Factorization
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Howto find factors of 2^1024 and 2^2048 in Matlab??
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回答(5 个)
Walter Roberson
2011-5-20
With some difficulty, seeing that both numbers are larger than any number that can be directly represented in MATLAB except by using the Fixed Point Toolbox or the Symbolic Toolbox.
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John D'Errico
2011-5-20
No problem. Just use a tool designed to solve the problem.
>> factor(vpi(2)^1024)
ans =
Columns 1 through 16
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
Columns 17 through 32
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
Columns 33 through 48
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
...
And so on for all 1024 factors of the number.
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Andy
2011-5-20
For any prime p and positive integer n, the factors of p^n are all of the form p^k for 0 <= k <= n. So the factors of 2^1024 are just 2^k for 0 <= k <= 1024, and similarly for 2^2048. No need to use MATLAB.
2 个评论
Walter Roberson
2011-5-20
But those numbers are not necessarily directly representable in MATLAB, especially if p is not a power of 2.
Andy
2011-5-20
That's true. But the OP was unclear as to whether he wanted to represent the factors of these numbers in MATLAB, or whether he simply wanted to solve this problem (and thought of MATLAB as a tool for solving it). Since all of the other answers were of the first sort, I thought I'd throw in an answer of the second sort.
Andrew Newell
2011-5-20
The Symbolic Toolbox has a really nice way of handling this:
two = sym(2);
factor(two^1024)
ans =
2^1024
2 个评论
Walter Roberson
2011-5-20
factor() for the symbolic toolbox does algebraic factoring and does not touch numbers.
I do not know at the moment how to produce factors in MuPad; in Maple it would be by using ifactor() or numtheory[divisors]() or one of the related numtheory package members.
Walter Roberson
2011-5-21
In Maple,
numtheory[divisors](2^512-5) %a smaller problem
yields an error,
Error, (in ifactor/QuadraticSieve) object too large
It thus seems unlikely that Maple would be able to factor 2^2048-5 using the built-in routines.
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