def s(p,S): '''return the number of primes when we replace all letters at positions corresponding to the set S ''' sp=str(p) count = 0 for s in S: if sp[s]!=sp[S[0]]:return -1 if 0 in S: lower=1 else: lower=0 for i in range(lower,10): sp1="" for j in range(0,len(sp)): if j in S: sp1 += str(i) else: sp1+=sp[j] if functions.isPrime(int(sp1)) : # print sp1 count +=1 return count
import functions as funcs
from time import time
primes1 = [x**2 for x in range(2, 7071) if funcs.isPrime(x)]
primes2 = [x**3 for x in range(2, 368) if funcs.isPrime(x)]
primes3 = [x**4 for x in range(2, 84) if funcs.isPrime(x)]
ctr = []
for i in primes1:
# if i % 1000 == 0:
# print i
for j in primes2:
for k in primes3:
num = i + j + k
if num >= 50000000:
break
ctr.append(num) #
# else:
# if num not in ctr:
# ctr.append(num)
a = time()
len(ctr)
print time() - a
a = time()
len(set(ctr))
print time() - a
a = time()
set(ctr).__len__()
print time() - a
def circularPrime(p): if '0' in str(p): return False for n in range(0, len(str(p))-1): p = nextCircularNumber(p) if not f.isPrime(p): return False return True
40 19 6 1 2 11 28
41 20 7 8 9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49
It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?
"""
import sys
sys.path.append("../")
import functions as f
primes = 0
nonePrimes = 1
ac = 1
p = 1
i = 0
while p > 0.1:
ac += 2
i += 1
a = pow(ac, 2)
for j in range(0,4):
n = a-(i*2)*j
if f.isPrime(n):
primes += 1
else:
nonePrimes += 1
p = primes / (nonePrimes+primes)
print(ac)
def isPrimePairs(p, q): n = int(str(p) + str(q)) m = int(str(q) + str(p)) if not f.isPrime(n) or not f.isPrime(m): return False; return True
def check(n):
n_str = str(n)
for i in range(len(n_str)):
if not isPrime(int(n_str[i:len(n_str)])):
return False
for i in range(1, len(n_str)):
if not isPrime(int(n_str[0:i])):
return False
return True
def truncatable(p):
s = str(p)
while p > 0:
if p > 2 and p % 2 == 0:
return False
p //= 10
if any(not isPrime(int(s[i:])) for i in range(0, len(s))):
return False
if any(not isPrime(int(s[:i])) for i in range(len(s), 0, -1)):
return False
return True
def fillP(): count, t = 0,2 while count < numberPrimes: if isPrime(t): P.append(t) count +=1 t+=1
def main():
start = t.time()
i = 9
while True:
if not f.isPrime(i) and not isComposite(i):
print(f"Found the answer, it is {i}")
print(f"This took {t.time() - start} seconds to run.")
return
i += 2
def main():
i = 9
while True:
if not isPrime(i):
if not goldbach_split(i):
print('-------->', i, '<--------')
break
else:
print(i)
i += 2
def main():
num = 2
index = 1
while True:
if isPrime(num):
print(index, '-->', num)
index += 1
if index == 10002:
break
num += 1
def countPrime(self):
count = 0
ptr = self.head
while ptr != None:
if isPrime(ptr.data):
count += 1
ptr = ptr.next
return count
def euler58():
i = 7
while True:
c = 0
d = diagonals(i)
for j in d:
if isPrime(j):
c+=1
per = 100.0*(c/(i*2-1))
print(f'{i}: {per}')
if per<10:
print(f'-->{i}<--')
print(i)
i+=2
def main():
start = t.time()
for n in range(1000, 10000):
if badNum(n):
continue
pg = f.permuteGen(n)
primes = []
for i in range(24):
perm = next(pg)
if f.isPrime(int(perm)):
primes.append(int(perm))
primes = set(primes)
if len(primes) >= 3 and 1487 not in primes and equiSpaced(primes):
print(f"This took {t.time() - start} seconds to run.")
