<< problem 71 - Ordered fractions | Counting fractions in a range - problem 73 >> |

# Problem 72: Counting fractions

(see projecteuler.net/problem=72)

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 21 elements in this set.

How many elements would be contained in the set of reduced proper fractions for d <= 1,000,000?

# My Algorithm

All reduced fraction have gcd(numerator, denominator) = 1 (gcd stands for "greatest common divisor")

therefore for a given denominator the number of suitable numerators is the same as the Euler totient of the denominator.

In project 70 I implemented phi(x) like this:

result = x * (1 - 1/prime1) * (1 - 1/prime2) * (1 - 1/prime3) * ...

I realized that something like a prime sieve can do the same job "in reverse" much faster:

- initialize phi(x) = x for all numbers

- for all multiples of a prime: phi_{new}(k * prime) = phi_{old}(k * prime) * (1 - 1/prime) = phi_{old}(k * prime) - phi_{old}(k * prime) / prime

- compute all sums phi(2) + phi(3) + ... + phi(i)

# 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>
int main()
{
// 1. initialize phi(x) = x for all numbers
unsigned int limit = 1000000;
std::vector<unsigned int> phi(limit + 1); // vectors are zero-based
for (size_t i = 0; i < phi.size(); i++)
phi[i] = i;
// 2. for all multiples of a prime:
// phi_new(k*prime) = phi_old(k*prime) * (1 - 1/prime)
// = phi_old(k*prime) - phi_old(k*prime) / prime
for (unsigned int i = 2; i <= limit; i++)
{
// prime number ? (because not modified yet by other primes)
if (phi[i] == i)
// adjust all multiples
for (unsigned int k = 1; k * i <= limit; k++)
phi[k * i] -= phi[k * i] / i;
}
// note: since we are only interested in 0 < fractions < 1
// we have to exclude phi(1) (which would yield 1/1 = 1)
// and start at phi(2)
std::vector<unsigned long long> sums(phi.size(), 0);
// 3. compute all sums phi(2) + phi(3) + ... + phi(i)
for (unsigned int i = 2; i <= limit; i++)
sums[i] = sums[i - 1] + phi[i];
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
std::cin >> limit;
std::cout << sums[limit] << std::endl;
}
}

This solution contains 5 empty lines, 10 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:

This is equivalent to`echo "1 8" | ./72`

Output:

*Note:* the original problem's input `1000000`

__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 0.03 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

Peak memory usage was about 13 MByte.

(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 13, 2017 submitted solution

May 2, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler072

My code solves **21** out of **21** test cases (score: **100%**)

# Difficulty

Project Euler ranks this problem at **20%** (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.

# Links

projecteuler.net/thread=72 - **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-72-reduced-proper-fractions/ (written by Kristian Edlund)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p072.java (written by Nayuki)

Go: github.com/frrad/project-euler/blob/master/golang/Problem072.go (written by Frederick Robinson)

Scala: github.com/samskivert/euler-scala/blob/master/Euler072.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 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.

red problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte.

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

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I scored 12,983 points (out of 15100 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

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

My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

# Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.

All of my solutions can be used for any purpose and I am in no way liable for any damages caused.

You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler. Thanks for all their endless effort.

<< problem 71 - Ordered fractions | Counting fractions in a range - problem 73 >> |