[ python-Bugs-942952 ] Weakness in tuple hash
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Wed May 19 12:15:30 EDT 2004
Bugs item #942952, was opened at 2004-04-27 07:41
Message generated for change (Comment added) made by tim_one
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Category: Python Interpreter Core
Group: None
Status: Open
Resolution: None
Priority: 6
Submitted By: Steve Tregidgo (smst)
Assigned to: Nobody/Anonymous (nobody)
Summary: Weakness in tuple hash
Initial Comment:
I've encountered some performance problems when
constructing dictionaries with keys of a particular
form, and have pinned the problem down to the hashing
function. I've reproduced the effect in Python 1.5.2,
Python 2.2 and Python 2.3.3. I came across this when
loading a marshalled dictionary with about 40000
entries. Loading other dictionaries of this size has
always been fast, but in this case I killed the
interpreter after a few minutes of CPU thrashing. The
performance problem was caused because every key in the
dictionary had the same hash value.
The problem is as follows: for hashable values X and Y,
hash( (X, (X, Y)) ) == hash(Y). This is always true
(except in the corner case where hash((X, Y)) is
internally calculated to be -1 (the error value) and so
-2 is forced as the return value). With data in this
form where X varies more than Y (Y is constant, or
chosen from relatively few values compared to X)
chances of collision become high as X is effectively
ignored.
The hash algorithm for tuples starts with a seed value,
then generates a new value for each item in the tuple
by multiplying the iteration's starting value by a
constant (keeping the lowest 32 bits) and XORing with
the hash of the item. The final value is then XORed
with the tuple's length. In Python (ignoring the
careful business with -1):
# assume 'my_mul' would multiply two numbers and
return the low 32 bits
value = seed
for item in tpl:
value = my_mul(const, value) ^ hash(item)
value = value ^ len(tpl)
The tuple (X, Y) therefore has hash value:
my_mul(const, my_mul(const, seed) ^ hash(X)) ^
hash(Y) ^ 2
...and the tuple (X, (X, Y)) has hash value:
my_mul(const, my_mul(const, seed) ^ hash(X)) ^
hash((X, Y)) ^ 2
The outer multiplication is repeated, and is XORed with
itself (cancelling both of them). The XORed 2s cancel
also, leaving just hash(Y). Note that this
cancellation property also means that the same hash
value is shared by (X, (X, (X, (X, Y)))), and (X, (X,
(X, (X, (X, (X, Y)))))), and so on, and (X, Z, (X, Z,
Y)) and so on.
I realise that choosing a hash function is a difficult
task, so it may be that the behaviour seen here is a
tradeoff against other desireable properties -- I don't
have the expertise to tell. My naive suggestion would
be that an extra multiplication is necessary, which
presumably has a performance impact (multiplication
being more expensive than XOR) but would reduce the
possibility of cancellation. On the other hand, perhaps
this particular case is rare enough that it's not worth
the effort.
For my own application I'm fortunate in that I can
probably rearrange the data structure to avoid this case.
----------------------------------------------------------------------
>Comment By: Tim Peters (tim_one)
Date: 2004-05-19 12:15
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If 250,050 balls are tossed into 2**32 bins at random, the
mean number of occupied bins is ~250042.7 (== expected
number of collisions is ~7.3), with std deviation ~2.7. While
we're happy to get "better than random"
distributions, "astronomically worse than random" distributions
usually predict more of the same kind of problem reported
later.
That said, I certainly agree that would be a major, easy, and
cheap improvement over the status quo.
----------------------------------------------------------------------
Comment By: Armin Rigo (arigo)
Date: 2004-05-19 11:10
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Tim:
> Most simple changes to the current algorithm don't
> do significantly better, and some do much worse.
The simple change suggested by ygale (put the xor before the
multiply) does a quite reasonable job in your test suite.
The number of collisions drop from 66966 to 1466. It also
solves the original problem with (x, (x,y)), which should
not be ignored because it could naturally occur if someone
is using d.items() to compute a hash out of a dict built by
d[x] = x,y.
With this change it seems more difficult to find examples of
regular families of tuples that all turn out to have the
same hash value. Such examples do exist: (x,
(1684537936,x)) yields 2000004 for any x. But the value
1684537936 has been carefully constructed for this purpose,
it's the only number to misbehave like this :-)
----------------------------------------------------------------------
Comment By: Tim Peters (tim_one)
Date: 2004-05-17 22:14
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Nothing wrong with being "better than random" -- it's not the
purpose of Python's hash() to randomize, but to minimize
hash collisions. That's why, e.g., hash(int) returns int
unchanged. Then when indexing a dict by any contiguous
range of ints (excepting -1), there are no collisions at all.
