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1fdf9883b8
Prepares some others. More than 35 problem solved ! Signed-off-by: Julien Lengrand-Lambert <julien@lengrand.fr>
99 lines
2.8 KiB
Python
99 lines
2.8 KiB
Python
#!/usr/bin/env python
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"""
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##---
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# Julien Lengrand-Lambert
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#Created on : Thu Jan 19 10:12:06 CET 2012
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#
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# DESCRIPTION : Solves problem 35 of Project Euler
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The number, 197, is called a circular prime because all rotations of the digits:
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197, 971, and 719, are themselves prime.
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There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
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How many circular primes are there below one million?
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##---
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"""
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import pickle
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def all_permutations(seq):
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"""permutate a sequence and return a list of the permutations"""
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if not seq:
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return [seq] # is an empty sequence
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else:
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temp = []
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for k in range(len(seq)):
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part = seq[:k] + seq[k+1:]
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for m in all_permutations(part):
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temp.append(seq[k:k+1] + m)
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return temp
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def is_prime(value):
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"""
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Returns True or False depending whether value is prime or not.
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"""
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start = 2
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while (start <= value / 2):
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if value % start == 0 :
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return False
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else :
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start += 1
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return True
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def prime_list(max_val):
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"""
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Returns a list of all primes below max_val
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"""
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plist = []
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cur = 2
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while (cur < max_val):
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if (cur % 1000) == 0:
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print "%d/%d" % (cur, max_val)
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if (is_prime(cur)):
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plist.append(cur)
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cur += 1
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return plist
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def perm_primes(max_val):
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"""
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Returns the number of circular primes below max_val
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"""
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circulars = 0
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circl = []
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# to avoid calculating primes
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plist = pickle.load(open("primes_list.dup", "rb"))
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#plist = prime_list(max_val)
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checks_list = [1] * len(plist) # all primes to be tested
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for ii in range(len(plist)):
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if (ii % 99 == 0):
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print "%d/%d" % ( ii + 1, len(plist))
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if checks_list[ii]: # if prime still to be checked
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prime = plist[ii]
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perms = all_permutations(str(prime)) # finds all permutations
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# counts how many permutations are in prime list given one prime
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p_cpt = 0
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for perm in perms:
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if int(perm) in plist:
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p_cpt += 1
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# if prime is circular
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if p_cpt == len(perms):
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# updating checks_list
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for perm in perms:
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if not(int(perm) in circl): # avoiding duplicates
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circulars += 1
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circl.append(int(perm))
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checks_list[plist.index(int(perm))] = 0
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else:
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checks_list[ii] = 0
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return circulars, circl
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if __name__ == '__main__' :
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answer, plist = perm_primes(1000000)
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print "Answer is : %d" % (answer) |