<< 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.

# My 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 in OEIS 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.

# 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:

(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 <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.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.10 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

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 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 77 - Prime summations Passcode derivation - problem 79 >>
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