Optimisation routines

Now that we are able to train the Neural Network, let's take a look at how we can optimise the training as in reality we will be dealing with deep (many hidden neurons) networks which will potentially take days to train. I'm going to take a step back and have a look at how we might generally attack a convex optimisation problem. Convex optimisation problems are very well researched and understood area, so it makes sense to have a look at these first.

The problem

We are going to fit a simple logistic regression to the following data 

• http://cs229.stanford.edu/ps/ps1/logistic_x.txt
• http://cs229.stanford.edu/ps/ps1/logistic_y.txt

There are two predictors (columns) in the first file and 1 outcome ($ y = \pm 1$) in the second. Note that I've borrowed these from the cs229 class that I'm currently working through. Our task is to minimise  the average empirical loss: $$ L(\theta) = \frac{1}{m} \sum_{i=1}^{m} \log (1 + \exp^{-y^{(i)} \theta^{T} x^{(i)}}) $$ where $y^{(i)}$ is the actual outcome for observation $x^{(i)}$. Note that $x^{(i)}$ is a vector containing the predictors for that observation.

Is this a convex optimisation problem?

It turns out that if we can show that the Hessian $H$ for this loss function satisfies $$z^{T} H z \ge 0$$ $ \forall z \in \mathbb{R}^{3}$ then $L(\theta) $ is a convex function (more generally if $H \in \mathbb{R}^{n \times n}$ then this result must hold true $\forall z \in \mathbb{R}^{n}$. Using the definition of $H$, we have $$ H_{pq} = \frac{\partial^2 L(\theta)}{\partial \theta_p \partial \theta_q}$$ The details of the calculation are quite straightforward, so I'll omit them but the result yields $$H_{pq} = \frac{1}{m} \sum_{i=1}^{m} \frac{1}{1 + \exp^{-\theta^T x^{(i)}}} \times \left( 1 - \frac{1}{1 + \exp^{-\theta^T x^{(i)}}} \right) \times x^{(i)}_p x^{(i)}_q $$ where the sum is over all of our training examples. We can write the last two terms as a matrix product (and subsequently drop the indices) as $x^{(i)} x^{(i)^{T}}$. Since the first two terms are $\in (0, 1]$ and $[0,1)$ respectively, then the product is $\in [0, 1]$, thus we can ignore this term when assessing if $z^T H z \ge 0 \hspace{.1in} \forall z$. Thus $$z^T H z \propto \sum_{i=1}^{m} z^T x^{(i)} x^{(i)^{T}} z = \sum_{i=1}^{m} (z^T x^{(i)})^2 \ge 0 \hspace{0.1in} \forall z$$ Hence $H$ is positive semidefinite which implies that $L(\theta) is a convex problem. This means it has a global optima, which makes our lives a lot easier. Since calculating the Hessian is rather easy in this setting, we can use Newton's method.

Newton's method

Newton's method utilises the Hessian and hence the curvature of the loss surface to find an optimal path to the minima. With this additional information, we can expect the algorithm to converge faster than gradient descent, which only uses first derivative information. The update rule is as follows $$ \theta \rightarrow \theta - H^{-1} \nabla_{\theta} L(\theta) $$

Let's see the performance of Newton's method vs Gradient descent:

I've performed 20 iterations for both Newton's method and gradient descent - clearly Newton's method converges a lot faster than gradient descent. Looking at the update step, it is obvious that this method won't scale well with more parameters since each step requires the calculation of the matrix of first derivatives and the Hessian. Furthermore, it's obvious that this method will fall apart if $H$ is singular. So in our toy example above which had a convex loss function, minimal parameters to estimate - Newton's method was king, but obviously in our neural network we couldn't apply such an algorithm. The beauty of backpropagation was that after a forward pass through the network, we had all of the first order derivatives we required in the update step. This fortune does not however extend to the second derivatives we would require to calculate $H$. It seems we're stuck with regular old batch gradient descent to train or neural network...or are we?

Stochastic Gradient Descent (SGD)

Recall our update step for batch gradient descent: $$\theta_j \rightarrow \theta_j - \eta \frac{\partial}{\partial \theta_j}L(\theta)$$ where  $$ L(\theta) = \frac{1}{m} \sum_{i=1}^{m} \log (1 + \exp^{-y^{(i)} \theta^{T} x^{(i)}}) $$$L(\theta)$ is a function of all m training examples. That is, for a single update to parameter $\theta_j$ we need to use all of our training data. What if we could calculate our loss on a subset of the training data? The hope is that this subset is "representative enough" of the entire dataset such that resulting update to $\theta_j$ is generally in the same direction to that of the update calculated on the entire dataset. This is the essence of Stochastic Gradient Descent.

Consider the fact when the training dataset has several hundred million rows, we may not even be able to fit all the data in memory to perform a batch gradient descent! 

I'll define the following terms commonly found when talking about SGD

• Epoch - once every datapoint in the training set has been used, one epoch has occurred.
• Mini batch - the subset of training data that is used in the parameter update
• Batch size - the size of the mini batch

So per epoch, we will be able to update the parameters $\frac{N}{\text{batch size}}$ times. Compare this to batch gradient descent which by definition, the parameters are updated once per epoch.

The pseudocode is as follows

for epoch in num_epochs:
    shuffle(data)
    for mini_batch in data:
        evaluate derivative of loss on mini_batch
        update parameters

Since we are taking a subset of our training data to update our parameters, the results may be volatile as each subset may contain slightly different information about our loss surface. This is slightly troublesome as we approach the optima, as in SGD the path of optimisation will tend to oscillate around the minimum we seek. In order to remedy this, we use a learning rate schedule which is simply a scaling (generally based on heuristics) of our learning rate at each iteration. The hope is that by the time the algorithm is near the minimum, the learning rate has been scaled down such that the successive parameter updates are relatively stable from this point forward. This process is also called annealing the learning rate. The trick is getting the balance between timing the schedule such that the learning rate is small enough when the algorithm is near the minimum - if you are too aggressive with the schedule, the learning rate will become too small too soon and you won't get near the minimum as the update to the parameters will tend to zero.

Let's have a look at the performance of SGD vs batch gradient descent on our neural network. If implemented SGD on the aforementioned neural network and have run the make moons dataset with 100000 data points and 10 neurons in the hidden layer. See below for a plot of the loss vs. epoch for both SGD and batch gradient descent.

Batch gradient descent:

• Epochs: 200 (i.e 200 updates to the parameters, each based on the full training set)
• Execution time: 17s

SGD:

• Epochs: 1
• Mini batch size: 500 (i.e 200 updates to the parameters, each based on mini batch of size 500)
• Execution time: 6s

We can see that SGD actually outperforms batch gradient descent here and takes about a third of the time to run! I haven't actually applied any learning rate schedule here, you can see an egregious spike in the loss from SGD at around the $160^{th}$ epoch. See what happens below when I simply set $\eta \rightarrow \frac{\eta}{10}$ at the $150^{th}$ epoch:

Notice that the aforementioned spike is now gone and the SGD results look very promising given the accuracy and performance (compared to batch gradient descent). In practice, batch gradient descent will never be used due to the memory constraints and the fact that the randomness of SGD can help the algorithm escape local minima that batch gradient descent would naturally get stuck in.

