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Coursera: Neural Networks and Deep Learning (Week 2) Quiz [MCQ Answers] - deeplearning.ai

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Coursera : Neural Networks and Deep Learning Week 2 MCQ Quiz Answers | APDaga | DumpBox

  1. What does a neuron compute?

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

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

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

    • 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, ...).




  2. Which of these is the "Logistic Loss"?



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






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  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((32*32*3,1))
      Correct

    • x = img.reshape((3,32*32))

    • x = img.reshape((32*32,3))




  4. Consider the two following random arrays "a" and "b":
    a = np.random.randn(2, 3) # a.shape = (2, 3)
    b = np.random.randn(2, 1) # b.shape = (2, 1)
    c = a + b
    What will be the shape of "c"?

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

    • c.shape = (2, 1)

    • c.shape = (3, 2)

    • 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.




  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 = (4, 3)

    • c.shape = (4,2)

    • c.shape = (3, 3)

    • 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. Suppose you have input features per example. Recall that . What is the dimension of X?



    • Correct






  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":

    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)
    What is the shape of c?

    • 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"!

    • c.shape = (150,150)




  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.T + b

    • c = a.T + b.T 

    • c = a + b

    • c = a + b.T
      Correct




  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)
      Correct

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

    • 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. Consider the following computation graph.
    computation graph | APDaga | DumpBox | Coursera
    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)


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