<< problem 25 - 1000-digit Fibonacci number Quadratic primes - problem 27 >>

# Problem 26: Reciprocal cycles

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

# Hackerrank

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.

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)
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.

Please click on a problem's number to open my solution to that problem:

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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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