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

# Problem 73: Counting fractions in a range

(see projecteuler.net/problem=73)

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 ?

# 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 (https://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.

# My code

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

The code contains `#ifdef`

s 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.

# 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.

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

Output:

*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)*

# Benchmark

The correct solution to the original Project Euler problem was found in **less than 0.01** 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

March 15, 2017 submitted solution

May 3, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler073

My code solved **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 never an option.

# Links

projecteuler.net/thread=73 - **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-73-sorted-reduced-proper-fractions/ (written by Kristian Edlund)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p073.java (written by Nayuki)

Go: github.com/frrad/project-euler/blob/master/golang/Problem073.go (written by Frederick Robinson)

Scala: github.com/samskivert/euler-scala/blob/master/Euler073.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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<< problem 72 - Counting fractions | Digit factorial chains - problem 74 >> |