from primes import mr
import time
from itertools import permutations
import string
START = time.time()
#sum from 1 to 8 = 36, sum from 1 to 9 is 45, so if we went from 1 to 8 or 1 to 9, it would be divisible by 3.
permutation = permutations(['1', '2', '3', '4', '5', '6', '7'], 7)
maxz = 0
for perm in permutation:
n = int(string.join(perm, ''))
if n > maxz and mr(n):
maxz = n
print maxz
print "Time Taken:", time.time() - START
import time
from primes import mr
start = time.time()
lst = [i**3 for i in xrange(1,578)]
candidates = set()
for i in xrange(1,len(lst)):
for j in xrange(i-1,-1,-2):
candidates.add(lst[i]-lst[j])
if lst[i] -lst[j] > 10**6:
break
print len(candidates)
count = 0
for val in candidates:
if mr(val):
count+= 1
print count
print "Time Taken: ", time.time() - start
"""
~/Desktop/python_projects/proj_euler $python prob131.py
4527
173
Time Taken: 0.131557941437
We only need to go up to 577
n^2(n+p)
n is cube (since n+p can't turn n^2 into a cube)
import time
START = time.time()
from primes import mr
mult = 1111111111
successes = set()
sumz = 0
for big_dig in xrange(9,0,-1):
n = big_dig * mult
for dig in xrange(0,10):
for diff in xrange(big_dig-9,big_dig+1):
m = n - diff*10**dig
if mr(m) and m > 10**9:
successes.add(big_dig)
sumz += m
print "Digits we've seen:", successes
print "Time taken:", time.time() - START
for big_dig in xrange(9,0,-1):
if big_dig in successes:
continue
n = big_dig * mult
for dig,dig2 in ((a,b) for a in xrange(10) for b in xrange(a)):
for diff,diff2 in ((d1,d2) for d1 in xrange(big_dig-9,big_dig+1) for d2 in xrange(big_dig-9,big_dig+1)):
m = n - diff*10**dig - diff2*10**dig2
if mr(m) and m > 10**9:
successes.add(big_dig)
sumz += m
print "Digits we've seen:", successes
print "Time taken:", time.time() - START
import time
START = time.time()
from primes import mr
sumz = 0
count = 0
SIZE = 150 * 10**6
not_p = [11, 17, 19, 21, 23]
for i in xrange(3, 28):
a = i**2
if all([mr(k) for k in [a + 1, a + 3, a + 7, a + 9, a + 13, a + 27]]):
count += 1
sumz += i
print i
for i in xrange(30, SIZE, 10):
if i % 3 == 0 or i % 7 == 0 or i % 13 == 0:
continue
a = i**2
if all([mr(k) for k in [a + 1, a + 3, a + 7, a + 9, a + 13, a + 27]]):
if not any([mr(a + j) for j in not_p]):
count += 1
sumz += i
print i
print sumz, count
print "Time Taken:", time.time() - START
"""
676333270 12
Time Taken: 226.555887938
from primes import mr
import time
START = time.time()
#the four diagonals are 4n^2-2n+1, 4n^2+1,4n^2+2n+1,
# and (2n+1)^2
count = 0
for i in range(1, 100000):
a = 4 * i * i + 1
j = 2 * i
count += mr(a + j) + mr(a) + mr(a - j)
if count * 1.0 / (4 * i + 1) < .1:
break
print j + 1, count / (j * 2.0 + 1)
print "Time taken:", time.time() - START
"""
LOL. I revisited this problem on 1/20/2013. I ran it with the original prime checking algorithm, then I ran it with the miller rabin primality test.
Original time:
Time taken: 31.7114129066
New time:
Time taken: 0.593687772751
looks like I cleaned up my code a bit more since then haha... 3/25/14
Time taken: 0.185644865036
"""
a_lst = range(-1, -1000, -2)
b_lst = primes[12:168]
a_lst2 = []
b_lst2 = []
maxz = 0
for a in a_lst:
for b in b_lst:
n = 1
while True:
if n**2 + n * a + b < 0: # We're no longer dealing with positive composite numbers
break
elif not mr(
n
): # Honestly, I don't remember why I had this condition...
n += 1
elif mr(n**2 + n * a + b): #If the result is a prime, keep going
n += 1
else:
break
if 40 < n:
maxz = n - 1
#print maxz, a, b
a_lst2 += [a]
b_lst2 += [b]
#print 'ON TO PHASE 2. YAY!'
