<< problem 327 - Rooms of Doom Special partitions - problem 333 >>

Problem 329: Prime Frog

Susan has a prime frog.
Her frog is jumping around over 500 squares numbered 1 to 500. He can only jump one square to the left or to the right, with equal probability,
and he cannot jump outside the range [1;500].
(if it lands at either end, it automatically jumps to the only available square on the next move.)

When he is on a square with a prime number on it, he croaks 'P' (PRIME) with probability 2/3 or 'N' (NOT PRIME) with probability 1/3 just before jumping to the next square.
When he is on a square with a number on it that is not a prime he croaks 'P' with probability 1/3 or 'N' with probability 2/3 just before jumping to the next square.

Given that the frog's starting position is random with the same probability for every square, and given that she listens to his first 15 croaks,
what is the probability that she hears the sequence PPPPNNPPPNPPNPN?

My Algorithm

A simple prime sieve finds all 95 primes between 1 and 500 in less than a millisecond.
The probability of reaching a square is 1/2 and it has the "right prime label" in 1/2 or 2/3 of all cases.
That means that there are 14 jumps and 15 "croak decisions". The denominator is always 500 * 2^14 * 3^15 (before reducing the fraction).

My program calculates the numerator recursively:

• each step checks whether the frog croaks in 1/3 or 2/3 of all cases (chance = 1 or chance = 2)
• then the frog jumps right or left (respecting the borders of square 1 and 500)
• memoization speeds up the whole process
The greatest common divisor of numerator and denominator is computed with the gcd function from my toolbox (this time I changed it to a template - only for the fun of it).
The result is displayed after dividing numerator and denominator by their gcd .

Alternative Approaches

There is no need to have the gcd function/template. The denominator has only prime factors 2, 3 and 5.
I could repeatedly divide the numerator and denominator by those numbers.

Note

Sequence has a dummy whitespace at position 0 because that position is unused.
Depth x queries Sequence[x] where x starts at 1 and finished at 15.

Memoization isn't strictly necessary because the result will be found without it in less than a second.
There are just 2^15 * 500 = 16384000 final states.

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 squares and the number of croaks

This is equivalent to
echo "10 3" | ./329

Output:

Note: the original problem's input 500 15 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 <vector>

// find greatest common divisor
template <typename T>
T gcd(T a, T b)
{
while (a != 0)
{
T c = a;
a = b % a;
b = c;
}
return b;
}

// croaked sequence
const char* Sequence = " PPPPNNPPPNPPNPN";
// length of croaked sequence
unsigned int maxDepth = 15;
// number of different starting positions
unsigned int limit = 500;

// prime numbers <= limit
static std::vector<char> isPrime; // I started with vector<bool> but vector<char> is a bit faster
// memoize intermediate results
static std::vector<unsigned int> cache;

// return only numerator, the denominator will be (2*3)^depth (before reducing)
unsigned int probability(unsigned int square, unsigned int depth = 1)
{
// either 1/3 or 2/3, discard the constant denominator 3 for now
unsigned int chance = 1;
if (isPrime[square] ^ (Sequence[depth] == 'N'))
chance = 2;

// done ?
if (depth == maxDepth)
return chance;

// memoize
auto id = square * maxDepth + depth;
if (cache.empty())
cache.resize((maxDepth + 1) * (limit + 1), 0);
if (cache[id] != 0)
return cache[id];

auto left  = square - 1;
if (left < 1) // avoid left border
left = 1 + 1;

auto right = square + 1;
if (right > limit) // avoid right border
right = limit - 1;

auto result = chance * (probability(left, depth + 1) + probability(right, depth + 1));
cache[id] = result;
return result;
}

int main()
{
std::cin >> limit >> maxDepth;

// prime sieve
isPrime.resize(limit + 1, true);
isPrime[1] = false;
for (unsigned int i = 2; i*i <= limit; i++)
if (isPrime[i])
for (unsigned int j = i*i; j <= limit; j += i)
isPrime[j] = false;

// let's croak !
unsigned long long sum = 0;
for (unsigned int i = 1; i <= limit; i++)
sum += probability(i);

// 500 * 3^15 * 2^14
unsigned long long denominator = limit;
for (unsigned int i = 1; i < maxDepth; i++)
denominator *= 3 * 2; // 3^14 * 2^14
denominator *= 3;

// reduce fraction
auto divide = gcd(sum, denominator);
std::cout << sum / divide << "/" << denominator / divide << std::endl;
return 0;
}


This solution contains 14 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

August 13, 2017 submitted solution

Difficulty

Project Euler ranks this problem at 25% (out of 100%).

Heatmap

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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 327 - Rooms of Doom Special partitions - problem 333 >>
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