<< problem 259 - Reachable Numbers Binary Circles - problem 265 >>

# Problem 260: Stone Game

A game is played with three piles of stones and two players.
At her turn, a player removes one or more stones from the piles.
However, if she takes stones from more than one pile, she must remove the same number of stones from each of the selected piles.

In other words, the player chooses some N>0 and removes:

N stones from any single pile; or
N stones from each of any two piles (2N total); or
N stones from each of the three piles (3N total).

The player taking the last stone(s) wins the game.

A winning configuration is one where the first player can force a win.
For example, (0,0,13), (0,11,11) and (5,5,5) are winning configurations because the first player can immediately remove all stones.

A losing configuration is one where the second player can force a win, no matter what the first player does.
For example, (0,1,2) and (1,3,3) are losing configurations: any legal move leaves a winning configuration for the second player.

Consider all losing configurations (x_i,y_i,z_i) where x_i <= y_i <= z_i <= 100.
We can verify that sum{x_i+y_i+z_i} = 173895 for these.

Find sum{x_i+y_i+z_i} where (x_i,y_i,z_i) ranges over the losing configurations with x_i <= y_i <= z_i <= 1000.

# My Algorithm

Every position (0,0,s) is won because the current player will take all s stones from the third pile and win immediately.
Every position (0,s,s) is won because the current player will take all s stones from the second and from the third pile and win immediately.
Every position (s,s,s) is won because the current player will take all s stones from all three piles and win immediately.
More specifically, these positions are won because there is a transition from the current state to the final state (0,0,0).

If the player to move has position (x_i,y_i,z_i) then he/she must avoid to take stones in such a way that his/her opponent has a won position.
For example, taking 3 stones from the largest pile of (0,2,5) will yield (0,2,2) and the opponent can will immediately.

A position is lost when all transitions lead to a position where the opponent wins.
A position is won when there is at least one transition leading to a position where the opponent loses.

I only need to consider moves where one pile becomes empty:

• if that is a winning position, then the current player will play that move
• if it turns out to be a losing position, then the opponent will enforce that move by mirroring
My program walks through all possible pile sizes x <= y <= z and looks at every potential move which "empties" one or more piles.
Only if all of them lead to a won position (→ the opponent wins), then count is incremented by x + y + z.

# 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):

This is equivalent to
echo 100 | ./260

Output:

Note: the original problem's input 1000 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>

unsigned int maxPileSize = 1000;

// compute linear index from two offsets
unsigned int id(unsigned int a, unsigned int b)
{
return a * (maxPileSize + 1) + b;
}

int main()
{
std::cin >> maxPileSize;

// the current position is only lost if every possible move leads to a position
// where the opponent wins
// that means: no matter how many stones I pick from one, two or three piles,
//             the follow-up state must not be lost

const bool Won  = false;
const bool Lost = true;

// assume each position is potentially won - unless proven otherwise
std::vector<bool> one((maxPileSize + 1) * (maxPileSize + 1), Won);
std::vector<bool> two((maxPileSize + 1) * (maxPileSize + 1), Won);
std::vector<bool> all((maxPileSize + 1) * (maxPileSize + 1), Won);

// count x+y+z for each lost position
unsigned int count = 0;

for (unsigned int x = 0; x <= maxPileSize; x++)
for (unsigned int y = x; y <= maxPileSize; y++)
{
// copy this condition from the inner loop (and keep there for the sake of completeness, though it's redundant)
if (one[id(x,y)] == Lost)
continue;

for (unsigned int z = y; z <= maxPileSize; z++)
{
// can current player win by clearing one pile ?
// check if (0,y,z) or (x,0,z) or (x,y,0) wins
// take x stones from first  pile: check (x-x,  y,   z ) = (0, y, z)
// take y stones from second pile: check ( x,  y-y,  z ) = (x, 0, z)
// take z stones from third  pile: check ( x,   y,  z-z) = (x, y, 0)
if (one[id(y,z)] == Lost ||
one[id(x,z)] == Lost ||
one[id(x,y)] == Lost)
continue; // opponent can be forced into a lost position

// can current player win by clearing one pile
// and removing the same number of stones from a second pile ?
// take x stones from first and second pile: check (x-x, y-x,  z ) = (0, y-x,  z )
// take y stones from first and second pile: check ( x,  y-y, z-y) = (x,  0,  z-y)
// take x stones from first and third  pile: check (x-x,  y,  z-x) = (0,  y,  z-x)
// (note that x <= y <= z, therefore can't take z stones)
if (two[id(y-x, z)] == Lost ||
two[id(z-y, x)] == Lost ||
two[id(z-x, y)] == Lost)
continue; // opponent can be forced into a lost position

// can current player win by clearing the first pile
// and removing the same number of stones from the second and third pile ?
// note that x <= y <= z
// take x stones from all piles: check (x-x, y-x, z-x) = (0, y-x, z-x)
if (all[id(y-x, z-x)] == Lost)
continue; // opponent can be forced into a lost position

// current player lost, opponents wins all follow-up positions
count += x + y + z;

// mark all reachable positions as lost
// (same indices and arrays as used in the if-conditions above)
one[id(y, z)] = Lost;
one[id(x, z)] = Lost;
one[id(x, y)] = Lost;

two[id(y-x, z)] = Lost;
two[id(z-y, x)] = Lost;
two[id(z-x, y)] = Lost;

all[id(y-x, z-x)] = Lost;
}
}

// and done !
std::cout << count << std::endl;
return 0;
}


This solution contains 16 empty lines, 27 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.5 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 4, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 70% (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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