A challenging question of ANN 
I ran into a challenge when see this solved old exam.



As seen in this image the author select $(D)$ as the best option with
minimum node. and in another page mentioned that if we use Bipolar
then $(E)$ is the answer. anyone can describe why $(E)$ is the answer
when we use bipolar?

if we have step function then if input of neuron $> 0$ then output of neuron is $1$ else $0$
if we have bipolar function then if input of neuron $> 0$ then output of neuron is $1$ else $-1$
 A: As long as we are talking only about additive neurons (i.e. all inputs to the neuron are summed together before being passed to the activation function), "unipolar" and "bipolar" can be used interchangeably. We can always transform a "unipolar" output to a "bipolar" one by multiplying by 2 and subtracting one:
$$
o_{bipolar} = 2 \cdot o_{unipolar} - 1
$$
To implement this in the network, we just need to double the weights and decrease the bias in by one for each input neuron:
$$
w_{ij}' = 2 \cdot w_{ij}
$$
$$
bias_j' = bias_{j} - N_{in[j]}
$$
where $N_{in[j]}$ is the number of neurons feeding their output as the input to the $j$-th neuron.
So the part

if we use Bipolar

can be safely ignored. Now, as Thomas points out in his comment, the first layer of the networks (D) and (E) simply map the continuous $(x, y)$-space onto $\{0, 1\}^2$ (or, alternatively, $\{-1, 1\}^2$, if you use "bipolar" neurons). With the given arrangement of the classes this becomes the classical XOR-problem, and you need two further layers to solve it.

A: If neuron had three outputs, say [-1,0,1] then it could draw three areas with linear boundaries as shown here for the first layer and solution would be (E).

The second layer simply picks the south and north region as one category, and west and east regions as another.
A neuron with two outputs, whether it's [0,1] or [-1,1] or any other pair of values, can only criss-cross. So the solution can only be (D)
Sideways
If you abstract yourself from the actual question, then it's clear that the variables are "wrong" :) This is asking for feature engineering (another buzzword!) - shift and rotate by 45 degrees would work beautifully. First you de-mean the data, then create new variables: S = x+y and V=x-y. Then your classification becomes simply a bit problem: L is (S*V<0).
No, this is not the solution of the problem, because it still requires four regions, and with binary neurons you still need D in this problem. I just thought it's an interesting twist to consider

