Given the following data observations a b 1 16 15 2 3 19 3 8...

Given the following data: Observations A B 1 16 15 2 3 19 3 8 12 4 9 8 5 5 4 Using K-means clustering and k=2. Assuming the starting points are observations 2 and 4. How many iterations will be performed before the clustering process stops? Also what will be the centroids of the final clusters that have been formed? Iterations performed 2, Centroid for cluster 1 (3, 19), Centroid for Cluster 2 (9.5, 9.75) Iterations performed 1, Centroid for cluster 1 (3, 19), Centroid for Cluster 2 (14.2, 11.65) Iterations performed 1, Centroid for cluster 1 (4, 18), Centroid for Cluster 2 (7, 8.5) Iterations performed 2, Centroid for cluster 1 (4, 18), Centroid for Cluster 2 (9.5, 8.75)

Transcript

00:01 Here it is given that in terms of a and b, 1, 2, 3, 4 and 5, that is a value it is 16, 3, 8, 9 and 5, whereas b value it is 15, 19, 12, 8 and 4.

00:23 Starting point is observation 2 and 4.

00:28 So observation 2 and 4 are the starting points of the concept.

00:35 So a and b, 2 and 4, it is nothing but 3, 19, 9, 18, 8.

00:46 The remaining terms are 1, 3 and 5.

00:49 So it will be in terms of 16, 15, 15, 8, 12, 5 and 4.

00:58 Now, the cluster will be in terms of cluster in terms of a and b.

01:08 It will be, cluster will be 24.

01:11 That is 3 plus 9 divided by 2 which is equal to 6 and 19 plus 8 divided by 2 which is equal to 6.

01:21 And next one it will be 135 which is nothing but 16 plus 8 plus 5 divided by 3 which gives you 9 .67 and 15 12 14 which gives you 10 .3p.

01:42 So the euclidean distance can be calculated in terms of euclidean distance can be calculated in terms of euclidean distance can be calculated in terms of.

01:53 Of d square of 1 comma 2 and 4 which is nothing but 3 minus 6 the whole square plus 19 minus 13 .5 the whole square which gives you 39 .25.

02:12 3 .25 similarly in terms of 2 .24 you will be getting 39 .25 then in terms of 2 .24 you will be getting 39 .25 then in terms of of 3 comma 24 it will be 1 or 2 .25 and this process continues for both the term and we will be obtaining the concept in terms of cluster 24 24 and for 135 now in terms of 1 .35 now in terms of 1 2 3 4 and 5 the solutions will be 39 .25 and 5.

02:53 The solutions will be 39 .25.

02:57 39 .25, 102 .25, 6 .25 and 91 .25.

03:08 Similarly, it will be 61 .88, 119 .66, 5 .58, again 5 .58, then 61 .83.

03:23 So here the minimum is the fourth one.

03:32 The total will be 101 .23, 158 .91, 107 .83, 13 .13 .13...

Question

Please give Ace some feedback