整数的因式分解 [英] Factorization of an integer

问题描述

在回答另一个问题时，我迷迷糊糊地问了一个问题:我如何才能真正找到所有没有符号数学工具箱的整数因子.

While answering another, I stumbled over the question how I actually could find all factors of an integer number without the Symbolic Math Toolbox.

例如:

factor(60)

返回:

 2     2     3     5

---

unique(factor(60))

因此将返回所有缺少的素数，"1" .

would therefore return all prime-factors, "1" missing.

 2     3     5

---

我正在寻找一个可以返回所有因子的函数( 1 和数字本身并不重要，但它们会很好)

---

And I'm looking for a function which would return all factors (1 and the number itself are not important, but they would be nice)

x = 60的预期输出:

Intended output for x = 60:

 1     2     3     4     5     6    10    12    15    20    30    60     

---

我想出了一个相当庞大的解决方案，除了可以将其向量化之外，还没有任何优雅的解决方案吗?

---

I came up with that rather bulky solution, apart from that it probably could be vectorized, isn't there any elegant solution?

x = 60;

P = perms(factor(x));
[n,m] = size(P);
Q = zeros(n,m);
for ii = 1:n
    for jj = 1:m
        Q(ii,jj) = prod(P(ii,1:jj));
    end
end

factors = unique(Q(:))'

我还认为，对于某些较大的数字，此解决方案将失败，因为烫发需要向量长度< 11.

Also I think, this solution will fail for certain big numbers, because perms requires a vector length < 11.

推荐答案

这里是六个不同实现的比较，用于查找整数因子:

Here is a comparison of six different implementations for finding factors of an integer:

function [t,v] = testFactors()
    % integer to factor
    %{45, 60, 2059, 3135, 223092870, 3491888400};
    n = 2*2*2*2*3*3*3*5*5*7*11*13*17*19;

    % functions to compare
    fcns = {
        @() factors1(n);
        @() factors2(n);
        @() factors3(n);
        @() factors4(n);
        %@() factors5(n);
        @() factors6(n);
    };

    % timeit
    t = cellfun(@timeit, fcns);

    % check results
    v = cellfun(@feval, fcns, 'UniformOutput',false);
    assert(isequal(v{:}));
end

function f = factors1(n)
    % vectorized implementation of factors2()
    f = find(rem(n, 1:floor(sqrt(n))) == 0);
    f = unique([1, n, f, fix(n./f)]);
end

function f = factors2(n)
    % factors come in pairs, the smaller of which is no bigger than sqrt(n)
    f = [1, n];
    for k=2:floor(sqrt(n))
        if rem(n,k) == 0
            f(end+1) = k;
            f(end+1) = fix(n/k);
        end
    end
    f = unique(f);
end

function f = factors3(n)
    % Get prime factors, and compute products of all possible subsets of size>1
    pf = factor(n);
    f = arrayfun(@(k) prod(nchoosek(pf,k),2), 2:numel(pf), ...
        'UniformOutput',false);
    f = unique([1; pf(:); vertcat(f{:})])'; %'
end

function f = factors4(n)
    % http://rosettacode.org/wiki/Factors_of_an_integer#MATLAB_.2F_Octave
    pf = factor(n);                    % prime decomposition
    K = dec2bin(0:2^length(pf)-1)-'0'; % all possible permutations
    f = ones(1,2^length(pf));
    for k=1:size(K)
      f(k) = prod(pf(~K(k,:)));        % compute products 
    end; 
    f = unique(f);                     % eliminate duplicates
end

function f = factors5(n)
    % @LuisMendo: brute-force implementation
    f = find(rem(n, 1:n) == 0);
end

function f = factors6(n)
    % Symbolic Math Toolbox
    f = double(evalin(symengine, sprintf('numlib::divisors(%d)',n)));
end

结果:

>> [t,v] = testFactors();
>> t
t =
    0.0019        % factors1()
    0.0055        % factors2()
    0.0102        % factors3()
    0.0756        % factors4()
    0.1314        % factors6()

>> numel(v{1})
ans =
        1920

尽管第一个矢量化版本是最快的，但是由于自动JIT优化，等效的基于循环的实现(factors2)也相差不远.

Although the first vectorized version is the fastest, the equivalent loop-based implementation (factors2) is not far behind, thanks to automatic JIT optimization.

请注意，我必须禁用蛮力实现(factors5())，因为它会引发内存不足错误(以双精度存储向量1:3491888400时需要超过26GB的内存！).对于大整数，无论​​是在空间上还是时间上，这种方法显然都不可行.

Note that I had to disable the brute-force implementation (factors5()) because it throws an out-of-memory error (storing the vector 1:3491888400 in double-precision requires over 26GB of memory!). This method is obviously not feasible for large integers, neither space- or time-wise.

结论:使用以下矢量化实现:)

Conclusion: use the following vectorized implementation :)

n = 3491888400;
f = find(rem(n, 1:floor(sqrt(n))) == 0);
f = unique([1, n, f, fix(n./f)]);

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