def main():
start = time()
primeFactorizationDict = {}
p = Prime()
firstOfFour = 0
candidate = 210 # 210 is the product of the smallest 4 unique primes
while firstOfFour == 0:
if p.factor(candidate):
if candidate not in primeFactorizationDict:
primeFactorizationDict[candidate] = len(set([prime for prime in p.factorize(candidate)]))
if primeFactorizationDict[candidate] == PRIME_LIMIT:
if candidate + 1 not in primeFactorizationDict:
primeFactorizationDict[candidate + 1] = len(set([prime for prime in p.factorize(candidate + 1)]))
if primeFactorizationDict[candidate + 1] == PRIME_LIMIT:
if candidate + 2 not in primeFactorizationDict:
primeFactorizationDict[candidate + 2] = len(set([prime for prime in p.factorize(candidate + 2)]))
if primeFactorizationDict[candidate + 2] == PRIME_LIMIT:
if candidate + 3 not in primeFactorizationDict:
primeFactorizationDict[candidate + 3] = len(set([prime for prime in p.factorize(candidate + 3)]))
if primeFactorizationDict[candidate + 3] == PRIME_LIMIT:
firstOfFour = candidate
candidate += 1
end = time()
print "First of four integers: ", firstOfFour
print "Runtime: ", end - start, " seconds. "
def getHash(x):
if type(x) is str:
s = x
elif type(x) is int:
s = str(x)
elif type(x) is float:
s = str(int(x * 1000000))
elif isinstance(x, Hashable):
return x.getHash()
else:
raise Exception('getHash: Hash of ' + str(x) + ' is unknown')
if "NewHash" in Params.params:
if (len(Prime.primes) == 0):
Prime.init()
S = 1
for i in range(len(s)):
S = addHash(S, Prime.getPrime(256 * i + ord(s[i])))
# S = addHash( S, Prime.getPrime(i+256) );
# print "NEW HASH("+s + ")="+str(S)
else:
# print "OLD HASH"
S = 1
for i in range(len(s)):
S = addHash(S, ord(s[i]))
return S
def getHash(x):
if type(x) is str:
s=x;
elif type(x) is int:
s=str(x);
elif type(x) is float:
s=str(int(x*1000000))
elif isinstance(x,Hashable):
return x.getHash();
else:
raise Exception('getHash: Hash of ' + str(x) +' is unknown')
if "NewHash" in Params.params:
if(len(Prime.primes)==0):
Prime.init();
S=1;
for i in range(len(s)):
S = addHash( S, Prime.getPrime(256*i + ord(s[i])) );
# S = addHash( S, Prime.getPrime(i+256) );
# print "NEW HASH("+s + ")="+str(S)
else:
# print "OLD HASH"
S=1;
for i in range(len(s)):
S = addHash( S, ord(s[i]) );
return S;
def main():
start = time()
primeFactorizationDict = {}
p = Prime()
firstOfFour = 0
candidate = 210 # 210 is the product of the smallest 4 unique primes
while firstOfFour == 0:
if p.factor(candidate):
if candidate not in primeFactorizationDict:
primeFactorizationDict[candidate] = len(
set([prime for prime in p.factorize(candidate)]))
if primeFactorizationDict[candidate] == PRIME_LIMIT:
if candidate + 1 not in primeFactorizationDict:
primeFactorizationDict[candidate + 1] = len(
set([prime for prime in p.factorize(candidate + 1)]))
if primeFactorizationDict[candidate + 1] == PRIME_LIMIT:
if candidate + 2 not in primeFactorizationDict:
primeFactorizationDict[candidate + 2] = len(
set([
prime for prime in p.factorize(candidate + 2)
]))
if primeFactorizationDict[candidate + 2] == PRIME_LIMIT:
if candidate + 3 not in primeFactorizationDict:
primeFactorizationDict[candidate + 3] = len(
set([
prime
for prime in p.factorize(candidate + 3)
]))
if primeFactorizationDict[candidate +
3] == PRIME_LIMIT:
firstOfFour = candidate
candidate += 1
end = time()
print "First of four integers: ", firstOfFour
print "Runtime: ", end - start, " seconds. "
def run(self):
prime = Prime(self.num)
count = prime.count()
endAt = int(round(time.time() * 1000)) - self.startAt
print('Python thread id #%d: executes %d prime numbers in %d ms' %
(self.threadId, count, endAt))
#!/usr/bin/env python
# Pravin Paratey (http://pravin.insanitybegins.com)
#
# 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?
#
# Answer: 55
from Prime import Prime
# Generate prime numbers under 1 million
primes_list = Prime.generate_primes(1000000)
primes_dict = {}
for p in primes_list:
primes_dict[p] = None
# Just brute force
# 13 primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97
count = 13
for p in primes_list:
if p < 100:
continue
is_circular = True
str_p = str(p)
for i in xrange(len(str_p)):
int_p = int(str_p)
if not primes_dict.has_key(int_p):
is_circular = False
def __init__(self):
Prime.__init__(self)
#!/usr/bin/env python
# Pravin Paratey (http://pravin.insanitybegins.com)
#
# 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?
#
# Answer: 55
from Prime import Prime
# Generate prime numbers under 1 million
primes_list = Prime.generate_primes(1000000)
primes_dict = {}
for p in primes_list:
primes_dict[p] = None
# Just brute force
# 13 primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97
count = 13
for p in primes_list:
if p < 100:
continue
is_circular = True
str_p = str(p)
for i in xrange(len(str_p)):
int_p = int(str_p)
if not primes_dict.has_key(int_p):
is_circular = False
from Prime import Prime
import math
import time
PAGE = 1_0000
REPEAT = 1_000
prime = Prime(PAGE * REPEAT)
def prime_by_eratosthenes_interval(pos, limit):
top = 0
num = [True] * limit
for i in range(0, prime.size()):
p = prime.get(i)
if p * p >= pos + limit: break
for j in range(math.ceil(pos / p), int((pos + limit - 1) / p) + 1):
num[j * p - pos] = False
for i in range(0, limit):
if num[i]:
prime.add(pos + i)
top = top + 1
return top
def prime_by_euler(limit):
top = 0
num = [True] * limit
for i in range(2, limit):
if num[i]:
from Prime import Prime
print ("Which number you want to test is prime or not")
num = int(input())
p1 = Prime(num)
print ("%d is prime?"% num)
print (p1.isPrime())
''' Write a code to check whether no is prime or not. Condition use function check() to find whether entered no is positive or negative ,if negative then enter the no, And if yes pas no as a parameter to prime() and check whether no is prime or not? Whether the number is positive or not, if it is negative then print the message “please enter the positive number” It is positive then call the function prime and check whether the take positive number is prime or not ''' from Prime import Prime n = int(input()) print(Prime.isPrime(n))
'''
Created on 2014. 4. 2.
Summation of primes
Problem 10
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million(2,000,000).
Solution: 142913828922
@author: Ungsik Yun
'''
if __name__ == '__main__':
from Prime import Prime
p = Prime()
l = p.list_under_improved(2000000)
print sum(l)
