我在这两种方法之间做了一些基准测试:
- 转置+弹出计数
- 逐行更新
我为这两种方法编写了一个简单的版本和一个 AVX2 版本。我为 AVX2“transpose+popcount”方法使用了一些函数(在 stackoverflow 或其他地方找到)。
在我的测试中,我假设输入是nbRowsx32
位压缩格式的矩阵(nbRows 本身是 32 的倍数);因此,矩阵存储为 的数组uint32_t
。
代码如下:
#include <cinttypes>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cassert>
#include <chrono>
#include <immintrin.h>
#include <asmlib.h>
using namespace std;
using namespace std::chrono;
// see https://stackoverflow.com/questions/24225786/fastest-way-to-unpack-32-bits-to-a-32-byte-simd-vector
static __m256i expand_bits_to_bytes (uint32_t x);
// see https://mischasan.wordpress.com/2011/10/03/the-full-sse2-bit-matrix-transpose-routine/
static void sse_trans(char const *inp, char *out);
static double deviation (double n, double sum2, double sum);
////////////////////////////////////////////////////////////////////////////////
// Naive approach (matrix transposition)
////////////////////////////////////////////////////////////////////////////////
void test_transpose_popcnt_naive (uint64_t nbRows, const uint32_t* bitmap, uint64_t* globalSums)
{
assert (nbRows%32==0);
uint8_t transpo[32][32]; memset (transpo, 0, sizeof(transpo));
for (uint64_t k=0; k<nbRows; k+=32)
{
// We unpack and transpose the input into a 32x32 bytes matrix
for (size_t row=0; row<32; row++)
{
for (size_t col=0; col<32; col++) { transpo[col][row] = (bitmap[k+row] >> col) & 1 ; }
}
for (size_t row=0; row<32; row++)
{
// We popcount the current row
u_int8_t sum=0;
for (size_t col=0; col<32; col++) { sum += transpo[row][col]; }
// We update the corresponding global sum
globalSums[row] += sum;
}
}
}
////////////////////////////////////////////////////////////////////////////////
// Naive approach (row by row)
////////////////////////////////////////////////////////////////////////////////
void test_update_row_by_row_naive (uint64_t nbRows, const uint32_t* bitmap, uint64_t* globalSums)
{
for (uint64_t row=0; row<nbRows; row++)
{
for (size_t col=0; col<32; col++)
{
globalSums[col] += (bitmap[row] >> col) & 1;
}
}
}
////////////////////////////////////////////////////////////////////////////////
// AVX2 (matrix transposition + popcount)
////////////////////////////////////////////////////////////////////////////////
void test_transpose_popcnt_avx2 (uint64_t nbRows, const uint32_t* bitmap, uint64_t* globalSums)
{
assert (nbRows%32==0);
uint32_t transpo[32];
const uint32_t* loop = bitmap;
for (uint64_t k=0; k<nbRows; loop+=32, k+=32)
{
// We transpose the input as a 32x32 bytes matrix
sse_trans ((const char*)loop, (char*)transpo);
// We update the global sums
for (size_t i=0; i<32; i++)
{
globalSums[i] += __builtin_popcount (transpo[i]);
}
}
}
////////////////////////////////////////////////////////////////////////////////
// AVX2 approach (update totals row by row)
////////////////////////////////////////////////////////////////////////////////
// Note: we use template specialization to unroll some portions of a loop
template<int N>
void UpdateLocalSums (__m256i& localSums, const uint32_t* bitmap, uint64_t& k)
{
// We update the local sums with the current row
localSums = _mm256_sub_epi8 (localSums, expand_bits_to_bytes (bitmap[k++]));
// Go recursively
UpdateLocalSums<N-1>(localSums, bitmap, k);
}
template<>
void UpdateLocalSums<0> (__m256i& localSums, const uint32_t* bitmap, uint64_t& k)
{
}
// Dillon Davis proposal: use 4 registers holding uint32_t values and update them from local sums with AVX2
#define USE_AVX2_FOR_GRAND_TOTALS 1
void test_update_row_by_row_avx2 (uint64_t nbRows, const uint32_t* bitmap, uint64_t* globalSums)
{
union U256i { __m256i v; uint8_t a[32]; uint32_t b[8]; };
