5

我需要以双精度执行以下操作：

在此处输入图像描述

数字表示值如何存储在内存中。我想用 AVX 来实现这个。如果我[QK]先将列填充到 8 个元素，然后执行矩阵向量乘法，然后再执行[x]点[QK]积，那会更好吗？

编辑：好的，所以我决定使用填充向量实现FLOAT 32 位版本，如下所示：

// Perform matrix vector multiplication of QK*x            
// Load first four columns QK into 4 ymm registers    
ymm0 = _mm256_load_ps((float *)(QK));     
ymm1 = _mm256_load_ps((float *)(QK+8));       
ymm2 = _mm256_load_ps((float *)(QK+16));    
ymm3 = _mm256_load_ps((float *)(QK+24));  

// Load first four values of x
ymm4 = _mm256_broadcast_ss((float *)(x));
ymm5 = _mm256_broadcast_ss((float *)(x+1));
ymm6 = _mm256_broadcast_ss((float *)(x+2));
ymm7 = _mm256_broadcast_ss((float *)(x+3));

// Multiply in place - frees up ymm4,ymm5,ymm6,ymm7
ymm0 = _mm256_mul_ps(ymm0, ymm4);
ymm1 = _mm256_mul_ps(ymm1, ymm5);
ymm2 = _mm256_mul_ps(ymm2, ymm6);
ymm3 = _mm256_mul_ps(ymm3, ymm7);

// Add together, this frees up ymm1,ymm2,and ymm3
ymm0 = _mm256_add_ps(ymm0, ymm1);
ymm2 = _mm256_add_ps(ymm2, ymm3);
ymm0 = _mm256_add_ps(ymm0, ymm2);

// Load next last columns of QK
ymm1 = _mm256_load_ps((float *)(QK+32));     
ymm2 = _mm256_load_ps((float *)(QK+40)); 

// Load last two values of x
ymm6 = _mm256_broadcast_ss((float *)(x+4));
ymm7 = _mm256_broadcast_ss((float *)(x+5));

// Multiply in place
ymm1 = _mm256_mul_ps(ymm1, ymm6);
ymm2 = _mm256_mul_ps(ymm2, ymm7);

// Add together, this frees up every register except for ymm0
ymm0 = _mm256_add_ps(ymm0, ymm1);
ymm0 = _mm256_add_ps(ymm0, ymm2);  

// Answer stored in ymm0 and ymm1   
// Calculate dot product of y*(QK*x)
// Load x
ymm1 = _mm256_load_ps((float *)(y));   

// Do dotproduct by using horizontal multiply followed by horizontal add
// Multiply in place
ymm0 = _mm256_mul_ps(ymm0, ymm1);  

// Do horizontal sum
__m128 xmm1 = _mm256_extractf128_ps(ymm0, 1);
__m128 xmm2 = _mm256_extractf128_ps(ymm0, 0);
xmm2 = _mm_add_ps(xmm1, xmm2);
xmm1 = _mm_movehl_ps(xmm1, xmm2);
xmm2 = _mm_add_ps(xmm1, xmm2);
xmm1 = _mm_shuffle_ps(xmm2, xmm2, 1);
xmm2 = _mm_add_ss(xmm1, xmm2);
ans[0] = _mm_cvtss_f32(xmm2);

就目前而言，它的运行速度大约是以下的 3 倍：

ans[0] = (QK[0]*x[0]+QK[8]*x[1]+QK[16]*x[2]+QK[24]*x[3]+QK[32]*x[4]+QK[40]*x[5])*y[0]+
         (QK[1]*x[0]+QK[9]*x[1]+QK[17]*x[2]+QK[25]*x[3]+QK[33]*x[4]+QK[41]*x[5])*y[1]+
         (QK[2]*x[0]+QK[10]*x[1]+QK[18]*x[2]+QK[26]*x[3]+QK[34]*x[4]+QK[42]*x[5])*y[2]+
         (QK[3]*x[0]+QK[11]*x[1]+QK[19]*x[2]+QK[27]*x[3]+QK[35]*x[4]+QK[43]*x[5])*y[3]+
         (QK[4]*x[0]+QK[12]*x[1]+QK[20]*x[2]+QK[28]*x[3]+QK[36]*x[4]+QK[44]*x[5])*y[4]+
         (QK[5]*x[0]+QK[13]*x[1]+QK[21]*x[2]+QK[29]*x[3]+QK[37]*x[4]+QK[45]*x[5])*y[5];

对于 5 亿次迭代，标准 C 版本运行时间约为 9 秒，单精度 AVX 版本运行时间约为 3.5 秒。如果我在最后注释掉水平总和，那么它会在大约 0.5 秒内运行。水平总和真的扼杀了性能......

