Week 43: Deep Learning: Constructing a Neural Network code and solving differential equations

Morten Hjorth-Jensen [1, 2]
[1] Department of Physics, University of Oslo
[2] Department of Physics and Astronomy and Facility for Rare Isotope Beams, Michigan State University

October 21, 2024


Plans for week 43

Exercises and lab session week 43

  • Exercise on writing your own neural network code
  • The exercises this week will be continued next week as well
  • Discussion of project 2

Mathematics of deep learning

  1. The Modern Mathematics of Deep Learning, by Julius Berner, Philipp Grohs, Gitta Kutyniok, Philipp Petersen at https://arxiv.org/abs/2105.04026, published as Mathematical Aspects of Deep Learning, pp. 1-111. Cambridge University Press, 2022
  2. Mathematical Introduction to Deep Learning: Methods, Implementations, and Theory, Arnulf Jentzen, Benno Kuckuck, Philippe von Wurstemberger at https://doi.org/10.48550/arXiv.2310.20360

Reminder on books with hands-on material and codes

Reading recommendations

  1. Rashkca et al., chapter 11, jupyter-notebook sent separately, from GitHub site at https://github.com/rasbt/machine-learning-book. See also chapters 12 and 13 on using Pytorch to make a Neural network code.
  2. Goodfellow et al, chapter 6 and 7 contain most of the neural network background.

Using Automatic differentiation

In our discussions of ordinary differential equations and neural network codes we will also study the usage of Autograd, see for example https://www.youtube.com/watch?v=fRf4l5qaX1M&ab_channel=AlexSmola in computing gradients for deep learning. For the documentation of Autograd and examples see the lectures slides from week 39 and the Autograd documentation at https://github.com/HIPS/autograd.

Back propagation and automatic differentiation

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

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

Lecture Monday October 21

Setting up the back propagation algorithm and algorithm for a feed forward NN, initalizations

This is a reminder from where we ended last week.

  1. Set up your inputs and outputs (scalars, vectors, matrices or higher-order arrays)
  2. Define the number of hidden layers and hidden nodes
  3. Define activation functions for hidden layers and output layers
  4. Define optimizer (plan learning rate, momentum, ADAgrad, RMSprop, ADAM etc) and array of initial learning rates
  5. Define cost function and possible regularization terms with hyperparameters
  6. Initialize weights and biases
  7. Fix number of iterations for the feed forward part and back propagation part

Setting up the back propagation algorithm, part 1

Let us write this out in the form of an algorithm.

First, we set up the input data \( \boldsymbol{x} \) and the activations \( \boldsymbol{z}_1 \) of the input layer and compute the activation function and the pertinent outputs \( \boldsymbol{a}^1 \).

Secondly, we perform then the feed forward till we reach the output layer and compute all \( \boldsymbol{z}_l \) of the input layer and compute the activation function and the pertinent outputs \( \boldsymbol{a}^l \) for \( l=1,2,3,\dots,L \).

Notation: The first hidden layer has \( l=1 \) as label and the final output layer has \( l=L \).

Setting up the back propagation algorithm, part 2

Thereafter we compute the ouput error \( \boldsymbol{\delta}^L \) by computing all

$$ \delta_j^L = \sigma'(z_j^L)\frac{\partial {\cal C}}{\partial (a_j^L)}. $$

Then we compute the back propagate error for each \( l=L-1,L-2,\dots,1 \) as

$$ \delta_j^l = \sum_k \delta_k^{l+1}w_{kj}^{l+1}\sigma'(z_j^l). $$

Setting up the Back propagation algorithm, part 3

Finally, we update the weights and the biases using gradient descent for each \( l=L-1,L-2,\dots,1 \) (the first hidden layer) and update the weights and biases according to the rules

$$ w_{ij}^l\leftarrow = w_{ij}^l- \eta \delta_j^la_i^{l-1}, $$ $$ b_j^l \leftarrow b_j^l-\eta \frac{\partial {\cal C}}{\partial b_j^l}=b_j^l-\eta \delta_j^l, $$

with \( \eta \) being the learning rate.

Updating the gradients

With the back propagate error for each \( l=L-1,L-2,\dots,1 \) as

$$ \delta_j^l = \sum_k \delta_k^{l+1}w_{kj}^{l+1}\sigma'(z_j^l), $$

we update the weights and the biases using gradient descent for each \( l=L-1,L-2,\dots,1 \) and update the weights and biases according to the rules

$$ w_{ij}^l\leftarrow = w_{ij}^l- \eta \delta_j^la_i^{l-1}, $$ $$ b_j^l \leftarrow b_j^l-\eta \frac{\partial {\cal C}}{\partial b_j^l}=b_j^l-\eta \delta_j^l, $$

Activation functions

A property that characterizes a neural network, other than its connectivity, is the choice of activation function(s). The following restrictions are imposed on an activation function for an FFNN to fulfill the universal approximation theorem

Activation functions, examples

Typical examples are the logistic Sigmoid

$$ \sigma(x) = \frac{1}{1 + e^{-x}}, $$

and the hyperbolic tangent function

$$ \sigma(x) = \tanh(x) $$

The RELU function family

The ReLU activation function suffers from a problem known as the dying ReLUs: during training, some neurons effectively die, meaning they stop outputting anything other than 0.

In some cases, you may find that half of your network’s neurons are dead, especially if you used a large learning rate. During training, if a neuron’s weights get updated such that the weighted sum of the neuron’s inputs is negative, it will start outputting 0. When this happen, the neuron is unlikely to come back to life since the gradient of the ReLU function is 0 when its input is negative.

ELU function

To solve this problem, nowadays practitioners use a variant of the ReLU function, such as the leaky ReLU discussed above or the so-called exponential linear unit (ELU) function

$$ ELU(z) = \left\{\begin{array}{cc} \alpha\left( \exp{(z)}-1\right) & z < 0,\\ z & z \ge 0.\end{array}\right. $$

Which activation function should we use?

In general it seems that the ELU activation function is better than the leaky ReLU function (and its variants), which is better than ReLU. ReLU performs better than \( \tanh \) which in turn performs better than the logistic function.

If runtime performance is an issue, then you may opt for the leaky ReLU function over the ELU function If you don’t want to tweak yet another hyperparameter, you may just use the default \( \alpha \) of \( 0.01 \) for the leaky ReLU, and \( 1 \) for ELU. If you have spare time and computing power, you can use cross-validation or bootstrap to evaluate other activation functions.

More on activation functions, output layers

In most cases you can use the ReLU activation function in the hidden layers (or one of its variants).

It is a bit faster to compute than other activation functions, and the gradient descent optimization does in general not get stuck.

For the output layer:

Setting up a Multi-layer perceptron model for classification

We are now gong to develop an example based on the MNIST data base. This is a classification problem and we need to use our cross-entropy function we discussed in connection with logistic regression. The cross-entropy defines our cost function for the classificaton problems with neural networks.

In binary classification with two classes \( (0, 1) \) we define the logistic/sigmoid function as the probability that a particular input is in class \( 0 \) or \( 1 \). This is possible because the logistic function takes any input from the real numbers and inputs a number between 0 and 1, and can therefore be interpreted as a probability. It also has other nice properties, such as a derivative that is simple to calculate.

For an input \( \boldsymbol{a} \) from the hidden layer, the probability that the input \( \boldsymbol{x} \) 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 \( x \) represents our activation values \( z \). We have

$$ P(y = 0 \mid \boldsymbol{x}, \boldsymbol{\theta}) = \frac{1}{1 + \exp{(- \boldsymbol{x}})} , $$

and

$$ P(y = 1 \mid \boldsymbol{x}, \boldsymbol{\theta}) = 1 - P(y = 0 \mid \boldsymbol{x}, \boldsymbol{\theta}) , $$

where \( y \in \{0, 1\} \) and \( \boldsymbol{\theta} \) represents the weights and biases of our network.

Defining the cost function

Our cost function is given as (see the Logistic regression lectures)

$$ \mathcal{C}(\boldsymbol{\theta}) = - \ln P(\mathcal{D} \mid \boldsymbol{\theta}) = - \sum_{i=1}^n y_i \ln[P(y_i = 0)] + (1 - y_i) \ln [1 - P(y_i = 0)] = \sum_{i=1}^n \mathcal{L}_i(\boldsymbol{\theta}) . $$

This last equality means that we can interpret our cost function as a sum over the loss function for each point in the dataset \( \mathcal{L}_i(\boldsymbol{\theta}) \). The negative sign is just so that we can think about our algorithm as minimizing a positive number, rather than maximizing a negative number.

In multiclass classification it is common to treat each integer label as a so called one-hot vector:

\( y = 5 \quad \rightarrow \quad \boldsymbol{y} = (0, 0, 0, 0, 0, 1, 0, 0, 0, 0) , \) and

\( y = 1 \quad \rightarrow \quad \boldsymbol{y} = (0, 1, 0, 0, 0, 0, 0, 0, 0, 0) , \)

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 \( 9 \))..

If \( \boldsymbol{x}_i \) is the \( i \)-th input (image), \( y_{ic} \) refers to the \( c \)-th component of the \( i \)-th output vector \( \boldsymbol{y}_i \). The probability of \( \boldsymbol{x}_i \) being in class \( c \) will be given by the softmax function:

$$ P(y_{ic} = 1 \mid \boldsymbol{x}_i, \boldsymbol{\theta}) = \frac{\exp{((\boldsymbol{a}_i^{hidden})^T \boldsymbol{w}_c)}} {\sum_{c'=0}^{C-1} \exp{((\boldsymbol{a}_i^{hidden})^T \boldsymbol{w}_{c'})}} , $$

which reduces to the logistic function in the binary case. The likelihood of this \( C \)-class classifier is now given as:

$$ P(\mathcal{D} \mid \boldsymbol{\theta}) = \prod_{i=1}^n \prod_{c=0}^{C-1} [P(y_{ic} = 1)]^{y_{ic}} . $$

Again we take the negative log-likelihood to define our cost function:

$$ \mathcal{C}(\boldsymbol{\theta}) = - \log{P(\mathcal{D} \mid \boldsymbol{\theta})}. $$

See the logistic regression lectures for a full definition of the cost function.

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!

Example: binary classification problem

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

$$ \mathcal{C}(\boldsymbol{\beta}) = - \sum_{i=1}^n \left(y_i\log{p(y_i \vert x_i,\boldsymbol{\beta})}+(1-y_i)\log{1-p(y_i \vert x_i,\boldsymbol{\beta})}\right), $$

where we had defined the logistic (sigmoid) function

$$ p(y_i =1\vert x_i,\boldsymbol{\beta})=\frac{\exp{(\beta_0+\beta_1 x_i)}}{1+\exp{(\beta_0+\beta_1 x_i)}}, $$

and

$$ p(y_i =0\vert x_i,\boldsymbol{\beta})=1-p(y_i =1\vert x_i,\boldsymbol{\beta}). $$

The parameters \( \boldsymbol{\beta} \) were defined using a minimization method like gradient descent or Newton-Raphson's method.

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 \( b_i^l \). We have then

$$ a_i^l = y_i = \frac{\exp{(z_i^l)}}{1+\exp{(z_i^l)}}, $$

with

$$ z_i^l = \sum_{j}w_{ij}^l a_j^{l-1}+b_i^l, $$

where the superscript \( l-1 \) indicates that these are the outputs from layer \( l-1 \). Our cost function at the final layer \( l=L \) is now

$$ \mathcal{C}(\boldsymbol{W}) = - \sum_{i=1}^n \left(t_i\log{a_i^L}+(1-t_i)\log{(1-a_i^L)}\right), $$

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

$$ \frac{\partial \mathcal{C}(\boldsymbol{W})}{\partial a_i^L} = \frac{a_i^L-t_i}{a_i^L(1-a_i^L)}. $$

In case we use another activation function than the logistic one, we need to evaluate other derivatives.

The Softmax function

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

$$ \frac{\partial f(z_i^l)}{\partial w_{jk}^l} = \frac{\partial f(z_i^l)}{\partial z_j^l} \frac{\partial z_j^l}{\partial w_{jk}^l}= \frac{\partial f(z_i^l)}{\partial z_j^l}a_k^{l-1}. $$

For the Softmax function we have

$$ f(z_i^l) = \frac{\exp{(z_i^l)}}{\sum_{m=1}^K\exp{(z_m^l)}}. $$

Its derivative with respect to \( z_j^l \) gives

$$ \frac{\partial f(z_i^l)}{\partial z_j^l}= f(z_i^l)\left(\delta_{ij}-f(z_j^l)\right), $$

which in case of the simply binary model reduces to having \( i=j \).

Developing a code for doing neural networks with back propagation

One can identify a set of key steps when using neural networks to solve supervised learning problems:

  1. Collect and pre-process data
  2. Define model and architecture
  3. Choose cost function and optimizer
  4. Train the model
  5. Evaluate model performance on test data
  6. Adjust hyperparameters (if necessary, network architecture)

Collect and pre-process data

Here we will be using the MNIST dataset, which is readily available through the scikit-learn package. You may also find it for example here. The MNIST (Modified National Institute of Standards and Technology) database is a large database of handwritten digits that is commonly used for training various image processing systems. The MNIST dataset consists of 70 000 images of size \( 28\times 28 \) pixels, each labeled from 0 to 9. 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.

To feed data into a feed-forward neural network we need to represent the inputs as a design/feature matrix \( X = (n_{inputs}, n_{features}) \). Each row represents an input, in this case a handwritten digit, and each column represents a feature, in this case a pixel. The correct answers, also known as labels or targets are represented as a 1D array of integers \( Y = (n_{inputs}) = (5, 3, 1, 8,...) \).

As an example, say we want to build a neural network using supervised learning to predict Body-Mass Index (BMI) from measurements of height (in m) and weight (in kg). If we have measurements of 5 people the design/feature matrix could be for example:

$$ X = \begin{bmatrix} 1.85 & 81\\ 1.71 & 65\\ 1.95 & 103\\ 1.55 & 42\\ 1.63 & 56 \end{bmatrix} ,$$

and the targets would be:

$$ Y = (23.7, 22.2, 27.1, 17.5, 21.1) $$

Since each input image is a 2D matrix, we need to flatten the image (i.e. "unravel" the 2D matrix into a 1D array) to turn the data into a design/feature matrix. This means we lose all spatial information in the image, such as locality and translational invariance. More complicated architectures such as Convolutional Neural Networks can take advantage of such information, and are most commonly applied when analyzing images.

# import necessary packages
import numpy as np
import matplotlib.pyplot as plt
from sklearn import datasets


# ensure the same random numbers appear every time
np.random.seed(0)

# display images in notebook
%matplotlib inline
plt.rcParams['figure.figsize'] = (12,12)


# download MNIST dataset
digits = datasets.load_digits()

# define inputs and labels
inputs = digits.images
labels = digits.target

print("inputs = (n_inputs, pixel_width, pixel_height) = " + str(inputs.shape))
print("labels = (n_inputs) = " + str(labels.shape))


# flatten the image
# the value -1 means dimension is inferred from the remaining dimensions: 8x8 = 64
n_inputs = len(inputs)
inputs = inputs.reshape(n_inputs, -1)
print("X = (n_inputs, n_features) = " + str(inputs.shape))


# choose some random images to display
indices = np.arange(n_inputs)
random_indices = np.random.choice(indices, size=5)

for i, image in enumerate(digits.images[random_indices]):
    plt.subplot(1, 5, i+1)
    plt.axis('off')
    plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest')
    plt.title("Label: %d" % digits.target[random_indices[i]])
plt.show()

Train and test datasets

Performing analysis before partitioning the dataset is a major error, that can lead to incorrect conclusions.

