Kombinatoryal Oyun Teorisi | Set 4 (Sprague - Grundy Teorem)
Önkoşul: Grundy Numaralar/Numaralar ve Mex
Set 2'de (https://www.geeksforgeeks.org/dsa/combinatorial-game-the--2-name-nim/) gördük.
Klasik NIM oyununu biraz değiştirdiğimizi varsayalım. Bu sefer her oyuncu sadece 1 2 veya 3 taşı çıkarabilir (ve NIM'in klasik oyununda olduğu gibi herhangi bir sayıda taş değil). Kimin kazanacağını tahmin edebilir miyiz?
Evet, Sprague-Grundy teoremini kullanarak kazananı tahmin edebiliriz.
Sprague-grundy teoremi nedir?
N alt oyunlarından ve iki oyuncu A ve B'den oluşan kompozit bir oyun (birden fazla alt oyun) olduğunu varsayalım. Daha sonra Sprague-grundy teoremi, hem a hem de B'nin en iyi şekilde oynarsa (yani herhangi bir hata yapmazlar), o zaman ilk başlayan oyuncunun her bir alt oyundaki hasta sayısının xor'larının her bir alt oyundaki konumun konumunun xor'unun, oyunun başlangıcında olmaması garanti edildiğini söylüyor. Aksi takdirde, Xor sıfır olarak değerlendirilirse, A oyuncusu ne olursa olsun kesinlikle kaybedecektir.
Sprague Grundy Teoremi nasıl uygulanır?
Sprague-grundy teoremini herhangi birine uygulayabiliriz tarafsız oyun ve çözün. Temel adımlar aşağıdaki gibi listelenmiştir:
- Kompozit oyunu alt oyunlara ayırın.
- Daha sonra her alt oyun için bu pozisyondaki grundy numarasını hesaplayın.
- Ardından, hesaplanan tüm Grundy sayılarının XOR'unu hesaplayın.
- XOR değeri sıfır değilse, dönüşü yapacak olan oyuncu (ilk oyuncu) kazanır, ne olursa olsun kaybetmeye mahkumdur.
Örnek Oyun: Oyun 3 4 ve 5 taşa sahip 3 kazıkla başlar ve hareket edecek oyuncu, sadece kazıklardan 3'e kadar pozitif sayıda taş alabilir [kazığın bu kadar taşa sahip olması şartıyla]. Taşıyan son oyuncu kazanır. Her iki oyuncunun da optimum oynadığını varsayarak hangi oyuncu oyunu kazanır?
Sprague-grundy teoremini uygulayarak kimin kazanacağını nasıl anlarsınız?
Görebildiğimiz gibi, bu oyunun kendisi birkaç alt oyundan oluşuyor.
İlk adım: Alt oyunlar her yığın olarak düşünülebilir.
İkinci adım: Aşağıdaki tablodan görüyoruz
Grundy(3) = 3 Grundy(4) = 0 Grundy(5) = 1
Bu oyunun grundy numaralarını nasıl hesaplayacağını zaten gördük. öncesi madde.
Üçüncü adım: 3 0 1 = 2
Dördüncü adım: Xor sıfır olmayan bir numara olduğundan, ilk oyuncunun kazanacağını söyleyebiliriz.
Aşağıda 4 adımın üzerinde uygulanan program verilmiştir.
C++ /* 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- 123) 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 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 < int > 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 ); }
Java import java.util.* ; /* 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- 123) 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. */ class GFG { /* 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*/ static int PLAYER1 = 1 ; static int PLAYER2 = 2 ; // A Function to calculate Mex of all the values in that set static int calculateMex ( HashSet < Integer > Set ) { int Mex = 0 ; while ( Set . contains ( Mex )) Mex ++ ; return ( Mex ); } // A function to Compute Grundy Number of 'n' static 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 ] ); // A Hash Table HashSet < Integer > Set = new HashSet < Integer > (); for ( int i = 1 ; i <= 3 ; i ++ ) Set . add ( calculateGrundy ( n - i Grundy )); // Store the result Grundy [ n ] = calculateMex ( Set ); return ( Grundy [ n ] ); } // A function to declare the winner of the game static 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 ) System . out . printf ( 'Player 1 will winn' ); else System . out . printf ( 'Player 2 will winn' ); } else { if ( whoseTurn == PLAYER1 ) System . out . printf ( 'Player 2 will winn' ); else System . out . printf ( 'Player 1 will winn' ); } return ; } // Driver code public static void main ( String [] args ) { // Test Case 1 int piles [] = { 3 4 5 }; int n = piles . length ; // Find the maximum element int maximum = Arrays . stream ( piles ). max (). getAsInt (); // An array to cache the sub-problems so that // re-computation of same sub-problems is avoided int Grundy [] = new int [ maximum + 1 ] ; Arrays . fill ( Grundy - 1 ); // 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); */ } } // This code is contributed by PrinciRaj1992
Python3 ''' 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- 123) 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. 