Optimal Strategy for the Divisor game using Dynamic Programming

Last Updated : 10 Mar, 2023

Given an integer N and two players, A and B are playing a game. On each playerâ€™s turn, that player makes a move by subtracting a divisor of current N (which is less than N) from current N, thus forming a new N for the next turn. The player who does not have any divisor left to subtract loses the game. The task is to tell which player wins the game if player A takes the first turn, assuming both players play optimally.
Examples:

Input : N = 2
Output : Player A wins
Explanation :
Player A chooses 1, and B has no more moves.
Input : N = 3
Output : Player B wins
Explanation :
Player A chooses 1, player B chooses 1, and A has no more moves.

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Approach :
This problem mentioned above can be solved using Dynamic Programming.

• We will take a DP having 2 states i.e.

N -> current number left
A -> boolean value to decide if it’s player A’s turn or not

•
• At each state, we will try to find all the divisors of N and will try to find the next state where the current player is winning. For player A, we will try to find the next state where the return value is true while for player B, we will try to find the next state where the return value is false (as false represents the loss of player A).
• The base cases will be for N=1 where always the player A will lose and N=2 where always the player B will lose.
• To find the answer, we just need to find the value of DP[ N ][ 1 ].

Below is the implementation of the above approach:

C++

 `// C++ program for implementation of` `// Optimal Strategy for the Divisor` `// Game using Dynamic Programming` `#include ` `using` `namespace` `std;`   `// Recursive function to find the winner` `bool` `divisorGame(``int` `N, ``bool` `A, ``int` `dp[][2])` `{`   `    ``// check if N=1 or N=3 then player B wins` `    ``if` `(N == 1 or N == 3)` `        ``return` `false``;`   `    ``// check if N=2 then player A wins` `    ``if` `(N == 2)` `        ``return` `true``;`   `    ``// check if current state already visited` `    ``// then return the previously obtained ans` `    ``if` `(dp[N][A] != -1)` `        ``return` `dp[N][A];`   `    ``// check if currently it is player A's turn` `    ``// then initialise the ans to 0` `    ``int` `ans = (A == 1) ? 0 : 1;`   `    ``// Traversing across all the divisors of N` `    ``// which are less than N` `    ``for` `(``int` `i = 1; i * i <= N; i++) {`   `        ``// check if current value of` `        ``// i is a divisor of N` `        ``if` `(N % i == 0) {`   `            ``// check if it is player A's turn` `            ``// then we need at least one true` `            ``if` `(A)` `                ``ans |= divisorGame(N - i, 0, dp);`   `            ``// Else if it is player B's turn` `            ``// then we need at least one false` `            ``else` `                ``ans &= divisorGame(N - i, 1, dp);` `        ``}` `    ``}`   `    ``// Return the current ans` `    ``return` `dp[N][A] = ans;` `}`   `// Driver code` `int` `main()` `{` `    ``// initialise N` `    ``int` `N = 3;`   `    ``int` `dp[N + 1][2];`   `    ``memset``(dp, -1, ``sizeof``(dp));`   `    ``if` `(divisorGame(N, 1, dp) == ``true``)` `        ``cout << ``"Player A wins"``;` `    ``else` `        ``cout << ``"Player B wins"``;`   `    ``return` `0;` `}`

Java

 `// Java program for implementation of` `// optimal strategy for the divisor` `// game using dynamic programming`   `import` `java.util.*;`   `class` `GFG {`   `    ``// Recursive function to find the winner` `    ``static` `int` `divisorGame(``int` `N, ``int` `A, ``int` `dp[][]) {`   `        ``// Check if N = 1 or N = 3 then player B wins` `        ``if` `(N == ``1` `|| N == ``3``)` `            ``return` `0``;`   `        ``// Check if N = 2 then player A wins` `        ``if` `(N == ``2``)` `            ``return` `1``;`   `        ``// Check if current state already visited` `        ``// then return the previously obtained ans` `        ``if` `(dp[N][A] != -``1``)` `            ``return` `dp[N][A];`   `        ``// Check if currently it is player A's turn` `        ``// then initialise the ans to 0` `        ``int` `ans = (A == ``1``) ? ``0` `: ``1``;`   `        ``// Traversing across all the divisors of N` `        ``// which are less than N` `        ``for` `(``int` `i = ``1``; i * i <= N; i++) {`   `            ``// Check if current value of` `            ``// i is a divisor of N` `            ``if` `(N % i == ``0``) {`   `                ``// Check if it is player A's turn` `                ``// then we need at least one true` `                ``if` `(A == ``1``)` `                    ``ans |= divisorGame(N - i, ``0``, dp);`   `                ``// Else if it is player B's turn` `                ``// then we need at least one false` `                ``else` `                   ``ans &= divisorGame(N - i, ``1``, dp);` `            ``}` `        ``}`   `        ``// Return the current ans` `        ``return` `dp[N][A] = ans;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args) {` `        `  `        ``// Initialise N` `        ``int` `N = ``3``;`   `        ``int``[][] dp = ``new` `int``[N + ``1``][``2``];`   `        ``for` `(``int` `i = ``0``; i < N + ``1``; i++) {` `             ``for` `(``int` `j = ``0``; j < ``2``; j++) {` `                  ``dp[i][j] = -``1``;` `            ``}` `        ``}`   `        ``if` `(divisorGame(N, ``1``, dp) == ``1``)` `            ``System.out.print(``"Player A wins"``);` `        ``else` `            ``System.out.print(``"Player B wins"``);`   `    ``}` `}`   `// This code contributed by sapnasingh4991`