We will reserve \( 80 \% \) of our dataset for training and \( 20 \% \) for testing.

It is important that the train and test datasets are drawn randomly from our dataset, to ensure no bias in the sampling. Say you are taking measurements of weather data to predict the weather in the coming 5 days. 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 collected from 12.00 to 24.00.

from sklearn.model_selection import train_test_split

# one-liner from scikit-learn library
train_size = 0.8
test_size = 1 - train_size
X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size,
                                                    test_size=test_size)

# equivalently in numpy
def train_test_split_numpy(inputs, labels, train_size, test_size):
    n_inputs = len(inputs)
    inputs_shuffled = inputs.copy()
    labels_shuffled = labels.copy()
    
    np.random.shuffle(inputs_shuffled)
    np.random.shuffle(labels_shuffled)
    
    train_end = int(n_inputs*train_size)
    X_train, X_test = inputs_shuffled[:train_end], inputs_shuffled[train_end:]
    Y_train, Y_test = labels_shuffled[:train_end], labels_shuffled[train_end:]
    
    return X_train, X_test, Y_train, Y_test

#X_train, X_test, Y_train, Y_test = train_test_split_numpy(inputs, labels, train_size, test_size)

print("Number of training images: " + str(len(X_train)))
print("Number of test images: " + str(len(X_test)))

Define model and architecture

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

$$ z = \sum_{i=1}^n w_i a_i ,$$

$$ y = f(z) ,$$

where \( f \) is the activation function, \( a_i \) represents input from neuron \( i \) in the preceding layer and \( w_i \) is the weight to input \( i \). The activation of the neurons in the input layer is just the features (e.g. a pixel value).

The simplest activation function for a neuron is the Heaviside function:

$$ f(z) = \begin{cases} 1, & z > 0\\ 0, & \text{otherwise} \end{cases} $$

A feed-forward neural network with this activation is known as a perceptron. For a binary classifier (i.e. two classes, 0 or 1, dog or not-dog) we can also use this in our output layer. This activation can be generalized to \( k \) classes (using e.g. the one-against-all strategy), and we call these architectures multiclass perceptrons.

However, it is now common to use the terms Single Layer Perceptron (SLP) (1 hidden layer) and Multilayer Perceptron (MLP) (2 or more hidden layers) to refer to feed-forward neural networks with any activation function.

Typical choices for activation functions include the sigmoid function, hyperbolic tangent, and Rectified Linear Unit (ReLU). We will be using the sigmoid function \( \sigma(x) \):

$$ f(x) = \sigma(x) = \frac{1}{1 + e^{-x}} ,$$

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.

Layers

Since each input image has 8x8 = 64 pixels or features, we have an input layer of 64 neurons.

We will use 50 neurons in the hidden layer receiving input from the neurons in the input layer. Since each neuron in the hidden layer is connected to the 64 inputs we have 64x50 = 3200 weights to the hidden layer.

If we were building a binary classifier, it would be sufficient with a single neuron in the output layer, 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.

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.

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:

$$ P(\text{class \( j \)} \mid \text{input \( \boldsymbol{a} \)}) = \frac{\exp{(\boldsymbol{a}^T \boldsymbol{w}_j)}} {\sum_{c=0}^{9} \exp{(\boldsymbol{a}^T \boldsymbol{w}_c)}} ,$$

i.e. each neuron \( j \) outputs the probability of being in class \( j \) given an input from the hidden layer \( \boldsymbol{a} \), with \( \boldsymbol{w}_j \) the weights of neuron \( j \) to the inputs. The denominator is a normalization factor to ensure the outputs (probabilities) sum up to 1. The exponent is just the weighted sum of inputs as before:

$$ z_j = \sum_{i=1}^n w_ {ij} a_i+b_j.$$

Since each neuron in the output layer is connected to the 50 inputs from the hidden layer we have 50x10 = 500 weights to the output layer.

Weights and biases

Typically weights are initialized with small values distributed around zero, drawn from a uniform or normal distribution. Setting all weights to zero means all neurons give the same output, making the network useless.

Adding a bias value to the weighted sum of inputs allows the neural network to represent a greater range 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 \( j \), \( b_j \):

$$ z_j = \sum_{i=1}^n w_ {ij} a_i + b_j.$$

The bias weights \( \boldsymbol{b} \) 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.

# building our neural network

n_inputs, n_features = X_train.shape
n_hidden_neurons = 50
n_categories = 10

# we make the weights normally distributed using numpy.random.randn

# weights and bias in the hidden layer
hidden_weights = np.random.randn(n_features, n_hidden_neurons)
hidden_bias = np.zeros(n_hidden_neurons) + 0.01

# weights and bias in the output layer
output_weights = np.random.randn(n_hidden_neurons, n_categories)
output_bias = np.zeros(n_categories) + 0.01

Feed-forward pass

Denote \( F \) the number of features, \( H \) the number of hidden neurons and \( C \) the number of categories. For each input image we calculate a weighted sum of input features (pixel values) to each neuron \( j \) in the hidden layer \( l \):

$$ z_{j}^{l} = \sum_{i=1}^{F} w_{ij}^{l} x_i + b_{j}^{l},$$

this is then passed through our activation function

$$ a_{j}^{l} = f(z_{j}^{l}) .$$

We calculate a weighted sum of inputs (activations in the hidden layer) to each neuron \( j \) in the output layer:

$$ z_{j}^{L} = \sum_{i=1}^{H} w_{ij}^{L} a_{i}^{l} + b_{j}^{L}.$$

Finally we calculate the output of neuron \( j \) in the output layer using the softmax function:

$$ a_{j}^{L} = \frac{\exp{(z_j^{L})}} {\sum_{c=0}^{C-1} \exp{(z_c^{L})}} .$$

Matrix multiplications

Since our data has the dimensions \( X = (n_{inputs}, n_{features}) \) and our weights to the hidden layer have the dimensions \( W_{hidden} = (n_{features}, n_{hidden}) \), we can easily feed the network all our training data in one go by taking the matrix product

$$ X W^{h} = (n_{inputs}, n_{hidden}),$$

and obtain a matrix that holds the weighted sum of inputs to the hidden layer for each input image and each hidden neuron. We also add the bias to obtain a matrix of weighted sums to the hidden layer \( Z^{h} \):

$$ \boldsymbol{z}^{l} = \boldsymbol{X} \boldsymbol{W}^{l} + \boldsymbol{b}^{l} ,$$

meaning the same bias (1D array with size equal number of hidden neurons) is added to each input image. This is then passed through the activation:

$$ \boldsymbol{a}^{l} = f(\boldsymbol{z}^l) .$$

This is fed to the output layer:

$$ \boldsymbol{z}^{L} = \boldsymbol{a}^{L} \boldsymbol{W}^{L} + \boldsymbol{b}^{L} .$$

Finally we receive our output values for each image and each category by passing it through the softmax function:

$$ output = softmax (\boldsymbol{z}^{L}) = (n_{inputs}, n_{categories}) .$$

# setup the feed-forward pass, subscript h = hidden layer

def sigmoid(x):
    return 1/(1 + np.exp(-x))

def feed_forward(X):
    # weighted sum of inputs to the hidden layer
    z_h = np.matmul(X, hidden_weights) + hidden_bias
    # activation in the hidden layer
    a_h = sigmoid(z_h)
    
    # weighted sum of inputs to the output layer
    z_o = np.matmul(a_h, output_weights) + output_bias
    # softmax output
    # axis 0 holds each input and axis 1 the probabilities of each category
    exp_term = np.exp(z_o)
    probabilities = exp_term / np.sum(exp_term, axis=1, keepdims=True)
    
    return probabilities

probabilities = feed_forward(X_train)
print("probabilities = (n_inputs, n_categories) = " + str(probabilities.shape))
print("probability that image 0 is in category 0,1,2,...,9 = \n" + str(probabilities[0]))
print("probabilities sum up to: " + str(probabilities[0].sum()))
print()

# we obtain a prediction by taking the class with the highest likelihood
def predict(X):
    probabilities = feed_forward(X)
    return np.argmax(probabilities, axis=1)

predictions = predict(X_train)
print("predictions = (n_inputs) = " + str(predictions.shape))
print("prediction for image 0: " + str(predictions[0]))
print("correct label for image 0: " + str(Y_train[0]))

Choose cost function and optimizer

To measure how well our neural network is doing we need to introduce a cost function. We will call the function that gives the error of a single sample output the loss function, and the function that gives the total error of our network across all samples the cost function. A typical choice for multiclass classification is the cross-entropy loss, also known as the negative log likelihood.

In multiclass classification it is common to treat each integer label as a so called one-hot vector:

$$ y = 5 \quad \rightarrow \quad \boldsymbol{y} = (0, 0, 0, 0, 0, 1, 0, 0, 0, 0) ,$$

$$ y = 1 \quad \rightarrow \quad \boldsymbol{y} = (0, 1, 0, 0, 0, 0, 0, 0, 0, 0) ,$$

i.e. a binary bit string of length \( C \), where \( C = 10 \) is the number of classes in the MNIST dataset.

Let \( y_{ic} \) denote the \( c \)-th component of the \( i \)-th one-hot vector. We define the cost function \( \mathcal{C} \) as a sum over the cross-entropy loss for each point \( \boldsymbol{x}_i \) in the dataset.

In the one-hot representation only one of the terms in the loss function is non-zero, namely the probability of the correct category \( c' \) (i.e. the category \( c' \) such that \( y_{ic'} = 1 \)). This means that the cross entropy loss only punishes you for how wrong you got the correct label. The probability of category \( c \) is given by the softmax function. The vector \( \boldsymbol{\theta} \) represents the parameters of our network, i.e. all the weights and biases.

Optimizing the cost function

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 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. Each parameter \( \theta \) is iteratively adjusted according to the rule

$$ \theta_{i+1} = \theta_i - \eta \nabla \mathcal{C}(\theta_i) ,$$

where \( \eta \) is known as the learning rate, which controls how big a step we take towards the minimum. This update can be repeated for any number of iterations, or until we are satisfied with the result.

A simple and effective improvement is a variant called Batch Gradient Descent. Instead of calculating the gradient on the whole dataset, we calculate an approximation of the gradient on a subset of the data called a minibatch. If there are \( N \) data points and we have a minibatch size of \( M \), the total number of batches is \( N/M \). We denote each minibatch \( B_k \), with \( k = 1, 2,...,N/M \). The gradient then becomes:

$$ \nabla \mathcal{C}(\theta) = \frac{1}{N} \sum_{i=1}^N \nabla \mathcal{L}_i(\theta) \quad \rightarrow \quad \frac{1}{M} \sum_{i \in B_k} \nabla \mathcal{L}_i(\theta) ,$$

i.e. instead of averaging the loss over the entire dataset, we average over a minibatch.

This has two important benefits:

  1. Introducing stochasticity decreases the chance that the algorithm becomes stuck in a local minima.
  2. It significantly speeds up the calculation, since we do not have to use the entire dataset to calculate the gradient.

The various optmization methods, with codes and algorithms, are discussed in our lectures on Gradient descent approaches.

Regularization

It is common to add an extra term to the cost function, proportional to the size of the weights. This is equivalent to constraining the size of the weights, so that they do not grow out of control. Constraining the size of the weights means that the weights cannot grow arbitrarily large to fit the training data, and in this way reduces overfitting.

We will measure the size of the weights using the so called L2-norm, meaning our cost function becomes:

$$ \mathcal{C}(\theta) = \frac{1}{N} \sum_{i=1}^N \mathcal{L}_i(\theta) \quad \rightarrow \quad \frac{1}{N} \sum_{i=1}^N \mathcal{L}_i(\theta) + \lambda \lvert \lvert \boldsymbol{w} \rvert \rvert_2^2 = \frac{1}{N} \sum_{i=1}^N \mathcal{L}(\theta) + \lambda \sum_{ij} w_{ij}^2,$$

i.e. we sum up all the weights squared. The factor \( \lambda \) is known as a regularization parameter.

In order to train the model, we need to calculate the derivative of the cost function with respect to every bias and weight in the network. In total our network has \( (64 + 1)\times 50=3250 \) weights in the hidden layer and \( (50 + 1)\times 10=510 \) weights to the output layer (\( +1 \) for the bias), and the gradient must be calculated for every parameter. We use the backpropagation algorithm discussed above. This is a clever use of the chain rule that allows us to calculate the gradient efficently.

Matrix multiplication

To more efficently train our network these equations are implemented using matrix operations. The error in the output layer is calculated simply as, with \( \boldsymbol{t} \) being our targets,

$$ \delta_L = \boldsymbol{t} - \boldsymbol{y} = (n_{inputs}, n_{categories}) .$$

The gradient for the output weights is calculated as

$$ \nabla W_{L} = \boldsymbol{a}^T \delta_L = (n_{hidden}, n_{categories}) ,$$

where \( \boldsymbol{a} = (n_{inputs}, n_{hidden}) \). This simply means that we are summing up the gradients for each input. Since we are going backwards we have to transpose the activation matrix.

The gradient with respect to the output bias is then

$$ \nabla \boldsymbol{b}_{L} = \sum_{i=1}^{n_{inputs}} \delta_L = (n_{categories}) .$$

The error in the hidden layer is

$$ \Delta_h = \delta_L W_{L}^T \circ f'(z_{h}) = \delta_L W_{L}^T \circ a_{h} \circ (1 - a_{h}) = (n_{inputs}, n_{hidden}) ,$$

where \( f'(a_{h}) \) is the derivative of the activation in the hidden layer. The matrix products mean that we are summing up the products for each neuron in the output layer. The symbol \( \circ \) denotes the Hadamard product, meaning element-wise multiplication.

This again gives us the gradients in the hidden layer:

$$ \nabla W_{h} = X^T \delta_h = (n_{features}, n_{hidden}) ,$$

$$ \nabla b_{h} = \sum_{i=1}^{n_{inputs}} \delta_h = (n_{hidden}) .$$

# to categorical turns our integer vector into a onehot representation
from sklearn.metrics import accuracy_score

# one-hot in numpy
def to_categorical_numpy(integer_vector):
    n_inputs = len(integer_vector)
    n_categories = np.max(integer_vector) + 1
    onehot_vector = np.zeros((n_inputs, n_categories))
    onehot_vector[range(n_inputs), integer_vector] = 1
    
    return onehot_vector

#Y_train_onehot, Y_test_onehot = to_categorical(Y_train), to_categorical(Y_test)
Y_train_onehot, Y_test_onehot = to_categorical_numpy(Y_train), to_categorical_numpy(Y_test)

def feed_forward_train(X):
    # weighted sum of inputs to the hidden layer
    z_h = np.matmul(X, hidden_weights) + hidden_bias
    # activation in the hidden layer
    a_h = sigmoid(z_h)
    
    # weighted sum of inputs to the output layer
    z_o = np.matmul(a_h, output_weights) + output_bias
    # softmax output
    # axis 0 holds each input and axis 1 the probabilities of each category
    exp_term = np.exp(z_o)
    probabilities = exp_term / np.sum(exp_term, axis=1, keepdims=True)
    
    # for backpropagation need activations in hidden and output layers
    return a_h, probabilities

def backpropagation(X, Y):
    a_h, probabilities = feed_forward_train(X)
    
    # error in the output layer
    error_output = probabilities - Y
    # error in the hidden layer
    error_hidden = np.matmul(error_output, output_weights.T) * a_h * (1 - a_h)
    
    # gradients for the output layer
    output_weights_gradient = np.matmul(a_h.T, error_output)
    output_bias_gradient = np.sum(error_output, axis=0)
    
    # gradient for the hidden layer
    hidden_weights_gradient = np.matmul(X.T, error_hidden)
    hidden_bias_gradient = np.sum(error_hidden, axis=0)

    return output_weights_gradient, output_bias_gradient, hidden_weights_gradient, hidden_bias_gradient

print("Old accuracy on training data: " + str(accuracy_score(predict(X_train), Y_train)))

eta = 0.01
lmbd = 0.01
for i in range(1000):
    # calculate gradients
    dWo, dBo, dWh, dBh = backpropagation(X_train, Y_train_onehot)
    
    # regularization term gradients
    dWo += lmbd * output_weights
    dWh += lmbd * hidden_weights
    
    # update weights and biases
    output_weights -= eta * dWo
    output_bias -= eta * dBo
    hidden_weights -= eta * dWh
    hidden_bias -= eta * dBh

print("New accuracy on training data: " + str(accuracy_score(predict(X_train), Y_train)))

Improving performance

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. In order to obtain a network that does something useful, we will have to do a bit more work.

