Question

如何在Sage中获得给定位长的随机素数？

例如，要获得512位素数，我尝试了

p = random_prime(2^512)

但根据文件：

命令random_prime（a，True）将返回2和
之间的随机素数

因此我无法使用它，因为我需要精确长度的素数。

Answer 1

sage: random_prime?
Signature:      random_prime(n, proof=None, lbound=2)
Docstring:
   Returns a random prime p between lbound and n (i.e. lbound <= p <=
   n). The returned prime is chosen uniformly at random from the set
   of prime numbers less than or equal to n.

   INPUT:

   * "n" - an integer >= 2.

   * "proof" - bool or None (default: None) If False, the function
     uses a pseudo-primality test, which is much faster for really big
     numbers but does not provide a proof of primality. If None, uses
     the global default (see "sage.structure.proof.proof")

   * "lbound" - an integer >= 2 lower bound for the chosen primes

这样就足够了吗？

sage: random_prime(2^512-1,False,2^511)
7484165189517896027318192121767201416039872004910529422703501933303497309177247161202453673508851750059292999942026203470027056226694857512284815420448467
sage: is_prime(7484165189517896027318192121767201416039872004910529422703501933303497309177247161202453673508851750059292999942026203470027056226694857512284815420448467)
True

如何在Sage中获得给定长度的素数？

1 个答案: