Re: Fix object re-hashing

Linus Torvalds [off-list ref] writes:

On Sun, 12 Feb 2006, Linus Torvalds wrote:

quoted

I actually didn't see any of this trigger in real life, so maybe my 
analysis is wrong. Junio? Johannes?

Btw, if it does trigger, the behaviour would be that a subsequent object 
lookup will fail, because the last old slot would be NULL, and a few 
entries following it (likely just a couple - never mind that the event 
triggering in the first place is probably fairly rare) wouldn't have 
gotten re-hashed down.

As a result, we'd allocate a new object, and have _two_ "struct object"s 
that describe the same real object. I don't know what would get upset, but 
git-fsck-index certainly would be (one of them would likely be marked 
unreachable, because lookup wouldn't find it, but you might have other 
issues too).

This "fix" makes the symptom that me fire two (maybe three)
Grrrrr messages earlier this morning disappear.  I haven't had
my caffeine nor nicotine yet after my short sleep, so I need to
take some time understanding your explanation first, but I am
reasonably sure this must be it (not that I do not trust you,
not at all -- it is that I do not trust *me* applying a patch
without understanding when I have a bug reproducible).

Thanks.

On Sun, 12 Feb 2006, Junio C Hamano wrote:

This "fix" makes the symptom that me fire two (maybe three)
Grrrrr messages earlier this morning disappear.

Goodie. I assume that was the fixed fix, not my original "edit out the 
useless optimization and then break it totally" fix ;)

I haven't had my caffeine nor nicotine yet after my short sleep, so I 
need to take some time understanding your explanation first, but I am 
reasonably sure this must be it (not that I do not trust you, not at all 
-- it is that I do not trust *me* applying a patch without understanding 
when I have a bug reproducible).

The basic notion is that this hashing algorithm uses a normal "linear 
probing" overflow approach, which basically means that overflows in 
a hash bucket always just probe the next few buckets to find an empty one.

That's a really simple (and fairly cache-friendly) approach, and it makes 
tons of sense, especially since we always re-size the hash to guarantee 
that we'll have empty slots. It's a bit more subtle - especially when 
re-hashing - than the probably more common "collission chain" approach, 
though.

Now, when we re-hash, the important rule is:

 - the re-hashing has to walk in the same direction as the overflow.

This is important, because when we move a hashed entry, that automatically 
means that even otherwise _already_correctly_ hashed entries may need to 
be moved down (ie even if their "inherent hash" does not change, their 
_effective_ hash address changes because their overflow position needs to 
be fixed up).

There are two interesting cases:

 - the "overflow of the overflow": when the linear probing itself 
   overflows the size of the hash queue, it will "change direction" by 
   overflowing back to index zero.

   Happily, the re-hashing does not need to care about this case, because 
   the new hash is bigger: the rule we have when doing the re-hashing is 
   that as we re-hash, the "i" entries we have already re-hashed are all 
   valid in the new hash, so even if overflow occurs, it will occur the 
   right way (and if it overflows all the way past the current "i", we'll 
   re-hash the already re-hashed entry anyway).

 - the old/new border case. In particular, the trivial logic says that we 
   only need to re-hash entries that were hashed with the old hash. That's 
   what the broken code did: it only traversed "0..oldcount-1", because 
   any entries that had an index bigger than or equal to "oldcount" were 
   obviously _already_ re-hashed.

   That logic sounds obvious, but it falls down on exactly the fact that 
   we may indeed have to re-hash even entries that already were re-hashed 
   with the new algorithm, exactly because of the overflow changes.

So the boundary for old/new is really: "you need to rehash all entries 
that were old, but then you _also_ need to rehash the list of entries that 
you rehashed that might need to be moved down to an empty spot vacated by 
an old hash".

So the stop condition really ends up being: "stop when you have seen all 
old hash entries _and_ at least one empty entry after that", since an 
empty entry means that there was no overflow from earlier positions past 
that position. But it's just simpler to walk the whole damn new thing and 
not worry about it.

