def E49():
prime_perm = [ ]
for i in range(1487, 10000):
if is_prime(i) == False: continue
plist = list(itertools.permutations(str(i)))
for each_num in plist:
num = "".join(map(str, each_num))
if is_prime(num) == True:
prime_perm.append(num)
len_p = len(prime_perm)
for j in range(0, len_p-1):
for k in range(j+1, len_p):
if prime_perm[k] == prime_perm[j]: continue
diff = int(prime_perm[k]) - int(prime_perm[j])
next_num = int(prime_perm[k]) + diff
if (str(next_num) in prime_perm) == True :
print prime_perm[j], prime_perm[k], next_num
del prime_perm[:]
return 0
def CheckRight(num):
devide = 1
if (is_prime(num) != True):
return -1
for i in range(0, len(str(num)) - 1):
num = num / 10
if (is_prime(num) != True):
return -1
return 0
def is_prime_set(vals):
subsets = itertools.combinations(vals, 2)
failed_set = set([])
for a, b in subsets:
a = str(a)
b = str(b)
str1 = a + b
str2 = b + a
if not is_prime(int(str1)) or not is_prime(int(str2)):
return False
return True
def E70():
res = 0
min_res = 100
min_i = 100
p_set = []
for i in range (3, MAXNUM, 2):
if is_prime(i) == True:
continue
p_set = factor(i)
if len(p_set) != 2:
continue
if int(p_set[0][1])!=1 or int(p_set[1][1])!=1:
continue
# [(421, 1), (571, 1)] <- both are 1
cnt = i - (int(p_set[0][0])+int(p_set[1][0])-1)
if is_perm(i, cnt) == False:
continue
res = float(i)/float(cnt)
if min_res >= res:
min_res = res
min_i = i
#print min_res, min_i, cnt, (1, p_set[0][0], p_set[1][0], i)
print "result", min_res, min_i
return 0
def eight_prime_family(prime, rd):
c=0
for digit in '0123456789':
n = int(string.replace(prime, rd, digit))
if (n>100000 and is_prime(n)):
c=c+1
return c==8
def main():
best = int(1000000)
for i in range(11, 1000, 2):
if i % 5 == 0:
i += 2
patterns = deepcopy(patterns5)
if i > 100:
patterns = deepcopy(patterns6)
for j in range(len(patterns)):
pattern = fill_pattern(patterns[j], i)
for k in range(3):
if pattern[0] == -1 and k == 0:
continue
candidate = generate_number(k, pattern)
# print(i, j, k, patterns[j], pattern, candidate, is_prime(candidate), get_family_size(k, pattern))
if is_prime(candidate) and get_family_size(
k, pattern) >= 8 and candidate < best:
best = candidate
break
print(best)
def E70():
res = 0
min_res = 100
min_i = 100
p_set = []
for i in range(3, MAXNUM, 2):
if is_prime(i) == True:
continue
p_set = factor(i)
if len(p_set) != 2:
continue
if int(p_set[0][1]) != 1 or int(p_set[1][1]) != 1:
continue
# [(421, 1), (571, 1)] <- both are 1
cnt = i - (int(p_set[0][0]) + int(p_set[1][0]) - 1)
if is_perm(i, cnt) == False:
continue
res = float(i) / float(cnt)
if min_res >= res:
min_res = res
min_i = i
#print min_res, min_i, cnt, (1, p_set[0][0], p_set[1][0], i)
print "result", min_res, min_i
return 0
def get_family_size(repeated_num, pattern_fam):
family_size = 0
for i in range(repeated_num, 10):
if is_prime(generate_number(i, pattern_fam)):
family_size += 1
return int(family_size)
def test_make_pairs():
print(make_pairs([1, 3, 5]))
print(make_pairs([7, 9]))
print(make_pairs([75, 23, 18]))
test = make_pairs([3, 7, 109, 673])
for elem in iter(test):
print(elem, is_prime(elem))
def P(n):
all_sum = 0
i = 0
while i <= n:
if (is_prime(i)):
all_sum += i
i += 1
return all_sum
def P(n):
all_sum = 0
i = 0
while i <= n:
if (is_prime(i)):
all_sum += i
i += 1
return all_sum
def CheckLeft(num):
devide = 1
for i in range(0, len(str(num))):
devide *= 10
res = num % devide
if (is_prime(res) != True):
return -1
return 0
def main(limit): num = 1 vals = [] while num < limit: val = 2*num**2-1 vals.append(val) if not is_prime(val): print val, factor(val) num += 1 return vals
def Get_P(n):
i = 2
prime_th = 0
while (1):
if (prime_th == n):
return (i - 1)
break
if (is_prime(i)):
prime_th += 1
i += 1
def eight_prime_family(prime, rd):
c=0
for digit in '0123456789':
n = int(string.replace(prime, rd, digit))
if (n>100000 and is_prime(n)):
c=c+1
return c==8
def Get_P(n) :
i = 2
prime_th = 0
while (1):
if (prime_th == n):
return (i-1)
break
if (is_prime(i)):
prime_th += 1
i += 1
def project_euler_27():
"""Finds the product of a and b for that produces the most primes for
consecutive values of n, starting with n = 0. For n^2 + an + b where
|a| < 1000 and |b| < 1000."""
