（Python）质数函数无限循环

Question

I'm currently working on a supercomputer to distribute a task (find prime numbers) to other machines. Right now I'm hitting an infinite loop somewhere, and I think I've tracked it down to my prime number function, but can't figure exactly where it is. I'm only in into to compsci, so any help is greatly appreciated.

LB is the lower bound, UB upper bound of the range I want prime numbers from.

Thanks!

def do_work(self,LB,UB):

    msg = M_DATA
    initialdata = self.recv()
    n = UB
    p = 2
    total = 0


    for p in range(LB,UB):
        prime = True
        for i in range(2, p):
            if p % i == 0:
                prime = False
                break
        if prime == True:
            #print numbers found
            #print (p)
            total += 1

    return (total)

Answer 1

The distance between two primes increases and "tends towards infinity". Hence for quite big LB , the number of iterations necessary to find primes gets big enough to look like you're stuck in an infinite loop. That's why primes are hard to find.

You can improve your algorithm without much changes by looking at the definition of prime numbers. You don't need to check for the remainder of each integer less than prime , but only the primes lower or equal than sqrt(prime) to give a definitive answer. Take 2 for instance: if a number isn't divided by 2, it won't be devided by an even number (4, 6, 8, 10, etc.).

You can then halve the number of iterations necessary by using range(UB, LB, 2) in your outer loop, given that UB is odd (that is range(UB + (1 - (UB % 2), LB, 2) ).

For your inner loop, store a list of found primes and iterate over it while the it's lower than sqrt(prime_to_check) . That'll help a lot.

Answer 2

Have you tested the function? See below, for an example function extracted from your method. Without going into its efficiency and all, it doesn't seem to have an infinite loop, though you can see it as per the logic as well.

#! /usr/bin/env python

def do_work(LB,UB):
    p = 2
    total = 0
    primes = []

    for p in range(LB,UB):
        prime = True
        for i in range(2, p):
            if p % i == 0:
                prime = False
                break
        if prime == True:
            #print numbers found
            primes.append(p)
            total += 1

    return LB, UB, total, primes

for i in xrange(100): print do_work(2, i)

A couple of things to note: you have variables like msg, initialdata, n which are not being used. The return statement is somewhat strange. return (total,) would return a tuple containing total. return total would return just the total.

M_DATA and self.recv() are unknown here, so I venture that you might have an infinite loop related to those.