Generalizing LongestCommonSequence to 3 or more arguments?

Question

The function LongestCommonSequence finds a longest common subsequence between 2 lists. Apparently, this built-in function does not accept more than 2 arguments. How can I find a longest common subsequence between 3 or more lists using Mathematica? Or, better yet, all longest common subsequences?

In a response to a "close as a duplicate" vote: This is not a duplicate of Longest common substring for multiple strings? beacuse that question is concerned with substrings (contiguous subsequences), but my question is concerned with arbitrary (not necessarily contiguous) subsequences.

actually this answer http://mathematica.stackexchange.com/a/114987/2079 works by first converting strings to list and so contains exactly an answer to this question. — george2079, May 13 '16 at 21:40
@george2079 It is interesting, but seems to be slow as hell. Besides, I have a hunch that the complexity of even the best algorithm here would be proportional to the product of lists lengths, or something like that. So it is basically quadratic for 2 similarly sized lists, cubic for 3, and so on. Can't prove it though. — Leonid Shifrin, May 13 '16 at 21:48
@george2079 No, this is not a duplicate, as I explained in an addendum to the question. — Vladimir Reshetnikov, May 13 '16 at 21:54
Could you explain what do you mean by "not necessarily contiguous", or maybe give an example of what you want to achieve? — xslittlegrass, May 13 '16 at 22:30
@xslittlegrass The Wikipedia page I linked provides detailed definitions an examples. Let $S$ be a string or a list. A subsequence of $S$ is obtained by removing zero or more elements of $S$ at arbitrary positions (e.g. "tea", "aha" and "etc" are subsequences of "Mathematica"). A substring of $S$ is a prefix of a suffix of $S$ (e.g. "them" is a substring of "Mathematica", because it is a prefix of its suffix "thematica"). Every substring is also a subsequence. — Vladimir Reshetnikov, May 13 '16 at 22:46
you ask to "generalize" LongestCommonSequence to 3 or more arguments, but you also want to generalize to a different definition of sequence? Please give a clear definition of terms and an example to work with. — george2079, May 14 '16 at 04:06
Sorry, I don't see why my definition is different. Different from what? — Vladimir Reshetnikov, May 14 '16 at 19:08

score 8 · Answer 1 · answered May 14 '16 at 06:41

ClearAll[fuzzyLCS];
fuzzyLCS[strings__List] :=
 Module[
  {subsets, aligned, intersections},
  subsets = Subsets[strings, {2, Length@strings}];
  aligned = 
   Select[SequenceAlignment[#[[1]], #[[2]]], StringQ[#] &] & /@ 
    subsets;
  intersections = 
   Intersection @@ (Subsets[#, {1, 
         Length@#}] & /@ (Flatten[Characters[#]] & /@ aligned));
  StringJoin[SortBy[intersections, Length] // Last]
  ]

fuzzyLCS[{"theano", "mathematica", "matea"}] // AbsoluteTiming

{0.000150089, "tea"}

Generalizing LongestCommonSequence to 3 or more arguments?

1 Answers1