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 \( \lambda = 10^{-6},...,10^{-0} \).

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 going through the entire dataset (\( n/M \) batches) an epoch.

If this does not improve network performance, you may want to consider altering the network architecture, adding more neurons or hidden layers. Andrew Ng goes through some of these considerations in this video. You can find a summary of the video here.

Full object-oriented implementation

It is very natural to think of the network as an object, with specific instances of the network 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.

class NeuralNetwork:
    def __init__(
            self,
            X_data,
            Y_data,
            n_hidden_neurons=50,
            n_categories=10,
            epochs=10,
            batch_size=100,
            eta=0.1,
            lmbd=0.0):

        self.X_data_full = X_data
        self.Y_data_full = Y_data

        self.n_inputs = X_data.shape[0]
        self.n_features = X_data.shape[1]
        self.n_hidden_neurons = n_hidden_neurons
        self.n_categories = n_categories

        self.epochs = epochs
        self.batch_size = batch_size
        self.iterations = self.n_inputs // self.batch_size
        self.eta = eta
        self.lmbd = lmbd

        self.create_biases_and_weights()

    def create_biases_and_weights(self):
        self.hidden_weights = np.random.randn(self.n_features, self.n_hidden_neurons)
        self.hidden_bias = np.zeros(self.n_hidden_neurons) + 0.01

        self.output_weights = np.random.randn(self.n_hidden_neurons, self.n_categories)
        self.output_bias = np.zeros(self.n_categories) + 0.01

    def feed_forward(self):
        # feed-forward for training
        self.z_h = np.matmul(self.X_data, self.hidden_weights) + self.hidden_bias
        self.a_h = sigmoid(self.z_h)

        self.z_o = np.matmul(self.a_h, self.output_weights) + self.output_bias

        exp_term = np.exp(self.z_o)
        self.probabilities = exp_term / np.sum(exp_term, axis=1, keepdims=True)

    def feed_forward_out(self, X):
        # feed-forward for output
        z_h = np.matmul(X, self.hidden_weights) + self.hidden_bias
        a_h = sigmoid(z_h)

        z_o = np.matmul(a_h, self.output_weights) + self.output_bias
        
        exp_term = np.exp(z_o)
        probabilities = exp_term / np.sum(exp_term, axis=1, keepdims=True)
        return probabilities

    def backpropagation(self):
        error_output = self.probabilities - self.Y_data
        error_hidden = np.matmul(error_output, self.output_weights.T) * self.a_h * (1 - self.a_h)

        self.output_weights_gradient = np.matmul(self.a_h.T, error_output)
        self.output_bias_gradient = np.sum(error_output, axis=0)

        self.hidden_weights_gradient = np.matmul(self.X_data.T, error_hidden)
        self.hidden_bias_gradient = np.sum(error_hidden, axis=0)

        if self.lmbd > 0.0:
            self.output_weights_gradient += self.lmbd * self.output_weights
            self.hidden_weights_gradient += self.lmbd * self.hidden_weights

        self.output_weights -= self.eta * self.output_weights_gradient
        self.output_bias -= self.eta * self.output_bias_gradient
        self.hidden_weights -= self.eta * self.hidden_weights_gradient
        self.hidden_bias -= self.eta * self.hidden_bias_gradient

    def predict(self, X):
        probabilities = self.feed_forward_out(X)
        return np.argmax(probabilities, axis=1)

    def predict_probabilities(self, X):
        probabilities = self.feed_forward_out(X)
        return probabilities

    def train(self):
        data_indices = np.arange(self.n_inputs)

        for i in range(self.epochs):
            for j in range(self.iterations):
                # pick datapoints with replacement
                chosen_datapoints = np.random.choice(
                    data_indices, size=self.batch_size, replace=False
                )

                # minibatch training data
                self.X_data = self.X_data_full[chosen_datapoints]
                self.Y_data = self.Y_data_full[chosen_datapoints]

                self.feed_forward()
                self.backpropagation()

Evaluate model performance on test data

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. We measure the performance of the network using the accuracy score. 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 \( 1 \).

$$ \text{Accuracy} = \frac{\sum_{i=1}^n I(\tilde{y}_i = y_i)}{n} ,$$

where \( I \) is the indicator function, \( 1 \) if \( \tilde{y}_i = y_i \) and \( 0 \) otherwise.

epochs = 100
batch_size = 100

dnn = NeuralNetwork(X_train, Y_train_onehot, eta=eta, lmbd=lmbd, epochs=epochs, batch_size=batch_size,
                    n_hidden_neurons=n_hidden_neurons, n_categories=n_categories)
dnn.train()
test_predict = dnn.predict(X_test)

# accuracy score from scikit library
print("Accuracy score on test set: ", accuracy_score(Y_test, test_predict))

# equivalent in numpy
def accuracy_score_numpy(Y_test, Y_pred):
    return np.sum(Y_test == Y_pred) / len(Y_test)

#print("Accuracy score on test set: ", accuracy_score_numpy(Y_test, test_predict))

Adjust hyperparameters

We now perform a grid search to find the optimal hyperparameters for the network. Note that we are only using 1 layer with 50 neurons, and human performance is estimated to be around \( 98\% \) (\( 2\% \) error rate).

eta_vals = np.logspace(-5, 1, 7)
lmbd_vals = np.logspace(-5, 1, 7)
# store the models for later use
DNN_numpy = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)

# grid search
for i, eta in enumerate(eta_vals):
    for j, lmbd in enumerate(lmbd_vals):
        dnn = NeuralNetwork(X_train, Y_train_onehot, eta=eta, lmbd=lmbd, epochs=epochs, batch_size=batch_size,
                            n_hidden_neurons=n_hidden_neurons, n_categories=n_categories)
        dnn.train()
        
        DNN_numpy[i][j] = dnn
        
        test_predict = dnn.predict(X_test)
        
        print("Learning rate  = ", eta)
        print("Lambda = ", lmbd)
        print("Accuracy score on test set: ", accuracy_score(Y_test, test_predict))
        print()

Visualization

# visual representation of grid search
# uses seaborn heatmap, you can also do this with matplotlib imshow
import seaborn as sns

sns.set()

train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))

for i in range(len(eta_vals)):
    for j in range(len(lmbd_vals)):
        dnn = DNN_numpy[i][j]
        
        train_pred = dnn.predict(X_train) 
        test_pred = dnn.predict(X_test)

        train_accuracy[i][j] = accuracy_score(Y_train, train_pred)
        test_accuracy[i][j] = accuracy_score(Y_test, test_pred)

        
fig, ax = plt.subplots(figsize = (10, 10))
sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis")
ax.set_title("Training Accuracy")
ax.set_ylabel("$\eta$")
ax.set_xlabel("$\lambda$")
plt.show()

fig, ax = plt.subplots(figsize = (10, 10))
sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
ax.set_title("Test Accuracy")
ax.set_ylabel("$\eta$")
ax.set_xlabel("$\lambda$")
plt.show()

scikit-learn implementation

scikit-learn focuses more on traditional machine learning methods, such as regression, clustering, decision trees, etc. As such, it has only two types of neural networks: Multi Layer Perceptron outputting continuous values, MPLRegressor, and Multi Layer Perceptron outputting labels, MLPClassifier. We will see how simple it is to use these classes.

scikit-learn implements a few improvements from our neural network, such as early stopping, a varying learning rate, different optimization methods, etc. We would therefore expect a better performance overall.

from sklearn.neural_network import MLPClassifier
# store models for later use
DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)

for i, eta in enumerate(eta_vals):
    for j, lmbd in enumerate(lmbd_vals):
        dnn = MLPClassifier(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',
                            alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
        dnn.fit(X_train, Y_train)
        
        DNN_scikit[i][j] = dnn
        
        print("Learning rate  = ", eta)
        print("Lambda = ", lmbd)
        print("Accuracy score on test set: ", dnn.score(X_test, Y_test))
        print()

Visualization

# optional
# visual representation of grid search
# uses seaborn heatmap, could probably do this in matplotlib
import seaborn as sns

sns.set()

train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))

for i in range(len(eta_vals)):
    for j in range(len(lmbd_vals)):
        dnn = DNN_scikit[i][j]
        
        train_pred = dnn.predict(X_train) 
        test_pred = dnn.predict(X_test)

        train_accuracy[i][j] = accuracy_score(Y_train, train_pred)
        test_accuracy[i][j] = accuracy_score(Y_test, test_pred)

        
fig, ax = plt.subplots(figsize = (10, 10))
sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis")
ax.set_title("Training Accuracy")
ax.set_ylabel("$\eta$")
ax.set_xlabel("$\lambda$")
plt.show()

fig, ax = plt.subplots(figsize = (10, 10))
sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
ax.set_title("Test Accuracy")
ax.set_ylabel("$\eta$")
ax.set_xlabel("$\lambda$")
plt.show()

Building neural networks in Tensorflow and Keras

Now we want to build on the experience gained from our neural network implementation in NumPy and scikit-learn and use it to construct a neural network in Tensorflow. Once we have constructed a neural network in NumPy and Tensorflow, building one in Keras is really quite trivial, though the performance may suffer.

In our previous example we used only one hidden layer, and in this we will use two. From this it should be quite clear how to build one using an arbitrary number of hidden layers, using data structures such as Python lists or NumPy arrays.

Tensorflow

Tensorflow is an open source library machine learning library developed by the Google Brain team for internal use. It was released under the Apache 2.0 open source license in November 9, 2015.

Tensorflow is a computational framework that allows you to construct machine learning models at different levels of abstraction, from high-level, object-oriented APIs like Keras, down to the C++ kernels that Tensorflow is built upon. The higher levels of abstraction are simpler to use, but less flexible, and our choice of implementation should reflect the problems we are trying to solve.

Tensorflow uses so-called graphs to represent your computation in terms of the dependencies between individual operations, such that you first build a Tensorflow graph to represent your model, and then create a Tensorflow session to run the graph.

In this guide we will analyze the same data as we did in our NumPy and scikit-learn tutorial, gathered from the MNIST database of images. We will give an introduction to the lower level Python Application Program Interfaces (APIs), and see how we use them to build our graph. Then we will build (effectively) the same graph in Keras, to see just how simple solving a machine learning problem can be.

To install tensorflow on Unix/Linux systems, use pip as

pip3 install tensorflow

and/or if you use anaconda, just write (or install from the graphical user interface) (current release of CPU-only TensorFlow)

conda create -n tf tensorflow
conda activate tf

To install the current release of GPU TensorFlow

conda create -n tf-gpu tensorflow-gpu
conda activate tf-gpu

Using Keras

Keras is a high level neural network that supports Tensorflow, CTNK and Theano as backends. If you have Anaconda installed you may run the following command

conda install keras

You can look up the instructions here for more information.

We will to a large extent use keras in this course.

Collect and pre-process data

Let us look again at the MINST data set.

# import necessary packages
import numpy as np
import matplotlib.pyplot as plt
import tensorflow as tf
from sklearn import datasets


# ensure the same random numbers appear every time
np.random.seed(0)

# display images in notebook
%matplotlib inline
plt.rcParams['figure.figsize'] = (12,12)


# download MNIST dataset
digits = datasets.load_digits()

# define inputs and labels
inputs = digits.images
labels = digits.target

print("inputs = (n_inputs, pixel_width, pixel_height) = " + str(inputs.shape))
print("labels = (n_inputs) = " + str(labels.shape))


# flatten the image
# the value -1 means dimension is inferred from the remaining dimensions: 8x8 = 64
n_inputs = len(inputs)
inputs = inputs.reshape(n_inputs, -1)
print("X = (n_inputs, n_features) = " + str(inputs.shape))


# choose some random images to display
indices = np.arange(n_inputs)
random_indices = np.random.choice(indices, size=5)

for i, image in enumerate(digits.images[random_indices]):
    plt.subplot(1, 5, i+1)
    plt.axis('off')
    plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest')
    plt.title("Label: %d" % digits.target[random_indices[i]])
plt.show()
from tensorflow.keras.layers import Input
from tensorflow.keras.models import Sequential      #This allows appending layers to existing models
from tensorflow.keras.layers import Dense           #This allows defining the characteristics of a particular layer
from tensorflow.keras import optimizers             #This allows using whichever optimiser we want (sgd,adam,RMSprop)
from tensorflow.keras import regularizers           #This allows using whichever regularizer we want (l1,l2,l1_l2)
from tensorflow.keras.utils import to_categorical   #This allows using categorical cross entropy as the cost function

from sklearn.model_selection import train_test_split

# one-hot representation of labels
labels = to_categorical(labels)

# split into train and test data
train_size = 0.8
test_size = 1 - train_size
X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size,
                                                    test_size=test_size)
epochs = 100
batch_size = 100
n_neurons_layer1 = 100
n_neurons_layer2 = 50
n_categories = 10
eta_vals = np.logspace(-5, 1, 7)
lmbd_vals = np.logspace(-5, 1, 7)
def create_neural_network_keras(n_neurons_layer1, n_neurons_layer2, n_categories, eta, lmbd):
    model = Sequential()
    model.add(Dense(n_neurons_layer1, activation='sigmoid', kernel_regularizer=regularizers.l2(lmbd)))
    model.add(Dense(n_neurons_layer2, activation='sigmoid', kernel_regularizer=regularizers.l2(lmbd)))
    model.add(Dense(n_categories, activation='softmax'))
    
    sgd = optimizers.SGD(learning_rate=eta)
    model.compile(loss='categorical_crossentropy', optimizer=sgd, metrics=['accuracy'])
    
    return model
DNN_keras = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
        
for i, eta in enumerate(eta_vals):
    for j, lmbd in enumerate(lmbd_vals):
        DNN = create_neural_network_keras(n_neurons_layer1, n_neurons_layer2, n_categories,
                                         eta=eta, lmbd=lmbd)
        DNN.fit(X_train, Y_train, epochs=epochs, batch_size=batch_size, verbose=0)
        scores = DNN.evaluate(X_test, Y_test)
        
        DNN_keras[i][j] = DNN
        
        print("Learning rate = ", eta)
        print("Lambda = ", lmbd)
        print("Test accuracy: %.3f" % scores[1])
        print()
# optional
# visual representation of grid search
# uses seaborn heatmap, could probably do this in matplotlib
import seaborn as sns

sns.set()

train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))

for i in range(len(eta_vals)):
    for j in range(len(lmbd_vals)):
        DNN = DNN_keras[i][j]

        train_accuracy[i][j] = DNN.evaluate(X_train, Y_train)[1]
        test_accuracy[i][j] = DNN.evaluate(X_test, Y_test)[1]

        
fig, ax = plt.subplots(figsize = (10, 10))
sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis")
ax.set_title("Training Accuracy")
ax.set_ylabel("$\eta$")
ax.set_xlabel("$\lambda$")
plt.show()

fig, ax = plt.subplots(figsize = (10, 10))
sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
ax.set_title("Test Accuracy")
ax.set_ylabel("$\eta$")
ax.set_xlabel("$\lambda$")
plt.show()

