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## Neural Networks: Representation

TOTAL POINTS 5

EXPLANATION:
A two layer (one input layer, one output layer; no hidden layer) neural network can represent the XOR function. (False)
=>We must compose multiple logical operations by using a hidden layer to represent the XOR function.

Any logical function over binary-valued (0 or 1) inputs x1 and x2 can be (approximately) represented using some neural network.(True)
=>Since we can build the basic AND, OR, and NOT functions with a two layer network, we can (approximately) represent any logical function by composing these basic functions over multiple layers.

Suppose you have a multi-class classification problem with three classes, trained with a 3 layer network. Let a(3)1=(hΘ(x))1 be the activation of the first output unit, and similarly a(3)2=(hΘ(x))2 and a(3)3=(hΘ(x))3. Then for any input x, it must be the case that a(3)1+a(3)2+a(3)3=1. (False)
=>The outputs of a neural network are not probabilities, so their sum need not be 1.

The activation values of the hidden units in a neural network, with the sigmoid activation function applied at every layer, are always in the range (0, 1).
(True)
=>fact

Question 2

Consider the following neural network which takes two binary-valued inputs $x_1, x_2 \in \{0, 1\}$ and outputs $h_\Theta(x)$. Which of the following logical functions does it (approximately) compute?

VARIATION IN 2ND QUESTION:

EXPLANATION:
This network outputs approximately 1 when atleast one input is 1.

EXPLANATION:

$a_1^{(3)} = g(\Theta_{1,0}^{(2)}a_0^{(2)} + \Theta_{1,1}^{(2)}a_1^{(2)} + \Theta_{1,2}^{(2)}a_2^{(2)})$
This correctly uses the first row of Θ(2) and includes the "+1" term of a0(2)  This correctly uses the first
row of Θ(2) and includes the "+1" term of   a0(2)

EXPLANATION:
a(2) = g(Θ(1)x)=g(z)

EXPLANATION:
Swapping Θ(1) swaps the hidden layers output a^{(2)}. But the swap of Θ(2) cancels out the change, so the output will remain unchanged.

VARIATION IN 5 TH QUESTION:

EXPLANATION:
Swapping Θ(1) swaps the hidden layers output a^{(2)}. But the swap of Θ(2) cancels out the change, so the output will remain unchanged.

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reference : coursera

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