Question

Given then following data points

x(1) = (2, 8); x(2) = (2, 5); x(3) = (1, 2); x(4) = (5, 8)

x(5) = (7, 3); x(6) = (6, 4); x(7) = (8, 4); x(8) = (4, 7)

Compute 2 iterations of the K-Means algorithm by hand using the Forgy’s initialisation choosing x(3), x(4) and x(6). Calculate the loss function in each iteration.

Answer 1

Here the data points are given as x⁽¹⁾=(2,8) , x⁽²⁾= (2,5) , x⁽³⁾= (1,2) ,x⁽⁴⁾=(5,8) , x⁽⁵⁾=(7,3), x⁽⁶⁾=(6,4) , x⁽⁷⁾=(8,4) , x⁽⁸⁾=(4,7)

the initialized points are x⁽³⁾= (1,2) ,x⁽⁴⁾=(5,8) and x⁽⁶⁾=(6,4)

the distance or the loss can be calculated by the formula

L = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

	Distance from x(3)	Distance from x(4)	Distance from x(5)	Nearest point
(2,8)	6.08	3	5.65	x(4)
(2,5)	3.16	4.24	4.12	x(3)
(1,2)	0	7.12	5.38	x(3)
(5,8)	7.21	0	4.12	x(4)
(7,3)	6.08	5.38	1.41	x(6)
(6,4)	5.38	4.12	0	x(6)
(8,4)	7.28	5	2	x(6)
(4,7)	5.83	1.41	3.61	x(4)

x⁽³⁾ contains x⁽²⁾ and x⁽³⁾, so the new center of x⁽³⁾ is = $\frac{2+1}{2},\frac{5+2}{2}$ = (1.5 , 3.5)

x⁽⁴⁾ contains x⁽¹⁾ , x⁽⁴⁾ and x⁽⁸⁾ so the new center of x⁽³⁾ is = $\frac{2+5+4}{3},\frac{8+8+8}{3}$ = (3.67 , 8)

x⁽⁶⁾ contains x⁽⁵⁾ , x⁽⁶⁾ and x⁽⁷⁾ so the new center of x⁽³⁾ is = $\frac{7+6+8}{3},\frac{3+4+4}{3}$ = (7 , 3.67)

	Distance from (1.5,3.5)	Distance from (3.67, 8)	Distance from (7, 3.67)	Nearest point
(2,8)	4.52	1.67	6.61	(3.67, 8) = x(4)
(2,5)	1.58	3.43	5.17	(1.5,3.5) = x(3)
(1,2)	1.58	6.56	6.22	(1.5,3.5) = x(3)
(5,8)	5.7	1.33	4.77	(3.67, 8) = x(4)
(7,3)	5.52	6.01	0.67	(7, 3.67) = x(6)
(6,4)	4.52	4.63	1.05	(7, 3.67) = x(6)
(8,4)	6.51	5.89	1.05	(7, 3.67) = x(6)
(4,7)	4.30	1.05	4.48	(3.67, 8) = x(4)

so , the clusters

x⁽³⁾ contains x⁽²⁾ and x⁽³⁾ points

x⁽⁴⁾ contains x⁽¹⁾ , x⁽⁴⁾ and x⁽⁸⁾ points

x⁽⁶⁾ contains x⁽⁵⁾ , x⁽⁶⁾ and x⁽⁷⁾ points