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## Thursday, November 4, 2010

Next, lets examine whether using the complete-linkage criterion would result in a different clustering of this sample data set. Complete linkage seeks to minimize the distance among the records in two clusters that are farthest from each other. Figure 8.3 illustrates complete-linkage clustering for this data set.

* Step 1: Since each cluster contains a single record only, there is no difference between single linkage and complete linkage at step 1. The two clusters each containing 33 are again combined.
* Step 2: Just as for single linkage, the clusters containing values 15 and 16 are combined into a new cluster. Again, this is because there is no difference in the two criteria for single-record clusters.
* Step 3: At this point, complete linkage begins to diverge from its predecessor. In single linkage, cluster {15,16} was at this point combined with cluster {18}. But complete linkage looks at the farthest neighbors, not the nearest neighbors. The farthest neighbors for these two clusters are 15 and 18, for a distance of 3. This is the same distance separating clusters {2} and {5}. The completelinkage criterion is silent regarding ties, so we arbitrarily select the first such combination found, therefore combining the clusters {2} and {5} into a new cluster.
* Step 4: Now cluster {15,16} is combined with cluster {18}.
* Step 5: Cluster {2,5} is combined with cluster {9}, since the complete-linkage distance is 7, the smallest among remaining clusters.
* Step 6: Cluster {25} is combined with cluster {33,33}, with a complete-linkage distance of 8.
* Step 7: Cluster {2,5,9} is combined with cluster {15,16,18}, with a completelinkage distance of 16.
* Step 8: Cluster {25,33,33} is combined with cluster {45}, with a completelinkage distance of 20.
* Step 9: Cluster {2,5,9,15,16,18} is combined with cluster {25,33,33,45}. All records are now contained in this last large cluster.

Finally, with average linkage, the criterion is the average distance of all the records in cluster A from all the records in cluster B. Since the average of a single record is the records value itself, this method does not differ from the earlier methods in the early stages, where single-record clusters are being combined. At step 3, average linkage would be faced with the choice of combining clusters {2} and {5}, or combining the {15, 16} cluster with the single-record {18} cluster. The average distance between the {15, 16} cluster and the {18} cluster is the average of |18 ? 15| and |18?16|, which is 2.5, while the average distance between clusters {2} and {5} is of course 3. Therefore, average linkage would combine the {15, 16} cluster with cluster
{18} at this step, followed by combining cluster {2} with cluster {5}. The reader may verify that the average-linkage criterion leads to the same hierarchical structure for this example as the complete-linkage criterion. In general, average linkage leads to clusters more similar in shape to complete linkage than does single linkage.