Momentum

We can introduce the concept of momentum to our parameter updates. It is applied as follows:
Recall the usual update to our parameters is as follows $$\theta_j \rightarrow \theta_j - \eta \frac{\partial}{\partial \theta_j}L(\theta)$$ which we can write in two steps, define
$$\Delta \theta_j = \frac{\partial}{\partial \theta_j}L(\theta)$$ such that the update becomes $$\theta_j \rightarrow \theta_j - \eta \Delta \theta_j$$ We'll now modify our definition of $\Delta \theta_j$ as $$\Delta \theta_j = \gamma \Delta \theta_j - \eta \frac{\partial}{\partial \theta_j}L(\theta)$$ and our update is $$\theta_j \rightarrow \theta_j + \Delta \theta_j$$ What we've done is influenced our current update of $\theta_j$ by the amount it was updated by in the previous step. Yes we've introduced yet another hyperparameter, but the idea here is to give it some "momentum" such that the updates are influenced by previous updates and the optimsation continues towards the minimum. Consider the case when the loss surface is a long narrow ravine with steep walls - SGD will typically oscillate across the ravine as the update will point down the steep walls, the momentum term will help move the algorithm down the ravine towards to minimum we seek. See the effect of momentum on training our neural network below

We see that momentum allows the algorithm to descend very quickly and hits the minimum loss around 40 epochs in - compared to ~140 in our previous iteration.

Hopefully you've now got a good grasp of batch gradient descent vs SGD and some of the subtleties and hyperparameters that need to be considered when training a model with SGD. For a more in depth exploration of various optimisation techniques, including some more advanced methods see here. In production, we'll rely on the built in SGD routines which have been highly tuned for performance.

Feedforward Artificial Neural Network pt5: Additional analysis

Now that we've finally implemented our ANN, let's have a play around some of the parameters to get an understanding of how they affect our network and its results.
The tricky part about training ANNs is that the loss function isn't necessarily convex, which means that we can't use our usual optimisation routines. The fact that the loss function isn't necessarily convex means that just because we find a local minimum, it doesn't mean it's the global minimum. Thus non convex problems may converge on different local minima depending on the parameters of the optimisation routine. We'll explore some of these parameters below.

Learning Rate

Recall how the learning rate $\eta$ enters our optimisation procedure via the weight updates in gradient descent; $$ w \rightarrow w - \eta \frac{\partial L}{\partial w}$$ It essentially controls the step size at each update. Recall we had some funny bumps in our loss function at certain iterations, let's take a closer look: I've plotted the results of two different iterations of training the ANN below

The two lines correspond to the total loss as a function of the number of iterations in our training of the ANN. The blue line has $\eta = 0.001$ and the green line has $\eta = 0.01$. You can see that the green line has those funny bumps we witnessed before - this is the training example with a larger learning rate. The spikes occur when the step size is too large and we overshoot the minimum. Notice that the blue line doesn't have these overshoots, however it takes more iterations to approach the minimum. If we take a step size which is too large, then we consistently overshoot the minima - never converging on the minimum:

The key is finding a learning rate which will find the minimum within a reasonable timeframe. Although our selection ($\eta = 0.001$ vs $\eta = 0.01$) didn't make a huge difference in this case, consider an ANN with multiple hidden layers and thousands of neurons in each layer. This network may take hours (or days) to train depending on how we choose our learning rate.
Depending on the problem at hand, you may value accuracy more than efficiency or vice versa, this will dictate how you choose your learning rate, of which you will usually calculate using cross validation.

Regularisation / Weight Decay

Say we have our initial loss function (the cross entropy Loss) $L_0$ and we add a regularisation term such that we now have $$L = L_0 + \frac{\lambda}{2n} \sum_{w} w^2$$ where the sum is over all weights. Now if $\lambda$ is large then the second term will dominate $L$ and the task of optimising the entire expression will be reduced to minimising $\sum_w w^2$. If $\lambda$ is small then the first term dominates and there are less restrictions place on $w$. This regularisation term controls $w$ by preventing it from becoming overly large and helps us from overfitting the model. If we want to use gradient descent to minimise this regularised loss function we have $$ \frac{\partial L}{\partial w} = \frac{\partial L_0}{\partial w} + \frac{\lambda}{n} \sum_w w$$ so our update at each iteration is $$ w \rightarrow w - \eta \frac{\partial L}{\partial w}$$ becomes $$ w \rightarrow w - \eta \frac{\partial L_0}{\partial w} - \frac{\eta \lambda}{n} w$$ $$\implies w \rightarrow \left(1 - \frac{\eta \lambda}{n} \right) w - \eta \frac{\partial L_0}{\partial w}$$ That is at each update, the weight $w$ is rescaled by a factor of $\left( 1 - \frac{\eta \lambda}{n} \right)$ at each iteration; this is referred to as weight decay and as mentioned before, limits the magnitude of $w$.

Weight initialisation

In this section we'll take a look at why we chose to intialise our values as we did (from a normal distribution with specific parameters). Recall the definition of the weight update from our gradient descent algorithm $$ w \rightarrow w - \eta \frac{\partial L}{\partial w}$$, if the second term in this expression is small or zero, then there is effectively no (or very little) weight update to $w$. This causes our training to slow down incredibly, such that after each iteration our weight $w$ is only changing ever so slightly; obviously we would like to avoid this situation at the start of the procedure. Recall the backpropagation rules for $W^{(1)}$:

$\delta^{(1)} = (1-\tanh^{2}(Z^{(1)}) \odot \delta^{(2)}{W^{(2)}}^T$

$\frac{\partial L}{\partial W^{(1)}} = {x}^T \delta^{(1)}$

we see that $(1-\tanh^{2}(Z^{(1)})$ term enters the equation (more generally, this will be the derivative of the activation function). So we have our update to the weight $w$ as $$w \rightarrow w - \eta (1-\tanh^{2}(Z^{(1)})) \odot {x}^T  \delta^{(2)}{W^{(2)}}^T$$ That is the amount we update $w$ by is proportional to the derivative of our activation function. Thus we want to avoid initialising our weights in a region where this derivative is close to zero. Below is a plot of the function

We can see that this activation function has its derivative approach zero at both extremes: as $x \rightarrow \infty$ and as $x \rightarrow -\infty$. Let's think about a more general ANN for a moment - suppose we have an ANN with 1000 inputs and a single training example where each input is equal to $1$. We have as usual $$Z^{(1)} = x W^{(1)} +b^{(1)}$$ If we have intialised each entry $W^{(1)}_{ij}$ and $b^{(1)}_j$ as selected from a standard normal distribution (iid), then each entry $Z^{(1)}_{i}$ will be the sum of 1001 iid standard normal variables. Then since the sum of $N$ standard normal variables will have mean $0$ and standard deviation $\sqrt{1001}$ i.e a very wide distribution with a relatively high probability of giving a large negative or positive result (this almost looks like a uniform distribution), the derivative of the activation function will be very close to zero! This isn't what we want.

What about if we initialise with a random normal with mean $0$ and standard deviation $\frac{1}{\sqrt{1000}}$? Now we know that the variance of a sum of iid random normal variances is the sum of the variances so we now have each entry in $Z^{(1)}_{ij}$ has mean $0$ and standard deviation $$\sigma = \sqrt{\frac{1000}{1000}+1} = \sqrt{2}$$ which is a lot narrower than our distribution before - there is a lot smaller chance intialising at values where the derivative of the activation function is close to $0$. Below is a comparison of the resulting initialisation distributions from the toy example - the green line is the resulting distribution for the refined initialisation, where the red line results from initialisation by standard normal variables.