best = (0, -1001, -1001)
for a in range(-999, 1000):
for b in range(-999, 1000):
n = 0
while is_prime(abs(n*n + a*n + b)): n += 1
if n > best[0]: best = (n, a, b)
print(best[1]*best[2])
def pFact(num):
pfactors = []
while True:
j = 2
while j < num+1:
if num%j==0 and is_prime(j):
num = num/j
pfactors.append(j)
break
j+=1
if num==1:
break
return pfactors
def main():
answers = list()
for index_a in range(0, len(primes)):
if primes[index_a] * 5 >= result: break
if pairs[index_a] is None: pairs[index_a] = make_pairs([index_a])
for index_b in range(index_a, len(primes)):
if all([
is_prime(n)
for n in iter(make_pairs(answers + [primes[index_b]]))
]):
answers.append(primes[index_b])
for index_c in range(index_b, len(primes)):
if all([
is_prime(n) for n in iter(
make_pairs(answers + [primes[index_c]]))
]):
answers.append(primes[index_c])
for index_d in range(index_c, len(primes)):
if all([
is_prime(n) for n in iter(
make_pairs(answers +
[primes[index_d]]))
]):
answers.append(primes[index_d])
for index_e in range(index_d, len(primes)):
if all([
is_prime(n) for n in iter(
make_pairs(answers +
[primes[index_e]]))
]):
answers.append(primes[index_e])
return answers
return "Failed", answers
def sol_1():
def formula(i, j, k):
return k**2 + i * k + j
max_primes, resI, resJ = 0, 0, 0
for i in xrange(-1000, 1000):
for j in xrange(i, 1000):
k = 0
while is_prime(formula(i, j, k)):
k += 1
if k - 1 > max_primes:
max_primes = k - 1
resI, resJ = i, j
print resI * resJ
def project_euler_26():
"""Calculates the value of d < 1000 for which 1/d contains the longest
recurring cycle in its decimal fraction part.
Decimal fraction parts only repeat if d is relatively prime to 10,
so we filter out non-primes from consideration. Next, the number
of digits is equal to the multiplicative order of 10 mod d."""
best, index = 0, 0
for x in [i for i in range(10,1000) if is_prime(i)]:
y = mult_order_mod(x)
if y > best:
index = x
best = y
print(index)
def main ():
"""
"""
#
primes = [i for i in range(10**3,10**4) if is_prime(i) and '0' not in str(i)]
#print primes
for p in primes:
perms = sorted(list(set([int(''.join(perm)) for perm in permutations(str(p)) if is_prime(int(''.join(perm)))])))
if len(perms) < 3:continue
diffDiff = [b-a for a,b in pairwise([y-x for x,y in pairwise(perms)])]
if 0 in diffDiff:
print p,perms
print p,diffDiff
print
pass
def FindPrimeAndTwiceSquare(n):
max_twice = math.sqrt(n)
print n, max_twice
for i in range(1, int(max_twice)):
if (2 * (i **2)) > n:
break
prime = n - (2 * (i ** 2))
print prime, n
if (is_prime(prime)) == True:
print i, "=", prime, "+", "2 x ", i, "^2"
return True
#else :
# print "not....",i, "=", prime, "+", "2 x ", i, "^2"
return False
def FindPrimeAndTwiceSquare(n):
max_twice = math.sqrt(n)
print n, max_twice
for i in range(1, int(max_twice)):
if (2 * (i**2)) > n:
break
prime = n - (2 * (i**2))
print prime, n
if (is_prime(prime)) == True:
print i, "=", prime, "+", "2 x ", i, "^2"
return True
#else :
# print "not....",i, "=", prime, "+", "2 x ", i, "^2"
return False
def main ():
"""
"""
maxN = 10**6+1
maxConsecTerms = 0
maxPrime = 0
# make a list of primes
primes = [i for i in range(2,maxN) if is_prime(i)]
for j,p in enumerate(primes):
print j/780.