		Linus

On Sun, 12 Feb 2006, Linus Torvalds wrote:

 - the "overflow of the overflow": when the linear probing itself 
   overflows the size of the hash queue, it will "change direction" by 
   overflowing back to index zero.

   Happily, the re-hashing does not need to care about this case, because 
   the new hash is bigger: the rule we have when doing the re-hashing is 
   that as we re-hash, the "i" entries we have already re-hashed are all 
   valid in the new hash, so even if overflow occurs, it will occur the 
   right way (and if it overflows all the way past the current "i", we'll 
   re-hash the already re-hashed entry anyway).

Btw, this is only always true if the new hash is at least twice the size 
of the old hash, I think. Otherwise a re-hash can fill up the new entries 
and overflow entirely before we've actually even re-hashed all the old 
entries, and then we'd need to re-hash even the overflowed entries (which 
are now below "i").

If the new size is at least twice the old size, the "upper area" cannot 
overflow completely (there has to be empty room), and we cannot be in the 
situation that we need to move even the overflowed entries when we remove 
an old hash entry.

Anyway, if all this makes you nervous, the conceptually much simpler way 
to do the re-sizing is to not do the in-place re-hashing. Instead of doing 
the xrealloc(), just do a "xmalloc()" of the new area, do the re-hashing 
(which now _must_ re-hash in just the "0..oldcount-1" old area) into the 
new area, and then free the old area after rehashing.

That would make things more obviously correct, and perhaps simpler.

Johannes, do you want to try that?

Btw, as it currently stands, I worry a tiny tiny bit about the

	obj_allocs = (obj_allocs < 32 ? 32 : 2 * obj_allocs)

thing, because I think that second "32" needs to be a "64" to be really 
safe (ie guarantee that the new obj_allocs value is always at least twice 
the old one).

Anyway, I'm pretty sure people smarter than me have already codified 
exactly what needs to be done for a in-place rehash of a linear probe hash 
overflow algorithm. This must all be in some "hashing 101" book. I had to 
think it through from first principles rather than "knowing" what the 
right answer was (which probably means that I slept through some 
fundamental algorithms class in University ;)

		Linus

Hi,

On Sun, 12 Feb 2006, Linus Torvalds wrote:

[something about the overflow in another mail]

Thank you for thinking it through! I was soooo stuck with my original 
idea: Ideally (i.e. if there are no collisions), if the hashtable is 
doubled in size, then each offset should either stay the same, or be just 
incremented by the original size (since the index is the hash modulo the 
hashtable size).

So I wanted to be clever about resizing, and just increment the offset if 
necessary. As it turns out, it's more complicated than that. You have to 
make sure that those entries which collided  with another entry, but do no 
longer, are adjusted appropriately.

And the overflow problem eluded my attention entirely. (I feel quite silly 
about it, because I fixed so many buffer-overflow problems myself, and 
the cause of the problem is the same there.)

On Sun, 12 Feb 2006, Linus Torvalds wrote:

quoted

 - the "overflow of the overflow": when the linear probing itself 
   overflows the size of the hash queue, it will "change direction" by 
   overflowing back to index zero.

   Happily, the re-hashing does not need to care about this case, because 
   the new hash is bigger: the rule we have when doing the re-hashing is 
   that as we re-hash, the "i" entries we have already re-hashed are all 
   valid in the new hash, so even if overflow occurs, it will occur the 
   right way (and if it overflows all the way past the current "i", we'll 
   re-hash the already re-hashed entry anyway).

Btw, this is only always true if the new hash is at least twice the size 
of the old hash, I think. Otherwise a re-hash can fill up the new entries 
and overflow entirely before we've actually even re-hashed all the old 
entries, and then we'd need to re-hash even the overflowed entries (which 
are now below "i").

After thinking long and hard about it, I tend to agree.