More generally for a given network we will initialise from a Gaussian distribution with mean $0$ and standard deviation $\frac{1}{\sqrt{N_{in}}}$ where $N_{in}$ is the number of inputs into the neural network.
Next time we'll have a look at optimising our network using stochastic gradient descent and maybe play around with some different datasets.

Feedforward Artificial Neural Network pt4: Implementation!

So if you've survived this far, it's all worth it! We've done all the groundwork required to code up our ANN to classify our toy example (make moons dataset kindly provided by sklearn).

Recap

Recall that a motivation for exploring ANN's was that our standard Logistic Regression classification technique could only ever give us a linear separating boundary:

ANN structure and Matrices

In the last blog post we derived general expressions for a single hidden layer feedforward ANN. The results were;

For a network with the following features:

• 2 inputs, $x_1$, $x_2$
• 1 Hidden Layer with n neurons - we will treat n as a parameter and see how it affects the output classification
• 2 outputs, $\hat{y}_1$, $\hat{y}_2$
• Activation function: $\sigma(x) = \tanh(x)$
• Softmax output
• Loss function: Cross entropy loss $$L = - \frac{1}{N} \sum_{n \in N} \sum_{i \in C} y_{n,i} \ln{\hat{y}_{n,i}}$$

Forward Propagation Summary

• $Z^{(1)} = xW^{(1)} + b^{(1)}$
• $a^{(1)} = \tanh(Z^{(1)})$
• $Z^{(2)} = a^{(1)}W^{(2)} + b^{(2)}$
• $\hat{y} \equiv a^{(2)} = softmax(Z^{(2)})$

Backpropagation Summary

• $\delta^{(2)} = \hat{y} - y$
• $\delta^{(1)} = (1-\tanh^{2}(Z^{(1)}) \odot \delta^{(2)}{W^{(2)}}^T$
• $\frac{\partial L}{\partial W^{(2)}} = {a^{(1)}}^T \delta^{(2)}$
• $\frac{\partial L}{\partial b^{(2)}} = \sum_{rows} \delta^{(2)}$
• $\frac{\partial L}{\partial W^{(1)}} = {x}^T \delta^{(1)}$
• $\frac{\partial L}{\partial b^{(1)}} = \sum_{rows} \delta^{(1)}$

With the above, we have entirely defined how our ANN classifies training examples, how the weights and biases of the network affect the loss function and hence have our gradient descent update rule completely specified (up to the learning rate parameter).

Implementation in Python

Now instead of defining our network and neurons as a class, I've decided to closely follow this page (which actually inspired all of these ANN related posts) as it is more transparent and easier to grasp what is going on in the code. So without further delay, I present the code for an ANN with a single hidden layer of size 3:

Code

import numpy as np import sklearn.datasets
import pandas as pd import matplotlib.pyplot as plt 
np.random.seed(0) 
X, y = sklearn.datasets.make_moons(200, noise=0.20)
df = pd.DataFrame()
df['x1'] = X[:,0]
df['x2'] = X[:,1]
df['z'] = y


##define ANN which takes inputs X and y with parameters n_iter as number of iterations and learning_rating the rate used in gradient descent
def calculate(X ,y, num_hidden_neurons, n_iter=2000, learning_rate = 0.01, regularisation_rate = 0):
    loss = [];
    #initialise weights and biases
    X_size = len(X)
    np.random.seed(0)
    W1 = np.random.randn(2, num_hidden_neurons) / np.sqrt(2)
    b1 = np.zeros((1, num_hidden_neurons))
    W2 = np.random.randn(num_hidden_neurons, 2) / np.sqrt(num_hidden_neurons)
    b2 = np.zeros((1, 2))
    
    model = {}
    
    for i in xrange(n_iter):
    #feedforward
        Z1 = X.dot(W1)+b1
        a1 = np.tanh(Z1)
        Z2 = a1.dot(W2)+b2
        yhat = softmax(Z2)
    #backpropagation
        d2 = yhat
        d2[range(X_size),y] -= 1
        d1 = (1-a1**2)*d2.dot(W2.T)
        dW2 = a1.T.dot(d2)
        db2 = np.sum(d2, axis=0, keepdims = True)
        dW1 = X.T.dot(d1)
        db1 = np.sum(d1, axis=0)
    #regularisation
        dW1 += regularisation_rate * W1     
        dW2 += regularisation_rate * W2  
    #gradient descent    
        W1 = W1 - learning_rate * dW1
        b1 = b1 - learning_rate * db1
        W2 = W2 - learning_rate * dW2
        b2 = b2 - learning_rate * db2
        
        model = {'W1' : W1, 'b1' : b1, 'W2' : W2, 'b2' : b2}
        loss.append(predict(model,X,y,True))
                
    return loss, model

Comments

The code is fairly self explanatory, but I'll walk through the steps

1. Importing all the libraries we need
2. Initialisation of the weights and biases (more on this later)
3. Feedforward step - implemented via matrices exactly how we calculated them in the previous blog post
4. Backpropagation step, again - implemented just how we derived them before. Note as the $b^{(i)}$ matrices consist of a single unique row each, we let python broadcast them into the appropriate dimensions when it does the matrix addition in the calculation of $Z^{(1)}$ and $Z^{(2)}$.
5. Gradient descent to alter the weights after each iteration in order to minimize the loss function.
6. Regularisation rate - I haven't mentioned this before, I've put it in there but set it to $0$ as to have no effect on the results. I will discuss this term later.
7. Output a dictionary Model which contains our weight and bias matrices $W^{(1)}$, $b^{(1)}$, $W^{(2)}$, $b^{(2)}$.
8. Note that I call a function (that's not defined here) called predict which will make predictions based on the input $X$ or return the cross entropy loss of a set of weights and biases when applied to $X$. The calculation of the cross entropy loss is the reason that $y$ is an input into the overall calculate function; if we were just calculating the weights and biases we would only require $X$ as an input. See code for predict at the bottom of this post.

Results

#function call - 3 neurons in the hidden layer
loss, model = calculate(X, y, 3)
The decision boundary is plotted below

Notice the non-linear decision boundary, and how well it captures the separation of data points! This gives us great hope that if the data generating process is distinctly non-linear then even though the standard GLM (i.e logistic regression) decision boundaries are crude, with an ANN we can potentially get a good fit!
The corresponding cross entropy loss as a function of iterations is also plotted below:

Ok, the overall trend of the loss looks alright, but what the hell are the spikes??? Stay tuned...

What about if we have 10 neurons in the hidden layer?
#function call - 10 neurons in the hidden layer
loss, model = calculate(X, y, 10)

and the corresponding loss: 

Now the spikes are gone and the loss is lower than in the 3 neuron case. Whoa this is a really good fit, or is it? How good is too good? Obviously here, with the 10 neurons the ANN is not only fitting the overall general shape of the data, but its capturing the noise from the input too. Just looking at the data (ignoring the decision boundary) we can imagine a kind of smooth sinusoidal shape separating the classes (as in the 3 neuron case above). Notice the jagged edge near the long red dot amongst the blue dots at coordinates (-0.5, 0.7), you can see that the single red training example has 'pulled' the decision boundary towards it. This is an ominous sign as the decision boundary is far too sensitive to a single data point, remember the idea is to capture the essence of the data generating process excluding noise.

