Question

1. apply k-means clustering to a dataset Task Consider the following set of two-dimensional records: RID Dimension 1 Dimension2 1 00 8 4 5 4 N 3 2 4 4 6 N 5 2. 00 6 00 8 6 Use the k-means algorithm to cluster the data in the dataset with K=3. You can assume that the records with RIDS 1, 3, and 5 are used for the initial cluster centroids (means). You must include the intermediate results in each iteration of the algorithm, including distance calculation, placement of records in a cluster, and possible changes of centroids. Just showing the final clustering results will not be accepted.

Answer 1

the & we taking the points representation as (x, J.) new point (records and the centoid of each cluster and assign that point to the cluster Centroid for the cluster by the frormula /2 + x2 4 + 2) Here 13.5 that meon (8.4),(2,4), (2,8) are centroids of initial clusters. Here and (22,92). (2, 9,) is the distance between with minimum distance, than If a to a cluster then we have to calculate the a precord which we are newly assigning to cluster and (92 Y) !s existing centroid. we calcular new Point added new 요 2 Formula por distance Calculation -x,)+(42.9. end de cord is the newly adding point here we calculate the distance between the point and centroids of each cluster с 2 X2 92 C3 Newly Point - 22,8) C, = {(8,4)} Cq={(2,4)} cz = {(2,8)} S L (and record) di = [(aq-x)² + (92-9,² = √(8-5) + (x-4)² = √3²+0=5a = 3 de - (-2,77(4,-9.):12-5)+(u-u)' 143)+0=19 -3 to end ce stron dz = [(4-x} + (92-9.) -(8-5)+ (8-4) = / -33% + (0)2 = 19 + 16 = 25 25 GA (X2%₂] ² (2,4) The minimum distance is same for a cluster centroids So the record (5, 4) belongs to either C, or Cq. I am taking it in с. the

C = {(8,4), (5,4)} = {(2,4)} cz 22 (2,8)} New centroid for the c.. C2 (2, (5,4) wew c, - (85.444) =(6.5,4) . 2 y. next Point (2,6) (4th record) Here we calculate distance from new centroid for the cluster 1. di = √ (6.5-2) + (4-6)² (4.s)+(-2)20.25 +4 =124.25 = 4,924 da = √ (2-2)² + (6-4)² = Jo+q² = 5 = dz = 5(2-2)²+(8-6² = 0+qe = 5 = 2 Here also hor (2,6) we got two minimum distances so taking it into and cluster and claculating centroid. C;={8,4) (5,4)} Cz={(2,4).(2,6)} = {(2,3)} (242.4+5) =(2,5) (new cen wold for the C2 (2,6) centroid (6.5, u) New C₂ = second cluster) 29, d.-6.5 * 2 neat Point (86) (6th record) calculating distance Centroid forcat (25,4) Porcy 2,5) -6.5-8)+(4-6) (-1.5) (2) ² = 12.25 + 4 = 16.25 = 2.5 de 21 (2-8) + (5-6) (66)² + (-1)² = (36+1 2 37 2 6.082 d3 z 162-8) +(3-6) 108 +62) : 36+4 = 140 26.324

Here we get less distance for d, means the point (816) belongs to the cluster 1 so the Final (5,4) (816) clusters C, -18,4),(5,4),(8,6)} a = {(2,4), (2,6)} 3= {(2,8)} 2. (2.0 NOTE:- for the first distance calculation we got new Point two minimum distances. we may take the to any of the clusters. If we take it into cluster 1 we have to eupdate the centroid of cluster I. cor if we take it into the cluster 2. then we have to calculate centroid for the cluster 2. like this adding a new Point and calculating the new cennoid and use that centoid to calculate the distance and add the point, PR we have only one point in the cluster then point as centroid of the cluster. we take that RA In the last condition we have three points in cluster in this case to calculate centroid we apply mean formula! foo existing cenroid and newly point, means cluster 1 centroid is {6.5,u) by adding new Point (8,6) centroid is 6.5 +8 114.5 10 6,548,644)= (1.25,5) This is not red, if we have another new Point we have to use this centoid to calculate distance.