Ncol >= Nrow + 64

[ymym1234]

I see the iuput limits:Ncol >= Nrow + 64@SebWouters
So can this package only be used when the column size is 64 compared to the row size?

[SebWouters]

In the current state, the package can only be used when the column size is at least 64 larger than the row size.

For the linear system B * x = 0, with B of size Nrows x Ncols, it's guaranteed that when Nrows < Ncols that there are Ncols - Nrows linearly independent x which solve the system. blanczos uses a buffer of Ncols uint64_t to compute up to 64 nullspace vectors at once. It's typically used in algorithms for integer factorization. In such a context one can always generate additional columns. The requirement is not strict per se, but it's best practice to use the uint64_t to the max.

Feel free to clone blanczos and tweak src/blanczos.cpp so that the requirement is not strict anymore.

What do you want to use it for?

[SebWouters]

Feel free to clone blanczos and tweak src/blanczos.cpp so that the requirement is not strict anymore.

https://github.com/SebWouters/blanczos/blob/master/src/blanczos.cpp#L471

[ymym1234]

Thank you for your reply. I will try it immediately

[ymym1234]

I'm sorry to ask this question
According to your previous count, I randomly set the 64U, but there were other issues that arose:
Assuming that the columns are only 1U more than the rows, I directly arranged the matrix in the main as follows:
uint32_t matrix[] = {
0, 1, 0, 0,
1, 1, 0, 1,
1, 0, 1, 1
};
uint32_t Nrow = 3; // 矩阵的行数
uint32_t Ncol = 4; // 矩阵的列数
uint64_t N = Nrow * Ncol; // 矩阵的元素总数
The compilation was successful, but during runtime debugging, when stepping into sparsehpp, line 146 for (uint64_t elem=0U; elem<N;++elem) {++rowmap [B [elem<<1U];}: a segmentation fault occurred,
Is this error caused by my matrix being too small? Or is it due to other reasons?

[SebWouters]

B is in a different format, it contains the row and column indices of the non-zero elements, see https://github.com/SebWouters/blanczos?tab=readme-ov-file#blanczos-block-lanczos-for-gf2

[ymym1234]

Thank you for your previous answer. The program can now run successfully; @SebWouters

In fact, I am trying to use this program to solve linear equations on a binary domain, but I am only interested in the kernel space of matrices. Through researching relevant materials, I know that the block Lanczos method can be used for solving large sparse matrices, but the running results are not satisfactory. Therefore, are there any other restrictions on the use of this method in this program? Is it because my matrix is a bit sparse?

The running result is as follows:
rows:88 cols:92
Lanczos iteration 0
Lanczos iteration 1
Returning 64 of 107 non-zero nullspace vectors retrieved within 123 candidates.
Blanczos algorithm execution failed!

[SebWouters]

What goes wrong? Can you provide your entire test case?

[SebWouters]

Please note that if it fails in your own testing code, that the 64 returned null-space vectors are bit-packed. So one solution is formed by bit 0 of each uint64_t of the result array, another by bit 1 of each uint64_t etc.

[ymym1234]

I 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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@SebWouters This is a matrix with 48 rows and 54 columns，and its sparsity is very high

[SebWouters]

#include "blanczos.h"

#include <stdint.h>
#include <random>
#include <chrono>
#include <vector>
#include <array>
#include <iostream>

int main()
{
    constexpr bool verbose = true;

    constexpr uint Nrow = 48;
    constexpr uint Ncol = 54;
    const std::array<uint8_t, Nrow * Ncol> input = {
1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1
    };

    uint64_t number = 0;
    for (int i = 0; i < input.size(); ++i)
        if (input[i] == 1)
            ++number;
    uint32_t * b = new uint32_t[2 * number];

    uint64_t count = 0;
    for (uint32_t row = 0; row < Nrow; ++row)
        for (uint32_t col = 0; col < Ncol; ++col)
            if (input[row * Ncol + col] == 1)
            {
                b[count++] = row;
                b[count++] = col;
            }

    uint64_t * result = new uint64_t[Ncol];

    uint32_t Nsol = blanczos(b, number, Nrow, Ncol, result);

    if (verbose)
        std::cout << "Number of solutions = " << Nsol << std::endl;

    if (Nsol < 1U || Nsol > 64U)
        return -1;

    std::array<uint8_t, Ncol> vector = {};
    std::array<uint8_t, Nrow> matvec = {};
    for (uint32_t sol = 0; sol < Nsol; ++sol)
    {
        for (uint32_t col = 0; col < Ncol; ++col)
            vector[col] = static_cast<uint8_t>(result[col] >> sol) & static_cast<uint8_t>(1);
        
        uint32_t nonzero = 0;
        for (uint32_t col = 0; col < Ncol; ++col)
            if (vector[col] == 1)
                ++nonzero;

        if (verbose)
            std::cout << "    Solution " << sol << ":" << std::endl;
        if (verbose)
            std::cout << "        Vector has " << nonzero << " non-zero entries." << std::endl;
        
        if (nonzero < 1)
            return -1;

        for (uint32_t row = 0; row < Nrow; ++row)
        {
            matvec[row] = 0;
            for (uint32_t col = 0; col < Ncol; ++col)
                if (input[row * Ncol + col] == 1)
                    matvec[row] ^= vector[col];
        }

        nonzero = 0;
        for (uint32_t row = 0; row < Nrow; ++row)
            if (matvec[row] == 1)
                ++nonzero;

        if (verbose)
            std::cout << "        Matrix x Vector has " << nonzero << " non-zero entries." << std::endl;

        if (nonzero != 0)
            return -1;
    }

    delete[] b;
    delete[] result;

    return 0;
}

The above code makes everything very explicit, and works just fine (with the checking of Ncol >= Nrow + 64 in blanczos.cpp commented out, of course).

[ymym1234]

Thank you again for your patient answer. I will try it right away

[ymym1234]

I have completed the testing through your program and it was very successful. Thank you for your help!!!
I have conducted a more detailed examination of the output results and found that some of the solutions are the same, while others are combinations of other solutions. Since the number of iterations is sufficient, the obtained solutions should be sufficient?Therefore, if I want to obtain the basis of the solutions, perhaps I need to perform Gaussian elimination on all solutions for linear independence testing at the end? But I think the cost of Gaussian elimination is a bit high. Do you have any suggestions?@SebWouters

[SebWouters]

The algorithm can return at most 64 nullspace vectors, and it's indeed not guaranteed that they are linearly independent. That is because for integer factorization, you're typically interested in one (or only a couple) of nullspace vectors.

If you want a full basis of nullspace vectors, then you will indeed need to perform a systematic routine such as Guassian elimination.

How large is your true system?