Correct Answer: YES (Option 1)

Let's solve this using the truth table. The trick is to calculate the output of each neuron for all four inputs.

Step 1: Understand the network

We are given:

$h=ReLU(W^{(1)}x)$$h=ReLU(W^{(1)}x)$

$\hat{y}=W^{(2)}h$$\hat y=W^{(2)}h$

Weight matrices:

$W^{(1)}=\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$$W^{(1)}= \begin{bmatrix} 1&-1\\ -1&1 \end{bmatrix}$

$W^{(2)}=\left[\begin{array}{cc}1& 1\end{array}\right]$$W^{(2)}= \begin{bmatrix} 1&1 \end{bmatrix}$

Recall:

$ReLU(z)=max(0,z)$$ReLU(z)=\max(0,z)$

And classification threshold = 0.5.

Step 2: Calculate for all inputs

XOR truth table:

Case 1: (0,0)

$W^{(1)}\left[\begin{array}{c}0\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$W^{(1)} \begin{bmatrix}0\\0\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$

Apply ReLU:

$h=\left[\begin{array}{c}0\\ 0\end{array}\right]$$h=\begin{bmatrix}0\\0\end{bmatrix}$

Output:

$\hat{y}=0+0=0$$\hat y=0+0=0$

Since $0<0.5$$0<0.5$, prediction = 0.

Case 2: (0,1)

$W^{(1)}\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}-1\\ 1\end{array}\right]$$W^{(1)} \begin{bmatrix}0\\1\end{bmatrix} = \begin{bmatrix}-1\\1\end{bmatrix}$

Apply ReLU:

$h=\left[\begin{array}{c}0\\ 1\end{array}\right]$$h=\begin{bmatrix}0\\1\end{bmatrix}$

Output:

$\hat{y}=0+1=1$$\hat y=0+1=1$

Since $1>0.5$$1>0.5$, prediction = 1.

Case 3: (1,0)

$W^{(1)}\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ -1\end{array}\right]$$W^{(1)} \begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix}$

Apply ReLU:

$h=\left[\begin{array}{c}1\\ 0\end{array}\right]$$h=\begin{bmatrix}1\\0\end{bmatrix}$

Output:

$\hat{y}=1+0=1$$\hat y=1+0=1$

Since $1>0.5$$1>0.5$, prediction = 1.

Case 4: (1,1)

$W^{(1)}\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$W^{(1)} \begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$

Apply ReLU:

$h=\left[\begin{array}{c}0\\ 0\end{array}\right]$$h=\begin{bmatrix}0\\0\end{bmatrix}$

Output:

$\hat{y}=0+0=0$$\hat y=0+0=0$

Since $0<0.5$$0<0.5$, prediction = 0.

Step 3: Final comparison

All four predictions are correct!

Therefore:

$\boxed{{\displaystyle \text{YES}}}$$\boxed{\text{YES}}$

Exam shortcut: Understand what the hidden neurons are doing

The two hidden neurons calculate:

$h_1=ReLU(x_1-x_2)$$h_1=ReLU(x_1-x_2)$

$h_2=ReLU(x_2-x_1)$$h_2=ReLU(x_2-x_1)$

The output neuron adds them:

$\hat{y}=h_1+h_2$$\hat y=h_1+h_2$

For binary inputs, this is equivalent to:

$\boxed{{\displaystyle \hat{y}=|x_1-x_2|}}$$\boxed{\hat y=|x_1-x_2|}$

And that is exactly XOR!

Remember: The two hidden neurons detect the two different cases:

• First neuron fires when $x_1=1,x_2=0$$x_1=1,x_2=0$.

• Second neuron fires when $x_1=0,x_2=1$$x_1=0,x_2=1$.

The output neuron adds them, giving XOR.