Daniele Fucini
3a09f3d4f5
Added solutions for problem 66, 67, 68, 69 and 70 in C and python. Also, minor improvements to the code for a few older problems.
52 lines
1.7 KiB
Python
52 lines
1.7 KiB
Python
#!/usr/bin/python3
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# Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of positive numbers
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# less than or equal to n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and
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# relatively prime to nine, φ(9)=6.
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# The number 1 is considered to be relatively prime to every positive number, so φ(1)=1.
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#
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# Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180.
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#
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# Find the value of n, 1 < n < 10^7, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.
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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, is_semiprime, phi_semiprime
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def main():
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start = default_timer()
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N = 10000000
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n = -1
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min_ = float('inf')
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semi_p = False
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primes = sieve(N)
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for i in range(2, N):
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# When n is prime, phi(n)=(n-1), so to minimize n/phi(n) we should
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# use n prime. But n-1 can't be a permutation of n. The second best
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# bet is to use semiprimes. For a semiprime n=p*q, phi(n)=(p-1)(q-1).
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# So we check if a number is semiprime, if yes calculate phi, finally
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# check if phi(n) is a permutation of n and update the minimum if it's
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# smaller.
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semi_p, a, b = is_semiprime(i, primes)
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if semi_p == True:
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p = phi_semiprime(i, a, b)
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if ''.join(sorted(str(p))) == ''.join(sorted(str(i))) and i / p < min_:
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n = i
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min_ = i / p
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end = default_timer()
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print('Project Euler, Problem 70')
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print('Answer: {}'.format(n))
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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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