Is there a function to find the paired factors of an integer?
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Matt J
el 4 de Dic. de 2022
Comentada: John D'Errico
el 4 de Dic. de 2022
Is there a function in Matlab to find the pairs of factors that multiply together to give a specified integer, e.g,
F=somefunc(16)
F =
1 16
2 8
4 4
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John D'Errico
el 4 de Dic. de 2022
Editada: John D'Errico
el 4 de Dic. de 2022
allfactorpairs = @(n) [divisors(n)',n./divisors(n)'];
allfactorpairs(16)
If you want only those pairs where the first element is the smaller of the two, then you might write a little function, like this:
allfactorpairs2(16)
allfactorpairs2(210)
allfactorpairs2(137)
Finally, see that this code runs on both doubles as well as syms, and is efficient as long as the number is esily factored, so the limit is really in terms of how well factor works for huge numbers. That of course can be terribly difficult.
allfactorpairs2(sym(factorial(10)))
Your choice. Note that divisors is a function from the symbolic toolbox. So if someone finds it is too slow, or you lack that toolbox, it is easily replaced. And while I expect that @Matt J has the symbolic TB, I'll write it the replacement below to operate efficiently on doubles as well as syms.
function D = principaldivisors(N)
% Compute the principal divisors of the scalar variable N, so only those divisors < N.
% Note if N is prime, then this function returns ONLY 1, as the only factor less than N.
F = factor(N);
D = 1;
if numel(F) > 1
% N is composite, so it has more than one divisor less than N
% if N were prime, then 1 and N are the only factors.
for Fi = F
D = unique([D,D*Fi]);
end
D(D == N) = [];
end
end
function pairs = allfactorpairs2(N)
D = principaldivisors(N)';
D(D > sqrt(N)) = [];
pairs = [D,N./D];
end
2 comentarios
John D'Errico
el 4 de Dic. de 2022
Arguably there should be a utiltiy in MATLAB itself to do this, or even just to find a list of all proper divisors of an integer. But there are relatively few utilities in MATLAB for working with integers, not much more than primes, factor, gcd, lcm, and a few others.
Should there be an expansion of those tools? Perhaps.
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