<< problem 99 - Largest exponential Optimum polynomial - problem 101 >>

# Problem 100: Arranged probability

If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs,
and two discs were taken at random, it can be seen that the probability of taking two blue discs, P(BB) = dfrac{15}{21} * dfrac{14}{20} = dfrac{1}{2}.

The next such arrangement, for which there is exactly 50% chance of taking two blue discs at random,
is a box containing eighty-five blue discs and thirty-five red discs.

By finding the first arrangement to contain over 10^12 = 1,000,000,000,000 discs in total, determine the number of blue discs that the box would contain.

# My Algorithm

This was a crude hack: I wrote a brute-force program to find the first solutions (ignoring the 10^12 limit) and looked at the numbers.
For p / q = 1 / 2 I saw that
red_{n+1} = 2 * blue_n + red_n - 1
blue_{n+1} = blue_n + 2 * red_{n+1}

The initial values red_0 = 6 and blue_0 = 15 were already given in the problem statement.
It takes just a few iterations of these simple operations to find the correct solution. Program runtime is close to 0ms !

## Modifications by HackerRank

A loop analyzes all values 2 <= blue < 100000 (I chose that limit to prevent timeouts).
The left side of blue * (blue - 1) * q / p = su m * (su m - 1) is known and if an integer solution for su m = blue + red exists
such that su m > minimum then a valid solution was found.

[TODO] My code fails most test cases because of my arbitrary limit of 1000000.
[TODO] So far I haven't implemented a proper detection whether a solution exists at all.
[TODO] It looks a lot like a Pell equation solver might help.

# 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):
Note: Enter numerator and denominator of a fraction and then the minimum sum of blue and red discs

This is equivalent to
echo "1 1 2 100" | ./100

Output:

Note: the original problem's input 1 2 1000000000000 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++ 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 <cmath>

//#define ORIGINAL

int main()
{
unsigned int tests;
std::cin >> tests;

while (tests--)
{
unsigned long long minimum = 1000000000000ULL;

unsigned long long p = 1;
unsigned long long q = 2;
std::cin >> p >> q >> minimum;

unsigned long long blue = 0;
unsigned long long red  = 0;

// special code for p/q = 1/2
if (p == 1 && q == 2)
{
blue = 15;
red  =  6;

// keep going until limit is reached
while (blue + red < minimum)
{
// at first I brute-forced the first solutions and wrote them down
// then I saw the following relationship for p/q = 1/2:
//  red(i+1) = 2 * blue(i) + red(i) - 1;
// blue(i+1) = 2 * red(i+1)
red   = 2 * blue + red - 1; // seems to be true for most p/q
blue += 2 * red;            // but this line is not correct for p/q != 1/2
}
#ifdef ORIGINAL
std::cout << blue << std::endl;
#else
std::cout << blue << " " << (red + blue) << std::endl;
#endif
continue;
}

// brute-force smallest solution
bool found = false;
for (blue = 2; blue < 100000; blue++)
{
// sum = red + blue
// blue * (blue - 1) / (sum * (sum - 1)) = p / q
// blue * (blue - 1) * q / p = sum * (sum - 1)
unsigned long long b2 = blue * (blue - 1);
b2 *= q;
// right side must be an integer
if (b2 % p != 0)
continue;
unsigned long long sum2 = b2 / p; // sum2 = sum * (sum - 1)

// (sum-1)^2 < sum2 < sum^2
unsigned long long sum  = std::sqrt(sum2) + 1;
// sqrt may have returned a floating-point number
if (sum * (sum - 1) != sum2)
continue;

// now we have the correct solution if minimum is small (< 100000)
red = sum - blue;
if (blue + red >= minimum)
{
found = true;
break;
}
}

// failed ? TODO: this means just that my simple search aborted
if (!found)
{
std::cout << "No solution" << std::endl;
continue;
}

// show solution
std::cout << blue << " " << (red + blue) << std::endl;
}

return 0;
}


This solution contains 13 empty lines, 17 comments and 5 preprocessor commands.

# Benchmark

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

March 15, 2017 submitted solution

# Hackerrank

My code solves 4 out of 22 test cases (score: 22.22%)

I failed 18 test cases due to wrong answers and 0 because of timeouts

# Difficulty

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

Hackerrank describes this problem as advanced.

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
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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 99 - Largest exponential Optimum polynomial - problem 101 >>
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