to come back to life since the gradient\n", + "of the ReLU function is 0 when its input is negative.\n", + "\n", + "To solve this problem, nowadays practitioners use a variant of the ReLU\n", + "function, such as the leaky ReLU discussed above or the so-called\n", + "exponential linear unit (ELU) function" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "ELU(z) = \\left\\{\\begin{array}{cc} \\alpha\\left( \\exp{(z)}-1\\right) & z < 0,\\\\ z & z \\ge 0.\\end{array}\\right.\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In general it seems that the ELU activation function is better than\n", + "the leaky ReLU function (and its variants), which is better than\n", + "ReLU. ReLU performs better than $\\tanh$ which in turn performs better\n", + "than the logistic function. \n", + "\n", + "If runtime\n", + "performance is an issue, then you may opt for the leaky ReLU function over the \n", + "ELU function If you don’t\n", + "want to tweak yet another hyperparameter, you may just use the default\n", + "$\\alpha$ of $0.01$ for the leaky ReLU, and $1$ for ELU. If you have\n", + "spare time and computing power, you can use cross-validation or\n", + "bootstrap to evaluate other activation functions.\n", + "\n", + "\n", + "\n", + "In most cases you can use the ReLU activation function in the hidden layers (or one of its variants).\n", + "\n", + "It is a bit faster to compute than other activation functions, and the gradient descent optimization does in general not get stuck.\n", + "\n", + "**For the output layer:**\n", + "\n", + "* For classification the softmax activation function is generally a good choice for classification tasks (when the classes are mutually exclusive).\n", + "\n", + "* For regression tasks, you can simply use no activation function at all.\n", + "\n", + "## Batch Normalization\n", + "\n", + "Batch Normalization\n", + "aims to address the vanishing/exploding gradients problems, and more generally the problem that the\n", + "distribution of each layer’s inputs changes during training, as the parameters of the previous layers change.\n", + "\n", + "The technique consists of adding an operation in the model just before the activation function of each\n", + "layer, simply zero-centering and normalizing the inputs, then scaling and shifting the result using two new\n", + "parameters per layer (one for scaling, the other for shifting). In other words, this operation lets the model\n", + "learn the optimal scale and mean of the inputs for each layer.\n", + "In order to zero-center and normalize the inputs, the algorithm needs to estimate the inputs’ mean and\n", + "standard deviation. It does so by evaluating the mean and standard deviation of the inputs over the current\n", + "mini-batch, from this the name batch normalization.\n", + "\n", + "## Dropout\n", + "\n", + "It is a fairly simple algorithm: at every training step, every neuron (including the input neurons but\n", + "excluding the output neurons) has a probability $p$ of being temporarily dropped out, meaning it will be\n", + "entirely ignored during this training step, but it may be active during the next step.\n", + "\n", + "The\n", + "hyperparameter $p$ is called the dropout rate, and it is typically set to 50%. After training, the neurons are not dropped anymore.\n", + " It is viewed as one of the most popular regularization techniques.\n", + "\n", + "## Gradient Clipping\n", + "\n", + "A popular technique to lessen the exploding gradients problem is to simply clip the gradients during\n", + "backpropagation so that they never exceed some threshold (this is mostly useful for recurrent neural\n", + "networks).\n", + "\n", + "This technique is called Gradient Clipping.\n", + "\n", + "In general however, Batch\n", + "Normalization is preferred.\n", + "\n", + "\n", + "## A top-down perspective on Neural networks\n", + "\n", + "\n", + "The first thing we would like to do is divide the data into two or three\n", + "parts. A training set, a validation or dev (development) set, and a\n", + "test set. The test set is the data on which we want to make\n", + "predictions. The dev set is a subset of the training data we use to\n", + "check how well we are doing out-of-sample, after training the model on\n", + "the training dataset. We use the validation error as a proxy for the\n", + "test error in order to make tweaks to our model. It is crucial that we\n", + "do not use any of the test data to train the algorithm. This is a\n", + "cardinal sin in ML. Then:\n", + "\n", + "\n", + "* Estimate optimal error rate\n", + "\n", + "* Minimize underfitting (bias) on training data set.\n", + "\n", + "* Make sure you are not overfitting.\n", + "\n", + "If the validation and test sets are drawn from the same distributions,\n", + "then a good performance on the validation set should lead to similarly\n", + "good performance on the test set. \n", + "\n", + "However, sometimes\n", + "the training data and test data differ in subtle ways because, for\n", + "example, they are collected using slightly different methods, or\n", + "because it is cheaper to collect data in one way versus another. In\n", + "this case, there can be a mismatch between the training and test\n", + "data. This can lead to the neural network overfitting these small\n", + "differences between the test and training sets, and a poor performance\n", + "on the test set despite having a good performance on the validation\n", + "set. To rectify this, Andrew Ng suggests making two validation or dev\n", + "sets, one constructed from the training data and one constructed from\n", + "the test data. The difference between the performance of the algorithm\n", + "on these two validation sets quantifies the train-test mismatch. This\n", + "can serve as another important diagnostic when using DNNs for\n", + "supervised learning.\n", + "\n", + "\n", + "## Limitations of supervised learning with deep networks\n", + "\n", + "Like all statistical methods, supervised learning using neural\n", + "networks has important limitations. This is especially important when\n", + "one seeks to apply these methods, especially to physics problems. Like\n", + "all tools, DNNs are not a universal solution. Often, the same or\n", + "better performance on a task can be achieved by using a few\n", + "hand-engineered features (or even a collection of random\n", + "features). \n", + "\n", + "Here we list some of the important limitations of supervised neural network based models. \n", + "\n", + "\n", + "\n", + "* **Need labeled data**. All supervised learning methods, DNNs for supervised learning require labeled data. Often, labeled data is harder to acquire than unlabeled data (e.g. one must pay for human experts to label images).\n", + "\n", + "* **Supervised neural networks are extremely data intensive.** DNNs are data hungry. They perform best when data is plentiful. This is doubly so for supervised methods where the data must also be labeled. The utility of DNNs is extremely limited if data is hard to acquire or the datasets are small (hundreds to a few thousand samples). In this case, the performance of other methods that utilize hand-engineered features can exceed that of DNNs.\n", + "\n", + "* **Homogeneous data.** Almost all DNNs deal with homogeneous data of one type. It is very hard to design architectures that mix and match data types (i.e. some continuous variables, some discrete variables, some time series). In applications beyond images, video, and language, this is often what is required. In contrast, ensemble models like random forests or gradient-boosted trees have no difficulty handling mixed data types.\n", + "\n", + "* **Many problems are not about prediction.** In natural science we are often interested in learning something about the underlying distribution that generates the data. In this case, it is often difficult to cast these ideas in a supervised learning setting. While the problems are related, it is possible to make good predictions with a *wrong* model. The model might or might not be useful for understanding the underlying science.\n", + "\n", + "Some of these remarks are particular to DNNs, others are shared by all supervised learning methods. This motivates the use of unsupervised methods which in part circumvent these problems." + ] + } + ], + "metadata": {}, + "nbformat": 4, + "nbformat_minor": 4 +} diff --git a/doc/BookChapters/chapter11.ipynb b/doc/BookChapters/chapter11.ipynb new file mode 100644 index 000000000..c03e4a368 --- /dev/null +++ b/doc/BookChapters/chapter11.ipynb @@ -0,0 +1,3023 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Solving Differential Equations with Deep Learning\n", + "\n", + "The Universal Approximation Theorem states that a neural network can\n", + "approximate any function at a single hidden layer along with one input\n", + "and output layer to any given precision. \n", + "\n", + "\n", + "An ordinary differential equation (ODE) is an equation involving functions having one variable.\n", + "\n", + "In general, an ordinary differential equation looks like" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{ode} \\tag{1}\n", + "f\\left(x, \\, g(x), \\, g'(x), \\, g''(x), \\, \\dots \\, , \\, g^{(n)}(x)\\right) = 0\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$.\n", + "\n", + "The $f\\left(x, g(x), g'(x), g''(x), \\, \\dots \\, , g^{(n)}(x)\\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \\ g'(x), \\ g''(x), \\, \\dots \\, , \\text{ and } g^{(n)}(x)$ on the left side of the equality sign in ([1](#ode)).\n", + "The highest order of derivative, that is the value of $n$, determines to the order of the equation.\n", + "The equation is referred to as a $n$-th order ODE.\n", + "Along with ([1](#ode)), some additional conditions of the function $g(x)$ are typically given\n", + "for the solution to be unique.\n", + "\n", + "\n", + "\n", + "Let the trial solution $g_t(x)$ be" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + "\tg_t(x) = h_1(x) + h_2(x,N(x,P))\n", + "\\label{_auto1} \\tag{2}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set\n", + "of conditions, $N(x,P)$ a neural network with weights and biases\n", + "described by $P$ and $h_2(x, N(x,P))$ some expression involving the\n", + "neural network. The role of the function $h_2(x, N(x,P))$, is to\n", + "ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is\n", + "evaluated at the values of $x$ where the given conditions must be\n", + "satisfied. The function $h_1(x)$ should alone make $g_t(x)$ satisfy\n", + "the conditions.\n", + "\n", + "But what about the network $N(x,P)$?\n", + "\n", + "\n", + "As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation.\n", + "\n", + "\n", + "\n", + "For the minimization to be defined, we need to have a cost function at hand to minimize.\n", + "\n", + "It is given that $f\\left(x, \\, g(x), \\, g'(x), \\, g''(x), \\, \\dots \\, , \\, g^{(n)}(x)\\right)$ should be equal to zero in ([1](#ode)).\n", + "We can choose to consider the mean squared error as the cost function for an input $x$.\n", + "Since we are looking at one input, the cost function is just $f$ squared.\n", + "The cost function $c\\left(x, P \\right)$ can therefore be expressed as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(x, P\\right) = \\big(f\\left(x, \\, g(x), \\, g'(x), \\, g''(x), \\, \\dots \\, , \\, g^{(n)}(x)\\right)\\big)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "If $N$ inputs are given as a vector $\\boldsymbol{x}$ with elements $x_i$ for $i = 1,\\dots,N$,\n", + "the cost function becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{cost} \\tag{3}\n", + "\tC\\left(\\boldsymbol{x}, P\\right) = \\frac{1}{N} \\sum_{i=1}^N \\big(f\\left(x_i, \\, g(x_i), \\, g'(x_i), \\, g''(x_i), \\, \\dots \\, , \\, g^{(n)}(x_i)\\right)\\big)^2\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The neural net should then find the parameters $P$ that minimizes the cost function in\n", + "([3](#cost)) for a set of $N$ training samples $x_i$.\n", + "\n", + "\n", + "\n", + "To perform the minimization using gradient descent, the gradient of $C\\left(\\boldsymbol{x}, P\\right)$ is needed.\n", + "It might happen so that finding an analytical expression of the gradient of $C(\\boldsymbol{x}, P)$ from ([3](#cost)) gets too messy, depending on which cost function one desires to use.\n", + "\n", + "Luckily, there exists libraries that makes the job for us through automatic differentiation.\n", + "Automatic differentiation is a method of finding the derivatives numerically with very high precision.\n", + "\n", + "\n", + "### Example: Exponential decay\n", + "\n", + "An exponential decay of a quantity $g(x)$ is described by the equation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{solve_expdec} \\tag{4}\n", + " g'(x) = -\\gamma g(x)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $g(0) = g_0$ for some chosen initial value $g_0$.\n", + "\n", + "The analytical solution of ([4](#solve_expdec)) is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " g(x) = g_0 \\exp\\left(-\\gamma x\\right)\n", + "\\label{_auto2} \\tag{5}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of ([4](#solve_expdec)).\n", + "\n", + "\n", + "\n", + "The program will use a neural network to solve" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{solveode} \\tag{6}\n", + "g'(x) = -\\gamma g(x)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(0) = g_0$ with $\\gamma$ and $g_0$ being some chosen values.\n", + "\n", + "In this example, $\\gamma = 2$ and $g_0 = 10$.\n", + "\n", + "\n", + "To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g_t(x, P) = h_1(x) + h_2(x, N(x, P))\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer.\n", + "\n", + "\n", + "\n", + "In this network, there are no weights and bias at the input layer, so $P = \\{ P_{\\text{hidden}}, P_{\\text{output}} \\}$.\n", + "If there are $N_{\\text{hidden} }$ neurons in the hidden layer, then $P_{\\text{hidden}}$ is a $N_{\\text{hidden} } \\times (1 + N_{\\text{input}})$ matrix, given that there are $N_{\\text{input}}$ neurons in the input layer.\n", + "\n", + "The first column in $P_{\\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer.\n", + "If there are $N_{\\text{output} }$ neurons in the output layer, then $P_{\\text{output}} $ is a $N_{\\text{output} } \\times (1 + N_{\\text{hidden} })$ matrix.\n", + "\n", + "Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron.\n", + "\n", + "It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of ([6](#solveode)). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \\cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{trial} \\tag{7}\n", + "g_t(x, P) = g_0 + x \\cdot N(x, P)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Reformulating the problem\n", + "\n", + "We wish that our neural network manages to minimize a given cost function.\n", + "\n", + "A reformulation of out equation, ([6](#solveode)), must therefore be done,\n", + "such that it describes the problem a neural network can solve for.\n", + "\n", + "The neural network must find the set of weights and biases $P$ such that the trial solution in ([7](#trial)) satisfies ([6](#solveode)).\n", + "\n", + "The trial solution" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g_t(x, P) = g_0 + x \\cdot N(x, P)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{nnmin} \\tag{8}\n", + "g_t'(x, P) = - \\gamma g_t(x, P)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "is fulfilled as *best as possible*.\n", + "\n", + "\n", + "The left hand side and right hand side of ([8](#nnmin)) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible.\n", + "This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero.\n", + "In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network.\n", + "\n", + "This gives the following cost function our neural network must solve for:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\min_{P}\\Big\\{ \\big(g_t'(x, P) - ( -\\gamma g_t(x, P) \\big)^2 \\Big\\}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "(the notation $\\min_{P}\\{ f(x, P) \\}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$)\n", + "\n", + "or, in terms of weights and biases for the hidden and output layer in our network:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\min_{P_{\\text{hidden} }, \\ P_{\\text{output} }}\\Big\\{ \\big(g_t'(x, \\{ P_{\\text{hidden} }, P_{\\text{output} }\\}) - ( -\\gamma g_t(x, \\{ P_{\\text{hidden} }, P_{\\text{output} }\\}) \\big)^2 \\Big\\}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for an input value $x$.\n", + "\n", + "\n", + "\n", + "If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \\dots, N$, then the *total* error to minimize becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{min} \\tag{9}\n", + "\\min_{P}\\Big\\{\\frac{1}{N} \\sum_{i=1}^N \\big(g_t'(x_i, P) - ( -\\gamma g_t(x_i, P) \\big)^2 \\Big\\}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Letting $\\boldsymbol{x}$ be a vector with elements $x_i$ and $C(\\boldsymbol{x}, P) = \\frac{1}{N} \\sum_i \\big(g_t'(x_i, P) - ( -\\gamma g_t(x_i, P) \\big)^2$ denote the cost function, the minimization problem that our network must solve, becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\min_{P} C(\\boldsymbol{x}, P)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In terms of $P_{\\text{hidden} }$ and $P_{\\text{output} }$, this could also be expressed as\n", + "\n", + "$$\n", + "\\min_{P_{\\text{hidden} }, \\ P_{\\text{output} }} C(\\boldsymbol{x}, \\{P_{\\text{hidden} }, P_{\\text{output} }\\})\n", + "$$\n", + "\n", + "\n", + "For simplicity, it is assumed that the input is an array $\\boldsymbol{x} = (x_1, \\dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills ([9](#min)).\n", + "\n", + "First, the neural network must feed forward the inputs.\n", + "This means that $\\boldsymbol{x}s$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further.\n", + "The input layer will consist of $N_{\\text{input} }$ neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be $N_{\\text{hidden} }$.\n", + "\n", + "\n", + "For the $i$-th in the hidden layer with weight $w_i^{\\text{hidden} }$ and bias $b_i^{\\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "z_{i,j}^{\\text{hidden}} &= b_i^{\\text{hidden}} + w_i^{\\text{hidden}}x_j \\\\\n", + "&=\n", + "\\begin{pmatrix}\n", + "b_i^{\\text{hidden}} & w_i^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 \\\\\n", + "x_j\n", + "\\end{pmatrix}\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "\\boldsymbol{z}_{i}^{\\text{hidden}} &= \\Big( b_i^{\\text{hidden}} + w_i^{\\text{hidden}}x_1 , \\ b_i^{\\text{hidden}} + w_i^{\\text{hidden}} x_2, \\ \\dots \\, , \\ b_i^{\\text{hidden}} + w_i^{\\text{hidden}} x_N\\Big) \\\\\n", + "&=\n", + "\\begin{pmatrix}\n", + " b_i^{\\text{hidden}} & w_i^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 & 1 & \\dots & 1 \\\\\n", + "x_1 & x_2 & \\dots & x_N\n", + "\\end{pmatrix} \\\\\n", + "&= \\boldsymbol{p}_{i, \\text{hidden}}^T X\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The vector $\\boldsymbol{p}_{i, \\text{hidden}}^T$ constitutes each row in $P_{\\text{hidden} }$, which contains the weights for the neural network to minimize according to ([9](#min)).