Finding the center of a cluster

Question

I have the following problem - made abstract to bring out the key issues.

I have 10 points each which is some distance from the other. I want to

be able to find the center of the cluster i.e. the point for which the pairwise distance to each other point is minimised,
let p(j) ~ p(k) represent the pairwise distance beteen points j and k
p(i) is center-point of the cluster iff p(i) s.t. min[sum(p(j)~p(k))] for all 0 < j,k <= n where we have n points in the cluster
determine how to split the cluster in to two clusters once the number of data points in the cluster goes above some threshold t.

This is not euclidean space. But the distances can be summarised as follows - p(i) is point i:

       p(1)    p(2)    p(3)    p(4)    p(5)    p(6)    p(7)    p(8)    p(9)    p(10)
p(1)    0       2       1       3       2       3       3       2       3        4
p(2)    2       0       1       3       2       3       3       2       3        4
p(3)    1       1       0       2       0       1       2       1       2        3
p(4)    3       3       2       0       1       2       3       2       3        4      
p(5)    2       2       1       1       0       1       2       1       2        3   
p(6)    3       3       2       2       1       0       3       2       3        4   
p(7)    3       3       2       3       2       3       0       1       2        3  
p(8)    2       2       1       2       1       2       1       0       1        2 
p(9)    3       3       2       3       2       3       2       1       0        1
p(10)   4       4       3       4       3       4       3       2       1        0

How would I calculate which is the center point of this cluster?

The application has to do with clustering concepts - my application is a semantic data store - the points represent abstract objects. I want to cluster objects to be able to determine "concepts" — Ankur, Aug 10 '09 at 09:07
Do you mean: choose i that minimizes [sum(p(i)~p(j)] for 0 < j <= n, so that p(i) is the center? If so, I think that's your answer to part 1; otherwise I'm not sure what you mean. — Jim Lewis, Aug 10 '09 at 09:23
@Jim - in words, we want to minimise the pairwise distances for all pairs of points in a cluster. The point that gives the minimum is the center point. — Ankur, Aug 10 '09 at 17:34

score 9 · Accepted Answer · edited Feb 23 '18 at 04:56

9

As far as I understand this looks like K Means Clustering, and what you are looking for is usually known as 'Medoids'.

See here: http://en.wikipedia.org/wiki/Medoids or here: http://en.wikipedia.org/wiki/K-medoids

edited Feb 23 '18 at 04:56

hirse

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answered Aug 10 '09 at 09:05

Darknight

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Upvoted: this indeed looks like the way to go for non-Euclidean metrics. But it still requires at least that the triangle inequality holds; I confess I'm not patient enough to verify that for Ankur's example. – Jim Lewis Aug 10 '09 at 09:38
@ Jim, the triangle inequality does hold in my "metric space", so then this should work. – Ankur Aug 10 '09 at 17:35

score 4 · Answer 2 · answered Aug 11 '09 at 14:02

I may be about to have that frisson that comes just before displaying utter stupidity. But doesn't this yield easily to brute force? In Python:

distances = [
[ 0 , 2 , 1 , 3 , 2 , 3 , 3 , 2 , 3 , 4 , ],
[ 2 , 0 , 1 , 3 , 2 , 3 , 3 , 2 , 3 , 4 , ],
[ 1 , 1 , 0 , 2 , 0 , 1 , 2 , 1 , 2 , 3 , ],
[ 3 , 3 , 2 , 0 , 1 , 2 , 3 , 2 , 3 , 4 , ],
[ 2 , 2 , 1 , 1 , 0 , 1 , 2 , 1 , 2 , 3 , ],
[ 3 , 3 , 2 , 2 , 1 , 0 , 3 , 2 , 3 , 4 , ],
[ 3 , 3 , 2 , 3 , 2 , 3 , 0 , 1 , 2 , 3 , ],
[ 2 , 2 , 1 , 2 , 1 , 2 , 1 , 0 , 1 , 2 , ],
[ 3 , 3 , 2 , 3 , 2 , 3 , 2 , 1 , 0 , 1 , ],
[ 4 , 4 , 3 , 4 , 3 , 4 , 3 , 2 , 1 , 0 , ],
]

currentMinimum = 99999

for point in range ( 10 ) :
    distance_sum = 0
    for second_point in range ( 10 ) :
        if point == second_point : continue
        distance_sum += distances [ point ] [ second_point ]
    print '>>>>>', point, distance_sum 

    if distance_sum < currentMinimum :
        currentMinimum = distance_sum 
        centre = point

print centre

Hi Bill, I have come to a similar conclusion. Once you have your cluster and you want to chose the centre of a cluster, this is pretty much the way to do it. What I am doing is this: I start with a single cluster because I start with 1 data point, as more are added and my cluster becomes greater than some threshold t, I split it. I then choose the two furtherest points as centres for the new clusters. Each point is then allocated to one of the two points depending on which is closer. Then the actual centre in each cluster is computed. — Ankur, Aug 12 '09 at 07:53

score 1 · Answer 3 · answered Aug 10 '09 at 08:59

a)

find median or average values of all distances. = avgAll
For each p, find average distance to other machines. = avgP(i)
Pick the closer one as center. avgAll ~= avgP(i)

b) no idea for now..

maybe for each p, find the closer machine.

by this logic make a graph.

than somehow (i dont know yet) divide the graph

score 1 · Answer 4 · answered Aug 10 '09 at 14:23

What you're trying to do, or at least (b) belongs to Cluster Analysis. A branch of mathematics / statistics / econometrics where datapoints (e.g. points in n-dimensional space) are divided among groups or clusters. How to do this is not a trivial questions, there are many, many possible ways.

Read more at the wikipedia article on cluster analysis.

Finding the center of a cluster

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