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The naive algorithm would compare n segments to (n-1) neighbours, so would be O(n^2) |
m →References: Various citation cleanup + WP:AWB fixes . Report errors and suggestions at User talk:CitationCleanerBot |
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*{{citation|last=Brown|first=K. Q.|title=Comments on
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*{{citation|last1=Chen|first1=E. Y.|last2=Chan|first2=T. M.|author2-link=Timothy M. Chan|contribution=A space-efficient algorithm for segment intersection|title=Proc. 15th Canadian Conference on Computational Geometry|year=2003|url=http://www.cccg.ca/proceedings/2003/31.pdf}}.
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