我需要实现前缀和算法，并且需要它尽可能快。 例如：
[3, 1, 7, 0, 4, 1, 6, 3]
should give
[3, 4, 11, 11, 15, 16, 22, 25]
有没有办法使用SSE / mmx / SIMD cpu指令执行此操作？
我的第一个想法是递归并行地对每一对求和，直到所有的总和都计算如下！
//in parallel do
for (int i = 0; i<z.length; i++){
z[i] = x[i<<1] + x[(i<<1)+1];
}
为了使算法更清晰，“z”不是最终输出
但是用来计算输出
int[] w = computePrefixSum(z);
for (int i = 1; i<ouput.length; i++){
ouput[i] = (i%2==0) ? (x[i] + ouput[i-1]) : w[(i-1)>>1];
}
答案 0 :(得分：11)
对于大寄存器长度和小总和，您可以利用一些小的并行性。例如，将1个字节的16个值（恰好适合一个sse寄存器）相加只需要log 2 16个加法和相同数量的移位。
不多，但比15个依赖的添加和额外的内存访问快。
__m128i x = _mm_set_epi8(3,1,7,0,4,1,6,3,3,1,7,0,4,1,6,3);
x = _mm_add_epi8(x, _mm_srli_si128(x, 1));
x = _mm_add_epi8(x, _mm_srli_si128(x, 2));
x = _mm_add_epi8(x, _mm_srli_si128(x, 4));
x = _mm_add_epi8(x, _mm_srli_si128(x, 8));
// x == 3, 4, 11, 11, 15, 16, 22, 25, 28, 29, 36, 36, 40, 41, 47, 50
如果你有更长的总和，可以通过利用指令级并行性并利用指令重新排序来隐藏依赖关系。
编辑：类似
__m128i x0 = _mm_set_epi8(3,1,7,0,4,1,6,3,3,1,7,0,4,1,6,3);
__m128i x1 = _mm_set_epi8(3,1,7,0,4,1,6,3,3,1,7,0,4,1,6,3);
__m128i x2 = _mm_set_epi8(3,1,7,0,4,1,6,3,3,1,7,0,4,1,6,3);
__m128i x3 = _mm_set_epi8(3,1,7,0,4,1,6,3,3,1,7,0,4,1,6,3);
__m128i mask = _mm_set_epi8(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0);
x0 = _mm_add_epi8(x0, _mm_srli_si128(x0, 1));
x1 = _mm_add_epi8(x1, _mm_srli_si128(x1, 1));
x2 = _mm_add_epi8(x2, _mm_srli_si128(x2, 1));
x3 = _mm_add_epi8(x3, _mm_srli_si128(x3, 1));
x0 = _mm_add_epi8(x0, _mm_srli_si128(x0, 2));
x1 = _mm_add_epi8(x1, _mm_srli_si128(x1, 2));
x2 = _mm_add_epi8(x2, _mm_srli_si128(x2, 2));
x3 = _mm_add_epi8(x3, _mm_srli_si128(x3, 2));
x0 = _mm_add_epi8(x0, _mm_srli_si128(x0, 4));
x1 = _mm_add_epi8(x1, _mm_srli_si128(x1, 4));
x2 = _mm_add_epi8(x2, _mm_srli_si128(x2, 4));
x3 = _mm_add_epi8(x3, _mm_srli_si128(x3, 4));
x0 = _mm_add_epi8(x0, _mm_srli_si128(x0, 8));
x1 = _mm_add_epi8(x1, _mm_srli_si128(x1, 8));
x2 = _mm_add_epi8(x2, _mm_srli_si128(x2, 8));
x3 = _mm_add_epi8(x3, _mm_srli_si128(x3, 8));
x1 = _mm_add_epi8(_mm_shuffle_epi8(x0, mask), x1);
x2 = _mm_add_epi8(_mm_shuffle_epi8(x1, mask), x2);
x3 = _mm_add_epi8(_mm_shuffle_epi8(x2, mask), x3);
答案 1 :(得分：8)
我所知道的最快的并行前缀和算法是在两个并行中并行运行总和，并在第二遍中使用SSE。
在第一遍中，您并行计算部分和，并存储每个部分和的总和。在第二遍中，将前一部分和的总和加到下一部分和。您可以使用多个线程并行运行两个传递（例如，使用OpenMP）。第二遍你也可以使用SIMD，因为每个部分和都会加一个常数值。
假设数组的n个元素，m核心以及SIMD宽度w，则时间成本应为
n/m + n/(m*w) = (n/m)*(1+1/w)
由于第一次通过不使用SIMD，因此时间成本将始终大于n/m
例如，对于SIMD_width为4的四个内核（四个32位浮点数和SSE），成本为5n/16。或者比顺序代码快3.2倍，时间成本为n。使用超线程，速度会更快。
在特殊情况下，也可以在第一次使用时使用SIMD。那么时间成本就是
2*n/(m*w)
我发布了一般情况的代码，该代码使用OpenMP作为SSE代码的线程和内在函数，并在以下链接中讨论有关特殊情况的详细信息 parallel-prefix-cumulative-sum-with-sse