If we were to assess the performance of our ANN purely on the cross entropy loss - it would seem that the network with 10 neurons in the hidden layer outperforms the one with 3. However we always need to test the ANN on data it hasn't seen before - cross validation is usually employed to do so.

Well there it is, if you've followed the posts from pt.1 of this series knowing the bare minimum about ANNs only armed with a pen, paper and a desire to learn, then you should be able to implement your own feedforward ANN from scratch! It was a laborious task (all the matrix algebra, etc) but I hope it provided insight into the construction and architecture of ANNs.

Useful functions

# Predict y_hat from input X or calculate loss

def predict(model, X, y, calc_loss = False):

    W1, b1, W2, b2 = model['W1'], model['b1'], model['W2'], model['b2']

    # Forward propagation

    Z1 = X.dot(W1) + b1

    a1 = np.tanh(Z1)

    Z2 = a1.dot(W2) + b2

    probs = softmax(Z2)

    if calc_loss:

        loss = -np.log(probs[range(len(X)), y])

        tot_loss = np.sum(loss)

        return 1./len(X) * tot_loss

    return np.argmax(probs, axis=1)  

#Compute softmax values for each sets of scores in x.

def softmax(x):

    return np.exp(x) / np.sum(np.exp(x), axis=1, keepdims = True)

Given a model, the predict function returns a classification (0 or 1) for our input data X or the cross entropy loss. This is used to keep track of the loss at each iteration of the weight updates. Note that this is just for illustrative purposes, since it is an expensive calculation in reality, it isn't wise to call it every iteration - maybe every 100, or 1000 or so.

Softmax is used to calculate the softmax of the entries of a vector - used to calculate our $\hat{y}$.

Next steps

Currently I'm working my way through the CS229 Machine Learning course by Andrew Ng which takes up the majority of my spare time. It's a fantastic course which has accompanying lectures available on youtube and is a great introduction to ML techniques.

However, in the future I am to continue with this implementation but tweak it to explore the following

• Why the weights were initialised in such a peculiar way (why not just set them all to 0?)
• Extending to more than 1 hidden layer
• How to prevent overfitting
• Regularisation (I mentioned it above, but did not explain what/how/why)
• Drop out techniques
• Stochastic gradient descent
• A variant of our batch gradient descent which is industry practise when training an ANN (due to massive datasets)
• Different applications - when I get around to it, I'll implement an ANN for the classic MNIST handwritten digit recognition problem.

Feedforward Artificial Neural Network pt3: Matrices!

So far we've looked at the building blocks of an ANN - the different layers, weights, activation functions and error functions. We know have a decent understanding of what the objects are and how they relate, but so far we've only looked at relatively small parts of the network in isolation for individual training samples. We'll look at implementing the aforementioned ideas using matrices, which will be very helpful when we build our full network and have hundreds (or thousands) of training samples. This is a very natural thing to do - if we can set up the propagation of our network and training samples through matrix multiplication then the computer can do the work in optimising the calculations. The other option is the old fashioned way of using for loops to iterate through each neuron and each training example in our network, this way is markedly slower and inefficient.

Sample ANN - using matrices

We'll use the below network to demonstrate the power of the matrix implementation. Note that we will do this step by step, so it can be rather laborious/repetitive but it's important to understand how each element of the matrix relates to the inputs and is propagated through the network. 

The network has the following features:

• 2 inputs, $x_1$, $x_2$
• 1 Hidden Layer with 3 neurons
• 2 outputs, $\hat{y}_1$, $\hat{y}_2$
• Activation function: $\sigma(x) = \tanh(x)$
• Softmax output
• Loss function: Cross entropy loss $$L = - \frac{1}{N} \sum_{n \in N} \sum_{i \in C} y_{n,i} \ln{\hat{y}_{n,i}}$$

Inputs

We define our input matrix based on two training samples: $$x = \left[ \begin{array}{cc} x_{11} & x_{12} \\ x_{21} & x_{22} \\ \end{array} \right]$$ 

where the matrix is defined such that $x_{ij}$ is the $j^{th}$ input of the $i^{th}$ training sample.

Weights

Now we define a matrix $W^{(1)}$ of weights for the neurons in the Hidden Layer;

$$W^{(1)} = \left[ \begin{array}{ccc} w_{11}^{(1)} w_{12}^{(1)} w_{13}^{(1)} \\ w_{21}^{(1)} w_{22}^{(1)} w_{23}^{(1)} \end{array} \right]$$ where $w_{ij}^{(1)}$ is the weight for the $i^{th}$ input in the $j^{th}$ neuron in the hidden layer. Thus each column of $W^{(1)}$ represents the weight each input receives from each neuron in the hidden layer.

Bias

We introduce the concept of bias in an activation function. The bias is a translation of the entire activation function and is implemented as the following $$\sigma(w \cdot x +b)$$ where $w$ is the weight, $x$ is the input into the particular neuron and $b$ is the bias. The bias adds flexibility to the activation function; we can achieve any output from any input. 

For example consider an activation function without bias $$\sigma(x) = \tanh(w \cdot x)$$ and say we want to achieve an output of $0.5$ when $x=0$. There is no such $w$ that will allow us to achieve this result, as $\tanh(0) = 0$. However if we introduce bias we have $$\sigma(x) = \tanh(w \cdot x + b)$$ we can set $b \approx 0.549$ to achieve the desired result. Now we introduce the bias matrix - the matrix of biases for the hidden layer $$b^{(1)} = \left[ \begin{array}{ccc} b_{11}^{(1)} b_{12}^{(1)} b_{13}^{(1)} \\ b_{21}^{(1)} b_{22}^{(1)} b_{23}^{(1)} \end{array} \right]$$ with the $b^{(1)}_{ij} = b^{(1)}_{kj}$ for all $k$ i.e $b^{(1)}$ has a single unique row; all other rows in the matrix are a copy of this one. This is just to ensure that the matrix multiplication works as intended.

Forward Propagation

We'll now work our way through the network, starting at the input, traversing through the hidden layer and arriving at the output.

Using the notation described in this previous blog post, we get our matrix $Z^{(1)}$ which contains the results of applying the weights and biases to our inputs. Thus $$Z^{(1)} = x W^{(1)}+b^{(1)}$$ is a $2 \times 3$ matrix with the following structure
$$Z^{(1)} = \left[ \begin{array}{ccc} x_{11}w_{11}^{(1)}+x_{12}w_{21}^{(1)}+b_{11}^{(1)} \hspace{.3in} \cdots \hspace{.3in} \cdots \\ x_{21}w_{11}^{(1)}+x_{22}w_{21}^{(1)}+b_{21}^{(1)} \hspace{.3in} \cdots \hspace{.3in} \cdots \end{array} \right]$$ where I've only included the first column of the matrix. In matrix index notation we have $$Z_{ij}^{(1)} = \sum_{k} x_{ik} W^{(1)}_{kj} + b_{ij}^{(1)}$$We can interpret $Z_{ij}^{(1)}$ as the input the the $j^{th}$ hidden neuron receives from the $i^{th}$ test sample.

