Let us assume that we have five elements and the following matrix of pairwise distances between them:
a
b
c
d
e
a
0
17
21
31
23
b
17
0
30
34
21
c
21
30
0
28
39
d
31
34
28
0
43
e
23
21
39
43
0
In this example, is the lowest value of , so we join elements and .
First branch length estimation
Let denote the node to which and are now connected. Setting ensures that elements and are equidistant from . This corresponds to the expectation of the ultrametricity hypothesis.
The branches joining and to then have lengths (see the final dendrogram)
First distance matrix update
We then proceed to update the initial proximity matrix into a new proximity matrix (see below), reduced in size by one row and one column because of the clustering of with .
Bold values in correspond to the new distances, calculated by retaining the minimum distance between each element of the first cluster and each of the remaining elements:
Italicized values in are not affected by the matrix update as they correspond to distances between elements not involved in the first cluster.
Second step
Second clustering
We now reiterate the three previous steps, starting from the new distance matrix :
(a,b)
c
d
e
(a,b)
0
21
31
21
c
21
0
28
39
d
31
28
0
43
e
21
39
43
0
Here, and are the lowest values of , so we join cluster with element and with element .
Second branch length estimation
Let denote the node to which , and are now connected. Because of the ultrametricity constraint, the branches joining or to , and to , and also to are equal and have the following total length:
We then proceed to update the matrix into a new distance matrix (see below), reduced in size by two rows and two columns because of the clustering of with and with :
Final step
The final matrix is:
((a,b),c,e)
d
((a,b),c,e)
0
28
d
28
0
So we join clusters and .
Let denote the (root) node to which and are now connected.
The branches joining and to then have lengths:
We deduce the remaining branch length:
The single-linkage dendrogram
Single Linkage Dendrogram 5S data
The dendrogram is now complete. It is ultrametric because all tips (, , , , and ) are equidistant from :
The dendrogram is therefore rooted by , its deepest node.