修改： 我设法找到第一遍的SIMD版本，大约是顺序代码的两倍。现在，我的四核常春藤桥系统总共提升了7个左右。
修改
对于较大的阵列，一个问题是在第一次传递之后，大多数值已从缓存中逐出。我提出了一个在块内并行运行的解决方案，但是按顺序运行每个块。 chunk_size是应该调整的值。例如，我将其设置为1MB = 256K浮点数。现在，第二次传递完成，而值仍在二级缓存中。这样做可以为大型阵列带来很大的改进。
以下是SSE的代码。 AVX代码速度大致相同，所以我没有在这里发布。执行前缀和的函数是scan_omp_SSEp2_SSEp1_chunk。传递一个数组a的浮点数，它用累积的总和填充数组s。
__m128 scan_SSE(__m128 x) {
x = _mm_add_ps(x, _mm_castsi128_ps(_mm_slli_si128(_mm_castps_si128(x), 4)));
x = _mm_add_ps(x, _mm_shuffle_ps(_mm_setzero_ps(), x, 0x40));
return x;
}
float pass1_SSE(float *a, float *s, const int n) {
__m128 offset = _mm_setzero_ps();
#pragma omp for schedule(static) nowait
for (int i = 0; i < n / 4; i++) {
__m128 x = _mm_load_ps(&a[4 * i]);
__m128 out = scan_SSE(x);
out = _mm_add_ps(out, offset);
_mm_store_ps(&s[4 * i], out);
offset = _mm_shuffle_ps(out, out, _MM_SHUFFLE(3, 3, 3, 3));
}
float tmp[4];
_mm_store_ps(tmp, offset);
return tmp[3];
}
void pass2_SSE(float *s, __m128 offset, const int n) {
#pragma omp for schedule(static)
for (int i = 0; i<n/4; i++) {
__m128 tmp1 = _mm_load_ps(&s[4 * i]);
tmp1 = _mm_add_ps(tmp1, offset);
_mm_store_ps(&s[4 * i], tmp1);
}
}
void scan_omp_SSEp2_SSEp1_chunk(float a[], float s[], int n) {
float *suma;
const int chunk_size = 1<<18;
const int nchunks = n%chunk_size == 0 ? n / chunk_size : n / chunk_size + 1;
//printf("nchunks %d\n", nchunks);
#pragma omp parallel
{
const int ithread = omp_get_thread_num();
const int nthreads = omp_get_num_threads();
#pragma omp single
{
suma = new float[nthreads + 1];
suma[0] = 0;
}
float offset2 = 0.0f;
for (int c = 0; c < nchunks; c++) {
const int start = c*chunk_size;
const int chunk = (c + 1)*chunk_size < n ? chunk_size : n - c*chunk_size;
suma[ithread + 1] = pass1_SSE(&a[start], &s[start], chunk);
#pragma omp barrier
#pragma omp single
{
float tmp = 0;
for (int i = 0; i < (nthreads + 1); i++) {
tmp += suma[i];
suma[i] = tmp;
}
}
__m128 offset = _mm_set1_ps(suma[ithread]+offset2);
pass2_SSE(&s[start], offset, chunk);
#pragma omp barrier
offset2 = s[start + chunk-1];
}
}
delete[] suma;
}
答案 2 :(得分：7)
prefix-sum可以并行计算，它实际上是GPU编程中的基础算法之一。如果你在英特尔处理器上使用SIMD扩展，我不确定并行执行它实际上会对你有多大帮助，但是请看看nvidia关于实现并行前缀和的这篇论文（只看一下算法并忽略） CUDA）：http://http.developer.nvidia.com/GPUGems3/gpugems3_ch39.html
答案 3 :(得分：6)
对于1000个32位整数的数组，我能够在Intel Sandybridge的循环中使用@ hirschhornsalz的方法获得大约1.4x单线程的小加速。使用60kB的int缓冲区，加速比约为1.37。凭借8MiB的整数，加速仍为1.13。 （i3-2500k，3.8GHz turbo，DDR3-1600。）
较小的元素（int16_t或uint8_t或无符号版本）将为每个向量的每个元素数量的每倍增加一个额外的移位/加法阶段。溢出是不好的，所以不要试图使用不能保存所有元素总和的数据类型，即使它给SSE带来了更大的优势。
#include <immintrin.h>
// In-place rewrite an array of values into an array of prefix sums.
