Exercise 10.4

To solve this exercise, there are some definitions that we should take into account:


We don't have enough information to tell. If all observations in cluster {1,2,3} and cluster {4,5} have the same pairwise distance, the fusion between these clusters occur at the same height on the tree. An example would be d(1,4)=d(1,5)=d(2,4)=d(2,5)=d(3,4)=d(2,5)=1, where d(x,y) denotes the distance between observation x and observation y.

In contrast, if observations between clusters have different distances, the fusion will occur higher on the complete linkage. For example, if d(1,4)=1, d(1,5)=2, d(2,4)=3, d(2,5)=4, d(3,4)=5, and d(3,5)=6, single linkage would fuse at height 1 and complete linkage at height 6.


They fuse at the same height. The distance between two observations is unique. Thus, the smallest and largest pairwise distance are the same and the will fuse at the same height.