Christian Couder <chriscool@xxxxxxxxxxxxx> writes: > Yes, I assume C is also bad. > >> ... which means you are dealing with *two* breakages. That's outside >> what bisect deals with. > > Sorry, I don't understand why I am assuming 2 breakages. Your topology is >> > A-B-C-D >> > \E-F and you start with "D F" so D is bad and F is good, right? bisect operates on a history with a single breakage that separates older good ones and newer bad ones. Now D is bad, so there must be somewhere in A--B--C--D that broke D. It could be D itself, or it may have been broken at A already. But if A or B were bad, then, because F is good, this part of the picture: A--B--E--F does not deal with "old good followed by new bad sparated by a single breakage point". Maybe E or F have fixed it, but then before that is bad and after that is good. That's not what bisect deals with. Going back to the original picture, then: >> > A-B-C-D >> > \E-F For F to be good, all of its ancestors are by definition good (that is the basic assumption of bisection). A and B are good. If D is broken, C or D must be broken. So we do not decend into B nor its ancestor. The logic is the same if the history is linear. A--B--C--D--E--F If B is good and F is bad, and if you check D and find out that D is still good, then we can ignore everything behind D, and that is what lets us limit the search space and suspect only E or F. This is exactly the same "Everything that leads to a good one is good" principle upon which bisection works. -- To unsubscribe from this list: send the line "unsubscribe git" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
