吴恩达deeplearning.ai课程作业,自己写的答案。
补充说明:
1. 评论中总有人问为什么直接复制这些notebook运行不了?请不要直接复制粘贴,不可能运行通过的,这个只是notebook中我们要自己写的那部分,要正确运行还需要其他py文件,请自己到GitHub上下载完整的。这里的部分仅仅是参考用的,建议还是自己按照提示一点一点写,如果实在卡住了再看答案。个人觉得这样才是正确的学习方法,况且作业也不算难。
2. 关于评论中有人说我是抄袭,注释还没别人详细,复制下来还运行不过。答复是:做伸手党之前,请先搞清这个作业是干什么的。大家都是从GitHub上下载原始的作业,然后根据代码前面的提示(通常会指定函数和公式)来编写代码,而且后面还有expected output供你比对,如果程序正确,结果一般来说是一样的。请不要无脑喷,说什么跟别人的答案一样的。说到底,我们要做的就是,看他的文字部分,根据提示在代码中加入部分自己的代码。我们自己要写的部分只有那么一小部分代码。
3. 由于实在很反感无脑喷子,故禁止了下面的评论功能,请见谅。如果有问题,请私信我,在力所能及的范围内会尽量帮忙。
Welcome to Course 4’s first assignment! In this assignment, you will implement convolutional (CONV) and pooling (POOL) layers in numpy, including both forward propagation and (optionally) backward propagation.
Notation:
- Superscript [l] [ l ] denotes an object of the lth l t h layer.
- Example: a[4] a [ 4 ] is the 4th 4 t h layer activation. W[5] W [ 5 ] and b[5] b [ 5 ] are the 5th 5 t h layer parameters.
Superscript (i) ( i ) denotes an object from the ith i t h example.
Lowerscript i i denotes the ith i t h entry of a vector.
nH n H , nW n W and nC n C denote respectively the height, width and number of channels of a given layer. If you want to reference a specific layer l l , you can also write n[l]H n H [ l ] , n[l]W n W [ l ] , n[l]C n C [ l ] .
We assume that you are already familiar with numpy
and/or have completed the previous courses of the specialization. Let’s get started!
Let’s first import all the packages that you will need during this assignment.
- numpy is the fundamental package for scientific computing with Python.
- matplotlib is a library to plot graphs in Python.
- np.random.seed(1) is used to keep all the random function calls consistent. It will help us grade your work.
import numpy as np
import h5py
import matplotlib.pyplot as plt
%matplotlib inline
plt.rcParams['figure.figsize'] = (5.0, 4.0) # set default size of plots
plt.rcParams['image.interpolation'] = 'nearest'
plt.rcParams['image.cmap'] = 'gray'
%load_ext autoreload
%autoreload 2
np.random.seed(1)
You will be implementing the building blocks of a convolutional neural network! Each function you will implement will have detailed instructions that will walk you through the steps needed:
This notebook will ask you to implement these functions from scratch in numpy
. In the next notebook, you will use the TensorFlow equivalents of these functions to build the following model:
Note that for every forward function, there is its corresponding backward equivalent. Hence, at every step of your forward module you will store some parameters in a cache. These parameters are used to compute gradients during backpropagation.
Although programming frameworks make convolutions easy to use, they remain one of the hardest concepts to understand in Deep Learning. A convolution layer transforms an input volume into an output volume of different size, as shown below.
In this part, you will build every step of the convolution layer. You will first implement two helper functions: one for zero padding and the other for computing the convolution function itself.
Zero-padding adds zeros around the border of an image:
The main benefits of padding are the following:
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. Use np.pad. 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:
a = np.pad(a, ((0,0), (1,1), (0,0), (3,3), (0,0)), 'constant', constant_values = (..,..))
# GRADED FUNCTION: zero_pad
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)
"""
### START CODE HERE ### (≈ 1 line)
X_pad = np.pad(X, ((0, 0), (pad, pad), (pad, pad), (0, 0)), 'constant')
### END CODE HERE ###
return X_pad
np.random.seed(1)
x = np.random.randn(4, 3, 3, 2)
x_pad = zero_pad(x, 2)
print ("x.shape =", x.shape)
print ("x_pad.shape =", x_pad.shape)
print ("x[1,1] =", x[1,1])
print ("x_pad[1,1] =", x_pad[1,1])
fig, axarr = plt.subplots(1, 2)
axarr[0].set_title('x')
axarr[0].imshow(x[0,:,:,0])
axarr[1].set_title('x_pad')
axarr[1].imshow(x_pad[0,:,:,0])
x.shape = (4, 3, 3, 2)
x_pad.shape = (4, 7, 7, 2)
x[1,1] = [[ 0.90085595 -0.68372786]
[-0.12289023 -0.93576943]
[-0.26788808 0.53035547]]
x_pad[1,1] = [[ 0. 0.]
[ 0. 0.]