The Breast Cancer Data, now with Keras

import tensorflow as tf
from tensorflow.keras.layers import Input
from tensorflow.keras.models import Sequential      #This allows appending layers to existing models
from tensorflow.keras.layers import Dense           #This allows defining the characteristics of a particular layer
from tensorflow.keras import optimizers             #This allows using whichever optimiser we want (sgd,adam,RMSprop)
from tensorflow.keras import regularizers           #This allows using whichever regularizer we want (l1,l2,l1_l2)
from tensorflow.keras.utils import to_categorical   #This allows using categorical cross entropy as the cost function
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
from sklearn.model_selection import train_test_split as splitter
from sklearn.datasets import load_breast_cancer
import pickle
import os 


"""Load breast cancer dataset"""

np.random.seed(0)        #create same seed for random number every time

cancer=load_breast_cancer()      #Download breast cancer dataset

inputs=cancer.data                     #Feature matrix of 569 rows (samples) and 30 columns (parameters)
outputs=cancer.target                  #Label array of 569 rows (0 for benign and 1 for malignant)
labels=cancer.feature_names[0:30]

print('The content of the breast cancer dataset is:')      #Print information about the datasets
print(labels)
print('-------------------------')
print("inputs =  " + str(inputs.shape))
print("outputs =  " + str(outputs.shape))
print("labels =  "+ str(labels.shape))

x=inputs      #Reassign the Feature and Label matrices to other variables
y=outputs

#%% 

# Visualisation of dataset (for correlation analysis)

plt.figure()
plt.scatter(x[:,0],x[:,2],s=40,c=y,cmap=plt.cm.Spectral)
plt.xlabel('Mean radius',fontweight='bold')
plt.ylabel('Mean perimeter',fontweight='bold')
plt.show()

plt.figure()
plt.scatter(x[:,5],x[:,6],s=40,c=y, cmap=plt.cm.Spectral)
plt.xlabel('Mean compactness',fontweight='bold')
plt.ylabel('Mean concavity',fontweight='bold')
plt.show()


plt.figure()
plt.scatter(x[:,0],x[:,1],s=40,c=y,cmap=plt.cm.Spectral)
plt.xlabel('Mean radius',fontweight='bold')
plt.ylabel('Mean texture',fontweight='bold')
plt.show()

plt.figure()
plt.scatter(x[:,2],x[:,1],s=40,c=y,cmap=plt.cm.Spectral)
plt.xlabel('Mean perimeter',fontweight='bold')
plt.ylabel('Mean compactness',fontweight='bold')
plt.show()


# Generate training and testing datasets

#Select features relevant to classification (texture,perimeter,compactness and symmetery) 
#and add to input matrix

temp1=np.reshape(x[:,1],(len(x[:,1]),1))
temp2=np.reshape(x[:,2],(len(x[:,2]),1))
X=np.hstack((temp1,temp2))      
temp=np.reshape(x[:,5],(len(x[:,5]),1))
X=np.hstack((X,temp))       
temp=np.reshape(x[:,8],(len(x[:,8]),1))
X=np.hstack((X,temp))       

X_train,X_test,y_train,y_test=splitter(X,y,test_size=0.1)   #Split datasets into training and testing

y_train=to_categorical(y_train)     #Convert labels to categorical when using categorical cross entropy
y_test=to_categorical(y_test)

del temp1,temp2,temp

# %%

# Define tunable parameters"

eta=np.logspace(-3,-1,3)                    #Define vector of learning rates (parameter to SGD optimiser)
lamda=0.01                                  #Define hyperparameter
n_layers=2                                  #Define number of hidden layers in the model
n_neuron=np.logspace(0,3,4,dtype=int)       #Define number of neurons per layer
epochs=100                                   #Number of reiterations over the input data
batch_size=100                              #Number of samples per gradient update

# %%

"""Define function to return Deep Neural Network model"""

def NN_model(inputsize,n_layers,n_neuron,eta,lamda):
    model=Sequential()      
    for i in range(n_layers):       #Run loop to add hidden layers to the model
        if (i==0):                  #First layer requires input dimensions
            model.add(Dense(n_neuron,activation='relu',kernel_regularizer=regularizers.l2(lamda),input_dim=inputsize))
        else:                       #Subsequent layers are capable of automatic shape inferencing
            model.add(Dense(n_neuron,activation='relu',kernel_regularizer=regularizers.l2(lamda)))
    model.add(Dense(2,activation='softmax'))  #2 outputs - ordered and disordered (softmax for prob)
    sgd=optimizers.SGD(learning_rate=eta)
    model.compile(loss='categorical_crossentropy',optimizer=sgd,metrics=['accuracy'])
    return model

    
Train_accuracy=np.zeros((len(n_neuron),len(eta)))      #Define matrices to store accuracy scores as a function
Test_accuracy=np.zeros((len(n_neuron),len(eta)))       #of learning rate and number of hidden neurons for 

for i in range(len(n_neuron)):     #run loops over hidden neurons and learning rates to calculate 
    for j in range(len(eta)):      #accuracy scores 
        DNN_model=NN_model(X_train.shape[1],n_layers,n_neuron[i],eta[j],lamda)
        DNN_model.fit(X_train,y_train,epochs=epochs,batch_size=batch_size,verbose=1)
        Train_accuracy[i,j]=DNN_model.evaluate(X_train,y_train)[1]
        Test_accuracy[i,j]=DNN_model.evaluate(X_test,y_test)[1]
               

def plot_data(x,y,data,title=None):

    # plot results
    fontsize=16


    fig = plt.figure()
    ax = fig.add_subplot(111)
    cax = ax.matshow(data, interpolation='nearest', vmin=0, vmax=1)
    
    cbar=fig.colorbar(cax)
    cbar.ax.set_ylabel('accuracy (%)',rotation=90,fontsize=fontsize)
    cbar.set_ticks([0,.2,.4,0.6,0.8,1.0])
    cbar.set_ticklabels(['0%','20%','40%','60%','80%','100%'])

    # put text on matrix elements
    for i, x_val in enumerate(np.arange(len(x))):
        for j, y_val in enumerate(np.arange(len(y))):
            c = "${0:.1f}\\%$".format( 100*data[j,i])  
            ax.text(x_val, y_val, c, va='center', ha='center')

    # convert axis vaues to to string labels
    x=[str(i) for i in x]
    y=[str(i) for i in y]


    ax.set_xticklabels(['']+x)
    ax.set_yticklabels(['']+y)

    ax.set_xlabel('$\\mathrm{learning\\ rate}$',fontsize=fontsize)
    ax.set_ylabel('$\\mathrm{hidden\\ neurons}$',fontsize=fontsize)
    if title is not None:
        ax.set_title(title)

    plt.tight_layout()

    plt.show()
    
plot_data(eta,n_neuron,Train_accuracy, 'training')
plot_data(eta,n_neuron,Test_accuracy, 'testing')

Building a neural network code

Here we present a flexible object oriented codebase for a feed forward neural network, along with a demonstration of how to use it. Before we get into the details of the neural network, we will first present some implementations of various schedulers, cost functions and activation functions that can be used together with the neural network.

The codes here were developed by Eric Reber and Gregor Kajda during spring 2023.

Learning rate methods

The code below shows object oriented implementations of the Constant, Momentum, Adagrad, AdagradMomentum, RMS prop and Adam schedulers. All of the classes belong to the shared abstract Scheduler class, and share the update_change() and reset() methods allowing for any of the schedulers to be seamlessly used during the training stage, as will later be shown in the fit() method of the neural network. Update_change() only has one parameter, the gradient (\( δ^l_ja^{l−1}_k \)), and returns the change which will be subtracted from the weights. The reset() function takes no parameters, and resets the desired variables. For Constant and Momentum, reset does nothing.

import autograd.numpy as np

class Scheduler:
    """
    Abstract class for Schedulers
    """

    def __init__(self, eta):
        self.eta = eta

    # should be overwritten
    def update_change(self, gradient):
        raise NotImplementedError

    # overwritten if needed
    def reset(self):
        pass


class Constant(Scheduler):
    def __init__(self, eta):
        super().__init__(eta)

    def update_change(self, gradient):
        return self.eta * gradient
    
    def reset(self):
        pass


class Momentum(Scheduler):
    def __init__(self, eta: float, momentum: float):
        super().__init__(eta)
        self.momentum = momentum
        self.change = 0

    def update_change(self, gradient):
        self.change = self.momentum * self.change + self.eta * gradient
        return self.change

    def reset(self):
        pass


class Adagrad(Scheduler):
    def __init__(self, eta):
        super().__init__(eta)
        self.G_t = None

    def update_change(self, gradient):
        delta = 1e-8  # avoid division ny zero

        if self.G_t is None:
            self.G_t = np.zeros((gradient.shape[0], gradient.shape[0]))

        self.G_t += gradient @ gradient.T

        G_t_inverse = 1 / (
            delta + np.sqrt(np.reshape(np.diagonal(self.G_t), (self.G_t.shape[0], 1)))
        )
        return self.eta * gradient * G_t_inverse

    def reset(self):
        self.G_t = None


class AdagradMomentum(Scheduler):
    def __init__(self, eta, momentum):
        super().__init__(eta)
        self.G_t = None
        self.momentum = momentum
        self.change = 0

    def update_change(self, gradient):
        delta = 1e-8  # avoid division ny zero

        if self.G_t is None:
            self.G_t = np.zeros((gradient.shape[0], gradient.shape[0]))

        self.G_t += gradient @ gradient.T

        G_t_inverse = 1 / (
            delta + np.sqrt(np.reshape(np.diagonal(self.G_t), (self.G_t.shape[0], 1)))
        )
        self.change = self.change * self.momentum + self.eta * gradient * G_t_inverse
        return self.change

    def reset(self):
        self.G_t = None


class RMS_prop(Scheduler):
    def __init__(self, eta, rho):
        super().__init__(eta)
        self.rho = rho
        self.second = 0.0

    def update_change(self, gradient):
        delta = 1e-8  # avoid division ny zero
        self.second = self.rho * self.second + (1 - self.rho) * gradient * gradient
        return self.eta * gradient / (np.sqrt(self.second + delta))

    def reset(self):
        self.second = 0.0


class Adam(Scheduler):
    def __init__(self, eta, rho, rho2):
        super().__init__(eta)
        self.rho = rho
        self.rho2 = rho2
        self.moment = 0
        self.second = 0
        self.n_epochs = 1

    def update_change(self, gradient):
        delta = 1e-8  # avoid division ny zero

        self.moment = self.rho * self.moment + (1 - self.rho) * gradient
        self.second = self.rho2 * self.second + (1 - self.rho2) * gradient * gradient

        moment_corrected = self.moment / (1 - self.rho**self.n_epochs)
        second_corrected = self.second / (1 - self.rho2**self.n_epochs)

        return self.eta * moment_corrected / (np.sqrt(second_corrected + delta))

    def reset(self):
        self.n_epochs += 1
        self.moment = 0
        self.second = 0

Usage of the above learning rate schedulers

To initalize a scheduler, simply create the object and pass in the necessary parameters such as the learning rate and the momentum as shown below. As the Scheduler class is an abstract class it should not called directly, and will raise an error upon usage.

momentum_scheduler = Momentum(eta=1e-3, momentum=0.9)
adam_scheduler = Adam(eta=1e-3, rho=0.9, rho2=0.999)

Here is a small example for how a segment of code using schedulers could look. Switching out the schedulers is simple.

weights = np.ones((3,3))
print(f"Before scheduler:\n{weights=}")

epochs = 10
for e in range(epochs):
    gradient = np.random.rand(3, 3)
    change = adam_scheduler.update_change(gradient)
    weights = weights - change
    adam_scheduler.reset()

print(f"\nAfter scheduler:\n{weights=}")

Cost functions

Here we discuss cost functions that can be used when creating the neural network. Every cost function takes the target vector as its parameter, and returns a function valued only at \( x \) such that it may easily be differentiated.

import autograd.numpy as np

def CostOLS(target):
    
    def func(X):
        return (1.0 / target.shape[0]) * np.sum((target - X) ** 2)

    return func


def CostLogReg(target):

    def func(X):
        
        return -(1.0 / target.shape[0]) * np.sum(
            (target * np.log(X + 10e-10)) + ((1 - target) * np.log(1 - X + 10e-10))
        )

    return func


def CostCrossEntropy(target):
    
    def func(X):
        return -(1.0 / target.size) * np.sum(target * np.log(X + 10e-10))

    return func

Below we give a short example of how these cost function may be used to obtain results if you wish to test them out on your own using AutoGrad's automatics differentiation.

from autograd import grad

target = np.array([[1, 2, 3]]).T
a = np.array([[4, 5, 6]]).T

cost_func = CostCrossEntropy
cost_func_derivative = grad(cost_func(target))

valued_at_a = cost_func_derivative(a)
print(f"Derivative of cost function {cost_func.__name__} valued at a:\n{valued_at_a}")

Activation functions

Finally, before we look at the neural network, we will look at the activation functions which can be specified between the hidden layers and as the output function. Each function can be valued for any given vector or matrix X, and can be differentiated via derivate().

import autograd.numpy as np
from autograd import elementwise_grad

def identity(X):
    return X


def sigmoid(X):
    try:
        return 1.0 / (1 + np.exp(-X))
    except FloatingPointError:
        return np.where(X > np.zeros(X.shape), np.ones(X.shape), np.zeros(X.shape))


def softmax(X):
    X = X - np.max(X, axis=-1, keepdims=True)
    delta = 10e-10
    return np.exp(X) / (np.sum(np.exp(X), axis=-1, keepdims=True) + delta)


def RELU(X):
    return np.where(X > np.zeros(X.shape), X, np.zeros(X.shape))


def LRELU(X):
    delta = 10e-4
    return np.where(X > np.zeros(X.shape), X, delta * X)


def derivate(func):
    if func.__name__ == "RELU":

        def func(X):
            return np.where(X > 0, 1, 0)

        return func

    elif func.__name__ == "LRELU":

        def func(X):
            delta = 10e-4
            return np.where(X > 0, 1, delta)

        return func

    else:
        return elementwise_grad(func)

Below follows a short demonstration of how to use an activation function. The derivative of the activation function will be important when calculating the output delta term during backpropagation. Note that derivate() can also be used for cost functions for a more generalized approach.

z = np.array([[4, 5, 6]]).T
print(f"Input to activation function:\n{z}")

act_func = sigmoid
a = act_func(z)
print(f"\nOutput from {act_func.__name__} activation function:\n{a}")

act_func_derivative = derivate(act_func)
valued_at_z = act_func_derivative(a)
print(f"\nDerivative of {act_func.__name__} activation function valued at z:\n{valued_at_z}")

The Neural Network

Now that we have gotten a good understanding of the implementation of some important components, we can take a look at an object oriented implementation of a feed forward neural network. The feed forward neural network has been implemented as a class named FFNN, which can be initiated as a regressor or classifier dependant on the choice of cost function. The FFNN can have any number of input nodes, hidden layers with any amount of hidden nodes, and any amount of output nodes meaning it can perform multiclass classification as well as binary classification and regression problems. Although there is a lot of code present, it makes for an easy to use and generalizeable interface for creating many types of neural networks as will be demonstrated below.

import math
import autograd.numpy as np
import sys
import warnings
from autograd import grad, elementwise_grad
from random import random, seed
from copy import deepcopy, copy
from typing import Tuple, Callable
from sklearn.utils import resample

warnings.simplefilter("error")


class FFNN:
    """
    Description:
    ------------
        Feed Forward Neural Network with interface enabling flexible design of a
        nerual networks architecture and the specification of activation function
        in the hidden layers and output layer respectively. This model can be used
        for both regression and classification problems, depending on the output function.