Once at the hidden layer, the activation function is applied (elementwise) to $Z^{(1)}$ as $$a^{(1)} = \left[ \begin{array}{ccc} a_{11}^{(1)} a_{12}^{(1)} a_{13}^{(1)}  \\ a_{21}^{(1)} a_{22}^{(1)} a_{23}^{(1)} \end{array} \right] \equiv \tanh{Z^{(1)}} $$

We now propagate these values through the hidden layer, applying the weights and biases resulting in a $2 \times 2$ matrix $Z^{(2)}$ defined as $$Z^{(2)} = a^{(1)}W^{(2)}+b^{(2)}$$ where $$W^{(2)} = \left[ \begin{array}{cc} w_{11}^{(2)} w_{12}^{(2)} \\ w_{21}^{(2)} w_{22}^{(2)} \\ w_{31}^{(2)} w_{32}^{(2)} \end{array} \right]$$ and $$b^{(2)} = \left[ \begin{array}{cc} b_{11}^{(1)} b_{12}^{(2)} \\ b_{21}^{(2)} b_{22}^{(2)} \end{array} \right]$$ Similarly to $b^{(1)}$, this matrix has a unique single row, with the rest of the rows of the matrix being exact copies.  Explicitly, $Z^{(2)}$ has the following entries
$$\left[ \begin{array}{cc} a_{11}^{(1)}w_{11}^{(2)}+a_{12}^{(1)}w_{21}^{(2)}+a_{13}^{(1)}w_{31}^{(2)}+b_{11}^{(2)} \hspace{.3in} a_{11}^{(1)}w_{12}^{(2)}+a_{12}^{(1)}w_{22}^{(2)}+a_{13}^{(1)}w_{23}^{(2)}+b_{12}^{(2)} \\ a_{21}^{(1)}w_{11}^{(2)}+a_{22}^{(1)}w_{21}^{(2)}+a_{23}^{(1)}w_{31}^{(2)}+b_{21}^{(2)} \hspace{.3in} a_{21}^{(1)}w_{12}^{(2)}+a_{22}^{(1)}w_{22}^{(2)}+a_{23}^{(1)}w_{23}^{(2)}+b_{22}^{(2)} \end{array} \right]$$ More compactly, we have $$Z_{ij}^{(2)} = \sum_{k} a^{(1)}_{ik} W^{(2)}_{kj} + b^{(2)}_{ij} $$ where $Z_{ij}^{(2)}$ can be considered the input to the $j^{th}$ output neuron from the $i^{th}$ training sample.

Finally, applying the softmax function in the output neurons we end up with $$a^{(2)} = \text{softmax}(Z^{(2)}) \equiv \left[ \begin{array}{cc} \hat{y}_{11} \hspace{.15in} \hat{y}_{12} \\ \hat{y}_{21} \hspace{.15in} \hat{y}_{22} \end{array} \right]$$ where $a_{ij}^{(2)}$ is the output of the $j^{th}$ output neuron of the $i^{th}$ training sample. Hopefully the explicit matrix multiplication helps illuminate how the input values are propagated through the network and how their values are affected by the various weights in the input/hidden/output layers. Next we'll take an explicit look at backpropagation and establish similar matrix results which will allow us to train the network in a relatively efficient manner.

Backpropagation

In this section we'll derive the backpropagation rules for the network. Make sure you understand matrix index notation as we'll use it here to succinctly write the results.

Recall our definition $$\delta_j \equiv \frac{\partial L}{\partial z_j} =  \sigma ' (z_j) \sum_{k \in path(j)} \delta_k w_{j \rightarrow k}$$ We'll slightly alter our notation to make it easier to track all of the indices floating around $ \delta^{(j)} \equiv \delta_j$ Starting at our output neurons and using the result from the previous blog post we have $$\delta^{(2)}_{ij} \equiv  \frac{\partial L}{\partial Z^{(2)}_{ij}} = \hat{y}_{ij} - y_{ij}$$ or in full matrix form
$$\left[ \begin{array}{cc} \hat{y}_{11}-y_{11} \hspace{.3in} \hat{y}_{12}-y_{12} \\ \hat{y}_{21}-y_{21} \hspace{.3in} \hat{y}_{22}-y_{22} \end{array} \right]$$ Now $$\delta^{(1)}_{ij} \equiv  \frac{\partial L}{\partial Z^{(1)}_{ij}}$$ using the chain rule we can decompose this into $$ \frac{\partial L}{\partial Z^{(1)}_{ij}} = \sum_{m,n,p,q} \frac{\partial L}{\partial{Z^{(2)}_{mn}}} \times \frac{\partial{Z^{(2)}_{mn}}}{\partial{a^{(1)}_{pq}}}\times \frac{\partial{a^{(1)}_{pq}}}{\partial {Z^{(1)}_{ij}}}$$ The first term is simply $\delta^{(2)}_{mn}$. For the second term, recall above $$Z^{(2)}_{mn} = \sum_{k} a^{(1)}_{mk} W^{(2)}_{kn} + b^{(2)}_{mn} $$ $$\implies \frac{\partial Z^{(1)}_{mn}}{\partial a^{(1)}_{pq}} = W^{(2)}_{qn}$$ where the sum over k has been collapsed since the derivative is only non-zero when $m=p$ and $k=q$. Now, the third term $$\frac{\partial a^{(1)}_{pq}}{\partial Z^{(1)}_{ij}} = \frac{\partial}{\partial Z^{(1)}_{ij}} \left( \tanh(Z^{(1)}_{pq}) \right) = 1 - \tanh^{2}(Z^{(1)}_{ij})$$ where the only non-zero elements of the above expression are when $p=i$ and $q=j$. Putting this all together we have $$\frac{\partial L}{\partial Z^{(1)}_{ij}} = \left( 1 - \tanh^{2}(Z^{(1)}_{ij}) \right)  \sum_{n} \delta^{(2)}_{in} W^{(2)}_{jn} $$ note the sum is only over $n$ now, instead of $m, n, p, q$ make sure you understand why this is. We can now write this in full matrix form (as we will use in the implementation) as
$$\delta^{(1)} = \left( 1 - \tanh^{2}(Z^{(1)}) \right) \odot \delta^{(2)} {W^{(2)}}^T$$ where $\odot$ is the elemetwise multiplication of the matrices (Hadamard product) and $T$ indicates the transpose of the matrix.
Now we can use the above results for $\delta^{(1)}$ and $\delta^{(2)}$ to calculate how the loss function changes with respect to the weights - recall this is the value we use to alter our weights at each iteration in our gradient descent routine. Now $$\frac{\partial L}{\partial W^{(2)}_{ij}} = \sum_{p,q} \frac{\partial L}{\partial Z^{(2)}_{pq}} \times \frac{\partial Z^{(2)}_{pq}}{\partial W^{(2)}_{ij}}$$ The first term is simply $\delta^{(2)}_{pq}$ and the second term is $$\frac{\partial}{\partial W^{(2)}_{ij}} \left[ \sum_{k} a^{(1)}_{pk} W^{(2)}_{kq} + b^{(2)}_{pq} \right]$$ $\implies k=i$ and $q=j$ hence the sum collapses and the term evaluates to $a^{(1)}_{pi}$. Hence $$\frac{\partial L}{\partial W^{(2)}_{ij}} = \sum_{p} \delta^{(2)}_{pj} a^{(1)}_{pi}$$ or in matrix form, this is represented as $$ \frac{\partial L}{\partial W^{(2)}} = {a^{(1)}}^T \delta^{(2)}$$ Similarly $$\frac{\partial L}{ \partial b^{(2)}_{ij}} = \sum_{p,q} \frac{\partial L}{\partial Z^{(2)}_{pq}} \times \frac{\partial Z^{(2)}_{pq}}{\partial b^{(2)}_{ij}} $$ the second term forces $q=j$ resulting in $$\frac{\partial L}{ \partial b^{(2)}_{ij}} = \sum_p \delta^{(2)}_{pj}$$ i.e a sum over the rows of $\delta^{(2)}$ note the right hand side of the expression is indendent of $i$ this implies that every row in $\frac{\partial L}{ \partial b^{(2)}}$ is the same. Similar analysis yields $$\frac{\partial L}{\partial W^{(1)}_{ij}} = \sum_{p} \delta^{(1)}_{pj} x_{pi} $$ and $$\frac{\partial L}{\partial b^{(1)}_{ij}} = \sum_p \delta^{(1)}_{pj}$$

