Could someone please tell me what I’m doing wrong with this code? It is just printing ‘count’ anyway. I just want a very simple prime generator (nothing fancy).
import math def main(): count = 3 one = 1 while one == 1: for x in range(2, int(math.sqrt(count) + 1)): if count % x == 0: continue if count % x != 0: print count count += 1
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Answer
There are some problems:
- Why do you print out count when it didn’t divide by x? It doesn’t mean it’s prime, it means only that this particular x doesn’t divide it
continue
moves to the next loop iteration – but you really want to stop it usingbreak
Here’s your code with a few fixes, it prints out only primes:
import math def main(): count = 3 while True: isprime = True for x in range(2, int(math.sqrt(count) + 1)): if count % x == 0: isprime = False break if isprime: print count count += 1
For much more efficient prime generation, see the Sieve of Eratosthenes, as others have suggested. Here’s a nice, optimized implementation with many comments:
# Sieve of Eratosthenes # Code by David Eppstein, UC Irvine, 28 Feb 2002 # http://code.activestate.com/recipes/117119/ def gen_primes(): """ Generate an infinite sequence of prime numbers. """ # Maps composites to primes witnessing their compositeness. # This is memory efficient, as the sieve is not "run forward" # indefinitely, but only as long as required by the current # number being tested. # D = {} # The running integer that's checked for primeness q = 2 while True: if q not in D: # q is a new prime. # Yield it and mark its first multiple that isn't # already marked in previous iterations # yield q D[q * q] = [q] else: # q is composite. D[q] is the list of primes that # divide it. Since we've reached q, we no longer # need it in the map, but we'll mark the next # multiples of its witnesses to prepare for larger # numbers # for p in D[q]: D.setdefault(p + q, []).append(p) del D[q] q += 1
Note that it returns a generator.