WPGMA

Algorithm

The WPGMA algorithm constructs a rooted tree (dendrogram) that reflects the structure present in a pairwise distance matrix (or a similarity matrix). At each step, the nearest two clusters, say i and j , are combined into a higher-level cluster i ∪ j . Then, its distance to another cluster k is simply the arithmetic mean of the distances between k and members of i ∪ j  :

d ( i ∪ j ) , k = d i , k + d j , k 2

The WPGMA algorithm produces rooted dendrograms and requires a constant-rate assumption: it produces an ultrametric tree in which the distances from the root to every branch tip are equal. This ultrametricity assumption is called the molecular clock when the tips involve DNA, RNA and protein data.

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 :

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 δ ( a , u ) = δ ( 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 δ ( a , u ) = δ ( b , u ) = 17 / 2 = 8.5 (see the final dendrogram)

We then proceed to update the initial distance matrix D 1 into a new distance 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 averaging distances between the first cluster ( a , b ) and each of the remaining elements:

D 2 ( ( a , b ) , c ) = ( D 1 ( a , c ) + D 1 ( b , c ) ) / 2 = ( 21 + 30 ) / 2 = 25.5

D 2 ( ( a , b ) , d ) = ( D 1 ( a , d ) + D 1 ( b , d ) ) / 2 = ( 31 + 34 ) / 2 = 32.5

D 2 ( ( a , b ) , e ) = ( D 1 ( a , e ) + D 1 ( b , e ) ) / 2 = ( 23 + 21 ) / 2 = 22

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  :

Here, D 2 ( ( a , b ) , e ) = 22 is the smallest value of D 2 , so we join cluster ( a , b ) and 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 length: δ ( a , v ) = δ ( b , v ) = δ ( e , v ) = 22 / 2 = 11

We deduce the missing branch length: δ ( u , v ) = δ ( e , v ) − δ ( a , u ) = δ ( e , v ) − δ ( b , u ) = 11 − 8.5 = 2.5 (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  : D 3 ( ( ( a , b ) , e ) , c ) = ( D 2 ( ( a , b ) , c ) + D 2 ( e , c ) ) / 2 = ( 25.5 + 39 ) / 2 = 32.25

Of note, this average calculation of the new distance does not account for the larger size of the ( a , b ) cluster (two elements) with respect to e (one element). Similarly:

D 3 ( ( ( a , b ) , e ) , d ) = ( D 2 ( ( a , b ) , d ) + D 2 ( e , d ) ) / 2 = ( 32.5 + 43 ) / 2 = 37.75

The averaging procedure therefore gives differential weight to the initial distances of matrix D 1 . This is the reason why the method is weighted, not with respect to the mathematical procedure but with respect to the initial distances.

Third step

We again reiterate the three previous steps, starting from the updated distance matrix D 3 .

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 δ ( c , w ) = δ ( d , w ) = 28 / 2 = 14 (see the final dendrogram)

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 ) ) ) / 2 = ( 32.25 + 37.75 ) / 2 = 35

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:

δ ( ( ( a , b ) , e ) , r ) = δ ( ( c , d ) , r ) = 35 / 2 = 17.5

We deduce the two remaining branch lengths:

δ ( v , r ) = δ ( ( ( a , b ) , e ) , r ) − δ ( e , v ) = 17.5 − 11 = 6.5

δ ( w , r ) = δ ( ( c , d ) , r ) − δ ( c , w ) = 17.5 − 14 = 3.5

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

δ ( a , r ) = δ ( b , r ) = δ ( e , r ) = δ ( c , r ) = δ ( d , r ) = 17.5

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