<< problem 235 - An Arithmetic Geometric sequence | Top Dice - problem 240 >> |

# Problem 239: Twenty-two Foolish Primes

(see projecteuler.net/problem=239)

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:

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.

#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 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

September 9, 2017 submitted solution

September 9, 2017 added comments

# Difficulty

Project Euler ranks this problem at **65%** (out of 100%).

# Links

projecteuler.net/thread=239 - **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 235 - An Arithmetic Geometric sequence | Top Dice - problem 240 >> |