Note: I chose the factor 2 because hashtables tend to have *awful* 
performance when space becomes scarce. So, 2 is not only a wise choice for 
rehashing, but for the operation in general.

Anyway, if all this makes you nervous, the conceptually much simpler way 
to do the re-sizing is to not do the in-place re-hashing. Instead of doing 
the xrealloc(), just do a "xmalloc()" of the new area, do the re-hashing 
(which now _must_ re-hash in just the "0..oldcount-1" old area) into the 
new area, and then free the old area after rehashing.

That would make things more obviously correct, and perhaps simpler.

Johannes, do you want to try that?

I do not particularly like it, since doubling the hashtable size is not 
particularly space efficient, and this makes it worse. Anyway, see below.

Btw, as it currently stands, I worry a tiny tiny bit about the

	obj_allocs = (obj_allocs < 32 ? 32 : 2 * obj_allocs)

thing, because I think that second "32" needs to be a "64" to be really 
safe (ie guarantee that the new obj_allocs value is always at least twice 
the old one).

As Junio already pointed out: obj_allocs is initially set to 0. But you're 
right, it is conceptually wrong.

Anyway, I'm pretty sure people smarter than me have already codified 
exactly what needs to be done for a in-place rehash of a linear probe hash 
overflow algorithm. This must all be in some "hashing 101" book. I had to 
think it through from first principles rather than "knowing" what the 
right answer was (which probably means that I slept through some 
fundamental algorithms class in University ;)

Well, it seems like a long time, doesn't it? But I always liked the 
Fibonacci numbers, and therefore the Fibonacci heap.

---

Make hashtable resizing more robust AKA do not resize in-place

diff --git a/object.c b/object.c
index c9ca481..94f0f5d 100644
--- a/object.c
+++ b/object.c

@@ -56,18 +56,14 @@ void created_object(const unsigned char 
 
 	if (obj_allocs - 1 <= nr_objs * 2) {
 		int i, count = obj_allocs;
-		obj_allocs = (obj_allocs < 32 ? 32 : 2 * obj_allocs);
-		objs = xrealloc(objs, obj_allocs * sizeof(struct object *));
-		memset(objs + count, 0, (obj_allocs - count)
-				* sizeof(struct object *));
-		for (i = 0; i < obj_allocs; i++)
-			if (objs[i]) {
-				int j = find_object(objs[i]->sha1);
-				if (j != i) {
-					j = -1 - j;
-					objs[j] = objs[i];
-					objs[i] = NULL;
-				}
+		struct object** old_objs = objs;
+		obj_allocs = (obj_allocs < 32 ? 64 : 2 * obj_allocs);
+		objs = xcalloc(obj_allocs, sizeof(struct object *));
+		for (i = 0; i < count; i++)
+			if (old_objs[i]) {
+				/* it is guaranteed to be new */
+				int j = -1 - find_object(old_objs[i]->sha1);
+				objs[j] = old_objs[i];
 			}
 	}
 

On Mon, 13 Feb 2006, Johannes Schindelin wrote:

Make hashtable resizing more robust AKA do not resize in-place

You forgot to release the old array afterwards.

Anyway, I think the in-place version is fine now, even if it has a few 
subtleties. So this isn't needed, but keep it in mind if we find another 
bug, or if somebody wants to shrink the hash table less aggressively than 
with doubling it every time.

		Linus

Hi,

On Sun, 12 Feb 2006, Linus Torvalds wrote:

On Mon, 13 Feb 2006, Johannes Schindelin wrote:

quoted

Make hashtable resizing more robust AKA do not resize in-place

You forgot to release the old array afterwards.

D'oh! I am going to bed now.

Anyway, I think the in-place version is fine now, even if it has a few 
subtleties. So this isn't needed, but keep it in mind if we find another 
bug, or if somebody wants to shrink the hash table less aggressively than 
with doubling it every time.

Sounds fine to me.

Good night,
Dscho