\n", + "\n", + "After having found $\\boldsymbol{z}_{i}^{\\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\\boldsymbol{z})$.\n", + "\n", + "In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "f(z) = \\frac{1}{1 + \\exp{(-z)}}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "It is possible to use other activations functions for the hidden layer also.\n", + "\n", + "The output $\\boldsymbol{x}_i^{\\text{hidden}}$ from each $i$-th hidden neuron is:\n", + "\n", + "$$\n", + "\\boldsymbol{x}_i^{\\text{hidden} } = f\\big( \\boldsymbol{z}_{i}^{\\text{hidden}} \\big)\n", + "$$\n", + "\n", + "The outputs $\\boldsymbol{x}_i^{\\text{hidden} } $ are then sent to the output layer.\n", + "\n", + "The output layer consists of one neuron in this case, and combines the\n", + "output from each of the neurons in the hidden layers. The output layer\n", + "combines the results from the hidden layer using some weights $w_i^{\\text{output}}$\n", + "and biases $b_i^{\\text{output}}$. In this case,\n", + "it is assumes that the number of neurons in the output layer is one.\n", + "\n", + "\n", + "\n", + "The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "z_{1,j}^{\\text{output}} & =\n", + "\\begin{pmatrix}\n", + "b_1^{\\text{output}} & \\boldsymbol{w}_1^{\\text{output}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 \\\\\n", + "\\boldsymbol{x}_j^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Expressing $z_{1,j}^{\\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\boldsymbol{z}_{1}^{\\text{output}} =\n", + "\\begin{pmatrix}\n", + "b_1^{\\text{output}} & \\boldsymbol{w}_1^{\\text{output}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 & 1 & \\dots & 1 \\\\\n", + "\\boldsymbol{x}_1^{\\text{hidden}} & \\boldsymbol{x}_2^{\\text{hidden}} & \\dots & \\boldsymbol{x}_N^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\\boldsymbol{z}_{1}^{\\text{output}}$ the neural network has finished its feed forward step, and $\\boldsymbol{z}_{1}^{\\text{output}}$ is the final output of the network.\n", + "\n", + "\n", + "The next step is to decide how the parameters should be changed such that they minimize the cost function.\n", + "\n", + "The chosen cost function for this problem is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C(\\boldsymbol{x}, P) = \\frac{1}{N} \\sum_i \\big(g_t'(x_i, P) - ( -\\gamma g_t(x_i, P) \\big)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In order to minimize the cost function, an optimization method must be chosen.\n", + "\n", + "Here, gradient descent with a constant step size has been chosen.\n", + "\n", + "### Gradient descent\n", + "\n", + "The idea of the gradient descent algorithm is to update parameters in\n", + "a direction where the cost function decreases goes to a minimum.\n", + "\n", + "In general, the update of some parameters $\\boldsymbol{\\omega}$ given a cost\n", + "function defined by some weights $\\boldsymbol{\\omega}$, $C(\\boldsymbol{x},\n", + "\\boldsymbol{\\omega})$, goes as follows:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\boldsymbol{\\omega}_{\\text{new} } = \\boldsymbol{\\omega} - \\lambda \\nabla_{\\boldsymbol{\\omega}} C(\\boldsymbol{x}, \\boldsymbol{\\omega})\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for a number of iterations or until $ \\big|\\big| \\boldsymbol{\\omega}_{\\text{new} } - \\boldsymbol{\\omega} \\big|\\big|$ becomes smaller than some given tolerance.\n", + "\n", + "The value of $\\lambda$ decides how large steps the algorithm must take\n", + "in the direction of $ \\nabla_{\\boldsymbol{\\omega}} C(\\boldsymbol{x}, \\boldsymbol{\\omega})$.\n", + "The notation $\\nabla_{\\boldsymbol{\\omega}}$ express the gradient with respect\n", + "to the elements in $\\boldsymbol{\\omega}$.\n", + "\n", + "In our case, we have to minimize the cost function $C(\\boldsymbol{x}, P)$ with\n", + "respect to the two sets of weights and biases, that is for the hidden\n", + "layer $P_{\\text{hidden} }$ and for the output layer $P_{\\text{output}\n", + "}$ .\n", + "\n", + "This means that $P_{\\text{hidden} }$ and $P_{\\text{output} }$ is updated by" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "P_{\\text{hidden},\\text{new}} &= P_{\\text{hidden}} - \\lambda \\nabla_{P_{\\text{hidden}}} C(\\boldsymbol{x}, P) \\\\\n", + "P_{\\text{output},\\text{new}} &= P_{\\text{output}} - \\lambda \\nabla_{P_{\\text{output}}} C(\\boldsymbol{x}, P)\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### The code for solving the ODE" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "%matplotlib inline\n", + "\n", + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "# Assuming one input, hidden, and output layer\n", + "def neural_network(params, x):\n", + "\n", + " # Find the weights (including and biases) for the hidden and output layer.\n", + " # Assume that params is a list of parameters for each layer.\n", + " # The biases are the first element for each array in params,\n", + " # and the weights are the remaning elements in each array in params.\n", + "\n", + " w_hidden = params[0]\n", + " w_output = params[1]\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " ## Hidden layer:\n", + "\n", + " # Add a row of ones to include bias\n", + " x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_input)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " ## Output layer:\n", + "\n", + " # Include bias:\n", + " x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_hidden)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "# The trial solution using the deep neural network:\n", + "def g_trial(x,params, g0 = 10):\n", + " return g0 + x*neural_network(params,x)\n", + "\n", + "# The right side of the ODE:\n", + "def g(x, g_trial, gamma = 2):\n", + " return -gamma*g_trial\n", + "\n", + "# The cost function:\n", + "def cost_function(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the neural network\n", + " d_net_out = elementwise_grad(neural_network,1)(P,x)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d_g_t = elementwise_grad(g_trial,0)(x,P)\n", + "\n", + " # The right side of the ODE\n", + " func = g(x, g_t)\n", + "\n", + " err_sqr = (d_g_t - func)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum / np.size(err_sqr)\n", + "\n", + "# Solve the exponential decay ODE using neural network with one input, hidden, and output layer\n", + "def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb):\n", + " ## Set up initial weights and biases\n", + "\n", + " # For the hidden layer\n", + " p0 = npr.randn(num_neurons_hidden, 2 )\n", + "\n", + " # For the output layer\n", + " p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included\n", + "\n", + " P = [p0, p1]\n", + "\n", + " print('Initial cost: %g'%cost_function(P, x))\n", + "\n", + " ## Start finding the optimal weights using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_grad = grad(cost_function,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of two arrays;\n", + " # one for the gradient w.r.t P_hidden and\n", + " # one for the gradient w.r.t P_output\n", + " cost_grad = cost_function_grad(P, x)\n", + "\n", + " P[0] = P[0] - lmb * cost_grad[0]\n", + " P[1] = P[1] - lmb * cost_grad[1]\n", + "\n", + " print('Final cost: %g'%cost_function(P, x))\n", + "\n", + " return P\n", + "\n", + "def g_analytic(x, gamma = 2, g0 = 10):\n", + " return g0*np.exp(-gamma*x)\n", + "\n", + "# Solve the given problem\n", + "if __name__ == '__main__':\n", + " # Set seed such that the weight are initialized\n", + " # with same weights and biases for every run.\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " N = 10\n", + " x = np.linspace(0, 1, N)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = 10\n", + " num_iter = 10000\n", + " lmb = 0.001\n", + "\n", + " # Use the network\n", + " P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " # Print the deviation from the trial solution and true solution\n", + " res = g_trial(x,P)\n", + " res_analytical = g_analytic(x)\n", + "\n", + " print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical)))\n", + "\n", + " # Plot the results\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, res_analytical)\n", + " plt.plot(x, res[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('x')\n", + " plt.ylabel('g(x)')\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## The network with one input layer, specified number of hidden layers, and one output layer\n", + "\n", + "It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers.\n", + "\n", + "The number of neurons within each hidden layer are given as a list of integers in the program below." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "# The neural network with one input layer and one output layer,\n", + "# but with number of hidden layers specified by the user.\n", + "def deep_neural_network(deep_params, x):\n", + " # N_hidden is the number of hidden layers\n", + "\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consists of\n", + " # parameters to all the hidden\n", + " # layers AND the output layer.\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "# The trial solution using the deep neural network:\n", + "def g_trial_deep(x,params, g0 = 10):\n", + " return g0 + x*deep_neural_network(params, x)\n", + "\n", + "# The right side of the ODE:\n", + "def g(x, g_trial, gamma = 2):\n", + " return -gamma*g_trial\n", + "\n", + "# The same cost function as before, but calls deep_neural_network instead.\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the neural network\n", + " d_net_out = elementwise_grad(deep_neural_network,1)(P,x)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d_g_t = elementwise_grad(g_trial_deep,0)(x,P)\n", + "\n", + " # The right side of the ODE\n", + " func = g(x, g_t)\n", + "\n", + " err_sqr = (d_g_t - func)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum / np.size(err_sqr)\n", + "\n", + "# Solve the exponential decay ODE using neural network with one input and one output layer,\n", + "# but with specified number of hidden layers from the user.\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # The number of elements in the list num_hidden_neurons thus represents\n", + " # the number of hidden layers.\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weights and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weights using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "def g_analytic(x, gamma = 2, g0 = 10):\n", + " return g0*np.exp(-gamma*x)\n", + "\n", + "# Solve the given problem\n", + "if __name__ == '__main__':\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " N = 10\n", + " x = np.linspace(0, 1, N)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = np.array([10,10])\n", + " num_iter = 10000\n", + " lmb = 0.001\n", + "\n", + " P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " res = g_trial_deep(x,P)\n", + " res_analytical = g_analytic(x)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, res_analytical)\n", + " plt.plot(x, res[0,:])\n", + " plt.legend(['analytical','dnn'])\n", + " plt.ylabel('g(x)')\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example: Population growth\n", + "\n", + "A logistic model of population growth assumes that a population converges toward an equilibrium.\n", + "The population growth can be modeled by" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{log} \\tag{10}\n", + "\tg'(t) = \\alpha g(t)(A - g(t))\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(t)$ is the population density at time $t$, $\\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment.\n", + "Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant.\n", + "\n", + "In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability\n", + "and high execution time (this might be more apparent in the examples solving PDEs),\n", + "using a library like TensorFlow is recommended.\n", + "Here, we stay with a more simple approach and implement for comparison, the simple forward Euler method.\n", + "\n", + "\n", + "\n", + "Here, we will model a population $g(t)$ in an environment having carrying capacity $A$.\n", + "The population follows the model" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{solveode_population} \\tag{11}\n", + "g'(t) = \\alpha g(t)(A - g(t))\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(0) = g_0$.\n", + "\n", + "In this example, we let $\\alpha = 2$, $A = 1$, and $g_0 = 1.2$.\n", + "\n", + "\n", + "We will get a slightly different trial solution, as the boundary conditions are different\n", + "compared to the case for exponential decay.\n", + "\n", + "A possible trial solution satisfying the condition $g(0) = g_0$ could be\n", + "\n", + "$$\n", + "h_1(t) = g_0 + t \\cdot N(t,P)\n", + "$$\n", + "\n", + "with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$.\n", + "\n", + "The analytical solution is\n", + "\n", + "$$\n", + "g(t) = \\frac{Ag_0}{g_0 + (A - g_0)\\exp(-\\alpha A t)}\n", + "$$\n", + "\n", + "\n", + "\n", + "The network will be the similar as for the exponential decay example, but with some small modifications for our problem." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "# Function to get the parameters.\n", + "# Done such that one can easily change the paramaters after one's liking.\n", + "def get_parameters():\n", + " alpha = 2\n", + " A = 1\n", + " g0 = 1.2\n", + " return alpha, A, g0\n", + "\n", + "def deep_neural_network(P, x):\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = P[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = P[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d_g_t = elementwise_grad(g_trial_deep,0)(x,P)\n", + "\n", + " # The right side of the ODE\n", + " func = f(x, g_t)\n", + "\n", + " err_sqr = (d_g_t - func)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum / np.size(err_sqr)\n", + "\n", + "# The right side of the ODE:\n", + "def f(x, g_trial):\n", + " alpha,A, g0 = get_parameters()\n", + " return alpha*g_trial*(A - g_trial)\n", + "\n", + "# The trial solution using the deep neural network:\n", + "def g_trial_deep(x, params):\n", + " alpha,A, g0 = get_parameters()\n", + " return g0 + x*deep_neural_network(params,x)\n", + "\n", + "# The analytical solution:\n", + "def g_analytic(t):\n", + " alpha,A, g0 = get_parameters()\n", + " return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t))\n", + "\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weigths using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nt = 10\n", + " T = 1\n", + " t = np.linspace(0,T, Nt)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [100, 50, 25]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(t,P)\n", + " g_analytical = g_analytic(t)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + " diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " print(\"The max absolute difference between the solutions is: %g\"%diff_ag)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(t, g_analytical)\n", + " plt.plot(t, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('t')\n", + " plt.ylabel('g(t)')\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Using forward Euler to solve the ODE\n", + "\n", + "A straightforward way of solving an ODE numerically, is to use Euler's method.\n", + "\n", + "Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\\Delta x$ from $x$:\n", + "\n", + "$$\n", + "f(x + \\Delta x) \\approx f(x) + \\Delta x f'(x)\n", + "$$\n", + "\n", + "In our case, using Euler's method to approximate the value of $g$ at a step $\\Delta t$ from $t$ yields" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + " g(t + \\Delta t) &\\approx g(t) + \\Delta t g'(t) \\\\\n", + " &= g(t) + \\Delta t \\big(\\alpha g(t)(A - g(t))\\big)\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "along with the condition that $g(0) = g_0$.\n", + "\n", + "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 $i = 0, \\dots, N_t-1$.\n", + "\n", + "For $i \\geq 1$, we have that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "t_i &= i\\Delta t \\\\\n", + "&= (i - 1)\\Delta t + \\Delta t \\\\\n", + "&= t_{i-1} + \\Delta t\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Now, if $g_i = g(t_i)$ then" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " \\begin{aligned}\n", + " g_i &= g(t_i) \\\\\n", + " &= g(t_{i-1} + \\Delta t) \\\\\n", + " &\\approx g(t_{i-1}) + \\Delta t \\big(\\alpha g(t_{i-1})(A - g(t_{i-1}))\\big) \\\\\n", + " &= g_{i-1} + \\Delta t \\big(\\alpha g_{i-1}(A - g_{i-1})\\big)\n", + " \\end{aligned}\n", + "\\end{equation} \\label{odenum} \\tag{12}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for $i \\geq 1$ and $g_0 = g(t_0) = g(0) = g_0$.\n", + "\n", + "Equation ([12](#odenum)) could be implemented in the following way,\n", + "extending the program that uses the network using Autograd:" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# Assume that all function definitions from the example program using Autograd\n", + "# are located here.\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nt = 10\n", + " T = 1\n", + " t = np.linspace(0,T, Nt)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [100,50,25]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(t,P)\n", + " g_analytical = g_analytic(t)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + " diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " print(\"The max absolute difference between the solutions is: %g\"%diff_ag)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(t, g_analytical)\n", + " plt.plot(t, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('t')\n", + " plt.ylabel('g(t)')\n", + "\n", + " ## Find an approximation to the funtion using forward Euler\n", + "\n", + " alpha, A, g0 = get_parameters()\n", + " dt = T/(Nt - 1)\n", + "\n", + " # Perform forward Euler to solve the ODE\n", + " g_euler = np.zeros(Nt)\n", + " g_euler[0] = g0\n", + "\n", + " for i in range(1,Nt):\n", + " g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1]))\n", + "\n", + " # Print the errors done by each method\n", + " diff1 = np.max(np.abs(g_euler - g_analytical))\n", + " diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical))\n", + "\n", + " print('Max absolute difference between Euler method and analytical: %g'%diff1)\n", + " print('Max absolute difference between deep neural network and analytical: %g'%diff2)\n", + "\n", + " # Plot results\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.plot(t,g_euler)\n", + " plt.plot(t,g_analytical)\n", + " plt.plot(t,g_dnn_ag[0,:])\n", + "\n", + " plt.legend(['euler','analytical','dnn'])\n", + " plt.xlabel('Time t')\n", + " plt.ylabel('g(t)')\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Solving the one dimensional Poisson equation\n", + "\n", + "The Poisson equation for $g(x)$ in one dimension is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{poisson} \\tag{13}\n", + " -g''(x) = f(x)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $f(x)$ is a given function for $x \\in (0,1)$.