On Wed, Jul 09, 2014 at 03:18:43PM -0700, Junio C Hamano wrote:

> Jeff King <p...@peff.net> writes:
> > I think the hash here does not collide in that way. It really is just
> > the last sixteen characters shoved into a uint32_t.
> All bytes overlap with their adjacent byte because they are shifted
> by only 2 bits, not 8 bits, when a new byte is brought in.  We can
> say that the topmost two bits of the result must have come from the
> last character, but other than these, there are more than one input
> byte for each bit position to be set/unset by, so two names that human
> would not consider "similar" would be given the same hash, no?

Yeah, you're right. I didn't look at the algorithm closely enough.

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