Problem Statement

Introduction Define a subsequence to be any output string obtained by deleting zero or more symbols from an input string. The Longest Common Subsequence (LCS) is a subsequence of maximum length common to two or more strings. Let A ≡ A[0]… A[m - 1] and B ≡ B[0]… B[n - 1], m < n be strings drawn from an alphabet Σ of size s, containing every distinct symbol in A + B. An ordered pair (i, j) will be referred to as a match if A[i] = B[j], where 0 ≤ i < m and 0 ≤ j < n. The set of matches M defines a relation over matches: M[i, j] ⇔ (i, j) ∈ M. Define a non-strict product-order (≤) over ordered pairs, such that (i1, j1) ≤ (i2, j2) ⇔ i1 ≤ i2 and j1 ≤ j2. We define (≥) similarly. We say ordered pairs p1 and p2 are comparable if either p1 ≤ p2 or p1 ≥ p2 holds. If i1 < i2 and j2 < j1 (or i2 < i1 and j1 < j2) then neither p1 ≤ p2 nor p1 ≥ p2 are possible, and we say p1 and p2 are incomparable. Define the strict product-order (<) over ordered pairs, such that (i1, j1) < (i2, j2) ⇔ i1 < i2 and j1 < j2. We define (>) similarly. A chain C is a subset of M consisting of at least one element m; and where either m1 < m2 or m1 > m2 for every pair of distinct elements m1 and m2. An antichain D is any subset of M in which every pair of distinct elements m1 and m2 are incomparable. A chain can be visualized as a strictly increasing curve that passes through matches (i, j) in the mn coordinate space of M[i, j]. Every Common Sequence of length q corresponds to a chain of cardinality q, over the set of matches M. Thus, finding an LCS can be restated as the problem of finding a chain of maximum cardinality p. According to [Dilworth 1950], this cardinality p equals the minimum number of disjoint antichains into which M can be decomposed. Note that such a decomposition into the minimal number p of disjoint antichains may not be unique. Background Where the number of symbols appearing in matches is small relative to the length of the input strings, reuse of the symbols increases; and the number of matches will tend towards O(mn) quadratic growth. This occurs, for example, in the Bioinformatics application of nucleotide and protein sequencing. The divide-and-conquer approach of [Hirschberg 1975] limits the space required to O(n). However, this approach requires O(mn) time even in the best case. This quadratic time dependency may become prohibitive, given very long input strings. Thus, heuristics are often favored over optimal Dynamic Programming solutions. In the application of comparing file revisions, records from the input files form a large symbol space; and the number of symbols approaches the length of the LCS. In this case the number of matches reduces to linear, O(n) growth. A binary search optimization due to [Hunt and Szymanski 1977] can be applied to the basic Dynamic Programming approach, resulting in an expected performance of O(n log m). Performance can degrade to O(mn log m) time in the worst case, as the number of matches grows to O(mn). Note [Rick 2000] describes a linear-space algorithm with a time bound of O(ns + p*min(m, n - p)). Legend References [Dilworth 1950] "A decomposition theorem for partially ordered sets" by Robert P. Dilworth, published January 1950, Annals of Mathematics [Volume 51, Number 1, pp. 161-166] [Goeman and Clausen 2002] "A New Practical Linear Space Algorithm for the Longest Common Subsequence Problem" by Heiko Goeman and Michael Clausen, published 2002, Kybernetika [Volume 38, Issue 1, pp. 45-66] [Hirschberg 1975] "A linear space algorithm for computing maximal common subsequences" by Daniel S. Hirschberg, published June 1975 Communications of the ACM [Volume 18, Number 6, pp. 341-343] [Hunt and McIlroy 1976] "An Algorithm for Differential File Comparison" by James W. Hunt and M. Douglas McIlroy, June 1976 Computing Science Technical Report, Bell Laboratories 41 [Hunt and Szymanski 1977] "A Fast Algorithm for Computing Longest Common Subsequences" by James W. Hunt and Thomas G. Szymanski, published May 1977 Communications of the ACM [Volume 20, Number 5, pp. 350-353] [Rick 2000] "Simple and fast linear space computation of longest common subsequences" by Claus Rick, received 17 March 2000, Information Processing Letters, Elsevier Science [Volume 75, pp. 275–281]

Examples The sequences "1234" and "1224533324" have an LCS of "1234": For a string example, consider the sequences "thisisatest" and "testing123testing". An LCS would be "tsitest": In this puzzle, your code only needs to deal with strings. Write a function which returns an LCS of two strings (case-sensitive). You don't need to show multiple LCS's. For more information on this problem please see Wikipedia.

Let's start with the solution:

lcs_wiki(X,Y) = V => 
  [C, _Len] = lcs_length(X,Y),
  V = backTrace(C,X,Y,X.length+1,Y.length+1).

lcs_length(X, Y) = V=>
  M = X.length, 
  N = Y.length,
  C = [[0 : J in 1..N+1]  : I in 1..N+1],
  foreach(I in 2..M+1,J in 2..N+1)
     if X[I-1] == Y[J-1] then
        C[I,J] := C[I-1,J-1] + 1
     else
        C[I,J] := max([C[I,J-1], C[I-1,J]])
     end
  end,
  V = [C, C[M+1,N+1]].

backTrace(C, X, Y, I, J) = V =>
  if I == 1; J == 1 then
    V = ""
  elseif X[I-1] == Y[J-1] then
    V = backTrace(C, X, Y, I-1, J-1) ++ [X[I-1]]
  else 
    if C[I,J-1] > C[I-1,J] then
      V = backTrace(C, X, Y, I, J-1)
    else 
      V = backTrace(C, X, Y, I-1, J)
    end
  end.

table
lcs_rule(A, B) = "", (A == ""; B == "") => true.
lcs_rule(A, B) = [A[1]] ++ lcs_rule(butfirst(A), butfirst(B)), A[1] == B[1] => true.
lcs_rule(A, B) = longest(lcs_rule(butfirst(A), B), lcs_rule(A, butfirst(B))) => true.

% Return the longest string of A and B
longest(A, B) = cond(A.length > B.length, A, B).
    
% butfirst (everything except first element)
butfirst(A) = [A[I] : I in 2..A.length].

go =>
   Tests = [["thisisatest","testing123testing"],
            ["XMJYAUZ", "MZJAWXU"],
            ["1234", "1224533324"],
            ["beginning-middle-ending","beginning-diddle-dum-ending"]
            ],
   Funs = [lcs_wiki,lcs_rule],

   foreach(Fun in Funs)
     println(fun=Fun),
     foreach(Test in Tests)
        printf("%w : %w\n", Test, apply(Fun,Test[1],Test[2]))
     end,
     nl
   end,

   nl.