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72 lines
2.2 KiB
Python
72 lines
2.2 KiB
Python
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"""Find the value of d < 1000 for which 1/d contains the longest recurring cycle.A unit fraction contains 1 in the numerator.
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The decimal representation of the unit fractions with denominators 2 to 10 are given:
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1/2 = 0.5
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1/3 = 0.(3)
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1/4 = 0.25
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1/5 = 0.2
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1/6 = 0.1(6)
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1/7 = 0.(142857)
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1/8 = 0.125
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1/9 = 0.(1)
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1/10 = 0.1
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Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
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Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
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"""
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from decimal import Decimal
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from e003 import pfactor
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def coprime(a, b):
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"""Are integers a and b coprime?"""
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primes_a = pfactor(a)
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primes_b = pfactor(b)
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primes = [p for p in primes_a if p in primes_b]
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return bool(primes)
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def period_length(numerator, denominator):
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"""Determine the period length of a non-terminating decimal"""
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length = None
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numerator = Decimal(numerator)
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denominator = Decimal(denominator)
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if numerator == 1:
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if denominator == 1:
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return 1
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primes = pfactor(int(denominator))
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"""
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From Wikipedia:
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Terminating decimals represent rational numbers of the form
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k/2^n5^m
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However, a terminating decimal also has a representation as a
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repeating decimal, obtained by decreasing the final (nonzero)
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digit by one and appending an infinitely repeating sequence of
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nines. 1 = 0.999999... and 1.585 = 1.584999999... are two
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examples of this.
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"""
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if len([p for p in primes if p not in [2, 5]]) == 0:
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return 1
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if coprime(denominator, 10):
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pass
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else:
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raise Exception('Non-reciprocal period detection not yet implemented')
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return length
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def main():
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MAX = 10
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longest_cycle = 0
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for i in xrange((MAX), 1, -1):
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# The period of 1/k for integer k is always <20> k ? 1.
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if longest_cycle >= i:
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break
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cycle = period_length(1, i)
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fraction = Decimal(1) / Decimal(i)
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print '1/{0} = {1} ({2})'.format(i, fraction, cycle), fraction.is_subnormal()
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if __name__ == '__main__':
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main()
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