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 ($$e$$).

First step
Let us assume that we have five elements $$(a,b,c,d,e)$$ and the following matrix $$D_1$$ of pairwise distances between them:
 * First clustering

In this example, $$D_1 (a,b)=17$$ is the smallest value of $$D_1$$, so we join elements $$a$$ and $$b$$.

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 branch length estimation

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:
 * First distance matrix update

$$D_2((a,b),c)=max(D_1(a,c),D_1(b,c))=max(21,30)=30$$

$$D_2((a,b),d)=max(D_1(a,d),D_1(b,d))=max(31,34)=34$$

$$D_2((a,b),e)=max(D_1(a,e),D_1(b,e))=max(23,21)=23$$

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
We now reiterate the three previous steps, starting from the new distance matrix $$D_2$$ :
 * Second clustering

Here, $$D_2 ((a,b),e)=23$$ is the lowest value of $$D_2$$, so we join cluster $$(a,b)$$ with element $$e$$.

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: $$\delta(a,v)=\delta(b,v)=\delta(e,v)=23/2=11.5$$
 * Second branch length estimation

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)

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 $$e$$ :
 * Second distance matrix update

$$D_3(((a,b),e),c)=max(D_2((a,b),c),D_2(e,c))=max(30,39)=39$$

$$D_3(((a,b),e),d)=max(D_2((a,b),d),D_2(e,d))=max(34,43)=43$$

Third step
We again reiterate the three previous steps, starting from the updated distance matrix $$D_3$$.
 * Third clustering

Here, $$D_3 (c,d)=28$$ is the smallest value of $$D_3$$, so we join elements $$c$$ and $$d$$.

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 branch length estimation

There is a single entry to update: $$D_4((c,d),((a,b),e))=(D_3(c,((a,b),e)), D_3(d,((a,b),e)))=(39, 43)=43$$
 * Third distance matrix update

Final step
The final $$D_4$$ matrix is:

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:

$$\delta(((a,b),e),r)=\delta((c,d),r)=43/2=21.5$$

We deduce the two remaining branch lengths:

$$\delta(v,r)=\delta(((a,b),e),r)-\delta(e,v)=21.5-11.5=10$$

$$\delta(w,r)=\delta((c,d),r)-\delta(c,w)=21.5-14=7.5$$

The complete-linkage dendrogram


The dendrogram is now complete. It is ultrametric because all tips ($$a$$ to $$e$$) are equidistant from $$r$$ :

$$\delta(a,r)=\delta(b,r)=\delta(e,r)=\delta(c,r)=\delta(d,r)=21.5$$

The dendrogram is therefore rooted by $$r$$, its deepest node.