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Re: reordering segments to form a polygon

by Random_Walk (Prior)
on Aug 13, 2004 at 10:42 UTC ( #382605=note: print w/replies, xml ) Need Help??

in reply to reordering segments to form a polygon

someone beat me to it but I coded it so I might as well post it.
#!/usr/local/bin/perl -w use strict; my @seg = ( [ 1, 1, 2, 1 ], [ 5, 5, 1, 5 ], [ 2, 3, 5, 3 ], [ 5, 3, 5, 5 ], [ 2, 1, 2, 3 ], [ 1, 5, 1, 1 ]); my ( $i, $end, $start, %polygon )=(-1, ""); foreach (@seg) { $i++; $start = $_->[0] ." ". $_->[1]; $polygon{$start} = $i; } while ( ($end=$seg[$i]->[2]." ".$seg[$i]->[3]) ne $start) { print "$seg[$i]->[0] $seg[$i]->[1] $end\n"; $i = $polygon{$end}; } print "$seg[$i]->[0] $seg[$i]->[1] $end\n";
This will not play nice if your polygon is not closed.

code seriously tidied up, posted too hastily. The print can be replaced with a push of $seg[$i] onto a new ordered array for use as a sub of other code.

Thought I'd add a quick comment on how it works

The first loop reads each element of the array and records its index number in a hash keyed by starting point.
The second loop starts with the last array element processed, reads it's end from the array and then looks into the hash to find the index of the array element starting there, This index it uses to find the end of that line
Rinse repeat until the end is where we started.

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