Problem 26: Reciprocal cycles

(see projecteuler.net/problem=26)

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

frac{1}{2} = 0.5
frac{1}{3} = 0.overline{3}
frac{1}{4} = 0.25
frac{1}{5} = 0.2
frac{1}{6} = 0.1\overline{6}
frac{1}{7} = 0.overline{142857}
frac{1}{8} = 0.125
frac{1}{9} = 0.overline{1}
frac{1}{10} = 0.1

Where 0.1\overline{6} means 0.166666..., and has a 1-digit recurring cycle.
It can be seen that frac{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.

Algorithm

I implemented the same basic division algorithm I learnt in school (3rd grade ?!)
but of course extended it to operate with fractionals instead of integers, e.g. 1/7 = 0.overline{142857}

Let's do it the good old-fashioned way:
1 / 7 = 0 (and 1 remains, multiply it by 10 and it becomes our next dividend)
1*10 / 7 = 1 (and 3 remains, multiply as above)
3*10 / 7 = 4 (and 2 remains, you know the game)
2*10 / 7 = 2 (and 6 remains)
6*10 / 7 = 8 (and 4 remains)
4*10 / 7 = 5 (and 5 remains)
5*10 / 7 = 7 (and 1 remains)

Now we have the same remainder 1 we had in the first step and the circle begins again.
Moreover, the digits after the equation sign produce the recurring cycle 142857.
Their length is 6 because it took six steps until we saw the same remainder again.

More or less the same algorithm can be found in the Wikipedia, too: en.wikipedia.org/wiki/Repeating_decimal

Similar to many problems before, the modified Hackerrank problem forced me to precompute the results
and then just look up those numbers for each test case.

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include <iostream>
#include <vector>
 
// find length of the recurring cycle in the decimal fraction part of 1/x
unsigned int cycleLength(unsigned int x)
{
// catch invalid fractions
if (x == 0)
return 0;
 
// [remainder] => [pos]
const unsigned int NotSeenYet = 0;
std::vector<unsigned int> lastPos(x, NotSeenYet);
 
// start at first digit after the decimal dot
unsigned int position = 1;
// 1/x => initial dividend is 1
unsigned int dividend = 1;
 
// exit-conditions are inside the loop
while (true)
{
// find remainder
unsigned int remainder = dividend % x;
 
// if remainder becomes zero then stop immediately
// because the fraction doesn't have a recurring cycle
if (remainder == 0)
return 0;
 
// same remainder ? => abort
if (lastPos[remainder] != NotSeenYet)
// length of recurring cycle
return position - lastPos[remainder];
 
// remember position of current remainder
lastPos[remainder] = position;
 
// next step
position++;
dividend = remainder * 10;
}
}
 
int main()
{
// Hackerrank's upper limit
const unsigned int MaxDenominator = 10000;
 
// cache results to speed up running tons of test cases
std::vector<unsigned int> cache = { 0 }; // no cycles for 1/0
 
unsigned int longestDenominator = 0;
unsigned int longestCycle = 0;
for (unsigned int denominator = 1; denominator <= MaxDenominator; denominator++)
{
// found a longer circle ?
auto length = cycleLength(denominator);
if (longestCycle < length)
{
longestCycle = length;
longestDenominator = denominator;
}
 
// cache result
cache.push_back(longestDenominator);
}
 
// plain look up
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int x;
std::cin >> x;
// find "best" denominator smaller (!) than the input value, therefore minus one
std::cout << cache[x - 1] << std::endl;
}
return 0;
}

This solution contains 13 empty lines, 19 comments and 2 preprocessor commands.

Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 3" | ./26

Output:

(please click 'Go !')

Note: the original problem's input 1000 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

Benchmark

The correct solution to the original Project Euler problem was found in 0.1 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

Changelog

February 24, 2017 submitted solution
April 4, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler026

My code solved 4 out of 4 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is never an option.

Links

projecteuler.net/thread=26 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-26-find-the-value-of-d-1000-for-which-1d-contains-the-longest-recurring-cycle/ (written by Kristian Edlund)
Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p026.hs (written by Nayuki)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p026.java (written by Nayuki)
Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p026.mathematica (written by Nayuki)
C: github.com/eagletmt/project-euler-c/blob/master/20-29/problem26.c (written by eagletmt)
Go: github.com/frrad/project-euler/blob/master/golang/Problem026.go (written by Frederick Robinson)
Javascript: github.com/dsernst/ProjectEuler/blob/master/26 Reciprocal cycles.js (written by David Ernst)
Scala: github.com/samskivert/euler-scala/blob/master/Euler026.scala (written by Michael Bayne)

Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are already solved and there wasn't a Hackerrank version of it (at the time I solved it) or I didn't care about it because it differed too much.

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The 133 solved problems had an average difficulty of 16.9% at Project Euler and I scored 11,174 points (out of 12300) at Hackerrank's Project Euler+.
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