<< problem 43 - Sub-string divisibility Triangular, pentagonal, and hexagonal - problem 45 >>

# Problem 44: Pentagon numbers

Pentagonal numbers are generated by the formula, P_n=n(3n-1)/2. The first ten pentagonal numbers are:
1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...

It can be seen that P_4 + P_7 = 22 + 70 = 92 = P_8. However, their difference, 70 - 22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, P_j and P_k, for which their sum and difference are pentagonal and D = |P_k - P_j| is minimised; what is the value of D?

# My Algorithm

The function isPentagonal(x) plays a core role is my solution:
under the assumption that x is pentagonal, that means x = P(n):
P(n) = n (3n - 1) / 2 → pentagonal formula from problem statement
P(n) = frac{3}{2} n^2 - frac{n}{2}
2 P(n) = 3n^2 - n
0 = 3n^2 - n - 2P(n)

Solving this quadratic equation (coefficients a=3, b=-1, c=-2P(n)):
n_{1,2} = dfrac{-b \pm sqrt{b^2 - 4ac}}{2a}

n_{1,2} = dfrac{1 \pm sqrt{1 + 24P(n)}}{6}

n must be a positive integer, therefore the only possible solution is
n = dfrac{1 + sqrt{1 + 24P(n)}}{6}

= dfrac{1 + sqrt{1 + 24x}}{6}

When generating pentagonal numbers, my code checks P(n)-P(n-d) and P(n)+P(n-d) for all d < n whether both are pentagonal, too.

## Modifications by HackerRank

The Hackerrank problem is a bit easier because the distance is predefined by the user.

# 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.

Input data (separated by spaces or newlines):
Note: This live test requires Hackerrank-style input: enter a max. pentagonal index and a distance, then the program will display all pentagonal indices

This is equivalent to
echo 8 | ./44

Output:

Note: the original problem's input 9 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. Or just jump to my GitHub repository.

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 <cmath>

bool isPentagonal(unsigned long long x)
{
// see explanation above
unsigned long long n = (1 + sqrt(24*x + 1)) / 6;

// if x was indeed a pentagonal number then our assumption P(n) = x must be true
auto p_n = n * (3 * n - 1) / 2;
return p_n == x;
}

int main()
{
//#define ORIGINAL
#ifdef ORIGINAL

const unsigned int HugeNumber = 999999999;
unsigned int best = HugeNumber;
unsigned int last = 1; // P(1)

while (best == HugeNumber)
{
// next pentagonal number
auto p_n  = n * (3 * n - 1) / 2;
// difference to closest pentagonal number larger than our best result ?
if (p_n - last > best)
break;

// check all pairs P(n) and P(n - distance) where 1 <= distance < n
for (unsigned int distance = 1; distance < n; distance++)
{
// compute P(n - distance) pentagonal number
auto x   = n - distance;
auto p_x = x * (3 * x - 1) / 2;

// their sum and difference
auto sum        = p_n + p_x;
auto difference = p_n - p_x;

// too far away ?
if (difference > best)
break;

// yes, found something
if (isPentagonal(sum) && isPentagonal(difference))
best = difference;
}

// check next pentagonal number
last = p_n;
n++;
}
std::cout << best << std::endl;

#else

unsigned int maxIndex;
unsigned int distance;
std::cin >> maxIndex >> distance;

// iterate over all pairs at a given distance
for (unsigned long long n = distance + 1; n <= maxIndex; n++)
{
auto p_n = n * (3 * n - 1) / 2;

auto x   = n - distance;
auto p_x = x * (3 * x - 1) / 2;

// check sum and difference
auto sum        = p_n + p_x;
auto difference = p_n - p_x;

// yes, found something
if (isPentagonal(sum) || isPentagonal(difference))
std::cout << p_n << std::endl;
}
#endif

return 0;
}


This solution contains 17 empty lines, 14 comments and 5 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.05 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

February 27, 2017 submitted solution

# Hackerrank

My code solves 6 out of 6 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 5% (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.

# Similar problems at Project Euler

Problem 45: Triangular, pentagonal, and hexagonal

Note: I'm not even close to solving all problems at Project Euler. Chances are that similar problems do exist and I just haven't looked at them.

# 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 43 - Sub-string divisibility Triangular, pentagonal, and hexagonal - problem 45 >>
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