\n", + "\n", + "The conditions that $g(x)$ is chosen to fulfill, are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + " g(0) &= 0 \\\\\n", + " g(1) &= 0\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used.\n", + "The results from the networks can then be compared to the analytical solution.\n", + "In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks.\n", + "\n", + "\n", + "Here, the function $g(x)$ to solve for follows the equation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "-g''(x) = f(x),\\qquad x \\in (0,1)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $f(x)$ is a given function, along with the chosen conditions" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{aligned}\n", + "g(0) = g(1) = 0\n", + "\\end{aligned}\\label{cond} \\tag{14}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this example, we consider the case when $f(x) = (3x + x^2)\\exp(x)$.\n", + "\n", + "For this case, a possible trial solution satisfying the conditions could be" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g_t(x) = x \\cdot (1-x) \\cdot N(P,x)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The analytical solution for this problem is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g(x) = x(1 - x)\\exp(x)\n", + "$$" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weigths using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "## Set up the cost function specified for this Poisson equation:\n", + "\n", + "# The right side of the ODE\n", + "def f(x):\n", + " return (3*x + x**2)*np.exp(x)\n", + "\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)\n", + "\n", + " right_side = f(x)\n", + "\n", + " err_sqr = (-d2_g_t - right_side)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum/np.size(err_sqr)\n", + "\n", + "# The trial solution:\n", + "def g_trial_deep(x,P):\n", + " return x*(1-x)*deep_neural_network(P,x)\n", + "\n", + "# The analytic solution;\n", + "def g_analytic(x):\n", + " return x*(1-x)*np.exp(x)\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10\n", + " x = np.linspace(0,1, Nx)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [200,100]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(x,P)\n", + " g_analytical = g_analytic(x)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + " max_diff = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " print(\"The max absolute difference between the solutions is: %g\"%max_diff)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, g_analytical)\n", + " plt.plot(x, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('x')\n", + " plt.ylabel('g(x)')\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Comparing with a numerical scheme\n", + "\n", + "The Poisson equation is possible to solve using Taylor series to approximate the second derivative.\n", + "\n", + "Using Taylor series, the second derivative can be expressed as\n", + "\n", + "$$\n", + "g''(x) = \\frac{g(x + \\Delta x) - 2g(x) + g(x-\\Delta x)}{\\Delta x^2} + E_{\\Delta x}(x)\n", + "$$\n", + "\n", + "where $\\Delta x$ is a small step size and $E_{\\Delta x}(x)$ being the error term.\n", + "\n", + "Looking away from the error terms gives an approximation to the second derivative:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{approx} \\tag{15}\n", + "g''(x) \\approx \\frac{g(x + \\Delta x) - 2g(x) + g(x-\\Delta x)}{\\Delta x^2}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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](#approx)) becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "g''(x_i) &\\approx \\frac{g(x_i + \\Delta x) - 2g(x_i) + g(x_i -\\Delta x)}{\\Delta x^2} \\\\\n", + "&= \\frac{g_{i+1} - 2g_i + g_{i-1}}{\\Delta x^2}\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Since we know from our problem that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "-g''(x) &= f(x) \\\\\n", + "&= (3x + x^2)\\exp(x)\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "along with the conditions $g(0) = g(1) = 0$,\n", + "the following scheme can be used to find an approximate solution for $g(x)$ numerically:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " \\begin{aligned}\n", + " -\\Big( \\frac{g_{i+1} - 2g_i + g_{i-1}}{\\Delta x^2} \\Big) &= f(x_i) \\\\\n", + " -g_{i+1} + 2g_i - g_{i-1} &= \\Delta x^2 f(x_i)\n", + " \\end{aligned}\n", + "\\end{equation} \\label{odesys} \\tag{16}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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.\n", + "\n", + "The equation can be rewritten into a matrix equation:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "\\begin{pmatrix}\n", + "2 & -1 & 0 & \\dots & 0 \\\\\n", + "-1 & 2 & -1 & \\dots & 0 \\\\\n", + "\\vdots & & \\ddots & & \\vdots \\\\\n", + "0 & \\dots & -1 & 2 & -1 \\\\\n", + "0 & \\dots & 0 & -1 & 2\\\\\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "g_1 \\\\\n", + "g_2 \\\\\n", + "\\vdots \\\\\n", + "g_{N_x - 3} \\\\\n", + "g_{N_x - 2}\n", + "\\end{pmatrix}\n", + "&=\n", + "\\Delta x^2\n", + "\\begin{pmatrix}\n", + "f(x_1) \\\\\n", + "f(x_2) \\\\\n", + "\\vdots \\\\\n", + "f(x_{N_x - 3}) \\\\\n", + "f(x_{N_x - 2})\n", + "\\end{pmatrix} \\\\\n", + "\\boldsymbol{A}\\boldsymbol{g} &= \\boldsymbol{f},\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "which makes it possible to solve for the vector $\\boldsymbol{g}$.\n", + "\n", + "\n", + "We can then compare the result from this numerical scheme with the output from our network using Autograd:" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weigths using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "## Set up the cost function specified for this Poisson equation:\n", + "\n", + "# The right side of the ODE\n", + "def f(x):\n", + " return (3*x + x**2)*np.exp(x)\n", + "\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)\n", + "\n", + " right_side = f(x)\n", + "\n", + " err_sqr = (-d2_g_t - right_side)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum/np.size(err_sqr)\n", + "\n", + "# The trial solution:\n", + "def g_trial_deep(x,P):\n", + " return x*(1-x)*deep_neural_network(P,x)\n", + "\n", + "# The analytic solution;\n", + "def g_analytic(x):\n", + " return x*(1-x)*np.exp(x)\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10\n", + " x = np.linspace(0,1, Nx)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [200,100]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(x,P)\n", + " g_analytical = g_analytic(x)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, g_analytical)\n", + " plt.plot(x, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('x')\n", + " plt.ylabel('g(x)')\n", + "\n", + " ## Perform the computation using the numerical scheme\n", + "\n", + " dx = 1/(Nx - 1)\n", + "\n", + " # Set up the matrix A\n", + " A = np.zeros((Nx-2,Nx-2))\n", + "\n", + " A[0,0] = 2\n", + " A[0,1] = -1\n", + "\n", + " for i in range(1,Nx-3):\n", + " A[i,i-1] = -1\n", + " A[i,i] = 2\n", + " A[i,i+1] = -1\n", + "\n", + " A[Nx - 3, Nx - 4] = -1\n", + " A[Nx - 3, Nx - 3] = 2\n", + "\n", + " # Set up the vector f\n", + " f_vec = dx**2 * f(x[1:-1])\n", + "\n", + " # Solve the equation\n", + " g_res = np.linalg.solve(A,f_vec)\n", + "\n", + " g_vec = np.zeros(Nx)\n", + " g_vec[1:-1] = g_res\n", + "\n", + " # Print the differences between each method\n", + " max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " max_diff2 = np.max(np.abs(g_vec - g_analytical))\n", + " print(\"The max absolute difference between the analytical solution and DNN Autograd: %g\"%max_diff1)\n", + " print(\"The max absolute difference between the analytical solution and numerical scheme: %g\"%max_diff2)\n", + "\n", + " # Plot the results\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.plot(x,g_vec)\n", + " plt.plot(x,g_analytical)\n", + " plt.plot(x,g_dnn_ag[0,:])\n", + "\n", + " plt.legend(['numerical scheme','analytical','dnn'])\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Partial Differential Equations\n", + "\n", + "A partial differential equation (PDE) has a solution here the function\n", + "is defined by multiple variables. The equation may involve all kinds\n", + "of combinations of which variables the function is differentiated with\n", + "respect to.\n", + "\n", + "In general, a partial differential equation for a function $g(x_1,\\dots,x_N)$ with $N$ variables may be expressed as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{PDE} \\tag{17}\n", + " f\\left(x_1, \\, \\dots \\, , x_N, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1}, \\dots , \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_N}, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(x_1,\\dots,x_N) }{\\partial x_N^n} \\right) = 0\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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.\n", + "\n", + "### Type of problem\n", + "\n", + "The problem our network must solve for, is similar to the ODE case.\n", + "We must have a trial solution $g_t$ at hand.\n", + "\n", + "For instance, the trial solution could be expressed as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + " g_t(x_1,\\dots,x_N) = h_1(x_1,\\dots,x_N) + h_2(x_1,\\dots,x_N,N(x_1,\\dots,x_N,P))\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $h_1(x_1,\\dots,x_N)$ is a function that ensures $g_t(x_1,\\dots,x_N)$ satisfies some given conditions.\n", + "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.\n", + "\n", + "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.\n", + "\n", + "\n", + "\n", + "### Network requirements\n", + "\n", + "The network tries then the minimize the cost function following the\n", + "same ideas as described for the ODE case, but now with more than one\n", + "variables to consider. The concept still remains the same; find a set\n", + "of parameters $P$ such that the expression $f$ in ([17](#PDE)) is as\n", + "close to zero as possible.\n", + "\n", + "As for the ODE case, the cost function is the mean squared error that\n", + "the network must try to minimize. The cost function for the network to\n", + "minimize is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(x_1, \\dots, x_N, P\\right) = \\left( f\\left(x_1, \\, \\dots \\, , x_N, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1}, \\dots , \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_N}, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(x_1,\\dots,x_N) }{\\partial x_N^n} \\right) \\right)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(\\boldsymbol{x}, P\\right) = f\\left( \\left( \\boldsymbol{x}, \\frac{\\partial g(\\boldsymbol{x}) }{\\partial x_1}, \\dots , \\frac{\\partial g(\\boldsymbol{x}) }{\\partial x_N}, \\frac{\\partial g(\\boldsymbol{x}) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(\\boldsymbol{x}) }{\\partial x_N^n} \\right) \\right)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(X, P \\right) = \\sum_{i=1}^M f\\left( \\left( \\boldsymbol{x}_i, \\frac{\\partial g(\\boldsymbol{x}_i) }{\\partial x_1}, \\dots , \\frac{\\partial g(\\boldsymbol{x}_i) }{\\partial x_N}, \\frac{\\partial g(\\boldsymbol{x}_i) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(\\boldsymbol{x}_i) }{\\partial x_N^n} \\right) \\right)^2.\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example: The diffusion equation\n", + "\n", + "In one spatial dimension, the equation reads" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\frac{\\partial g(x,t)}{\\partial t} = \\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where a possible choice of conditions are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "g(0,t) &= 0 ,\\qquad t \\geq 0 \\\\\n", + "g(1,t) &= 0, \\qquad t \\geq 0 \\\\\n", + "g(x,0) &= u(x),\\qquad x\\in [0,1]\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $u(x)$ being some given function.\n", + "\n", + "\n", + "\n", + "For this case, we want to find $g(x,t)$ such that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " \\frac{\\partial g(x,t)}{\\partial t} = \\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "\\end{equation} \\label{diffonedim} \\tag{18}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "and" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "g(0,t) &= 0 ,\\qquad t \\geq 0 \\\\\n", + "g(1,t) &= 0, \\qquad t \\geq 0 \\\\\n", + "g(x,0) &= u(x),\\qquad x\\in [0,1]\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $u(x) = \\sin(\\pi x)$.\n", + "\n", + "First, let us set up the deep neural network.\n", + "The deep neural network will follow the same structure as discussed in the examples solving the ODEs.\n", + "First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions.\n", + "\n", + "\n", + "\n", + "\n", + "The only change to do here, is to extend our network such that\n", + "functions of multiple parameters are correctly handled. In this case\n", + "we have two variables in our function to solve for, that is time $t$\n", + "and position $x$. The variables will be represented by a\n", + "one-dimensional array in the program. The program will evaluate the\n", + "network at each possible pair $(x,t)$, given an array for the desired\n", + "$x$-values and $t$-values to approximate the solution at." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # x is now a point and a 1D numpy array; make it a column vector\n", + " num_coordinates = np.size(x,0)\n", + " x = x.reshape(num_coordinates,-1)\n", + "\n", + " num_points = np.size(x,1)\n", + "\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output[0][0]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The cost function must then iterate through the given arrays\n", + "containing values for $x$ and $t$, defines a point $(x,t)$ the deep\n", + "neural network and the trial solution is evaluated at, and then finds\n", + "the Jacobian of the trial solution.\n", + "\n", + "A possible trial solution for this PDE is\n", + "\n", + "$$\n", + "g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P)\n", + "$$\n", + "\n", + "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 $P$.\n", + "\n", + "To fulfill the conditions, $A(x,t)$ could be:\n", + "\n", + "$$\n", + "h_1(x,t) = (1-t)\\Big(u(x) - \\big((1-x)u(0) + x u(1)\\big)\\Big) = (1-t)u(x) = (1-t)\\sin(\\pi x)\n", + "$$\n", + "since $(0) = u(1) = 0$ and $u(x) = \\sin(\\pi x)$.\n", + "\n", + "\n", + "\n", + "The Jacobian is used because the program must find the derivative of\n", + "the trial solution with respect to $x$ and $t$.\n", + "\n", + "This gives the necessity of computing the Jacobian matrix, as we want\n", + "to evaluate the gradient with respect to $x$ and $t$ (note that the\n", + "Jacobian of a scalar-valued multivariate function is simply its\n", + "gradient).\n", + "\n", + "In Autograd, the differentiation is by default done with respect to\n", + "the first input argument of your Python function. Since the points is\n", + "an array representing $x$ and $t$, the Jacobian is calculated using\n", + "the values of $x$ and $t$.\n", + "\n", + "To find the second derivative with respect to $x$ and $t$, the\n", + "Jacobian can be found for the second time. The result is a Hessian\n", + "matrix, which is the matrix containing all the possible second order\n", + "mixed derivatives of $g(x,t)$." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# Set up the trial function:\n", + "def u(x):\n", + " return np.sin(np.pi*x)\n", + "\n", + "def g_trial(point,P):\n", + " x,t = point\n", + " return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)\n", + "\n", + "# The right side of the ODE:\n", + "def f(point):\n", + " return 0.\n", + "\n", + "# The cost function:\n", + "def cost_function(P, x, t):\n", + " cost_sum = 0\n", + "\n", + " g_t_jacobian_func = jacobian(g_trial)\n", + " g_t_hessian_func = hessian(g_trial)\n", + "\n", + " for x_ in x:\n", + " for t_ in t:\n", + " point = np.array([x_,t_])\n", + "\n", + " g_t = g_trial(point,P)\n", + " g_t_jacobian = g_t_jacobian_func(point,P)\n", + " g_t_hessian = g_t_hessian_func(point,P)\n", + "\n", + " g_t_dt = g_t_jacobian[1]\n", + " g_t_d2x = g_t_hessian[0][0]\n", + "\n", + " func = f(point)\n", + "\n", + " err_sqr = ( (g_t_dt - g_t_d2x) - func)**2\n", + " cost_sum += err_sqr\n", + "\n", + " return cost_sum" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Setting up the network using Autograd; The full program\n", + "\n", + "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.\n", + "\n", + "The analytical solution of our problem is\n", + "\n", + "$$\n", + "g(x,t) = \\exp(-\\pi^2 t)\\sin(\\pi x)\n", + "$$\n", + "\n", + "A possible way to implement a neural network solving the PDE, is given below.\n", + "Be aware, though, that it is fairly slow for the parameters used.\n", + "A better result is possible, but requires more iterations, and thus longer time to complete.\n", + "\n", + "\n", + "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.\n", + "Using TensorFlow results in a much better execution time. Try it!" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import jacobian,hessian,grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import cm\n", + "from matplotlib import pyplot as plt\n", + "from mpl_toolkits.mplot3d import axes3d\n", + "\n", + "## Set up the network\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # x is now a point and a 1D numpy array; make it a column vector\n", + " num_coordinates = np.size(x,0)\n", + " x = x.reshape(num_coordinates,-1)\n", + "\n", + " num_points = np.size(x,1)\n", + "\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output[0][0]\n", + "\n", + "## Define the trial solution and cost function\n", + "def u(x):\n", + " return np.sin(np.pi*x)\n", + "\n", + "def g_trial(point,P):\n", + " x,t = point\n", + " return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)\n", + "\n", + "# The right side of the ODE:\n", + "def f(point):\n", + " return 0.