// We use 1 register for updating local totals
__m256i localSums = _mm256_setzero_si256();
#ifdef USE_AVX2_FOR_GRAND_TOTALS
// Dillon Davis proposal: use 4 registers holding uint32_t values and update them from local sums with AVX2
__m256i globalSumsReg[4]; for (size_t r=0; r<4; r++) { globalSumsReg[r] = _mm256_setzero_si256(); }
#endif
uint64_t steps = nbRows / 255;
uint64_t k=0;
const int divisorOf255 = 5;
// We iterate over all rows
for (uint64_t i=0; i<steps; i++)
{
// we update the local totals (255*32=8160 additions)
for (int j=0; j<255/divisorOf255; j++)
{
// unroll some portion of the 255 loop through template specialization
UpdateLocalSums<divisorOf255>(localSums, bitmap, k);
}
#ifdef USE_AVX2_FOR_GRAND_TOTALS
// Dillon Davis proposal: use 4 registers holding uint32_t values and update them from local sums
// We take the 128 high bits of the local sums
__m256i localSums2 = _mm256_broadcastsi128_si256(_mm256_extracti128_si256(localSums,1));
globalSumsReg[0] = _mm256_add_epi32 (globalSumsReg[0],
_mm256_cvtepu8_epi32 (_mm256_castsi256_si128 (_mm256_srli_si256(localSums, 0)))
);
globalSumsReg[1] = _mm256_add_epi32 (globalSumsReg[1],
_mm256_cvtepu8_epi32 (_mm256_castsi256_si128 (_mm256_srli_si256(localSums, 8)))
);
globalSumsReg[2] = _mm256_add_epi32 (globalSumsReg[2],
_mm256_cvtepu8_epi32 (_mm256_castsi256_si128 (_mm256_srli_si256(localSums2, 0)))
);
globalSumsReg[3] = _mm256_add_epi32 (globalSumsReg[3],
_mm256_cvtepu8_epi32 (_mm256_castsi256_si128 (_mm256_srli_si256(localSums2, 8)))
);
#else
// we update the global totals
U256i tmp = { localSums };
for (size_t k=0; k<32; k++) { globalSums[k] += tmp.a[k]; }
#endif
// we reset the local totals
localSums = _mm256_setzero_si256();
}
#ifdef USE_AVX2_FOR_GRAND_TOTALS
// We update the global totals into the final uint32_t array
for (size_t r=0; r<4; r++)
{
U256i tmp = { globalSumsReg[r] };
for (size_t k=0; k<8; k++) { globalSums[r*8+k] += tmp.b[k]; }
}
#endif
// we update the remaining local totals
for (uint64_t i=steps*255; i<nbRows; i++)
{
UpdateLocalSums<1>(localSums, bitmap, k);
}
// we update the global totals
U256i tmp = { localSums };
for (size_t k=0; k<32; k++) { globalSums[k] += tmp.a[k]; }
}
////////////////////////////////////////////////////////////////////////////////
void execute (
const char* name,
void (*fct)(uint64_t nbRows, const uint32_t* bitmap, uint64_t* globalSums),
size_t nbRuns,
uint64_t nbRows,
u_int32_t* bitmap
)
{
uint64_t sums[32];
double timeTotal=0;
double cycleTotal=0;
double timeTotal2=0;
double cycleTotal2=0;
uint64_t check=0;
for (size_t n=0; n<nbRuns; n++)
{
memset(sums,0,sizeof(sums));
// We want both time and cpu cycles information
milliseconds t0 = duration_cast< milliseconds >(system_clock::now().time_since_epoch());
uint64_t c0 = ReadTSC();
// We run the test
(*fct) (nbRows, bitmap, sums);
uint64_t c1 = ReadTSC();
milliseconds t1 = duration_cast< milliseconds >(system_clock::now().time_since_epoch());
timeTotal += (t1-t0).count();
cycleTotal += (double)(c1-c0) / nbRows;
timeTotal2 += (t1-t0).count() * (t1-t0).count();
cycleTotal2 += ((double)(c1-c0) / nbRows) * ((double)(c1-c0) / nbRows);
// We compute some dummy checksum
for (size_t k=0; k<32; k++) { check += sums[k]; }
}
printf ("%-21s | %5.0lf (%5.1lf) | %5.2lf (%4.2lf) | %.3lf | 0x%lx\n",
name,
timeTotal / nbRuns,
deviation (nbRuns, timeTotal2, timeTotal),
cycleTotal/nbRuns,
deviation (nbRuns, cycleTotal2, cycleTotal),
check,
nbRows * cycleTotal / timeTotal / 1000000.0
);
}
////////////////////////////////////////////////////////////////////////////////
int main(int argc, char **argv)