Recap

Relax. Take a breath. The derivations are over (for now). I think going through all the derivations in detail at least once is paramount to understanding the inner workings of the neural network. Once you get your head around the notation, there actually isn't anything that fancy going on - just repeated applications of the chain rule. They key for me was attacking this problem via matrix index notation which illustrated exactly how values are propagated (backwards and forwards) in the network via matrix multiplication.

Forward Propagation Summary

• $Z^{(1)} = xW^{(1)} + b^{(1)}$
• $a^{(1)} = \tanh(Z^{(1)})$
• $Z^{(2)} = a^{(1)}W^{(2)} + b^{(2)}$
• $\hat{y} \equiv a^{(2)} = softmax(Z^{(2)})$

Backpropagation Summary

$\delta^{(2)} = \hat{y} - y$

$\delta^{(1)} = (1-\tanh^{2}(Z^{(1)}) \odot \delta^{(2)}{W^{(2)}}^T$

$\frac{\partial L}{\partial W^{(2)}} = {a^{(1)}}^T \delta^{(2)}$

$\frac{\partial L}{\partial b^{(2)}} = \sum_{rows} \delta^{(2)}$

$\frac{\partial L}{\partial W^{(1)}} = {x}^T \delta^{(1)}$

$\frac{\partial L}{\partial b^{(1)}} = \sum_{rows} \delta^{(1)}$

Note that the $\frac{\partial L}{\partial W^{(i)}}$ and $\frac{\partial L}{\partial b^{(i)}}$ are not actually derivatives with respect to matrices, these are just labels for the matrices containing the derivatives with respect to the various weights and biases. The $\sum_rows \delta^{(i)}$ indicates that we sum over the rows of the matrix $\delta^{(i)}$ such that the matrices $$\frac{\partial L}{\partial b^{(i)}}$$ consist of a single unique row repeated $m$ times, where $n$ is the number of training examples.

With these matrices we are now in a position to finally implement our ANN to classify our dataset! The next blog will detail the implementation in python and the associated results.

Derivative of cross entropy loss with softmax

Derivative of the softmax function

I've dedicated a separate post for the derivation of the derivative of the cross entropy loss function with softmax as the activation function in the output layer, which I'll reference in the future. It's nothing groundbreaking but sometimes it's nice to work through some of the results which are often quoted without derivation. 

The softmax function

The softmax function is often used as the activation function for neurons in the output layer of an ANN. We borrow the notation from a previous blog post, where $\hat{y}$ is the output of a particular neuron in the output layer and $$z_o = \sum_{j \in input(o)} z_j w_{j \rightarrow o}$$ i.e. we sum over all paths $j$ which are an input into our output neuron $o$.

It is defined as the following:

$$ \hat{y}_o = \frac{e^{z_o}}{\sum_k e^{z_k}}$$ where the sum is over the $k$ output neurons in the network. The softmax essentially takes an input $z_o \in \mathbb{R}$ and maps it to the interval $(0,1)$ such that the sum of all outputs is 1 - it gives us a probability distribution over our $k$ outputs (i.e. $k$ different classifications). 

Cross entropy loss function

Recalling the definition of the cross entropy loss function from a previous blog post we have

$$L = - \frac{1}{N} \sum_{n \in N} \sum_{i \in C} y_{n,i} \ln{\hat{y}_{n,i}}$$

where $N$ is the number of training examples and $C$ is the number of classifications (i.e 2 for binary classification, etc...). Let us assume $N = 1$ without loss of generality, thus 

$$L = - \sum_{o} y_{o} \ln{\hat{y}_{o}}$$ where the sum is over the output neurons in our network (given that we have 1 output neuron for each class, the sum over $i \in C$ and $o$ are equivalent). Note that $y_o$ is the actual classification of our data, i.e. $y_o \in \{0,1\}$ and that if $y_o=1$ then all other $y_j =0$ where $ i \neq o$ see a description of one hot encoding. Now our object of interest is $$ \frac{ \partial L}{\partial z_i} = \frac{\partial L}{\partial \hat{y}_o} \times \frac{\partial \hat{y}_o}{\partial z_i}$$ Let's focus on the second term $\frac{\partial \hat{y}_o}{\partial z_i}$ for now and split it into two cases.

• Case 1: $ i = o$

$$\frac{\partial \hat{y}_o}{\partial z_i} = \frac{\partial \hat{y}_o}{\partial z_o} =  \frac{\partial}{\partial z_o} \left( \frac{e^{z_o}}{\sum_k e^{z_k}} \right) = \frac{e^{z_o}}{\sum_k e^{z_k}} - e^{z_o} (\sum_k e^{z_k})^{-2}e^{z_o}$$

$$= \frac{e^{z_o}}{\sum_k e^{z_k}} \left ( 1 - \frac{e^{z_o}}{\sum_k e^{z_k}} \right) = \hat{y}_o (1 - \hat{y}_o)$$

