Я не знаю, будет ли кому-нибудь интересно, но мне удалось найти решение.Код решения является частью fl-core1.0, пакета нечеткой логики для Octave, и выпущен под лицензией LGPL.(Код опирается на некоторые октавные функции)
// Calculate the S-Norm/T-Norm composition of sparse matrices (single thread)
void sparse_compose(octave_value_list args)
{
// Create constant versions of the input matrices to prevent them to be filled by zeros on reading.
// a is the const reference to the transpose of a because octave sparse matrices are column compressed
// (to cycle on the rows, we cycle on the columns of the transpose).
SparseMatrix atmp = args(0).sparse_matrix_value();
const SparseMatrix a = atmp.transpose();
const SparseMatrix b = args(1).sparse_matrix_value();
// Declare variables for the T-Norm and S-Norm values
float snorm_val;
float tnorm_val;
// Initialize the result sparse matrix
sparseC = SparseMatrix((int)colsB, (int)rowsA, (int)(colsB*rowsA));
// Initialize the number of nonzero elements in the sparse matrix c
int nel = 0;
sparseC.xcidx(0) = 0;
// Calculate the composition for each element
for (int i = 0; i < rowsC; i++)
{
for(int j = 0; j < colsC; j++)
{
// Get the index of the first element of the i-th column of a transpose (i-th row of a)
// and the index of the first element of the j-th column of b
int ka = a.cidx(i);
int kb = b.cidx(j);
snorm_val = 0;
// Check if the values of the matrix are really not 0 (it happens if the column of a or b hasn't any value)
// because otherwise the cidx(i) or cidx(j) returns the first nonzero element of the previous column
if(a(a.ridx(ka),i)!=0 && b(b.ridx(kb),j)!=0)
{
// Cicle on the i-th column of a transpose (i-th row of a) and j-th column of b
// From a.cidx(i) to a.cidx(i+1)-1 there are all the nonzero elements of the column i of a transpose (i-th row of a)
// From b.cidx(j) to b.cidx(j+1)-1 there are all the nonzero elements of the column j of b
while ((ka <= (a.cidx(i+1)-1)) && (kb <= (b.cidx(j+1)-1)))
{
// If a.ridx(ka) == b.ridx(kb) is true, then there's a nonzero value on the same row
// so there's a k for that a'(k, i) (equals to a(i, k)) and b(k, j) are both nonzero
if (a.ridx(ka) == b.ridx(kb))
{
tnorm_val = calc_tnorm(a.data(ka), b.data(kb));
snorm_val = calc_snorm(snorm_val, tnorm_val);
ka++;
kb++;
}
// If a.ridx(ka) == b.ridx(kb) ka should become the index of the next nonzero element on the i column of a
// transpose (i row of a)
else if (a.ridx(ka) < b.ridx(kb))
ka++;
// If a.ridx(ka) > b.ridx(kb) kb should become the index of the next nonzero element on the j column of b
else
kb++;
}
}
if (snorm_val != 0)
{
// Equivalent to sparseC(i, j) = snorm_val;
sparseC.xridx(nel) = j;
sparseC.xdata(nel++) = snorm_val;
}
}
sparseC.xcidx(i+1) = nel;
}
// Compress the result sparse matrix because it is initialized with a number of nonzero element probably greater than the real one
sparseC.maybe_compress();
// Transpose the result
sparseC = sparseC.transpose();
}