<< problem 109 - Darts Primes with runs - problem 111 >>

# Problem 110: Diophantine reciprocals II

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

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

It can be verified that when n = 1260 there are 113 distinct solutions and this is the least value of n
for which the total number of distinct solutions exceeds one hundred.

What is the least value of n for which the number of distinct solutions exceeds four million?

NOTE: This problem is a much more difficult version of Problem 108 and as it is well beyond the limitations of a brute force approach it requires a clever implementation.

# My Algorithm

As mentioned in my solution for problem 108, a prime factorization can find the number of solutions.
This time I "invert" the algorithm: I generate prime factorizations in ascending order.
If e_0, e_1, e_2, ... are the exponents of the first primes 2, 3, 5, ... then:
value = 2^{e_0} * 3^{e_1} * 5^{e_2} ...
The container exponents stores these exponents, while prime obviously holds the prime numbers.
Empirically I found that the first 12 prime numbers (2 ... 37) are sufficient for the original problem.

This gives ((2e_0 + 1)(2e_1 + 1)(2e_2 + 1) ... + 1) factorizations. However, x and y are interchangeable, therefore I must divide the result by 2 as well.

A nice property of std::map is that its elements are always sorted by their keys.
Therefore todo.begin() always refers to the smallest unprocessed number.
If it has too few prime factorizations then I increment its components and re-insert them into todo.

## Modifications by HackerRank

Hackerrank has a broader input ranges which requires a few more prime numbers.
Even more, the result may be too big for 64 bit integer. That's why I opt for long double instead.
The failed test case is probably due to exceeding the precision of long double.

## Note

I noticed that the exponents of larger prime numbers are at most 1 (when processing numbers <= 4000000).
My code runs about 3x faster and needs less memory when I avoid incrementing e_4, e_5, ...

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):

This is equivalent to
echo 100 | ./110

Output:

Note: the original problem's input 4000000 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)

# 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>
#include <iomanip>
#include <vector>
#include <map>

int main()
{
unsigned long long limit = 4000000;
std::cin >> limit;

#define ORIGINAL
#ifdef ORIGINAL
// 12 primes are enough for the original problem, I added a few more for the Hackerrank version
const unsigned int NumPrimes = 12;
// a short list of primes
unsigned int primes[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 };
#else
const unsigned int NumPrimes = 18;
// a short list of primes
unsigned int primes[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61 };
#endif

// use 1 as a "seed" value (all its primes' exponents are zero)
typedef std::vector<unsigned char> Exponents;
std::map<long double, Exponents> todo = { { 1, Exponents(NumPrimes, 0) } };

while (true)
{
// pick the smallest value and its exponents
auto current   = todo.begin();
auto value     = current->first;
auto exponents = current->second;

// remove from todo list
todo.erase(current);

// find number of factorizations
unsigned long long uniqueFactors = 1;
for (auto x : exponents)
uniqueFactors *= 2 * x + 1;
// a*b = b*a => divide number of factorizations by 2
uniqueFactors++;
uniqueFactors /= 2;

// stop ?
if (uniqueFactors >= limit)
{
std::cout << std::fixed << std::setprecision(0) << value << std::endl;
break;
}

// increment each primes' exponent by 1 and append new value to todo list
for (size_t i = 0; i < exponents.size(); i++)
{
#ifdef ORIGINAL
// speed optimization: exponents of larger primes can be at most 1
if (exponents[i] == 1 && i > 3) // 3 was found by trial-n-error
break;
#endif

exponents[i]++;
value *= primes[i];

todo[value] = exponents;
// note: std::map prunes duplicates automatically
}
}

return 0;
}


This solution contains 11 empty lines, 12 comments and 10 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.02 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 8 MByte.

(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

July 28, 2017 submitted solution

# Hackerrank

My code solves 12 out of 20 test cases (score: 60%)

I failed 1 test cases due to wrong answers and 7 because of timeouts

# Difficulty

Project Euler ranks this problem at 40% (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 the flashing problem is the one I solved most recently

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The 289 solved problems (level 11) had an average difficulty of 32.1% at Project Euler and
I scored 13,486 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.

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