What is the fastest way to transpose a matrix in C++?
Some details about transposing 4x4 square float (I will discuss 32-bit integer later) matrices with x86 hardware. It's helpful to start here in order to transpose larger square matrices such as 8x8 or 16x16.
_MM_TRANSPOSE4_PS(r0, r1, r2, r3) is implemented differently by different compilers. GCC and ICC (I have not checked Clang) use unpcklps, unpckhps, unpcklpd, unpckhpd whereas MSVC uses only shufps. We can actually combine these two approaches together like this.
t0 = _mm_unpacklo_ps(r0, r1);
t1 = _mm_unpackhi_ps(r0, r1);
t2 = _mm_unpacklo_ps(r2, r3);
t3 = _mm_unpackhi_ps(r2, r3);
r0 = _mm_shuffle_ps(t0,t2, 0x44);
r1 = _mm_shuffle_ps(t0,t2, 0xEE);
r2 = _mm_shuffle_ps(t1,t3, 0x44);
r3 = _mm_shuffle_ps(t1,t3, 0xEE);
One interesting observation is that two shuffles can be converted to one shuffle and two blends (SSE4.1) like this.
t0 = _mm_unpacklo_ps(r0, r1);
t1 = _mm_unpackhi_ps(r0, r1);
t2 = _mm_unpacklo_ps(r2, r3);
t3 = _mm_unpackhi_ps(r2, r3);
v = _mm_shuffle_ps(t0,t2, 0x4E);
r0 = _mm_blend_ps(t0,v, 0xC);
r1 = _mm_blend_ps(t2,v, 0x3);
v = _mm_shuffle_ps(t1,t3, 0x4E);
r2 = _mm_blend_ps(t1,v, 0xC);
r3 = _mm_blend_ps(t3,v, 0x3);
This effectively converted 4 shuffles into 2 shuffles and 4 blends. This uses 2 more instructions than the implementation of GCC, ICC, and MSVC. The advantage is that it reduces port pressure which may have a benefit in some circumstances. Currently all the shuffles and unpacks can go only to one particular port whereas the blends can go to either of two different ports.
I tried using 8 shuffles like MSVC and converting that into 4 shuffles + 8 blends but it did not work. I still had to use 4 unpacks.
I used this same technique for a 8x8 float transpose (see towards the end of that answer). https://stackoverflow.com/a/25627536/2542702. In that answer I still had to use 8 unpacks but I manged to convert the 8 shuffles into 4 shuffles and 8 blends.
For 32-bit integers there is nothing like shufps (except for 128-bit shuffles with AVX512) so it can only be implemented with unpacks which I don't think can be convert to blends (efficiently). With AVX512 vshufi32x4 acts effectively like shufps except for 128-bit lanes of 4 integers instead of 32-bit floats so this same technique might be possibly with vshufi32x4 in some cases. With Knights Landing shuffles are four times slower (throughput) than blends.
This is going to depend on your application but in general the fastest way to transpose a matrix would be to invert your coordinates when you do a look up, then you do not have to actually move any data.
This is a good question. There are many reason you would want to actually transpose the matrix in memory rather than just swap coordinates, e.g. in matrix multiplication and Gaussian smearing.
First let me list one of the functions I use for the transpose (EDIT: please see the end of my answer where I found a much faster solution)
void transpose(float *src, float *dst, const int N, const int M) {
#pragma omp parallel for
for(int n = 0; n<N*M; n++) {
int i = n/N;
int j = n%N;
dst[n] = src[M*j + i];
}
}
Now let's see why the transpose is useful. Consider matrix multiplication C = A*B. We could do it this way.
for(int i=0; i<N; i++) {
for(int j=0; j<K; j++) {
float tmp = 0;
for(int l=0; l<M; l++) {
tmp += A[M*i+l]*B[K*l+j];
}
C[K*i + j] = tmp;
}
}
That way, however, is going to have a lot of cache misses. A much faster solution is to take the transpose of B first
transpose(B);
for(int i=0; i<N; i++) {
for(int j=0; j<K; j++) {
float tmp = 0;
for(int l=0; l<M; l++) {
tmp += A[M*i+l]*B[K*j+l];
}
C[K*i + j] = tmp;
}
}
transpose(B);
Matrix multiplication is O(n^3) and the transpose is O(n^2), so taking the transpose should have a negligible effect on the computation time (for large n). In matrix multiplication loop tiling is even more effective than taking the transpose but that's much more complicated.