return
primes.clear()
def euler58_():
lst = [3,5,7]
c = 3
for i in range(5,30001,2):
a = lst[-3]+(len(lst)//3*8+2)
b = len(lst)//3+1
lst.extend([a,a+2*b,a+4*b])
c+=len([
i
for i in [a,a+2*b,a+4*b]
if isPrime(i)
])
if (c/(i*2-1))<0.1:
print(f'-->{i}<--')
break
print(i)
def main():
num_diags = 1 # start with 1 cuz 1 is in the middle
num_primes = 0
side_len = 2
curr = 1
while True:
for i in range(4):
curr += side_len
num_diags += 1
if f.isPrime(curr):
num_primes += 1
if num_primes / num_diags < 0.1:
print(f"answer is {side_len+1}")
return side_len
side_len += 2
print(f"num diags is {num_diags}")
print(f"num primes is {num_primes}")
"""
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
"""
from functions import isPrime
arr = []
i = 1
while len(arr) < 10001:
if isPrime(i):
arr.append(i)
i += 1
print(arr.pop())
from functions import isPrime P = [] for a in range(2,400): if isPrime(a): P.append(a) print P def smallestPrime(pN): '''return the smallest prime divisor of pN''' for p in P: if pN % p == 0: return p return pN limit = 10**5 L = [0,1] #L[n] = rad(n) def fillL(): n = 1 while n < limit: n+=1 p = smallestPrime(n) t = n/p # print n,p,t if t % p == 0: L.append(L[t]) else: L.append(L[t]*p) print 'L is filled' fillL() def test(): for i in range(limit) : # print i,L[i]
return True
def no0(n):
return str(n).count("0") == 0
primes = list(i for i in primesUnder(10000) if noRepeats(i) and no0(i))
total = 0
for i in primes:
digits = "123456789"
for j in str(i):
digits = digits.replace(j, "")
for j in ["".join(p) for p in permutations(digits)]:
if isPrime(int(j)):
total += 1
print(i, int(j), total)
for i in range(len(primes) - 1):
for j in range(i + 1, len(primes)):
digits = "123456789"
p1 = primes[i]
p2 = primes[j]
digitCounts = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
valid = True
for k in str(p1) + str(p2):
digitCounts[int(k)] += 1
if max(digitCounts) > 1:
valid = False
break
def main():
n = 0
for i in range(1, int(600851475143**(1 / 2)) + 1):
if isPrime(i) and 600851475143 % i == 0:
print(i)
from functions import isPrime, primeSieve n=1000 #list all the primes up to n primeList=primeSieve(n) maxNumber=0 # a and b have a range of abs(a) and abs(b) < 1000 # this means we have to go from -999 to 999 for a in range(-n, n, 1): for b in primeList: x=1 while isPrime(x**2 + a*x + b): x+=1 if x>maxNumber: maxNumber=x prime=a*b print "Project Euler - Problem 27" print "Quadratic Primes Answer : ",prime
return newN
pl = f.primeList(1000000)
low = 999999999
for p in pl:
#print("Testing prime: %d" % p)
for i in range(1, len(str(p))):
lists = perm(i, p)
for l in lists:
#print("### %d ###" % p)
c = 0
cc = 0
for j in range(0,10):
cc += 1
#print(j,l,p)
primeTest = replace(j,l,p)
if len(str(primeTest)) == len(str(p)) and f.isPrime(primeTest):
c += 1
# print("%d:%d. Prime found: %d" % (c, cc, replace(j,l,p)))
if cc-c > 2:
break
if c == 8:
#print("PRIME: %d, nr: %d" % (p, c))
for j in range(0,10):
#print(j,l,p)
primeTest = replace(j,l,p)
if f.isPrime(primeTest):
print(primeTest)
exit(0)
#!/bin/py
"""
We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
What is the largest n-digit pandigital prime that exists?
"""
import sys
sys.path.append("../")
import functions as f
import itertools
t = 0
n = list("1")
while len(n) < 10:
perm = itertools.permutations(n)
for nr in perm:
t = int(''.join(map(str,nr)))
if f.isPrime(t):
answer = t
n.append(len(n)+1)
print(answer)
import math import functions # Modules in Python contains a list of related functions. # In Python, import can import a module. For example, math is a module. # Importing math module allows Python program to use various functions and variables in it. print(math.sqrt(81)) # sqrt returns square root of a number print(math.pi) # prints value of variable pi in math module print(math.sin(math.radians(30))) # Returns sine value of x. x must be in radians. radians() converts degrees to radians # Python files of same project can also be imported. print(functions.isPrime(23)) # Note the output of functions is printed when you import functions. # Python runs the imported modules when import runs. # Make sure not to import unused modules. import os print(os.getcwd()) # import can be used anywhere in program. But, it is recommended to import at the top of program to avoid any confusion.
def subPrimes(p):
out = []
for i in subString(str(p)):
if isPrime(int(i)):
out.append(int(i))
return out
# 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?
from functions import isPrime
import itertools
import string
def rotate(string):
x = string[1:]
y = string[:1]
return x + y
count = 1
for x in range(3, 1000000):
if isPrime(x):
length = len(str(x))
prime = True
a = str(x)
for i in range(1, length):
a = rotate(a)
if not isPrime(int(a)):
prime = False
if prime:
print x
count += 1
print "Project Euler"
print "Answer to Problem 35 : ", count
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
Which prime, below one-million, can be written as the sum of the most consecutive primes?