# make a list of the candidate primes
cand = [c for c in primes[:j]]
# starting at each prime in the candidate list...
consecTerms=0
for ind,c in enumerate(cand):
candSum = 0
i = ind
while candSum < p and i < len(cand):
candSum += cand[i]
i += 1
# aggregate terms until they equal or exceed the target prime
#print p,c,candSum
consecTerms = i-ind
if p == candSum:
if consecTerms > maxConsecTerms:
maxPrime = p
maxConsecTerms = consecTerms
#print p,i-ind #,sum(cand[ind:i]),cand[ind:i]
break
continue
print '****',maxPrime,maxConsecTerms
print 'And the answer is:',maxPrime,'with a total of',maxConsecTerms,'consecutive terms.'
pass
def CirPrime(num):
buf = str(num)
len_buf = len(buf)
devide = 1
for i in range (0, len_buf-1):
devide *= 10
new_num = num
for i in range(0, len_buf):
#print new_num
if (is_prime(new_num)) :
d1 = new_num / devide
d2 = new_num % devide
new_num = d2 * 10 + d1
else :
return -1
return 0
def E69():
res = 0
max_res = 0
max_i = 0
for i in range (2, MAXNUM):
if i % 2 == 1:
continue;
if is_prime(i) == True:
continue
cnt = 1
for j in range (1, i):
if gcd(i, j) == 1:
cnt+=1
if cnt != 0:
res = float(i)/float(cnt)
if max_res < res:
max_res = res
max_i = i
print max_res, max_i
return 0
def E69():
res = 0
max_res = 0
max_i = 0
for i in range(2, MAXNUM):
if i % 2 == 1:
continue
if is_prime(i) == True:
continue
cnt = 1
for j in range(1, i):
if gcd(i, j) == 1:
cnt += 1
if cnt != 0:
res = float(i) / float(cnt)
if max_res < res:
max_res = res
max_i = i
print max_res, max_i
return 0
chart[(i, j)] = result
return result
else:
result = sum_of_primes(i - 1, j) - primes[i - 1]
chart[(i, j)] = result
return result
(highest, num_of_terms) = (41, 6)
starting_prime_index = 0
for i in xrange(0, 4): #index of the starting prime
m = num_of_terms
for j in xrange(i + m, 550): #index of the ending prime
the_sum = sum_of_primes(i, j)
if the_sum > 1000000: break #increment here to test
if is_prime(the_sum):
highest = the_sum
num_of_terms = j - i
starting_prime_index = i
print highest, "is the sum"
print num_of_terms, "terms in the sum"
print "The sum starts at the prime", primes[starting_prime_index]
print time.clock() - t_0, " seconds to complete"
##k=1000000
##while not is_prime(k):
## k -= 1
##
##print k, "has index", primes.index(k)
##
chart[(i,j)] = result
return result
else:
result = sum_of_primes(i-1,j) - primes[i-1]
chart[(i,j)] = result
return result
(highest, num_of_terms) = (41,6)
starting_prime_index = 0
for i in xrange(0,4): #index of the starting prime
m = num_of_terms
for j in xrange(i+m,550): #index of the ending prime
the_sum = sum_of_primes(i,j)
if the_sum > 1000000: break #increment here to test
if is_prime(the_sum):
highest = the_sum
num_of_terms = j-i
starting_prime_index = i
print highest, "is the sum"
print num_of_terms, "terms in the sum"
print "The sum starts at the prime", primes[starting_prime_index]
print time.clock() - t_0, " seconds to complete"
##k=1000000
##while not is_prime(k):
## k -= 1
##
#27. Come up with a formula n^2 + an + b for |a|,|b|<=1000 which
#generates the most number of consecutive primes starting from n=0.
