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

# Problem 493: Under The Rainbow

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?

# 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:

Input data (separated by spaces or newlines):
Note: Enter the the numbers of colors, then balls per color and finally the number of picks

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 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

August 21, 2017 submitted solution

# Difficulty

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

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