<< problem 491 - Double pandigital number divisible by 11 | Problem 500!!! - problem 500 >> |

# Problem 493: Under The Rainbow

(see projecteuler.net/problem=493)

70 colored balls are placed in an urn, 10 for each of the seven rainbow colors.

What is the expected number of distinct colors in 20 randomly picked balls?

Give your answer with nine digits after the decimal point (a.bcdefghij).

# My Algorithm

My standard procedure for probability problems is to write a simple Monte-Carlo simulation.

The first four or five digits are correct - but still a far cry from the nine digits requested by the problem statement.

A much smarter approach is to turn the problem upside-down:

for each color I compute the probability that this color is *not* among the 20 picked balls.

That's the probability that 20 out of 70 balls are actually taken from a set of only 70 - 10 = 60 balls:

{{60}choose{20}} div {{70}choose{20}}

Then the opposite is the probability that this color was picked:

1 - {{60}choose{20}} div {{70}choose{20}}

Computing that equation for all 7 colors gives the result:

7 * (1 - {{60}choose{20}} div {{70}choose{20}} )

## Note

The main problem is to have a proper `choose`

function. The naive way to compute `choose(70, 20)`

is to find 70! and 20!.

However, 70! causes serious problems because of its size (about 100 digits, way more than a 64 bit integer can handle).

Lucky for me, I already wrote such a `choose`

function for problem 116. Thus the code for this problem was actually pretty simple.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This is equivalent to`echo "4 5 7" | ./493`

Output:

*Note:* the original problem's input `7 10 20`

__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.

#include <iostream>
#include <iomanip>
#include <vector>
// ---------- Monte-Carlo simulation to get a rough estimation ----------
// note: not used anymore

// a simple pseudo-random number generator
// (produces the same result no matter what compiler you have - unlike rand() from math.h)

unsigned int myrand()
{
static unsigned long long seed = 0;
seed = 6364136223846793005ULL * seed + 1;
return (unsigned int)(seed >> 30);
}
// Monte Carlo simulation

double monteCarlo(unsigned int numColors, unsigned int numBallsPerColor, unsigned int picks, unsigned int iterations)
{
// count different colors of all iterations
unsigned int sum = 0;
auto numBalls = numColors * numBallsPerColor;
for (unsigned int i = 0; i < iterations; i++)
{
std::vector<bool> current(numBalls, true);
for (unsigned int j = 0; j < picks; j++)
{
unsigned int id;
// pick a ball (if still available)
do
id = myrand() % numBalls;
while (!current[id]);
// remove that ball
current[id] = false;
}
// for each color check if at least one ball was picked
for (unsigned int c = 0; c < numColors; c++)
for (unsigned int b = 0; b < numBallsPerColor; b++)
{
auto id = c * numBallsPerColor + b;
if (!current[id])
{
sum++;
break;
}
}
}
return sum / double(iterations);
}
// ---------- my solution ----------

// number of ways to choose n elements from k available
// taken from problem 116

unsigned long long choose(unsigned long long n, unsigned long long k)
{
// n! / (n-k)!k!
unsigned long long result = 1;
// reduce overflow by dividing as soon as possible to keep numbers small
for (unsigned long long invK = 1; invK <= k; invK++)
{
result *= n;
result /= invK;
n--;
}
return result;
}
int main()
{
unsigned int colors = 7;
unsigned int ballsPerColor = 10;
unsigned int picks = 20;
std::cin >> colors >> ballsPerColor >> picks;
const auto numBalls = colors * ballsPerColor;
std::cout << std::fixed << std::setprecision(9);
// run Monte-Carlo simulation to estimate the result
//while (true)
//std::cout << monteCarlo(colors, ballsPerColor, picks, 10000000) << std::endl;
// and now the precise computation
double result = 0;
for (unsigned int current = 0; current < colors; current++)
{
// count combinations where any ball is picked - but not from a special color
auto dontPickThatColor = choose(numBalls - ballsPerColor, picks);
// count combinations where any ball can be picked, no matter what color
auto total = choose(numBalls, picks);
// probability that the current color was not picked
auto absent = dontPickThatColor / double(total);
// probability that the current color was picked
auto hasColor = 1 - absent;
// expected value: 0 if absent, 1 if present
result += 0 * absent + 1 * hasColor; // which is the same as result += hasColor
}
// obviously all 7 iterations don't depend on "current" and could be merged into a single equation:
//result = colors * (1 - choose(numBalls - ballsPerColor, picks) / (double)choose(numBalls, picks));
// and we're done !
std::cout << result << std::endl;
return 0;
}

This solution contains 18 empty lines, 26 comments and 3 preprocessor commands.

# Benchmark

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

August 21, 2017 submitted solution

August 21, 2017 added comments

# Difficulty

Project Euler ranks this problem at **15%** (out of 100%).

# Links

projecteuler.net/thread=493 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

# 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 |

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I scored 13,386 points (out of 15600 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.

# Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.

All of my solutions can be used for any purpose and I am in no way liable for any damages caused.

You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.

Thanks for all their endless effort !!!

<< problem 491 - Double pandigital number divisible by 11 | Problem 500!!! - problem 500 >> |