## Here's the output:
##-59231 is the product of a and b
##70 is the maximum consecutive number of primes
##8.589944 seconds to complete
from Euler import is_prime
import time
import cProfile
t_0 = time.clock()
max_consecutive = 0
for a in xrange(-1000,1001):
for b in xrange(-1000,1001):
for n in xrange(80):
if not is_prime(n**2+a*n+b):
break
elif n > max_consecutive:
max_consecutive = n+1
if max_consecutive==70:
print a*b, " is the product of a and b"
print max_consecutive, " is the maximum consecutive number of primes"
print time.clock()-t_0, " seconds to complete"
from Euler import is_prime
import math
def FindPrimeAndTwiceSquare(n):
max_twice = math.sqrt(n)
print n, max_twice
for i in range(1, int(max_twice)):
if (2 * (i **2)) > n:
break
prime = n - (2 * (i ** 2))
print prime, n
if (is_prime(prime)) == True:
print i, "=", prime, "+", "2 x ", i, "^2"
return True
#else :
# print "not....",i, "=", prime, "+", "2 x ", i, "^2"
return False
MAX_LIMIT = 1000000
count = 0
for i in range(3, MAX_LIMIT, 2):
if (is_prime(i) == True):
continue
if (FindPrimeAndTwiceSquare(i) == False):
print "(DONE) result = [",i,"]"
exit(0)
#!/usr/bin/python -tt
import sys
sys.path.append('/Users/admin/Desktop/python-work/Project_Euler/primes')
from Euler import prime_sieve, is_prime
nmax = 0;
primes = prime_sieve(1000)
for a in range(-999,999,2):
for b in primes:
n = 1
while is_prime(n**2 + a*n + b): n += 1
if n>nmax: nmax, p = n, a*b
print "Answer to PE27 = ",p
def main():
print 'hello'
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
from Euler import prime_sieve, is_prime
max = 1000000
primes = prime_sieve(max)
prime_sum = [0]
sum = 0
count = 0
while (sum<terms and is_prime(n)):
(terms, max_prime) = (j-i, n)
print "Project Euler 50 Solution = ", max_prime, " with ", terms, " terms"
def right_prime(n): if len(str(n)) ==1:return is_prime(n) return is_prime(n) and right_prime(int(str(n)[1:]))
from Euler import is_prime
num_diag_primes = 3
step_length = 2
corner = 9
while float(num_diag_primes) / (2 * step_length + 1) > 0.1:
step_length += 2
for i in range(0, 3):
corner += step_length
if is_prime(corner):
num_diag_primes += 1
corner += step_length
print(step_length + 1)
# SOLVED : 26241
def genPrimes(maxN):
primes = []
for n in range(2, maxN + 1):
if is_prime(n):
primes.append(n)
return primes
def left_prime(n): if len(str(n)) ==1:return is_prime(n) return is_prime(n) and left_prime(int(str(n)[:-1]))
BUF[a4] = 1
for a5 in range(1, 8):
if (BUF[a5] == 1):
continue
BUF[a5] = 1
for a6 in range(1, 8):
if (BUF[a6] == 1):
continue
BUF[a6] = 1
for a7 in range(1, 8):
if (BUF[a7] == 1):
continue
count += 1
buf = "%d%d%d%d%d%d%d" % (a1, a2, a3, a4, a5, a6,
a7)
if (is_prime(int(buf)) == True):
print "prime = ", buf
BUF[a7] = 0
BUF[a6] = 0
BUF[a5] = 0
BUF[a4] = 0
BUF[a3] = 0
BUF[a2] = 0
'''
for a1 in range(1, 10):
BUF = [0,0,0,0,0,0,0,0,0,0]
BUF[a1] = 1
for a2 in range(1, 10):
if (BUF[a2] == 1):
continue
BUF[a2] = 1
#!/usr/bin/python
from Euler import is_prime
from time import clock
# start a process timer
clock()
# setup a counter
max = 0
# get a list of primes
z = [n for n in range(10**6+1) if is_prime(n)]
lenght = len(z)
for a in range(0, lenght-1):
#
t = 0
sum = 0
wyrazy = 0
#
for i in range(a, lenght-1):
sum = sum + z[i]
if sum > 10**6:
break
wyrazy += 1
if is_prime(sum) and sum != z[i] and wyrazy > max:
r = sum
max = wyrazy
print(r)
#!/usr/bin/python
from Euler import is_prime
from sys import stdout
print "{ ",
for n in range(2,10**6):
if is_prime(n) :
stdout.write("'%s': 1, " % (str(n)))
print "}"
def test(n):
for factor in divisors(n):
if not is_prime(factor + (n / factor)):
return False
return True
'''
What is the side length of the square spiral for which the ratio of primes
along both diagonals first falls below 10%?