I wish I knew a faster way to do the transpose (Edit: I found a faster solution, see the end of my answer). When Haswell/AVX2 comes out in a few weeks it will have a gather function. I don't know if that will be helpful in this case but I could image gathering a column and writing out a row. Maybe it will make the transpose unnecessary.
For Gaussian smearing what you do is smear horizontally and then smear vertically. But smearing vertically has the cache problem so what you do is
Smear image horizontally
transpose output
Smear output horizontally
transpose output
Here is a paper by Intel explaining that http://software.intel.com/en-us/articles/iir-gaussian-blur-filter-implementation-using-intel-advanced-vector-extensions
Lastly, what I actually do in matrix multiplication (and in Gaussian smearing) is not take exactly the transpose but take the transpose in widths of a certain vector size (e.g. 4 or 8 for SSE/AVX). Here is the function I use
void reorder_matrix(const float* A, float* B, const int N, const int M, const int vec_size) {
#pragma omp parallel for
for(int n=0; n<M*N; n++) {
int k = vec_size*(n/N/vec_size);
int i = (n/vec_size)%N;
int j = n%vec_size;
B[n] = A[M*i + k + j];
}
}
EDIT:
I tried several function to find the fastest transpose for large matrices. In the end the fastest result is to use loop blocking with block_size=16 (Edit: I found a faster solution using SSE and loop blocking - see below). This code works for any NxM matrix (i.e. the matrix does not have to be square).
inline void transpose_scalar_block(float *A, float *B, const int lda, const int ldb, const int block_size) {
#pragma omp parallel for
for(int i=0; i<block_size; i++) {
for(int j=0; j<block_size; j++) {
B[j*ldb + i] = A[i*lda +j];
}
}
}
inline void transpose_block(float *A, float *B, const int n, const int m, const int lda, const int ldb, const int block_size) {
#pragma omp parallel for
for(int i=0; i<n; i+=block_size) {
for(int j=0; j<m; j+=block_size) {
transpose_scalar_block(&A[i*lda +j], &B[j*ldb + i], lda, ldb, block_size);
}
}
}
The values lda and ldb are the width of the matrix. These need to be multiples of the block size. To find the values and allocate the memory for e.g. a 3000x1001 matrix I do something like this
#define ROUND_UP(x, s) (((x)+((s)-1)) & -(s))
const int n = 3000;
const int m = 1001;
int lda = ROUND_UP(m, 16);
int ldb = ROUND_UP(n, 16);
float *A = (float*)_mm_malloc(sizeof(float)*lda*ldb, 64);
float *B = (float*)_mm_malloc(sizeof(float)*lda*ldb, 64);
For 3000x1001 this returns ldb = 3008 and lda = 1008
Edit:
I found an even faster solution using SSE intrinsics:
inline void transpose4x4_SSE(float *A, float *B, const int lda, const int ldb) {
__m128 row1 = _mm_load_ps(&A[0*lda]);
__m128 row2 = _mm_load_ps(&A[1*lda]);
__m128 row3 = _mm_load_ps(&A[2*lda]);
__m128 row4 = _mm_load_ps(&A[3*lda]);
_MM_TRANSPOSE4_PS(row1, row2, row3, row4);
_mm_store_ps(&B[0*ldb], row1);
_mm_store_ps(&B[1*ldb], row2);
_mm_store_ps(&B[2*ldb], row3);
_mm_store_ps(&B[3*ldb], row4);
}
inline void transpose_block_SSE4x4(float *A, float *B, const int n, const int m, const int lda, const int ldb ,const int block_size) {
#pragma omp parallel for
for(int i=0; i<n; i+=block_size) {
for(int j=0; j<m; j+=block_size) {
int max_i2 = i+block_size < n ? i + block_size : n;
int max_j2 = j+block_size < m ? j + block_size : m;
for(int i2=i; i2<max_i2; i2+=4) {
for(int j2=j; j2<max_j2; j2+=4) {
transpose4x4_SSE(&A[i2*lda +j2], &B[j2*ldb + i2], lda, ldb);
}
}
}
}
}
If the size of the arrays are known prior then we could use the union to our help. Like this-
#include <bits/stdc++.h>
using namespace std;
union ua{
int arr[2][3];
int brr[3][2];
};
int main() {
union ua uav;
int karr[2][3] = {{1,2,3},{4,5,6}};
memcpy(uav.arr,karr,sizeof(karr));
for (int i=0;i<3;i++)
{
for (int j=0;j<2;j++)
cout<<uav.brr[i][j]<<" ";
cout<<'\n';
}
return 0;
}