Problem 26 at Project Euler asks

A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:

^(1)/_(2) = 0.5

^(1)/_(3) = 0.(3)

^(1)/_(4) = 0.25

^(1)/_(5) = 0.2

^(1)/_(6) = 0.1(6)

^(1)/_(7) = 0.(142857)

^(1)/_(8) = 0.125

^(1)/_(9) = 0.(1)

^(1)/_(10) = 0.1

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.

Find the value of d < 1000 for which ^(1)/_(d) contains the longest recurring cycle in its decimal fraction part.

This is a problem about repeated division by the same number – how many divisions can you do before you start repeating yourself? The division algorithm can be expressed as follows:

def div_by(rem, div): y = min([x for x in range(5) if rem*(10**x) > div]) return rem*(10**y) % div

So then I need to use this to repeatedly divide by the test value d – len_pattern calls div_by until it detects a repeated remainder:

def len_pattern(init, rems, d): rem=div_by(init, d) if rem in rems: return len(rems) - rems.index(rem) else: rems.append(rem) return len_pattern(rem, rems, d)

I then have to use this to find out which d < 1000 has the longest cycle. However, I don't need to call it on all the numbers under 1000 – the only ones I really need to test are the primes. The upper bound on the length of the cycle is n-1 for any number n (for example the length of the cycle generated by 7 is 6), and primes are the only numbers which generate cycles of this length. So I need to use the list generated by the primes_under() function, then call len_pattern on each of these, returning the maximum length.

def get_longest(): from EulerFunc import primes_under divs = dict([(div, len_pattern(1.0, [], div)) \ for div in primes_under(1000) if div > 5]) longest=max(divs, key=lambda x: divs[x]) print "Longest cycle is for %d with length %d" % \ (longest, divs[longest]) if __name__ == '__main__': from timeit import Timer t = Timer("get_longest()", "from __main__ import get_longest") print t.timeit(number=100) >>> Longest cycle is for 983 with length 884 Longest cycle is for 983 with length 884 Longest cycle is for 983 with length 884 . . . Longest cycle is for 983 with length 884 151.166881084

So it takes on average around 1.5 seconds to run. Not bad, I see some people had Python code which ran in under a second, so I’m sure this could be optimised…