• Case 2: $ i \neq o $

$$ \frac{\partial \hat{y}_i}{\partial z_i} = \frac{\partial}{\partial z_i} \left( \frac{e^{z_o}}{\sum_k e^{z_k}} \right) = -e^{z_o}(\sum_k e^{z_k})^{-2} e^{z_i}$$

$$= -\hat{y}_o \hat{y}_i$$ Now $$\frac{\partial L}{\partial \hat{y}_o} = - \sum_o \frac{1}{\hat{y}_o} y_o$$ Therefore $$ \frac{\partial L}{\partial z_i} = -\sum_o \frac{1}{\hat{y}_o} y_o  \times \frac{\partial \hat{y}_o}{\partial z_i}$$ Substituting in our above computations and splitting out the $i=o$ case from the sum we get

$$\frac{\partial L}{\partial z_i} = -\frac{1}{\hat{y}_i} y_i \times \hat{y}_i(1-\hat{y}_i) - \sum_{o \neq i} \frac{1}{\hat{y}_o} y_o \times (- \hat{y}_o \hat{y}_i)$$

$$ = -y_i(1-\hat{y}_i) + \sum_{o \neq i} y_o \hat{y}_i$$ $$= y_i \hat{y}_i -y_i + \sum_{o \neq i} y_o \hat{y}_i$$ Moving the first term inside the sum and summing over all $o$ we get 

$$\frac{\partial L}{\partial z_i} = -y_i + \sum_{o} y_o \hat{y}_i$$ $$ = \hat{y}_i \sum_o y_o - y_i$$ But $\sum_o y_o$ is the sum over all the actual labels of our data which is one hot encoded (as mentioned above). Thus $y_o = 1$ for exactly one class and $0$ for all else - hence $\sum_o y_o = 1$. We finally arrive at a result we'll use in the future in constructing and implementing our ANN to classify the make moons dataset

$$ \frac{\partial L}{\partial z_i}  = \hat{y}_i - y_i$$

Feedforward Artificial Neural Network pt2

Feedforward Artificial Neural Network pt2

In the previous post, we briefly looked at the general structure of a Feedforward ANN and why such a construction should be useful. Now that we have a general handle of the idea behind a Feedforward ANN, we need to train the network - find a method to calculate the appropriate weights for our hidden layer neurons. In the context of our classification case, we want to train the network by providing it a dataset with labels so that it can predict the labels of new data once trained - a classic supervised learning scenario.

Backpropagation

Backpropagation is the algorithm we'll use to train (i.e minimize the error function of) our ANN. It is essentially repeated applications of the chain rule - chaining together all possible paths in the network to establish how the error changes as a function of the inputs and outputs of each neuron. This may not seem like a very profound approach, nor may it be clear that this use of the chain rule should deserve it's own name, however the power from this method comes from the implementation. The set up of backpropagation allows the derivatives we use in the chain rule to be calculated using intermediate results in our network. This becomes especially important when the size of our network grows - it is efficient in that it saves the recalculation of derivatives.

Some notation

Suppose we have two neurons labelled by $i$ and $j$ respectively, as depicted below:

where $\sigma_j$ is a label for the neuron with the following meaning:

• We define the activation function for neuron $i$ as $a_j(w_{i \rightarrow j} \cdot a_i)$, where $w_{i \rightarrow j}$ is the weight applied to the input $a_i$ of neuron $j$. 

Gradient Descent

In order to minimize the Loss function, one might think to simply take the derivatives and throw them into a prebuilt optimisation routine from your favourite library. This approach is fine when the network is small, but when it scales to millions of neurons it becomes computationally intractable. We will in fact use a standard optimisation routine, but it is the clever implementation which makes it so attractive. The optimisation method we'll be using is called the batch gradient descent method, this is in contrast to the stochastic gradient descent method. It is a fairly standard iterative method to minmize a loss function $L\left(w_{i \rightarrow j} \right)$ by updating the target variables by

$$ w_{i\rightarrow j} \rightarrow w_{i\rightarrow j} - \eta \frac{\partial L}{\partial w_{i\rightarrow j}} $$

at each iteration. $\eta$ is the so called learning rate which determines how big/small the steps are at each iteration. Intuitively small steps will take a longer time for the algorithm to converge, while larger steps cover more ground, they may overshoot the true minimum - the ideal value is dependent on the loss function and is somewhere in between.

The Loss Function

Once we have the architecture of our ANN set up, we need to measure its performance. In order to do so, we borrow from information theory, a loss function which is commonly found in ANN classification tasks: the cross entropy loss
$$ L = -\frac{1}{N} \sum_{n \in N} \sum_{ i \in C} y_{n,i} \ln \hat{y}_{n,i} $$
where we sum over N, the number of training examples and C, the number of classes in our classification problem. The hat $\hat{y}$ indicates an output classification from our network, where $y$ is the actual classification of the training sample. This is the Loss function we will use once we begin to classify the examples from the previous blog post. Note that if we were building the ANN with a continuous output (i.e. regression) we could use the usual Residual Sum of Squares

$$L_{reg} = \sum_{i=1}^{N} \frac{1}{2} \left( \hat{y_{i}} - y_{i} \right)^{2} $$

where the factor of a half is chosen for convenience (it prevents stray factors of $\frac{1}{2}$ floating around when we differentiate).
Thus the task of training our neural network then becomes an optimisation problem; minimize the classification error function $L$ with respect to the weights $w_{i \rightarrow j}$ of each neuron in the hidden layer -  see below for an explanation of the notation. Note this is ignoring any potential problems of overfitting.

Illuminating example - a sample ANN

I'm still working my way around TikZ which will allow me to make nice diagrams in LaTeX, but in the interest of time I've borrowed the below image from here. The network pictured has a single input $x_i$, 2 hidden layers and a single output. We will not assume the functional form of the loss function $L$ or the activation functions $\sigma$. 

We want to understand how the output $\hat{y}_i$ changes with respect to the input weight $w_{in \rightarrow i}$, we can use the chain rule to get this relationship. Noting that there are two paths from the input layer to the output, we need to consider each path separately. 

We define the following in the network, where $i$, $j$, $k$, $o$ index the neurons 

$$z_i = w_{in \rightarrow i} x_i,  \quad a_i = \sigma(z_i)$$

$$z_j = w_{i \rightarrow j} a_i, \quad a_j = \sigma(z_j)$$

$$z_k = w_{i \rightarrow k} a_i, \quad a_j = \sigma(z_k)$$

$$z_o = w_{k \rightarrow o} a_k + w_{j \rightarrow o} a_j, \quad \hat{y} \equiv a_j = output(z_o)$$

where $output(x)$ is the output function, which for now we will keep generic and $\sigma(x)$ is the activation function, which is the same for each neuron in the network.

We decompose $\frac{\partial L}{\partial w_{in \rightarrow j}}$: starting at the right side of the graph and moving through $o \rightarrow k \rightarrow i$ as below:

$$\frac{\partial L}{\partial z_o} \times \frac{\partial z_o}{\partial z_k} \times \frac{\partial z_k}{\partial z_i} \times \frac{\partial z_i}{\partial w_{in \rightarrow j}}$$

Let's look at an individual component $$\frac{\partial z_o}{ \partial z_k} = \frac{\partial}{\partial z_k} \left( w_{k \rightarrow o} a_k + w_{j \rightarrow o} a_j \right)$$ noting the second term does not depend on $z_k$

$$\frac{\partial}{\partial z_k}  w_{k \rightarrow o} \sigma(z_k)  = w_{k \rightarrow o} \sigma ' (z_k)$$With a similar analysis we see that the product of derivatives can be expressed as

$$\frac{\partial L}{\partial z_o} \times w_{k \rightarrow o} \sigma ' (z_k) \times w_{i \rightarrow k} \sigma ' (z_i) \times x_i$$

and now moving through $o \rightarrow j \rightarrow i$

$$\frac{\partial L}{\partial z_o} \times \frac{\partial z_o}{\partial z_j} \times \frac{\partial z_j}{\partial z_i} \times \frac{\partial z_i}{\partial w_{in \rightarrow j}}$$

$$=\frac{\partial L}{\partial z_o} \times w_{j \rightarrow o} \sigma ' (z_j) \times w_{i \rightarrow k} \sigma ' (z_i) \times x_i$$

Putting this all together we have $$\frac{\partial L}{\partial w_{in \rightarrow i}} = \frac{\partial L}{\partial z_o} \left[ w_{k \rightarrow o} \sigma ' (z_k) \times w_{i \rightarrow k} \sigma '(z_i) + w_{j \rightarrow o} \sigma ' (z_j) \times w_{i \rightarrow j} \sigma '(z_i) \right] x_i$$

where we have yet to specify the exact form of the loss function $L$ and activation function $\sigma(x)$ in this example. Once these are specified, we can just plug them in to the above formula. Hopefully this simplified example illuminates how the derivatives of the loss function wrt to each neuron is calculated. In terms of backpropagation, we can think of starting out the output and working backwards through each neuron in the path leading to it and understanding how the error changes at each point. In the next section I will generalise these results so they are easily extendable to a larger network.

