[coursera/dl&nn/week2]Basics of Neural Network programming(quiz)

This blog helps me review the course on coursera.

Wrong answer:

3.reshape to a column vector

9."*" means the elementwise product 

".dot" means matrix multiplication operation



1. Question 1

What does a neuron compute?

A neuron
computes a linear function (z = Wx + b) followed by an activation function

Correct 

Correct, we generally say that the output of a neuron is a = g(Wx + b) where g is the activation function (sigmoid, tanh, ReLU, ...).

A neuron
computes a function g that scales the input x linearly (Wx + b)

A neuron
computes the mean of all features before applying the output to an activation function

A neuron
computes an activation function followed by a linear function (z = Wx + b)

2. Question 2

Which of these is the "Logistic Loss"?

L(i)(ˆy(i),y(i))=∣y(i)−ˆy(i)∣2

L(i)(ˆy(i),y(i))=−(y(i)log(ˆy(i))+(1−y(i))log(1−ˆy(i)))

Correct 

Correct, this is the logistic loss you've seen in lecture!

L(i)(ˆy(i),y(i))=∣y(i)−ˆy(i)∣

L(i)(ˆy(i),y(i))=max(0,y(i)−ˆy(i))

3. Question 3

Suppose img is a (32,32,3) array, representing a 32x32 image with 3 color channels red, green and blue. How do you reshape this into a column vector?

x = img.reshape((1,32*32,*3))

x = img.reshape((3,32*32))

x = img.reshape((32*32*3,1))

x = img.reshape((32*32,3))

This should not be selected 

4. Question 4

Consider the two following random arrays "a" and "b":What will be the shape of "c"?

a = np.random.randn(2, 3) # a.shape = (2, 3)
b = np.random.randn(2, 1) # b.shape = (2, 1)
c = a + b

c.shape
= (2, 1)

c.shape
= (2, 3)

Correct 

Yes! This is broadcasting. b (column vector) is copied 3 times so that it can be summed to each column of a.

c.shape
= (3, 2)

The computation
cannot happen because the sizes don't match. It's going to be "Error"!

5. Question 5

Consider the two following random arrays "a" and "b":

a = np.random.randn(4, 3) # a.shape = (4, 3)
b = np.random.randn(3, 2) # b.shape = (3, 2)
c = a*b

What will be the shape of "c"?

c.shape
= (3, 3)

c.shape
= (4, 3)

c.shape =
(4,2)

The computation
cannot happen because the sizes don't match. It's going to be "Error"!

Correct 

Indeed! In numpy the "*" operator indicates element-wise multiplication. It is different from "np.dot()". If you would try "c = np.dot(a,b)" you would get c.shape = (4, 2).

6. Question 6

Suppose you have nxinput
features per example. Recall that X=[x(1)x(2)...x(m)].
What is the dimension of X?

(1,m)

(m,1)

(m,nx)

(nx,m)

Correct 

7. Question 7

Recall that "np.dot(a,b)" performs a matrix multiplication on a and b, whereas "a*b" performs an element-wise multiplication.Consider the two following random arrays "a" and "b":What
is the shape of c?a = np.random.randn(12288, 150) # a.shape = (12288, 150)
b = np.random.randn(150, 45) # b.shape = (150, 45)
c = np.dot(a,b)

c.shape
= (150, 150)

c.shape
= (12288, 150)

c.shape =
(12288, 45)

Correct 

Correct, remember that a np.dot(a, b) has shape (number of rows of a, number of columns of b). The sizes match because :"number of columns of a = 150 = number of rows of b"

The
computation cannot happen because the sizes don't match. It's going to be "Error"!

8. Question 8

Consider the following code snippet:

# a.shape = (3,4)
# b.shape = (4,1)
for i in range(3):
for j in range(4):
c[i][j] = a[i][j] + b[j]

How do you vectorize this?

c = a + b

c = a.T +
b.T

c = a.T +
b

c = a +
b.T

Correct 

9. Question 9

Consider the following code:

a = np.random.randn(3, 3)
b = np.random.randn(3, 1)
c = a*b

What will be c? (If you’re not sure, feel free to run this in python to find out).

This will
invoke broadcasting, so b is copied three times to become (3,3), and ∗is
an element-wise product so c.shape will be (3, 3)

This will
invoke broadcasting, so b is copied three times to become (3, 3), and ∗[Math
Processing Error] invokes a matrix multiplication operation of two 3x3 matrices so c.shape will be (3, 3)

This should not be selected

This will
multiply a 3x3 matrix a with a 3x1 vector, thus resulting in a 3x1 vector. That is, c.shape = (3,1).

It will lead
to an error since you cannot use “*” to operate on these two matrices. You need to instead use np.dot(a,b)

10. Question 10

Consider the following computation graph.


What is the output J?

J = (c -
1)*(b + a)

J = (a -
1) * (b + c)

Correct 

Yes. J = u + v - w = a*b + a*c - (b + c) = a * (b + c) - (b + c) = (a - 1) * (b + c).

J = a*b
+ b*c + a*c

J = (b
- 1) * (c + a)