Reorder simple 2D matrix in-place

Question

I have a simple 2D (row, column) matrix which I currently reorder according to the algorithm below, using another array as final container to swap items.

The problem is that I need to save memory (the code is running on a very low end device), and thus I need to figure a way to reorder the array in-place.

The algorithm is as follows:

for (int iRHS = 0; iRHS < NUM_VARS; iRHS++)
    for (int iRow = 0; iRow < _numEquations; iRow++) {
        coef[iRHS][iRow] = _matrixCoef(iRow, iRHS); 
    }

Note: coef is a pointer to double accessed via subscript, _matrixCoef is a matrix helper class and uses a vector of double accessed by operator(row,col). Here I want to eliminate coef, so that all values are reordered in _matrixCoef in-place instead.

Edit: NUM_VARS is a define set to 2.

Is this possible in-place after all?

Edit 2:

Here is the matrix class which is accessed above via operator overload (row, col):

struct Matrix
{
    /// Creates a matrix with zero rows and columns.
    Matrix() = default;
    /// Creates a matrix with \a rows rows and \a col columns
    /// Its elements are initialized to 0.
    Matrix(int rows, int cols) : n_rows(rows), n_cols(cols), v(rows * cols, 0.) {}
    /// Returns the number or rows of the matrix
    inline int getNumRows() const { return n_rows; }
    /// Returns the number or columns of the matrix.
    inline int getNumCols() const { return n_cols; }
    /// Returns the reference to the element at the position \a row, \a col.
    inline double & operator()(int row, int col) { return v[row + col * n_rows]; }
    /// Returns the element at the position \a row, \a col by value.
    inline double operator()(int row, int col) const { return  v[row + col * n_rows]; }
    /// Returns the values of the matrix in column major order.
    double const * data() const { return v.data(); }
    /// Returns the values of the matrix in column major order.
    double * data() { return v.data(); }
    /// Initialize the matrix with given size. All values are set to zero.
    void initialize(int iRows, int iCols)
    {
        n_rows = iRows;
        n_cols = iCols;
        v.clear();
        v.resize(iRows * iCols);
    }
    
    void resize(int iRows, int iCols)
    {
        n_rows = iRows;
        n_cols = iCols;
        v.resize(iRows * iCols);
    }
private:
    int n_rows = 0;
    int n_cols = 0;
    std::vector<double> v;
};

Answer 1

After you posted code, I will suggest another solution, that's rather simple and quick to implement.

In your current Matrix class:

struct Matrix
{
    // ...

    // add this:
       void transpose()
       {
           is_transposed = !is_transposed;
       }
    // ...

    // modify these:

    /// Returns the number or rows of the matrix
    inline int getNumRows() const { return (is_transposed) ? n_cols : n_rows; }
    /// Returns the number or columns of the matrix.
    inline int getNumCols() const { return (is_transposed) ? n_rows : n_cols; }
    /// Returns the reference to the element at the position \a row, \a col.
    inline double & operator()(int row, int col) 
    {
        if (is_transposed) 
            return v[col + row * n_rows]; 
        return v[row + col * n_rows]; 
    }
    /// Returns the element at the position \a row, \a col by value.
    inline double operator()(int row, int col) const 
    { 
        if (is_transposed) 
            return  v[col + row * n_rows]; 
        return  v[row + col * n_rows]; 
    }

private:
    // ...

    // add this:
    bool is_transposed = false;
};

You may want to modify other member functions, depending on your application.

Answer 2

Well, assuming your matrix is a square matrix, that is NUM_VARS == _numEquations, it is possible. Otherwise the size of the resulting matrix will not allow in-place transposition. In that case, a solution would be to modify the downstream computations to swap the row/column indices while doing the operations.

But if it's square, you can try this:

for (size_t r = 0; r < NUM_VARS; ++r)
    for (size_t c = 0; c < NUM_VARS; ++c)
    {
         if (&_matrixCoef[r][c] < &_matrixCoef[c][r])
             std::swap(_matrixCoef[r][c], _matrixCoef[c][r]);
    }

Answer 3

If you want to explicitly re-order the data array in-place:

Consider this post that details in-place re-ordering of an array given a permutation. I have modified it only slightly. Algorithm to apply permutation in constant memory space

The expression for the permutation can be deduced as follows:

For index k = i + j * rows, j = k / rows and i = k - j * rows (integer division). The index of the transposed matrix is k_transpose = j + i * cols. Now for the code:

int perm(int k, int rows, int cols)
{
    int j = k / rows;
    int i = k - j * rows;
    return j + i * cols;
}
template<typename Scalar>
void transpose_colmajor(Scalar* A, int rows, int cols)
{
//tiny optimization: for (int i = 1; i < rows * cols - 1; i++)
    for (int i = 0; i < rows * cols; i++) {
        int ind = perm(i, rows, cols);
        while (ind > i)
            ind = perm(ind, rows, cols);
        std::swap(A[i], A[ind]);
    }
}

Example with a symbolic matrix:

    int rows = 4;
    int cols = 2;
    char entry = 'a';
    std::vector<char> symbolic_matrix;
    for (int col = 0; col < cols; col++)
    {
        for (int row = 0; row < rows; row++)
        {
            symbolic_matrix.push_back(entry);
            entry++;
        }
    }
    for (char coefficient : symbolic_matrix)
        std::cout << coefficient << ",";
    std::cout << "\n\n";

    transpose_colmajor(symbolic_matrix.data(), rows, cols);

    for (char coefficient : symbolic_matrix)
        std::cout << coefficient << ",";
    std::cout << "\n\n";

Output:

a,b,c,d,e,f,g,h,

a,e,b,f,c,g,d,h,

Of course, you will also have to update your struct with the correct rows and columns. The asymptotic computational complexity is not guaranteed to be linear, but there are no auxiliary data structures that consume your precious memory!

EDIT: I have to admit, I learned a lot from this question. I was wondering why transposing a square matrix is so intuitive while the rectangular case is not. It has to do with the cycles of the permutation. This algorithm https://www.geeksforgeeks.org/minimum-number-swaps-required-sort-array/ calculates the minimum swaps required for a given permutation. If you plug in the permutation expression above, you can see that the minimum number of swaps increases dramatically when rows != cols. For example, a 4 x 4 matrix requires 6 swaps, whereas a 4 x 3 matrix requires 8, which to me is very counter-intuitive! The algorithm I posted requires rows * cols - 2 swaps, which is not optimal. However, based on random experimentation of rectangular matrices, it appears that the gap between the minimum swaps is pretty small and is less important for larger matrices. It is definitely worth specializing the transposition routine for square matrices, as the minimum swaps is always n * (n - 1)/2.