"""
import sys
sys.path.append("../")
sys.stdout.flush()
import functions as f
limit = 1000000
pl = f.primeList(limit)
t = []
i = 0
j = 0
mL = 0
mV = 0
for i in range(0, len(pl)):
for j in range(i+1, len(pl)):
s = sum(pl[i:j])
if s > limit:
break
if f.isPrime(s) and s < limit and j-i > mL:
t.append(sum(pl[i:j]))
mL = j-i
mV = sum(pl[i:j])
print(mL)
print(mV) #answer
def test(n):
for j in range(1, int(sqrt(n / 2)) + 1):
val = n - 2 * pow(j, 2)
if isPrime(val):
return True
return False
for j in range(10):
add = 0
for x in range(7):
if segments[i][x] == "1" and segments[j][x] == "1":
add += 1
new.append(add)
common.append(new)
def difference(n):
r = 0
lastNum = n
x = digitSum(n)
while len(str(lastNum)) > 1:
temp = str(lastNum)
temp = temp[0 - len(str(x)):]
for a in range(len(temp)):
r += common[int(str(x)[a])][int(temp[a])] * 2
lastNum = x
x = digitSum(x)
return r
tot = 0
for a in range(10**7 + 1, 2 * 10**7, 2):
if isPrime(a):
d = difference(a)
tot += d
print(a, d, tot)
print(tot)
def isCircularPrime(n): n = str(n) for start in range(len(n)): if not isPrime(int(n[start:len(n)]+n[0:start])): return False return True
n² + an + b, where |a| < 1000 and |b| < 1000
where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |−4| = 4
Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.
"""
import sys
sys.path.append("../")
import functions as f
import math as m
answer = [0, 0, 0]
for a in range(-1000, 1000):
for b in range(-1000, 1000):
n = 0
while f.isPrime(m.pow(n,2) + a*n + b):
n += 1
if n > answer[2]:
answer = (a, b, n)
print(answer[0] * answer[1])
return True
def subString(s):
n = len(s)
out = set()
for Len in range(1, n + 1):
for i in range(n - Len + 1):
j = i + Len - 1
sub = ''
for k in range(i, j + 1):
sub += s[k]
out.add(sub)
return out
def subPrimes(p):
out = []
for i in subString(str(p)):
if isPrime(int(i)):
out.append(int(i))
return out
for i in range(1, 100000, 2):
if len(subPrimes(i)) > 4:
if all(isPrime(n[0]) for n in getPermutations(subPrimes(i))):
print(i)
print(sum(subPrimes(i)))
print()
def fillP(): '''filling P''' for i in range(2,limit): if functions.isPrime(i): P.append(i) print "filling P complete"
def fillP(): for i in range(2,limit2): if functions.isPrime(i): P.append(i) print 'P is filled', len(P)
from functions import sieve, isPrime
primes = sieve(1000)[1:]
def sum_of_digits(n):
return sum([int(i) for i in str(n)])
for idx, i in enumerate(primes):
# if idx == 1:
# break
s = set()
for j in primes:
if isPrime(int(str(j) + str(i))) and isPrime(int(str(i) + str(j))):
s.add(j)
print(s)
# s = set([j for j in primes if isPrime(int(str(j)+str(i))) and isPrime(int(str(i)+str(j)))])
# s = set(lst)
# print(lst)
# for j in s:
# if len(set([k for k in primes if isPrime(int(str(k)+str(j))) and isPrime(int(str(j)+str(k)))]))>len(lst):
# lst = [j]+[k for k in primes if isPrime(int(str(k)+str(j))) and isPrime(int(str(j)+str(k)))]
def fillP() : for i in range(2, 10**3): if functions.isPrime(i): P.append(i) print 'P is filled'
def main():
for i in range(10001, 10000000, 2):
if not same_digits(i) and i % 10 != 0:
if isPrime(i) and isPadnigital(i):
print(i)
n = str(n)
for i in range(1, len(n)):
if direction == 'r':
r.append(int(n[:-i]))
elif direction == 'l':
r.append(int(n[i:]))
elif direction == 'b':
r.append(int(n[i:]))
r.append(int(n[:-i]))
return r
solution = 0
primes = file('Primzahlen.txt', 'r')
count = 11
for n_ in primes:
n = int(n_[:-1])
if n < 10:
continue
if all([isPrime(i) for i in truncatable(n, 'b')]):
print n
solution += n
count -= 1
if count == 0:
break
print(solution)
def find_ns(a, b):
n = 0
while isPrime(n**2 + a * n + b):
n += 1
return n