4

1 回答 1

1

我创建了代码来有效地做到这一点。将 AVX 用于单线程时，我得到了近 4 倍的速度提升（最好是在 AVX 上使用双打）。以下是在 6x6 矩阵上运行 2000、32000 和 4000000 个 x 和 y 六分量向量的结果。这些大致对应于我系统上的 L2、L3 和 >>L3 缓存大小（每个向量使用 48 个字节）。

编辑：我在最后添加了文本（和代码）以使用浮动来执行此操作。在单线程上使用 AVX 的加速比接近 8 倍。

i5-3317U (2 core ivy bridge)
compiled with: g++ m6.cpp -o m6 -O3 -mavx -fopenmp 
nvec 2000, repeat 10000, 1 thread : time scalar/SIMD 3.95
nvec 32000, repeat 1000, 1 thread : time scalar/SIMD 3.53
nvec 4000000, repeat 10, 1 thread : time scalar/SIMD 3.28
1 thread for both the SIMD and non-SIMD functions 

nvec 2000, repeat 10000, 2 threads: time scalar/SIMD 5.96
nvec 32000, repeat 1000, 2 threads: time scalar/SIMD 5.88
nvec 4000000, repeat 10, 2 threads: time scalar/SIMD 4.52
2 threads on the SIMD function vs. 1 thread on the non-SIMD function

compiled with g++ m6.cpp -o m6 -O3 -msse4.2 -fopenmp
nvec 2000, repeat 10000, 1 thread: time scalar/SIMD 2.15
nvec 32000, repeat 1000, 1 thread: time scalar/SIMD 2.06 
nvec 4000000, repeat 10, 1 thread: time scalar/SIMD 2.00

基本算法在 x 和 y 向量上执行 SIMD，而不是在 6x6 矩阵上。一个关键点是 x 和 y 向量必须是数组结构块的数组。这也称为数组结构数组 (AoSoA) 或数组混合结构。您可以在此处阅读有关它的更多信息“为便携式 SIMD 编程扩展类 C 语言” http://www.sci.utah.edu/~wald/Publications/2012///ppopp/download//vecimp.pdf

下面是代码。该函数aos2aosoa将六个分量向量的数组转换为 SoA 数组。如果您想获得 SIMD 的全部好处，您的应用程序应该以这种形式生成向量（不进行转换）。此代码使用 Agner Fog 的矢量类http://www.agner.org/optimize/#vectorclass。这只是一些头文件。此代码也适用于 SSE（但提升仅为预期的 2 倍左右）。

一个小警告，x 和 y 向量的数组和结果必须是 4 的倍数。但是，向量的数量不是。

像这样编译

g++ m6.cpp -o m6 -O3 -mavx -fopenmp  //system with AVX
g++ m6.cpp -o m6 -O3 -msse4.2 -fopenmp //system with SSE

在 Visual Studio 中，将 /arch:AVX 放入编译器命令行选项中

编码：

#include <stdio.h>
#include <omp.h>
#include "vectorclass.h"
#include <stdlib.h>

double prod_scalar(double *x, double *M, double *y) {
    double sum = 0.0f;
    for(int i=0; i<6; i++) {
        for(int j=0; j<6; j++) {
            sum += x[i]*M[i*6 + j]*y[j];
        }
    }
    return sum;
}

void prod_block4(double *x, double *M, double *y, double *result) {
    Vec4d sum4 = 0.0f;
    for(int i=0; i<6; i++) {
        Vec4d x4 = Vec4d().load(&x[4*i]);
        for(int j=0; j<6; j++) {
            Vec4d y4 = Vec4d().load(&y[4*j]);
            sum4 += x4*M[i*6 + j]*y4;
        }
    }
    sum4.store(result);
}

