<< problem 179 - Consecutive positive divisors RSA encryption - problem 182 >>

# Problem 181: Investigating in how many ways objects of two different colours can be grouped

Having three black objects B and one white object W they can be grouped in 7 ways like this:
(BBBW), (B,BBW), (B,B,BW), (B,B,B,W), (B,BB,W), (BBB,W), (BB,BW)

In how many ways can sixty black objects B and forty white objects W be thus grouped?

# My Algorithm

My solution is somehow similar to the coin-change algorithm:

• each group can be treated as a sequence of black and white objects, BWBW is the same as BBWW
• all groups can be sorted by their size and, if multiple groups have the same size, by their lexicographical order
For maxBlack = 3 and maxWhite = 1 these groups exist:
(B,B,B,W), (B,B,BW), (B,W,BB), (B, BBW), (W, BBB), (BB, BW), (BBBW)
These are exactly the same as in the problem statement but in a different order.

Two outer loops iterate over all possible sequences of black and white objects.
The inner loops place them at every possible positions (until the number of available objects is exhausted).

The result will be found in current[60][40].

## Alternative Approaches

You can solve this problem with Dynamic Programming, too.
I wrote a simple prototype but it turned out to be much slower (8 seconds vs. 0.04 seconds).

# 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):
Note: Enter the number of black and the number of white objects

This is equivalent to
echo "3 1" | ./181

Output:

Note: the original problem's input 60 40 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>

// limits
const unsigned int MaxBlack = 160;
const unsigned int MaxWhite = 160;
const unsigned int Modulo   = 1000000007; // Hackerrank only

int main()
{
#define ORIGINAL
#ifndef ORIGINAL
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
#endif
{
unsigned int maxBlack = MaxBlack;
unsigned int maxWhite = MaxWhite;
std::cin >> maxBlack >> maxWhite;

// static array size: actually it would be sufficient to use maxBlack instead of MaxBlack
// (and maxWhite instead of MaxWhite)
unsigned long long previous[MaxBlack + 1][MaxWhite + 1];
unsigned long long current [MaxBlack + 1][MaxWhite + 1];

// initialize
for (unsigned int i = 0; i <= maxBlack; i++)
for (unsigned int j = 0; j <= maxWhite; j++)
previous[i][j] = 0;
previous[0][0] = 1;

// all possible subsets
for (unsigned int useBlack = 0; useBlack <= maxBlack; useBlack++)
for (unsigned int useWhite = 0; useWhite <= maxWhite; useWhite++)
{
// skip empty subset
if (useBlack == 0 && useWhite == 0)
continue;

// put subset at every possible position
for (unsigned int i = 0; i <= maxBlack; i++)
for (unsigned int j = 0; j <= maxWhite; j++)
{
current[i][j] = 0;

// place it repeatedly
unsigned int k = 0;
while (i >= k * useBlack && j >= k * useWhite)
{
current[i][j] += previous[i - k * useBlack][j - k *  useWhite];
k++;
}
#ifndef ORIGINAL
current[i][j] %= Modulo;
#endif
}

// copy for next iteration
for (unsigned int i = 0; i <= maxBlack; i++)
for (unsigned int j = 0; j <= maxWhite; j++)
previous[i][j] = current[i][j];
}

// print result
std::cout << current[maxBlack][maxWhite] << std::endl;
}

return 0;
}


This solution contains 10 empty lines, 10 comments and 6 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.01 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

June 28, 2017 submitted solution
July 28, 2017 modified to solve Hackerrank, too

# Hackerrank

My code solves 2 out of 3 test cases (score: 9.09%)

I failed 0 test cases due to wrong answers and 1 because of timeouts

# Difficulty

Project Euler ranks this problem at 70% (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 179 - Consecutive positive divisors RSA encryption - problem 182 >>
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