\n", + "\n", + "# The cost function:\n", + "def cost_function(P, x, t):\n", + " cost_sum = 0\n", + "\n", + " g_t_jacobian_func = jacobian(g_trial)\n", + " g_t_hessian_func = hessian(g_trial)\n", + "\n", + " for x_ in x:\n", + " for t_ in t:\n", + " point = np.array([x_,t_])\n", + "\n", + " g_t = g_trial(point,P)\n", + " g_t_jacobian = g_t_jacobian_func(point,P)\n", + " g_t_hessian = g_t_hessian_func(point,P)\n", + "\n", + " g_t_dt = g_t_jacobian[1]\n", + " g_t_d2x = g_t_hessian[0][0]\n", + "\n", + " func = f(point)\n", + "\n", + " err_sqr = ( (g_t_dt - g_t_d2x) - func)**2\n", + " cost_sum += err_sqr\n", + "\n", + " return cost_sum /( np.size(x)*np.size(t) )\n", + "\n", + "## For comparison, define the analytical solution\n", + "def g_analytic(point):\n", + " x,t = point\n", + " return np.exp(-np.pi**2*t)*np.sin(np.pi*x)\n", + "\n", + "## Set up a function for training the network to solve for the equation\n", + "def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):\n", + " ## Set up initial weigths and biases\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: ',cost_function(P, x, t))\n", + "\n", + " cost_function_grad = grad(cost_function,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " cost_grad = cost_function_grad(P, x , t)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_grad[l]\n", + "\n", + " print('Final cost: ',cost_function(P, x, t))\n", + "\n", + " return P\n", + "\n", + "if __name__ == '__main__':\n", + " ### Use the neural network:\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10; Nt = 10\n", + " x = np.linspace(0, 1, Nx)\n", + " t = np.linspace(0,1,Nt)\n", + "\n", + " ## Set up the parameters for the network\n", + " num_hidden_neurons = [100, 25]\n", + " num_iter = 250\n", + " lmb = 0.01\n", + "\n", + " P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " ## Store the results\n", + " g_dnn_ag = np.zeros((Nx, Nt))\n", + " G_analytical = np.zeros((Nx, Nt))\n", + " for i,x_ in enumerate(x):\n", + " for j, t_ in enumerate(t):\n", + " point = np.array([x_, t_])\n", + " g_dnn_ag[i,j] = g_trial(point,P)\n", + "\n", + " G_analytical[i,j] = g_analytic(point)\n", + "\n", + " # Find the map difference between the analytical and the computed solution\n", + " diff_ag = np.abs(g_dnn_ag - G_analytical)\n", + " print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag))\n", + "\n", + " ## Plot the solutions in two dimensions, that being in position and time\n", + "\n", + " T,X = np.meshgrid(t,x)\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))\n", + " s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Analytical solution')\n", + " s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Difference')\n", + " s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + " ## Take some slices of the 3D plots just to see the solutions at particular times\n", + " indx1 = 0\n", + " indx2 = int(Nt/2)\n", + " indx3 = Nt-1\n", + "\n", + " t1 = t[indx1]\n", + " t2 = t[indx2]\n", + " t3 = t[indx3]\n", + "\n", + " # Slice the results from the DNN\n", + " res1 = g_dnn_ag[:,indx1]\n", + " res2 = g_dnn_ag[:,indx2]\n", + " res3 = g_dnn_ag[:,indx3]\n", + "\n", + " # Slice the analytical results\n", + " res_analytical1 = G_analytical[:,indx1]\n", + " res_analytical2 = G_analytical[:,indx2]\n", + " res_analytical3 = G_analytical[:,indx3]\n", + "\n", + " # Plot the slices\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t1)\n", + " plt.plot(x, res1)\n", + " plt.plot(x,res_analytical1)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t2)\n", + " plt.plot(x, res2)\n", + " plt.plot(x,res_analytical2)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t3)\n", + " plt.plot(x, res3)\n", + " plt.plot(x,res_analytical3)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Solving the wave equation with Neural Networks\n", + "\n", + "The wave equation is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\frac{\\partial^2 g(x,t)}{\\partial t^2} = c^2\\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $c$ being the specified wave speed.\n", + "\n", + "Here, the chosen conditions are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "\tg(0,t) &= 0 \\\\\n", + "\tg(1,t) &= 0 \\\\\n", + "\tg(x,0) &= u(x) \\\\\n", + "\t\\frac{\\partial g(x,t)}{\\partial t} \\Big |_{t = 0} &= v(x)\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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.\n", + "\n", + "\n", + "The wave equation to solve for, is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{wave} \\tag{19}\n", + "\\frac{\\partial^2 g(x,t)}{\\partial t^2} = c^2 \\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $c$ is the given wave speed.\n", + "The chosen conditions for this equation are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{aligned}\n", + "g(0,t) &= 0, &t \\geq 0 \\\\\n", + "g(1,t) &= 0, &t \\geq 0 \\\\\n", + "g(x,0) &= u(x), &x\\in[0,1] \\\\\n", + "\\frac{\\partial g(x,t)}{\\partial t}\\Big |_{t = 0} &= v(x), &x \\in [0,1]\n", + "\\end{aligned} \\label{condwave} \\tag{20}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this example, let $c = 1$ and $u(x) = \\sin(\\pi x)$ and $v(x) = -\\pi\\sin(\\pi x)$.\n", + "\n", + "\n", + "\n", + "Setting up the network is done in similar matter as for the example of solving the diffusion equation.\n", + "The only things we have to change, is the trial solution such that it satisfies the conditions from ([20](#condwave)) and the cost function.\n", + "\n", + "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\n", + "\n", + "$$\n", + "g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P)\n", + "$$\n", + "\n", + "where\n", + "\n", + "$$\n", + "h_1(x,t) = (1-t^2)u(x) + tv(x)\n", + "$$\n", + "\n", + "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.\n", + "\n", + "\n", + "The analytical solution for our specific problem, is\n", + "\n", + "$$\n", + "g(x,t) = \\sin(\\pi x)\\cos(\\pi t) - \\sin(\\pi x)\\sin(\\pi t)\n", + "$$" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import hessian,grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import cm\n", + "from matplotlib import pyplot as plt\n", + "from mpl_toolkits.mplot3d import axes3d\n", + "\n", + "## Set up the trial function:\n", + "def u(x):\n", + " return np.sin(np.pi*x)\n", + "\n", + "def v(x):\n", + " return -np.pi*np.sin(np.pi*x)\n", + "\n", + "def h1(point):\n", + " x,t = point\n", + " return (1 - t**2)*u(x) + t*v(x)\n", + "\n", + "def g_trial(point,P):\n", + " x,t = point\n", + " return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point)\n", + "\n", + "## Define the cost function\n", + "def cost_function(P, x, t):\n", + " cost_sum = 0\n", + "\n", + " g_t_hessian_func = hessian(g_trial)\n", + "\n", + " for x_ in x:\n", + " for t_ in t:\n", + " point = np.array([x_,t_])\n", + "\n", + " g_t_hessian = g_t_hessian_func(point,P)\n", + "\n", + " g_t_d2x = g_t_hessian[0][0]\n", + " g_t_d2t = g_t_hessian[1][1]\n", + "\n", + " err_sqr = ( (g_t_d2t - g_t_d2x) )**2\n", + " cost_sum += err_sqr\n", + "\n", + " return cost_sum / (np.size(t) * np.size(x))\n", + "\n", + "## The neural network\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # x is now a point and a 1D numpy array; make it a column vector\n", + " num_coordinates = np.size(x,0)\n", + " x = x.reshape(num_coordinates,-1)\n", + "\n", + " num_points = np.size(x,1)\n", + "\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output[0][0]\n", + "\n", + "## The analytical solution\n", + "def g_analytic(point):\n", + " x,t = point\n", + " return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t)\n", + "\n", + "def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):\n", + " ## Set up initial weigths and biases\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: ',cost_function(P, x, t))\n", + "\n", + " cost_function_grad = grad(cost_function,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " cost_grad = cost_function_grad(P, x , t)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_grad[l]\n", + "\n", + "\n", + " print('Final cost: ',cost_function(P, x, t))\n", + "\n", + " return P\n", + "\n", + "if __name__ == '__main__':\n", + " ### Use the neural network:\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10; Nt = 10\n", + " x = np.linspace(0, 1, Nx)\n", + " t = np.linspace(0,1,Nt)\n", + "\n", + " ## Set up the parameters for the network\n", + " num_hidden_neurons = [50,20]\n", + " num_iter = 1000\n", + " lmb = 0.01\n", + "\n", + " P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " ## Store the results\n", + " res = np.zeros((Nx, Nt))\n", + " res_analytical = np.zeros((Nx, Nt))\n", + " for i,x_ in enumerate(x):\n", + " for j, t_ in enumerate(t):\n", + " point = np.array([x_, t_])\n", + " res[i,j] = g_trial(point,P)\n", + "\n", + " res_analytical[i,j] = g_analytic(point)\n", + "\n", + " diff = np.abs(res - res_analytical)\n", + " print(\"Max difference between analytical and solution from nn: %g\"%np.max(diff))\n", + "\n", + " ## Plot the solutions in two dimensions, that being in position and time\n", + "\n", + " T,X = np.meshgrid(t,x)\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))\n", + " s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Analytical solution')\n", + " s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Difference')\n", + " s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + " ## Take some slices of the 3D plots just to see the solutions at particular times\n", + " indx1 = 0\n", + " indx2 = int(Nt/2)\n", + " indx3 = Nt-1\n", + "\n", + " t1 = t[indx1]\n", + " t2 = t[indx2]\n", + " t3 = t[indx3]\n", + "\n", + " # Slice the results from the DNN\n", + " res1 = res[:,indx1]\n", + " res2 = res[:,indx2]\n", + " res3 = res[:,indx3]\n", + "\n", + " # Slice the analytical results\n", + " res_analytical1 = res_analytical[:,indx1]\n", + " res_analytical2 = res_analytical[:,indx2]\n", + " res_analytical3 = res_analytical[:,indx3]\n", + "\n", + " # Plot the slices\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t1)\n", + " plt.plot(x, res1)\n", + " plt.plot(x,res_analytical1)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t2)\n", + " plt.plot(x, res2)\n", + " plt.plot(x,res_analytical2)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t3)\n", + " plt.plot(x, res3)\n", + " plt.plot(x,res_analytical3)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Resources on differential equations and deep learning\n", + "\n", + "1. [Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al](https://pdfs.semanticscholar.org/d061/df393e0e8fbfd0ea24976458b7d42419040d.pdf)\n", + "\n", + "2. [Neural networks for solving differential equations by A. Honchar](https://becominghuman.ai/neural-networks-for-solving-differential-equations-fa230ac5e04c)\n", + "\n", + "3. [Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener](http://cs229.stanford.edu/proj2013/ChiaramonteKiener-SolvingDifferentialEquationsUsingNeuralNetworks.pdf)\n", + "\n", + "4. [Introduction to Partial Differential Equations by A. Tveito, R. Winther](https://www.springer.com/us/book/9783540225515)" + ] + } + ], + "metadata": {}, + "nbformat": 4, + "nbformat_minor": 4 +} diff --git a/doc/LectureNotes/chapter11.ipynb b/doc/LectureNotes/chapter11.ipynb new file mode 100644 index 000000000..c03e4a368 --- /dev/null +++ b/doc/LectureNotes/chapter11.ipynb @@ -0,0 +1,3023 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Solving Differential Equations with Deep Learning\n", + "\n", + "The Universal Approximation Theorem states that a neural network can\n", + "approximate any function at a single hidden layer along with one input\n", + "and output layer to any given precision. \n", + "\n", + "\n", + "An ordinary differential equation (ODE) is an equation involving functions having one variable.\n", + "\n", + "In general, an ordinary differential equation looks like" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{ode} \\tag{1}\n", + "f\\left(x, \\, g(x), \\, g'(x), \\, g''(x), \\, \\dots \\, , \\, g^{(n)}(x)\\right) = 0\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$.\n", + "\n", + "The $f\\left(x, g(x), g'(x), g''(x), \\, \\dots \\, , g^{(n)}(x)\\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \\ g'(x), \\ g''(x), \\, \\dots \\, , \\text{ and } g^{(n)}(x)$ on the left side of the equality sign in ([1](#ode)).\n", + "The highest order of derivative, that is the value of $n$, determines to the order of the equation.\n", + "The equation is referred to as a $n$-th order ODE.\n", + "Along with ([1](#ode)), some additional conditions of the function $g(x)$ are typically given\n", + "for the solution to be unique.\n", + "\n", + "\n", + "\n", + "Let the trial solution $g_t(x)$ be" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + "\tg_t(x) = h_1(x) + h_2(x,N(x,P))\n", + "\\label{_auto1} \\tag{2}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set\n", + "of conditions, $N(x,P)$ a neural network with weights and biases\n", + "described by $P$ and $h_2(x, N(x,P))$ some expression involving the\n", + "neural network. The role of the function $h_2(x, N(x,P))$, is to\n", + "ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is\n", + "evaluated at the values of $x$ where the given conditions must be\n", + "satisfied. The function $h_1(x)$ should alone make $g_t(x)$ satisfy\n", + "the conditions.\n", + "\n", + "But what about the network $N(x,P)$?\n", + "\n", + "\n", + "As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation.\n", + "\n", + "\n", + "\n", + "For the minimization to be defined, we need to have a cost function at hand to minimize.\n", + "\n", + "It is given that $f\\left(x, \\, g(x), \\, g'(x), \\, g''(x), \\, \\dots \\, , \\, g^{(n)}(x)\\right)$ should be equal to zero in ([1](#ode)).\n", + "We can choose to consider the mean squared error as the cost function for an input $x$.\n", + "Since we are looking at one input, the cost function is just $f$ squared.\n", + "The cost function $c\\left(x, P \\right)$ can therefore be expressed as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(x, P\\right) = \\big(f\\left(x, \\, g(x), \\, g'(x), \\, g''(x), \\, \\dots \\, , \\, g^{(n)}(x)\\right)\\big)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "If $N$ inputs are given as a vector $\\boldsymbol{x}$ with elements $x_i$ for $i = 1,\\dots,N$,\n", + "the cost function becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{cost} \\tag{3}\n", + "\tC\\left(\\boldsymbol{x}, P\\right) = \\frac{1}{N} \\sum_{i=1}^N \\big(f\\left(x_i, \\, g(x_i), \\, g'(x_i), \\, g''(x_i), \\, \\dots \\, , \\, g^{(n)}(x_i)\\right)\\big)^2\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The neural net should then find the parameters $P$ that minimizes the cost function in\n", + "([3](#cost)) for a set of $N$ training samples $x_i$.\n", + "\n", + "\n", + "\n", + "To perform the minimization using gradient descent, the gradient of $C\\left(\\boldsymbol{x}, P\\right)$ is needed.\n", + "It might happen so that finding an analytical expression of the gradient of $C(\\boldsymbol{x}, P)$ from ([3](#cost)) gets too messy, depending on which cost function one desires to use.\n", + "\n", + "Luckily, there exists libraries that makes the job for us through automatic differentiation.\n", + "Automatic differentiation is a method of finding the derivatives numerically with very high precision.\n", + "\n", + "\n", + "### Example: Exponential decay\n", + "\n", + "An exponential decay of a quantity $g(x)$ is described by the equation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{solve_expdec} \\tag{4}\n", + " g'(x) = -\\gamma g(x)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $g(0) = g_0$ for some chosen initial value $g_0$.\n", + "\n", + "The analytical solution of ([4](#solve_expdec)) is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " g(x) = g_0 \\exp\\left(-\\gamma x\\right)\n", + "\\label{_auto2} \\tag{5}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of ([4](#solve_expdec)).\n", + "\n", + "\n", + "\n", + "The program will use a neural network to solve" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{solveode} \\tag{6}\n", + "g'(x) = -\\gamma g(x)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(0) = g_0$ with $\\gamma$ and $g_0$ being some chosen values.\n", + "\n", + "In this example, $\\gamma = 2$ and $g_0 = 10$.\n", + "\n", + "\n", + "To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g_t(x, P) = h_1(x) + h_2(x, N(x, P))\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer.\n", + "\n", + "\n", + "\n", + "In this network, there are no weights and bias at the input layer, so $P = \\{ P_{\\text{hidden}}, P_{\\text{output}} \\}$.\n", + "If there are $N_{\\text{hidden} }$ neurons in the hidden layer, then $P_{\\text{hidden}}$ is a $N_{\\text{hidden} } \\times (1 + N_{\\text{input}})$ matrix, given that there are $N_{\\text{input}}$ neurons in the input layer.\n", + "\n", + "The first column in $P_{\\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer.\n", + "If there are $N_{\\text{output} }$ neurons in the output layer, then $P_{\\text{output}} $ is a $N_{\\text{output} } \\times (1 + N_{\\text{hidden} })$ matrix.\n", + "\n", + "Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron.\n", + "\n", + "It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of ([6](#solveode)). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \\cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{trial} \\tag{7}\n", + "g_t(x, P) = g_0 + x \\cdot N(x, P)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Reformulating the problem\n", + "\n", + "We wish that our neural network manages to minimize a given cost function.\n", + "\n", + "A reformulation of out equation, ([6](#solveode)), must therefore be done,\n", + "such that it describes the problem a neural network can solve for.\n", + "\n", + "The neural network must find the set of weights and biases $P$ such that the trial solution in ([7](#trial)) satisfies ([6](#solveode)).\n", + "\n", + "The trial solution" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g_t(x, P) = g_0 + x \\cdot N(x, P)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{nnmin} \\tag{8}\n", + "g_t'(x, P) = - \\gamma g_t(x, P)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "is fulfilled as *best as possible*.\n", + "\n", + "\n", + "The left hand side and right hand side of ([8](#nnmin)) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible.\n", + "This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero.\n", + "In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network.\n", + "\n", + "This gives the following cost function our neural network must solve for:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\min_{P}\\Big\\{ \\big(g_t'(x, P) - ( -\\gamma g_t(x, P) \\big)^2 \\Big\\}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "(the notation $\\min_{P}\\{ f(x, P) \\}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$)\n", + "\n", + "or, in terms of weights and biases for the hidden and output layer in our network:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\min_{P_{\\text{hidden} }, \\ P_{\\text{output} }}\\Big\\{ \\big(g_t'(x, \\{ P_{\\text{hidden} }, P_{\\text{output} }\\}) - ( -\\gamma g_t(x, \\{ P_{\\text{hidden} }, P_{\\text{output} }\\}) \\big)^2 \\Big\\}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for an input value $x$.