Myers-Diff-in-c-

c++ Myers Algorithm implementation and some extra proof.

According to Myers Diff Algorithm(http://www.xmailserver.org/diff2.pdf), there are 3 parts:

1. Find the end of the furthest reaching reverse D-path in diagonal k+∆.
2. Find the middle snake and length of an optimal path for A and B.
3. Get the SES(which is an exercise left to the reader).

which were not given by a pseudocode, this code gives a implementation of them.

The reason to use c++ is that with c++'s operator override feature, the code can be similar to the pseudocode while can be excicuted.

Note: Since the example strings is put in the graph, so the index is start from 1 in the paper, but when implement it, there's no such graph thing, so the start index is from 0, this will lead the code slightly different from the pseudocode.

Besides, I think the paper missed a proof, which is very simple, but I think the missing will lead to some confusion (for me), so I gives this proof here:

in page 6, line 4, the code is as below:

 if k = −D or k ≠ D and V[k − 1] < V[k + 1] Then
    x ← V[k + 1]
 Else
    x ← V[k − 1]+1

I don't think lemma2 will lead to this, the purpose of lemma2 is to proof this problem can be solved with a greedy strategy. And according to Given the endpoints of the furthest reaching (D − 1)-paths in diagonal k+1 and k−1, say (x’,y’) and (x",y") respectively, Lemma 2 gives a procedure for computing the endpoint of the furthest reaching D-path in diagonal k. Namely, take the further reaching of (x’,y’+1) and (x"+1,y") in diagonal k and then follow diagonal edges until it is no longer possible to do so or until the boundary of the edit graph is reached. The normal way is to get both points which extends from v[k - 1] and v[k + 1] and figure out which is a further reach path. But this will lead a double-culculate. Here's proof why only check V[k − 1] < V[k + 1] works:

lemma4: if v[k - 1] < v[k + 1], then path followed by v[k + 1] with a vertical edge with a snake will be the further rearched path in diagonal k.

proof: let's make v[k - 1] to x1 and v[k + 1] to x2. so the point followed by x1 in diagonal k is (x1 + 1, x1 + 1 - k)(go right), the point followed by x2 is (x2, x2 - k)(go down); the y position is useless here. then the problem changes to: if x1 < x2, then x1 + 1 <= x2 since x1 < x2 -> x2 - x1 > 0 assume that x1 + 1 > x2, then x2 - x1 < 1, so 0 < x2 - x1 < 1. But in edit graph, the basic move is 1, the 0 < x2 - x1 < 1 will never happen, so the assumption is wrong, so x1 + 1 <= x2.