    Attributes:
    ------------
        I   dimensions (tuple[int]): A list of positive integers, which specifies the
            number of nodes in each of the networks layers. The first integer in the array
            defines the number of nodes in the input layer, the second integer defines number
            of nodes in the first hidden layer and so on until the last number, which
            specifies the number of nodes in the output layer.
        II  hidden_func (Callable): The activation function for the hidden layers
        III output_func (Callable): The activation function for the output layer
        IV  cost_func (Callable): Our cost function
        V   seed (int): Sets random seed, makes results reproducible
    """

    def __init__(
        self,
        dimensions: tuple[int],
        hidden_func: Callable = sigmoid,
        output_func: Callable = lambda x: x,
        cost_func: Callable = CostOLS,
        seed: int = None,
    ):
        self.dimensions = dimensions
        self.hidden_func = hidden_func
        self.output_func = output_func
        self.cost_func = cost_func
        self.seed = seed
        self.weights = list()
        self.schedulers_weight = list()
        self.schedulers_bias = list()
        self.a_matrices = list()
        self.z_matrices = list()
        self.classification = None

        self.reset_weights()
        self._set_classification()

    def fit(
        self,
        X: np.ndarray,
        t: np.ndarray,
        scheduler: Scheduler,
        batches: int = 1,
        epochs: int = 100,
        lam: float = 0,
        X_val: np.ndarray = None,
        t_val: np.ndarray = None,
    ):
        """
        Description:
        ------------
            This function performs the training the neural network by performing the feedforward and backpropagation
            algorithm to update the networks weights.

        Parameters:
        ------------
            I    X (np.ndarray) : training data
            II   t (np.ndarray) : target data
            III  scheduler (Scheduler) : specified scheduler (algorithm for optimization of gradient descent)
            IV   scheduler_args (list[int]) : list of all arguments necessary for scheduler

        Optional Parameters:
        ------------
            V    batches (int) : number of batches the datasets are split into, default equal to 1
            VI   epochs (int) : number of iterations used to train the network, default equal to 100
            VII  lam (float) : regularization hyperparameter lambda
            VIII X_val (np.ndarray) : validation set
            IX   t_val (np.ndarray) : validation target set

        Returns:
        ------------
            I   scores (dict) : A dictionary containing the performance metrics of the model.
                The number of the metrics depends on the parameters passed to the fit-function.

        """

        # setup 
        if self.seed is not None:
            np.random.seed(self.seed)

        val_set = False
        if X_val is not None and t_val is not None:
            val_set = True

        # creating arrays for score metrics
        train_errors = np.empty(epochs)
        train_errors.fill(np.nan)
        val_errors = np.empty(epochs)
        val_errors.fill(np.nan)

        train_accs = np.empty(epochs)
        train_accs.fill(np.nan)
        val_accs = np.empty(epochs)
        val_accs.fill(np.nan)

        self.schedulers_weight = list()
        self.schedulers_bias = list()

        batch_size = X.shape[0] // batches

        X, t = resample(X, t)

        # this function returns a function valued only at X
        cost_function_train = self.cost_func(t)
        if val_set:
            cost_function_val = self.cost_func(t_val)

        # create schedulers for each weight matrix
        for i in range(len(self.weights)):
            self.schedulers_weight.append(copy(scheduler))
            self.schedulers_bias.append(copy(scheduler))

        print(f"{scheduler.__class__.__name__}: Eta={scheduler.eta}, Lambda={lam}")

        try:
            for e in range(epochs):
                for i in range(batches):
                    # allows for minibatch gradient descent
                    if i == batches - 1:
                        # If the for loop has reached the last batch, take all thats left
                        X_batch = X[i * batch_size :, :]
                        t_batch = t[i * batch_size :, :]
                    else:
                        X_batch = X[i * batch_size : (i + 1) * batch_size, :]
                        t_batch = t[i * batch_size : (i + 1) * batch_size, :]

                    self._feedforward(X_batch)
                    self._backpropagate(X_batch, t_batch, lam)

                # reset schedulers for each epoch (some schedulers pass in this call)
                for scheduler in self.schedulers_weight:
                    scheduler.reset()

                for scheduler in self.schedulers_bias:
                    scheduler.reset()

                # computing performance metrics
                pred_train = self.predict(X)
                train_error = cost_function_train(pred_train)

                train_errors[e] = train_error
                if val_set:
                    
                    pred_val = self.predict(X_val)
                    val_error = cost_function_val(pred_val)
                    val_errors[e] = val_error

                if self.classification:
                    train_acc = self._accuracy(self.predict(X), t)
                    train_accs[e] = train_acc
                    if val_set:
                        val_acc = self._accuracy(pred_val, t_val)
                        val_accs[e] = val_acc

                # printing progress bar
                progression = e / epochs
                print_length = self._progress_bar(
                    progression,
                    train_error=train_errors[e],
                    train_acc=train_accs[e],
                    val_error=val_errors[e],
                    val_acc=val_accs[e],
                )
        except KeyboardInterrupt:
            # allows for stopping training at any point and seeing the result
            pass

        # visualization of training progression (similiar to tensorflow progression bar)
        sys.stdout.write("\r" + " " * print_length)
        sys.stdout.flush()
        self._progress_bar(
            1,
            train_error=train_errors[e],
            train_acc=train_accs[e],
            val_error=val_errors[e],
            val_acc=val_accs[e],
        )
        sys.stdout.write("")

        # return performance metrics for the entire run
        scores = dict()

        scores["train_errors"] = train_errors

        if val_set:
            scores["val_errors"] = val_errors

        if self.classification:
            scores["train_accs"] = train_accs

            if val_set:
                scores["val_accs"] = val_accs

        return scores

    def predict(self, X: np.ndarray, *, threshold=0.5):
        """
         Description:
         ------------
             Performs prediction after training of the network has been finished.

         Parameters:
        ------------
             I   X (np.ndarray): The design matrix, with n rows of p features each

         Optional Parameters:
         ------------
             II  threshold (float) : sets minimal value for a prediction to be predicted as the positive class
                 in classification problems

         Returns:
         ------------
             I   z (np.ndarray): A prediction vector (row) for each row in our design matrix
                 This vector is thresholded if regression=False, meaning that classification results
                 in a vector of 1s and 0s, while regressions in an array of decimal numbers

        """

        predict = self._feedforward(X)

        if self.classification:
            return np.where(predict > threshold, 1, 0)
        else:
            return predict

    def reset_weights(self):
        """
        Description:
        ------------
            Resets/Reinitializes the weights in order to train the network for a new problem.

        """
        if self.seed is not None:
            np.random.seed(self.seed)

        self.weights = list()
        for i in range(len(self.dimensions) - 1):
            weight_array = np.random.randn(
                self.dimensions[i] + 1, self.dimensions[i + 1]
            )
            weight_array[0, :] = np.random.randn(self.dimensions[i + 1]) * 0.01

            self.weights.append(weight_array)

    def _feedforward(self, X: np.ndarray):
        """
        Description:
        ------------
            Calculates the activation of each layer starting at the input and ending at the output.
            Each following activation is calculated from a weighted sum of each of the preceeding
            activations (except in the case of the input layer).

        Parameters:
        ------------
            I   X (np.ndarray): The design matrix, with n rows of p features each

        Returns:
        ------------
            I   z (np.ndarray): A prediction vector (row) for each row in our design matrix
        """

        # reset matrices
        self.a_matrices = list()
        self.z_matrices = list()

        # if X is just a vector, make it into a matrix
        if len(X.shape) == 1:
            X = X.reshape((1, X.shape[0]))

        # Add a coloumn of zeros as the first coloumn of the design matrix, in order
        # to add bias to our data
        bias = np.ones((X.shape[0], 1)) * 0.01
        X = np.hstack([bias, X])

        # a^0, the nodes in the input layer (one a^0 for each row in X - where the
        # exponent indicates layer number).
        a = X
        self.a_matrices.append(a)
        self.z_matrices.append(a)

        # The feed forward algorithm
        for i in range(len(self.weights)):
            if i < len(self.weights) - 1:
                z = a @ self.weights[i]
                self.z_matrices.append(z)
                a = self.hidden_func(z)
                # bias column again added to the data here
                bias = np.ones((a.shape[0], 1)) * 0.01
                a = np.hstack([bias, a])
                self.a_matrices.append(a)
            else:
                try:
                    # a^L, the nodes in our output layers
                    z = a @ self.weights[i]
                    a = self.output_func(z)
                    self.a_matrices.append(a)
                    self.z_matrices.append(z)
                except Exception as OverflowError:
                    print(
                        "OverflowError in fit() in FFNN\nHOW TO DEBUG ERROR: Consider lowering your learning rate or scheduler specific parameters such as momentum, or check if your input values need scaling"
                    )

        # this will be a^L
        return a

    def _backpropagate(self, X, t, lam):
        """
        Description:
        ------------
            Performs the backpropagation algorithm. In other words, this method
            calculates the gradient of all the layers starting at the
            output layer, and moving from right to left accumulates the gradient until
            the input layer is reached. Each layers respective weights are updated while
            the algorithm propagates backwards from the output layer (auto-differentation in reverse mode).

        Parameters:
        ------------
            I   X (np.ndarray): The design matrix, with n rows of p features each.
            II  t (np.ndarray): The target vector, with n rows of p targets.
            III lam (float32): regularization parameter used to punish the weights in case of overfitting

        Returns:
        ------------
            No return value.

        """
        out_derivative = derivate(self.output_func)
        hidden_derivative = derivate(self.hidden_func)

        for i in range(len(self.weights) - 1, -1, -1):
            # delta terms for output
            if i == len(self.weights) - 1:
                # for multi-class classification
                if (
                    self.output_func.__name__ == "softmax"
                ):
                    delta_matrix = self.a_matrices[i + 1] - t
                # for single class classification
                else:
                    cost_func_derivative = grad(self.cost_func(t))
                    delta_matrix = out_derivative(
                        self.z_matrices[i + 1]
                    ) * cost_func_derivative(self.a_matrices[i + 1])

            # delta terms for hidden layer
            else:
                delta_matrix = (
                    self.weights[i + 1][1:, :] @ delta_matrix.T
                ).T * hidden_derivative(self.z_matrices[i + 1])

            # calculate gradient
            gradient_weights = self.a_matrices[i][:, 1:].T @ delta_matrix
            gradient_bias = np.sum(delta_matrix, axis=0).reshape(
                1, delta_matrix.shape[1]
            )

            # regularization term
            gradient_weights += self.weights[i][1:, :] * lam

            # use scheduler
            update_matrix = np.vstack(
                [
                    self.schedulers_bias[i].update_change(gradient_bias),
                    self.schedulers_weight[i].update_change(gradient_weights),
                ]
            )

            # update weights and bias
            self.weights[i] -= update_matrix

    def _accuracy(self, prediction: np.ndarray, target: np.ndarray):
        """
        Description:
        ------------
            Calculates accuracy of given prediction to target

        Parameters:
        ------------
            I   prediction (np.ndarray): vector of predicitons output network
                (1s and 0s in case of classification, and real numbers in case of regression)
            II  target (np.ndarray): vector of true values (What the network ideally should predict)

        Returns:
        ------------
            A floating point number representing the percentage of correctly classified instances.
        """
        assert prediction.size == target.size
        return np.average((target == prediction))
    def _set_classification(self):
        """
        Description:
        ------------
            Decides if FFNN acts as classifier (True) og regressor (False),
            sets self.classification during init()
        """
        self.classification = False
        if (
            self.cost_func.__name__ == "CostLogReg"
            or self.cost_func.__name__ == "CostCrossEntropy"
        ):
            self.classification = True

    def _progress_bar(self, progression, **kwargs):
        """
        Description:
        ------------
            Displays progress of training
        """
        print_length = 40
        num_equals = int(progression * print_length)
        num_not = print_length - num_equals
        arrow = ">" if num_equals > 0 else ""
        bar = "[" + "=" * (num_equals - 1) + arrow + "-" * num_not + "]"
        perc_print = self._format(progression * 100, decimals=5)
        line = f"  {bar} {perc_print}% "

        for key in kwargs:
            if not np.isnan(kwargs[key]):
                value = self._format(kwargs[key], decimals=4)
                line += f"| {key}: {value} "
        sys.stdout.write("\r" + line)
        sys.stdout.flush()
        return len(line)

    def _format(self, value, decimals=4):
        """
        Description:
        ------------
            Formats decimal numbers for progress bar
        """
        if value > 0:
            v = value
        elif value < 0:
            v = -10 * value
        else:
            v = 1
        n = 1 + math.floor(math.log10(v))
        if n >= decimals - 1:
            return str(round(value))
        return f"{value:.{decimals-n-1}f}"

Before we make a model, we will quickly generate a dataset we can use for our linear regression problem as shown below

import autograd.numpy as np
from sklearn.model_selection import train_test_split

def SkrankeFunction(x, y):
    return np.ravel(0 + 1*x + 2*y + 3*x**2 + 4*x*y + 5*y**2)

def create_X(x, y, n):
    if len(x.shape) > 1:
        x = np.ravel(x)
        y = np.ravel(y)

    N = len(x)
    l = int((n + 1) * (n + 2) / 2)  # Number of elements in beta
    X = np.ones((N, l))

    for i in range(1, n + 1):
        q = int((i) * (i + 1) / 2)
        for k in range(i + 1):
            X[:, q + k] = (x ** (i - k)) * (y**k)

    return X

step=0.5
x = np.arange(0, 1, step)
y = np.arange(0, 1, step)
x, y = np.meshgrid(x, y)
target = SkrankeFunction(x, y)
target = target.reshape(target.shape[0], 1)

poly_degree=3
X = create_X(x, y, poly_degree)

X_train, X_test, t_train, t_test = train_test_split(X, target)

Now that we have our dataset ready for the regression, we can create our regressor. Note that with the seed parameter, we can make sure our results stay the same every time we run the neural network. For inititialization, we simply specify the dimensions (we wish the amount of input nodes to be equal to the datapoints, and the output to predict one value).

input_nodes = X_train.shape[1]
output_nodes = 1

linear_regression = FFNN((input_nodes, output_nodes), output_func=identity, cost_func=CostOLS, seed=2023)

We then fit our model with our training data using the scheduler of our choice.

linear_regression.reset_weights() # reset weights such that previous runs or reruns don't affect the weights

scheduler = Constant(eta=1e-3)
scores = linear_regression.fit(X_train, t_train, scheduler)

Due to the progress bar we can see the MSE (train_error) throughout the FFNN's training. Note that the fit() function has some optional parameters with defualt arguments. For example, the regularization hyperparameter can be left ignored if not needed, and equally the FFNN will by default run for 100 epochs. These can easily be changed, such as for example:

linear_regression.reset_weights() # reset weights such that previous runs or reruns don't affect the weights

scores = linear_regression.fit(X_train, t_train, scheduler, lam=1e-4, epochs=1000)

We see that given more epochs to train on, the regressor reaches a lower MSE.

