written 7.0 years ago by | modified 2.6 years ago by |
Step 1: Consider A and B be the main clusters.
Let c1 and c2 denotes the co-ordinate of the centroid.
So, c1= (2, 2), c2= (3, 2)
I | X1 | X2 |
---|---|---|
A | 2 | 2 |
B | 3 | 2 |
C | 1 | 1 |
D | 3 | 1 |
E | 1.5 | 0.5 |
Step 2:
Distance from first centroid c1 = (2, 2) to B = (3, 2)
c1 (2, 2) is $\sqrt{(3-2)^2+(2-2)^2}=1$
Distance from first centroid c1 = (2, 2) to C = (1, 1)
c1 (2, 2) is $\sqrt{(2-1)^2+(2-1)^2}=1.41$
Distance from first centroid c1 = (2, 2) to D = (3, 1)
c1 (2, 2) is $\sqrt{(3-2)^2+(1-2)^2}=1.41$
Distance from first centroid c2 = (2, 2) to E = (1.5, 0.5)
c1 (3, 2) is $\sqrt{(2-1.5)^2+(2-0.5)^2}=1.58$
Distance from first centroid c2 = (3, 2) to C = (1, 1)
c1 (3, 2) is $\sqrt{(3-1)^2+(2-1)^2}=2.24$
Distance from first centroid c2 = (3, 2) to D = (3, 1)
c1 (3, 2) is $\sqrt{(3-3)^2+(2-1)^2}=1$
Distance from first centroid c2 = (3, 2) to E = (1.5, 0.5)
c1 (3, 2) is $\sqrt{(3-1.5)^2+(2-0.5)^2}=2.12$
$ D0=\begin{matrix} \ 0 & 1 & 1.41 & 1.41 & 1.58\\ 1& 0 & 2.24 & 1 & 2.12\\ \end{matrix}$
We assign objects based on min distance to G0.
$ G0=\begin{matrix} \ 1 & 0 & 1 & 0 & 1\\ 0 & 1 & 0 & 1 & 0\\ \end{matrix}$
Now we compute the new centroid of each group as shown below.
c1 = (((2+1+1.5)/3), ((2+1+0.5)/3)) = (1.5, 1.167)
c2 = (((3+3)/2), ((2+1)/2)) = (3, 1.5)
The next step is to compute the distance of all objects to the new centroid.
Similar to above steps we have distance matrix at iteration 1 is
$ D1=\begin{matrix} \ 0.97 & 1.71 & 0.53 & 1.51 & 0.67\\ 1.12 & 0.5 & 2.06 & 0.5 & 1.80\\ \end{matrix} $
$ G1=\begin{matrix} \ 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0\\ \end{matrix} $
We obtain the result that G0=G1.
Comparing the grouping of last iteration and this iteration reveals that the object does not move group anymore.
Thus, we get final grouping as the result
I | X1 | X2 | Group |
---|---|---|---|
A | 2 | 2 | 1 |
B | 3 | 2 | 2 |
C | 1 | 1 | 1 |
D | 3 | 1 | 2 |
E | 1.5 | 0.5 | 1 |