User:Manudouz/sandbox/Complete-linkage clustering

Working example

This working example is based on a JC69 genetic distance matrix computed from the 5S ribosomal RNA sequence alignment of five bacteria: Bacillus subtilis ( $a$ ), Bacillus stearothermophilus ( $b$ ), Lactobacillus viridescens ( $c$ ), Acholeplasma modicum ( $d$ ), and Micrococcus luteus ( $e$ ).^[1]^[2]

First step

First clustering

Let us assume that we have five elements $(a,b,c,d,e)$ and the following matrix $D_{1}$ 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, $D_{1}(a,b)=17$ is the smallest value of $D_{1}$ , so we join elements $a$ and $b$ .

First branch length estimation

Let $u$ denote the node to which $a$ and $b$ are now connected. Setting $\delta (a,u)=\delta (b,u)=D_{1}(a,b)/2$ ensures that elements $a$ and $b$ are equidistant from $u$ . This corresponds to the expectation of the ultrametricity hypothesis. The branches joining $a$ and $b$ to $u$ then have lengths $\delta (a,u)=\delta (b,u)=17/2=8.5$ (see the final dendrogram)

First distance matrix update

We then proceed to update the initial proximity matrix $D_{1}$ into a new proximity matrix $D_{2}$ (see below), reduced in size by one row and one column because of the clustering of $a$ with $b$ . Bold values in $D_{2}$ correspond to the new distances, calculated by retaining the maximum distance between each element of the first cluster $(a,b)$ and each of the remaining elements:

$D_{2}((a,b),c)=max(D_{1}(a,c),D_{1}(b,c))=max(21,30)=30$

$D_{2}((a,b),d)=max(D_{1}(a,d),D_{1}(b,d))=max(31,34)=34$

$D_{2}((a,b),e)=max(D_{1}(a,e),D_{1}(b,e))=max(23,21)=23$

Italicized values in $D_{2}$ 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 $D_{2}$ :

	(a,b)	c	d	e
(a,b)	0	30	34	23
c	30	0	28	39
d	34	28	0	43
e	23	39	43	0

Here, $D_{2}((a,b),e)=23$ is the lowest value of $D_{2}$ , so we join cluster $(a,b)$ with element $e$ .

Second branch length estimation

Let $v$ denote the node to which $(a,b)$ and $e$ are now connected. Because of the ultrametricity constraint, the branches joining $a$ or $b$ to $v$ , and $e$ to $v$ , are equal and have the following total length: $\delta (a,v)=\delta (b,v)=\delta (e,v)=23/2=11.5$

We deduce the missing branch length: $\delta (u,v)=\delta (e,v)-\delta (a,u)=\delta (e,v)-\delta (b,u)=11.5-8.5=3$ (see the final dendrogram)

Second distance matrix update

We then proceed to update the $D_{2}$ matrix into a new distance matrix $D_{3}$ (see below), reduced in size by one row and one column because of the clustering of $(a,b)$ with $e$ :

$D_{3}(((a,b),e),c)=max(D_{2}((a,b),c),D_{2}(e,c))=max(30,39)=39$

$D_{3}(((a,b),e),d)=max(D_{2}((a,b),d),D_{2}(e,d))=max(34,43)=43$

Third step

Third clustering

We again reiterate the three previous steps, starting from the updated distance matrix $D_{3}$ .

	((a,b),e)	c	d
((a,b),e)	0	39	43
c	39	0	28
d	43	28	0

Here, $D_{3}(c,d)=28$ is the smallest value of $D_{3}$ , so we join elements $c$ and $d$ .

Third branch length estimation

Let $w$ denote the node to which $c$ and $d$ are now connected. The branches joining $c$ and $d$ to $w$ then have lengths $\delta (c,w)=\delta (d,w)=28/2=14$ (see the final dendrogram)

Third distance matrix update

There is a single entry to update: $D_{4}((c,d),((a,b),e))=(D_{3}(c,((a,b),e)),D_{3}(d,((a,b),e)))=(39,43)=43$

Final step

The final $D_{4}$ matrix is:

	((a,b),e)	(c,d)
((a,b),e)	0	43
(c,d)	43	0

So we join clusters $((a,b),e)$ and $(c,d)$ .

Let $r$ denote the (root) node to which $((a,b),e)$ and $(c,d)$ are now connected. The branches joining $((a,b),e)$ and $(c,d)$ to $r$ then have lengths: