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Improve C function performance with cache locality?

开发者 https://www.devze.com 2023-04-12 22:10 出处:网络
I have to find a diagonal difference in a matrix represented as 2d array and the function prototype is

I have to find a diagonal difference in a matrix represented as 2d array and the function prototype is

int diagonal_diff(int x[512][512])

I have to use a 2d array, and the data is 512x512. This is tested on a SPARC machine: my current timing is 6ms but I need to be under 2ms.

Sample data:

[3][4][5][9]
[2][8][9][4]
[6][9][7][3]
[5][8][8][2]

The difference is:

|4-2| + |5-6| 开发者_如何转开发+ |9-5| + |9-9| + |4-8| + |3-8| = 2 + 1 + 4 + 0 + 4 + 5 = 16

In order to do that, I use the following algorithm:

int i,j,result=0;
for(i=0; i<4; i++)
    for(j=0; j<4; j++)
        result+=abs(array[i][j]-[j][i]);

return result;

But this algorithm keeps accessing the column, row, column, row, etc which make inefficient use of cache.

Is there a way to improve my function?


EDIT: Why is a block oriented approach faster? We are taking advantage of the CPU's data cache by ensuring that whether we iterate over a block by row or by column, we guarantee that the entire block fits into the cache.

For example, if you have a cache line of 32-bytes and an int is 4 bytes, you can fit a 8x8 int matrix into 8 cache lines. Assuming you have a big enough data cache, you can iterate over that matrix either by row or by column and be guaranteed that you do not thrash the cache. Another way to think about it is if your matrix fits in the cache, you can traverse it any way you want.

If you have a matrix that is much bigger, say 512x512, then you need to tune your matrix traversal such that you don't thrash the cache. For example, if you traverse the matrix in the opposite order of the layout of the matrix, you will almost always miss the cache on every element you visit.

A block oriented approach ensures that you only have a cache miss for data you will eventually visit before the CPU has to flush that cache line. In other words, a block oriented approach tuned to the cache line size will ensure you don't thrash the cache.

So, if you are trying to optimize for the cache line size of the machine you are running on, you can iterate over the matrix in block form and ensure you only visit each matrix element once:

int sum_diagonal_difference(int array[512][512], int block_size)
{
    int i,j, block_i, block_j,result=0;

     // sum diagonal blocks
    for (block_i= 0; block_i<512; block_i+= block_size)
        for (block_j= block_i + block_size; block_j<512; block_j+= block_size)
            for(i=0; i<block_size; i++)
                for(j=0; j<block_size; j++)
                    result+=abs(array[block_i + i][block_j + j]-array[block_j + j][block_i + i]);

    result+= result;

     // sum diagonal
    for (int block_offset= 0; block_offset<512; block_offset+= block_size)
    {
        for (i= 0; i<block_size; ++i)
        {
            for (j= i+1; j<block_size; ++j)
            {
                int value= abs(array[block_offset + i][block_offset + j]-array[block_offset + j][block_offset + i]);
                result+= value + value;
            }
        }
    }

    return result;
}

You should experiment with various values for block_size. On my machine, 8 lead to the biggest speed up (2.5x) compared to a block_size of 1 (and ~5x compared to the original iteration over the entire matrix). The block_size should ideally be cache_line_size_in_bytes/sizeof(int).


If you have a good vector/matrix library like intel MKL, also try the vectorized way.

very simple in matlab: result = sum(sum(abs(x-x')));

I reproduced Hans's method and MSN's method in matlab too, and the results are:

Elapsed time is 0.211480 seconds.  (Hans)

Elapsed time is 0.009172 seconds.  (MSN)

Elapsed time is 0.002193 seconds.  (Mine)


With one minor change you can have your loops only operate on the desired indices. I just changed the j loop initialization.

int i, j, result = 0;
for (i = 0; i < 4; ++i) {
    for (j = i + 1; j < 4; ++j) {
        result += abs(array[i][j] - array[j][i]);
    }
}
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