Subject: Re: Reachability lists in git

From: Junio C Hamano <[email-masked]>
Date: Tue, 18 Nov 2014 13:11:51 -0800

Junio C Hamano <[email-masked]> writes:

> Alan Stern <[email-masked]> writes:
>
>> On Tue, 18 Nov 2014, Jonathan Nieder wrote:
>>
>>> Alan Stern wrote:
>>> 
>>> > Tracking down regressions.  Bisection isn't perfect.  Suppose a
>>> > bisection run ends up saying that B is the first bad commit.  It's easy
>>> > enough to build B and test it, to verify that it really is bad.
>>> >
>>> > But to be sure that B introduced the fault, it would help to find the
>>> > latest commit that doesn't include B's changes -- that is, the latest
>>> > commit that B isn't reachable from (or the maximal elements in the set
>>> > of all such commits).
>>> 
>>> Isn't that B^ (or B^ and B^2, if B is a merge)?
>>
>> No.  Here's a simple example:
>>
>>             Y
>>            /
>>           /
>>          X--B
>>
>> In this diagram, X = B^.  But B isn't reachable from either X or Y, 
>> whereas it is reachable from one of X's children (namely Y).
> ...
>
> Why do you say B is reachable from Y?
>
> If you mean that B is a merge between X and Y, then that is already
> covered by what Jonathan wrote "(or B^ and B^2 if B is a merge)".
>
>     X----Y
>      \    \
>       .----B
> ...

No, that cannot be what you meant.  I was confused.  The above
picture does not make B reachable from Y (it is the other way
around: B reaches Y).  The topology where B isn't reachable from
either X or Y and is reachable from Y would be

	X---Y	i.e. B = Y^2, X = Y^1 = B^1 
         \ /  
          B

If B is broken, and X is not, then Y would be contaminated by the
breakage B introduces relative to X, unless Y is an evil merge and
fixed that breakage while merging.

In any case, even if Y is found to be broken, its parent B is
already broken, so that does not place the blame on Y, does it?

Still confused why you feel Y is any significant...