Question

Gi 1D: 11-10,22 = 8,13-6,24 clustering to obtain k 2 clusters by hand. Specifically, 4,15-3,16 2, perform k-means ven the following six iteims in 1. Start from initial cluster centers c0,2 9. Show your steps for all iterations: (1 the cluster assignments i.... ys: (2) the updated cluster centers at the end of that iteration; (3) the energy at the end of that iteration 2. Repeat the above but start from initial cluster centers c 3. Which k-means solution is better? Why? 8, 2 9.

Answer 1

x_1 = 10, x_2 = 8, x_3 = 6, x_4 = 4, x_5 = 3, x_6 = 2 {10, 8, 6, 4, 3, 2} c_1 = 0 c_2 = 9 k_1 = {4, 3, 42} k_2 = {10, 8, 6} c_1 = 9/3 = 3 c_2 = 24/3 = 8 therefore k_1 = {4, 3, 2} k_2 = {10, 8, 6} therefore c_1 = 3 c_2 = 8 since we are getting same mean we must stop therefore k_1 = {4, 3, 2} k_2 = {10, 8, 6} therefore y_1 = 4, y_2 = 3, y_3 = 2, y_4 = 10, y_5 = 8, y_6 = 6 updated cluster centers are c_1 = 3 and c_2 = 8 Energy = sum of all clusters = 10 + 8 + 6 + 4 + 3 + 2 = 33 c_1 = 8 c_2 = 9 k_1 = {6, 4, 3, 2} k_2 = {10, 8} c_1 = 15/4 = 3.7 almostequalto

The second k means solution is better as it has cluster center values close to each other which help to calculate the mean cluster faster.