# Backpropagation algorithm NN with Rectified Linear Unit (ReLU) activation

I am trying to follow a great example in R by Peng Zhao of a simple, "manually"-composed NN to classify the iris dataset into the three different species (setosa, virginica and versicolor), based on $4$ features. The initial input matrix in the training set (excluding the species column) is $[90 \times 4]$ (90 examples and 4 features - of note, the number of rows may already be different because I have made some changes - here is the entire code ready to copy, paste and run).

This is the architecture of the NN:

My question is not about the code, which I follow, but about the mathematical expression (and possibly brief explanation of the backpropagation algorithm followed in the code). The entire function train.dnn is in the link provided above (and the entire code conveniently available here); however, I want to place the magnifying glass on the backpropagation lines in that function that are the object of my question (my annotations hopefully convey the problems, and make the actual code irrelevant to the understanding of what's being done):

# backward ....
# probs below is a [90 x 3] matrix corresponding to the softmax output:

dscores <- probs

#        [,1]      [,2]      [,3]
# [1,] 0.3332967 0.3327828 0.3339205
# [2,] 0.3332917 0.3327686 0.3339397
# [3,] 0.3332868 0.3328144 0.3338988

# what follows is probably THE most confusing part...
# effectively the matrix of softmax scores (now in dscores) is modified...
# so that for each example, the probability corresponding to the true species...
# is the value that had been calculated MINUS 1, rendering negative values.
# Update: It is the DERIVATIVE OF THE COST FUNCTION OF SOFTMAX wrt output logit z.

dscores[Y.index] <- dscores[Y.index] - 1

#         [,1]       [,2]       [,3]
# [1,] 0.3332967 -0.6672172  0.3339205
# [2,] 0.3332917 -0.6672314  0.3339397
# [3,] 0.3332868 -0.6671856  0.3338988

######################################################

# Regarding Y.index it is a [90 x 2] matrix like this:
#       [,1] [,2]
# [1,]    1    2
# [2,]    2    2
# [3,]    3    2
# [4,]    4    3
# The second column is the codified species {1, 2, 3}
# dscores[Y.index] selects the Pr assigned to the true species in each example.

#####################################################

# Now we divide each value by batchsize,
# which is the size of the training set (90). Why?
# I presume this is part of the BATCH GRADIENT DESCENT ALGORITHM?

dscores <- dscores / batchsize

#          [,1]         [,2]         [,3]
# [1,] 0.003703297 -0.007413524  0.003710227
# [2,] 0.003703241 -0.007413682  0.003710441
# [3,] 0.003703187 -0.007413173  0.003709987

# For the next step, hidden.layer is a [90 x 6] matrix, corresponding to the
# activated values of the hidden layer (ReLU activation function):

#         [,1]    [,2]       [,3]      [,4]    [,5]       [,6]
# [1,] 0.1583907    0     0.09240448    0       0     0.01297566
# [2,] 0.1629522    0     0.09634972    0       0     0.01663396
# [3,] 0.1556569    0     0.08518394    0       0     0.01390234

dW2 <- t(hidden.layer) %*% dscores # [6 x 3] matrix

#             [,1]          [,2]          [,3]
# [1,] 2.143109e-02 -7.865665e-03 -1.356543e-02
# [2,] 0.000000e+00  0.000000e+00  0.000000e+00
# [3,] 3.267824e-03 -1.908723e-03 -1.359102e-03
# [4,] 0.000000e+00  0.000000e+00  0.000000e+00
# [5,] 2.085304e-06 -4.173916e-06  2.088612e-06
# [6,] 2.761881e-03 -5.875996e-04 -2.174282e-03

db2 <- colSums(dscores)            # a vector of 3 numbers (sums for each species)

dhidden <- dscores %*% t(W2) # [90 x 6] matrix
# W2 is a [6 x 3] matrix of weights.
# dhidden is [90 x 6]

dhidden[hidden.layer <= 0] <- 0 # We get rid of negative values

dW1 <- t(X) %*% dhidden # X is the input matrix [90 x 4]
db1 <- colSums(dhidden) # 6-element vector (sum columns across examples)

# update ....
dW2 <- dW2  + reg * W2 # reg is regularization rate reg = 1e-3
dW1 <- dW1  + reg * W1

W1  <- W1 - lr * dW1    # lr is the learning rate lr = 1e-2
b1  <- b1 - lr * db1    # b1 is the first bias

W2  <- W2 - lr * dW2
b2  <- b2 - lr * db2    # b2 is the second bias


QUESTION:

Following the R code (or my English notes on what the code is doing), can I get the mathematical expressions and explanation for the backpropagation algorithm being applied?