maxz = 0
max_values = (0, 0)
import time, string
START = time.time()
from itertools import permutations
from primes import mr
digits = [str(i) for i in range(1, 10)]
prime_sets = [set() for i in range(10)]
SIZE_LIM = 9
for num_dig in range(1, SIZE_LIM):
for perm in permutations(digits, num_dig):
next_num = int(string.join(perm, ''))
if mr(next_num):
prime_sets[num_dig].add(next_num)
print num_dig, len(prime_sets[num_dig])
prime_lst = sorted(prime_sets)
print "Time Taken:", time.time() - START
def get_next(seen, max_size, last_seen):
if max_size == 0 and len(seen) == 9:
return 1
elif max_size == 0:
return 0
sumz = 0
for prime in prime_sets[max_size]:
if prime > last_seen: #Since we're not requiring that the num_digs goes down, we still need some ordering to avoid double counting.
continue
new_set = set(seen)
import time, string
START = time.time()
from itertools import permutations
from primes import mr
digits = [str(i) for i in range(1,10)]
prime_sets = [set() for i in range(10)]
SIZE_LIM = 9
for num_dig in range(1,SIZE_LIM):
for perm in permutations(digits,num_dig):
next_num = int(string.join(perm,''))
if mr(next_num):
prime_sets[num_dig].add(next_num)
print num_dig, len(prime_sets[num_dig])
prime_lst = sorted(prime_sets)
print "Time Taken:", time.time() - START
def get_next(seen,max_size, last_seen):
if max_size == 0 and len(seen) == 9:
return 1
elif max_size == 0:
return 0
sumz = 0
for prime in prime_sets[max_size]:
if prime > last_seen: #Since we're not requiring that the num_digs goes down, we still need some ordering to avoid double counting.
continue
new_set = set(seen)
import time
from primes import mr
start = time.time()
lst = [i**3 for i in xrange(1, 578)]
candidates = set()
for i in xrange(1, len(lst)):
for j in xrange(i - 1, -1, -2):
candidates.add(lst[i] - lst[j])
if lst[i] - lst[j] > 10**6:
break
print len(candidates)
count = 0
for val in candidates:
if mr(val):
count += 1
print count
print "Time Taken: ", time.time() - start
"""
~/Desktop/python_projects/proj_euler $python prob131.py
4527
173
Time Taken: 0.131557941437
We only need to go up to 577
n^2(n+p)
n is cube (since n+p can't turn n^2 into a cube)
p+n is cube
from primes import mr
import time
START = time.time()
#the four diagonals are 4n^2-2n+1, 4n^2+1,4n^2+2n+1,
# and (2n+1)^2
count = 0
for i in range(1,100000):
a = 4 * i*i +1
j = 2* i
count += mr(a+j) + mr(a) + mr(a-j)
if count * 1.0 / (4*i+1) < .1:
break
print j+1, count / (j*2.0+1)
print "Time taken:", time.time() - START
"""
LOL. I revisited this problem on 1/20/2013. I ran it with the original prime checking algorithm, then I ran it with the miller rabin primality test.
Original time:
Time taken: 31.7114129066
New time:
Time taken: 0.593687772751
looks like I cleaned up my code a bit more since then haha... 3/25/14
Time taken: 0.185644865036
"""
a_lst = range(-1,-1000,-2)
b_lst = primes[12:168]
a_lst2 = []
b_lst2 = []
maxz = 0
for a in a_lst:
for b in b_lst:
n = 1
while True:
if n**2 + n*a + b < 0: # We're no longer dealing with positive composite numbers
break
elif not mr(n): # Honestly, I don't remember why I had this condition...
n+=1
elif mr(n**2 + n*a + b): #If the result is a prime, keep going
n+=1
else: break
if 40 < n:
maxz = n-1
#print maxz, a, b
a_lst2 += [a]
b_lst2 += [b]
#print 'ON TO PHASE 2. YAY!'
maxz = 0
max_values = (0,0)
for i in range(0,len(a_lst2)):
n = 1
import sys
size = 10**6
if len(sys.argv) > 1:
size = int(sys.argv[1])
lst = bitarray('0' * (size+1))
count = 0
for i in xrange(2,size+1):
if i%10240 == 0:
print i
if lst[i] == 0:
if not mr(2*i*i - 1):
if i < size//500:
a = set(factor(2*i**2-1)[:-1])
for j in a:
for k in xrange((-i)%j+j,size+1,j):
lst[k] = 1
for k in xrange(i,size+1,j):
lst[k] = 1
count +=1
else:
count +=1
print size -1 - count
print "Time Taken:", time.time() - start
from primes import mr
import time
from itertools import permutations
import string
START = time.time()
#sum from 1 to 8 = 36, sum from 1 to 9 is 45, so if we went from 1 to 8 or 1 to 9, it would be divisible by 3.
permutation = permutations(['1','2','3','4','5','6','7'],7)
maxz = 0
for perm in permutation:
n = int(string.join(perm, ''))
if n > maxz and mr(n):
maxz = n
print maxz
print "Time Taken:", time.time() -START