'''
from Euler import is_prime, itertools
# consider a layer N
# for example, N = 3
# ....... .......
# .NNNNN. S .C...C.
# .N...N. S ..XXXX.
# .N...N. => S => ..XXXX.
# .N...N. S ..XXXX.
# .NNNNN. S .C.....
# ....... .......
count = 0
for N in itertools.count(start=2):
S = 2*N-1
X = (S-1)*(S-2)
i = X+1
count += sum(is_prime(i+j*(S-1)) for j in range(3))
total = 4*N-3
if count/total < 0.1:
break
ans = S
print(ans)
def project_euler_3():
'''Prints the largest prime factor of 600851475143.
See the is_prime method in Euler.py for more details.'''
print(max(i for i in range(1, int(600851475143**.5) +1) if 600851475143 % i == 0 and is_prime(i)))
<i>n</i>² + <i>an</i> + <i>b</i>, where |<i>a</i>| < 1000 and |<i>b</i>| < 1000<br /><br />
<div class="info" style="text-align:left;">where |<i>n</i>| is the modulus/absolute value of <i>n</i><br />e.g. |11| = 11 and |−4| = 4</div>
</blockquote>
<p>Find the product of the coefficients, <i>a</i> and <i>b</i>, for the quadratic expression that produces the maximum number of primes for consecutive values of <i>n</i>, starting with <i>n</i> = 0.</p>
</div><br />
<br /></div>
"""
from Euler import is_prime
max_cnt = 0
max_a = 0
max_b = 0
for a in range(-1000, 1001):
for b in range(-1000, 1001):
n = 0
while (True):
result = (n * n) + (a * n) + b
if (is_prime(result)):
n += 1
else:
if (max_cnt < n):
max_cnt = n
max_a = a
max_b = b
break
print max_cnt, max_a, max_b
print "result=", (max_a * max_b)
#!/usr/bin/python -Wall
# -*- coding: utf-8 -*-
"""
<div id="content">
<div style="text-align:center;" class="print"><img src="images/print_page_logo.png" alt="projecteuler.net" style="border:none;" /></div>
<h2>Largest prime factor</h2><div id="problem_info" class="info"><h3>Problem 3</h3><span>Published on Friday, 2nd November 2001, 06:00 pm; Solved by 281868; Difficulty rating: 5%</span></div>
<div class="problem_content" role="problem">
<p>The prime factors of 13195 are 5, 7, 13 and 29.</p>
<p>What is the largest prime factor of the number 600851475143 ?</p>
<!--
Note: This problem has been changed recently, please check that you are using the right number.