PROGRESS NOTES:

Thanks to the help from Lukasz it seems as though the key operation dscores[Y.index] -1 is the derivative of the softmax cost function:

$$\text{Cost}=-\sum_j t_j \log y_j$$

where $t_j$ is the target value.

$$\frac{\partial \text{Cost}}{\partial z_i}=\sum_j \frac{\partial \text{Cost}}{\partial y_j}\frac{\partial y_j}{\partial z_i}= y_i - t_i$$

1. We want to modify the weights backpropagating the error or cost ($J$):

$$\frac{\partial J}{\partial z_i}=\sum_j \frac{\partial J}{\partial y_j}\frac{\partial y_j}{\partial z_i}=y_i - t_i$$

This corresponds to dscores[Y.index] <- dscores[Y.index] - 1.

However, the weights have not been updated, yet.

1. Division by number of examples in training set:

This is dscores <- dscores / batchsize, and probably it has to do with formulas like the one given by Andrew Ng for gradient descent for linear regression:

$$\theta_1:=\theta_1-\alpha\color{red}{\frac{1}{m}}\sum_{i=1}^m\left(\left(h_\theta(x_i)-y_i\right)x_i\right)$$

The $\sum$ part only seems to come into play in db2 <- colSums(dscores), which only updates the bias.

Instead, the update of the second set of weights $\mathbf w_2$ will be directly related to dW2 <- t(hidden.layer) %*% dscores.

And for the first set of weights, $\mathbf w_1$, dW1 <- t(X) %*% (dscores %*% t(W2)).

• I'm guessing that $Y.index \in \{1, 2, 3\}$ so the line that confuses you is a quick formula to apply derivative of softmax $t_i - y_i$, when $y_i$ is not in one-hot encoding – Łukasz Grad Mar 17 '17 at 22:50
• As for the backpropagation, try to get a grasp on what is going on with a single training example. In this code we apply it to a whole dataset, so it's certainly not easy to follow. – Łukasz Grad Mar 17 '17 at 23:01
• @Lukasz Grad Y.index is a [90 x 2] matrix. I'll edit OP. – Antoni Parellada Mar 17 '17 at 23:04
• I checked the link: Y.index <- cbind(1:N, match(Y, Y.set)) # Matching example number and "answer" (1, 2, 3) – Łukasz Grad Mar 17 '17 at 23:11
• @ŁukaszGrad Right! This is what it does, but I wonder what is actually happening with the "- 1"... And then the division by 90. – Antoni Parellada Mar 17 '17 at 23:13

As for the confusing part. Softmax derivative is simply

$$\frac{\partial L}{\partial t_i} = t_i - y_i$$

where $t_i$ is predicted output. Now, in this case $t_i, y_i \in \Re^3$, but $y_i$ has to be in the one-hot encoding form which looks like this

$$y_i = (0, \dots, \overset{\text{k'th}}{1}, \dots, 0)$$

So, for example class 2 is represented as $(0, 1, 0)$

And in your code we have $y.index$ be the position of $1$ in above example. So the line

dscores[Y.index] <- dscores[Y.index] - 1


Is a short and clever way for applying derivative to a whole dataset, i.e.

$$t_i - y_i \equiv t_i[k] = t_i[k] - 1, \text{ for } y_i = (0, \dots, \overset{\text{k'th}}{1}, \dots, 0)$$

## Backpropagation derivatives

Notation:

• $i^k$ - input vector of $k$th layer
• $o^k$ - output vector of $k$th layer
• $W^k$ - transition matrix of $k$th layer
• $b^k$ - biases of $k$th layer
• $X = o^0$ - network input
• $T$ - predicted output
• $T = o^n$ assuming we have $n$ layers

We have a transition between layers

$$i^{k} = W^ko^{k-1} + b^k$$

$ReLU$ as activation function

$$o^k = max(0, i^k)$$

where we assume element-wise application.

