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

▸ Neural Network Basics : 

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1. Coursera: Machine Learning  
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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, ...).





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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))





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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.





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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).





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


• 


• 
  Correct


• 


• 





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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)





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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





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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)





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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)





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