It allows you to use a CONV layer without necessarily shrinking the height and width of the volumes. This is important for building deeper networks, since otherwise the height/width would shrink as you go to deeper layers. An important special case is the "same" convolution, in which the height/width is exactly preserved after one layer.
It helps us keep more of the information at the border of an image. Without padding, very few values at the next layer would be affected by pixels as the edges of an image.
Exercise: Implement the following function, which pads all the images of a batch of examples X with zeros. . Note if you want to pad the array "a" of shape (5,5,5,5,5)(5,5,5,5,5) with pad = 1 for the 2nd dimension, pad = 3 for the 4th dimension and pad = 0 for the rest, you would do:
def zero_pad(X, pad):
"""
Pad with zeros all images of the dataset X. The padding is applied to the height and width of an image,
as illustrated in Figure 1.
Argument:
X -- python numpy array of shape (m, n_H, n_W, n_C) representing a batch of m images
pad -- integer, amount of padding around each image on vertical and horizontal dimensions
Returns:
X_pad -- padded image of shape (m, n_H + 2*pad, n_W + 2*pad, n_C)
"""
X_pad = np.pad(X, ((0, 0),(pad, pad),(pad, pad),(0, 0)), 'constant', constant_values=0)
return X_pad
Single step of convolution
In this part, implement a single step of convolution, in which you apply the filter to a single position of the input. This will be used to build a convolutional unit, which:
Takes an input volume
Applies a filter at every position of the input
Outputs another volume (usually of different size)
def conv_single_step(a_slice_prev, W, b):
"""
Apply one filter defined by parameters W on a single slice (a_slice_prev) of the output activation
of the previous layer.
Arguments:
a_slice_prev -- slice of input data of shape (f, f, n_C_prev)
W -- Weight parameters contained in a window - matrix of shape (f, f, n_C_prev)
b -- Bias parameters contained in a window - matrix of shape (1, 1, 1)
Returns:
Z -- a scalar value, result of convolving the sliding window (W, b) on a slice x of the input data
"""
# Element-wise product between a_slice and W. Add bias.
s = np.multiply(a_slice_prev, W) + b
# Sum over all entries of the volume s
Z = np.sum(s)
return Z
# test function conv_single_step()
np.random.seed(1)
a_slice_prev = np.random.randn(4, 4, 3)
W = np.random.randn(4, 4, 3)
b = np.random.randn(1, 1, 1)
Z = conv_single_step(a_slice_prev, W, b)
print("Z =", Z)
Convolutional Neural Networks - Forward pass
In the forward pass, you will take many filters and convolve them on the input. Each 'convolution' gives you a 2D matrix output. You will then stack these outputs to get a 3D volume
Exercise: Implement the function below to convolve the filters W on an input activation A_prev. This function takes as input A_prev, the activations output by the previous layer (for a batch of m inputs), F filters/weights denoted by W, and a bias vector denoted by b, where each filter has its own (single) bias. Finally you also have access to the hyperparameters dictionary which contains the stride and the padding.
For this exercise, we won't worry about vectorization, and will just implement everything with for-loops.