\n", + "\n", + "\n", + "\n", + "If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \\dots, N$, then the *total* error to minimize becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{min} \\tag{9}\n", + "\\min_{P}\\Big\\{\\frac{1}{N} \\sum_{i=1}^N \\big(g_t'(x_i, P) - ( -\\gamma g_t(x_i, P) \\big)^2 \\Big\\}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Letting $\\boldsymbol{x}$ be a vector with elements $x_i$ and $C(\\boldsymbol{x}, P) = \\frac{1}{N} \\sum_i \\big(g_t'(x_i, P) - ( -\\gamma g_t(x_i, P) \\big)^2$ denote the cost function, the minimization problem that our network must solve, becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\min_{P} C(\\boldsymbol{x}, P)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In terms of $P_{\\text{hidden} }$ and $P_{\\text{output} }$, this could also be expressed as\n", + "\n", + "$$\n", + "\\min_{P_{\\text{hidden} }, \\ P_{\\text{output} }} C(\\boldsymbol{x}, \\{P_{\\text{hidden} }, P_{\\text{output} }\\})\n", + "$$\n", + "\n", + "\n", + "For simplicity, it is assumed that the input is an array $\\boldsymbol{x} = (x_1, \\dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills ([9](#min)).\n", + "\n", + "First, the neural network must feed forward the inputs.\n", + "This means that $\\boldsymbol{x}s$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further.\n", + "The input layer will consist of $N_{\\text{input} }$ neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be $N_{\\text{hidden} }$.\n", + "\n", + "\n", + "For the $i$-th in the hidden layer with weight $w_i^{\\text{hidden} }$ and bias $b_i^{\\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "z_{i,j}^{\\text{hidden}} &= b_i^{\\text{hidden}} + w_i^{\\text{hidden}}x_j \\\\\n", + "&=\n", + "\\begin{pmatrix}\n", + "b_i^{\\text{hidden}} & w_i^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 \\\\\n", + "x_j\n", + "\\end{pmatrix}\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "\\boldsymbol{z}_{i}^{\\text{hidden}} &= \\Big( b_i^{\\text{hidden}} + w_i^{\\text{hidden}}x_1 , \\ b_i^{\\text{hidden}} + w_i^{\\text{hidden}} x_2, \\ \\dots \\, , \\ b_i^{\\text{hidden}} + w_i^{\\text{hidden}} x_N\\Big) \\\\\n", + "&=\n", + "\\begin{pmatrix}\n", + " b_i^{\\text{hidden}} & w_i^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 & 1 & \\dots & 1 \\\\\n", + "x_1 & x_2 & \\dots & x_N\n", + "\\end{pmatrix} \\\\\n", + "&= \\boldsymbol{p}_{i, \\text{hidden}}^T X\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The vector $\\boldsymbol{p}_{i, \\text{hidden}}^T$ constitutes each row in $P_{\\text{hidden} }$, which contains the weights for the neural network to minimize according to ([9](#min)).\n", + "\n", + "After having found $\\boldsymbol{z}_{i}^{\\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\\boldsymbol{z})$.\n", + "\n", + "In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "f(z) = \\frac{1}{1 + \\exp{(-z)}}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "It is possible to use other activations functions for the hidden layer also.\n", + "\n", + "The output $\\boldsymbol{x}_i^{\\text{hidden}}$ from each $i$-th hidden neuron is:\n", + "\n", + "$$\n", + "\\boldsymbol{x}_i^{\\text{hidden} } = f\\big( \\boldsymbol{z}_{i}^{\\text{hidden}} \\big)\n", + "$$\n", + "\n", + "The outputs $\\boldsymbol{x}_i^{\\text{hidden} } $ are then sent to the output layer.\n", + "\n", + "The output layer consists of one neuron in this case, and combines the\n", + "output from each of the neurons in the hidden layers. The output layer\n", + "combines the results from the hidden layer using some weights $w_i^{\\text{output}}$\n", + "and biases $b_i^{\\text{output}}$. In this case,\n", + "it is assumes that the number of neurons in the output layer is one.\n", + "\n", + "\n", + "\n", + "The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "z_{1,j}^{\\text{output}} & =\n", + "\\begin{pmatrix}\n", + "b_1^{\\text{output}} & \\boldsymbol{w}_1^{\\text{output}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 \\\\\n", + "\\boldsymbol{x}_j^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Expressing $z_{1,j}^{\\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\boldsymbol{z}_{1}^{\\text{output}} =\n", + "\\begin{pmatrix}\n", + "b_1^{\\text{output}} & \\boldsymbol{w}_1^{\\text{output}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 & 1 & \\dots & 1 \\\\\n", + "\\boldsymbol{x}_1^{\\text{hidden}} & \\boldsymbol{x}_2^{\\text{hidden}} & \\dots & \\boldsymbol{x}_N^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\\boldsymbol{z}_{1}^{\\text{output}}$ the neural network has finished its feed forward step, and $\\boldsymbol{z}_{1}^{\\text{output}}$ is the final output of the network.\n", + "\n", + "\n", + "The next step is to decide how the parameters should be changed such that they minimize the cost function.\n", + "\n", + "The chosen cost function for this problem is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C(\\boldsymbol{x}, P) = \\frac{1}{N} \\sum_i \\big(g_t'(x_i, P) - ( -\\gamma g_t(x_i, P) \\big)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In order to minimize the cost function, an optimization method must be chosen.\n", + "\n", + "Here, gradient descent with a constant step size has been chosen.\n", + "\n", + "### Gradient descent\n", + "\n", + "The idea of the gradient descent algorithm is to update parameters in\n", + "a direction where the cost function decreases goes to a minimum.\n", + "\n", + "In general, the update of some parameters $\\boldsymbol{\\omega}$ given a cost\n", + "function defined by some weights $\\boldsymbol{\\omega}$, $C(\\boldsymbol{x},\n", + "\\boldsymbol{\\omega})$, goes as follows:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\boldsymbol{\\omega}_{\\text{new} } = \\boldsymbol{\\omega} - \\lambda \\nabla_{\\boldsymbol{\\omega}} C(\\boldsymbol{x}, \\boldsymbol{\\omega})\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for a number of iterations or until $ \\big|\\big| \\boldsymbol{\\omega}_{\\text{new} } - \\boldsymbol{\\omega} \\big|\\big|$ becomes smaller than some given tolerance.\n", + "\n", + "The value of $\\lambda$ decides how large steps the algorithm must take\n", + "in the direction of $ \\nabla_{\\boldsymbol{\\omega}} C(\\boldsymbol{x}, \\boldsymbol{\\omega})$.\n", + "The notation $\\nabla_{\\boldsymbol{\\omega}}$ express the gradient with respect\n", + "to the elements in $\\boldsymbol{\\omega}$.\n", + "\n", + "In our case, we have to minimize the cost function $C(\\boldsymbol{x}, P)$ with\n", + "respect to the two sets of weights and biases, that is for the hidden\n", + "layer $P_{\\text{hidden} }$ and for the output layer $P_{\\text{output}\n", + "}$ .\n", + "\n", + "This means that $P_{\\text{hidden} }$ and $P_{\\text{output} }$ is updated by" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "P_{\\text{hidden},\\text{new}} &= P_{\\text{hidden}} - \\lambda \\nabla_{P_{\\text{hidden}}} C(\\boldsymbol{x}, P) \\\\\n", + "P_{\\text{output},\\text{new}} &= P_{\\text{output}} - \\lambda \\nabla_{P_{\\text{output}}} C(\\boldsymbol{x}, P)\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### The code for solving the ODE" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "%matplotlib inline\n", + "\n", + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "# Assuming one input, hidden, and output layer\n", + "def neural_network(params, x):\n", + "\n", + " # Find the weights (including and biases) for the hidden and output layer.\n", + " # Assume that params is a list of parameters for each layer.\n", + " # The biases are the first element for each array in params,\n", + " # and the weights are the remaning elements in each array in params.\n", + "\n", + " w_hidden = params[0]\n", + " w_output = params[1]\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " ## Hidden layer:\n", + "\n", + " # Add a row of ones to include bias\n", + " x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_input)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " ## Output layer:\n", + "\n", + " # Include bias:\n", + " x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_hidden)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "# The trial solution using the deep neural network:\n", + "def g_trial(x,params, g0 = 10):\n", + " return g0 + x*neural_network(params,x)\n", + "\n", + "# The right side of the ODE:\n", + "def g(x, g_trial, gamma = 2):\n", + " return -gamma*g_trial\n", + "\n", + "# The cost function:\n", + "def cost_function(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the neural network\n", + " d_net_out = elementwise_grad(neural_network,1)(P,x)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d_g_t = elementwise_grad(g_trial,0)(x,P)\n", + "\n", + " # The right side of the ODE\n", + " func = g(x, g_t)\n", + "\n", + " err_sqr = (d_g_t - func)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum / np.size(err_sqr)\n", + "\n", + "# Solve the exponential decay ODE using neural network with one input, hidden, and output layer\n", + "def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb):\n", + " ## Set up initial weights and biases\n", + "\n", + " # For the hidden layer\n", + " p0 = npr.randn(num_neurons_hidden, 2 )\n", + "\n", + " # For the output layer\n", + " p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included\n", + "\n", + " P = [p0, p1]\n", + "\n", + " print('Initial cost: %g'%cost_function(P, x))\n", + "\n", + " ## Start finding the optimal weights using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_grad = grad(cost_function,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of two arrays;\n", + " # one for the gradient w.r.t P_hidden and\n", + " # one for the gradient w.r.t P_output\n", + " cost_grad = cost_function_grad(P, x)\n", + "\n", + " P[0] = P[0] - lmb * cost_grad[0]\n", + " P[1] = P[1] - lmb * cost_grad[1]\n", + "\n", + " print('Final cost: %g'%cost_function(P, x))\n", + "\n", + " return P\n", + "\n", + "def g_analytic(x, gamma = 2, g0 = 10):\n", + " return g0*np.exp(-gamma*x)\n", + "\n", + "# Solve the given problem\n", + "if __name__ == '__main__':\n", + " # Set seed such that the weight are initialized\n", + " # with same weights and biases for every run.\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " N = 10\n", + " x = np.linspace(0, 1, N)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = 10\n", + " num_iter = 10000\n", + " lmb = 0.001\n", + "\n", + " # Use the network\n", + " P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " # Print the deviation from the trial solution and true solution\n", + " res = g_trial(x,P)\n", + " res_analytical = g_analytic(x)\n", + "\n", + " print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical)))\n", + "\n", + " # Plot the results\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, res_analytical)\n", + " plt.plot(x, res[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('x')\n", + " plt.ylabel('g(x)')\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## The network with one input layer, specified number of hidden layers, and one output layer\n", + "\n", + "It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers.\n", + "\n", + "The number of neurons within each hidden layer are given as a list of integers in the program below." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "# The neural network with one input layer and one output layer,\n", + "# but with number of hidden layers specified by the user.\n", + "def deep_neural_network(deep_params, x):\n", + " # N_hidden is the number of hidden layers\n", + "\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consists of\n", + " # parameters to all the hidden\n", + " # layers AND the output layer.\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "# The trial solution using the deep neural network:\n", + "def g_trial_deep(x,params, g0 = 10):\n", + " return g0 + x*deep_neural_network(params, x)\n", + "\n", + "# The right side of the ODE:\n", + "def g(x, g_trial, gamma = 2):\n", + " return -gamma*g_trial\n", + "\n", + "# The same cost function as before, but calls deep_neural_network instead.\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the neural network\n", + " d_net_out = elementwise_grad(deep_neural_network,1)(P,x)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d_g_t = elementwise_grad(g_trial_deep,0)(x,P)\n", + "\n", + " # The right side of the ODE\n", + " func = g(x, g_t)\n", + "\n", + " err_sqr = (d_g_t - func)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum / np.size(err_sqr)\n", + "\n", + "# Solve the exponential decay ODE using neural network with one input and one output layer,\n", + "# but with specified number of hidden layers from the user.\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # The number of elements in the list num_hidden_neurons thus represents\n", + " # the number of hidden layers.\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weights and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weights using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "def g_analytic(x, gamma = 2, g0 = 10):\n", + " return g0*np.exp(-gamma*x)\n", + "\n", + "# Solve the given problem\n", + "if __name__ == '__main__':\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " N = 10\n", + " x = np.linspace(0, 1, N)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = np.array([10,10])\n", + " num_iter = 10000\n", + " lmb = 0.001\n", + "\n", + " P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " res = g_trial_deep(x,P)\n", + " res_analytical = g_analytic(x)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, res_analytical)\n", + " plt.plot(x, res[0,:])\n", + " plt.legend(['analytical','dnn'])\n", + " plt.ylabel('g(x)')\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example: Population growth\n", + "\n", + "A logistic model of population growth assumes that a population converges toward an equilibrium.\n", + "The population growth can be modeled by" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{log} \\tag{10}\n", + "\tg'(t) = \\alpha g(t)(A - g(t))\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(t)$ is the population density at time $t$, $\\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment.\n", + "Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant.\n", + "\n", + "In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability\n", + "and high execution time (this might be more apparent in the examples solving PDEs),\n", + "using a library like TensorFlow is recommended.\n", + "Here, we stay with a more simple approach and implement for comparison, the simple forward Euler method.\n", + "\n", + "\n", + "\n", + "Here, we will model a population $g(t)$ in an environment having carrying capacity $A$.\n", + "The population follows the model" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{solveode_population} \\tag{11}\n", + "g'(t) = \\alpha g(t)(A - g(t))\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(0) = g_0$.\n", + "\n", + "In this example, we let $\\alpha = 2$, $A = 1$, and $g_0 = 1.2$.\n", + "\n", + "\n", + "We will get a slightly different trial solution, as the boundary conditions are different\n", + "compared to the case for exponential decay.\n", + "\n", + "A possible trial solution satisfying the condition $g(0) = g_0$ could be\n", + "\n", + "$$\n", + "h_1(t) = g_0 + t \\cdot N(t,P)\n", + "$$\n", + "\n", + "with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$.\n", + "\n", + "The analytical solution is\n", + "\n", + "$$\n", + "g(t) = \\frac{Ag_0}{g_0 + (A - g_0)\\exp(-\\alpha A t)}\n", + "$$\n", + "\n", + "\n", + "\n", + "The network will be the similar as for the exponential decay example, but with some small modifications for our problem." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "# Function to get the parameters.\n", + "# Done such that one can easily change the paramaters after one's liking.\n", + "def get_parameters():\n", + " alpha = 2\n", + " A = 1\n", + " g0 = 1.2\n", + " return alpha, A, g0\n", + "\n", + "def deep_neural_network(P, x):\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = P[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = P[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d_g_t = elementwise_grad(g_trial_deep,0)(x,P)\n", + "\n", + " # The right side of the ODE\n", + " func = f(x, g_t)\n", + "\n", + " err_sqr = (d_g_t - func)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum / np.size(err_sqr)\n", + "\n", + "# The right side of the ODE:\n", + "def f(x, g_trial):\n", + " alpha,A, g0 = get_parameters()\n", + " return alpha*g_trial*(A - g_trial)\n", + "\n", + "# The trial solution using the deep neural network:\n", + "def g_trial_deep(x, params):\n", + " alpha,A, g0 = get_parameters()\n", + " return g0 + x*deep_neural_network(params,x)\n", + "\n", + "# The analytical solution:\n", + "def g_analytic(t):\n", + " alpha,A, g0 = get_parameters()\n", + " return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t))\n", + "\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weigths using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nt = 10\n", + " T = 1\n", + " t = np.linspace(0,T, Nt)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [100, 50, 25]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(t,P)\n", + " g_analytical = g_analytic(t)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + " diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " print(\"The max absolute difference between the solutions is: %g\"%diff_ag)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(t, g_analytical)\n", + " plt.plot(t, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('t')\n", + " plt.ylabel('g(t)')\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Using forward Euler to solve the ODE\n", + "\n", + "A straightforward way of solving an ODE numerically, is to use Euler's method.\n", + "\n", + "Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\\Delta x$ from $x$:\n", + "\n", + "$$\n", + "f(x + \\Delta x) \\approx f(x) + \\Delta x f'(x)\n", + "$$\n", + "\n", + "In our case, using Euler's method to approximate the value of $g$ at a step $\\Delta t$ from $t$ yields" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + " g(t + \\Delta t) &\\approx g(t) + \\Delta t g'(t) \\\\\n", + " &= g(t) + \\Delta t \\big(\\alpha g(t)(A - g(t))\\big)\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "along with the condition that $g(0) = g_0$.\n", + "\n", + "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 $i = 0, \\dots, N_t-1$.