Question

SECTION B. This part is about using KNN algorithms. In this section we will consider the following points and we will use the Euclidean distance. D = (2). D;= 0) (1.D = (). 0,= (*) Y3 = x, Y4 = x. And their labels are yı = 0, y2 = 0, Now we are given a test point T: 1 = () B1. What is the label of T if we use 1NN algorithm to C? (a) 50% o and 50% x (b) No decision (c) o (d) x B2. What is the label of T if we use 2NN algorithm to C/? (a) 50% o and 50% x (b) No decision (c) o (d) x B3. What is the label of T if we use 3NN algorithm to C ? (a) 50% o and 50% x (b) No decision (c) o (d) x

Answer 1

d1=(2,2)          
T =(3,3)            

dis=sqrt(1+1)
    =1.14

d2=(1,3)
T= (3,3)

dis=sqrt(4)
    =2


d3=(4,3)
T= (3,3)

dis=sqrt(1)
    =1


d4=(5,4)
T= (3,3)

dis=sqrt(4+1)
    =2.23



B1)1NN Algorithm

near point to T is d3
d3 lable is x
So Lable of T is x

B2)2NN Algorithm

near point to T is d3,d1
d3 lable is x
d1 lable is 0
So Lable of T is 50% 0 and 50% x


B3)3NN Algorithm

near point to T is d1,d2,d3
d3 lable is x
d1 lable is 0
d2 lable is 0
So Lable of T is 0