# Generalizing LongestCommonSequence to 3 or more arguments?

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.

• Related: mathematica.stackexchange.com/a/114987/9490 – Jason B. May 13 '16 at 21:32
• actually this answer 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
• @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 4:06

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

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


{0.000150089, "tea"}