SIMD optimization puzzle

I Want to optimize the following function using SIMD (SSE2 & such):

int64_t fun(int64_t N, int size, int* p)
{
    int64_t sum = 0;
    for(int i=1; i<size; i++)
       sum += (N/i)*p[i];
    return sum;
}

This seems like an eminently vectorizable task, except that the needed instructions just aren't there ... 开发者_开发百科

We can assume that N is very large (10^12 to 10^18) and size~sqrt(N). We can also assume that p can only take values of -1, 0, and 1; so we don't need a real multiplication, the (N/i)*p[i] can be done with four instructions (pcmpgt, pxor, psub, pand), if we could just somehow compute N/i.

This is as close as I could get to vectorizing that code. I don't really expect it to be faster. I was just trying my hand at writting SIMD code.

#include <stdint.h>

int64_t fun(int64_t N, int size, const int* p)
{
    int64_t sum = 0;
    int i;
    for(i=1; i<size; i++) {
        sum += (N/i)*p[i];
    }
    return sum;
}

typedef int64_t v2sl __attribute__ ((vector_size (2*sizeof(int64_t))));

int64_t fun_simd(int64_t N, int size, const int* p)
{
    int64_t sum = 0;
    int i;
    v2sl v_2 = { 2, 2 };
    v2sl v_N = { N, N };
    v2sl v_i = { 1, 2 };
    union { v2sl v; int64_t a[2]; } v_sum;

    v_sum.a[0] = 0;
    v_sum.a[1] = 0;
    for(i=1; i<size-1; i+=2) {
        v2sl v_p = { p[i], p[i+1] };
        v_sum.v += (v_N / v_i) * v_p;
        v_i += v_2;
    }
    sum = v_sum.a[0] + v_sum.a[1];
    for(; i<size; i++) {
        sum += (N/i)*p[i];
    }
    return sum;
}

typedef double v2df __attribute__ ((vector_size (2*sizeof(double))));

int64_t fun_simd_double(int64_t N, int size, const int* p)
{
    int64_t sum = 0;
    int i;
    v2df v_2 = { 2, 2 };
    v2df v_N = { N, N };
    v2df v_i = { 1, 2 };
    union { v2df v; double a[2]; } v_sum;

    v_sum.a[0] = 0;
    v_sum.a[1] = 0;
    for(i=1; i<size-1; i+=2) {
        v2df v_p = { p[i], p[i+1] };
        v_sum.v += (v_N / v_i) * v_p;
        v_i += v_2;
    }
    sum = v_sum.a[0] + v_sum.a[1];
    for(; i<size; i++) {
        sum += (N/i)*p[i];
    }
    return sum;
}

#include <stdio.h>

static const int test_array[] = {
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0
};
#define test_array_len (sizeof(test_array)/sizeof(int))

#define big_N (1024 * 1024 * 1024)

int main(int argc, char *argv[]) {
    int64_t res1;
    int64_t res2;
    int64_t res3;
    v2sl a = { 123, 456 };
    v2sl b = { 100, 200 };
    union { v2sl v; int64_t a[2]; } tmp;

    a = a + b;
    tmp.v = a;
    printf("a = { %ld, %ld }\n", tmp.a[0], tmp.a[1]);

    printf("test_array size = %zd\n", test_array_len);

    res1 = fun(big_N, test_array_len, test_array);
    printf("fun() = %ld\n", res1);

    res2 = fun_simd(big_N, test_array_len, test_array);
    printf("fun_simd() = %ld\n", res2);

    res3 = fun_simd_double(big_N, test_array_len, test_array);
    printf("fun_simd_double() = %ld\n", res3);

    return 0;
}

The derivative of 1/x is -1/x^2, which means as x gets bigger, N/x==N/(x + 1).

For a known value of N/x (let's call that value r), we can determine the next value of x (let's call that value x' such that N/x'<r:

x'= N/(r - 1)

And since we are dealing with integers:

x'= ceiling(N/(r - 1))

So, the loop becomes something like this:

int64_t sum = 0;   
int i=1; 
int r= N;
while (i<size)
{
    int s= (N + r - 1 - 1)/(r - 1);

    while (i<s && i<size)
    {
        sum += (r)*p[i];
        ++i;
    }

    r= N/s;
}
return sum;   

For sufficiently large N, you will have many many runs of identical values for N/i. Granted, you will hit a divide by zero if you aren't careful.

I suggest you do this with floating point SIMD operations - either single or double precision depending on your accuracy requirements. Conversion from int to float or double is relatively fast using SSE.

The cost is concentrated in computing the divisions. There is no opcode in SSE2 for integral divisions, so you would have to implement a division algorithm yourself, bit by bit. I do not think it would be worth the effort: SSE2 allow you to perform two instances in parallel (you use 64-bit numbers, and SSE2 registers are 128-bit) but I find it likely that a handmade division algorithm would be at least twice as slow as the CPU idiv opcode.

(By the way, do you compile in 32-bit or 64-bit mode ? The latter will be more comfortable with 64-bit integers.)

Reducing the overall number of divisions looks like a more promising way. One may note that for positive integers x and y, then floor(x/(2y)) = floor(floor(x/y)/2). In C terminology, once you have computed N/i (truncated division) then you just have to shift it right by one bit to obtain N/(2*i). Used properly, this makes half of your divisions almost free (that "properly" also includes accessing the billions of p[i] values in a way which does not wreak havoc with the caches, so it does not seem very easy).