// This makes the code simpler, and minimizes cache effects.
int prefix_sum_sse(int data[], int n)
{
// const int elemsz = sizeof(data[0]);
#define elemsz sizeof(data[0]) // clang-3.5 doesn't allow compile-time-const int as an imm8 arg to intrinsics
__m128i *datavec = (__m128i*)data;
const int vec_elems = sizeof(*datavec)/elemsz;
// to use this for int8/16_t, you still need to change the add_epi32, and the shuffle
const __m128i *endp = (__m128i*) (data + n - 2*vec_elems); // don't start an iteration beyond this
__m128i carry = _mm_setzero_si128();
for(; datavec <= endp ; datavec += 2) {
IACA_START
__m128i x0 = _mm_load_si128(datavec + 0);
__m128i x1 = _mm_load_si128(datavec + 1); // unroll / pipeline by 1
// __m128i x2 = _mm_load_si128(datavec + 2);
// __m128i x3;
x0 = _mm_add_epi32(x0, _mm_slli_si128(x0, elemsz)); // for floats, use shufps not bytewise-shift
x1 = _mm_add_epi32(x1, _mm_slli_si128(x1, elemsz));
x0 = _mm_add_epi32(x0, _mm_slli_si128(x0, 2*elemsz));
x1 = _mm_add_epi32(x1, _mm_slli_si128(x1, 2*elemsz));
// more shifting if vec_elems is larger
x0 = _mm_add_epi32(x0, carry); // this has to go after the byte-shifts, to avoid double-counting the carry.
_mm_store_si128(datavec +0, x0); // store first to allow destructive shuffle (non-avx pshufb if needed)
x1 = _mm_add_epi32(_mm_shuffle_epi32(x0, _MM_SHUFFLE(3,3,3,3)), x1);
_mm_store_si128(datavec +1, x1);
carry = _mm_shuffle_epi32(x1, _MM_SHUFFLE(3,3,3,3)); // broadcast the high element for next vector
}
// FIXME: scalar loop to handle the last few elements
IACA_END
return data[n-1];
#undef elemsz
}
int prefix_sum_simple(int data[], int n)
{
int sum=0;
for (int i=0; i<n ; i++) {
IACA_START
sum += data[i];
data[i] = sum;
}
IACA_END
return sum;
}
// perl -we '$n=1000; sub rnlist($$) { return map { int rand($_[1]) } ( 1..$_[0] );} @a=rnlist($n,127); $"=", "; print "$n\n@a\n";'
int data[] = { 51, 83, 126, 11, 20, 63, 113, 102,
126,67, 83, 113, 86, 123, 30, 109,
97, 71, 109, 86, 67, 60, 47, 12,
/* ... */ };
int main(int argc, char**argv)
{
const int elemsz = sizeof(data[0]);
const int n = sizeof(data)/elemsz;
const long reps = 1000000 * 1000 / n;
if (argc >= 2 && *argv[1] == 'n') {
for (int i=0; i < reps ; i++)
prefix_sum_simple(data, n);
}else {
for (int i=0; i < reps ; i++)
prefix_sum_sse(data, n);
}
return 0;
}
使用n = 1000进行测试，并将列表编译为二进制文件。 （是的，我检查它实际上是循环的，没有采用任何使矢量或非矢量测试无意义的编译时快捷方式。）
请注意，使用AVX进行编译以获得3操作数非破坏性向量指令可以节省大量movdqa指令，但只能节省很少的周期。这是因为shuffle和vector-int-add都只能在SnB / IvB上的端口1和5上运行，因此port0有足够的备用周期来运行mov指令。 uop-cache吞吐量瓶颈可能是非AVX版本稍慢的原因。 （所有那些额外的mov指令将我们推到3.35 insn / cycle）。前端只剩下4.54％的周期，所以几乎没有跟上。
gcc -funroll-loops -DIACA_MARKS_OFF -g -std=c11 -Wall -march=native -O3 prefix-sum.c -mno-avx -o prefix-sum-noavx
# gcc 4.9.2
################# SSE (non-AVX) vector version ############
$ ocperf.py stat -e task-clock,cycles,instructions,uops_issued.any,uops_dispatched.thread,uops_retired.all,uops_retired.retire_slots,stalled-cycles-frontend,stalled-cycles-backend ./prefix-sum-noavx