Now the important part

$$\frac{\partial L}{\partial o^{k-1}} = W^k\frac{\partial L}{\partial i^k} \ \ \ \ \text{ (1)}$$

$$\frac{\partial L}{\partial i^k} = \frac{\partial L}{\partial o^k} \circ I[o^k > 0] \ \ \ \ \text{ (2)}$$

where $\circ$ means Hadamard product (element-wise)

$$\frac{\partial L}{\partial W^k} = \frac{\partial L}{\partial i^k}(o^{k-1})^T$$

$$\frac{\partial L}{\partial b^k} = \frac{\partial L}{\partial i^k} \ \ \ \ \text{ (3)}$$

Since $\frac{\partial i^k}{\partial b^k} = I$ (identity matrix) and we use chain rule

You can check that above equations hold when $i, o$ are matrices, i.e. we process whole batches at once. Now we can also update weights for a whole batch as follows

\begin{align}\frac{\partial \sum_{l=1}^d L_{(l)}}{\partial W^k} &= \sum_{l=1}^d \frac{\partial L_{(l)}}{\partial W^k} \\ &= \sum_{l=1}^d \frac{\partial L_{(l)}}{\partial i_{(l)}^k}(o_{(l)}^{k-1})^T \\ &= \begin{pmatrix} \frac{\partial L_{(1)}}{\partial i_{(1)}^k} & \dots & \frac{\partial L_{(d)}}{\partial i_{(d)}^k} \end{pmatrix} \begin{pmatrix} o_{(1)}^{k-1})^T \\ \dots \\ o_{(d)}^{k-1})^T \end{pmatrix} \\ &= \frac{\partial \sum_{l=1}^d L_{(l)}}{\partial I^k} (O^{k-1})^T \ \ \ \ \text{ (4)} \end{align}

dscores <- dscores / batchsize


Since we process a whole batch at once and derivative of ReLU is constant we can divide $T - Y$ at start

dW2 <- t(hidden.layer) %*% dscores # [6 x 3] matrix


Using (4)

db2 <- colSums(dscores)            # a vector of 3 numbers (sums for each species)


Apply (3) to a whole batch

dhidden <- dscores %*% t(W2) # [90 x 6] matrix]
# W2 is a [6 x 3] matrix of weights.
# dhidden is [90 x 6]


Applying (1)

dhidden[hidden.layer <= 0] <- 0 # We get rid of negative values


Using (2)

dW1 <- t(X) %*% dhidden # X is the input matrix [90 x 4]
db1 <- colSums(dhidden) # 6-element vector (sum columns across examples)


Again using (4) and (3) to a whole batch for layer 1

# update ....
dW2 <- dW2  + reg * W2 # reg is regularization rate reg = 1e-3
dW1 <- dW1  + reg * W1

W1  <- W1 - lr * dW1    # lr is the learning rate lr = 1e-2
b1  <- b1 - lr * db1    # b1 is the first bias

W2  <- W2 - lr * dW2
b2  <- b2 - lr * db2    # b2 is the second bias


• While you were writing your answer, I was updating my OP with the tip you gave me in the comments. Can you take a look at the edit of the OP? Three questions: 1. To prevent a basic misunderstanding t_i is simply 1 (if the species is correct) or 0, am I right? 2. Can you elaborate beyond that line of code into the next steps, and specifically the division by 90? Also, I am not sure what you mean by "one-hot encoding." – Antoni Parellada Mar 17 '17 at 23:39
• @AntoniParellada $t_i$ is a single row in dscores, one hot encoding en.wikipedia.org/wiki/One-hot . I can update my answer tomorrow – Łukasz Grad Mar 17 '17 at 23:45
• In the print out in the original post head(dscores) every line is the result of softmax for one example. Now only one of the columns is correct for each example. In the derivative of the cost, $y_i - t_i$ applied to the first row (0.3142438 0.3395843 0.3461719) is y_1 - t_1 = 0.3142438 - 1 because the first column is the correct species. Right? If so, $t_1 = 1.$, and it would be $1$ if it had happened to be the second or the third column. That's why it is coded as $dscores[Y.index] -1$. Am I right? – Antoni Parellada Mar 17 '17 at 23:51
• We use swapped notations. In my answer $t_i = (0.3142438, 0.3395843 ,0.3461719)$ and $y_i = (1,0,0)$ , but you got it right i guess. – Łukasz Grad Mar 18 '17 at 0:05
• So $y_i$ can be $y_i=(1,0,0)$, $y_i=(0,1,0)$ or $y_i=(0,0,1)$ depending on the example... – Antoni Parellada Mar 18 '17 at 0:07

If anyone is concerned, this "answer" will never become the accepted answer. It's more like some notes on the issue, possibly helping other people as well.

