<< problem 107 - Minimal network | Darts - problem 109 >> |

# Problem 108: Diophantine reciprocals I

(see projecteuler.net/problem=108)

In the following equation x, y, and n are positive integers.

dfrac{1}{x} + dfrac{1}{y} = dfrac{1}{n}

For n = 4 there are exactly three distinct solutions:

dfrac{1}{5} + dfrac{1}{20} = dfrac{1}{4}

dfrac{1}{6} + dfrac{1}{12} = dfrac{1}{4}

dfrac{1}{8} + dfrac{1}{8} = dfrac{1}{4}

What is the least value of n for which the number of distinct solutions exceeds one-thousand?

NOTE: This problem is an easier version of Problem 110; it is strongly advised that you solve this one first.

# Algorithm

It's safe to assume x <= y. Moreover, x > n because frac{1}{x} < frac{1}{n}. That means y > n, too.

There must be some a and b such that x = n + a and y = n + b.

The original equation becomes:

dfrac{1}{n + a} + dfrac{1}{n + b} = dfrac{1}{n}

Multiply both sides by n:

dfrac{n}{n + a} + dfrac{n}{n + b} = dfrac{n}{n} = 1

Multiply by n+a:

dfrac{n(n+a)}{n + a} + dfrac{n(n+a)}{n + b} = n + a

n + dfrac{n(n+a)}{n + b} = n + a

And the same for n+b:

n(n+b) + dfrac{n(n+a)(n+b)}{n + b} = (n + a)(n + b)

n(n+b) + n(n+a) = (n + a)(n + b)

Simplify:

n^2 + nb + n^2 + na = n^2 + na + nb + ab

n^2 = ab

In the example where n=4 you can compute x = n + a → 5 = 4 + a → a = 1 and because of y = n + b → b = 16.

Indeed, ab = 1 * 16 = 16 = 4^2 = n^2.

(And for the other two solutions: ab = (6-4) * (12-4) = 16 = n^2 and ab = (8-4) * (8-4) = 16 = n^2).

What does it mean ? Well, finding all ab = n^2 produces all solutions.

The number of solutions is the number of divisors of n^2.

First first attempt was based on brute force and is at the end of the code (it's not used anymore).

A smarter approach is to perform a prime factorization.

A high number of divisors means that prime factors must be small.

To speed up the program I abort when a prime factor > 100 is left. This is kind of cheating ... and roughly 100x faster.

## Modifications by HackerRank

I didn't notice that prime factorization of n^2 can be reduced to a slightly modified prime factorization of n.

Thanks to www.mathblog.dk/project-euler-108-diophantine-equation/ where I found that trick (see `numSquareDivisors`

)

The code becomes much faster but I kept my old code `numDivisors`

for the original problem.

However, I still time-out on two thirds of the test cases (my old code timed out in 80% of all cases).

# 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>
//#define ORIGINAL

// count divisors

unsigned long long numDivisors(unsigned long long n)
{
unsigned int result = 1;
auto reduce = n;
// trial division by all prime numbers
// => I don't precomputed a sieve, therefore divide by 2 and all odd numbers
for (unsigned long long divisor = 2; divisor <= reduce; divisor++)
{
// 2 is the only even prime number
if (divisor % 2 == 0 && divisor > 2)
divisor++;
if (divisor > 100) // WARNING: unsafe speed optimization !
break; // returns correct values for original problem but fails for some Hackerrank test cases
unsigned int exponent = 0;
while (reduce % divisor == 0)
{
exponent++;
reduce /= divisor;
}
result *= exponent + 1;
}
return result;
}
// count divisors of n^2, note: parameter is n, not n^2 (this is different from my old code in numDivisors)

unsigned long long numSquareDivisors(unsigned long long n)
{
unsigned int result = 1;
auto reduce = n;
// trial division by all prime numbers
// => I don't precomputed a sieve, therefore divide by 2 and all odd numbers
for (unsigned long long divisor = 2; divisor <= reduce; divisor++)
{
// 2 is the only even prime number
if (divisor % 2 == 0 && divisor > 2)
divisor++;
unsigned int exponent = 0;
while (reduce % divisor == 0)
{
exponent++;
reduce /= divisor;
}
result *= 2*exponent + 1; // changed vs. my code: times 2
}
return result;
}
int main()
{
#ifdef ORIGINAL
unsigned long long n = 1;
unsigned long long threshold = 1000;
while (true)
{
auto divisors = numDivisors(n * n);
// a and b are interchangeable therefore only half of the solutions are "unique"
auto half = (divisors + 1) / 2; // plus 1 because n^2 is obviously a square number and
if (half >= threshold)
{
std::cout << n << std::endl;
break;
}
// check next square number
n++;
}
#else
unsigned int tests;
std::cin >> tests;
while (tests--)
{
// find the number of solutions
unsigned long long n;
std::cin >> n;
auto divisors = numSquareDivisors(n);
// a and b are interchangeable therefore only half of the solutions are "unique"
auto half = (divisors + 1) / 2; // plus 1 because n^2 is obviously a square number and
std::cout << half << std::endl;
}
#endif
return 0;
}
// my initial brute force solution

int bruteForce(unsigned int threshold)
{
unsigned long long n = 4;
while (true)
{
unsigned int solutions = 0;
// try all values between n+1 ... 2n
for (unsigned long long x = n + 1; x <= 2*n; x++)
{
// the same as 1/x + 1/y = 1/n
auto y = n*x / (x - n);
// integer arithmetic might produce a wrong result, re-compute the formula backwards
if (y * (x - n) != n*x)
continue;
// exhausted ?
if (y < x)
break;
// valid solution
solutions++;
}
if (solutions > threshold)
break;
n++;
}
std::cout << n << std::endl;
return 0;
}

This solution contains 31 empty lines, 19 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 "1 3" | ./108`

Output:

*(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 **0.14** 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

May 12, 2017 submitted solution

May 16, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler108

My code solved **5** out of **12** test cases (score: **36.36%**)

I failed **7** test cases due to wrong answers and **0** because of timeouts

# Difficulty

Project Euler ranks this problem at **30%** (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=108 - **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-108-diophantine-equation/ (written by Kristian Edlund)

Scala: github.com/samskivert/euler-scala/blob/master/Euler108.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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