Error propagation

We will now generalise the above computation, in order to get the derivative of the loss function with respect to any $z_j$ in any of the layers of the network. We define the error signal $$\delta_j = \frac{\partial L}{\partial z_j}$$ Now, using the chain rule as in the example above we have $$\delta_j = \frac{\partial L}{\partial \hat{y}} \times \frac{\partial \hat{y}}{\partial z_j}$$ Once we specify our loss function $L$ we'll have a nice expression for the first term in the expression above, for now let's keep it generic. Now using the definitions of the $z_i$ and $a_i$ above: $$\frac{\partial \hat{y}}{\partial z_j} = \frac{\partial \hat{y}}{\partial a_j} \times \frac{\partial a_j}{\partial z_j} = \sigma ' (z_j) \sum_{k \in path(j)} \frac{\partial \hat{y}}{\partial z_k} \frac{\partial z_k}{\partial a_j}$$ where the sum is over the $k$ paths which are connected to the output of neuron $j$. We have $z_k = a_j w_{j \rightarrow k}$ so our expression becomes $$\frac{ \partial \hat{y}}{\partial z_j} = \sigma '(z_j) \sum_{k \in path(j)} \frac{\partial \hat{y}}{\partial z_k} w_{j \rightarrow k}$$ $$\implies \delta_j = \sigma '(z_j) \sum_{k \in path(j)} \frac{\partial L}{\partial \hat{y}} \frac{\partial \hat{y}}{\partial z_k} w_{j \rightarrow k}$$ Note the first two terms in the summation are actually $\delta_k$ in disguise! Thus we have a recursive definition for our error propagation: $$\delta_j =  \sigma ' (z_j) \sum_{k \in path(j)} \delta_k w_{j \rightarrow k}$$
Thus if we have $\delta_\text{output}$, we can work our way backwards through the neural network, calculating the error for each neuron as a function of the error of the neurons that its output is connected to. This is the key concept of back propagation - although it is just an application of the chain rule, because we can propagate the error at each point in the network through subsequent neurons, the calculations can be done on the fly by utilising calculations that have already been performed.

Next time we'll describe the architecture of the actual ANN we'll use to classify the make moons example detailted in pt1. To be honest, these concepts didn't fully sink in until the first time I implemented an ANN, going through the details of the calculations, I suggest you work through this and the next blog post by hand in order to ensure you have a handle on how the feedforward and error propagations are affected by different topologies of ANNs.

Feedforward Artificial Neural Network pt1

Architecture of an ANN

The history of ANNs can be traced back to Warren McCulloch, who proposed a mechanism of how neurons might work by modelling a simple neural network with an electrical circuit. This line of thinking was reinforced by Donald Hebb, who discovered that neural pathways between neurons were strengthened through repetitive use. Understanding the motivation behind ANNs will help us understand why they are structured the way they are, particularly the action potential of a neuron.

An ANN consists of three major components;

• Input Layer
• This layer takes the inputs as is - they do not modify the data at all; they push it into the hidden layers.
• Hidden Layer(s)
• In line with the aforementioned action potential of a neuron in the brain, the hidden layer neurons take an input, applies an activation function and pushes the result to the next layer in the network.
• The activation function $\phi :  \mathbb{R} \rightarrow \mathbb{R}$ takes $\sum_{i=1}^{n} x_i w_i + b$ as an argument, where $n$ is the number of inputs flowing into the particular hidden layer neuron, $w_{i}$ is the associated weight with input $x_{i}$ and $b$ is called the bias. Our task is to find the $w_{i}$ and $b$ for each neuron in the hidden layer which minimize our error function (more on this later).
• The choice of activation function is entirely up to you. Obviously there are some standard choices for certain tasks. See below for a (by no means exhaustive) list of activation functions
  
  
   
• The activation functions are they key to capturing non-linearities in our decision boundary.
• Output Layer
• The output of the neural network. In our case (classification) it will be a label classifying our input data to one of two classes.

These three components together with the topology of the network will define the entire structure of our ANN. Below is a stock image of a typical ANN

On the left there are 3 inputs which flow into the hidden layers (more on these below) which then in turn flow into the outputs. There are two hidden layers in this setup, each with 4 neurons - these are hyper parameters we can tune to increase performance and reduce overfitting.

Note that the flow of the network is strictly left to right, none of the neurons in the hidden layer are connected to one another and none of the neurons in the network loop back into themselves. This single direction of information flow and the fact that there are no loops or cycles in the network define a Feedforward Artificial Neural Network. Also note that the number of inputs and outputs are fixed once we define the network under this architecture. Other types of neural networks exist that are able to handle inputs with multiple lengths, but our focus for now will be Feedforward ANNs.

By why this structure?

The above structure of a neural network may seem like an arbitrary construction - how can we guarantee that it can be trained in such a way to produce our desired output?

Well..luckily for us there exists a theorem called the Universal Approximation Theorem which states that under reasonable circumstances, a feedforward network containing a finite number of neurons can approximate continuous functions on compact subsets of $\mathbb{R}^{n}$.

Statement

Let $\phi \left( \cdot \right)$ be a nonconstant, bounded and monotonically-increasing continuous function. Let $I_{m}$ denote the $m$-dimensional unit hypercube $\left[ 0, 1 \right]$. The space of continuous functions on $I_{m}$ is $C \left( I_{m} \right)$. Then $\forall f \in C \left( I_{m} \right)$ and $ \epsilon > 0$ $\exists N \in \mathbb{Z}$, $v_{i}, b_{i} \in \mathbb{R}$ and $w_i \in \mathbb{R}^{m}$ where $ i = 1, \dots, N$ such that we may define

$$F(x) \equiv \sum_{i=1}^{N} v_{i} \phi \left( w_{i}^{T}x + b_{i} \right)$$

such that $$\left| F(x) - f(x) \right| < \epsilon \quad \forall x \in I_{m}$$

Which is quite fascinating really, that we can essentially approximate any continuous function to an arbitrary degree using a linear combination of our activation functions. Please note that this tells us nothing of actually how to approximate the function; what the parameters $N, v_{i}, b_{i}$ or $w_{i}$ are - only that they exist.

Summary

So there it is, a whirlwind introduction to Feedforward Artificial Networks. By no means is this supposed to be a comprehensive introduction, but to set the scene for the next few blog posts. In my next blog I'll define the error functions and consequently the Backpropogation algorithm which we will use to train our Feedforward ANN to find the weights and biases of our hidden layer neurons.