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K-means on six 2-D instances (K=2)
Short answer
Using K=2 and initial centroids (1,2), K-means after two iterations converges to Cluster 1: instances 1,3,4, centroid (1.50, 1.83); Cluster 2: instances 2,5,6, centroid (3.00, 5.07).
Identify the cluster of the following instances using K-Means clustering algorithm (Take K=2).
| Instance | X | Y |
|---|---|---|
| 1 | 1.0 | 1.5 |
| 2 | 2.5 | 5.5 |
| 3 | 1.5 | 1.0 |
| 4 | 2.0 | 3.0 |
| 5 | 2.5 | 3.5 |
| 6 | 4.0 | 6.2 |
Data
| Instance | X | Y |
|---|---|---|
| 1 | 1.0 | 1.5 |
| 2 | 2.5 | 5.5 |
| 3 | 1.5 | 1.0 |
| 4 | 2.0 | 3.0 |
| 5 | 2.5 | 3.5 |
| 6 | 4.0 | 6.2 |
Assumption: K = 2, and the first two instances are the initial centroids: C1 = (1.0, 1.5) and C2 = (2.5, 5.5). Euclidean distance d = √[(x − cx)² + (y − cy)²].
Iteration 1 (centroids C1 = (1.0, 1.5), C2 = (2.5, 5.5))
| Instance | (X, Y) | d to C1 | d to C2 | Cluster |
|---|---|---|---|---|
| 1 | (1.0, 1.5) | 0.00 | 4.27 | C1 |
| 2 | (2.5, 5.5) | 4.27 | 0.00 | C2 |
| 3 | (1.5, 1.0) | 0.71 | 4.61 | C1 |
| 4 | (2.0, 3.0) | 1.80 | 2.55 | C1 |
| 5 | (2.5, 3.5) | 2.50 | 2.00 | C2 |
| 6 | (4.0, 6.2) | 5.58 | 1.66 | C2 |
Clusters: C1 = {1, 3, 4}, C2 = {2, 5, 6}. New centroids: C1 = ((1.0 + 1.5 + 2.0)/3, (1.5 + 1.0 + 3.0)/3) = (1.50, 1.83); C2 = ((2.5 + 2.5 + 4.0)/3, (5.5 + 3.5 + 6.2)/3) = (3.00, 5.07).
Iteration 2 (centroids C1 = (1.50, 1.83), C2 = (3.00, 5.07))
| Instance | d to C1 | d to C2 | Cluster |
|---|---|---|---|
| 1 | 0.60 | 4.09 | C1 |
| 2 | 3.80 | 0.66 | C2 |
| 3 | 0.83 | 4.33 | C1 |
| 4 | 1.27 | 2.30 | C1 |
| 5 | 1.94 | 1.64 | C2 |
| 6 | 5.03 | 1.51 | C2 |
The assignment is unchanged, so the algorithm has converged (the centroids would stay the same).
Answer: Cluster 1 = {instances 1, 3, 4} with centroid (1.50, 1.83); Cluster 2 = {instances 2, 5, 6} with centroid (3.00, 5.07).
✓ Answer checked by Claude Opus 5.5 (high effort)
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