\n", + "\n", + "For $i \\geq 1$, we have that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "t_i &= i\\Delta t \\\\\n", + "&= (i - 1)\\Delta t + \\Delta t \\\\\n", + "&= t_{i-1} + \\Delta t\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Now, if $g_i = g(t_i)$ then" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " \\begin{aligned}\n", + " g_i &= g(t_i) \\\\\n", + " &= g(t_{i-1} + \\Delta t) \\\\\n", + " &\\approx g(t_{i-1}) + \\Delta t \\big(\\alpha g(t_{i-1})(A - g(t_{i-1}))\\big) \\\\\n", + " &= g_{i-1} + \\Delta t \\big(\\alpha g_{i-1}(A - g_{i-1})\\big)\n", + " \\end{aligned}\n", + "\\end{equation} \\label{odenum} \\tag{12}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for $i \\geq 1$ and $g_0 = g(t_0) = g(0) = g_0$.\n", + "\n", + "Equation ([12](#odenum)) could be implemented in the following way,\n", + "extending the program that uses the network using Autograd:" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# Assume that all function definitions from the example program using Autograd\n", + "# are located here.\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nt = 10\n", + " T = 1\n", + " t = np.linspace(0,T, Nt)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [100,50,25]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(t,P)\n", + " g_analytical = g_analytic(t)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + " diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " print(\"The max absolute difference between the solutions is: %g\"%diff_ag)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(t, g_analytical)\n", + " plt.plot(t, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('t')\n", + " plt.ylabel('g(t)')\n", + "\n", + " ## Find an approximation to the funtion using forward Euler\n", + "\n", + " alpha, A, g0 = get_parameters()\n", + " dt = T/(Nt - 1)\n", + "\n", + " # Perform forward Euler to solve the ODE\n", + " g_euler = np.zeros(Nt)\n", + " g_euler[0] = g0\n", + "\n", + " for i in range(1,Nt):\n", + " g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1]))\n", + "\n", + " # Print the errors done by each method\n", + " diff1 = np.max(np.abs(g_euler - g_analytical))\n", + " diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical))\n", + "\n", + " print('Max absolute difference between Euler method and analytical: %g'%diff1)\n", + " print('Max absolute difference between deep neural network and analytical: %g'%diff2)\n", + "\n", + " # Plot results\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.plot(t,g_euler)\n", + " plt.plot(t,g_analytical)\n", + " plt.plot(t,g_dnn_ag[0,:])\n", + "\n", + " plt.legend(['euler','analytical','dnn'])\n", + " plt.xlabel('Time t')\n", + " plt.ylabel('g(t)')\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Solving the one dimensional Poisson equation\n", + "\n", + "The Poisson equation for $g(x)$ in one dimension is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{poisson} \\tag{13}\n", + " -g''(x) = f(x)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $f(x)$ is a given function for $x \\in (0,1)$.\n", + "\n", + "The conditions that $g(x)$ is chosen to fulfill, are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + " g(0) &= 0 \\\\\n", + " g(1) &= 0\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used.\n", + "The results from the networks can then be compared to the analytical solution.\n", + "In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks.\n", + "\n", + "\n", + "Here, the function $g(x)$ to solve for follows the equation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "-g''(x) = f(x),\\qquad x \\in (0,1)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $f(x)$ is a given function, along with the chosen conditions" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{aligned}\n", + "g(0) = g(1) = 0\n", + "\\end{aligned}\\label{cond} \\tag{14}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this example, we consider the case when $f(x) = (3x + x^2)\\exp(x)$.\n", + "\n", + "For this case, a possible trial solution satisfying the conditions could be" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g_t(x) = x \\cdot (1-x) \\cdot N(P,x)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The analytical solution for this problem is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g(x) = x(1 - x)\\exp(x)\n", + "$$" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weigths using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "## Set up the cost function specified for this Poisson equation:\n", + "\n", + "# The right side of the ODE\n", + "def f(x):\n", + " return (3*x + x**2)*np.exp(x)\n", + "\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)\n", + "\n", + " right_side = f(x)\n", + "\n", + " err_sqr = (-d2_g_t - right_side)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum/np.size(err_sqr)\n", + "\n", + "# The trial solution:\n", + "def g_trial_deep(x,P):\n", + " return x*(1-x)*deep_neural_network(P,x)\n", + "\n", + "# The analytic solution;\n", + "def g_analytic(x):\n", + " return x*(1-x)*np.exp(x)\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10\n", + " x = np.linspace(0,1, Nx)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [200,100]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(x,P)\n", + " g_analytical = g_analytic(x)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + " max_diff = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " print(\"The max absolute difference between the solutions is: %g\"%max_diff)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, g_analytical)\n", + " plt.plot(x, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('x')\n", + " plt.ylabel('g(x)')\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Comparing with a numerical scheme\n", + "\n", + "The Poisson equation is possible to solve using Taylor series to approximate the second derivative.\n", + "\n", + "Using Taylor series, the second derivative can be expressed as\n", + "\n", + "$$\n", + "g''(x) = \\frac{g(x + \\Delta x) - 2g(x) + g(x-\\Delta x)}{\\Delta x^2} + E_{\\Delta x}(x)\n", + "$$\n", + "\n", + "where $\\Delta x$ is a small step size and $E_{\\Delta x}(x)$ being the error term.\n", + "\n", + "Looking away from the error terms gives an approximation to the second derivative:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{approx} \\tag{15}\n", + "g''(x) \\approx \\frac{g(x + \\Delta x) - 2g(x) + g(x-\\Delta x)}{\\Delta x^2}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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](#approx)) becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "g''(x_i) &\\approx \\frac{g(x_i + \\Delta x) - 2g(x_i) + g(x_i -\\Delta x)}{\\Delta x^2} \\\\\n", + "&= \\frac{g_{i+1} - 2g_i + g_{i-1}}{\\Delta x^2}\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Since we know from our problem that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "-g''(x) &= f(x) \\\\\n", + "&= (3x + x^2)\\exp(x)\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "along with the conditions $g(0) = g(1) = 0$,\n", + "the following scheme can be used to find an approximate solution for $g(x)$ numerically:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " \\begin{aligned}\n", + " -\\Big( \\frac{g_{i+1} - 2g_i + g_{i-1}}{\\Delta x^2} \\Big) &= f(x_i) \\\\\n", + " -g_{i+1} + 2g_i - g_{i-1} &= \\Delta x^2 f(x_i)\n", + " \\end{aligned}\n", + "\\end{equation} \\label{odesys} \\tag{16}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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.\n", + "\n", + "The equation can be rewritten into a matrix equation:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "\\begin{pmatrix}\n", + "2 & -1 & 0 & \\dots & 0 \\\\\n", + "-1 & 2 & -1 & \\dots & 0 \\\\\n", + "\\vdots & & \\ddots & & \\vdots \\\\\n", + "0 & \\dots & -1 & 2 & -1 \\\\\n", + "0 & \\dots & 0 & -1 & 2\\\\\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "g_1 \\\\\n", + "g_2 \\\\\n", + "\\vdots \\\\\n", + "g_{N_x - 3} \\\\\n", + "g_{N_x - 2}\n", + "\\end{pmatrix}\n", + "&=\n", + "\\Delta x^2\n", + "\\begin{pmatrix}\n", + "f(x_1) \\\\\n", + "f(x_2) \\\\\n", + "\\vdots \\\\\n", + "f(x_{N_x - 3}) \\\\\n", + "f(x_{N_x - 2})\n", + "\\end{pmatrix} \\\\\n", + "\\boldsymbol{A}\\boldsymbol{g} &= \\boldsymbol{f},\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "which makes it possible to solve for the vector $\\boldsymbol{g}$.\n", + "\n", + "\n", + "We can then compare the result from this numerical scheme with the output from our network using Autograd:" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weigths using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "## Set up the cost function specified for this Poisson equation:\n", + "\n", + "# The right side of the ODE\n", + "def f(x):\n", + " return (3*x + x**2)*np.exp(x)\n", + "\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)\n", + "\n", + " right_side = f(x)\n", + "\n", + " err_sqr = (-d2_g_t - right_side)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum/np.size(err_sqr)\n", + "\n", + "# The trial solution:\n", + "def g_trial_deep(x,P):\n", + " return x*(1-x)*deep_neural_network(P,x)\n", + "\n", + "# The analytic solution;\n", + "def g_analytic(x):\n", + " return x*(1-x)*np.exp(x)\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10\n", + " x = np.linspace(0,1, Nx)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [200,100]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(x,P)\n", + " g_analytical = g_analytic(x)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, g_analytical)\n", + " plt.plot(x, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('x')\n", + " plt.ylabel('g(x)')\n", + "\n", + " ## Perform the computation using the numerical scheme\n", + "\n", + " dx = 1/(Nx - 1)\n", + "\n", + " # Set up the matrix A\n", + " A = np.zeros((Nx-2,Nx-2))\n", + "\n", + " A[0,0] = 2\n", + " A[0,1] = -1\n", + "\n", + " for i in range(1,Nx-3):\n", + " A[i,i-1] = -1\n", + " A[i,i] = 2\n", + " A[i,i+1] = -1\n", + "\n", + " A[Nx - 3, Nx - 4] = -1\n", + " A[Nx - 3, Nx - 3] = 2\n", + "\n", + " # Set up the vector f\n", + " f_vec = dx**2 * f(x[1:-1])\n", + "\n", + " # Solve the equation\n", + " g_res = np.linalg.solve(A,f_vec)\n", + "\n", + " g_vec = np.zeros(Nx)\n", + " g_vec[1:-1] = g_res\n", + "\n", + " # Print the differences between each method\n", + " max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " max_diff2 = np.max(np.abs(g_vec - g_analytical))\n", + " print(\"The max absolute difference between the analytical solution and DNN Autograd: %g\"%max_diff1)\n", + " print(\"The max absolute difference between the analytical solution and numerical scheme: %g\"%max_diff2)\n", + "\n", + " # Plot the results\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.plot(x,g_vec)\n", + " plt.plot(x,g_analytical)\n", + " plt.plot(x,g_dnn_ag[0,:])\n", + "\n", + " plt.legend(['numerical scheme','analytical','dnn'])\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Partial Differential Equations\n", + "\n", + "A partial differential equation (PDE) has a solution here the function\n", + "is defined by multiple variables. The equation may involve all kinds\n", + "of combinations of which variables the function is differentiated with\n", + "respect to.\n", + "\n", + "In general, a partial differential equation for a function $g(x_1,\\dots,x_N)$ with $N$ variables may be expressed as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{PDE} \\tag{17}\n", + " f\\left(x_1, \\, \\dots \\, , x_N, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1}, \\dots , \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_N}, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(x_1,\\dots,x_N) }{\\partial x_N^n} \\right) = 0\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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.\n", + "\n", + "### Type of problem\n", + "\n", + "The problem our network must solve for, is similar to the ODE case.\n", + "We must have a trial solution $g_t$ at hand.\n", + "\n", + "For instance, the trial solution could be expressed as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + " g_t(x_1,\\dots,x_N) = h_1(x_1,\\dots,x_N) + h_2(x_1,\\dots,x_N,N(x_1,\\dots,x_N,P))\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $h_1(x_1,\\dots,x_N)$ is a function that ensures $g_t(x_1,\\dots,x_N)$ satisfies some given conditions.\n", + "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.\n", + "\n", + "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.\n", + "\n", + "\n", + "\n", + "### Network requirements\n", + "\n", + "The network tries then the minimize the cost function following the\n", + "same ideas as described for the ODE case, but now with more than one\n", + "variables to consider. The concept still remains the same; find a set\n", + "of parameters $P$ such that the expression $f$ in ([17](#PDE)) is as\n", + "close to zero as possible.\n", + "\n", + "As for the ODE case, the cost function is the mean squared error that\n", + "the network must try to minimize. The cost function for the network to\n", + "minimize is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(x_1, \\dots, x_N, P\\right) = \\left( f\\left(x_1, \\, \\dots \\, , x_N, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1}, \\dots , \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_N}, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(x_1,\\dots,x_N) }{\\partial x_N^n} \\right) \\right)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(\\boldsymbol{x}, P\\right) = f\\left( \\left( \\boldsymbol{x}, \\frac{\\partial g(\\boldsymbol{x}) }{\\partial x_1}, \\dots , \\frac{\\partial g(\\boldsymbol{x}) }{\\partial x_N}, \\frac{\\partial g(\\boldsymbol{x}) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(\\boldsymbol{x}) }{\\partial x_N^n} \\right) \\right)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(X, P \\right) = \\sum_{i=1}^M f\\left( \\left( \\boldsymbol{x}_i, \\frac{\\partial g(\\boldsymbol{x}_i) }{\\partial x_1}, \\dots , \\frac{\\partial g(\\boldsymbol{x}_i) }{\\partial x_N}, \\frac{\\partial g(\\boldsymbol{x}_i) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(\\boldsymbol{x}_i) }{\\partial x_N^n} \\right) \\right)^2.\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example: The diffusion equation\n", + "\n", + "In one spatial dimension, the equation reads" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\frac{\\partial g(x,t)}{\\partial t} = \\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where a possible choice of conditions are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "g(0,t) &= 0 ,\\qquad t \\geq 0 \\\\\n", + "g(1,t) &= 0, \\qquad t \\geq 0 \\\\\n", + "g(x,0) &= u(x),\\qquad x\\in [0,1]\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $u(x)$ being some given function.\n", + "\n", + "\n", + "\n", + "For this case, we want to find $g(x,t)$ such that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " \\frac{\\partial g(x,t)}{\\partial t} = \\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "\\end{equation} \\label{diffonedim} \\tag{18}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "and" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "g(0,t) &= 0 ,\\qquad t \\geq 0 \\\\\n", + "g(1,t) &= 0, \\qquad t \\geq 0 \\\\\n", + "g(x,0) &= u(x),\\qquad x\\in [0,1]\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $u(x) = \\sin(\\pi x)$.\n", + "\n", + "First, let us set up the deep neural network.\n", + "The deep neural network will follow the same structure as discussed in the examples solving the ODEs.\n", + "First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions.\n", + "\n", + "\n", + "\n", + "\n", + "The only change to do here, is to extend our network such that\n", + "functions of multiple parameters are correctly handled. In this case\n", + "we have two variables in our function to solve for, that is time $t$\n", + "and position $x$. The variables will be represented by a\n", + "one-dimensional array in the program. The program will evaluate the\n", + "network at each possible pair $(x,t)$, given an array for the desired\n", + "$x$-values and $t$-values to approximate the solution at." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # x is now a point and a 1D numpy array; make it a column vector\n", + " num_coordinates = np.size(x,0)\n", + " x = x.reshape(num_coordinates,-1)\n", + "\n", + " num_points = np.size(x,1)\n", + "\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output[0][0]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The cost function must then iterate through the given arrays\n", + "containing values for $x$ and $t$, defines a point $(x,t)$ the deep\n", + "neural network and the trial solution is evaluated at, and then finds\n", + "the Jacobian of the trial solution.\n", + "\n", + "A possible trial solution for this PDE is\n", + "\n", + "$$\n", + "g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P)\n", + "$$\n", + "\n", + "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 $P$.\n", + "\n", + "To fulfill the conditions, $A(x,t)$ could be:\n", + "\n", + "$$\n", + "h_1(x,t) = (1-t)\\Big(u(x) - \\big((1-x)u(0) + x u(1)\\big)\\Big) = (1-t)u(x) = (1-t)\\sin(\\pi x)\n", + "$$\n", + "since $(0) = u(1) = 0$ and $u(x) = \\sin(\\pi x)$.\n", + "\n", + "\n", + "\n", + "The Jacobian is used because the program must find the derivative of\n", + "the trial solution with respect to $x$ and $t$.\n", + "\n", + "This gives the necessity of computing the Jacobian matrix, as we want\n", + "to evaluate the gradient with respect to $x$ and $t$ (note that the\n", + "Jacobian of a scalar-valued multivariate function is simply its\n", + "gradient).\n", + "\n", + "In Autograd, the differentiation is by default done with respect to\n", + "the first input argument of your Python function. Since the points is\n", + "an array representing $x$ and $t$, the Jacobian is calculated using\n", + "the values of $x$ and $t$.\n", + "\n", + "To find the second derivative with respect to $x$ and $t$, the\n", + "Jacobian can be found for the second time. The result is a Hessian\n", + "matrix, which is the matrix containing all the possible second order\n", + "mixed derivatives of $g(x,t)$." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# Set up the trial function:\n", + "def u(x):\n", + " return np.sin(np.pi*x)\n", + "\n", + "def g_trial(point,P):\n", + " x,t = point\n", + " return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)\n", + "\n", + "# The right side of the ODE:\n", + "def f(point):\n", + " return 0.