def conv_forward(A_prev, W, b, hparameters):
"""
Implements the forward propagation for a convolution function
Arguments:
A_prev -- output activations of the previous layer, numpy array of shape (m, n_H_prev, n_W_prev, n_C_prev)
W -- Weights, numpy array of shape (f, f, n_C_prev, n_C)
b -- Biases, numpy array of shape (1, 1, 1, n_C)
hparameters -- python dictionary containing "stride" and "pad"
Returns:
Z -- conv output, numpy array of shape (m, n_H, n_W, n_C)
cache -- cache of values needed for the conv_backward() function
"""
# Retrieve dimensions from A_prev's shape (≈1 line)
(m, n_H_prev, n_W_prev, n_C_prev) = A_prev.shape
# Retrieve dimensions from W's shape (≈1 line)
(f, f, n_C_prev, n_C) = W.shape
# Retrieve information from "hparameters" (≈2 lines)
stride = hparameters['stride']
pad = hparameters['pad']
# Compute the dimensions of the CONV output volume using the formula given above. Hint: use int() to floor. (≈2 lines)
n_H = 1 + int((n_H_prev + 2 * pad - f) / stride)
n_W = 1 + int((n_W_prev + 2 * pad - f) / stride)
# Initialize the output volume Z with zeros. (≈1 line)
Z = np.zeros((m, n_H, n_W, n_C))
# Create A_prev_pad by padding A_prev
A_prev_pad = zero_pad(A_prev, pad)
for i in range(m): # loop over the batch of training examples
a_prev_pad = A_prev_pad[i] # Select ith training example's padded activation
for h in range(n_H): # loop over vertical axis of the output volume
for w in range(n_W): # loop over horizontal axis of the output volume
for c in range(n_C): # loop over channels (= #filters) of the output volume
# Find the corners of the current "slice" (≈4 lines)
# Note: there is a stride, therefore it is not wise to assign vert_start a value of h and it is the same with horiz_start
vert_start = h * stride
vert_end = vert_start + f
horiz_start = w * stride
horiz_end = horiz_start + f
# Use the corners to define the (3D) slice of a_prev_pad (See Hint above the cell). (≈1 line)
a_slice_prev = a_prev_pad[vert_start:vert_end, horiz_start:horiz_end, :]
# Convolve the (3D) slice with the correct filter W and bias b, to get back one output neuron. (≈1 line)
Z[i, h, w, c] = np.sum(np.multiply(a_slice_prev, W[:, :, :, c]) + b[:, :, :, c])
# Making sure your output shape is correct
assert(Z.shape == (m, n_H, n_W, n_C))
# Save information in "cache" for the backprop
cache = (A_prev, W, b, hparameters)
return Z, cache
# test
np.random.seed(1)
A_prev = np.random.randn(10,4,4,3)
W = np.random.randn(2,2,3,8)
b = np.random.randn(1,1,1,8)
hparameters = {"pad" : 2,
"stride": 1}
Z, cache_conv = conv_forward(A_prev, W, b, hparameters)
print("Z's mean =", np.mean(Z))
print("cache_conv[0][1][2][3] =", cache_conv[0][1][2][3])
Finally, CONV layer should also contain an activation, in which case we would add the following line of code:
# Convolve the window to get back one output neuron
Z[i, h, w, c] = ...
# Apply activation
A[i, h, w, c] = activation(Z[i, h, w, c])
Pooling layer
The pooling (POOL) layer reduces the height and width of the input. It helps reduce computation, as well as helps make feature detectors more invariant to its position in the input. The two types of pooling layers are:
Max-pooling layer: slides an (f,f) window over the input and stores the max value of the window in the output.
Average-pooling layer: slides an (f,f) window over the input and stores the average value of the window in the output.
These pooling layers have no parameters for backpropagation to train. However, they have hyperparameters such as the window size f. This specifies the height and width of the fxf window you would compute a max or average over.
