⚠️ Warning: This is a draft ⚠️
This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.
If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.
== More specific? ==
What happens when there are more than one way to match? Say you have string "aBaBaBa" and want to find non-overlapping "aBa"s, you could say both (aBa)B(aBa) and aB(aBa)Ba, where brackets denote the found matches, are valid. --[[User:Ledrug|Ledrug]] 17:07, 16 June 2011 (UTC)
:My guess would be to specify to match left-to-right, but that might be too restrictive of possible methods. Maybe say "give the highest possible number of non-overlapping matches"? Can you think of any situations where that isn't the same as the count when matching from left-to-right (or even right-to-left)? --[[User:Mwn3d|Mwn3d]] 17:20, 16 June 2011 (UTC)
::Earliest match always produces most matches. Proof: suppose a string is matched in two ways:
:: where m1 and m2 overlap, and suppose both are giving highest possible number of matches at their starting location; further suppose second way matches more times: but this can't be, because we can then give up m2 and use m1 while keeping the rest, and it would have more matches than first way, which is a contradiction. Hence, highest num of matches can be obtained by starting earliest. :: Since reversing both pattern and string makes left-to-right problem into a right-to-left one, this also proves highest matches can be obtained from either end. :: That exercise aside, I just thought the task should be worded so that any given input would produce an unambiguous result. --[[User:Ledrug|Ledrug]] 17:45, 16 June 2011 (UTC) :::For completeness' sake (and because it wouldn't affect the existing examples), I added that clarification and referenced this proof. --[[User:Mwn3d|Mwn3d]] 17:52, 16 June 2011 (UTC) Yes, the problem is stated ambiguously and was probably inspired by some particular language's implementation of a built in function. It would be more appropriate to count all occurrences of a substring. Counting the maximum number of non-overlapping substrings can turn into a very complicated problem requiring backtracking. : Yes, some languages' BIF may treat it as caseless, still others may treat all whitespace equally (as blank or blanks). It would've been interesting to make a case-sensative and a caseless version. The whitespace issue can be more complicated. -- [[User:Gerard Schildberger|Gerard Schildberger]] 18:33, 12 June 2012 (UTC)