<< problem 239 - Twenty-two Foolish Primes Resilience - problem 243 >>

# Problem 240: Langton's ant

There are 1111 ways in which five 6-sided dice (sides numbered 1 to 6) can be rolled so that the top three sum to 15.
Some examples are:

D_1,D_2,D_3,D_4,D_5 = 4,3,6,3,5
D_1,D_2,D_3,D_4,D_5 = 4,3,3,5,6
D_1,D_2,D_3,D_4,D_5 = 3,3,3,6,6
D_1,D_2,D_3,D_4,D_5 = 6,6,3,3,3

In how many ways can twenty 12-sided dice (sides numbered 1 to 12) be rolled so that the top ten sum to 70?

# My Algorithm

Another Dynamic Programming solution:

• generate dices in descending order, e.g. 655311
• when all 10 top dices are generated, check whether their sum is 70, abort if not
• when all 20 dices are generated, compute the number of permutations:
dfrac{20!}{s_1! s_2! s_3! ... s_12!}
... where s_i is the number of dices where the visible side is i, e.g. s_2 counts all 2.

## Alternative Approaches

You can probably solve this with a pure analytic (on a piece of paper).

## Note

• precomputing the factorial in main is about twice as fast as computing it on-the-fly in search.
• re-using the parameter dices avoids a huge number of memory allocations / deallocations and makes the code about 5x faster.
• re-using the container howOften (in count) has a less significant effect: about 30% faster.

# 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 dices, followed by the maximum points of a dice, the number of top dices and their sum.

This is equivalent to
echo "5 6 3 15" | ./240

Output:

Note: the original problem's input 20 12 10 70 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>

// configure environment
unsigned int numDice   = 20;// 5;
unsigned int maxPoints = 12;// 6;
unsigned int numTop    = 10;// 3;
unsigned int sumTop    = 70;//15;

// 0! ... 20!
unsigned long long factorial[21];

// return number of permutations of dices
unsigned long long count(const std::vector<unsigned int>& dices)
{
static std::vector<unsigned int> howOften(maxPoints + 1, 0); // re-use the same vector
// re-initialize
for (auto& x : howOften)
x = 0;

// count how often we see each side (1,2,3,...,6 or ...,12)
for (auto x : dices)
howOften[x]++;

// n! / (ones! * twos! * threes! * ...)
unsigned long long result = factorial[numDice];
for (auto x : howOften)
if (x > 1) // factorial[0] = factorial[1] = 1 ==> no need to divide
result /= factorial[x];

return result;
}

// generate all dices in descending order
// then compute how many permutations  exist for each combination
unsigned long long search(std::vector<unsigned int>& dices)
{
// enough dices ?
if (dices.size() == numDice)
return count(dices);

// check top dices: "hit the spot" ?
if (dices.size() == numTop)
{
unsigned int sum = 0;
for (auto x : dices)
sum += x;

// no, top dices' sum is either too high or too low => abort
if (sum != sumTop)
return 0;
}

// add one more dice - always in descending order, never higher than the previous dice
unsigned int maxDice = maxPoints;
if (!dices.empty())
maxDice = dices.back();

// all possible dices ...
unsigned long long result = 0;
for (unsigned int dice = 1; dice <= maxDice; dice++)
{
dices.push_back(dice);
result += search(dices);
dices.pop_back(); // re-use the same vector over and over again (for performance reasons)
}

return result;
}

int main()
{
std::cin >> numDice >> maxPoints >> numTop >> sumTop;
// catch invalid input
if (numTop > numDice)
return 1;

// precompute factorials 0! .. 20!
factorial[0] = 1;
unsigned long long current = 1;
for (unsigned int i = 1; i <= numDice; i++)
{
current *= i;
factorial[i] = current;
}

// and go !
std::vector<unsigned int> dices;
std::cout << search(dices) << std::endl;
return 0;
}


This solution contains 15 empty lines, 16 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.12 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

July 20, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 60% (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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