Improve solution for problem 47
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@ -14,65 +14,46 @@
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# Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?
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from timeit import default_timer
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from projecteuler import sieve
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def count_distinct_factors(n):
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global primes
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count = 0
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# Function using a modified sieve of Eratosthenes to count
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# the prime factors of each number.
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def count_factors(n):
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factors = [0] * n
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i = 2
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# Start checking if 2 is a prime factor of n. Then remove
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# all 2s factore.
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if n % 2 == 0:
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count = count + 1
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while i < n // 2:
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if factors[i] == 0:
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for j in range(i, n, i):
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factors[j] = factors[j] + 1
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i = i + 1
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while True:
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n = n // 2
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if n % 2 != 0:
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break
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i = 3
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# Check all odd numbers i, if they're prime and they're a factor
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# of n, count them and then divide n for by i until all factors i
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# are eliminated. Stop the loop when n=1, i.e. all factors have
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# been found.
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while n > 1:
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if primes[i] == 1 and n % i == 0:
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count = count + 1
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while True:
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n = n // i
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if n % i != 0:
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break
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i = i + 2
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return count
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return factors
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def main():
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start = default_timer()
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global primes
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N = 150000
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primes = sieve(N)
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found = 0
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i = 647
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factors = count_factors(N)
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# Starting from 647, count the distinct prime factors of n, n+1, n+2 and n+3.
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# If they all have 4, the solution is found.
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while not found and i < N - 3:
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if primes[i] == 0 and primes[i+1] == 0 and primes[i+2] == 0 and primes[i+3] == 0:
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if count_distinct_factors(i) == 4 and count_distinct_factors(i+1) == 4 and count_distinct_factors(i+2) == 4 and count_distinct_factors(i+3) == 4:
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found = 1
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i = i + 1
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count = 0
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# Find the first instance of four consecutive numbers
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# having four distinct prime factors.
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for i in range(N):
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if factors[i] == 4:
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count = count + 1
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else:
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count = 0
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if count == 4:
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res = i - 3
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break
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end = default_timer()
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print('Project Euler, Problem 47')
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print('Answer: {}'.format(i-1))
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print('Answer: {}'.format(res))
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print('Elapsed time: {:.9f} seconds'.format(end - start))
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