Let us then switch to a binary classification. We use a binary classification dataset, and follow a similar setup to the regression case.

from sklearn.datasets import load_breast_cancer
from sklearn.preprocessing import MinMaxScaler

wisconsin = load_breast_cancer()
X = wisconsin.data
target = wisconsin.target
target = target.reshape(target.shape[0], 1)

X_train, X_val, t_train, t_val = train_test_split(X, target)

scaler = MinMaxScaler()
scaler.fit(X_train)
X_train = scaler.transform(X_train)
X_val = scaler.transform(X_val)
input_nodes = X_train.shape[1]
output_nodes = 1

logistic_regression = FFNN((input_nodes, output_nodes), output_func=sigmoid, cost_func=CostLogReg, seed=2023)

We will now make use of our validation data by passing it into our fit function as a keyword argument

logistic_regression.reset_weights() # reset weights such that previous runs or reruns don't affect the weights

scheduler = Adam(eta=1e-3, rho=0.9, rho2=0.999)
scores = logistic_regression.fit(X_train, t_train, scheduler, epochs=1000, X_val=X_val, t_val=t_val)

Finally, we will create a neural network with 2 hidden layers with activation functions.

input_nodes = X_train.shape[1]
hidden_nodes1 = 100
hidden_nodes2 = 30
output_nodes = 1

dims = (input_nodes, hidden_nodes1, hidden_nodes2, output_nodes)

neural_network = FFNN(dims, hidden_func=RELU, output_func=sigmoid, cost_func=CostLogReg, seed=2023)
neural_network.reset_weights() # reset weights such that previous runs or reruns don't affect the weights

scheduler = Adam(eta=1e-4, rho=0.9, rho2=0.999)
scores = neural_network.fit(X_train, t_train, scheduler, epochs=1000, X_val=X_val, t_val=t_val)

Multiclass classification

Finally, we will demonstrate the use case of multiclass classification using our FFNN with the famous MNIST dataset, which contain images of digits between the range of 0 to 9.

from sklearn.datasets import load_digits

def onehot(target: np.ndarray):
    onehot = np.zeros((target.size, target.max() + 1))
    onehot[np.arange(target.size), target] = 1
    return onehot

digits = load_digits()

X = digits.data
target = digits.target
target = onehot(target)

input_nodes = 64
hidden_nodes1 = 100
hidden_nodes2 = 30
output_nodes = 10

dims = (input_nodes, hidden_nodes1, hidden_nodes2, output_nodes)

multiclass = FFNN(dims, hidden_func=LRELU, output_func=softmax, cost_func=CostCrossEntropy)

multiclass.reset_weights() # reset weights such that previous runs or reruns don't affect the weights

scheduler = Adam(eta=1e-4, rho=0.9, rho2=0.999)
scores = multiclass.fit(X, target, scheduler, epochs=1000)

Testing the XOR gate and other gates

Let us now use our code to test the XOR gate.

X = np.array([ [0, 0], [0, 1], [1, 0],[1, 1]],dtype=np.float64)

# The XOR gate
yXOR = np.array( [[ 0], [1] ,[1], [0]])

input_nodes = X.shape[1]
output_nodes = 1

logistic_regression = FFNN((input_nodes, output_nodes), output_func=sigmoid, cost_func=CostLogReg, seed=2023)
logistic_regression.reset_weights() # reset weights such that previous runs or reruns don't affect the weights
scheduler = Adam(eta=1e-1, rho=0.9, rho2=0.999)
scores = logistic_regression.fit(X, yXOR, scheduler, epochs=1000)

Not bad, but the results depend strongly on the learning reate. Try different learning rates.

Solving differential equations with Deep Learning

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

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

Ordinary Differential Equations first

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

In general, an ordinary differential equation looks like

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

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

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

The trial solution

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

$$ \begin{equation} g_t(x) = h_1(x) + h_2(x,N(x,P)) \label{_auto1} \end{equation} $$

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

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

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

Minimization process

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

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

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

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

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

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

Minimizing the cost function using gradient descent and automatic differentiation

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

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

Example: Exponential decay

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

$$ \begin{equation} \label{solve_expdec} g'(x) = -\gamma g(x) \end{equation} $$

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

The analytical solution of \eqref{solve_expdec} is

$$ \begin{equation} g(x) = g_0 \exp\left(-\gamma x\right) \label{_auto2} \end{equation} $$

Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of \eqref{solve_expdec}.

The function to solve for

The program will use a neural network to solve

$$ \begin{equation} \label{solveode} g'(x) = -\gamma g(x) \end{equation} $$

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

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

The trial solution

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

$$ g_t(x, P) = h_1(x) + h_2(x, N(x, P)) $$

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

Setup of Network

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

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

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

It is given that \( g(0) = g_0 \). The trial solution must fulfill this condition to be a proper solution of \eqref{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:

$$ \begin{equation} \label{trial} g_t(x, P) = g_0 + x \cdot N(x, P) \end{equation} $$

Reformulating the problem

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

A reformulation of out equation, \eqref{solveode}, must therefore be done, such that it describes the problem a neural network can solve for.

The neural network must find the set of weights and biases \( P \) such that the trial solution in \eqref{trial} satisfies \eqref{solveode}.

The trial solution

$$ g_t(x, P) = g_0 + x \cdot N(x, P) $$

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

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

is fulfilled as best as possible.

More technicalities

The left hand side and right hand side of \eqref{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. This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to \( P \) of the neural network.

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

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

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

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

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

for an input value \( x \).

More details

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

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

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

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

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

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

A possible implementation of a neural network

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 \eqref{min}.

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

Technicalities

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

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

Final technicalities I

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

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

Final technicalities II

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 \eqref{min}.

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

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

$$ f(z) = \frac{1}{1 + \exp{(-z)}} $$

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

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

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

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

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

Final technicalities III

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.

$$ \begin{aligned} z_{1,j}^{\text{output}} & = \begin{pmatrix} b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}} \end{pmatrix} \begin{pmatrix} 1 \\ \boldsymbol{x}_j^{\text{hidden}} \end{pmatrix} \end{aligned} $$

Final technicalities IV

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

$$ \boldsymbol{z}_{1}^{\text{output}} = \begin{pmatrix} b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}} \end{pmatrix} \begin{pmatrix} 1 & 1 & \dots & 1 \\ \boldsymbol{x}_1^{\text{hidden}} & \boldsymbol{x}_2^{\text{hidden}} & \dots & \boldsymbol{x}_N^{\text{hidden}} \end{pmatrix} $$

In 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.

Back propagation

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

The chosen cost function for this problem is

$$ C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 $$

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

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

Gradient descent

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

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

$$ \boldsymbol{\omega}_{\text{new} } = \boldsymbol{\omega} - \lambda \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega}) $$

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

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

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

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

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

The code for solving the ODE

import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt

def sigmoid(z):
    return 1/(1 + np.exp(-z))

# Assuming one input, hidden, and output layer
def neural_network(params, x):

    # Find the weights (including and biases) for the hidden and output layer.
    # Assume that params is a list of parameters for each layer.
    # The biases are the first element for each array in params,
    # and the weights are the remaning elements in each array in params.

    w_hidden = params[0]
    w_output = params[1]

    # Assumes input x being an one-dimensional array
    num_values = np.size(x)
    x = x.reshape(-1, num_values)

    # Assume that the input layer does nothing to the input x
    x_input = x

    ## Hidden layer:

    # Add a row of ones to include bias
    x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0)

    z_hidden = np.matmul(w_hidden, x_input)
    x_hidden = sigmoid(z_hidden)

    ## Output layer:

    # Include bias:
    x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0)

    z_output = np.matmul(w_output, x_hidden)
    x_output = z_output

    return x_output

# The trial solution using the deep neural network:
def g_trial(x,params, g0 = 10):
    return g0 + x*neural_network(params,x)

# The right side of the ODE:
def g(x, g_trial, gamma = 2):
    return -gamma*g_trial

# The cost function:
def cost_function(P, x):

    # Evaluate the trial function with the current parameters P
    g_t = g_trial(x,P)

    # Find the derivative w.r.t x of the neural network
    d_net_out = elementwise_grad(neural_network,1)(P,x)

    # Find the derivative w.r.t x of the trial function
    d_g_t = elementwise_grad(g_trial,0)(x,P)

    # The right side of the ODE
    func = g(x, g_t)

    err_sqr = (d_g_t - func)**2
    cost_sum = np.sum(err_sqr)

    return cost_sum / np.size(err_sqr)

# Solve the exponential decay ODE using neural network with one input, hidden, and output layer
def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb):
    ## Set up initial weights and biases

    # For the hidden layer
    p0 = npr.randn(num_neurons_hidden, 2 )

    # For the output layer
    p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included

    P = [p0, p1]

    print('Initial cost: %g'%cost_function(P, x))

    ## Start finding the optimal weights using gradient descent

    # Find the Python function that represents the gradient of the cost function
    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
    cost_function_grad = grad(cost_function,0)

    # Let the update be done num_iter times
    for i in range(num_iter):
        # Evaluate the gradient at the current weights and biases in P.
        # The cost_grad consist now of two arrays;
        # one for the gradient w.r.t P_hidden and
        # one for the gradient w.r.t P_output
        cost_grad =  cost_function_grad(P, x)

        P[0] = P[0] - lmb * cost_grad[0]
        P[1] = P[1] - lmb * cost_grad[1]

    print('Final cost: %g'%cost_function(P, x))

    return P

def g_analytic(x, gamma = 2, g0 = 10):
    return g0*np.exp(-gamma*x)

# Solve the given problem
if __name__ == '__main__':
    # Set seed such that the weight are initialized
    # with same weights and biases for every run.
    npr.seed(15)

    ## Decide the vales of arguments to the function to solve
    N = 10
    x = np.linspace(0, 1, N)

    ## Set up the initial parameters
    num_hidden_neurons = 10
    num_iter = 10000
    lmb = 0.001

    # Use the network
    P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb)

    # Print the deviation from the trial solution and true solution
    res = g_trial(x,P)
    res_analytical = g_analytic(x)

    print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical)))

    # Plot the results
    plt.figure(figsize=(10,10))

    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
    plt.plot(x, res_analytical)
    plt.plot(x, res[0,:])
    plt.legend(['analytical','nn'])
    plt.xlabel('x')
    plt.ylabel('g(x)')
    plt.show()

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

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

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

import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt

def sigmoid(z):
    return 1/(1 + np.exp(-z))

# The neural network with one input layer and one output layer,
# but with number of hidden layers specified by the user.
def deep_neural_network(deep_params, x):
    # N_hidden is the number of hidden layers
    # deep_params is a list, len() should be used
    N_hidden = len(deep_params) - 1 # -1 since params consists of
                                        # parameters to all the hidden
                                        # layers AND the output layer.

    # Assumes input x being an one-dimensional array
    num_values = np.size(x)
    x = x.reshape(-1, num_values)

    # Assume that the input layer does nothing to the input x
    x_input = x

    # Due to multiple hidden layers, define a variable referencing to the
    # output of the previous layer:
    x_prev = x_input

    ## Hidden layers:

    for l in range(N_hidden):
        # From the list of parameters P; find the correct weigths and bias for this layer
        w_hidden = deep_params[l]

        # Add a row of ones to include bias
        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)

        z_hidden = np.matmul(w_hidden, x_prev)
        x_hidden = sigmoid(z_hidden)

        # Update x_prev such that next layer can use the output from this layer
        x_prev = x_hidden

    ## Output layer:

    # Get the weights and bias for this layer
    w_output = deep_params[-1]

    # Include bias:
    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)

    z_output = np.matmul(w_output, x_prev)
    x_output = z_output

    return x_output

# The trial solution using the deep neural network:
def g_trial_deep(x,params, g0 = 10):
    return g0 + x*deep_neural_network(params, x)

# The right side of the ODE:
def g(x, g_trial, gamma = 2):
    return -gamma*g_trial

# The same cost function as before, but calls deep_neural_network instead.
def cost_function_deep(P, x):

    # Evaluate the trial function with the current parameters P
    g_t = g_trial_deep(x,P)

    # Find the derivative w.r.t x of the neural network
    d_net_out = elementwise_grad(deep_neural_network,1)(P,x)

    # Find the derivative w.r.t x of the trial function
    d_g_t = elementwise_grad(g_trial_deep,0)(x,P)

    # The right side of the ODE
    func = g(x, g_t)

    err_sqr = (d_g_t - func)**2
    cost_sum = np.sum(err_sqr)

    return cost_sum / np.size(err_sqr)

# Solve the exponential decay ODE using neural network with one input and one output layer,
# but with specified number of hidden layers from the user.
def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
    # num_hidden_neurons is now a list of number of neurons within each hidden layer

    # The number of elements in the list num_hidden_neurons thus represents
    # the number of hidden layers.

    # Find the number of hidden layers:
    N_hidden = np.size(num_neurons)

    ## Set up initial weights and biases

    # Initialize the list of parameters:
    P = [None]*(N_hidden + 1) # + 1 to include the output layer

    P[0] = npr.randn(num_neurons[0], 2 )
    for l in range(1,N_hidden):
        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias

    # For the output layer
    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included

    print('Initial cost: %g'%cost_function_deep(P, x))

    ## Start finding the optimal weights using gradient descent

    # Find the Python function that represents the gradient of the cost function
    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
    cost_function_deep_grad = grad(cost_function_deep,0)

    # Let the update be done num_iter times
    for i in range(num_iter):
        # Evaluate the gradient at the current weights and biases in P.
        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
        # in the hidden layers and output layers evaluated at x.
        cost_deep_grad =  cost_function_deep_grad(P, x)

        for l in range(N_hidden+1):
            P[l] = P[l] - lmb * cost_deep_grad[l]

    print('Final cost: %g'%cost_function_deep(P, x))

    return P

def g_analytic(x, gamma = 2, g0 = 10):
    return g0*np.exp(-gamma*x)

# Solve the given problem
if __name__ == '__main__':
    npr.seed(15)

    ## Decide the vales of arguments to the function to solve
    N = 10
    x = np.linspace(0, 1, N)

    ## Set up the initial parameters
    num_hidden_neurons = np.array([10,10])
    num_iter = 10000
    lmb = 0.001

    P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)

    res = g_trial_deep(x,P)
    res_analytical = g_analytic(x)

    plt.figure(figsize=(10,10))

    plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution')
    plt.plot(x, res_analytical)
    plt.plot(x, res[0,:])
    plt.legend(['analytical','dnn'])
    plt.ylabel('g(x)')
    plt.show()

Example: Population growth

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

$$ \begin{equation} \label{log} g'(t) = \alpha g(t)(A - g(t)) \end{equation} $$

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

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

Setting up the problem

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

$$ \begin{equation} \label{solveode_population} g'(t) = \alpha g(t)(A - g(t)) \end{equation} $$

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

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

The trial solution

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

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

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

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

The analytical solution is

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

The program using Autograd

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

import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt

def sigmoid(z):
    return 1/(1 + np.exp(-z))

# Function to get the parameters.
# Done such that one can easily change the paramaters after one's liking.
def get_parameters():
    alpha = 2
    A = 1
    g0 = 1.2
    return alpha, A, g0

def deep_neural_network(deep_params, x):
    # N_hidden is the number of hidden layers
    # deep_params is a list, len() should be used
    N_hidden = len(deep_params) - 1 # -1 since params consists of
                                        # parameters to all the hidden
                                        # layers AND the output layer.

