2020-5-31 吳恩達-NN&DL-w2 NN基礎(課後作業)

參考https://zhuanlan.zhihu.com/p/31268885

1、What does a neuron compute?

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

1、神經元計算什麼？

• 神經元先計算激活函數，再計算線性函數(z = Wx + b)
• 神經元先計算線性函數（z = Wx + b），再計算激活函數。（正確）
• 神經元計算函數g，函數g計算(Wx + b)。
• 在將輸出應用於激活函數之前，神經元計算所有特徵的平均值

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2、Which of these is the “Logistic Loss”?以下哪個是邏輯迴歸損失函數？

• $L^{(i)}(\hat y^{(i)},y^{(i)})=|y^{(i)}-\hat y^{(i)}|$
• $L^{(i)}(\hat y^{(i)},y^{(i)})=max(0,y^{(i)}-\hat y^{(i)})$
• $L^{(i)}(\hat y^{(i)},y^{(i)})=-(y^{(i)}log(\hat y^{(i)})+(1+y^{(i)})log(1-\hat y^{(i)}))$。（正確）
• $L^{(i)}(\hat y^{(i)},y^{(i)})=|y^{(i)}-\hat y^{(i)}|^2$

參見2.3 logistic 迴歸損失函數 。採用平方誤差可能導致優化非凸（局部最優，不是全局最優），而上面定義的損失函數可以得到全局最優結果。

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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?

假設img是一個（32,32,3）數組，具有3個顏色通道：紅色、綠色和藍色的32x32像素的圖像。 如何將其轉換爲列向量？

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

參見2.16 關於 python / numpy 向量的說明

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4、Consider the two following random arrays “a” and “b”: 有2個隨機數組a和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”? 請問數組c的維度是怎麼樣？

• c.shape = (3, 2)
• c.shape = (2, 1)
• c.shape = (2, 3)。（正確）
• The computation cannot happen because the sizes don’t match.It’s going to be “Error”!

參見2.15 Python 中的廣播。
B（列向量）複製3次，然後和A的每一列相加。

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5、Consider the two following random arrays “a” and “b”:有2個隨機數組a和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”? 請問數組c的維度是怎麼樣？

• The computation cannot happen because the sizes don’t match.It’s going to be “Error”!（正確）
• c.shape = (4, 3)
• c.shape = (3, 3)
• c.shape = (4, 2)

數組按照元素相乘需要兩個矩陣之間的維數相同，但是a和b維度不同，所以這將報錯，無法計算。

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6、Suppose you have n_x input features per example. Recall that $X=[x^{(1)}, x^{(2)}…x^{(m)}]$. What is the dimension of X?
假設你的每一個實例有$n_x$個輸入特徵，想一下在$X=[x^{(1)}, x^{(2)}…x^{(m)}]$中，X的維度是多少？

• (m,1)
• (m,$n_x$)
• ($n_x$,m)。（正確）
• (1,m)

m個x向量橫向堆疊。x是包含$n_x$個元素的列向量。

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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.
回想一下，np.dot（a，b）在a和b上執行矩陣乘法，而`a * b’執行元素方式的乘法。

Consider the two following random arrays “a” and “b”:
有2個隨機數組“a”和“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? 請問數組c的維度是怎麼樣？

• c.shape = (150, 150)
• The computation cannot happen because the sizes don’t match.It’s going to be “Error”!
• c.shape = (12288, 150)
• c.shape = (12288, 45)。（正確）

矩陣乘法，沒什麼好說的。

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8、Consider the following code snippet: 觀察下面代碼

# 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? 如何向量化？

• c = a + b.T。（正確）
• c = a.T + b
• c = a + b
• c = a.T + b.T

c的維度（3，4），a維度無需轉置，b需要轉置並廣播。

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9、Consider the following code: 觀察下面代碼

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

What will be c?

• This will invoke broadcasting, so b is copied three times to become (3,3), and * is an element-wise product so c.shape = (3, 3).（正確）
• This will invoke broadcasting, so b is copied three times to become (3,3), and * invokes a matrix multiplication operation of 3x3 matrices so c 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).

數組按照元素相乘。使用廣播機制，b會被複制三次。

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10、Consider the following computation graph. 觀察下面計算圖

What is the output J? J是什麼？

• J= (c - 1) * (b + a)
• J= (a - 1) * (b + c)。（正確）
• J= a * b + b * c + a * c
• J= (b - 1) * (c + a)

推導過程如下

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