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

# Problem 260: Stone Game

(see projecteuler.net/problem=260)

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`

.

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#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.

# 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 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)*

# Benchmark

The correct solution to the original Project Euler problem was found in 0.5 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

August 4, 2017 submitted solution

August 4, 2017 added comments

# Difficulty

Project Euler ranks this problem at **70%** (out of 100%).

# Links

projecteuler.net/thread=260 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

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My username at Project Euler is

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# 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 259 - Reachable Numbers | Binary Circles - problem 265 >> |