def permutePrimes(prime, character):
# Replaces each instance of character in prime with 0-9 and
# returns a list of each resulting number that is also prime
digits = [x for x in range(10)]
digits = list(filter(lambda x: x != character, digits))
answers = [prime]
for x in digits:
testnum = replaceDigit(prime, character, x)
if Euler.isPrime(testnum) == True:
if len(str(testnum)) == len(str(prime)):
answers.append(testnum)
return answers
def main(limit): # Iterates through each number less than limit testing for primeness. # Counter is multiplied by each prime number for the a number of times # equal to the greatest number of times that prime divides a number less # than or equal to the limit. For example, 2 divides 8 3 times (8 = 2*2*2), # so when limit == 10 counter is multiplied by 2^3. counter = 1 for x in range(2, limit+1): if Euler.isPrime(x): counter *= (x ** math.floor(math.log(limit, x))) return counter
def naiveResilience(num):
if(num==1): return 0/num
if(Euler.isPrime(num,Euler.currPrimes)): return (num-1)/num
count = 1
for i in range(2,num):
hit = False
b = math.sqrt(num)
for p in Euler.currPrimes:
if(p>b): break
if(i%p==0 and num%p==0):
hit = True
break
if(not hit): count+=1
return count/num
# Problem 58: Spiral primes
import time
import Euler
t1 = time.clock()
diagonal = 1
primes = []
non_primes = [1]
side_length = 3
ratio = 1
while ratio > 1/10: # Change to: while found = False
for x in range(4):
diagonal += side_length - 1
if Euler.isPrime(diagonal) == True:
primes.append(diagonal)
else:
non_primes.append(diagonal)
ratio = len(primes) / (len(primes) + len(non_primes))
side_length += 2
t2 = time.clock()
print("Side length = ", side_length - 2)
print(str(t2-t1)[:9])
def is_pair(x,y):
# Returns True if xy and yx are each prime, else returns False
test1 = int(str(x) + str(y))
test2 = int(str(y) + str(x))
return Euler.isPrime(test1) == True and Euler.isPrime(test2) == True
if is_pair(prime, candidate) == True:
pairs.append(candidate)
candidate -= 2
if len(pairs) > 1:
answer.append((len(pairs), pairs))
candidate = Euler.previous_prime(pairs[-1])
pairs = pairs[:-1]
if len(pairs) < 2:
finished = True
if len(answer) == 0:
prime_table[prime] = False
else:
prime_table[prime] = answer
for x in range(8, 700):
if Euler.isPrime(x) == True:
evaluate_prime(x)
print(prime_table[13])
t2 = time.clock()
print(str(t2-t1)[:5])
# Problem 50: Consecutive Prime Sum
import time
t1 = time.clock()
import Euler
limit = 4000
primes = [x for x in Euler.primeSieve(limit)]
answer = [(0,[])]
for x, prime1 in enumerate(primes):
startprime = x
sequence = []
counter = 0
for y, prime2 in enumerate(primes):
if y >= x:
counter += prime2
if Euler.isPrime(counter) == True:
if counter < 1000000:
sequentials = [x for x in primes[x:y+1]]
if len(answer[0][1]) < len(sequentials):
answer[0] = ((counter, sequentials))
print("Prime = " + str(answer[0][0]) + " ; Length of sequence: " + str(len(answer[0][1])))
print("Sequence start = " + str(answer[0][1][0]) + " ; Sequence end = " + str(answer[0][1][-1]))
t2 = time.clock()
print("Runtime = " + str(t2-t1)[:4] + " seconds")