This post is very useful:

$f$ is the array of class scores for a single example (e.g. array of 3 numbers here):

$f= X\cdot W + b$,

then the Softmax classifier computes the loss for that example as:

$L_i = -\log\left(\frac{e^{f_{y_i}}}{ \sum_j e^{f_j} }\right)$

Recall also that the full Softmax classifier loss is then defined as the average cross-entropy loss over the training examples and the regularization:

$L = \underbrace{ \frac{1}{N} \sum_i L_i }_\text{data loss} + > \underbrace{ \frac{1}{2} \lambda \sum_k\sum_l W_{k,l}^2 > }_\text{regularization loss} \\\\$

Lets introduce the intermediate variable $p$, which is a vector of the (normalized) probabilities. The loss for one example is:

$$p_k = \frac{e^{f_k}}{ \sum_j e^{f_j} } \hspace{1in} L_i =-\log\left(p_{y_i}\right)$$

We now wish to understand how the computed scores inside $f$ should change to decrease the loss $L_i$ that this example contributes to the full objective. In other words, we want to derive the gradient $\partial L_i/\partial f_k$. The loss $L_i$ is computed from $p$, which in turn depends on $f$. It’s a fun exercise to the reader to use the chain rule to derive the gradient, but it turns out to be extremely simple and interpretible [sic] in the end, after a lot of things cancel out:

$$\frac{\partial L_i }{ \partial f_k } = p_k - \mathbb{1}(y_i = k)$$

Notice how elegant and simple this expression is. Suppose the probabilities we computed were $p = [0.2, 0.3, 0.5]$, and that the correct class was the middle one (with probability $0.3$). According to this derivation the gradient on the scores would be $\text{df} = [0.2, -0.7, 0.5]$. Recalling what the interpretation of the gradient, we see that this result is highly intuitive: increasing the first or last element of the score vector f (the scores of the incorrect classes) leads to an increased loss (due to the positive signs $+0.2$ and $+0.5$) - and increasing the loss is bad, as expected. However, increasing the score of the correct class has negative influence on the loss. The gradient of $-0.7$ is telling us that increasing the correct class score would lead to a decrease of the loss $L_i$, which makes sense.

The code implementation of scores in Python (the example in this linked post has no hidden layer) is:

# compute class scores for a linear classifier
scores = np.dot(X, W) + b


This corresponds to the line providing the net input for the outer layer on the example in the OP:

score <- sweep(hidden.layer %*% W2, 2, b2, '+')


Given the array of scores we’ve computed above, we can compute the loss. First, the way to obtain the probabilities is straight forward:

# get unnormalized probabilities
exp_scores = np.exp(scores)
# normalize them for each example
probs = exp_scores / np.sum(exp_scores, axis=1, keepdims=True)


This corresponds to the lines in the OP:

# softmax
score.exp <- exp(score) # This is a [90 x 3] matrix
probs <- sweep(score.exp, 1, rowSums(score.exp), '/')
# Calculated probability mass function for every example


All of this boils down to the following code. Recall that probs stores the probabilities of all classes (as rows) for each example. To get the gradient on the scores, which we call dscores, we proceed as follows:

dscores = probs
dscores[range(num_examples),y] -= 1
dscores /= num_examples


This is the equivalent in the OP to:

dscores <- probs #[90 x 3] matrix of PMF'S (ONE PER EXAMPLE)
#dscores[Y.index] will pick up the probability associated with the
#column that contains the true value (species).
# subtracted "1", which is the gradient of partial L / partial f
dscores[Y.index] <- dscores[Y.index] - 1
dscores <- dscores / batchsize


Lastly, we had that scores = np.dot(X, W) + b, so armed with the gradient on scores (stored in dscores), we can now backpropagate into W and b:

dW = np.dot(X.T, dscores)
db = np.sum(dscores, axis=0, keepdims=True)
dW += reg * W # don't forget the regularization gradient


Where we see that we have backpropped through the matrix multiply operation, and also added the contribution from the regularization. Note that the regularization gradient has the very simple form reg*W since we used the constant $0.5$ for its loss contribution (i.e. $\frac{d}{dw} ( \frac{1}{2} \lambda w^2) = \lambda w$). This is a common convenience trick that simplifies the gradient expression.

Here the matrix of weights is updated, and that's it, because the example quoted has no hidden layer. However, in the OP, there is a hidden layer, which explains that the first matrix of weights being updated are $W_2$ connecting the output of the hidden layer to the net input of the outer layer:

Remembering that:

hidden.layer <- pmax(hidden.layer, 0)


is the activation function or the output of the hidden layer...

dW2 <- t(hidden.layer) %*% dscores # [6 x 3] matrix
dW2 <- dW2  + reg * W2 # reg is regularization rate
W2 <- W2 - lr * dW2 # lr is the learning rate.