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Finally solves problem 35. Problem of . . . definition .

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julien Lengrand-Lambert
2012-05-01 22:03:10 +02:00
parent 200d812bc4
commit f09f83e0a4
3 changed files with 81 additions and 174 deletions
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2
.pydevproject
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@@ -3,7 +3,7 @@
<pydev_project>
<pydev_pathproperty name="org.python.pydev.PROJECT_SOURCE_PATH">
<path>/project_euler</path>
<path>/peuler</path>
</pydev_pathproperty>
<pydev_property name="org.python.pydev.PYTHON_PROJECT_VERSION">python 2.7</pydev_property>
<pydev_property name="org.python.pydev.PYTHON_PROJECT_INTERPRETER">Default</pydev_property>

105
e_35.py
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@@ -13,8 +13,25 @@
"""
import pickle
def shift(seq, n):
"""
Shifts a sequence by n rotations
"""
seq = str(seq)
return int(seq[n:]+seq[:n])
def all_rotations(seq):
"""
Returns a list of all possible rotations of a number
"""
out = []
for n in range(len(str(seq))):
out.append(int(shift(str(seq), n)))
return out
def all_permutations(seq):
"""permutate a sequence and return a list of the permutations"""
"""permutates a sequence and returns a list of the permutations"""
if not seq:
return [seq] # is an empty sequence
else:
@@ -53,47 +70,75 @@ def prime_list(max_val):
return plist
def contains_even(val):
"""
Returns True if val contains an even number (ex : 998)
"""
return ((str(2) in str(val)) or (str(4) in str(val)) or (str(6) in str(val)) or (str(8) in str(val)))
def divisible_by_5(val):
"""
Returns True if any circular permutation of val is divisible by (ex : 907)
"""
return ((str(5) in str(val)) or (str(0) in str(val)))
def divisible_by_3(val):
"""
Returns True if any circular permutation of val is divisible by 3 (ex : 12)
"""
temp = sum([int(p) for p in str(val)])
if len(str(temp)) > 1:
divisible_by_3(temp)
else:
return ((temp % 3) == 0)
def check_easy_out(plist):
"""
Returns the number of circular primes below max_val
TODO : offer cleaner method
"""
pred = list(plist)
for p in plist:
if p > 11: # my filter does not work for values under 10
# ugly method
if (contains_even(p) or divisible_by_5(p)): # no prime divisible by 3 . . .
pred.remove(p)
return pred
def perm_primes(max_val):
"""
Returns the number of circular primes below max_val
"""
circulars = 0
circl = []
plist = prime_list(max_val)
#plist = pickle.load(open("primes_list.dup", "rb"))
p_reduced = check_easy_out(plist)
# to avoid calculating primes
plist = pickle.load(open("primes_list.dup", "rb"))
#plist = prime_list(max_val)
checks_list = [1] * len(plist) # all primes to be tested
p_out = []
for ii in range(len(plist)):
if (ii % 99 == 0):
print "%d/%d" % ( ii + 1, len(plist))
print len(plist)
print len(p_reduced)
if checks_list[ii]: # if prime still to be checked
prime = plist[ii]
perms = all_permutations(str(prime)) # finds all permutations
print p_reduced
# counts how many permutations are in prime list given one prime
p_cpt = 0
for perm in perms:
if int(perm) in plist:
p_cpt += 1
for p in p_reduced:
perms = all_rotations(p)
perms = list(set(perms)) # removes duplicates
perms = [int(p) for p in perms]
# if prime is circular
if p_cpt == len(perms):
# updating checks_list
for perm in perms:
if not(int(perm) in circl): # avoiding duplicates
circulars += 1
circl.append(int(perm))
checks_list[plist.index(int(perm))] = 0
circl = 0
for pp in perms:
if pp in p_reduced: # in list of primes
circl += 1
else:
checks_list[ii] = 0
if circl == len(perms):
p_out = p_out + perms
return circulars, circl
circl = 0
p_out = list(set(p_out))
return len(p_out)
if __name__ == '__main__' :
answer, plist = perm_primes(1000000)
answer = perm_primes(1000000)
print "Answer is : %d" % (answer)

138
e_35_2.py
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@@ -1,138 +0,0 @@
#!/usr/bin/env python
"""
##---
# Julien Lengrand-Lambert
#Created on : Thu Jan 19 10:12:06 CET 2012
#
# DESCRIPTION : Solves problem 35 of Project Euler
The number, 197, is called a circular prime because all rotations of the digits:
197, 971, and 719, are themselves prime.
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
##---
"""
import pickle
def all_permutations(seq):
"""permutate a sequence and return a list of the permutations"""
if not seq:
return [seq] # is an empty sequence
else:
temp = []
for k in range(len(seq)):
part = seq[:k] + seq[k+1:]
for m in all_permutations(part):
temp.append(seq[k:k+1] + m)
return temp
def is_prime(value):
"""
Returns True or False depending whether value is prime or not.
"""
start = 2
while (start <= value / 2):
if value % start == 0 :
return False
else :
start += 1
return True
def prime_list(max_val):
"""
Returns a list of all primes below max_val
"""
plist = []
cur = 2
while (cur < max_val):
if (cur % 1000) == 0:
print "%d/%d" % (cur, max_val)
if (is_prime(cur)):
plist.append(cur)
cur += 1
return plist
def contains_even(val):
"""
Returns True f val contains an even number (ex : 998)
"""
return (str(2) in str(val) or str(4) in str(val) or str(6) in str(val) or str(8) in str(val))
def divisible_by_3(val):
"""
Returns True of val is divisible by 3 (ex : 12)
"""
temp = sum([int(p) for p in str(val)])
if len(str(temp)) > 1:
divisible_by_3(temp)
else:
return ((temp % 3) == 0)
def check_easy_out(plist):
"""
Returns the number of circular primes below max_val
TODO : offer cleaner method
"""
pred = list(plist)
for p in plist:
if p > 11: # my filter does not work for values under 10
# ugly method
#if (contains_even(p) or divisible_by_3(p)):
if (contains_even(p)): # no prime divisible by 3 . . .
# could also test for 3 and 9
pred.remove(p)
return pred
def perm_primes(max_val):
"""
Returns the number of circular primes below max_val
"""
circulars = 0
circl = []
#plist = prime_list(max_val)
plist = pickle.load(open("primes_list.dup", "rb"))
p_reduced = check_easy_out(plist)
print len(plist)
print len(p_reduced)
print p_reduced
checks_list = [1] * len(p_reduced) # all primes to be tested
for ii in range(len(p_reduced)):
if (ii % 99 == 0):
print "%d/%d" % ( ii + 1, len(p_reduced))
if checks_list[ii]: # if prime still to be checked
prime = p_reduced[ii]
perms = all_permutations(str(prime)) # finds all permutations
# counts how many permutations are in prime list given one prime
p_cpt = 0
for perm in perms:
if int(perm) in p_reduced:
p_cpt += 1
# if prime is circular
if p_cpt == len(perms):
# updating checks_list
for perm in perms:
if not(int(perm) in circl): # avoiding duplicates
circulars += 1
circl.append(int(perm))
checks_list[p_reduced.index(int(perm))] = 0
else:
checks_list[ii] = 0
return circulars, circl
if __name__ == '__main__' :
#perm_primes(1000000)
answer, plist = perm_primes(1000000)
print "Answer is : %d" % (answer)