void prod_block4_unroll2(double *x, double *M, double *y, double *result) {
    Vec4d sum4_1 = 0.0f;
    Vec4d sum4_2 = 0.0f;
    Vec4d yrow[6];
    for(int i=0; i<6; i++) {
        yrow[i] = Vec4d().load(&y[4*i]);
    }
    for(int i=0; i<6; i++) {
        Vec4d x4 = Vec4d().load(&x[4*i]);
        for(int j=0; j<6; j+=2) {
            sum4_1 += x4*M[i*6 + j]*yrow[j];
            sum4_2 += x4*M[i*6 + j+1]*yrow[j+1];
        }
    }
    sum4_1 += sum4_2;
    sum4_1.store(result);
}
void loop_scalar(double *x, double *M, double *y, double *result, const int nvec) {
//  #pragma omp parallel for
    for(int i=0; i<nvec; i++) {
        result[i] = prod_scalar(&x[6*i], M, &y[6*i]);
    }   
}

void loop_block4(double *x, double *M, double *y, double *result, const int nvec) {
//  #pragma omp parallel for
    for(int i=0; i<(nvec/4); i++) {
//      prod_block4(&x[i*24], M, &y[i*24], &result[4*i]);
        prod_block4_unroll2(&x[i*24], M, &y[i*24], &result[4*i]);
    }
}


void aos2soa(double *in, double *out) {
    int cnt = 0;
    for(int i=0; i<6; i++) {
        for(int j=0; j<4; j++) {
            out[cnt++] = in[6*j + i];
        }
    }
}

void aos2aosoa(double *in, double *out, const int nvec) {
    for(int i=0; i<(nvec/4); i++) {
        aos2soa(&in[i*24], &out[i*24]);
    }
}

double compare_results(double *r1, double * r2, const int nvec) {
    double out = 0.0f;
    for(int i=0; i<nvec; i++) {
        out += r1[i] -r2[i];
        //if(out!=0) printf("%f %f\n",r1[i], r2[i]);
    }
    return out;
}

void loop(const int nvec, const int repeat) {
    double *M = (double*)_mm_malloc(6*6*sizeof(double),32);
    double *x_aos = (double*)_mm_malloc(nvec*6*sizeof(double),32);
    double *y_aos = (double*)_mm_malloc(nvec*6*sizeof(double),32);
    double *x_aosoa = (double*)_mm_malloc(nvec*6*sizeof(double),32);
    double *y_aosoa = (double*)_mm_malloc(nvec*6*sizeof(double),32);
    double *results_scalar = (double*)_mm_malloc(nvec*sizeof(double),32);
    double *results_vector = (double*)_mm_malloc(nvec*sizeof(double),32);

    for(int i=0; i<6; i++) {
        for(int j=0; j<6; j++) {
            M[j*6 + i] = i*6 + j; 
        }
    }
    for(int i=0; i<(6*nvec); i++) {
        double r1 = (double)rand()/RAND_MAX;
        double r2 = (double)rand()/RAND_MAX;
        //x_aos[i] = i;
        x_aos[i] = r1;
        //y_aos[i] = i;
        y_aos[i] = r2;

    } 

    aos2aosoa(x_aos, x_aosoa, nvec);
    aos2aosoa(y_aos, y_aosoa, nvec);

    double dtime;
    dtime = omp_get_wtime();
    for(int i=0; i<repeat; i++) {
        loop_scalar(x_aos, M, y_aos, results_scalar, nvec);
    }
    dtime = omp_get_wtime() - dtime;
    printf("time scalar %f\n", dtime);

    double dtime_old = dtime;
    dtime = omp_get_wtime();
    for(int i=0; i<repeat; i++) {
        loop_block4(x_aosoa, M, y_aosoa, results_vector, nvec);
    }
    dtime = omp_get_wtime() - dtime;
    printf("time vector %f\n", dtime);
    printf("time scalar/vector %f\n", dtime_old/dtime);

    printf("difference %f\n", compare_results(results_scalar, results_vector, nvec));

