Question

(b) Apply the perceptron algorithm to the following pattern classes 5 Wi (0,0,0)T, (1,0,0)7, (1,0,1)T, (1,1,0)T\ W2 ((0,0,1)T, (0,1,1)T, (0,1,0)T, (1,1,1)T Let C 1 and W(1) = (-1, -2, -2, 0)1. Sketch the decision surface

Answer 1

For class ω1 we let y(1) = (0, 0, 0,1)^T, y(2) = (1, 0, 0,1)^T , y(3) = (1, 0, 1,1)^T ,
y(4) = (1, 1, 0,1)^T .

1 WI Figure of DECISION SURFACE fAmber Aauy

Similarly, for class ω2, y(5) = (0, 0, 1,1)^T , y(6) = (0, 1, 1,1)^T,
y(7) = (0, 1, 0,1)^T , y(8) = (1, 1, 1,1)^T .

Then,using c = 1 and
w(1) = (−1,−2,−2,0)^T

Going through percepton theorem we can say that :-

w(1)^T y(1) = 0, w(2) = w(1)+y(1) = (−1,−2,−2,1)^T ;
w(2)^T y(2) = 0, w(3) = w(2)+y(2) = (0,−2,−2,2)^T ;
w(3)^T y(3) = 0, w(4) = w(3)+y(3) = (1,−2,−1,3)^T ;
w(4)^T y(4) = 2, w(5) = w(4) = (1,−2,−1,3)^T ;
w(5)^T y(5) = 2, w(6) = w(5)−y(5) = (−1,−2,−2,2)^T ;
w(6)^T y(6) =−2, w(7) = w(6) = (−1,−2,−2,2)^T;
w(7)^T y(7) = 0, w(8) = w(7)−y(7) = (1,−3,−2,1)^T ;
w(8)^T y(8) =−3, w(9) = w(8) = (1,−3,−2,1)^T .

Once more, a total emphasis over all examples without a blunder was not accomplished,

so the examples are reused by letting

y(17) = y(1), y(18) = y(2), and so on, which
gives:
w(17)^T y(17) = 1, w(18) =w(17) = (2,−3,−2,1)^T ;
w(18)^Ty(18) = 3, w(19) =w(18) = (2,−3,−2,1)^T;
w(19)^T y(19) = 1, w(20) =w(19) = (2,−3,−2,1)^T ;
w(20)^Ty(20) = 0, w(21) =w(20)+y(20) = (3,−2,−2,2)^T ;
w(21)^Ty(21) = 0, w(22) =w(21)−y(21) = (3,−2,−3,1)^T .

It is effectively checked that no more rectifications happen after this progression, so

w(22) = (3,−2,−3,1)^T is a solution weight vector.

SKETCH THE DECISION SURFACE:

The decision surface is given by the equation
w^T y= 3y1 −2y2 −3y3 +1 =0.

P.S. -THE FIGURE IS ATTACHED