| written 6.6 years ago by | • modified 5.8 years ago |
The table below comprises sample data items indicting the distance between the elements.
| Item | E | A | C | B | D |
|---|---|---|---|---|---|
| E | 0 | 1 | 2 | 2 | 3 |
| A | 1 | 0 | 2 | 5 | 3 |
| C | 2 | 2 | 0 | 1 | 6 |
| B | 2 | 5 | 1 | 0 | 3 |
| D | 3 | 3 | 6 | 3 | 0 |
| written 6.6 years ago by | • modified 5.8 years ago |
The table below comprises sample data items indicting the distance between the elements.
| Item | E | A | C | B | D |
|---|---|---|---|---|---|
| E | 0 | 1 | 2 | 2 | 3 |
| A | 1 | 0 | 2 | 5 | 3 |
| C | 2 | 2 | 0 | 1 | 6 |
| B | 2 | 5 | 1 | 0 | 3 |
| D | 3 | 3 | 6 | 3 | 0 |
| written 6.6 years ago by | • modified 6.6 years ago |
Agglomerative Clustering Technique:
In Hierarchical clustering algorithms, either top down or bottom up approach is followed. In Bottom up approach, every object is considered to be a cluster and in subsequent iterations they are merged into single cluster. Therefore it is also called as Hierarchical Agglomerative Clustering.
An HAC clustering is typically visualised as a dendrogram , where each merge is represented by a horizontal line.
Flowchart:

Numerical:
Given:
Distance matrix:

Step 1: From above given distance matrix, E and A clusters are having minimum distance, so merge them together to form cluster(E,A).

Distance matrix:
$\begin{aligned} \text{dist((E A), C)} &= \text{MIN(dist(E,C), dist(A,C))} \\ &= \text{MIN(2,2)} \\ &= 2 \\ \text{} \\ \text{dist((E A), B)} &= \text{MIN(dist(E,B), dist(A,B))} \\ &= \text{MIN(2,5)} \\ &= 2 \\ \text{} \\ \text{dist((E A), D)} &= \text{MIN(dist(E,D), dist(A,D))} \\ &= \text{MIN(3,3)} \\ &= 3 \\ \end{aligned}$

Step 2: Consider the distance matrix obtained in step 1. Since B,C distance is minimum, we combine B and C.

$\begin{aligned} \text{dist((B C), (E A))} &= \text{MIN(dist(B,E), dist(B,A), dist(C E), dist(C A))} \\ &= \text{MIN(2,5, 2, 2)} \\ &= 2 \\ \text{} \\ \text{dist((B C), D)} &= \text{MIN(dist(B, D), dist(C,D))} \\ &= \text{MIN(3,6)} \\ &= 3 \\ \end{aligned}$

Step 3: Consider the distance matrix obtained in step 2. Since (E,A) and (B,C) distance is minimum, we combine them

$\begin{aligned} \text{dist((E A), (B C))} &= \text{MIN(dist(E,B), dist(E,C), dist(A B), dist(A C))} \\ &= \text{MIN(2, 2, 2, 5, 2)} \\ &= 2 \\ \end{aligned}$

Step 4: Finally combine D with (E A B C)
