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Despite what I just said about NP-completeness, the following algorithm might give a reasonable solution (certainly not optimal).
  1. Let C be the set of all complaints.
  2. For each complaint c in C find the set S(c) of all complaints d in C such that the distance between c and d is less than X (X is user defined).
  3. Find complaint e in C such that |S(e)| <= |S(f)| for all f in C; that is, find the complaint who has the most other complaints nearby. Pick a random one in case of a tie.
  4. Make a clump out of e and S(e).
  5. Remove all complaints g in S(e) from C. For all h remaining in C, remove from S(h) all g in S(e).
  6. If C is empty, we're done. Else, goto 3.
Some pseudo code:
# Get set of all complaints. my @C = get_all_complaints; # Find all the associated sets. my %D = map {my $c = $_; $c => {map {$_ => 1} grep {$c ne $_ && distance ($c, $_) < $X} @C}} @C; while (%D) { # Find complaint with the most nearby. my ($complaint, $size) = (undef, 0); while (my ($c, $set) = each %D) { ($complaint = $c, $size = keys %$set) if keys %$set > $size; } # Found largest, make a clump. make_clump ($complaint, @{$D {$complaint}}); # Delete largest from set. my $set = delete $D {$complaint}; # Delete associated set from set. delete @D {keys %$set}; # Delete associated set from associated sets. delete @{$_} {keys %$set} for values %D; }
The performance will be quadratic, I'm afraid.

Abigail


In reply to Re: Sorting by geographical proximity / clumping groups of items based on X and Y by Abigail-II
in thread Sorting by geographical proximity / clumping groups of items based on X and Y by vroom

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