> Hi Cristian. First, let me see if I understand you. You are talking about > any graph consisting of a sequence of 'complete' merges[1] between the > two branches A and B. Is this a > concrete example, where n=2 and m=3? > > / -- p ---- q ----- A1 -- s ----- A2 > O \ \ / \ / > \ --- B1 --- B2 -- r ----- B3 -- t > > Here, p/q/r/s/t means a change that is not a merge. The p/q/r/s/t states are > the base candidates, not Ab and Bb.
Ah... my mistake. But that is the idea, yes. > Did you mean, "(merge B into A produces Ax, where 1 <= x <= n)" and > "(merge A into B produces By, where 1 <= y <= m)"? Yes. B into A will produce in this example A3, A into B will produce B4 (n=2, m=3) and my goal is to find bases which will have at least a way (path) for which A3 == B4 (symmetry). > [1] <http://wiki.apache.org/subversion/SymmetricMerge#Terminology> >> >> we get the possible paths of walking Ax->An and By->Bm. > > By "paths of walking" you mean following the "merge arrows" (the '/' and '\' > in my graph) and also following the "natural history" of a branch (e.g. B1 -> > B2)? > > So we get the possible paths of walking A1 -> A2, which are: > > A1 -> s -> A2 > A1 -> s -> B3 -> t -> A2 > > and the possible paths of walking (let's choose y=2) B2 -> B3, which are: > > B2 -> r -> B3 > B2 -> r -> A1 -> s -> B3 > > I'll stop here, for now. You got my idea exactly (despite the clumsy description...). Also B2 -> r -> A1 -> B3 -> s -> t -> A2 ==> A3 (end in A) B2 -> r -> A1 -> B3 -> s -> A2 -> t ==> B4 (end in B) can be possible paths, if A3 == B4 is verified. - Cristian