Forward Pooling
def pool_forward(A_prev, hparameters, mode = "max"):
"""
Implements the forward pass of the pooling layer
Arguments:
A_prev -- Input data, numpy array of shape (m, n_H_prev, n_W_prev, n_C_prev)
hparameters -- python dictionary containing "f" and "stride"
mode -- the pooling mode you would like to use, defined as a string ("max" or "average")
Returns:
A -- output of the pool layer, a numpy array of shape (m, n_H, n_W, n_C)
cache -- cache used in the backward pass of the pooling layer, contains the input and hparameters
"""
# Retrieve dimensions from the input shape
(m, n_H_prev, n_W_prev, n_C_prev) = A_prev.shape
# Retrieve hyperparameters from "hparameters"
f = hparameters["f"]
stride = hparameters["stride"]
# Define the dimensions of the output
n_H = int(1 + (n_H_prev - f) / stride)
n_W = int(1 + (n_W_prev - f) / stride)
n_C = n_C_prev
# Initialize output matrix A
A = np.zeros((m, n_H, n_W, n_C))
for i in range(m): # loop over the training examples
for h in range(n_H): # loop on the vertical axis of the output volume
for w in range(n_W): # loop on the horizontal axis of the output volume
for c in range (n_C): # loop over the channels of the output volume
# Find the corners of the current "slice" (≈4 lines)
vert_start = h * stride
vert_end = vert_start + f
horiz_start = w * stride
horiz_end = horiz_start + f
# Use the corners to define the current slice on the ith training example of A_prev, channel c. (≈1 line)
a_prev_slice = A_prev[i, vert_start:vert_end, horiz_start:horiz_end, c]
# Compute the pooling operation on the slice. Use an if statment to differentiate the modes. Use np.max/np.mean.
if mode == "max":
A[i, h, w, c] = np.max(a_prev_slice)
elif mode == "average":
A[i, h, w, c] = np.mean(a_prev_slice)
# Store the input and hparameters in "cache" for pool_backward()
cache = (A_prev, hparameters)
# Making sure your output shape is correct
assert(A.shape == (m, n_H, n_W, n_C))
return A, cache
# test
np.random.seed(1)
A_prev = np.random.randn(2, 4, 4, 3)
hparameters = {"stride" : 1, "f": 4}
A, cache = pool_forward(A_prev, hparameters)
print("mode = max")
print("A =", A)
print()
A, cache = pool_forward(A_prev, hparameters, mode = "average")
print("mode = average")
print("A =", A)
Backpropagation in convolutional neural networks
Convolutional layer backward pass
Computing dA:
This is the formula for computing $dA$ with respect to the cost for a certain filter $W_c$ and a given training example:
Where $Wc$ is a filter and $dZ{hw}$ is a scalar corresponding to the gradient of the cost with respect to the output of the conv layer Z at the hth row and wth column (corresponding to the dot product taken at the ith stride left and jth stride down). Note that at each time, we multiply the the same filter $W_c$ by a different dZ when updating dA. We do so mainly because when computing the forward propagation, each filter is dotted and summed by a different a_slice. Therefore when computing the backprop for dA, we are just adding the gradients of all the a_slices.
In code, inside the appropriate for-loops, this formula translates into:
Where $a{slice}$ corresponds to the slice which was used to generate the acitivation $Z{ij}$. Hence, this ends up giving us the gradient for $W$ with respect to that slice. Since it is the same $W$, we will just add up all such gradients to get $dW$.
In code, inside the appropriate for-loops, this formula translates into:
dW[:,:,:,c] += a_slice * dZ[i, h, w, c]
Computing db:
This is the formula for computing $db$ with respect to the cost for a certain filter $W_c$:
db = \sum_h \sum_w dZ_{hw} \tag{3}
As you have previously seen in basic neural networks, db is computed by summing $dZ$. In this case, you are just summing over all the gradients of the conv output (Z) with respect to the cost.
In code, inside the appropriate for-loops, this formula translates into:
db[:,:,:,c] += dZ[i, h, w, c]
Exercise: Implement the conv_backward function below. You should sum over all the training examples, filters, heights, and widths. You should then compute the derivatives using formulas 1, 2 and 3 above.
