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

# Problem 108: Diophantine reciprocals I

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.

# My 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 + a5 = 4 + aa = 1 and because of y = n + bb = 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.

My 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).

# 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.

Number of test cases (1-5):

Input data (separated by spaces or newlines):
Note: Enter n and the program will compute the number of solutions

This is equivalent to
echo "1 3" | ./108

Output:

(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.

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>

//#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 didn't precompute 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 didn't precompute 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
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
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 28 empty lines, 19 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.12 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

May 12, 2017 submitted solution

# Hackerrank

My code solves 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 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 107 - Minimal network Darts - problem 109 >>
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