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''' PLAYER1 = 1 PLAYER2 = 2 # A Function to calculate Mex of all # the values in that set def calculateMex ( Set ): Mex = 0 ; while ( Mex in Set ): Mex += 1 return ( Mex ) # A function to Compute Grundy Number of 'n' def calculateGrundy ( n Grundy ): Grundy [ 0 ] = 0 Grundy [ 1 ] = 1 Grundy [ 2 ] = 2 Grundy [ 3 ] = 3 if ( Grundy [ n ] != - 1 ): return ( Grundy [ n ]) # A Hash Table Set = set () for i in range ( 1 4 ): Set . add ( calculateGrundy ( n - i Grundy )) # Store the result Grundy [ n ] = calculateMex ( Set ) return ( Grundy [ n ]) # A function to declare the winner of the game def declareWinner ( whoseTurn piles Grundy n ): xorValue = Grundy [ piles [ 0 ]]; for i in range ( 1 n ): xorValue = ( xorValue ^ Grundy [ piles [ i ]]) if ( xorValue != 0 ): if ( whoseTurn == PLAYER1 ): print ( 'Player 1 will win n ' ); else : print ( 'Player 2 will win n ' ); else : if ( whoseTurn == PLAYER1 ): print ( 'Player 2 will win n ' ); else : print ( 'Player 1 will win n ' ); # Driver code if __name__ == '__main__' : # Test Case 1 piles = [ 3 4 5 ] n = len ( piles ) # Find the maximum element maximum = max ( piles ) # An array to cache the sub-problems so that # re-computation of same sub-problems is avoided Grundy = [ - 1 for i in range ( maximum + 1 )]; # Calculate Grundy Value of piles[i] and store it for i in range ( n ): 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); ''' # This code is contributed by rutvik_56
C# using System ; using System.Linq ; using System.Collections.Generic ; /* 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- 123) 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. */ class GFG { /* 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*/ static int PLAYER1 = 1 ; //static int PLAYER2 = 2; // A Function to calculate Mex of all the values in that set static int calculateMex ( HashSet < int > Set ) { int Mex = 0 ; while ( Set . Contains ( Mex )) Mex ++ ; return ( Mex ); } // A function to Compute Grundy Number of 'n' static 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 ]); // A Hash Table HashSet < int > Set = new HashSet < int > (); for ( int i = 1 ; i <= 3 ; i ++ ) Set . Add ( calculateGrundy ( n - i Grundy )); // Store the result Grundy [ n ] = calculateMex ( Set ); return ( Grundy [ n ]); } // A function to declare the winner of the game static 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 ) Console . Write ( 'Player 1 will winn' ); else Console . Write ( 'Player 2 will winn' ); } else { if ( whoseTurn == PLAYER1 ) Console . Write ( 'Player 2 will winn' ); else Console . Write ( 'Player 1 will winn' ); } return ; } // Driver code static void Main () { // Test Case 1 int [] piles = { 3 4 5 }; int n = piles . Length ; // Find the maximum element int maximum = piles . Max (); // An array to cache the sub-problems so that // re-computation of same sub-problems is avoided int [] Grundy = new int [ maximum + 1 ]; Array . Fill ( Grundy - 1 ); // 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); */ } } // This code is contributed by mits
JavaScript < script > /* 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- 123) 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. */ /* 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*/ let PLAYER1 = 1 ; let PLAYER2 = 2 ; // A Function to calculate Mex of all the values in that set function calculateMex ( Set ) { let Mex = 0 ; while ( Set . has ( Mex )) Mex ++ ; return ( Mex ); } // A function to Compute Grundy Number of 'n' function calculateGrundy ( n Grundy ) { Grundy [ 0 ] = 0 ; Grundy [ 1 ] = 1 ; Grundy [ 2 ] = 2 ; Grundy [ 3 ] = 3 ; if ( Grundy [ n ] != - 1 ) return ( Grundy [ n ]); // A Hash Table let Set = new Set (); for ( let i = 1 ; i <= 3 ; i ++ ) Set . add ( calculateGrundy ( n - i Grundy )); // Store the result Grundy [ n ] = calculateMex ( Set ); return ( Grundy [ n ]); } // A function to declare the winner of the game function declareWinner ( whoseTurn piles Grundy n ) { let xorValue = Grundy [ piles [ 0 ]]; for ( let i = 1 ; i <= n - 1 ; i ++ ) xorValue = xorValue ^ Grundy [ piles [ i ]]; if ( xorValue != 0 ) { if ( whoseTurn == PLAYER1 ) document . write ( 'Player 1 will win
' ); else document . write ( 'Player 2 will win
' ); } else { if ( whoseTurn == PLAYER1 ) document . write ( 'Player 2 will win
' ); else document . write ( 'Player 1 will win
' ); } return ; } // Driver code // Test Case 1 let piles = [ 3 4 5 ]; let n = piles . length ; // Find the maximum element let maximum = Math . max (... piles ) // An array to cache the sub-problems so that // re-computation of same sub-problems is avoided let Grundy = new Array ( maximum + 1 ); for ( let i = 0 ; i < maximum + 1 ; i ++ ) Grundy [ i ] = 0 ; // Calculate Grundy Value of piles[i] and store it for ( let 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); */ // This code is contributed by avanitrachhadiya2155 < /script>
Çıktı:
Player 1 will win
Zaman karmaşıklığı: O (n^2) burada n, bir yığın içindeki maksimum taş sayısıdır.
Uzay Karmaşıklığı: O (n) Grundy dizisi, yedek hesaplamaları önlemek için alt problemlerin sonuçlarını saklamak için kullanılır ve O (n) boşluğu alır.
Referanslar:
https://en.wikipedia.org/wiki/sprague%e2%93grundy_theorem
Okuyuculara egzersiz: Aşağıdaki oyunu düşünün.
Bir oyun N tamponu A1 A2 ile iki oyuncu tarafından oynanır. An. Onun dönüşünde bir oyuncu bir tamsayı seçer 2 3 veya 6 ile bölünür ve sonra zemini alır. Tamsayı 0 olursa kaldırılır. Taşıyan son oyuncu kazanır. Her iki oyuncu da optimum oynarsa hangi oyuncu oyunu kazanır?
İpucu: Örnek 3'e bakın öncesi madde.