def conv_backward(dZ, cache):
"""
Implement the backward propagation for a convolution function
Arguments:
dZ -- gradient of the cost with respect to the output of the conv layer (Z), numpy array of shape (m, n_H, n_W, n_C)
cache -- cache of values needed for the conv_backward(), output of conv_forward()
Returns:
dA_prev -- gradient of the cost with respect to the input of the conv layer (A_prev),
numpy array of shape (m, n_H_prev, n_W_prev, n_C_prev)
dW -- gradient of the cost with respect to the weights of the conv layer (W)
numpy array of shape (f, f, n_C_prev, n_C)
db -- gradient of the cost with respect to the biases of the conv layer (b)
numpy array of shape (1, 1, 1, n_C)
"""
# Retrieve information from "cache"
(A_prev, W, b, hparameters) = cache
# Retrieve dimensions from A_prev's shape
(m, n_H_prev, n_W_prev, n_C_prev) = A_prev.shape
# Retrieve dimensions from W's shape
(f, f, n_C_prev, n_C) = W.shape
# Retrieve information from "hparameters"
stride = hparameters['stride']
pad = hparameters['pad']
# Retrieve dimensions from dZ's shape
(m, n_H, n_W, n_C) = dZ.shape
# Initialize dA_prev, dW, db with the correct shapes
dA_prev = np.zeros((m, n_H_prev, n_W_prev, n_C_prev))
dW = np.zeros((f, f, n_C_prev, n_C))
db = np.zeros((1, 1, 1, n_C))
# Pad A_prev and dA_prev
A_prev_pad = zero_pad(A_prev, pad)
dA_prev_pad = zero_pad(dA_prev, pad)
for i in range(m): # loop over the training examples
# select ith training example from A_prev_pad and dA_prev_pad
a_prev_pad = A_prev_pad[i]
da_prev_pad = dA_prev_pad[i]
for h in range(n_H): # loop over vertical axis of the output volume
for w in range(n_W): # loop over horizontal axis of the output volume
for c in range(n_C): # loop over the channels of the output volume
# Find the corners of the current "slice"
vert_start = h * stride
vert_end = vert_start + f
horiz_start = w * stride
horiz_end = horiz_start + f
# Use the corners to define the slice from a_prev_pad
a_slice = a_prev_pad[vert_start:vert_end, horiz_start:horiz_end, :]
# Update gradients for the window and the filter's parameters using the code formulas given above
da_prev_pad[vert_start:vert_end, horiz_start:horiz_end, :] += W[:,:,:,c] * dZ[i, h, w, c]
dW[:,:,:,c] += a_slice * dZ[i, h, w, c]
db[:,:,:,c] += dZ[i, h, w, c]
# Set the ith training example's dA_prev to the unpaded da_prev_pad (Hint: use X[pad:-pad, pad:-pad, :])
dA_prev[i, :, :, :] = dA_prev_pad[i, pad:-pad, pad:-pad, :]
# Making sure your output shape is correct
assert(dA_prev.shape == (m, n_H_prev, n_W_prev, n_C_prev))
return dA_prev, dW, db
# test
np.random.seed(1)
dA, dW, db = conv_backward(Z, cache_conv)
print("dA_mean =", np.mean(dA))
print("dW_mean =", np.mean(dW))
print("db_mean =", np.mean(db))
Pooling layer - backward pass
Even though a pooling layer has no parameters for backprop to update, you still need to backpropagation the gradient through the pooling layer in order to compute gradients for layers that came before the pooling layer.
Max pooling - backward pass
Before jumping into the backpropagation of the pooling layer, you are going to build a helper function called create_mask_from_window() which does the following:
X = \begin{bmatrix}1 && 3 \\4 && 2\end{bmatrix} \quad \rightarrow \quad M =\begin{bmatrix}0 && 0 \\1 && 0\end{bmatrix}\tag{4}
As you can see, this function creates a "mask" matrix which keeps track of where the maximum of the matrix is. True (1) indicates the position of the maximum in X, the other entries are False (0). You'll see later that the backward pass for average pooling will be similar to this but using a different mask.
Exercise: Implement create_mask_from_window(). This function will be helpful for pooling backward. Hints:
If you have a matrix X and a scalar x: A = (X == x) will return a matrix A of the same size as X such that:
A[i,j] = True if X[i,j] = x
A[i,j] = False if X[i,j] != x
Here, you don't need to consider cases where there are several maxima in a matrix.
def create_mask_from_window(x):
"""
Creates a mask from an input matrix x, to identify the max entry of x.