Python3

 `# Python3 program for implementation of` `# Optimal Strategy for the Divisor` `# Game using Dynamic Programming`   `from` `math ``import` `sqrt`   `# Recursive function to find the winner` `def` `divisorGame(N,A,dp):` `    ``# check if N=1 or N=3 then player B wins` `    ``if` `(N ``=``=` `1` `or` `N ``=``=` `3``):` `        ``return` `False`   `    ``# check if N=2 then player A wins` `    ``if` `(N ``=``=` `2``):` `        ``return` `True`   `    ``# check if current state already visited` `    ``# then return the previously obtained ans` `    ``if` `(dp[N][A] !``=` `-``1``):` `        ``return` `dp[N][A]`   `    ``# check if currently it is player A's turn` `    ``# then initialise the ans to 0` `    ``if``(A ``=``=` `1``):` `        ``ans ``=` `0` `    ``else``:` `        ``ans ``=` `1`   `    ``# Traversing across all the divisors of N` `    ``# which are less than N` `    ``for` `i ``in` `range``(``1``,``int``(sqrt(N))``+``1``,``1``):` `        ``# check if current value of` `        ``# i is a divisor of N` `        ``if` `(N ``%` `i ``=``=` `0``):` `            ``# check if it is player A's turn` `            ``# then we need at least one true` `            ``if` `(A):` `                ``ans |``=` `divisorGame(N ``-` `i, ``0``, dp)`   `            ``# Else if it is player B's turn` `            ``# then we need at least one false` `            ``else``:` `                ``ans &``=` `divisorGame(N ``-` `i, ``1``, dp) `   `    ``dp[N][A] ``=` `ans`     `    ``# Return the current ans` `    ``return` `dp[N][A]`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``# initialise N` `    ``N ``=` `3`   `    ``dp ``=` `[[``-``1` `for` `i ``in` `range``(``2``)] ``for` `j ``in` `range``(N``+``1``)]`   `    ``if` `(divisorGame(N, ``1``, dp) ``=``=` `True``):` `        ``print``(``"Player A wins"``)` `    ``else``:` `        ``print``(``"Player B wins"``)`   `# This code is contributed by Surendra_Gangwar`

C#

 `// C# program for implementation of` `// optimal strategy for the divisor` `// game using dynamic programming` `using` `System;`   `class` `GFG {`   `// Recursive function to find the winner` `static` `int` `divisorGame(``int` `N, ``int` `A, ` `                       ``int` `[,]dp)` `{`   `    ``// Check if N = 1 or N = 3 ` `    ``// then player B wins` `    ``if` `(N == 1 || N == 3)` `        ``return` `0;`   `    ``// Check if N = 2 then player A wins` `    ``if` `(N == 2)` `        ``return` `1;`   `    ``// Check if current state already visited` `    ``// then return the previously obtained ans` `    ``if` `(dp[N, A] != -1)` `        ``return` `dp[N, A];`   `    ``// Check if currently it is player A's turn` `    ``// then initialise the ans to 0` `    ``int` `ans = (A == 1) ? 0 : 1;`   `    ``// Traversing across all the divisors of N` `    ``// which are less than N` `    ``for``(``int` `i = 1; i * i <= N; i++)` `    ``{` `        `  `       ``// Check if current value of` `       ``// i is a divisor of N` `       ``if` `(N % i == 0)` `       ``{` `           `  `           ``// Check if it is player A's turn` `           ``// then we need at least one true` `           ``if` `(A == 1)` `               ``ans |= divisorGame(N - i, 0, dp);` `               `  `           ``// Else if it is player B's turn` `           ``// then we need at least one false` `           ``else` `              ``ans &= divisorGame(N - i, 1, dp);` `       ``}` `    ``}` `    `  `    ``// Return the current ans` `    ``return` `dp[N, A] = ans;` `}`   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    `  `    ``// Initialise N` `    ``int` `N = 3;` `    ``int``[,] dp = ``new` `int``[N + 1, 2];` `    `  `    ``for``(``int` `i = 0; i < N + 1; i++) ` `    ``{` `       ``for``(``int` `j = 0; j < 2; j++) ` `       ``{` `          ``dp[i, j] = -1;` `       ``}` `    ``}` `    `  `    ``if` `(divisorGame(N, 1, dp) == 1)` `    ``{` `        ``Console.Write(``"Player A wins"``);` `    ``}` `    ``else` `    ``{` `        ``Console.Write(``"Player B wins"``);` `    ``}` `}` `}`   `// This code is contributed by amal kumar choubey`

Javascript

 ``

Output:

`Player B wins`

Time Complexity: O(N*log(N))
Auxiliary Space: O(N)

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