Hamming Distance
Jared Grubb
jared.grubb at gmail.com
Fri Jun 27 19:54:45 CEST 2008
Matimus,
I was surprised that "lazy" was the algorithm that won your time tests, and
I saw a way to improve it even better (algorithm is O(# ones in number)
rather than O(# bits in number))
def lazy2(a, b, bits=32):
x = (a ^ b) & ((1 << bits) - 1)
tot = 0
while x:
tot += 1
x &= x-1
return tot
# times on my system (run a few times just to check for sure)
python -mtimeit -s"from ham import *" "test(lazy)"
10000 loops, best of 3: 121 usec per loop
python -mtimeit -s"from ham import *" "test(lazy2)"
10000 loops, best of 3: 62.4 usec per loop
Check my math, but I think that's correct. Here's my derivation (but it's
been a while since my Boolean algebra days, so I may've made a mistake!) It
sounds right, though, since subtracting one in two's complement flips the
rightmost one and inverts the zeros to the right of it. So, x & (x-1) would
remove the rightmost one. Right?
Long Derivation: The "trick" to finding the rightmost one in a number:
pos = x ^ (x-1)
It has to do with how two's-complement works. In our algorithm above, we are
trying to count them, so we want to flip off the bits one by one from the
right. So in each loop:
x = x & ~pos
But, then you notice you can simplify it even more (let y=x-1) and use
mult/add syntax for & and | and use X=~x and Y=~y
x * ~(x ^ y)
x * ~(xY+Xy) [ def of ^ ]
x * (~(xY)*~(Xy)) [ DeMoires Law ]
x * ( (X+y)*(x+Y) ) [ inversion]
x * (X+y) * (x+Y) [ associative]
(xX+xy)*(x+Y) [ distributive ]
xy*(x+Y) [ xX = 0 ]
xy+xyY [ distrib ]
xy [yY = 0]
So,
x &= x-1
On 19 Jun 2008, at 17:37, Matimus wrote:
On Jun 19, 4:27 pm, godavemon <davefow... at gmail.com> wrote:
I need to calculate the Hamming Distance of two integers. The hamming
distance is the number of bits in two integers that don't match. I
thought there'd be a function in math or scipy but i haven't been able
to find one. This is my function but it seems like there should be a
faster way. I do this computation many times and speed up is
important.
def hamdist( a, b , bits = 32):
def _hamdist( x, bits):
if bits:
return (x & 1) + _hamdist(x >> 1, bits-1)
return x & 1
return _hamdist( a ^ b, bits)
Another alternative would be to convert the XOR to a binary string and
count the # of 1's.
Which would be fastest? Are there better alternatives?
Thanks!
I see no good reason to use recursion for this type of thing. Here are
some of my attempts:
[code]
from math import log
def yours(a, b , bits = 32):
def _hamdist( x, bits):
if bits:
return (x & 1) + _hamdist(x >> 1, bits-1)
return x & 1
return _hamdist(a ^ b, bits)
def simple(a, b, bits=32):
x = a ^ b
return sum((x >> i & 1) for i in xrange(bits))
def lazy(a, b, bits=32):
x = (a ^ b) & ((1 << bits) - 1)
tot = 0
while x:
tot += x & 1
x >>= 1
return tot
def fancy(a, b, bits=32):
x = (a ^ b) & ((1 << bits) - 1)
tot = 0
while x:
tot += 1
x ^= 1 << int(log(x, 2))
return tot
test_vals = (
((0xffffffff, 0), 32),
((0,0), 0),
((1,0), 1),
((0x80000000, 0), 1),
((0x55555555, 0), 16)
)
def test(f):
test_vals = (
((0xffffffff, 0), 32), # ALL
((0,0), 0), # None
((1,0), 1), # First
((0x80000000, 0), 1), # Last
((0x55555555, 0), 16), # Every Other
((0xffff, 0), 16), # First Half
((0xffff0000, 0), 16), # Last Half
)
for i, (args, exp) in enumerate(test_vals):
if f(*args) != exp:
return 0
return 1
if __name__ == "__main__":
for f in (yours, simple, lazy, fancy):
if not test(f):
print "%s failed"%f.__name__
[/code]
The python module `timeit` is handy for testing speed:
python -mtimeit -s"from hamdist import *" "test(yours)"
10000 loops, best of 3: 95.1 usec per loop
python -mtimeit -s"from hamdist import *" "test(simple)"
10000 loops, best of 3: 65.3 usec per loop
python -mtimeit -s"from hamdist import *" "test(lazy)"
10000 loops, best of 3: 59.8 usec per loop
python -mtimeit -s"from hamdist import *" "test(fancy)"
10000 loops, best of 3: 77.2 usec per loop
Even the ridiculous `fancy` version beat the recursive version.
Matt
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