降低Eratosthenes筛的空间复杂度，以生成一定范围内的质数 [英] Reducing space complexity of Sieve of Eratosthenes to generate primes in a range

问题描述

经过一些 SO帖子，我发现 Eratosthenes筛子是最好的&生成质数的最快方法.

After getting through some of the SO posts, I found Sieve of Eratosthenes is the best & fastest way of generating prime numbers.

我想生成两个数字之间的质数，例如 a 和 b .

I want to generate the prime numbers between two numbers, say a and b.

AFAIK，按照Sieve的方法，空间复杂度为 O(b).

AFAIK, in Sieve's method, the space complexity is O(b).

PS:我写的是Big-O而不是Theta，因为我不知道是否可以减少空间需求.

PS: I wrote Big-O and not Theta, because I don't know whether the space requirement can be reduced.

我们可以减少筛网筛网的空间复杂度吗?

Can we reduce the space complexity in Sieve of Eratosthenes ?

推荐答案

如果您有足够的空间将所有素数存储到sqrt(b)，则可以使用额外的空间O来筛选a到b范围内的素数(ba).

If you have enough space to store all the primes up to sqrt(b) then you can sieve for the primes in the range a to b using additional space O(b-a).

在Python中，它可能类似于:

In Python this might look like:

def primesieve(ps,start,n):
  """Sieve the interval [start,start+n) for primes.

     Returns a list P of length n.  
     P[x]==1 if the number start+x is prime.  
     Relies on being given a list of primes in ps from 2 up to sqrt(start+n)."""
  P=[1]*n
  for p in ps:
    for k in range((-start)%p,n,p):
      if k+start<=p: continue
      P[k]=0
  return P

只需筛分奇数，就可以轻松地占用一半的空间.

You could easily make this take half the space by only sieving the odd numbers.

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