70 lines
2.0 KiB
Python
70 lines
2.0 KiB
Python
#!/usr/bin/python3
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# The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime,
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# and, (ii) each of the 4-digit numbers are permutations of one another.
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#
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# There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit
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# increasing sequence.
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#
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# What 12-digit number do you form by concatenating the three terms in this sequence?
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from numpy import zeros
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from timeit import default_timer
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from projecteuler import sieve
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def check_digits(a, b):
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digits1 = zeros(10, int)
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digits2 = zeros(10, int)
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while a > 0:
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digits1[a%10] = digits1[a%10] + 1
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a = a // 10
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while b > 0:
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digits2[b%10] = digits2[b%10] + 1
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b = b // 10
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for i in range(10):
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if digits1[i] != digits2[i]:
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return False
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return True
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def main():
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start = default_timer()
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N = 10000
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primes = sieve(N)
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found = 0
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i = 1489
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# Starting from i=1489 (bigger than the first number in the sequence given in the problem),
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# check odd numbers. If they're prime, loop on even numbers j (odd+even=odd, odd+odd=even and
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# we need odd numbers because we're looking for primes) up to 4254 (1489+2*4256=10001 which has
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# 5 digits.
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while i < N:
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if primes[i] == 1:
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for j in range(1, 4255):
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# If i, i+j and i+2*j are all primes and they have
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# all the same digits, the result has been found.
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if i + 2 * j < N and primes[i+j] == 1 and primes[i+2*j] == 1 and\
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check_digits(i, i+j) and check_digits(i, i+2*j):
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found = 1
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break
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if(found):
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break
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i = i + 2
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end = default_timer()
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print('Project Euler, Problem 49')
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print('Answer: {}'.format(str(i)+str(i+j)+str(i+2*j)))
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print('Elapsed time: {:.9f} seconds'.format(end - start))
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if __name__ == '__main__':
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main()
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