[ 0. 0.]
[ 0. 0.]
[ 0. 0.]
[ 0. 0.]
[ 0. 0.]]
Expected Output:
x.shape: | (4, 3, 3, 2) |
x_pad.shape: | (4, 7, 7, 2) |
x[1,1]: | [[ 0.90085595 -0.68372786] [-0.12289023 -0.93576943] [-0.26788808 0.53035547]] |
x_pad[1,1]: | [[ 0. 0.] [ 0. 0.] [ 0. 0.] [ 0. 0.] [ 0. 0.] [ 0. 0.] [ 0. 0.]] |
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:
In a computer vision application, each value in the matrix on the left corresponds to a single pixel value, and we convolve a 3x3 filter with the image by multiplying its values element-wise with the original matrix, then summing them up. In this first step of the exercise, you will implement a single step of convolution, corresponding to applying a filter to just one of the positions to get a single real-valued output.
Later in this notebook, you’ll apply this function to multiple positions of the input to implement the full convolutional operation.
Exercise: Implement conv_single_step(). Hint.
# GRADED FUNCTION: conv_single_step
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
"""
### START CODE HERE ### (≈ 2 lines of code)
# Element-wise product between a_slice and W. Add bias.
s = a_slice_prev * W + b
# Sum over all entries of the volume s
Z = np.sum(s)
### END CODE HERE ###
return Z
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)
Z = -23.1602122025
Expected Output:
Z | -23.1602122025 |
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.
Hint:
1. To select a 2x2 slice at the upper left corner of a matrix “a_prev” (shape (5,5,3)), you would do:
a_slice_prev = a_prev[0:2,0:2,:]
This will be useful when you will define a_slice_prev
below, using the start/end
indexes you will define.
2. To define a_slice you will need to first define its corners vert_start
, vert_end
, horiz_start
and horiz_end
. This figure may be helpful for you to find how each of the corner can be defined using h, w, f and s in the code below.
Reminder:
The formulas relating the output shape of the convolution to the input shape is:
For this exercise, we won’t worry about vectorization, and will just implement everything with for-loops.
# GRADED FUNCTION: conv_forward
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
"""
### START CODE HERE ###
# 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 = np.int((n_H_prev + 2 * pad - f) / stride) + 1
n_W = np.int((n_W_prev + 2 * pad - f) / stride) + 1
# 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)
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] = conv_single_step(a_slice_prev, W[:, :, :, c], b[:, :, :, c])
### END CODE HERE ###
# 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
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])
Z's mean = 0.155859324889
cache_conv[0][1][2][3] = [-0.20075807 0.18656139 0.41005165]
Expected Output:
Z’s mean | 0.155859324889 |
cache_conv[0][1][2][3] | [-0.20075807 0.18656139 0.41005165] |
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])
You don’t need to do it here.
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 f , f ) window over the input and stores the max value of the window in the output.
Average-pooling layer: slides an ( f,f 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 f . This specifies the height and width of the fxf window you would compute a max or average over.
Now, you are going to implement MAX-POOL and AVG-POOL, in the same function.
Exercise: Implement the forward pass of the pooling layer. Follow the hints in the comments below.
Reminder:
As there’s no padding, the formulas binding the output shape of the pooling to the input shape is:
# GRADED FUNCTION: pool_forward
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))
### START CODE HERE ###
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)
### END CODE HERE ###
# 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
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)
mode = max
A = [[[[ 1.74481176 1.6924546 2.10025514]]]
[[[ 1.19891788 1.51981682 2.18557541]]]]
mode = average
A = [[[[-0.09498456 0.11180064 -0.14263511]]]
[[[-0.09525108 0.28325018 0.33035185]]]]
Expected Output:
A = | [[[[ 1.74481176 1.6924546 2.10025514]]] [[[ 1.19891788 1.51981682 2.18557541]]]] |
A = | [[[[-0.09498456 0.11180064 -0.14263511]]] [[[-0.09525108 0.28325018 0.33035185]]]] |
Congratulations! You have now implemented the forward passes of all the layers of a convolutional network.
The remainer of this notebook is optional, and will not be graded.
In modern deep learning frameworks, you only have to implement the forward pass, and the framework takes care of the backward pass, so most deep learning engineers don’t need to bother with the details of the backward pass. The backward pass for convolutional networks is complicated. If you wish however, you can work through this optional portion of the notebook to get a sense of what backprop in a convolutional network looks like.
When in an earlier course you implemented a simple (fully connected) neural network, you used backpropagation to compute the derivatives with respect to the cost to update the parameters. Similarly, in convolutional neural networks you can to calculate the derivatives with respect to the cost in order to update the parameters. The backprop equations are not trivial and we did not derive them in lecture, but we briefly presented them below.
Let’s start by implementing the backward pass for a CONV layer.
