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I am writing a program that computes and displays diffs. I implemented Meyers algorithm that computes the LCS between 2 subsequences (seq1 and seq2); its output is one of the possible LCS and a partition of seq1 and seq2, one projection of which is lcs.

I want to improve it so that the LCS displayed minimizes the number of breaks; to do so, I implemented a function f(lcs, seq):

  • seq is a sequence of characters
  • lcs is a subsequence of seq
  • the output is a partition of seq p0, p1, p2, ... pn such as
    • either p0 + p2 + or p1 + p3 + ... is lcs
    • and n is minimal

I did so using some sort of BFS, at each step finding the next element non-covered in lcs in seq and greedily expanding the common part.

The resulting algorithm is quite slow: on a typical input, ~3x slower that the Myers algorithm which seems to compute something way more complex. See code here: https://github.com/mookid/diffr/blob/c9ed7746193fd9833ddce1237d6e5005e91deaf4/diffr-lib/src/best_projection.rs

Am I missing a better algorithm?

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  • $\begingroup$ One way to do this is by using Gotoh's sequence alignment algorithm to compute affine gap costs -- this is where you pay an extra cost for opening a new gap (vs. extending an existing gap). But this takes $\Theta(nm)$ time and memory. Myers's algorithm takes $O(nd)$ time which is quadratic too, but usually $d \ll n$ (or the diff would be unhelpful) so it's much faster, and it can also work with just $O(n)$ memory. This page gives more details. $\endgroup$ Nov 26 '19 at 2:17
  • $\begingroup$ BTW, any greedy approach is likely to get stuck. E.g. if seq1 = "RED.APPLEREDAPPLE.R.E.D" and lcs = "REDAPPLERED", an approach that tries to make the longest initial unbroken match will produce "red.appleREDAPPLE.R.E.D" (positions matched to lcs are capitalised) with 3 breaks instead of "RED.APPLEREDapple.r.e.d" with just 1 break. $\endgroup$ Nov 26 '19 at 2:26
  • $\begingroup$ thanks for the pointer! I will have a look. Here greediness means that once I match lcs[0] = R with seq1[0], i keep matching E = seq1[1] rather than trying to match E at some later index. $\endgroup$
    – mookid
    Nov 26 '19 at 3:07
  • $\begingroup$ You're welcome :) I believe Myers's algorithm is already greedy in the way you describe -- it always tries to extend "snakes" as far as they can go. $\endgroup$ Nov 28 '19 at 1:21
  • $\begingroup$ Another approach that I would recommend (a variant of which is used by the DNA sequence alignment tool MUMmer to great effect) is to look for a maximum-length unique match (MUM) between seq1 and lcs -- that is, the longest substring that appears in both strings exactly once. This can be found in $O(n)$ time and space with a generalised suffix tree. This leaves you with 2 smaller subproblems to solve, one on each "side" of the MUM. Requires some careful programming if you need to implement the suffix tree yourself, but should be fast in practice. $\endgroup$ Nov 28 '19 at 1:31
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Your applications may be different from mine, but I think want you want is to input two strings $a$ and $b$ and output a diff $d$ such that no diff is shorter than $d$ and no shortest diff has fewer breaks than $d$. By a diff I mean the algebraic data type List of ((CommonToBoth | FromLeftString | FromRightString) * Char), best expressed in Rust as a sequence of a three-variant Enum, each storing a character (or a T, if generic). Minimizing breaks amounts roughly minimizing the number of runs of CommonToBoth. [Concretely I call mine Common, Insert and Delete.]

I'm not sure that computing a break-minimizing matching $m_a$ between $d$ and $a$ and a break-minimizing matching $m_b$ between $d$ and $b$ will let you compute a break-minimizing matching $m_{ab} = g(m_a, m_b)$ between $d$ and [$a$ and $b$ simultaneously], which is the thing I think you really want.

I've managed to compute a shortest diff with the smallest number of breaks, using an adaptation of Wagner-Fischer (dynamic programming, $O(n^2)$ space for a big table).

Normally you store the LCS length in the Wagner-Fischer table. Instead of doing that, you can choose a score of your own and store it instead. I've used triples similar to (lcsLength, numberOfBreaks, isCurrentRunSharedOrEdits). There's a trick: when characters don't match you need to do something more complicated than max, and when they do it's not always dominant to include the particular match you're looking at: table entry (i, j) is the best among adjustScoreInMatchingCase(table[i-1, j-1]) and combineMismatchingScore(table[i-1, j], table[i, j-1]), where the two functions depend on the particulars of how you score things.

This works, but takes $O(n^2)$ time and space. I can't do much about asymptotic time (provably so if the exponential time hypothesis is true, IINM). I think you can improve the space requirements by only storing the previous and current row of the table; you don't refer back more than a single row in the Wagner-Fischer algorithm.

Incidentally, I think the logic in the Myers algorithm which cleverly constructs and (re)uses the v array doesn't work in this case; the fact that if a[i] == b[j] you don't always extend from LCS(a[..i-1], b[..j-1]) is rooted in the reason why we can't trivially extend the v array. But I might be wrong, I haven't fully explored the problem.

Postscript: Unless smake is a new build tool I haven't heard about, you probably want to rename max_sMake_len ($m \rightarrow n$). If you like property testing you may want to steal my tests at https://github.com/jonaskoelker/equate/blob/master/test/EquateProperties.scala; but beware, one or two of them have slightly wrong labels.

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