That's a good thing.
There's nothing in the structure of my example function
geared toward the specific test instances used. It *should*
do a good job of "spreading out" inputs that are "close
together", and that's an important case, and the test inputs
have a lot of that. It's a good thing that it avoids all
collisions across them.
About speed, the only answer is to time both, and across
several major platforms. An extra mult may or may not cost
more than blowing extra cache lines to suck up a table of ints.
About the table of ints, it's always a dubious idea to multiply
by an even number. Because the high bits of the product are
thrown away, multiplying by an even number throws away
information needlessly (it throws away as many useful bits as
there are trailing zero bits in the multiplier). odd_number *
69069**i is always odd, so the multiplier in the table-free
way is never even (or, worse, divisible by 4, 8, 16, etc).
----------------------------------------------------------------------
Comment By: Yitz Gale (ygale)
Date: 2004-05-17 09:39
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Here is the Python code for the hash function
that I am describing:
def hx(x):
if isinstance(x, tuple):
result = 0x345678L
for n, elt in enumerate(x):
y = (tbl[n & 0x3f] + (n >> 6)) & 0xffffffffL
result = (y * (result ^ hx(elt))) & 0xffffffffL
result ^= len(x)
else:
result = hash(x)
return result
# 64 constants taken from MD5
# (see Modules/md5c.c in the Python sources)
tbl = (
-0x28955b88, -0x173848aa, 0x242070db, -0x3e423112,
-0xa83f051, 0x4787c62a, -0x57cfb9ed, -0x2b96aff,
0x698098d8, -0x74bb0851, -0xa44f, -0x76a32842,
0x6b901122, -0x2678e6d, -0x5986bc72, 0x49b40821,
-0x9e1da9e, -0x3fbf4cc0, 0x265e5a51, -0x16493856,
-0x29d0efa3, 0x2441453, -0x275e197f, -0x182c0438,
0x21e1cde6, -0x3cc8f82a, -0xb2af279, 0x455a14ed,
-0x561c16fb, -0x3105c08, 0x676f02d9, -0x72d5b376,
-0x5c6be, -0x788e097f, 0x6d9d6122, -0x21ac7f4,
-0x5b4115bc, 0x4bdecfa9, -0x944b4a0, -0x41404390,
0x289b7ec6, -0x155ed806, -0x2b10cf7b, 0x4881d05,
-0x262b2fc7, -0x1924661b, 0x1fa27cf8, -0x3b53a99b,
-0xbd6ddbc, 0x432aff97, -0x546bdc59, -0x36c5fc7,
0x655b59c3, -0x70f3336e, -0x100b83, -0x7a7ba22f,
0x6fa87e4f, -0x1d31920, -0x5cfebcec, 0x4e0811a1,
-0x8ac817e, -0x42c50dcb, 0x2ad7d2bb, -0x14792c6f
)
----------------------------------------------------------------------
Comment By: Yitz Gale (ygale)
Date: 2004-05-17 09:09
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Tim, your test fixture is very nice.
I verified your result that there are no collisions
over the fixture set for the algorithm you specified.
Based on that result, I would recommend that we
NOT use your algorithm.
The probability of no collisions is near zero for
250000 random samples taken from a set of 2^32
elements. So the result shows that your algorithm
is far from a good hash function, though it is
constructed to be great for that specific fixture set.
I ran my algorithm on your fixture set using the
table of 64 constants taken from MD5. I got 13
collisions, a reasonable result.
It is not true that I repeat multipliers (very often) -
note that I increment the elements of the table each
time around, so the sequence repeats itself only
after 2^38 tuple elements. Also, my algorithm runs
at esentially the same speed as the current one:
no additional multiplies, only a few adds and increments
and such.
----------------------------------------------------------------------
Comment By: Tim Peters (tim_one)
Date: 2004-05-16 17:54
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I can't make more time for this, so unassigned myself.
This seems to make a good set of test inputs:
N = whatever
base = range(N)
xp = [(i, j) for i in base for j in base]
inps = base + [(i, j) for i in base for j in xp] + [(i, j) for i in xp for j in base]
When N == 50, inps contains 250,050 distinct elements, and
66,966 of them generate a duplicated hash code. Most
simple changes to the current algorithm don't do significantly
better, and some do much worse. For example, just replacing
the xor with an add zooms it to 119,666 duplicated hash
codes across that set.