This is the formula for computing dA d A with respect to the cost for a certain filter Wc W c and a given training example:
Where Wc W c is a filter and dZhw d Z h w 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 Wc 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:
da_prev_pad[vert_start:vert_end, horiz_start:horiz_end, :] += W[:,:,:,c] * dZ[i, h, w, c]
This is the formula for computing dWc d W c ( dWc d W c is the derivative of one filter) with respect to the loss:
Where aslice a s l i c e corresponds to the slice which was used to generate the acitivation Zij Z i j . Hence, this ends up giving us the gradient for W W with respect to that slice. Since it is the same W W , we will just add up all such gradients to get dW d W .
In code, inside the appropriate for-loops, this formula translates into:
dW[:,:,:,c] += a_slice * dZ[i, h, w, c]
This is the formula for computing db d b with respect to the cost for a certain filter Wc W c :
As you have previously seen in basic neural networks, db is computed by summing dZ d Z . 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)
"""
### START CODE HERE ###
# 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[pad:-pad, pad:-pad, :]
### END CODE HERE ###
# 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
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))
dA_mean = 9.60899067587
dW_mean = 10.5817412755
db_mean = 76.3710691956
* Expected Output: *
dA_mean | 9.60899067587 |
dW_mean | 10.5817412755 |
db_mean | 76.3710691956 |
Next, let’s implement the backward pass for the pooling layer, starting with the MAX-POOL layer. 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.
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:
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:
- np.max() may be helpful. It computes the maximum of an array.
- 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
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.
"""
### START CODE HERE ### (≈1 line)
mask = (x == np.max(x))
### END CODE HERE ###
return mask
np.random.seed(1)
x = np.random.randn(2,3)
mask = create_mask_from_window(x)
print('x = ', x)
print("mask = ", mask)
x = [[ 1.62434536 -0.61175641 -0.52817175]
[-1.07296862 0.86540763 -2.3015387 ]]
mask = [[ True False False]
[False False False]]
Expected Output:
x = | [[ 1.62434536 -0.61175641 -0.52817175] [-1.07296862 0.86540763 -2.3015387 ]] |
mask = | [[ True False False] [False False False]] |
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.
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:
This implies that each position in the dZ d Z matrix contributes equally to output because in the forward pass, we took an average.
Exercise: Implement the function below to equally distribute a value dz through a matrix of dimension shape. Hint
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
"""
### START CODE HERE ###
# 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 = average * np.ones(shape)
### END CODE HERE ###
return a
a = distribute_value(2, (2,2))
print('distributed value =', a)
distributed value = [[ 0.5 0.5]
[ 0.5 0.5]]
Expected Output:
distributed_value = | [[ 0.5 0.5] |
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
"""
### START CODE HERE ###
# 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((m, n_H_prev, n_W_prev, n_C_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)
# print(dA[i, h, w, c])
# print(mask.shape)
# print(dA_prev[i, vert_start: vert_end, horiz_start: horiz_end, c].shape)
dA_prev[i, vert_start: vert_end, horiz_start: horiz_end, c] += dA[i, h, w, c] * mask
elif mode == "average":
# Get the value a from dA (≈1 line)
da = dA[i, h, w, 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)
### END CODE ###
# Making sure your output shape is correct
assert(dA_prev.shape == A_prev.shape)
return dA_prev
np.random.seed(1)
A_prev = np.random.randn(5, 5, 3, 2)
hparameters = {"stride" : 1, "f": 2}
A, cache = pool_forward(A_prev, hparameters)
dA = np.random.randn(5, 4, 2, 2)
dA_prev = pool_backward(dA, cache, mode = "max")
print("mode = max")
print('mean of dA = ', np.mean(dA))
print('dA_prev[1,1] = ', dA_prev[1,1])
print()
dA_prev = pool_backward(dA, cache, mode = "average")
print("mode = average")
print('mean of dA = ', np.mean(dA))
print('dA_prev[1,1] = ', dA_prev[1,1])
mode = max
mean of dA = 0.145713902729
dA_prev[1,1] = [[ 0. 0. ]
[ 5.05844394 -1.68282702]
[ 0. 0. ]]
mode = average
mean of dA = 0.145713902729
dA_prev[1,1] = [[ 0.08485462 0.2787552 ]
[ 1.26461098 -0.25749373]
[ 1.17975636 -0.53624893]]
Expected Output:
mode = max:
mean of dA = | 0.145713902729 |
dA_prev[1,1] = | [[ 0. 0. ] [ 5.05844394 -1.68282702] [ 0. 0. ]] |
mode = average
mean of dA = | 0.145713902729 |
dA_prev[1,1] = | [[ 0.08485462 0.2787552 ] [ 1.26461098 -0.25749373] [ 1.17975636 -0.53624893]] |
Congratulation on completing this assignment. You now understand how convolutional neural networks work. You have implemented all the building blocks of a neural network. In the next assignment you will implement a ConvNet using TensorFlow.