perf stat -e task-clock,cycles,instructions,cpu/event=0xe,umask=0x1,name=uops_issued_any/,cpu/event=0xb1,umask=0x1,name=uops_dispatched_thread/,cpu/event=0xc2,umask=0x1,name=uops_retired_all/,cpu/event=0xc2,umask=0x2,name=uops_retired_retire_slots/,stalled-cycles-frontend,stalled-cycles-backend ./prefix-sum-noavx
Performance counter stats for './prefix-sum-noavx':
206.986720 task-clock (msec) # 0.999 CPUs utilized
777,473,726 cycles # 3.756 GHz
2,604,757,487 instructions # 3.35 insns per cycle
# 0.01 stalled cycles per insn
2,579,310,493 uops_issued_any # 12461.237 M/sec
2,828,479,147 uops_dispatched_thread # 13665.027 M/sec
2,829,198,313 uops_retired_all # 13668.502 M/sec (unfused domain)
2,579,016,838 uops_retired_retire_slots # 12459.818 M/sec (fused domain)
35,298,807 stalled-cycles-frontend # 4.54% frontend cycles idle
1,224,399 stalled-cycles-backend # 0.16% backend cycles idle
0.207234316 seconds time elapsed
------------------------------------------------------------
######### AVX (same source, but built with -mavx). not AVX2 #########
$ ocperf.py stat -e task-clock,cycles,instructions,uops_issued.any,uops_dispatched.thread,uops_retired.all,uops_retired.retire_slots,stalled-cycles-frontend,stalled-cycles-backend ./prefix-sum-avx
Performance counter stats for './prefix-sum-avx':
203.429021 task-clock (msec) # 0.999 CPUs utilized
764,859,441 cycles # 3.760 GHz
2,079,716,097 instructions # 2.72 insns per cycle
# 0.12 stalled cycles per insn
2,054,334,040 uops_issued_any # 10098.530 M/sec
2,303,378,797 uops_dispatched_thread # 11322.764 M/sec
2,304,140,578 uops_retired_all # 11326.509 M/sec
2,053,968,862 uops_retired_retire_slots # 10096.735 M/sec
240,883,566 stalled-cycles-frontend # 31.49% frontend cycles idle
1,224,637 stalled-cycles-backend # 0.16% backend cycles idle
0.203732797 seconds time elapsed
------------------------------------------------------------
################## scalar version (cmdline arg) #############
$ ocperf.py stat -e task-clock,cycles,instructions,uops_issued.any,uops_dispatched.thread,uops_retired.all,uops_retired.retire_slots,stalled-cycles-frontend,stalled-cycles-backend ./prefix-sum-avx n
Performance counter stats for './prefix-sum-avx n':
287.567070 task-clock (msec) # 0.999 CPUs utilized
1,082,611,453 cycles # 3.765 GHz
2,381,840,355 instructions # 2.20 insns per cycle
# 0.20 stalled cycles per insn
2,272,652,370 uops_issued_any # 7903.034 M/sec
4,262,838,836 uops_dispatched_thread # 14823.807 M/sec
4,256,351,856 uops_retired_all # 14801.249 M/sec
2,256,150,510 uops_retired_retire_slots # 7845.650 M/sec
465,018,146 stalled-cycles-frontend # 42.95% frontend cycles idle
6,321,098 stalled-cycles-backend # 0.58% backend cycles idle
0.287901811 seconds time elapsed
------------------------------------------------------------
Haswell应该大致相同，但每个时钟可能稍慢一些，因为shuffle只能在端口5上运行，而不是在端口1上运行。（在Haswell上，vector-int add仍然是p1 / 5。）