Here's a hash function that yields no collisions across that
set:
def hx(x):
if isinstance(x, tuple):
result = 0x345678L
mult = 1000003
for elt in x:
y = hx(elt)
result = (((result + y) & 0xffffffffL) * mult) &
0xffffffffL
mult = (mult * 69069) & 0xffffffffL
result ^= len(x)
if result == -1:
result = -2
else:
result = hash(x)
return result
In C, none of the "& 0xffffffffL" clauses are needed; in Python
code, they're needed to cut results back to 32 bits. 69069 is
a well-known multiplier for a better-than-most pure
multiplicative PRNG.
This does add a multiply per loop trip over what we have
today, but it can proceed in parallel with the existing
multiply. Unlike a canned table, it won't repeat multipliers,
(unless the tuple has more than a billion elements).
----------------------------------------------------------------------
Comment By: Tim Peters (tim_one)
Date: 2004-05-16 17:09
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Noting that
(X, (Y, Z))
and
(Y, (X, Z))
hash to the same thing today too (module -1 endcases). For
example,
>>> hash((3, (4, 8))) == hash((4, (3, 8)))
True
>>>
----------------------------------------------------------------------
Comment By: Yitz Gale (ygale)
Date: 2004-05-16 08:25
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Hmm, you are correct. This is appears to be
an off-by-one problem: the original seed always gets
multiplied by the constant (which is silly), and the last
item in the tuple does not get multiplied (which causes
the bug).
The correct solution is to change:
value = my_mul(const, value) ^ hash(item)
in Steve's pseudo-code to:
value = my_mul(const, value ^ hash(item))
Of course, you still get a lot more robustness
for almost no cost if you vary "const" across
the tuple via a table.
----------------------------------------------------------------------
Comment By: Jim Jewett (jimjjewett)
Date: 2004-05-03 09:40
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Note that in this case, the problem was because of nested
tuples. X was the first element of both tuples, and both
tuples had the same length -- so the X's would still have been
multiplied by the same constant. (Not the same constant as
Y, and hash(X, Y), but the same constant as each other.) A
non-linear function might work, but would do bad things to
other data.
----------------------------------------------------------------------
Comment By: Yitz Gale (ygale)
Date: 2004-05-01 22:50
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I suggest leaving the function as is, but replacing the
constant with a value that varies as we step through the
tuple. Take the values from a fixed table. When we reach the
end of the table, start again from the beginning and mung
the values slightly (e.g., increment them).
True, there will always be the possibilities of collisions,
but this will make it very unlikely except in very weird
cases. Using a table of constants is a standard technique
for hashes.
----------------------------------------------------------------------
Comment By: Jim Jewett (jimjjewett)
Date: 2004-04-28 13:45
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Wouldn't that just shift the pathological case from (x, (x, y))
to (x, (-x, y))? My point was that any hash function will act
badly on *some* pattern of input, and if a pattern must be
penalized, this might be a good pattern to choose.
----------------------------------------------------------------------
Comment By: Raymond Hettinger (rhettinger)
Date: 2004-04-28 12:36
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The OP was not referring to "some collisions"; his app
collapsed all entries to a single hash value. Changing XOR
to + would partially eliminate the self cancelling property
of this hash function.
Also, I am concerned about the tuple hash using the same
multiplier as the hash for other objects. In sets.py, a
naive combination of the component hash values caused many
distinct sets to collapse to a handful of possibilities --
while tuples do not have an identical issue, it does
highlight the risks involved.
----------------------------------------------------------------------
Comment By: Jim Jewett (jimjjewett)
Date: 2004-04-28 10:50
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Any hash will have some collisions. If there is going to be
predictably bad data, this is probably a good place to have it.
The obvious alternatives are a more complicated hash (slows
everything down), a different hash for embedded tuples (bad,
since hash can't be cached then) or ignoring some elements
when determining the hash (bad in the normal case of different
data).
I would also expect your workaround of data rearrangement to
be sensible almost any time (X, (X, Y)) is really a common
case. (The intuitive meaning for me is "X - then map X to Y",
which could be done as (X, Y) or at least (X, (None, Y)), or
perhaps d[X]=(X,Y).)
----------------------------------------------------------------------
Comment By: Steve Tregidgo (smst)
Date: 2004-04-27 07:45
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I'll repeat the tuple hashing algorithm in a fixed-width
font (with the first line following "for..." being the loop
body):
# assume 'my_mul' would multiply two numbers and
return the low 32 bits
value = seed
for item in tpl:
value = my_mul(const, value) ^ hash(item)
value = value ^ len(tpl)
----------------------------------------------------------------------
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