<< problem 72 - Counting fractions Digit factorial chains - problem 74 >>

Problem 73: Counting fractions in a range

Consider the fraction, dfrac{n}{d}, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d <= 8 in ascending order of size, we get:

dfrac{1}{8}, dfrac{1}{7}, dfrac{1}{6}, dfrac{1}{5}, dfrac{1}{4}, dfrac{2}{7}, dfrac{1}{3}, dfrac{3}{8}, dfrac{2}{5}, dfrac{3}{7}, dfrac{1}{2}, dfrac{4}{7}, dfrac{3}{5}, dfrac{5}{8}, dfrac{2}{3}, dfrac{5}{7}, dfrac{3}{4}, dfrac{4}{5}, dfrac{5}{6}, dfrac{6}{7}, dfrac{7}{8}

It can be seen that there are 3 fractions between 1/3 and 1/2.

How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for d <= 12,000 ?

My Algorithm

This is a problem where I had to try three different approaches until I found one that is fast enough to solve all Hackerrank test cases.
(all algorithms almost instantly solve the original problem, though).

All algorithms are related to the Farey sequence (en.wikipedia.org/wiki/Farey_sequence).
They ignore the numerator because it is actually not needed to solve the problem.

The most simple algorithm is based on recursion (look at my function recursion).
Starting with recursion(3, 2) (which means 1/3 and 1/2) the mediant m of 1/3 and 1/2 is found
and then the function calls itself with 1/3 and m and a second second with m a 1/2.
This continues until the denominator exceeds the limit 12000.
Each call returns the number of fractions which is 0 when the denominator turns out to be too big or 1 + leftSide + rightSide else.

The second algorithm (see iterative) computes the adjacent fraction of 1/3 (its "right neighbor").
Then, the denominator of the next fraction is nextD = maxD - \lfloor dfrac{maxD + prevD}{currentD} \rfloor.
We are done when nextD = toD.

The third and by far fastest algorithm computes the "rank" of fraction. Thereby rank(n, d) means: how many fractions are between 0 and dfrac{n}{d} ?
I found the idea/concept online: people.csail.mit.edu/mip/papers/farey/talk.pdf

Then the number of fractions between dfrac{1}{fromD} and dfrac{1}{toD} is rank(1, toD) - rank(1, fromD) - 1.
The algorithm is similar to a prime sieve.

Interactive test

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

This live test is based on the Hackerrank problem.

Input data (separated by spaces or newlines):
Note: Enter the denominator of the bigger fraction (e.g. 2 for 1/2) and then the maximum denominator (e.g. 12000)

This is equivalent to
echo "2 8" | ./73

Output:

(please click 'Go !')

Note: the original problem's input 2 12000 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)

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       #include <iostream>
#include <vector>

// maximum denominator
unsigned int maxD = 12000;

// algorithm I:
// count mediants between 1/fromD and 1/toD using recursion
unsigned int recursion(unsigned int fromD, unsigned int toD)
{
auto mediantD = fromD + toD;
// denominator too big ?
if (mediantD > maxD)
return 0;

// recursion
return 1 + recursion(fromD, mediantD) + recursion(mediantD, toD);
}

// algorithm II:
// iteratively enumerate all denominators
unsigned int iteration(unsigned int fromD, unsigned int toD)
{
// find denominator of closest mediant of "from"
// initial mediant
auto d = fromD + toD;
// is there a mediant closer to fromD ?
while (d + fromD <= maxD)
d += fromD;

// if prevD and d are denominators of adjacent fractions prevN/prevD and n/d
// then the next denominator is nextD = maxD - (maxD + prevD) % d
auto prevD = fromD;

unsigned int count = 0;
// until we reach the final denominator
while (d != toD)
{
// find next denominator
auto nextD = maxD - (maxD + prevD) % d;

// shift denominators, the current becomes the previous
prevD = d;
d = nextD;

count++;
}

return count;
}

// algorithm III:
// return numbers of irreducible fractions a/b < n/d where b is less than maxD
unsigned int rank(unsigned int n, unsigned int d)
{
// algorithm from "Computer Order Statistics in the Farey Sequence" by C. & M. Patrascu
// http://people.csail.mit.edu/mip/papers/farey/talk.pdf
std::vector<unsigned int> data(maxD + 1);
for (unsigned int i = 0; i < data.size(); i++)
data[i] = i * n / d; // n is always 1 but I wanted to keep the original algorithm

// remove all multiples of 2*i, 3*i, 4*i, ...
// similar to a prime sieve
for (unsigned int i = 1; i < data.size(); i++)
for (unsigned int j = 2*i; j < data.size(); j += i)
data[j] -= data[i];

// return sum of all elements
unsigned int sum = 0;
for (auto x : data)
sum += x;
return sum;
}

int main()
{
// denominators are abbreviated D
unsigned int toD = 2; // which means 1/2 (original problem)

//#define ORIGINAL
#ifndef ORIGINAL
std::cin >> toD >> maxD;
#endif

// the algorithm search from 1/fromD to 1/toD
auto fromD = toD + 1;

// algorithm 1
//std::cout << recursion(fromD, toD) << std::endl;
// algorithm 2
//std::cout << iteration(fromD, toD) << std::endl;
// algorithm 3
auto result = rank(1, toD) - rank(1, fromD) - 1;
std::cout << result << std::endl;

return 0;
}


This solution contains 17 empty lines, 30 comments and 4 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++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

March 15, 2017 submitted solution
May 3, 2017 added comments

Hackerrank

My code solves 14 out of 14 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 15% (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 rarely an option.

Heatmap

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

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

 << problem 72 - Counting fractions Digit factorial chains - problem 74 >>
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