    _mm_free(M);
    _mm_free(x_aos);
    _mm_free(y_aos);
    _mm_free(x_aosoa);
    _mm_free(y_aosoa);
    _mm_free(results_scalar);
    _mm_free(results_vector);

}

int main() {
    int nveca[3];
    nveca[0] = 2000; // 2000*2*6*8 = 192kb //L2
    nveca[1] = 32000; // 32000*2*6*8 = 3Mb //L3
    nveca[2] = 4*1000000; //366Mb
    int nrepeat[3];
    nrepeat[0] = 10000;
    nrepeat[1] = 1000;
    nrepeat[2] = 10;
    for(int i=0; i<3; i++) {
        printf("nvec %d, repeat %d\n", nveca[i], nrepeat[i]);
        loop(nveca[i], nrepeat[i]);
        printf("\n");
    }
}

这是浮动的结果。提升约8倍

nvec 2000, repeat 10000, 1 thread:  time scalar/SIMD 7.86
nvec 32000, repeat 1000, 1 thread:  time scalar/SIMD 7.63
nvec 5000000, repeat 100, 1 thread: time scalar/SIMD 6.63

这是 float 而不是 double 的代码

#include <stdio.h>
#include <omp.h>
#include "vectorclass.h"
#include <stdlib.h>

#define ROUND_DOWN(x, s) ((x) & ~((s)-1))

float prod_scalar(float *x, float *M, float *y) {
    float sum = 0.0f;
    for(int i=0; i<6; i++) {
        for(int j=0; j<6; j++) {
            sum += x[i]*M[i*6 + j]*y[j];
        }
    }
    return sum;
}

float prod_scalar_unroll2(float *x, float *M, float *y) {
    float sum_1 = 0.0f;
    float sum_2 = 0.0f;
    for(int i=0; i<6; i++) {
        for(int j=0; j<6; j+=2) {
            sum_1 += x[i]*M[i*6 + j]*y[j];
            sum_2 += x[i]*M[i*6 + j+1]*y[j+1];
        }
    }
    return sum_1 + sum_2;
}

void prod_block4(float *x, float *M, float *y, float *result) {
    Vec8f sum4 = 0.0f;
    for(int i=0; i<6; i++) {
        Vec8f x4 = Vec8f().load(&x[8*i]);
        for(int j=0; j<6; j++) {
            Vec8f y4 = Vec8f().load(&y[8*j]);
            sum4 += x4*M[i*6 + j]*y4;
        }
    }
    sum4.store(result);
}

void prod_block4_unroll2(float *x, float *M, float *y, float *result) {
    Vec8f sum4_1 = 0.0f;
    Vec8f sum4_2 = 0.0f;
    Vec8f yrow[6];
    for(int i=0; i<6; i++) {
        yrow[i] = Vec8f().load(&y[8*i]);
    }
    for(int i=0; i<6; i++) {
        Vec8f x4 = Vec8f().load(&x[8*i]);
        for(int j=0; j<6; j+=2) {
            sum4_1 += x4*M[i*6 + j]*yrow[j];
            sum4_2 += x4*M[i*6 + j+1]*yrow[j+1];
        }
    }
    sum4_1 += sum4_2;
    sum4_1.store(result);
}

void loop_scalar(float *x, float *M, float *y, float *result, const int nvec) {
//  #pragma omp parallel for
    for(int i=0; i<nvec; i++) {
        result[i] = prod_scalar(&x[6*i], M, &y[6*i]);
        //result[i] = prod_scalar_unroll2(&x[6*i], M, &y[6*i]);
    }   
}

void loop_SIMD(float *x, float *M, float *y, float *result, const int nvec) {
    //#pragma omp parallel for schedule(static,256)
    //#pragma omp parallel for schedule(static)
    const int N = nvec/8;
    //printf("chuck %d\n", N/2);
    //omp_set_num_threads(2);

    //#pragma omp parallel 
    {
        //int nthreads = omp_get_num_threads();
        //int ithread = omp_get_thread_num();

