Python:非常大的矩阵的简化行梯形形式(mod p) [英] Python: reduced row echelon form (mod p) of a very large matrix

问题描述

我想找到一个大矩阵的简化的行梯形形式(在字段F_q中). 我尝试了以下代码. 尽管我使用gmpy2库加快了速度，但该程序仍然内存不足.因为我的输入矩阵非常大(100 x 2 ^ 15)，而p也非常大(| p | = 256位).有人可以建议如何降低这种算法的复杂性.

I want to find find a reduced a row echelon form (in field F_q) of a big matrix. I tried the following code. Although I used gmpy2 library to speed up, the program was still out of memory. because my input matrix is very large (100 x 2^15) and p is also very large (|p|=256 bits). Can someone suggest how to reduce the complexity of this alg.

谢谢

def invmodp(a, p):
    return gmpy2.invert(a,p)

def division_mod(a, b, p): #a/b mod p
    invert = invmodp(b, p)
    return (a * invert) %p

def row_echelon_form(M, p):
   lead = 0
   rowCount = len(M)
   columnCount = len(M[0])
   for r in range(rowCount):
       if lead >= columnCount:
           return
       i = r
       while M[i][lead] == 0:
           i += 1
           if i == rowCount:
               i = r
               lead += 1
               if columnCount == lead:
                   return
    M[i],M[r] = M[r],M[i]
    lv = M[r][lead]
    M[r] = [ division_mod(mrx, lv, p) for mrx in M[r]]
    for i in range(rowCount):
        if i != r:
            lv = M[i][lead]
            M[i] = [ (iv - lv*rv)%p for rv,iv in zip(M[r],M[i])]
    lead += 1
return M

推荐答案

通过使用gmpy2.divm替换您的division_mod，我能够节省几秒钟的运行时间.我无法进行其他任何重大改进.下面的程序创建一个随机的100 x 2 ^ 15矩阵，并在大约3分钟内计算行梯形形式，并消耗425MB的内存.

I was able to save a few seconds of running time by using gmpy2.divm to replace your division_mod. I wasn't able to make any other significant improvements. The following program creates a random 100 x 2^15 matrix and calculates the row echelon form in approximately 3 minutes and consumes 425MB of memory.

import gmpy2

bits = 256
r = 100
c = 2**15

p = gmpy2.next_prime(2**bits - 1234)
seed = gmpy2.random_state(42)

M = []
for i in range(r):
    M.append([gmpy2.mpz_urandomb(seed, bits) for j in range(c)])

def row_echelon_form(M, p):
    lead = 0
    rowCount = len(M)
    columnCount = len(M[0])
    for r in range(rowCount):
        if lead >= columnCount:
            return
        i = r
        while M[i][lead] == 0:
            i += 1
            if i == rowCount:
                i = r
                lead += 1
                if columnCount == lead:
                    return

        M[i],M[r] = M[r],M[i]
        lv = M[r][lead]
        M[r] = [ gmpy2.divm(mrx, lv, p) for mrx in M[r]]
        for i in range(rowCount):
            if i != r:
                lv = M[i][lead]
                M[i] = [ (iv - lv*rv) % p for rv,iv in zip(M[r],M[i])]
        lead += 1
    return M

N = row_echelon_form(M, p)

如果您的内存使用量增长到超过500MB，则您的gmpy2版本可能存在内存泄漏.或者我没有正确解释您的要求，并且矩阵很大.

If your memory usage grows beyond about 500MB, there may be a memory leak in your version of gmpy2. Or I've interpreted your requirements incorrectly and the matrix is significantly larger.

免责声明:我维护gmpy2.

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