    # Assumes input x being an one-dimensional array
    num_values = np.size(x)
    x = x.reshape(-1, num_values)

    # Assume that the input layer does nothing to the input x
    x_input = x

    # Due to multiple hidden layers, define a variable referencing to the
    # output of the previous layer:
    x_prev = x_input

    ## Hidden layers:

    for l in range(N_hidden):
        # From the list of parameters P; find the correct weigths and bias for this layer
        w_hidden = deep_params[l]

        # Add a row of ones to include bias
        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)

        z_hidden = np.matmul(w_hidden, x_prev)
        x_hidden = sigmoid(z_hidden)

        # Update x_prev such that next layer can use the output from this layer
        x_prev = x_hidden

    ## Output layer:

    # Get the weights and bias for this layer
    w_output = deep_params[-1]

    # Include bias:
    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)

    z_output = np.matmul(w_output, x_prev)
    x_output = z_output

    return x_output




def cost_function_deep(P, x):

    # Evaluate the trial function with the current parameters P
    g_t = g_trial_deep(x,P)

    # Find the derivative w.r.t x of the trial function
    d_g_t = elementwise_grad(g_trial_deep,0)(x,P)

    # The right side of the ODE
    func = f(x, g_t)

    err_sqr = (d_g_t - func)**2
    cost_sum = np.sum(err_sqr)

    return cost_sum / np.size(err_sqr)

# The right side of the ODE:
def f(x, g_trial):
    alpha,A, g0 = get_parameters()
    return alpha*g_trial*(A - g_trial)

# The trial solution using the deep neural network:
def g_trial_deep(x, params):
    alpha,A, g0 = get_parameters()
    return g0 + x*deep_neural_network(params,x)

# The analytical solution:
def g_analytic(t):
    alpha,A, g0 = get_parameters()
    return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t))

def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
    # num_hidden_neurons is now a list of number of neurons within each hidden layer

    # Find the number of hidden layers:
    N_hidden = np.size(num_neurons)

    ## Set up initial weigths and biases

    # Initialize the list of parameters:
    P = [None]*(N_hidden + 1) # + 1 to include the output layer

    P[0] = npr.randn(num_neurons[0], 2 )
    for l in range(1,N_hidden):
        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias

    # For the output layer
    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included

    print('Initial cost: %g'%cost_function_deep(P, x))

    ## Start finding the optimal weigths using gradient descent

    # Find the Python function that represents the gradient of the cost function
    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
    cost_function_deep_grad = grad(cost_function_deep,0)

    # Let the update be done num_iter times
    for i in range(num_iter):
        # Evaluate the gradient at the current weights and biases in P.
        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
        # in the hidden layers and output layers evaluated at x.
        cost_deep_grad =  cost_function_deep_grad(P, x)

        for l in range(N_hidden+1):
            P[l] = P[l] - lmb * cost_deep_grad[l]

    print('Final cost: %g'%cost_function_deep(P, x))

    return P

if __name__ == '__main__':
    npr.seed(4155)

    ## Decide the vales of arguments to the function to solve
    Nt = 10
    T = 1
    t = np.linspace(0,T, Nt)

    ## Set up the initial parameters
    num_hidden_neurons = [100, 50, 25]
    num_iter = 1000
    lmb = 1e-3

    P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)

    g_dnn_ag = g_trial_deep(t,P)
    g_analytical = g_analytic(t)

    # Find the maximum absolute difference between the solutons:
    diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))
    print("The max absolute difference between the solutions is: %g"%diff_ag)

    plt.figure(figsize=(10,10))

    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
    plt.plot(t, g_analytical)
    plt.plot(t, g_dnn_ag[0,:])
    plt.legend(['analytical','nn'])
    plt.xlabel('t')
    plt.ylabel('g(t)')

    plt.show()

Using forward Euler to solve the ODE

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

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

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

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

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

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

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

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

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

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

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

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

Equation \eqref{odenum} could be implemented in the following way, extending the program that uses the network using Autograd:

# Assume that all function definitions from the example program using Autograd
# are located here.

if __name__ == '__main__':
    npr.seed(4155)

    ## Decide the vales of arguments to the function to solve
    Nt = 10
    T = 1
    t = np.linspace(0,T, Nt)

    ## Set up the initial parameters
    num_hidden_neurons = [100,50,25]
    num_iter = 1000
    lmb = 1e-3

    P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)

    g_dnn_ag = g_trial_deep(t,P)
    g_analytical = g_analytic(t)

    # Find the maximum absolute difference between the solutons:
    diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))
    print("The max absolute difference between the solutions is: %g"%diff_ag)

    plt.figure(figsize=(10,10))

    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
    plt.plot(t, g_analytical)
    plt.plot(t, g_dnn_ag[0,:])
    plt.legend(['analytical','nn'])
    plt.xlabel('t')
    plt.ylabel('g(t)')

    ## Find an approximation to the funtion using forward Euler

    alpha, A, g0 = get_parameters()
    dt = T/(Nt - 1)

    # Perform forward Euler to solve the ODE
    g_euler = np.zeros(Nt)
    g_euler[0] = g0

    for i in range(1,Nt):
        g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1]))

    # Print the errors done by each method
    diff1 = np.max(np.abs(g_euler - g_analytical))
    diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical))

    print('Max absolute difference between Euler method and analytical: %g'%diff1)
    print('Max absolute difference between deep neural network and analytical: %g'%diff2)

    # Plot results
    plt.figure(figsize=(10,10))

    plt.plot(t,g_euler)
    plt.plot(t,g_analytical)
    plt.plot(t,g_dnn_ag[0,:])

    plt.legend(['euler','analytical','dnn'])
    plt.xlabel('Time t')
    plt.ylabel('g(t)')

    plt.show()

Example: Solving the one dimensional Poisson equation

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

$$ \begin{equation} \label{poisson} -g''(x) = f(x) \end{equation} $$

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

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

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

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

The specific equation to solve for

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

$$ -g''(x) = f(x),\qquad x \in (0,1) $$

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

$$ \begin{aligned} g(0) = g(1) = 0 \end{aligned}\label{cond} $$

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

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

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

The analytical solution for this problem is

$$ g(x) = x(1 - x)\exp(x) $$

Solving the equation using Autograd

import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt

def sigmoid(z):
    return 1/(1 + np.exp(-z))

def deep_neural_network(deep_params, x):
    # N_hidden is the number of hidden layers
    # deep_params is a list, len() should be used
    N_hidden = len(deep_params) - 1 # -1 since params consists of
                                        # parameters to all the hidden
                                        # layers AND the output layer.

    # Assumes input x being an one-dimensional array
    num_values = np.size(x)
    x = x.reshape(-1, num_values)

    # Assume that the input layer does nothing to the input x
    x_input = x

    # Due to multiple hidden layers, define a variable referencing to the
    # output of the previous layer:
    x_prev = x_input

    ## Hidden layers:

    for l in range(N_hidden):
        # From the list of parameters P; find the correct weigths and bias for this layer
        w_hidden = deep_params[l]

        # Add a row of ones to include bias
        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)

        z_hidden = np.matmul(w_hidden, x_prev)
        x_hidden = sigmoid(z_hidden)

        # Update x_prev such that next layer can use the output from this layer
        x_prev = x_hidden

    ## Output layer:

    # Get the weights and bias for this layer
    w_output = deep_params[-1]

    # Include bias:
    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)

    z_output = np.matmul(w_output, x_prev)
    x_output = z_output

    return x_output


def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
    # num_hidden_neurons is now a list of number of neurons within each hidden layer

    # Find the number of hidden layers:
    N_hidden = np.size(num_neurons)

    ## Set up initial weigths and biases

    # Initialize the list of parameters:
    P = [None]*(N_hidden + 1) # + 1 to include the output layer

    P[0] = npr.randn(num_neurons[0], 2 )
    for l in range(1,N_hidden):
        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias

    # For the output layer
    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included

    print('Initial cost: %g'%cost_function_deep(P, x))

    ## Start finding the optimal weigths using gradient descent

    # Find the Python function that represents the gradient of the cost function
    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
    cost_function_deep_grad = grad(cost_function_deep,0)

    # Let the update be done num_iter times
    for i in range(num_iter):
        # Evaluate the gradient at the current weights and biases in P.
        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
        # in the hidden layers and output layers evaluated at x.
        cost_deep_grad =  cost_function_deep_grad(P, x)

        for l in range(N_hidden+1):
            P[l] = P[l] - lmb * cost_deep_grad[l]

    print('Final cost: %g'%cost_function_deep(P, x))

    return P

## Set up the cost function specified for this Poisson equation:

# The right side of the ODE
def f(x):
    return (3*x + x**2)*np.exp(x)

def cost_function_deep(P, x):

    # Evaluate the trial function with the current parameters P
    g_t = g_trial_deep(x,P)

    # Find the derivative w.r.t x of the trial function
    d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)

    right_side = f(x)

    err_sqr = (-d2_g_t - right_side)**2
    cost_sum = np.sum(err_sqr)

    return cost_sum/np.size(err_sqr)

# The trial solution:
def g_trial_deep(x,P):
    return x*(1-x)*deep_neural_network(P,x)

# The analytic solution;
def g_analytic(x):
    return x*(1-x)*np.exp(x)

if __name__ == '__main__':
    npr.seed(4155)

    ## Decide the vales of arguments to the function to solve
    Nx = 10
    x = np.linspace(0,1, Nx)

    ## Set up the initial parameters
    num_hidden_neurons = [200,100]
    num_iter = 1000
    lmb = 1e-3

    P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)

    g_dnn_ag = g_trial_deep(x,P)
    g_analytical = g_analytic(x)

    # Find the maximum absolute difference between the solutons:
    max_diff = np.max(np.abs(g_dnn_ag - g_analytical))
    print("The max absolute difference between the solutions is: %g"%max_diff)

    plt.figure(figsize=(10,10))

    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
    plt.plot(x, g_analytical)
    plt.plot(x, g_dnn_ag[0,:])
    plt.legend(['analytical','nn'])
    plt.xlabel('x')
    plt.ylabel('g(x)')
    plt.show()

Comparing with a numerical scheme

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

Using Taylor series, the second derivative can be expressed as

$$ g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x) $$

where \( \Delta x \) is a small step size and \( E_{\Delta x}(x) \) being the error term.

Looking away from the error terms gives an approximation to the second derivative:

$$ \begin{equation} \label{approx} g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} \end{equation} $$

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 \( x \), \eqref{approx} becomes

$$ \begin{aligned} g''(x_i) &\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\ &= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \end{aligned} $$

Since we know from our problem that

$$ \begin{aligned} -g''(x) &= f(x) \\ &= (3x + x^2)\exp(x) \end{aligned} $$

along with the conditions \( g(0) = g(1) = 0 \), the following scheme can be used to find an approximate solution for \( g(x) \) numerically:

$$ \begin{equation} \begin{aligned} -\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &= f(x_i) \\ -g_{i+1} + 2g_i - g_{i-1} &= \Delta x^2 f(x_i) \end{aligned} \end{equation} \label{odesys} $$

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.

The equation can be rewritten into a matrix equation:

$$ \begin{aligned} \begin{pmatrix} 2 & -1 & 0 & \dots & 0 \\ -1 & 2 & -1 & \dots & 0 \\ \vdots & & \ddots & & \vdots \\ 0 & \dots & -1 & 2 & -1 \\ 0 & \dots & 0 & -1 & 2\\ \end{pmatrix} \begin{pmatrix} g_1 \\ g_2 \\ \vdots \\ g_{N_x - 3} \\ g_{N_x - 2} \end{pmatrix} &= \Delta x^2 \begin{pmatrix} f(x_1) \\ f(x_2) \\ \vdots \\ f(x_{N_x - 3}) \\ f(x_{N_x - 2}) \end{pmatrix} \\ \boldsymbol{A}\boldsymbol{g} &= \boldsymbol{f}, \end{aligned} $$

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

Setting up the code

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

import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt

def sigmoid(z):
    return 1/(1 + np.exp(-z))

def deep_neural_network(deep_params, x):
    # N_hidden is the number of hidden layers
    # deep_params is a list, len() should be used
    N_hidden = len(deep_params) - 1 # -1 since params consists of
                                        # parameters to all the hidden
                                        # layers AND the output layer.

    # Assumes input x being an one-dimensional array
    num_values = np.size(x)
    x = x.reshape(-1, num_values)

    # Assume that the input layer does nothing to the input x
    x_input = x

    # Due to multiple hidden layers, define a variable referencing to the
    # output of the previous layer:
    x_prev = x_input

    ## Hidden layers:

    for l in range(N_hidden):
        # From the list of parameters P; find the correct weigths and bias for this layer
        w_hidden = deep_params[l]

        # Add a row of ones to include bias
        x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)

        z_hidden = np.matmul(w_hidden, x_prev)
        x_hidden = sigmoid(z_hidden)

        # Update x_prev such that next layer can use the output from this layer
        x_prev = x_hidden

    ## Output layer:

    # Get the weights and bias for this layer
    w_output = deep_params[-1]

    # Include bias:
    x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)

    z_output = np.matmul(w_output, x_prev)
    x_output = z_output

    return x_output


def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
    # num_hidden_neurons is now a list of number of neurons within each hidden layer

    # Find the number of hidden layers:
    N_hidden = np.size(num_neurons)

    ## Set up initial weigths and biases

    # Initialize the list of parameters:
    P = [None]*(N_hidden + 1) # + 1 to include the output layer

    P[0] = npr.randn(num_neurons[0], 2 )
    for l in range(1,N_hidden):
        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias

    # For the output layer
    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included

    print('Initial cost: %g'%cost_function_deep(P, x))

    ## Start finding the optimal weigths using gradient descent

    # Find the Python function that represents the gradient of the cost function
    # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
    cost_function_deep_grad = grad(cost_function_deep,0)

    # Let the update be done num_iter times
    for i in range(num_iter):
        # Evaluate the gradient at the current weights and biases in P.
        # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
        # in the hidden layers and output layers evaluated at x.
        cost_deep_grad =  cost_function_deep_grad(P, x)

        for l in range(N_hidden+1):
            P[l] = P[l] - lmb * cost_deep_grad[l]

    print('Final cost: %g'%cost_function_deep(P, x))

    return P

## Set up the cost function specified for this Poisson equation:

# The right side of the ODE
def f(x):
    return (3*x + x**2)*np.exp(x)

def cost_function_deep(P, x):

    # Evaluate the trial function with the current parameters P
    g_t = g_trial_deep(x,P)

    # Find the derivative w.r.t x of the trial function
    d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)

    right_side = f(x)

    err_sqr = (-d2_g_t - right_side)**2
    cost_sum = np.sum(err_sqr)

    return cost_sum/np.size(err_sqr)

# The trial solution:
def g_trial_deep(x,P):
    return x*(1-x)*deep_neural_network(P,x)

# The analytic solution;
def g_analytic(x):
    return x*(1-x)*np.exp(x)

if __name__ == '__main__':
    npr.seed(4155)

    ## Decide the vales of arguments to the function to solve
    Nx = 10
    x = np.linspace(0,1, Nx)

    ## Set up the initial parameters
    num_hidden_neurons = [200,100]
    num_iter = 1000
    lmb = 1e-3

    P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)

    g_dnn_ag = g_trial_deep(x,P)
    g_analytical = g_analytic(x)

    # Find the maximum absolute difference between the solutons:

    plt.figure(figsize=(10,10))

    plt.title('Performance of neural network solving an ODE compared to the analytical solution')
    plt.plot(x, g_analytical)
    plt.plot(x, g_dnn_ag[0,:])
    plt.legend(['analytical','nn'])
    plt.xlabel('x')
    plt.ylabel('g(x)')

    ## Perform the computation using the numerical scheme

    dx = 1/(Nx - 1)

    # Set up the matrix A
    A = np.zeros((Nx-2,Nx-2))

    A[0,0] = 2
    A[0,1] = -1

    for i in range(1,Nx-3):
        A[i,i-1] = -1
        A[i,i] = 2
        A[i,i+1] = -1

    A[Nx - 3, Nx - 4] = -1
    A[Nx - 3, Nx - 3] = 2

    # Set up the vector f
    f_vec = dx**2 * f(x[1:-1])

    # Solve the equation
    g_res = np.linalg.solve(A,f_vec)

    g_vec = np.zeros(Nx)
    g_vec[1:-1] = g_res

    # Print the differences between each method
    max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical))
    max_diff2 = np.max(np.abs(g_vec - g_analytical))
    print("The max absolute difference between the analytical solution and DNN Autograd: %g"%max_diff1)
    print("The max absolute difference between the analytical solution and numerical scheme: %g"%max_diff2)

    # Plot the results
    plt.figure(figsize=(10,10))

    plt.plot(x,g_vec)
    plt.plot(x,g_analytical)
    plt.plot(x,g_dnn_ag[0,:])

    plt.legend(['numerical scheme','analytical','dnn'])
    plt.show()

Partial Differential Equations

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

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

$$ \begin{equation} \label{PDE} f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0 \end{equation} $$

where \( f \) 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.