        //int start = (ithread*N)/nthreads;
        //int end = ((ithread+1)*N)/nthreads;
        //printf("ithread, start %d, end %d, chunk %d\n",start, end, end-start);
        //#pragma omp for 
        for(int i=0; i<(nvec/8); i++) {
        //for(int i=start; i<end; i++) {
    //      prod_block4(&x[i*24], M, &y[i*24], &result[4*i]);
        prod_block4_unroll2(&x[i*48], M, &y[i*48], &result[8*i]);
        }
    }
}

void aos2soa(float *in, float *out) {
    int cnt = 0;
    for(int i=0; i<6; i++) {
        for(int j=0; j<8; j++) {
            out[cnt++] = in[6*j + i];
        }
    }
}

void aos2aosoa(float *in, float *out, const int nvec) {
    for(int i=0; i<(nvec/8); i++) {
        aos2soa(&in[i*48], &out[i*48]);
    }
}

float compare_results(float *r1, float * r2, const int nvec) {
    float out = 0.0f;
    for(int i=0; i<nvec; i++) {
        out += r1[i] -r2[i];
        //if(out!=0) printf("%f %f\n",r1[i], r2[i]);
    }
    return out;
}

void loop(const int nvec, const int repeat) {
    float *M = (float*)_mm_malloc(6*6*sizeof(float),64);
    float *x_aos = (float*)_mm_malloc(nvec*6*sizeof(float),64);
    float *y_aos = (float*)_mm_malloc(nvec*6*sizeof(float),64);
    float *x_aosoa = (float*)_mm_malloc(nvec*6*sizeof(float),64);
    float *y_aosoa = (float*)_mm_malloc(nvec*6*sizeof(float),64);
    float *results_scalar = (float*)_mm_malloc(nvec*sizeof(float),64);
    float *results_SIMD = (float*)_mm_malloc(nvec*sizeof(float),64);

    for(int i=0; i<6; i++) {
        for(int j=0; j<6; j++) {
            M[j*6 + i] = i*6 + j; 
        }
    }
    for(int i=0; i<(6*nvec); i++) {
        float r1 = (float)rand()/RAND_MAX;
        float r2 = (float)rand()/RAND_MAX;
        //x_aos[i] = i;
        x_aos[i] = r1;
        //y_aos[i] = i;
        y_aos[i] = r2;

    } 

    aos2aosoa(x_aos, x_aosoa, nvec);
    aos2aosoa(y_aos, y_aosoa, nvec);

    float dtime;
    dtime = omp_get_wtime();
    for(int i=0; i<repeat; i++) {
        loop_scalar(x_aos, M, y_aos, results_scalar, nvec);
    }
    dtime = omp_get_wtime() - dtime;
    printf("time scalar %f\n", dtime);

    float dtime_old = dtime;
    dtime = omp_get_wtime();
    for(int i=0; i<repeat; i++) {
        loop_SIMD(x_aosoa, M, y_aosoa, results_SIMD, nvec);
    }
    dtime = omp_get_wtime() - dtime;
    printf("time SIMD %f\n", dtime);
    printf("time scalar/SIMD %f\n", dtime_old/dtime);

    printf("difference %f\n", compare_results(results_scalar, results_SIMD, nvec));

    _mm_free(M);
    _mm_free(x_aos);
    _mm_free(y_aos);
    _mm_free(x_aosoa);
    _mm_free(y_aosoa);
    _mm_free(results_scalar);
    _mm_free(results_SIMD);

}

int main() {
    int nveca[3];
    nveca[0] = 2000; // 2000*2*6*8 = 192kb //L2
    nveca[1] = 32000; // 32000*2*6*8 = 3Mb //L3
    nveca[2] = 5*1000000; //366Mb
    int nrepeat[3];
    nrepeat[0] = 10000;
    nrepeat[1] = 1000;
    nrepeat[2] = 100;
    for(int i=0; i<3; i++) {
        printf("nvec %d, repeat %d\n", nveca[i], nrepeat[i]);
        loop(nveca[i], nrepeat[i]);
        printf("\n");
    }
}
于 2013-05-20T08:55:36.223 回答