"""Eratosthenes.py Space-efficient version of sieve of Eratosthenes. D. Eppstein, May 2004. The main storage of the algorithm is a hash table D with sqrt(n) nonempty entries for a total of O(sqrt n) space. At any point in the algorithm, each prime p occupies a cell with key at most 2n. E.g. by Bertrand's postulate, there is another prime p' between n/p and 2n/p, and p' can not yet have been included because it is greater than sqrt n, so key pp' can not be used by any other prime; therefore p is placed at or before key pp'<2n. Thus, the number of times p can have been moved from its initial placement at p^2 is < n/p. The time for the algorithm, up to output n, is O(n) + sum_{prime p <= sqrt(n)} O(n/p) = O(n log log n). The algorithm also makes a recursive call, but the recursion only generates primes up to sqrt n so its time and space is negligible compared to the outer call. If efficiency is a significant concern it may be better to combine this idea with segmentation and bitvectors, as in the code by T. Oliveira e Silva at http://www.ieeta.pt/~tos/software/prime_sieve.html Thanks to Alex Martelli for the suggestion of keeping one prime per entry of D, rather than a list of all prime factors of D. We also include a variant of the sieve that produces a list of all integers, with their factorizations, and an application of this variant in the generation of practical numbers. """ import unittest def primes(): '''Yields the sequence of primes via the Sieve of Eratosthenes.''' yield 2 # Only even prime. Sieve only odd numbers. # Generate recursively the sequence of primes up to sqrt(n). # Each p from the sequence is used to initiate sieving at p*p. roots = primes() root = roots.next() square = root*root # The main sieving loop. # We use a hash table D such that D[n]=2p for p a prime factor of n. # Each prime p up to sqrt(n) appears once as a value in D, and is # moved to successive odd multiples of p as the sieve progresses. D = {} n = 3 while True: if n >= square: # Time to include another square? D[square] = root+root root = roots.next() square = root*root if n not in D: # Not witnessed, must be prime. yield n else: # Move witness p to next free multiple. p = D[n] q = n+p while q in D: q += p del D[n] D[q] = p n += 2 # Move on to next odd number. def FactoredIntegers(): """ Generate pairs n,F where F is the prime factorization of n. F is represented as a dictionary in which each prime factor of n is a key and the exponent of that prime is the corresponding value. """ yield 1,{} i = 2 factorization = {} while True: if i not in factorization: # prime F = {i:1} yield i,F factorization[2*i] = F elif len(factorization[i]) == 1: # prime power p,x = factorization[i].items()[0] F = {p:x+1} yield i,F factorization[2*i] = F factorization.setdefault(i+p**x,{})[p] = x del factorization[i] else: yield i,factorization[i] for p,x in factorization[i].iteritems(): q = p**x iq = i+q if iq in factorization and p in factorization[iq]: iq += p**x # skip higher power of p factorization.setdefault(iq,{})[p] = x del factorization[i] i += 1 def isPracticalFactorization(f): """Test whether f is the factorization of a practical number.""" f = f.items() f.sort() sigma = 1 for p,x in f: if sigma < p - 1: return False sigma *= (p**(x+1)-1)//(p-1) return True def PracticalNumbers(): """Generate the sequence of practical (or panarithmic) numbers.""" for x,f in FactoredIntegers(): if isPracticalFactorization(f): yield x # If run standalone, perform unit tests class SieveTest(unittest.TestCase): def testPrime(self): """Test that the first few primes are generated correctly.""" G = primes() for p in [2,3,5,7,11,13,17,19,23,29,31,37]: self.assertEqual(p,G.next()) def testPractical(self): """Test that the first few practical nos are generated correctly.""" G = PracticalNumbers() for p in [1,2,4,6,8,12,16,18,20,24,28,30,32,36]: self.assertEqual(p,G.next()) if __name__ == "__main__": unittest.main()