\n", + "\n", + "# The cost function:\n", + "def cost_function(P, x, t):\n", + " cost_sum = 0\n", + "\n", + " g_t_jacobian_func = jacobian(g_trial)\n", + " g_t_hessian_func = hessian(g_trial)\n", + "\n", + " for x_ in x:\n", + " for t_ in t:\n", + " point = np.array([x_,t_])\n", + "\n", + " g_t = g_trial(point,P)\n", + " g_t_jacobian = g_t_jacobian_func(point,P)\n", + " g_t_hessian = g_t_hessian_func(point,P)\n", + "\n", + " g_t_dt = g_t_jacobian[1]\n", + " g_t_d2x = g_t_hessian[0][0]\n", + "\n", + " func = f(point)\n", + "\n", + " err_sqr = ( (g_t_dt - g_t_d2x) - func)**2\n", + " cost_sum += err_sqr\n", + "\n", + " return cost_sum" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Setting up the network using Autograd; The full program\n", + "\n", + "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.\n", + "\n", + "The analytical solution of our problem is\n", + "\n", + "$$\n", + "g(x,t) = \\exp(-\\pi^2 t)\\sin(\\pi x)\n", + "$$\n", + "\n", + "A possible way to implement a neural network solving the PDE, is given below.\n", + "Be aware, though, that it is fairly slow for the parameters used.\n", + "A better result is possible, but requires more iterations, and thus longer time to complete.\n", + "\n", + "\n", + "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.\n", + "Using TensorFlow results in a much better execution time. Try it!" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import jacobian,hessian,grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import cm\n", + "from matplotlib import pyplot as plt\n", + "from mpl_toolkits.mplot3d import axes3d\n", + "\n", + "## Set up the network\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # x is now a point and a 1D numpy array; make it a column vector\n", + " num_coordinates = np.size(x,0)\n", + " x = x.reshape(num_coordinates,-1)\n", + "\n", + " num_points = np.size(x,1)\n", + "\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output[0][0]\n", + "\n", + "## Define the trial solution and cost function\n", + "def u(x):\n", + " return np.sin(np.pi*x)\n", + "\n", + "def g_trial(point,P):\n", + " x,t = point\n", + " return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)\n", + "\n", + "# The right side of the ODE:\n", + "def f(point):\n", + " return 0.\n", + "\n", + "# The cost function:\n", + "def cost_function(P, x, t):\n", + " cost_sum = 0\n", + "\n", + " g_t_jacobian_func = jacobian(g_trial)\n", + " g_t_hessian_func = hessian(g_trial)\n", + "\n", + " for x_ in x:\n", + " for t_ in t:\n", + " point = np.array([x_,t_])\n", + "\n", + " g_t = g_trial(point,P)\n", + " g_t_jacobian = g_t_jacobian_func(point,P)\n", + " g_t_hessian = g_t_hessian_func(point,P)\n", + "\n", + " g_t_dt = g_t_jacobian[1]\n", + " g_t_d2x = g_t_hessian[0][0]\n", + "\n", + " func = f(point)\n", + "\n", + " err_sqr = ( (g_t_dt - g_t_d2x) - func)**2\n", + " cost_sum += err_sqr\n", + "\n", + " return cost_sum /( np.size(x)*np.size(t) )\n", + "\n", + "## For comparison, define the analytical solution\n", + "def g_analytic(point):\n", + " x,t = point\n", + " return np.exp(-np.pi**2*t)*np.sin(np.pi*x)\n", + "\n", + "## Set up a function for training the network to solve for the equation\n", + "def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):\n", + " ## Set up initial weigths and biases\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: ',cost_function(P, x, t))\n", + "\n", + " cost_function_grad = grad(cost_function,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " cost_grad = cost_function_grad(P, x , t)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_grad[l]\n", + "\n", + " print('Final cost: ',cost_function(P, x, t))\n", + "\n", + " return P\n", + "\n", + "if __name__ == '__main__':\n", + " ### Use the neural network:\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10; Nt = 10\n", + " x = np.linspace(0, 1, Nx)\n", + " t = np.linspace(0,1,Nt)\n", + "\n", + " ## Set up the parameters for the network\n", + " num_hidden_neurons = [100, 25]\n", + " num_iter = 250\n", + " lmb = 0.01\n", + "\n", + " P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " ## Store the results\n", + " g_dnn_ag = np.zeros((Nx, Nt))\n", + " G_analytical = np.zeros((Nx, Nt))\n", + " for i,x_ in enumerate(x):\n", + " for j, t_ in enumerate(t):\n", + " point = np.array([x_, t_])\n", + " g_dnn_ag[i,j] = g_trial(point,P)\n", + "\n", + " G_analytical[i,j] = g_analytic(point)\n", + "\n", + " # Find the map difference between the analytical and the computed solution\n", + " diff_ag = np.abs(g_dnn_ag - G_analytical)\n", + " print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag))\n", + "\n", + " ## Plot the solutions in two dimensions, that being in position and time\n", + "\n", + " T,X = np.meshgrid(t,x)\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))\n", + " s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Analytical solution')\n", + " s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Difference')\n", + " s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + " ## Take some slices of the 3D plots just to see the solutions at particular times\n", + " indx1 = 0\n", + " indx2 = int(Nt/2)\n", + " indx3 = Nt-1\n", + "\n", + " t1 = t[indx1]\n", + " t2 = t[indx2]\n", + " t3 = t[indx3]\n", + "\n", + " # Slice the results from the DNN\n", + " res1 = g_dnn_ag[:,indx1]\n", + " res2 = g_dnn_ag[:,indx2]\n", + " res3 = g_dnn_ag[:,indx3]\n", + "\n", + " # Slice the analytical results\n", + " res_analytical1 = G_analytical[:,indx1]\n", + " res_analytical2 = G_analytical[:,indx2]\n", + " res_analytical3 = G_analytical[:,indx3]\n", + "\n", + " # Plot the slices\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t1)\n", + " plt.plot(x, res1)\n", + " plt.plot(x,res_analytical1)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t2)\n", + " plt.plot(x, res2)\n", + " plt.plot(x,res_analytical2)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t3)\n", + " plt.plot(x, res3)\n", + " plt.plot(x,res_analytical3)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Solving the wave equation with Neural Networks\n", + "\n", + "The wave equation is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\frac{\\partial^2 g(x,t)}{\\partial t^2} = c^2\\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $c$ being the specified wave speed.\n", + "\n", + "Here, the chosen conditions are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "\tg(0,t) &= 0 \\\\\n", + "\tg(1,t) &= 0 \\\\\n", + "\tg(x,0) &= u(x) \\\\\n", + "\t\\frac{\\partial g(x,t)}{\\partial t} \\Big |_{t = 0} &= v(x)\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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.\n", + "\n", + "\n", + "The wave equation to solve for, is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{wave} \\tag{19}\n", + "\\frac{\\partial^2 g(x,t)}{\\partial t^2} = c^2 \\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $c$ is the given wave speed.\n", + "The chosen conditions for this equation are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{aligned}\n", + "g(0,t) &= 0, &t \\geq 0 \\\\\n", + "g(1,t) &= 0, &t \\geq 0 \\\\\n", + "g(x,0) &= u(x), &x\\in[0,1] \\\\\n", + "\\frac{\\partial g(x,t)}{\\partial t}\\Big |_{t = 0} &= v(x), &x \\in [0,1]\n", + "\\end{aligned} \\label{condwave} \\tag{20}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this example, let $c = 1$ and $u(x) = \\sin(\\pi x)$ and $v(x) = -\\pi\\sin(\\pi x)$.\n", + "\n", + "\n", + "\n", + "Setting up the network is done in similar matter as for the example of solving the diffusion equation.\n", + "The only things we have to change, is the trial solution such that it satisfies the conditions from ([20](#condwave)) and the cost function.\n", + "\n", + "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\n", + "\n", + "$$\n", + "g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P)\n", + "$$\n", + "\n", + "where\n", + "\n", + "$$\n", + "h_1(x,t) = (1-t^2)u(x) + tv(x)\n", + "$$\n", + "\n", + "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.\n", + "\n", + "\n", + "The analytical solution for our specific problem, is\n", + "\n", + "$$\n", + "g(x,t) = \\sin(\\pi x)\\cos(\\pi t) - \\sin(\\pi x)\\sin(\\pi t)\n", + "$$" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import hessian,grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import cm\n", + "from matplotlib import pyplot as plt\n", + "from mpl_toolkits.mplot3d import axes3d\n", + "\n", + "## Set up the trial function:\n", + "def u(x):\n", + " return np.sin(np.pi*x)\n", + "\n", + "def v(x):\n", + " return -np.pi*np.sin(np.pi*x)\n", + "\n", + "def h1(point):\n", + " x,t = point\n", + " return (1 - t**2)*u(x) + t*v(x)\n", + "\n", + "def g_trial(point,P):\n", + " x,t = point\n", + " return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point)\n", + "\n", + "## Define the cost function\n", + "def cost_function(P, x, t):\n", + " cost_sum = 0\n", + "\n", + " g_t_hessian_func = hessian(g_trial)\n", + "\n", + " for x_ in x:\n", + " for t_ in t:\n", + " point = np.array([x_,t_])\n", + "\n", + " g_t_hessian = g_t_hessian_func(point,P)\n", + "\n", + " g_t_d2x = g_t_hessian[0][0]\n", + " g_t_d2t = g_t_hessian[1][1]\n", + "\n", + " err_sqr = ( (g_t_d2t - g_t_d2x) )**2\n", + " cost_sum += err_sqr\n", + "\n", + " return cost_sum / (np.size(t) * np.size(x))\n", + "\n", + "## The neural network\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # x is now a point and a 1D numpy array; make it a column vector\n", + " num_coordinates = np.size(x,0)\n", + " x = x.reshape(num_coordinates,-1)\n", + "\n", + " num_points = np.size(x,1)\n", + "\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output[0][0]\n", + "\n", + "## The analytical solution\n", + "def g_analytic(point):\n", + " x,t = point\n", + " return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t)\n", + "\n", + "def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):\n", + " ## Set up initial weigths and biases\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: ',cost_function(P, x, t))\n", + "\n", + " cost_function_grad = grad(cost_function,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " cost_grad = cost_function_grad(P, x , t)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_grad[l]\n", + "\n", + "\n", + " print('Final cost: ',cost_function(P, x, t))\n", + "\n", + " return P\n", + "\n", + "if __name__ == '__main__':\n", + " ### Use the neural network:\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10; Nt = 10\n", + " x = np.linspace(0, 1, Nx)\n", + " t = np.linspace(0,1,Nt)\n", + "\n", + " ## Set up the parameters for the network\n", + " num_hidden_neurons = [50,20]\n", + " num_iter = 1000\n", + " lmb = 0.01\n", + "\n", + " P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " ## Store the results\n", + " res = np.zeros((Nx, Nt))\n", + " res_analytical = np.zeros((Nx, Nt))\n", + " for i,x_ in enumerate(x):\n", + " for j, t_ in enumerate(t):\n", + " point = np.array([x_, t_])\n", + " res[i,j] = g_trial(point,P)\n", + "\n", + " res_analytical[i,j] = g_analytic(point)\n", + "\n", + " diff = np.abs(res - res_analytical)\n", + " print(\"Max difference between analytical and solution from nn: %g\"%np.max(diff))\n", + "\n", + " ## Plot the solutions in two dimensions, that being in position and time\n", + "\n", + " T,X = np.meshgrid(t,x)\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))\n", + " s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Analytical solution')\n", + " s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Difference')\n", + " s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + " ## Take some slices of the 3D plots just to see the solutions at particular times\n", + " indx1 = 0\n", + " indx2 = int(Nt/2)\n", + " indx3 = Nt-1\n", + "\n", + " t1 = t[indx1]\n", + " t2 = t[indx2]\n", + " t3 = t[indx3]\n", + "\n", + " # Slice the results from the DNN\n", + " res1 = res[:,indx1]\n", + " res2 = res[:,indx2]\n", + " res3 = res[:,indx3]\n", + "\n", + " # Slice the analytical results\n", + " res_analytical1 = res_analytical[:,indx1]\n", + " res_analytical2 = res_analytical[:,indx2]\n", + " res_analytical3 = res_analytical[:,indx3]\n", + "\n", + " # Plot the slices\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t1)\n", + " plt.plot(x, res1)\n", + " plt.plot(x,res_analytical1)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t2)\n", + " plt.plot(x, res2)\n", + " plt.plot(x,res_analytical2)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t3)\n", + " plt.plot(x, res3)\n", + " plt.plot(x,res_analytical3)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Resources on differential equations and deep learning\n", + "\n", + "1. [Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al](https://pdfs.semanticscholar.org/d061/df393e0e8fbfd0ea24976458b7d42419040d.pdf)\n", + "\n", + "2. [Neural networks for solving differential equations by A. Honchar](https://becominghuman.ai/neural-networks-for-solving-differential-equations-fa230ac5e04c)\n", + "\n", + "3. [Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener](http://cs229.stanford.edu/proj2013/ChiaramonteKiener-SolvingDifferentialEquationsUsingNeuralNetworks.pdf)\n", + "\n", + "4. [Introduction to Partial Differential Equations by A. Tveito, R. Winther](https://www.springer.com/us/book/9783540225515)" + ] + } + ], + "metadata": {}, + "nbformat": 4, + "nbformat_minor": 4 +} diff --git a/doc/LectureNotes/chapter12.ipynb b/doc/LectureNotes/chapter12.ipynb new file mode 100644 index 000000000..dfbc0e322 --- /dev/null +++ b/doc/LectureNotes/chapter12.ipynb @@ -0,0 +1,718 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Convolutional Neural Networks\n", + "\n", + "\n", + "Convolutional neural networks (CNNs) were developed during the last\n", + "decade of the previous century, with a focus on character recognition\n", + "tasks. Nowadays, CNNs are a central element in the spectacular success\n", + "of deep learning methods. The success in for example image\n", + "classifications have made them a central tool for most machine\n", + "learning practitioners.\n", + "\n", + "CNNs are very similar to ordinary Neural Networks.\n", + "They are made up of neurons that have learnable weights and\n", + "biases. Each neuron receives some inputs, performs a dot product and\n", + "optionally follows it with a non-linearity. The whole network still\n", + "expresses a single differentiable score function: from the raw image\n", + "pixels on one end to class scores at the other. And they still have a\n", + "loss function (for example Softmax) on the last (fully-connected) layer\n", + "and all the tips/tricks we developed for learning regular Neural\n", + "Networks still apply (back propagation, gradient descent etc etc).\n", + "\n", + "What is the difference? **CNN architectures make the explicit assumption that\n", + "the inputs are images, which allows us to encode certain properties\n", + "into the architecture. These then make the forward function more\n", + "efficient to implement and vastly reduce the amount of parameters in\n", + "the network.**\n", + "\n", + "\n", + "As an example, consider\n", + "an image of size $32\\times 32\\times 3$ (32 wide, 32 high, 3 color channels), so a\n", + "single fully-connected neuron in a first hidden layer of a regular\n", + "Neural Network would have $32\\times 32\\times 3 = 3072$ weights. This amount still\n", + "seems manageable, but clearly this fully-connected structure does not\n", + "scale to larger images. For example, an image of more respectable\n", + "size, say $200\\times 200\\times 3$, would lead to neurons that have \n", + "$200\\times 200\\times 3 = 120,000$ weights. \n", + "\n", + "We could have\n", + "several such neurons, and the parameters would add up quickly! Clearly,\n", + "this full connectivity is wasteful and the huge number of parameters\n", + "would quickly lead to possible overfitting.\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "Convolutional Neural Networks take advantage of the fact that the\n", + "input consists of images and they constrain the architecture in a more\n", + "sensible way. \n", + "\n", + "In particular, unlike a regular Neural Network, the\n", + "layers of a CNN have neurons arranged in 3 dimensions: width,\n", + "height, depth. (Note that the word depth here refers to the third\n", + "dimension of an activation volume, not to the depth of a full Neural\n", + "Network, which can refer to the total number of layers in a network.)\n", + "\n", + "To understand it better, the above example of an image \n", + "with an input volume of\n", + "activations has dimensions $32\\times 32\\times 3$ (width, height,\n", + "depth respectively). \n", + "\n", + "The neurons in a layer will\n", + "only be connected to a small region of the layer before it, instead of\n", + "all of the neurons in a fully-connected manner. Moreover, the final\n", + "output layer could for this specific image have dimensions $1\\times 1 \\times 10$, \n", + "because by the\n", + "end of the CNN architecture we will reduce the full image into a\n", + "single vector of class scores, arranged along the depth\n", + "dimension. \n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "A simple CNN is a sequence of layers, and every layer of a CNN\n", + "transforms one volume of activations to another through a\n", + "differentiable function. We use three main types of layers to build\n", + "CNN architectures: Convolutional Layer, Pooling Layer, and\n", + "Fully-Connected Layer (exactly as seen in regular Neural Networks). We\n", + "will stack these layers to form a full CNN architecture.\n", + "\n", + "A simple CNN for image classification could have the architecture:\n", + "\n", + "* **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.\n", + "\n", + "* **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.\n", + "\n", + "* **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 ($[32\\times 32\\times 12]$).\n", + "\n", + "* **POOL** (pooling) layer will perform a downsampling operation along the spatial dimensions (width, height), resulting in volume such as $[16\\times 16\\times 12]$.\n", + "\n", + "* **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.\n", + "\n", + "CNNs transform the original image layer by layer from the original\n", + "pixel values to the final class scores. \n", + "\n", + "Observe that some layers contain\n", + "parameters and other don’t. In particular, the CNN layers perform\n", + "transformations that are a function of not only the activations in the\n", + "input volume, but also of the parameters (the weights and biases of\n", + "the neurons). On the other hand, the RELU/POOL layers will implement a\n", + "fixed function. The parameters in the CONV/FC layers will be trained\n", + "with gradient descent so that the class scores that the CNN computes\n", + "are consistent with the labels in the training set for each image.