<< problem 237 - Tours on a 4 x n playing board Top Dice - problem 240 >>

# Problem 239: Twenty-two Foolish Primes

A set of disks numbered 1 through 100 are placed in a line in random order.

What is the probability that we have a partial derangement such that exactly 22 prime number discs are found away from their natural positions?
(Any number of non-prime disks may also be found in or out of their natural positions.)

Give your answer rounded to 12 places behind the decimal point in the form 0.abcdefghijkl.

# My Algorithm

Let's simplify this problem:

• there are 25 prime numbers below 100
• exactly 25 - 22 = 3 must remain at their position
• actually it doesn't matter whether those fixed numbers are prime numbers or not !
• therefore I just look at the first 25 numbers (1 .. 25) and require that three of them are fixed
The Wikipedia article on Derangement contains all the needed formulas: en.wikipedia.org/wiki/Derangement
The "subfactorial" (I haven't heard that name before !) is the core concept behind my derangements() function.

derangements() tells me the number of ways such that the first three numbers are fixed.
There are 2300 ways of choosing any three primes out of the 25 available (choose(25,3)).
Now I a total count of deranged sets - dividing it by the number of permutations (it's 100!) gives the probability.

## Alternative Approaches

A different approach can be found on www.numericana.com/answer/counting.htm although they don't go into detail
(except mentioning inclusion-exclusion principle, see en.wikipedia.org/wiki/Inclusionâ€“exclusion_principle)

## Note

Unlike most of my solutions, this time choose() and factorial() return double because their results are really big and their last digits don't matter.
There is an opportunity to pre-compute the factorials but my program terminates after less than 0.01 seconds even without this optimization.

I was a bit confused whether my result has to be the "raw" result or a "percentage" (thus multiplied by 100).
Strangely enough, my first attempt was correct - that's usually never happens ...

# 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 primes moved away from their original position.

This is equivalent to
echo 21 | ./239

Output:

Note: the original problem's input 22 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.

       #include <iostream>
#include <iomanip>

// ---------- based on similar code in my toolbox ----------

// factorial
// result is not accurate but supports large ranges
double factorial(unsigned int n)
{
double result = 1;
while (n > 1)
result *= n--;
return result;
}

// number of ways to choose n elements from k available
double choose(unsigned int n, unsigned int k)
{
// n! / (n-k)!k!
return factorial(n) / (factorial(n - k) * factorial(k));
}

// ---------- problem-specific code ----------

// count derangement
// note: need double as return type because results will be HUGE
double derangements(unsigned int move, unsigned int dontCare)
{
// don't need to move a prime away from its original position ?
if (move < 1)
return factorial(dontCare); // permutation of all remaining numbers

// recursion
move--;
auto result = dontCare * derangements(move,     dontCare);
if (move > 0)
result   += move     * derangements(move - 1, dontCare + 1);

return result;
}

int main()
{
unsigned int disks  = 100;
unsigned int primes =  25;
unsigned int moved  =  22;
std::cin >> moved;

// detect invalid input: for live test only
if (moved > primes)
return 1;

unsigned int unchanged = primes - moved;

// count ways
double result = derangements(moved, disks - primes);

// => 2300 ways to choose 3 primes
result *= choose(primes, unchanged);
// divide by total number of permutations
result /= factorial(disks);

// display result
std::cout << std::fixed << std::setprecision(12) << result << std::endl;
return 0;
}


This solution contains 13 empty lines, 15 comments and 2 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

September 9, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 65% (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.

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