Arguments:
x -- Array of shape (f, f)
Returns:
mask -- Array of the same shape as window, contains a True at the position corresponding to the max entry of x.
"""
mask = (x == np.max(x))
return mask
# test
np.random.seed(1)
x = np.random.randn(2,3)
mask = create_mask_from_window(x)
print('x = ', x)
print("mask = ", mask)
Why do we keep track of the position of the max? It's because this is the input value that ultimately influenced the output, and therefore the cost. Backprop is computing gradients with respect to the cost, so anything that influences the ultimate cost should have a non-zero gradient. So, backprop will "propagate" the gradient back to this particular input value that had influenced the cost.
Average pooling - backward pass
In max pooling, for each input window, all the "influence" on the output came from a single input value--the max. In average pooling, every element of the input window has equal influence on the output. So to implement backprop, you will now implement a helper function that reflects this.
For example if we did average pooling in the forward pass using a 2x2 filter, then the mask you'll use for the backward pass will look like: dZ = 1 \quad \rightarrow \quad dZ =\begin{bmatrix}1/4 && 1/4 \\1/4 && 1/4\end{bmatrix}\tag{5}
This implies that each position in the $dZ$ matrix contributes equally to output because in the forward pass, we took an average.
def distribute_value(dz, shape):
"""
Distributes the input value in the matrix of dimension shape
Arguments:
dz -- input scalar
shape -- the shape (n_H, n_W) of the output matrix for which we want to distribute the value of dz
Returns:
a -- Array of size (n_H, n_W) for which we distributed the value of dz
"""
# Retrieve dimensions from shape (≈1 line)
(n_H, n_W) = shape
# Compute the value to distribute on the matrix (≈1 line)
average = dz / (n_H * n_W)
# Create a matrix where every entry is the "average" value (≈1 line)
a = np.ones(shape) * average
return a
a = distribute_value(2, (2,2))
print('distributed value =', a)
Putting it together: Pooling backward
You now have everything you need to compute backward propagation on a pooling layer.
Exercise: Implement the pool_backward function in both modes ("max" and "average"). You will once again use 4 for-loops (iterating over training examples, height, width, and channels). You should use an if/elif statement to see if the mode is equal to 'max' or 'average'. If it is equal to 'average' you should use the distribute_value() function you implemented above to create a matrix of the same shape as a_slice. Otherwise, the mode is equal to 'max', and you will create a mask with create_mask_from_window() and multiply it by the corresponding value of dZ.
def pool_backward(dA, cache, mode = "max"):
"""
Implements the backward pass of the pooling layer
Arguments:
dA -- gradient of cost with respect to the output of the pooling layer, same shape as A
cache -- cache output from the forward pass of the pooling layer, contains the layer's input and hparameters
mode -- the pooling mode you would like to use, defined as a string ("max" or "average")
Returns:
dA_prev -- gradient of cost with respect to the input of the pooling layer, same shape as A_prev
"""
# Retrieve information from cache (≈1 line)
(A_prev, hparameters) = cache
# Retrieve hyperparameters from "hparameters" (≈2 lines)
stride = hparameters['stride']
f = hparameters['f']
# Retrieve dimensions from A_prev's shape and dA's shape (≈2 lines)
m, n_H_prev, n_W_prev, n_C_prev = A_prev.shape
m, n_H, n_W, n_C = dA.shape
# Initialize dA_prev with zeros (≈1 line)
dA_prev = np.zeros_like(A_prev)
for i in range(m): # loop over the training examples
# select training example from A_prev (≈1 line)
a_prev = A_prev[i]
for h in range(n_H): # loop on the vertical axis
for w in range(n_W): # loop on the horizontal axis
for c in range(n_C): # loop over the channels (depth)