Type of problem

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

For instance, the trial solution could be expressed as

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

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

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.

Network requirements

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

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

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

More details

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

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

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

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

Example: The diffusion equation

In one spatial dimension, the equation reads

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

where a possible choice of conditions are

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

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

Defining the problem

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

$$ \begin{equation} \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} \end{equation} \label{diffonedim} $$

and

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

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

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

Setting up the network using Autograd

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

def sigmoid(z):
    return 1/(1 + np.exp(-z))

def deep_neural_network(deep_params, x):
    # x is now a point and a 1D numpy array; make it a column vector
    num_coordinates = np.size(x,0)
    x = x.reshape(num_coordinates,-1)

    num_points = np.size(x,1)

    # N_hidden is the number of hidden layers
    N_hidden = len(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer

    # Assume that the input layer does nothing to the input x
    x_input = x
    x_prev = x_input

    ## Hidden layers:

    for l in range(N_hidden):
        # From the list of parameters P; find the correct weigths and bias for this layer
        w_hidden = deep_params[l]

        # Add a row of ones to include bias
        x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)

        z_hidden = np.matmul(w_hidden, x_prev)
        x_hidden = sigmoid(z_hidden)

        # Update x_prev such that next layer can use the output from this layer
        x_prev = x_hidden

    ## Output layer:

    # Get the weights and bias for this layer
    w_output = deep_params[-1]

    # Include bias:
    x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)

    z_output = np.matmul(w_output, x_prev)
    x_output = z_output

    return x_output[0][0]

Setting up the network using Autograd; The trial solution

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

A possible trial solution for this PDE is

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

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

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

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

Why the jacobian?

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

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

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

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

# Set up the trial function:
def u(x):
    return np.sin(np.pi*x)

def g_trial(point,P):
    x,t = point
    return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)

# The right side of the ODE:
def f(point):
    return 0.

# The cost function:
def cost_function(P, x, t):
    cost_sum = 0

    g_t_jacobian_func = jacobian(g_trial)
    g_t_hessian_func = hessian(g_trial)

    for x_ in x:
        for t_ in t:
            point = np.array([x_,t_])

            g_t = g_trial(point,P)
            g_t_jacobian = g_t_jacobian_func(point,P)
            g_t_hessian = g_t_hessian_func(point,P)

            g_t_dt = g_t_jacobian[1]
            g_t_d2x = g_t_hessian[0][0]

            func = f(point)

            err_sqr = ( (g_t_dt - g_t_d2x) - func)**2
            cost_sum += err_sqr

    return cost_sum

Setting up the network using Autograd; The full program

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

The analytical solution of our problem is

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

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

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

import autograd.numpy as np
from autograd import jacobian,hessian,grad
import autograd.numpy.random as npr
from matplotlib import cm
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import axes3d

## Set up the network

def sigmoid(z):
    return 1/(1 + np.exp(-z))

def deep_neural_network(deep_params, x):
    # x is now a point and a 1D numpy array; make it a column vector
    num_coordinates = np.size(x,0)
    x = x.reshape(num_coordinates,-1)

    num_points = np.size(x,1)

    # N_hidden is the number of hidden layers
    N_hidden = len(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer

    # Assume that the input layer does nothing to the input x
    x_input = x
    x_prev = x_input

    ## Hidden layers:

    for l in range(N_hidden):
        # From the list of parameters P; find the correct weigths and bias for this layer
        w_hidden = deep_params[l]

        # Add a row of ones to include bias
        x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)

        z_hidden = np.matmul(w_hidden, x_prev)
        x_hidden = sigmoid(z_hidden)

        # Update x_prev such that next layer can use the output from this layer
        x_prev = x_hidden

    ## Output layer:

    # Get the weights and bias for this layer
    w_output = deep_params[-1]

    # Include bias:
    x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)

    z_output = np.matmul(w_output, x_prev)
    x_output = z_output

    return x_output[0][0]

## Define the trial solution and cost function
def u(x):
    return np.sin(np.pi*x)

def g_trial(point,P):
    x,t = point
    return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)

# The right side of the ODE:
def f(point):
    return 0.

# The cost function:
def cost_function(P, x, t):
    cost_sum = 0

    g_t_jacobian_func = jacobian(g_trial)
    g_t_hessian_func = hessian(g_trial)

    for x_ in x:
        for t_ in t:
            point = np.array([x_,t_])

            g_t = g_trial(point,P)
            g_t_jacobian = g_t_jacobian_func(point,P)
            g_t_hessian = g_t_hessian_func(point,P)

            g_t_dt = g_t_jacobian[1]
            g_t_d2x = g_t_hessian[0][0]

            func = f(point)

            err_sqr = ( (g_t_dt - g_t_d2x) - func)**2
            cost_sum += err_sqr

    return cost_sum /( np.size(x)*np.size(t) )

## For comparison, define the analytical solution
def g_analytic(point):
    x,t = point
    return np.exp(-np.pi**2*t)*np.sin(np.pi*x)

## Set up a function for training the network to solve for the equation
def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):
    ## Set up initial weigths and biases
    N_hidden = np.size(num_neurons)

    ## Set up initial weigths and biases

    # Initialize the list of parameters:
    P = [None]*(N_hidden + 1) # + 1 to include the output layer

    P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias
    for l in range(1,N_hidden):
        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias

    # For the output layer
    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included

    print('Initial cost: ',cost_function(P, x, t))

    cost_function_grad = grad(cost_function,0)

    # Let the update be done num_iter times
    for i in range(num_iter):
        cost_grad =  cost_function_grad(P, x , t)

        for l in range(N_hidden+1):
            P[l] = P[l] - lmb * cost_grad[l]

    print('Final cost: ',cost_function(P, x, t))

    return P

if __name__ == '__main__':
    ### Use the neural network:
    npr.seed(15)

    ## Decide the vales of arguments to the function to solve
    Nx = 10; Nt = 10
    x = np.linspace(0, 1, Nx)
    t = np.linspace(0,1,Nt)

    ## Set up the parameters for the network
    num_hidden_neurons = [100, 25]
    num_iter = 250
    lmb = 0.01

    P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)

    ## Store the results
    g_dnn_ag = np.zeros((Nx, Nt))
    G_analytical = np.zeros((Nx, Nt))
    for i,x_ in enumerate(x):
        for j, t_ in enumerate(t):
            point = np.array([x_, t_])
            g_dnn_ag[i,j] = g_trial(point,P)

            G_analytical[i,j] = g_analytic(point)

    # Find the map difference between the analytical and the computed solution
    diff_ag = np.abs(g_dnn_ag - G_analytical)
    print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag))

    ## Plot the solutions in two dimensions, that being in position and time

    T,X = np.meshgrid(t,x)

    fig = plt.figure(figsize=(10,10))
    ax = fig.add_suplot(projection='3d')
    ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))
    s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis)
    ax.set_xlabel('Time $t$')
    ax.set_ylabel('Position $x$');


    fig = plt.figure(figsize=(10,10))
    ax = fig.add_suplot(projection='3d')
    ax.set_title('Analytical solution')
    s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)
    ax.set_xlabel('Time $t$')
    ax.set_ylabel('Position $x$');

    fig = plt.figure(figsize=(10,10))
    ax = fig.add_suplot(projection='3d')
    ax.set_title('Difference')
    s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis)
    ax.set_xlabel('Time $t$')
    ax.set_ylabel('Position $x$');

    ## Take some slices of the 3D plots just to see the solutions at particular times
    indx1 = 0
    indx2 = int(Nt/2)
    indx3 = Nt-1

    t1 = t[indx1]
    t2 = t[indx2]
    t3 = t[indx3]

    # Slice the results from the DNN
    res1 = g_dnn_ag[:,indx1]
    res2 = g_dnn_ag[:,indx2]
    res3 = g_dnn_ag[:,indx3]

    # Slice the analytical results
    res_analytical1 = G_analytical[:,indx1]
    res_analytical2 = G_analytical[:,indx2]
    res_analytical3 = G_analytical[:,indx3]

    # Plot the slices
    plt.figure(figsize=(10,10))
    plt.title("Computed solutions at time = %g"%t1)
    plt.plot(x, res1)
    plt.plot(x,res_analytical1)
    plt.legend(['dnn','analytical'])

    plt.figure(figsize=(10,10))
    plt.title("Computed solutions at time = %g"%t2)
    plt.plot(x, res2)
    plt.plot(x,res_analytical2)
    plt.legend(['dnn','analytical'])

    plt.figure(figsize=(10,10))
    plt.title("Computed solutions at time = %g"%t3)
    plt.plot(x, res3)
    plt.plot(x,res_analytical3)
    plt.legend(['dnn','analytical'])

    plt.show()

Example: Solving the wave equation with Neural Networks

The wave equation is

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

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

Here, the chosen conditions are

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

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.

The problem to solve for

The wave equation to solve for, is

$$ \begin{equation} \label{wave} \frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2} \end{equation} $$

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

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

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

The trial solution

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

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

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

where

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

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

The analytical solution

The analytical solution for our specific problem, is

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

Solving the wave equation - the full program using Autograd

import autograd.numpy as np
from autograd import hessian,grad
import autograd.numpy.random as npr
from matplotlib import cm
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import axes3d

## Set up the trial function:
def u(x):
    return np.sin(np.pi*x)

def v(x):
    return -np.pi*np.sin(np.pi*x)

def h1(point):
    x,t = point
    return (1 - t**2)*u(x) + t*v(x)

def g_trial(point,P):
    x,t = point
    return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point)

## Define the cost function
def cost_function(P, x, t):
    cost_sum = 0

    g_t_hessian_func = hessian(g_trial)

    for x_ in x:
        for t_ in t:
            point = np.array([x_,t_])

            g_t_hessian = g_t_hessian_func(point,P)

            g_t_d2x = g_t_hessian[0][0]
            g_t_d2t = g_t_hessian[1][1]

            err_sqr = ( (g_t_d2t - g_t_d2x) )**2
            cost_sum += err_sqr

    return cost_sum / (np.size(t) * np.size(x))

## The neural network
def sigmoid(z):
    return 1/(1 + np.exp(-z))

def deep_neural_network(deep_params, x):
    # x is now a point and a 1D numpy array; make it a column vector
    num_coordinates = np.size(x,0)
    x = x.reshape(num_coordinates,-1)

    num_points = np.size(x,1)

    # N_hidden is the number of hidden layers
    N_hidden = len(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer

    # Assume that the input layer does nothing to the input x
    x_input = x
    x_prev = x_input

    ## Hidden layers:

    for l in range(N_hidden):
        # From the list of parameters P; find the correct weigths and bias for this layer
        w_hidden = deep_params[l]

        # Add a row of ones to include bias
        x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)

        z_hidden = np.matmul(w_hidden, x_prev)
        x_hidden = sigmoid(z_hidden)

        # Update x_prev such that next layer can use the output from this layer
        x_prev = x_hidden

    ## Output layer:

    # Get the weights and bias for this layer
    w_output = deep_params[-1]

    # Include bias:
    x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)

    z_output = np.matmul(w_output, x_prev)
    x_output = z_output

    return x_output[0][0]

## The analytical solution
def g_analytic(point):
    x,t = point
    return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t)

def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):
    ## Set up initial weigths and biases
    N_hidden = np.size(num_neurons)

    ## Set up initial weigths and biases

    # Initialize the list of parameters:
    P = [None]*(N_hidden + 1) # + 1 to include the output layer

    P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias
    for l in range(1,N_hidden):
        P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias

    # For the output layer
    P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included

    print('Initial cost: ',cost_function(P, x, t))

    cost_function_grad = grad(cost_function,0)

    # Let the update be done num_iter times
    for i in range(num_iter):
        cost_grad =  cost_function_grad(P, x , t)

        for l in range(N_hidden+1):
            P[l] = P[l] - lmb * cost_grad[l]


    print('Final cost: ',cost_function(P, x, t))

    return P

if __name__ == '__main__':
    ### Use the neural network:
    npr.seed(15)

    ## Decide the vales of arguments to the function to solve
    Nx = 10; Nt = 10
    x = np.linspace(0, 1, Nx)
    t = np.linspace(0,1,Nt)

    ## Set up the parameters for the network
    num_hidden_neurons = [50,20]
    num_iter = 1000
    lmb = 0.01

    P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)

    ## Store the results
    res = np.zeros((Nx, Nt))
    res_analytical = np.zeros((Nx, Nt))
    for i,x_ in enumerate(x):
        for j, t_ in enumerate(t):
            point = np.array([x_, t_])
            res[i,j] = g_trial(point,P)

            res_analytical[i,j] = g_analytic(point)

    diff = np.abs(res - res_analytical)
    print("Max difference between analytical and solution from nn: %g"%np.max(diff))

    ## Plot the solutions in two dimensions, that being in position and time

    T,X = np.meshgrid(t,x)

    fig = plt.figure(figsize=(10,10))
    ax = fig.add_suplot(projection='3d')
    ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))
    s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis)
    ax.set_xlabel('Time $t$')
    ax.set_ylabel('Position $x$');


    fig = plt.figure(figsize=(10,10))
    ax = fig.add_suplot(projection='3d')
    ax.set_title('Analytical solution')
    s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)
    ax.set_xlabel('Time $t$')
    ax.set_ylabel('Position $x$');


    fig = plt.figure(figsize=(10,10))
    ax = fig.add_suplot(projection='3d')
    ax.set_title('Difference')
    s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis)
    ax.set_xlabel('Time $t$')
    ax.set_ylabel('Position $x$');

    ## Take some slices of the 3D plots just to see the solutions at particular times
    indx1 = 0
    indx2 = int(Nt/2)
    indx3 = Nt-1

    t1 = t[indx1]
    t2 = t[indx2]
    t3 = t[indx3]

    # Slice the results from the DNN
    res1 = res[:,indx1]
    res2 = res[:,indx2]
    res3 = res[:,indx3]

    # Slice the analytical results
    res_analytical1 = res_analytical[:,indx1]
    res_analytical2 = res_analytical[:,indx2]
    res_analytical3 = res_analytical[:,indx3]

    # Plot the slices
    plt.figure(figsize=(10,10))
    plt.title("Computed solutions at time = %g"%t1)
    plt.plot(x, res1)
    plt.plot(x,res_analytical1)
    plt.legend(['dnn','analytical'])

    plt.figure(figsize=(10,10))
    plt.title("Computed solutions at time = %g"%t2)
    plt.plot(x, res2)
    plt.plot(x,res_analytical2)
    plt.legend(['dnn','analytical'])

    plt.figure(figsize=(10,10))
    plt.title("Computed solutions at time = %g"%t3)
    plt.plot(x, res3)
    plt.plot(x,res_analytical3)
    plt.legend(['dnn','analytical'])

    plt.show()

Resources on differential equations and deep learning

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