1. Pehle question samjho

Perceptron ka output hai:

$\hat{y}=\{\begin{array}{ll}+1& \text{if }{w}^{T}x>0\\ -1& \text{if }{w}^{T}x<0\end{array}$$\hat y = \begin{cases} +1 & \text{if } w^Tx>0\\ -1 & \text{if } w^Tx<0 \end{cases}$

Yahan:

• $x$$x$ = input vector

• $w$$w$ = weight vector

• $w^{T}x$$w^Tx$ = dot product

• $y$$y$ = actual class label (+1 or -1)

Question bol raha hai ki perceptron ne data ko correctly separate kar diya hai.

Matlab prediction aur actual label same hone chahiye.

2. Ab dono possible cases dekho

Case 1: Positive class

Suppose:

$y=+1$$y=+1$

Correct classification ke liye:

$w^{T}x>0$$w^Tx>0$

Ab multiply kar do:

$y(w^{T}x)=(+1)(positive)$$y(w^Tx)= (+1)(positive)$

$\boxed{{\displaystyle y(w^{T}x)>0}}$$\boxed{y(w^Tx)>0}$

Case 2: Negative class

Suppose:

$y=-1$$y=-1$

Correct classification ke liye:

$w^{T}x<0$$w^Tx<0$

Ab multiply kar do:

$y(w^{T}x)=(-1)(negative)$$y(w^Tx)=(-1)(negative)$

Negative × Negative = Positive.

$\boxed{{\displaystyle y(w^{T}x)>0}}$$\boxed{y(w^Tx)>0}$

3. Dono cases ko combine karo

Notice karo, dono cases mein:

$\boxed{{\displaystyle y(w^{T}x)>0}}$$\boxed{y(w^Tx)>0}$

Isliye Option A correct hai.

4. Baaki options galat kyun hain?

5. Exam trick: Ye yaad rakhna

Jab bhi perceptron mein labels $+1,-1$$+1,-1$ diye hon aur data correctly classified ho:

$\boxed{{\displaystyle y_i(w^T{x}_i)>0}}$$\boxed{y_i(w^Tx_i)>0}$

Isko Perceptron classification condition bol sakte ho.

Ek chhota sa check:

Agar kisi data point ka:

$y=-1,w^{T}x=-5$$y=-1,\quad w^Tx=-5$

Toh kya point correctly classified hai?