OTOH，IACA认为Haswell在一次迭代中会比SnB略快，如果你编译没有-funroll-loops（这对SnB有帮助）。 Haswell可以在port6上做分支，但在SnB分支上是在port5上，我们已经饱和了。
# compile without -DIACA_MARKS_OFF
$ iaca -64 -mark 1 -arch HSW prefix-sum-avx
Intel(R) Architecture Code Analyzer Version - 2.1
Analyzed File - prefix-sum-avx
Binary Format - 64Bit
Architecture - HSW
Analysis Type - Throughput
*******************************************************************
Intel(R) Architecture Code Analyzer Mark Number 1
*******************************************************************
Throughput Analysis Report
--------------------------
Block Throughput: 6.20 Cycles Throughput Bottleneck: Port5
Port Binding In Cycles Per Iteration:
---------------------------------------------------------------------------------------
| Port | 0 - DV | 1 | 2 - D | 3 - D | 4 | 5 | 6 | 7 |
---------------------------------------------------------------------------------------
| Cycles | 1.0 0.0 | 5.8 | 1.4 1.0 | 1.4 1.0 | 2.0 | 6.2 | 1.0 | 1.3 |
---------------------------------------------------------------------------------------
N - port number or number of cycles resource conflict caused delay, DV - Divider pipe (on port 0)
D - Data fetch pipe (on ports 2 and 3), CP - on a critical path
F - Macro Fusion with the previous instruction occurred
* - instruction micro-ops not bound to a port
^ - Micro Fusion happened
# - ESP Tracking sync uop was issued
@ - SSE instruction followed an AVX256 instruction, dozens of cycles penalty is expected
! - instruction not supported, was not accounted in Analysis
| Num Of | Ports pressure in cycles | |
| Uops | 0 - DV | 1 | 2 - D | 3 - D | 4 | 5 | 6 | 7 | |
---------------------------------------------------------------------------------
| 1 | | | 1.0 1.0 | | | | | | | vmovdqa xmm2, xmmword ptr [rax]
| 1 | 1.0 | | | | | | | | | add rax, 0x20
| 1 | | | | 1.0 1.0 | | | | | | vmovdqa xmm3, xmmword ptr [rax-0x10]
| 1 | | | | | | 1.0 | | | CP | vpslldq xmm1, xmm2, 0x4
| 1 | | 1.0 | | | | | | | | vpaddd xmm2, xmm2, xmm1
| 1 | | | | | | 1.0 | | | CP | vpslldq xmm1, xmm3, 0x4
| 1 | | 1.0 | | | | | | | | vpaddd xmm3, xmm3, xmm1
| 1 | | | | | | 1.0 | | | CP | vpslldq xmm1, xmm2, 0x8
| 1 | | 1.0 | | | | | | | | vpaddd xmm2, xmm2, xmm1
| 1 | | | | | | 1.0 | | | CP | vpslldq xmm1, xmm3, 0x8
| 1 | | 1.0 | | | | | | | | vpaddd xmm3, xmm3, xmm1
| 1 | | 0.9 | | | | 0.2 | | | CP | vpaddd xmm1, xmm2, xmm0
| 2^ | | | | | 1.0 | | | 1.0 | | vmovaps xmmword ptr [rax-0x20], xmm1
| 1 | | | | | | 1.0 | | | CP | vpshufd xmm1, xmm1, 0xff
| 1 | | 0.9 | | | | 0.1 | | | CP | vpaddd xmm0, xmm1, xmm3
| 2^ | | | 0.3 | 0.3 | 1.0 | | | 0.3 | | vmovaps xmmword ptr [rax-0x10], xmm0
| 1 | | | | | | 1.0 | | | CP | vpshufd xmm0, xmm0, 0xff
| 1 | | | | | | | 1.0 | | | cmp rax, 0x602020
| 0F | | | | | | | | | | jnz 0xffffffffffffffa3
Total Num Of Uops: 20
load(datavec + i + 1)。这是最好的代码，尤其是。在SnB系列中，2寄存器寻址模式不能微熔合，因此为了clang的利益，我将源更改为该循环条件。