On Tue, Oct 02, 2007 at 08:47:48PM +0000, Pierre Habouzit wrote:
> On mar, oct 02, 2007 at 08:13:18 +0000, Pierre Habouzit wrote:
> > Ancestor: (aa*, aaa, bbb)
> > Left child: (aa*, bbb) <-- remove aaa because aa* covers it
> > Right child: (aaa, aabcd, bbb, cc*) <-- remove aa* and be explicit
> >
> > The proper result is probably: (aaa, aabcd, bbb, cc*) but is in fact a
> > case of conflict, because the "left" child could have used the fact that
> > aa* was present and hide say a aaXXX that the right child did not had,
> > and the merge would be wrong.
>
> Okay this example blows, I believe this one is better:
>
> (a*)
> / \
> (ab*) (ac*)
> \ /
> ????
>
> gitignore are subsets of the set of words. if S is the ancestor set,
> S1 and S2 the left and right sets. let Δ1 and Δ2 be S1 \ S and S2 \ S
I meant S \ S1 and S \ S2 in fact here ...
> respectively. I think there is a conflict if
> Δ1 n Δ2 != 0 and (Δ1 is not a subset of Δ2) and (Δ2 is not a subset of Δ1)
>
> If the condition holds, then I believe that the "merged" .gitignore
> would be: (S1 u S2) \ (Δ1 u Δ2)
after some more thoughts, as basically merging the complementary of
the sets I talk about here should yield the same "conflicts" (as it's
the dual problem), I suppose that the same restrictions should be
checked wrt the "added" deltas between S -> S1 (aka S1 \ S for real this
time) and S -> S2 (aka S2 \ S).
so if Δ(0,n) is Sn \ S and Δ(n, 0) is S \ Sn, it would mean that if:
{ Δ(0,1) n Δ(0,2) == 0 || ∃ i ∋ (1,2), Δ(0,i) ⊆ Δ(0, 3 - i) }
&& { Δ(1,0) n Δ(2,0) == 0 || ∃ i ∋ (1,2), Δ(i,0) ⊆ Δ(3 - i, 0) }
Then the correct merge (without conflicts) would be:
(S u Δ(1,0) u Δ(2,0)) \ (Δ(0,1) u Δ(0,2)) aka S + what was added -
what was removed.
in fact, I think that wrt the sets usual operations, there is a
conflict if the expression I just wrote does not commutes wrt the \ or
sth very similar.
Anyway, I'm going pretty off topic here, so I'll shut up now :)
--
·O· Pierre Habouzit
··O madcoder@xxxxxxxxxx
OOO http://www.madism.org
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