{
// We set rows number as 2^n where n is the provided argument
// For simplification, we assume that the rows number is a multiple of 32
uint64_t nbRows = 1ULL << (argc>1 ? atoi(argv[1]) : 28);
size_t nbRuns = argc>2 ? atoi(argv[2]) : 10;
// We build an bitmap of size nbRows*32
uint32_t* bitmap = new uint32_t[nbRows];
if (bitmap==nullptr)
{
fprintf(stderr, "unable to allocate the bitmap\n");
exit(1);
}
// We fill the bitmap with random values
srand(time(nullptr));
for (uint64_t i=0; i<nbRows; i++) { bitmap[i] = rand() & 0xFFFFFFFF; }
printf ("\n");
printf ("nbRows=%ld nbRuns=%ld\n", nbRows, nbRuns);
printf ("------------------------------------------------------------------------------------------------------------\n");
printf ("name | time in msec : mean (sd) | cycles/row : mean (sd) | frequency in GHz | checksum\n");
printf ("------------------------------------------------------------------------------------------------------------\n");
// We launch the benchmark
execute ("naive (transpo) ", test_transpose_popcnt_naive, nbRuns, nbRows, bitmap);
execute ("naive (row by row)", test_update_row_by_row_naive, nbRuns, nbRows, bitmap);
execute ("AVX2 (transpo) ", test_transpose_popcnt_avx2, nbRuns, nbRows, bitmap);
execute ("AVX2 (row by row)", test_update_row_by_row_avx2, nbRuns, nbRows, bitmap);
printf ("\n");
// Some clean up
delete[] bitmap;
return EXIT_SUCCESS;
}
////////////////////////////////////////////////////////////////////////////////
__m256i expand_bits_to_bytes(uint32_t x)
{
__m256i xbcast = _mm256_set1_epi32(x);
// Each byte gets the source byte containing the corresponding bit
__m256i shufmask = _mm256_set_epi64x(
0x0303030303030303, 0x0202020202020202,
0x0101010101010101, 0x0000000000000000);
__m256i shuf = _mm256_shuffle_epi8(xbcast, shufmask);
__m256i andmask = _mm256_set1_epi64x(0x8040201008040201); // every 8 bits -> 8 bytes, pattern repeats.
__m256i isolated_inverted = _mm256_and_si256(shuf, andmask);
// Avoid an _mm256_add_epi8 thanks to Peter Cordes's comment
return _mm256_cmpeq_epi8(isolated_inverted, andmask);
}
////////////////////////////////////////////////////////////////////////////////
void sse_trans(char const *inp, char *out)
{
#define INP(x,y) inp[(x)*4 + (y)/8]
#define OUT(x,y) out[(y)*4 + (x)/8]
int rr, cc, i, h;
union { __m256i x; uint8_t b[32]; } tmp;
for (cc = 0; cc < 32; cc += 8)
{
for (i = 0; i < 32; ++i)
tmp.b[i] = INP(i, cc);
for (i = 8; i--; tmp.x = _mm256_slli_epi64(tmp.x, 1))
*(uint32_t*)&OUT(0, cc + i) = _mm256_movemask_epi8(tmp.x);
}
}
////////////////////////////////////////////////////////////////////////////////
double deviation (double n, double sum2, double sum) { return sqrt (sum2/n - (sum/n)*(sum/n)); }
一些备注:
- 我使用 Agner Fog 的 asmlib 有一个返回 CPU 周期的函数
- 编译命令是
g++ -O3 -march=native ../Test.cpp -o ./Test -laelf64
- gcc 版本是 7.3.1
- CPU 是 Intel(R) Core(TM) i7-6700HQ CPU @ 2.60GHz
- 我计算了一些虚拟校验和来比较不同测试的结果
现在结果:
------------------------------------------------------------------------------------------------------------
name | time in msec : mean (sd) | cycles/row : mean (sd) | frequency in GHz | checksum
------------------------------------------------------------------------------------------------------------
naive (transpo) | 4548 ( 36.5) | 43.91 (0.35) | 2.592 | 0x9affeb5a6
naive (row by row) | 3033 ( 11.0) | 29.29 (0.11) | 2.592 | 0x9affeb5a6
AVX2 (transpo) | 767 ( 12.8) | 7.40 (0.12) | 2.592 | 0x9affeb5a6
AVX2 (row by row) | 130 ( 4.0) | 1.25 (0.04) | 2.591 | 0x9affeb5a6
所以看来AVX2中的“逐行”是目前为止最好的。
请注意,当我看到这个结果(每行少于 2 个周期)时,我没有再努力优化 AVX2“transpose+popcount”方法,这应该可以通过并行计算几个 popcount 来实现(我可能稍后测试它)。