-->
</div><br />
<br /></div>
"""
from Euler import is_prime
#limit = 13195
limit = 600851475143
i = 3
max_num = 0
for i in range(i, 10000):
if (is_prime(i)):
if (limit % i == 0):
temp = limit // i
max_num = i
print "max =", max_num
import math
def FindPrimeAndTwiceSquare(n):
max_twice = math.sqrt(n)
print n, max_twice
for i in range(1, int(max_twice)):
if (2 * (i**2)) > n:
break
prime = n - (2 * (i**2))
print prime, n
if (is_prime(prime)) == True:
print i, "=", prime, "+", "2 x ", i, "^2"
return True
#else :
# print "not....",i, "=", prime, "+", "2 x ", i, "^2"
return False
MAX_LIMIT = 1000000
count = 0
for i in range(3, MAX_LIMIT, 2):
if (is_prime(i) == True):
continue
if (FindPrimeAndTwiceSquare(i) == False):
print "(DONE) result = [", i, "]"
exit(0)
continue
BUF[a4] = 1
for a5 in range(1, 8):
if (BUF[a5] == 1):
continue
BUF[a5] = 1
for a6 in range(1, 8):
if (BUF[a6] == 1):
continue
BUF[a6] = 1
for a7 in range(1, 8):
if (BUF[a7] == 1):
continue
count += 1
buf = "%d%d%d%d%d%d%d" % (a1,a2,a3,a4,a5,a6,a7)
if (is_prime(int(buf)) == True):
print "prime = ", buf
BUF[a7] = 0
BUF[a6] = 0
BUF[a5] = 0
BUF[a4] = 0
BUF[a3] = 0
BUF[a2] = 0
'''
for a1 in range(1, 10):
BUF = [0,0,0,0,0,0,0,0,0,0]
BUF[a1] = 1
for a2 in range(1, 10):
if (BUF[a2] == 1):
def r_p(p):
for i in trunc_r(p):
if not is_prime(i):
return False
return True
from Euler import prime_sieve, is_prime
max = 1000000
primes = prime_sieve(max)
prime_sum = [0]
sum = 0
count = 0
while (sum < terms and is_prime(n)):
(terms, max_prime) = (j - i, n)
print "Project Euler 50 Solution = ", max_prime, " with ", terms, " terms"
def is_valid(my_list):
return all((is_prime(str(p[0]) + str(p[1]))) for p in perms(my_list, 2))
"""fact = lambda n: reduce (lambda x,y:x*y, range(1,n+1))
def mod10(n):
r = 0
for i in range(1,n+1):
f = fact(i)
r = (r + pow(i,f,10)) % 10
return r
while True:
a = raw_input()
if a == '#':break
print mod10(int(a))"""
from Euler import is_prime
if is_prime(23):
print 'hi'
else:
print 'no'
def genPrimes(maxN): primes = [] for n in range(2, maxN+1): if is_prime(n): primes.append(n) return primes
"""
<div id="content">
<div style="text-align:center;" class="print"><img src="images/print_page_logo.png" alt="projecteuler.net" style="border:none;" /></div>
<h2>Largest prime factor</h2><div id="problem_info" class="info"><h3>Problem 3</h3><span>Published on Friday, 2nd November 2001, 06:00 pm; Solved by 281868; Difficulty rating: 5%</span></div>
<div class="problem_content" role="problem">
<p>The prime factors of 13195 are 5, 7, 13 and 29.</p>
<p>What is the largest prime factor of the number 600851475143 ?</p>
<!--
Note: This problem has been changed recently, please check that you are using the right number.
-->
</div><br />
<br /></div>
"""
from Euler import is_prime
#limit = 13195
limit = 600851475143
i = 3
max_num = 0
for i in range (i, 10000):
if (is_prime(i)):
if (limit % i == 0):
temp = limit // i
max_num = i
print "max =", max_num
<div class="info" style="text-align:left;">where |<i>n</i>| is the modulus/absolute value of <i>n</i><br />e.g. |11| = 11 and |−4| = 4</div>
</blockquote>
<p>Find the product of the coefficients, <i>a</i> and <i>b</i>, for the quadratic expression that produces the maximum number of primes for consecutive values of <i>n</i>, starting with <i>n</i> = 0.</p>
</div><br />
<br /></div>
"""
from Euler import is_prime
max_cnt = 0
max_a = 0
max_b = 0
for a in range (-1000, 1001):
for b in range (-1000, 1001):
n = 0
while (True):
result = (n*n) + (a*n) + b
if (is_prime(result)):
n += 1
else:
if (max_cnt < n):
max_cnt = n
max_a = a
max_b = b
break
print max_cnt, max_a, max_b
print "result=", (max_a * max_b)