# Find the corners of the current "slice" (≈4 lines)
vert_start = h * stride
vert_end = vert_start + f
horiz_start = w * stride
horiz_end = horiz_start + f
# Compute the backward propagation in both modes.
if mode == "max":
# Use the corners and "c" to define the current slice from a_prev (≈1 line)
a_prev_slice = a_prev[vert_start:vert_end, horiz_start:horiz_end, c]
# Create the mask from a_prev_slice (≈1 line)
mask = create_mask_from_window(a_prev_slice)
# Set dA_prev to be dA_prev + (the mask multiplied by the correct entry of dA) (≈1 line)
dA_prev[i, vert_start: vert_end, horiz_start: horiz_end, c] += mask * dA[i, vert_start, horiz_start, c]
elif mode == "average":
# Get the value a from dA (≈1 line)
da = dA[i, vert_start, horiz_start, c]
# Define the shape of the filter as fxf (≈1 line)
shape = (f, f)
# Distribute it to get the correct slice of dA_prev. i.e. Add the distributed value of da. (≈1 line)
dA_prev[i, vert_start: vert_end, horiz_start: horiz_end, c] += distribute_value(da, shape)
# Making sure your output shape is correct
assert(dA_prev.shape == A_prev.shape)
return dA_prev
def create_placeholders(n_H0, n_W0, n_C0, n_y):
"""
Creates the placeholders for the tensorflow session.
Arguments:
n_H0 -- scalar, height of an input image
n_W0 -- scalar, width of an input image
n_C0 -- scalar, number of channels of the input
n_y -- scalar, number of classes
Returns:
X -- placeholder for the data input, of shape [None, n_H0, n_W0, n_C0] and dtype "float"
Y -- placeholder for the input labels, of shape [None, n_y] and dtype "float"
"""
X = tf.placeholder(tf.float32, shape=(None, n_H0, n_W0, n_C0))
Y = tf.placeholder(tf.float32,shape=(None,n_y))
return X, Y
Initialize parameters
You will initialize weights/filters $W_1$ and $W_2$ using tf.contrib.layers.xavier_initializer(seed = 0). You don't need to worry about bias variables as you will soon see that TensorFlow functions take care of the bias. Note also that you will only initialize the weights/filters for the conv2d functions. TensorFlow initializes the layers for the fully connected part automatically.
Reminder - to initialize a parameter W of shape [1,2,3,4] in Tensorflow, use:
W = tf.get_variable("W", [1,2,3,4], initializer = ...)
def initialize_parameters():
"""
Initializes weight parameters to build a neural network with tensorflow. The shapes are:
W1 : [4, 4, 3, 8]
W2 : [2, 2, 8, 16]
Returns:
parameters -- a dictionary of tensors containing W1, W2
"""
tf.set_random_seed(1) # so that your "random" numbers match ours
W1 =tf.get_variable('W1',[4,4,3,8],initializer=tf.contrib.layers.xavier_initializer(seed = 0))
W2 = tf.get_variable('W2',[2,2,8,16],initializer=tf.contrib.layers.xavier_initializer(seed = 0))
parameters = {"W1": W1,
"W2": W2}
return parameters
Reminder: The formulas relating the output shape of the convolution to the input shape is: nH=⌊stridenHprev−f+2×pad⌋+1nW=⌊stridenWprev−f+2×pad⌋+1nC=number of filters used in the convolution
Reminder: As there's no padding, the formulas binding the output shape of the pooling to the input shape is: nH=⌊stridenHprev−f⌋+1nW=⌊stridenWprev−f⌋+1nC=nCprev
may be helpful. It computes the maximum of an array.
Exercise: Implement the function below to equally distribute a value dz through a matrix of dimension shape.
Implement the function below to create placeholders for the input image X and the output Y. You should not define the number of training examples for the moment. To do so, you could use "None" as the batch size, it will give you the flexibility to choose it later. Hence X should be of dimension [None, n_H0, n_W0, n_C0] and Y should be of dimension [None, n_y]. .