qlear - 1 month ago 19

C++ Question

I am trying to implement Strassen's algorithm for matrix multiplication in C++, and I want to find a way to split two matrices into four parts each in constant time. Here is the current way I am doing so:

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

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

A11[i][j] = a[i][j];

A12[i][j] = a[i][j+n];

A21[i][j] = a[i+n][j];

A22[i][j] = a[i+n][j+n];

B11[i][j] = b[i][j];

B12[i][j] = b[i][j+n];

B21[i][j] = b[i+n][j];

B22[i][j] = b[i+n][j+n];

}

}

This approach is obviously O(n^2), and it adds n^2*log(n) to the runtime, as it is called for each recursive call.

It seems that the way to do this in constant time is to create pointers to the four sub-matrices, rather than copy over the values, but I am having a difficult time figuring out how to create those pointers. Any help would be appreciated.

Answer

Don't think of matrices, think of matrix views.

A matrix view has pointer to a buffer of `T`

, a width, a height, an offset, and a stride between columns (or rows).

We can start with an array view type.

```
template<class T>
struct array_view {
T* b = 0; T* e = 0;
T* begin() const{ return b; }
T* end() const{ return e; }
array_view( T* s, T* f ):b(s), e(f) {}
array_view( T* s, std::size_t l ):array_view(s, s+l) {}
std::size_t size() const { return end()-begin(); }
T& operator[]( std::size_t n ) const { return *(begin()+n); }
array_view slice( std::size_t start, std::size_t length ) const {
start = (std::min)(start, size());
length = (std::min)(size()-start, length);
return {b+start, length};
}
};
```

Now our matrix view:

```
temlpate<class T>
struct matrix_view {
std::size_t height, width;
std::size_t offset, stride;
array_view<T> buffer;
// TODO: Ctors
// one from a matrix that has offset and stirde set to 0.
// another that lets you create a sub-matrix
array_view<T> operator[]( std::size_t n ) const {
return buffer.slice( offset+stride*n, width ); // or width, depending on if row or column major
}
};
```

Now your code does work on `matrix_view`

s, not matrices.