Simple Python Challenge: Fastest Bitwise XOR on Data Buffers

algorithm, performance, python, xor

Solution

First Try

Using `scipy.weave` and SSE2 intrinsics gives a marginal improvement. The first invocation is a bit slower since the code needs to be loaded from the disk and cached, subsequent invocations are faster:

import numpy
import time
from os import urandom
from scipy import weave

SIZE = 2**20

def faster_slow_xor(aa,bb):
    b = numpy.fromstring(bb, dtype=numpy.uint64)
    numpy.bitwise_xor(numpy.frombuffer(aa,dtype=numpy.uint64), b, b)
    return b.tostring()

code = """
const __m128i* pa = (__m128i*)a;
const __m128i* pend = (__m128i*)(a + arr_size);
__m128i* pb = (__m128i*)b;
__m128i xmm1, xmm2;
while (pa < pend) {
  xmm1 = _mm_loadu_si128(pa); // must use unaligned access 
  xmm2 = _mm_load_si128(pb); // numpy will align at 16 byte boundaries
  _mm_store_si128(pb, _mm_xor_si128(xmm1, xmm2));
  ++pa;
  ++pb;
}
"""

def inline_xor(aa, bb):
    a = numpy.frombuffer(aa, dtype=numpy.uint64)
    b = numpy.fromstring(bb, dtype=numpy.uint64)
    arr_size = a.shape[0]
    weave.inline(code, ["a", "b", "arr_size"], headers = ['"emmintrin.h"'])
    return b.tostring()

Second Try

Taking into account the comments, I revisited the code to find out if the copying could be avoided. Turns out I read the documentation of the string object wrong, so here goes my second try:

support = """
#define ALIGNMENT 16
static void memxor(const char* in1, const char* in2, char* out, ssize_t n) {
    const char* end = in1 + n;
    while (in1 < end) {
       *out = *in1 ^ *in2;
       ++in1; 
       ++in2;
       ++out;
    }
}
"""

code2 = """
PyObject* res = PyString_FromStringAndSize(NULL, real_size);

const ssize_t tail = (ssize_t)PyString_AS_STRING(res) % ALIGNMENT;
const ssize_t head = (ALIGNMENT - tail) % ALIGNMENT;

memxor((const char*)a, (const char*)b, PyString_AS_STRING(res), head);

const __m128i* pa = (__m128i*)((char*)a + head);
const __m128i* pend = (__m128i*)((char*)a + real_size - tail);
const __m128i* pb = (__m128i*)((char*)b + head);
__m128i xmm1, xmm2;
__m128i* pc = (__m128i*)(PyString_AS_STRING(res) + head);
while (pa < pend) {
    xmm1 = _mm_loadu_si128(pa);
    xmm2 = _mm_loadu_si128(pb);
    _mm_stream_si128(pc, _mm_xor_si128(xmm1, xmm2));
    ++pa;
    ++pb;
    ++pc;
}
memxor((const char*)pa, (const char*)pb, (char*)pc, tail);
return_val = res;
Py_DECREF(res);
"""

def inline_xor_nocopy(aa, bb):
    real_size = len(aa)
    a = numpy.frombuffer(aa, dtype=numpy.uint64)
    b = numpy.frombuffer(bb, dtype=numpy.uint64)
    return weave.inline(code2, ["a", "b", "real_size"], 
                        headers = ['"emmintrin.h"'], 
                        support_code = support)

The difference is that the string is allocated inside the C code. It's impossible to have it aligned at a 16-byte-boundary as required by the SSE2 instructions, therefore the unaligned memory regions at the beginning and the end are copied using byte-wise access.

The input data is handed in using numpy arrays anyway, because `weave` insists on copying Python `str` objects to `std::string`s. `frombuffer` doesn't copy, so this is fine, but the memory is not aligned at 16 byte, so we need to use `_mm_loadu_si128` instead of the faster `_mm_load_si128`.

Instead of using `_mm_store_si128`, we use `_mm_stream_si128`, which will make sure that any writes are streamed to main memory as soon as possible---this way, the output array does not use up valuable cache lines.

Timings

As for the timings, the `slow_xor` entry in the first edit referred to my improved version (inline bitwise xor, `uint64`), I removed that confusion. `slow_xor` refers to the code from the original questions. All timings are done for 1000 runs.

- `slow_xor`: 1.85s (1x)

- `faster_slow_xor`: 1.25s (1.48x)

- `inline_xor`: 0.95s (1.95x)

- `inline_xor_nocopy`: 0.32s (5.78x)

The code was compiled using gcc 4.4.3 and I've verified that the compiler actually uses the SSE instructions.

Problem

Challenge: Perform a bitwise XOR on two equal sized buffers. The buffers will be required to be the python `str` type since this is traditionally the type for data buffers in python. Return the resultant value as a `str`. Do this as fast as possible. The inputs are two 1 megabyte (2**20 byte) strings. The challenge is to substantially beat my inefficient algorithm using python or existing third party python modules (relaxed rules: or create your own module.) Marginal increases are useless. ``` from os import urandom from numpy import frombuffer,bitwise_xor,byte def slow_xor(aa,bb): a=frombuffer(aa,dtype=byte) b=frombuffer(bb,dtype=byte) c=bitwise_xor(a,b) r=c.tostring() return r aa=urandom(2**20) bb=urandom(2**20) def test_it(): for x in xrange(1000): slow_xor(aa,bb) ```

Original source