my_rage's blog

By my_rage, history, 7 weeks ago, In English

Q. The game starts with 3 piles having 3, 4 and 5 stones, and the player to move may take any positive number of stones upto 3 only from any of the piles [Provided that the pile has that much amount of stones]. The last player to move wins. Which player wins the game assuming that both players play optimally?

ans )

/* Game Description- "A game is played between two players and there are N piles of stones such that each pile has certain number of stones. On his/her turn, a player selects a pile and can take any non-zero number of stones upto 3 (i.e- 1,2,3) The player who cannot move is considered to lose the game (i.e., one who take the last stone is the winner). Can you find which player wins the game if both players play optimally (they don't make any mistake)? "

A Dynamic Programming approach to calculate Grundy Number and Mex and find the Winner using Sprague — Grundy Theorem. */

include<bits/stdc++.h>

using namespace std;

/* piles[] -> Array having the initial count of stones/coins in each piles before the game has started. n -> Number of piles

Grundy[] -> Array having the Grundy Number corresponding to the initial position of each piles in the game

The piles[] and Grundy[] are having 0-based indexing*/

define PLAYER1 1

define PLAYER2 2

// A Function to calculate Mex of all the values in that set int calculateMex(unordered_set Set) { int Mex = 0;

while (Set.find(Mex) != Set.end())
    Mex++;

return (Mex);

}

// A function to Compute Grundy Number of 'n' int calculateGrundy(int n, int Grundy[]) { Grundy[0] = 0; Grundy[1] = 1; Grundy[2] = 2; Grundy[3] = 3;

if (Grundy[n] != -1)
    return (Grundy[n]);

unordered_set<int> Set; // A Hash Table

for (int i=1; i<=3; i++)
       Set.insert (calculateGrundy (n-i, Grundy));

// Store the result
Grundy[n] = calculateMex (Set);

return (Grundy[n]);

}

// A function to declare the winner of the game void declareWinner(int whoseTurn, int piles[], int Grundy[], int n) { int xorValue = Grundy[piles[0]];

for (int i=1; i<=n-1; i++)
    xorValue = xorValue ^ Grundy[piles[i]];

if (xorValue != 0)
{
    if (whoseTurn == PLAYER1)
       printf("Player 1 will win\n");
    else
       printf("Player 2 will win\n");
}
else
{
    if (whoseTurn == PLAYER1)
       printf("Player 2 will win\n");
    else
       printf("Player 1 will win\n");
}

return;

}

// Driver program to test above functions int main() { // Test Case 1 int piles[] = {3, 4, 5}; int n = sizeof(piles)/sizeof(piles[0]);

// Find the maximum element
int maximum = *max_element(piles, piles + n);

// An array to cache the sub-problems so that
// re-computation of same sub-problems is avoided
int Grundy[maximum + 1];
memset(Grundy, -1, sizeof (Grundy));

// Calculate Grundy Value of piles[i] and store it
for (int i=0; i<=n-1; i++)
    calculateGrundy(piles[i], Grundy);

declareWinner(PLAYER1, piles, Grundy, n);

/* Test Case 2
int piles[] = {3, 8, 2};
int n = sizeof(piles)/sizeof(piles[0]);


int maximum = *max_element (piles, piles + n);

// An array to cache the sub-problems so that
// re-computation of same sub-problems is avoided
int Grundy [maximum + 1];
memset(Grundy, -1, sizeof (Grundy));

// Calculate Grundy Value of piles[i] and store it
for (int i=0; i<=n-1; i++)
    calculateGrundy(piles[i], Grundy);

declareWinner(PLAYER2, piles, Grundy, n); */

return (0);

}

 
 
 
 
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7 weeks ago, # |
  Vote: I like it +14 Vote: I do not like it

Reading your article makes me feel very tired and terrible...

Could you use markdown and LaTeX please?

»
7 weeks ago, # |
  Vote: I like it +17 Vote: I do not like it

Atleast preview blog before posting.