\n", + "\n", + "\n", + "\n", + "### CNNs in brief\n", + "\n", + "In summary:\n", + "\n", + "* 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)\n", + "\n", + "* There are a few distinct types of Layers (e.g. CONV/FC/RELU/POOL are by far the most popular)\n", + "\n", + "* Each Layer accepts an input 3D volume and transforms it to an output 3D volume through a differentiable function\n", + "\n", + "* Each Layer may or may not have parameters (e.g. CONV/FC do, RELU/POOL don’t)\n", + "\n", + "* Each Layer may or may not have additional hyperparameters (e.g. CONV/FC/POOL do, RELU doesn’t)\n", + "\n", + "## CNNs in more detail, building convolutional neural networks in Tensorflow and Keras\n", + "\n", + "\n", + "As discussed above, CNNs are neural networks built from the assumption that the inputs\n", + "to the network are 2D images. This is important because the number of features or pixels in images\n", + "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", + "\n", + "As before, we still have our input, a hidden layer and an output. What's novel about convolutional networks\n", + "are the **convolutional** and **pooling** layers stacked in pairs between the input and the hidden layer.\n", + "In addition, the data is no longer represented as a 2D feature matrix, instead each input is a number of 2D\n", + "matrices, typically 1 for each color dimension (Red, Green, Blue). \n", + "\n", + "\n", + "\n", + "It means that to represent the entire\n", + "dataset of images, we require a 4D matrix or **tensor**. This tensor has the dimensions:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "(n_{inputs},\\, n_{pixels, width},\\, n_{pixels, height},\\, depth) .\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The MNIST dataset consists of grayscale images with a pixel size of\n", + "$28\\times 28$, meaning we require $28 \\times 28 = 724$ weights to each\n", + "neuron in the first hidden layer.\n", + "\n", + "If we were to analyze images of size $128\\times 128$ we would require\n", + "$128 \\times 128 = 16384$ weights to each neuron. Even worse if we were\n", + "dealing with color images, as most images are, we have an image matrix\n", + "of size $128\\times 128$ for each color dimension (Red, Green, Blue),\n", + "meaning 3 times the number of weights $= 49152$ are required for every\n", + "single neuron in the first hidden layer.\n", + "\n", + "\n", + "\n", + "Images typically have strong local correlations, meaning that a small\n", + "part of the image varies little from its neighboring regions. If for\n", + "example we have an image of a blue car, we can roughly assume that a\n", + "small blue part of the image is surrounded by other blue regions.\n", + "\n", + "Therefore, instead of connecting every single pixel to a neuron in the\n", + "first hidden layer, as we have previously done with deep neural\n", + "networks, we can instead connect each neuron to a small part of the\n", + "image (in all 3 RGB depth dimensions). The size of each small area is\n", + "fixed, and known as a [receptive](https://en.wikipedia.org/wiki/Receptive_field).\n", + "\n", + "\n", + "\n", + "The layers of a convolutional neural network arrange neurons in 3D: width, height and depth. \n", + "The input image is typically a square matrix of depth 3. \n", + "\n", + "A **convolution** is performed on the image which outputs\n", + "a 3D volume of neurons. The weights to the input are arranged in a number of 2D matrices, known as **filters**.\n", + "\n", + "\n", + "Each filter slides along the input image, taking the dot product\n", + "between each small part of the image and the filter, in all depth\n", + "dimensions. This is then passed through a non-linear function,\n", + "typically the **Rectified Linear (ReLu)** function, which serves as the\n", + "activation of the neurons in the first convolutional layer. This is\n", + "further passed through a **pooling layer**, which reduces the size of the\n", + "convolutional layer, e.g. by taking the maximum or average across some\n", + "small regions, and this serves as input to the next convolutional\n", + "layer.\n", + "\n", + "\n", + "\n", + "By systematically reducing the size of the input volume, through\n", + "convolution and pooling, the network should create representations of\n", + "small parts of the input, and then from them assemble representations\n", + "of larger areas. The final pooling layer is flattened to serve as\n", + "input to a hidden layer, such that each neuron in the final pooling\n", + "layer is connected to every single neuron in the hidden layer. This\n", + "then serves as input to the output layer, e.g. a softmax output for\n", + "classification.\n", + "\n", + "\n", + "\n", + "### Prerequisites: Collect and pre-process data" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "%matplotlib inline\n", + "\n", + "# import necessary packages\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "from sklearn import datasets\n", + "\n", + "\n", + "# ensure the same random numbers appear every time\n", + "np.random.seed(0)\n", + "\n", + "# display images in notebook\n", + "%matplotlib inline\n", + "plt.rcParams['figure.figsize'] = (12,12)\n", + "\n", + "\n", + "# download MNIST dataset\n", + "digits = datasets.load_digits()\n", + "\n", + "# define inputs and labels\n", + "inputs = digits.images\n", + "labels = digits.target\n", + "\n", + "# RGB images have a depth of 3\n", + "# our images are grayscale so they should have a depth of 1\n", + "inputs = inputs[:,:,:,np.newaxis]\n", + "\n", + "print(\"inputs = (n_inputs, pixel_width, pixel_height, depth) = \" + str(inputs.shape))\n", + "print(\"labels = (n_inputs) = \" + str(labels.shape))\n", + "\n", + "\n", + "# choose some random images to display\n", + "n_inputs = len(inputs)\n", + "indices = np.arange(n_inputs)\n", + "random_indices = np.random.choice(indices, size=5)\n", + "\n", + "for i, image in enumerate(digits.images[random_indices]):\n", + " plt.subplot(1, 5, i+1)\n", + " plt.axis('off')\n", + " plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest')\n", + " plt.title(\"Label: %d\" % digits.target[random_indices[i]])\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Importing Keras and Tensorflow" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "from tensorflow.keras import datasets, layers, models\n", + "from tensorflow.keras.layers import Input\n", + "from tensorflow.keras.models import Sequential #This allows appending layers to existing models\n", + "from tensorflow.keras.layers import Dense #This allows defining the characteristics of a particular layer\n", + "from tensorflow.keras import optimizers #This allows using whichever optimiser we want (sgd,adam,RMSprop)\n", + "from tensorflow.keras import regularizers #This allows using whichever regularizer we want (l1,l2,l1_l2)\n", + "from tensorflow.keras.utils import to_categorical #This allows using categorical cross entropy as the cost function\n", + "#from tensorflow.keras import Conv2D\n", + "#from tensorflow.keras import MaxPooling2D\n", + "#from tensorflow.keras import Flatten\n", + "\n", + "from sklearn.model_selection import train_test_split\n", + "\n", + "# representation of labels\n", + "labels = to_categorical(labels)\n", + "\n", + "# split into train and test data\n", + "# one-liner from scikit-learn library\n", + "train_size = 0.8\n", + "test_size = 1 - train_size\n", + "X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size,\n", + " test_size=test_size)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "def create_convolutional_neural_network_keras(input_shape, receptive_field,\n", + " n_filters, n_neurons_connected, n_categories,\n", + " eta, lmbd):\n", + " model = Sequential()\n", + " model.add(layers.Conv2D(n_filters, (receptive_field, receptive_field), input_shape=input_shape, padding='same',\n", + " activation='relu', kernel_regularizer=regularizers.l2(lmbd)))\n", + " model.add(layers.MaxPooling2D(pool_size=(2, 2)))\n", + " model.add(layers.Flatten())\n", + " model.add(layers.Dense(n_neurons_connected, activation='relu', kernel_regularizer=regularizers.l2(lmbd)))\n", + " model.add(layers.Dense(n_categories, activation='softmax', kernel_regularizer=regularizers.l2(lmbd)))\n", + " \n", + " sgd = optimizers.SGD(lr=eta)\n", + " model.compile(loss='categorical_crossentropy', optimizer=sgd, metrics=['accuracy'])\n", + " \n", + " return model\n", + "\n", + "epochs = 100\n", + "batch_size = 100\n", + "input_shape = X_train.shape[1:4]\n", + "receptive_field = 3\n", + "n_filters = 10\n", + "n_neurons_connected = 50\n", + "n_categories = 10\n", + "\n", + "eta_vals = np.logspace(-5, 1, 7)\n", + "lmbd_vals = np.logspace(-5, 1, 7)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "CNN_keras = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)\n", + " \n", + "for i, eta in enumerate(eta_vals):\n", + " for j, lmbd in enumerate(lmbd_vals):\n", + " CNN = create_convolutional_neural_network_keras(input_shape, receptive_field,\n", + " n_filters, n_neurons_connected, n_categories,\n", + " eta, lmbd)\n", + " CNN.fit(X_train, Y_train, epochs=epochs, batch_size=batch_size, verbose=0)\n", + " scores = CNN.evaluate(X_test, Y_test)\n", + " \n", + " CNN_keras[i][j] = CNN\n", + " \n", + " print(\"Learning rate = \", eta)\n", + " print(\"Lambda = \", lmbd)\n", + " print(\"Test accuracy: %.3f\" % scores[1])\n", + " print()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Final visualization" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# visual representation of grid search\n", + "# uses seaborn heatmap, could probably do this in matplotlib\n", + "import seaborn as sns\n", + "\n", + "sns.set()\n", + "\n", + "train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n", + "test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n", + "\n", + "for i in range(len(eta_vals)):\n", + " for j in range(len(lmbd_vals)):\n", + " CNN = CNN_keras[i][j]\n", + "\n", + " train_accuracy[i][j] = CNN.evaluate(X_train, Y_train)[1]\n", + " test_accuracy[i][j] = CNN.evaluate(X_test, Y_test)[1]\n", + "\n", + " \n", + "fig, ax = plt.subplots(figsize = (10, 10))\n", + "sns.heatmap(train_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n", + "ax.set_title(\"Training Accuracy\")\n", + "ax.set_ylabel(\"$\\eta$\")\n", + "ax.set_xlabel(\"$\\lambda$\")\n", + "plt.show()\n", + "\n", + "fig, ax = plt.subplots(figsize = (10, 10))\n", + "sns.heatmap(test_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n", + "ax.set_title(\"Test Accuracy\")\n", + "ax.set_ylabel(\"$\\eta$\")\n", + "ax.set_xlabel(\"$\\lambda$\")\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## The CIFAR01 data set\n", + "\n", + "The CIFAR10 dataset contains 60,000 color images in 10 classes, with\n", + "6,000 images in each class. The dataset is divided into 50,000\n", + "training images and 10,000 testing images. The classes are mutually\n", + "exclusive and there is no overlap between them." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import tensorflow as tf\n", + "\n", + "from tensorflow.keras import datasets, layers, models\n", + "import matplotlib.pyplot as plt\n", + "\n", + "# We import the data set\n", + "(train_images, train_labels), (test_images, test_labels) = datasets.cifar10.load_data()\n", + "\n", + "# Normalize pixel values to be between 0 and 1 by dividing by 255. \n", + "train_images, test_images = train_images / 255.0, test_images / 255.0" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "class_names = ['airplane', 'automobile', 'bird', 'cat', 'deer',\n", + " 'dog', 'frog', 'horse', 'ship', 'truck']\n", + "​\n", + "plt.figure(figsize=(10,10))\n", + "for i in range(25):\n", + " plt.subplot(5,5,i+1)\n", + " plt.xticks([])\n", + " plt.yticks([])\n", + " plt.grid(False)\n", + " plt.imshow(train_images[i], cmap=plt.cm.binary)\n", + " # The CIFAR labels happen to be arrays, \n", + " # which is why you need the extra index\n", + " plt.xlabel(class_names[train_labels[i][0]])\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The 6 lines of code below define the convolutional base using a common pattern: a stack of Conv2D and MaxPooling2D layers.\n", + "\n", + "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." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "model = models.Sequential()\n", + "model.add(layers.Conv2D(32, (3, 3), activation='relu', input_shape=(32, 32, 3)))\n", + "model.add(layers.MaxPooling2D((2, 2)))\n", + "model.add(layers.Conv2D(64, (3, 3), activation='relu'))\n", + "model.add(layers.MaxPooling2D((2, 2)))\n", + "model.add(layers.Conv2D(64, (3, 3), activation='relu'))\n", + "\n", + "# Let's display the architecture of our model so far.\n", + "\n", + "model.summary()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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.\n", + "\n", + "\n", + "\n", + "To complete our model, you will feed the last output tensor from the\n", + "convolutional base (of shape (4, 4, 64)) into one or more Dense layers\n", + "to perform classification. Dense layers take vectors as input (which\n", + "are 1D), while the current output is a 3D tensor. First, you will\n", + "flatten (or unroll) the 3D output to 1D, then add one or more Dense\n", + "layers on top. CIFAR has 10 output classes, so you use a final Dense\n", + "layer with 10 outputs and a softmax activation." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "model.add(layers.Flatten())\n", + "model.add(layers.Dense(64, activation='relu'))\n", + "model.add(layers.Dense(10))\n", + "Here's the complete architecture of our model.\n", + "\n", + "model.summary()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "As you can see, our (4, 4, 64) outputs were flattened into vectors of shape (1024) before going through two Dense layers.\n", + "\n", + "Compile and train the model." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "model.compile(optimizer='adam',\n", + " loss=tf.keras.losses.SparseCategoricalCrossentropy(from_logits=True),\n", + " metrics=['accuracy'])\n", + "​\n", + "history = model.fit(train_images, train_labels, epochs=10, \n", + " validation_data=(test_images, test_labels))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Finally, we evaluate the model." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "plt.plot(history.history['accuracy'], label='accuracy')\n", + "plt.plot(history.history['val_accuracy'], label = 'val_accuracy')\n", + "plt.xlabel('Epoch')\n", + "plt.ylabel('Accuracy')\n", + "plt.ylim([0.5, 1])\n", + "plt.legend(loc='lower right')\n", + "\n", + "test_loss, test_acc = model.evaluate(test_images, test_labels, verbose=2)\n", + "\n", + "print(test_acc)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Recurrent neural networks: Overarching view\n", + "\n", + "Till now our focus has been, including convolutional neural networks\n", + "as well, on feedforward neural networks. The output or the activations\n", + "flow only in one direction, from the input layer to the output layer.\n", + "\n", + "A recurrent neural network (RNN) looks very much like a feedforward\n", + "neural network, except that it also has connections pointing\n", + "backward. \n", + "\n", + "RNNs are used to analyze time series data such as stock prices, and\n", + "tell you when to buy or sell. In autonomous driving systems, they can\n", + "anticipate car trajectories and help avoid accidents. More generally,\n", + "they can work on sequences of arbitrary lengths, rather than on\n", + "fixed-sized inputs like all the nets we have discussed so far. For\n", + "example, they can take sentences, documents, or audio samples as\n", + "input, making them extremely useful for natural language processing\n", + "systems such as automatic translation and speech-to-text.\n", + "\n", + "\n", + "\n", + "\n", + "### A simple example" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# Start importing packages\n", + "import pandas as pd\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "import tensorflow as tf\n", + "from tensorflow.keras import datasets, layers, models\n", + "from tensorflow.keras.layers import Input\n", + "from tensorflow.keras.models import Model, Sequential \n", + "from tensorflow.keras.layers import Dense, SimpleRNN, LSTM, GRU\n", + "from tensorflow.keras import optimizers \n", + "from tensorflow.keras import regularizers \n", + "from tensorflow.keras.utils import to_categorical \n", + "\n", + "\n", + "\n", + "# convert into dataset matrix\n", + "def convertToMatrix(data, step):\n", + " X, Y =[], []\n", + " for i in range(len(data)-step):\n", + " d=i+step \n", + " X.append(data[i:d,])\n", + " Y.append(data[d,])\n", + " return np.array(X), np.array(Y)\n", + "\n", + "step = 4\n", + "N = 1000 \n", + "Tp = 800 \n", + "\n", + "t=np.arange(0,N)\n", + "x=np.sin(0.02*t)+2*np.random.rand(N)\n", + "df = pd.DataFrame(x)\n", + "df.head()\n", + "\n", + "plt.plot(df)\n", + "plt.show()\n", + "\n", + "values=df.values\n", + "train,test = values[0:Tp,:], values[Tp:N,:]\n", + "\n", + "# add step elements into train and test\n", + "test = np.append(test,np.repeat(test[-1,],step))\n", + "train = np.append(train,np.repeat(train[-1,],step))\n", + " \n", + "trainX,trainY =convertToMatrix(train,step)\n", + "testX,testY =convertToMatrix(test,step)\n", + "trainX = np.reshape(trainX, (trainX.shape[0], 1, trainX.shape[1]))\n", + "testX = np.reshape(testX, (testX.shape[0], 1, testX.shape[1]))\n", + "\n", + "model = Sequential()\n", + "model.add(SimpleRNN(units=32, input_shape=(1,step), activation=\"relu\"))\n", + "model.add(Dense(8, activation=\"relu\")) \n", + "model.add(Dense(1))\n", + "model.compile(loss='mean_squared_error', optimizer='rmsprop')\n", + "model.summary()\n", + "\n", + "model.fit(trainX,trainY, epochs=100, batch_size=16, verbose=2)\n", + "trainPredict = model.predict(trainX)\n", + "testPredict= model.predict(testX)\n", + "predicted=np.concatenate((trainPredict,testPredict),axis=0)\n", + "\n", + "trainScore = model.evaluate(trainX, trainY, verbose=0)\n", + "print(trainScore)\n", + "\n", + "index = df.index.values\n", + "plt.plot(index,df)\n", + "plt.plot(index,predicted)\n", + "plt.axvline(df.index[Tp], c=\"r\")\n", + "plt.show()" + ] + } + ], + "metadata": {}, + "nbformat": 4, + "nbformat_minor": 4 +} diff --git a/doc/LectureNotes/schedule.md b/doc/LectureNotes/schedule.md index dd159bf54..572474bbf 100644 --- a/doc/LectureNotes/schedule.md +++ b/doc/LectureNotes/schedule.md @@ -123,7 +123,7 @@ For the reading assignments we use the following abbreviations: ### Week 42 October 18-22 - Lab Wednesday: Work on project 2 -- Lecture Thursday: Convolutional Neural Networks and classification problems +- Lecture Thursday: Solving differential equations with neural networks and start Convolutional Neural Networks and classification problems - Lecture Friday: Convolutional Neural Networks and classification problems - Reading recommendations: - See lecture notes for week 42 at https://compphysics.github.io/MachineLearning/doc/web/course.html.