<< problem 77 - Prime summations Passcode derivation - problem 79 >>

# Problem 78: Coin partitions

Let p(n) represent the number of different ways in which n coins can be separated into piles.
For example, five coins can be separated into piles in exactly seven different ways, so p(5)=7.

OOOOO
OOOO O
OOO OO
OOO O O
OO OO O
OO O O O
O O O O O

Find the least value of n for which p(n) is divisible by one million.

# Algorithm

Brute-forcing the solution for small n yielded the sequence:
1,2,3,5,7,11,15,122,30,42,56,77,101,...
I searched the web and found these numbers at oeis.org/A000041

After that I read the problem description again and looked up "partition" on Wikipedia: en.wikipedia.org/wiki/Partition_(number_theory)#Recurrence_formula
And found a link to Euler's formula which is based on pentagonal numbers: en.wikipedia.org/wiki/Pentagonal_number_theorem
result(i) = result(pentagonal(+1)) + result(pentagonal(-1))
 - result(pentagonal(+2)) - result(pentagonal(-2))
 + result(pentagonal(+3)) + result(pentagonal(-3))
 - result(pentagonal(+4)) - result(pentagonal(-4))
 ...

result(417) is too big for a 64 bit integer (and there is no solution among the first 416 partitions).
Luckily, we need to find the first number that is divisible by one million, that means where result(x) % 1000000 = 0.
Hence I store the number of partitions modulo 1000000. Whenever it is zero, my program can abort.

## Modifications by HackerRank

The modified problem asks for the number of partitions of an input value (modulo 10^9 - 7).
Results from previous test cases are kept in partitions to speed up the process.

# 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 <vector>

int main()
{
// store result (modulo 10^6 or 10^9 + 7)
std::vector<unsigned long long> partitions;
// degenerated case, there's one partition for an empty pile
partitions.push_back(1);

//#define ORIGINAL
#ifdef ORIGINAL
const long long modulo =    1000000; // 10^6
#else
const long long modulo = 1000000007; // 10^9 + 7

unsigned int tests = 1;
std::cin >> tests;
while (tests--)
#endif

{
unsigned int limit = 100000; // the solution is < 100000, program ab
#ifndef ORIGINAL
std::cin >> limit;
#endif

// fill cache
for (unsigned int n = partitions.size(); n <= limit; n++)
{
// sum according to Euler's formula
long long sum = 0;

// all pentagonal numbers where pentagonal(i) <= n
for (unsigned int i = 0; ; i++) // abort inside loop
{
// generate alternating numbers +1,-1,+2,-2,+3,-3,...
int alternate = 1 + (i / 2); // generate the digit 1,1,2,2,3,3,...
if (i % 2 == 1)
alternate = -alternate;    // flip the sign for every second number

// pentagonal index, "how far we go back" in partitions[]
unsigned int offset = alternate * (3 * alternate - 1) / 2;
// can't go back that far ? (array index would be negative)
if (n < offset)
break;

// add two terms, subtract two terms, add two terms, subtract two terms, ...
if (i % 4 < 2)
sum += partitions[n - offset]; // i % 4 = { 0, 1 }
else
sum -= partitions[n - offset]; // i % 4 = { 2, 3 }

// only the last digits are relevant
sum %= modulo;

}

// note: sum can be temporarily negative
if (sum < 0)
sum += modulo;

#ifdef ORIGINAL
// "divisible by one million" => sum % 1000000 == 0
// last 6 digits (modulo was 10^6) are zero
if (sum == 0)
break;
#endif

partitions.push_back(sum);
}

// print (cached) result
#ifdef ORIGINAL
std::cout << partitions.size() << std::endl;
#else
std::cout << partitions[limit] << std::endl;
#endif
}

return 0;
}


This solution contains 15 empty lines, 15 comments and 12 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.

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 5" | ./78

Output:

(please click 'Go !')

(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 0.1 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

March 16, 2017 submitted solution

# Hackerrank

My code solves 8 out of 8 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 30% (out of 100%).

Hackerrank describes this problem as medium.

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=78 - 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-78-coin-piles/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p078.java (written by Nayuki)
Go: github.com/frrad/project-euler/blob/master/golang/Problem078.go (written by Frederick Robinson)
Scala: github.com/samskivert/euler-scala/blob/master/Euler078.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.
 << problem 77 - Prime summations Passcode derivation - problem 79 >>
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