A Cache Efficient Matrix Transpose Program?

Question

So the obvious way to transpose a matrix is to use :

  for( int i = 0; i < n; i++ )

    for( int j = 0; j < n; j++ )

      destination[j+i*n] = sourc         


        

Answer 1

Steve Jessop mentioned a cache oblivious matrix transpose algorithm. For the record, I want to share an possible implementation of a cache oblivious matrix transpose.

public class Matrix {
    protected double data[];
    protected int rows, columns;

    public Matrix(int rows, int columns) {
        this.rows = rows;
        this.columns = columns;
        this.data = new double[rows * columns];
    }

    public Matrix transpose() {
        Matrix C = new Matrix(columns, rows);
        cachetranspose(0, rows, 0, columns, C);
        return C;
    }

    public void cachetranspose(int rb, int re, int cb, int ce, Matrix T) {
        int r = re - rb, c = ce - cb;
        if (r <= 16 && c <= 16) {
            for (int i = rb; i < re; i++) {
                for (int j = cb; j < ce; j++) {
                    T.data[j * rows + i] = data[i * columns + j];
                }
            }
        } else if (r >= c) {
            cachetranspose(rb, rb + (r / 2), cb, ce, T);
            cachetranspose(rb + (r / 2), re, cb, ce, T);
        } else {
            cachetranspose(rb, re, cb, cb + (c / 2), T);
            cachetranspose(rb, re, cb + (c / 2), ce, T);
        }
    }
}

More details on cache oblivious algorithms can be found here.

Answer 2

Matrix multiplication comes to mind, but the cache issue there is much more pronounced, because each element is read N times.

With matrix transpose, you are reading in a single linear pass and there's no way to optimize that. But you can simultaneously process several rows so that you write several columns and so fill complete cache lines. You will only need three loops.

Or do it the other way around and read in columns while writing linearly.

Answer 3

Instead of transposing the matrix in memory, why not collapse the transposition operation into the next operation you're going to do on the matrix?

Answer 4

I had the exact same problem yesterday. I ended up with this solution:

void transpose(double *dst, const double *src, size_t n, size_t p) noexcept {
    THROWS();
    size_t block = 32;
    for (size_t i = 0; i < n; i += block) {
        for(size_t j = 0; j < p; ++j) {
            for(size_t b = 0; b < block && i + b < n; ++b) {
                dst[j*n + i + b] = src[(i + b)*p + j];
            }
        }
    }
}

This is 4 time faster than the obvious solution on my machine.

This solution takes care of a rectangular matrix with dimensions which are not a multiple of the block size.

if dst and src are the same square matrix an in place function should really be used instead:

void transpose(double*m,size_t n)noexcept{
    size_t block=0,size=8;
    for(block=0;block+size-1<n;block+=size){
        for(size_t i=block;i<block+size;++i){
            for(size_t j=i+1;j<block+size;++j){
                std::swap(m[i*n+j],m[j*n+i]);}}
        for(size_t i=block+size;i<n;++i){
            for(size_t j=block;j<block+size;++j){
                std::swap(m[i*n+j],m[j*n+i]);}}}
    for(size_t i=block;i<n;++i){
        for(size_t j=i+1;j<n;++j){
            std::swap(m[i*n+j],m[j*n+i]);}}}

I used C++11 but this could be easily translated in other languages.

Answer 5

You're probably going to want four loops - two to iterate over the blocks, and then another two to perform the transpose-copy of a single block. Assuming for simplicity a block size that divides the size of the matrix, something like this I think, although I'd want to draw some pictures on the backs of envelopes to be sure:

for (int i = 0; i < n; i += blocksize) {
    for (int j = 0; j < n; j += blocksize) {
        // transpose the block beginning at [i,j]
        for (int k = i; k < i + blocksize; ++k) {
            for (int l = j; l < j + blocksize; ++l) {
                dst[k + l*n] = src[l + k*n];
            }
        }
    }
}

An important further insight is that there's actually a cache-oblivious algorithm for this (see http://en.wikipedia.org/wiki/Cache-oblivious_algorithm, which uses this exact problem as an example). The informal definition of "cache-oblivious" is that you don't need to experiment tweaking any parameters (in this case the blocksize) in order to hit good/optimal cache performance. The solution in this case is to transpose by recursively dividing the matrix in half, and transposing the halves into their correct position in the destination.

Whatever the cache size actually is, this recursion takes advantage of it. I expect there's a bit of extra management overhead compared with your strategy, which is to use performance experiments to, in effect, jump straight to the point in the recursion at which the cache really kicks in, and go no further. On the other hand, your performance experiments might give you an answer that works on your machine but not on your customers' machines.

Answer 6

With a large matrix, possibly a large sparse matrix, it might be an idea to decompose it into smaller cache friendly chunks (Say, 4x4 sub matrices). You can also flag sub matrices as identity which will help you in creating optimized code paths.