Kombinatorinio žaidimo teorija | 4 rinkinys (Sprague - Grundy teorema)
Būtinos sąlygos: Grundy skaičiai/numeriai ir mex
Mes jau matėme 2 rinkinyje (https://www.geeksforgeeks.org/dsa/combinatoral-game-theory-set-2-game-nim/), kuriuos galime rasti, kas laimi NIM žaidimą, nežaidžiant žaidimo.
Tarkime, kad mes šiek tiek pakeisime klasikinį „Nim“ žaidimą. Šį kartą kiekvienas žaidėjas gali pašalinti tik 1 2 ar 3 akmenis (o ne bet kokį akmenų skaičių, kaip klasikiniame „Nim“ žaidime). Ar galime nuspėti, kas laimės?
Taip, mes galime numatyti nugalėtoją naudodami „Sprague-Grundy“ teoremą.
Kas yra „Sprague-Grundy“ teorema?
Tarkime, kad yra sudėtinis žaidimas (daugiau nei vienas pogrupis), sudarytas iš N sub-žaidimų ir dviejų žaidėjų A ir B., tada „Sprague-Grundy“ teorema sako, kad jei ir A, ir B žaidžia optimaliai (t. Y. Jie nepadaro jokių klaidų), tada žaidėjas pradedamas pirmiausia, jei laimės, jei laimės „Grundy“ pozicijos skaičių kiekvieno pobūdžio žaidimuose žaidimo pradžioje, o ne „Zero“. Priešingu atveju, jei XOR įvertins iki nulio, žaidėjas A praras tikrai, nesvarbu.
Kaip pritaikyti „Sprague Grundy“ teoremą?
Mes galime pritaikyti „Sprague-Grundy“ teoremą bet kurioje nešališkas žaidimas ir išspręsti. Pagrindiniai veiksmai yra išvardyti taip:
- Suskirstykite kompozicinį žaidimą į „Subbames“.
- Tada kiekvienam pogrupiui apskaičiuokite Grundy numerį toje padėtyje.
- Tada apskaičiuokite visų apskaičiuotų Grundy skaičių XOR.
- Jei XOR vertė nėra lygi nuliui, tada žaidėjas, kuris ketina pasukti (pirmasis žaidėjas), laimės kitą, jam lemta prarasti, nesvarbu.
Žaidimo pavyzdys: Žaidimas prasideda nuo 3 polių, turinčių 3 4 ir 5 akmenis, o žaidėjas gali judėti, bet kokį teigiamą akmenų skaičių iki 3 iš bet kurio iš polių gali paimti teigiamą skaičių iki 3 iki 3 [su sąlyga, kad krūva turi tiek daug akmenų]. Laimės paskutinis žaidėjas. Kuris žaidėjas laimi žaidimą, darant prielaidą, kad abu žaidėjai žaidžia optimaliai?
Kaip pasakyti, kas laimės pritaikydamas „Sprague-Grundy“ teoremą?
Kaip matome, kad šį žaidimą pats sudaro kelios pobūdžio.
Pirmasis žingsnis: Pogrupiai gali būti laikomi kiekvienu polių.
Antras žingsnis: Iš žemiau esančios lentelės matome
Grundy(3) = 3 Grundy(4) = 0 Grundy(5) = 1
Mes jau matėme, kaip apskaičiuoti šio žaidimo grūdų numerius ankstesnis straipsnis.
Trečias žingsnis: 3 0 1 = 2 xor
Ketvirtas žingsnis: Kadangi XOR yra ne nulinis skaičius, todėl galime pasakyti, kad pirmasis žaidėjas laimės.
Žemiau yra programa, kuri įgyvendina virš 4 žingsnių.
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>
Išvestis:
Player 1 will win
Laiko sudėtingumas: O (n^2), kur n yra maksimalus akmenų skaičius krūvoje.
Kosmoso sudėtingumas: O (n) Kadangi Grundy masyvas naudojamas subproblemų rezultatams saugoti, kad būtų išvengta nereikalingų skaičiavimų, ir jis užima o (n) erdvę.
Nuorodos:
https://en.wikipedia.org/wiki/sprague%E2%80%93GRUNDY_THEOREM
Mankšta skaitytojams: Apsvarstykite žemiau pateiktą žaidimą.
Žaidimą žaidžia du žaidėjai, turintys n sveikus skaičius A1 A2 .. an. Savo posūkyje žaidėjas pasirenka sveikąjį skaičių padalijus jį iš 2 3 arba 6, o tada paima grindis. Jei sveikasis skaičius tampa 0, jis pašalinamas. Laimės paskutinis žaidėjas. Kuris žaidėjas laimi žaidimą, jei abu žaidėjai žaidžia optimaliai?
Užuomina: žiūrėkite 3 pavyzdį ankstesnis straipsnis.
