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

# Problem 72: Counting fractions

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?

# 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)
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:

Number of test cases (1-5):

Input data (separated by spaces or newlines):

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

# Hackerrank

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 never an option.

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.

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

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The 163 solved problems had an average difficulty of 22.2% at Project Euler and I scored 11,907 points (out of 13200) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
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