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Re: Finding all connected nodes in an all-against-all comparisonby LanX (Saint) |
| on May 07, 2010 at 14:52 UTC ( [id://838921]=note: print w/replies, xml ) | Need Help?? |
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So which solution do you expect with this data?
There is no path from A to B, so do you consider them in the same or different groups? Or do you have a guaranty that these intersections are not possible?
Cheers Rolf
In Section
Seekers of Perl Wisdom
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