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Find winner in game of N balls, in which a player can remove any balls in range [A, B] in a single move

  • Last Updated : 02 Jul, 2021

Given two integers A and B, and also given that Alice and Bob are playing a game starting with a bag containing N balls, in which, in a single move, a player can remove any number of balls between the range [A, B] and if the player cannot remove any balls, then the player loses, the task is to find the winner of the game if Alice and Bob play the game alternatively and optimally and Alice starts the game.

Examples:

Input: N = 2, A = 1, B = 1
Output: Bob
Explanation: 
One way in which the game can be played is:

  1. Alice removes 1 ball, so the remaining balls are 1.
  2. Now, Bob removes the last ball.
  3. Alice cannot remove any balls, so she loses.

Input: N = 3, A = 1, B = 2
Output: Bob

Naive Approach: The simplest approach is to find the grundy number for every state and find the winning and losing states by using Sprague – Grundy Theorem.



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

Efficient Approach: The above approach can be optimized based on the following observations:

  1. Firstly, it can be observed that (A + B) is always losing state, because whatever X (A ≤ X ≤ B) number of balls the current player chooses, the opponent can always empty the bag as there will be (A + B – X) number of balls left where A ≤ (A + B – X) ≤ B.
  2. Also, from previous observations, one can observe that, for any multiple of (A + B) say, m*(A + B), the opponent can always reduce the current player to a state of (m – 1)*(A + B) and from (m – 1)*(A + B) to (m – 2)*(A + B) and so on.
  3. Thus, extending the above observation, one can say that for any multiple of (A + B), the opponent can always reduce the current player to a state of exactly (A + B), which is indeed a losing state. Therefore, all multiples of (A + B) are a losing state.
  4. So, the optimal choice for any player is to reduce the opponent to a multiple of (A + B), because after this, the player can always win, no matter what the opponent’s moves are.
  5. So, now losing states are the states from which one can never reduce the opponent to a multiple of (A + B).
  6. Therefore, any player with the state of the form: ((A + B)*m + y), where (0 ≤  y ≤ A-1) can never force the opponent to reduce to a multiple of (A + B), as any player can only pick at least A and at most B number of balls.

Follow the steps below to solve the problem:

  • If N%(A+B) is less than A, then print “Bob“.
  • Otherwise, print “Alice“.

  Below is the implementation of the above approach:

C++




// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the winner of the
// game
string NimGame(int N, int A, int B)
{
    // Stores sum of A and B
    int sum = A + B;
 
    // If N is of the form
    // m*(A+B)+y
    if (N % sum <= A - 1)
        return "Bob";
    // Otherwise,
    else
        return "Alice";
}
 
// Driver code
int main()
{
    // Input
    int N = 3, A = 1, B = 2;
    // Function call
    cout << NimGame(N, A, B) << endl;
 
    return 0;
}

Java




// Java Program for the above approach
import java.io.*;
 
class GFG
{
 
  // Function to find the winner of the
  // game
  public static String NimGame(int N, int A, int B)
  {
 
    // Stores sum of A and B
    int sum = A + B;
 
    // If N is of the form
    // m*(A+B)+y
    if (N % sum <= A - 1)
      return "Bob";
 
    // Otherwise,
    else
      return "Alice";
  }
 
  public static void main (String[] args)
  {
 
    // Input
    int N = 3, A = 1, B = 2;
 
    // Function call
    System.out.println(NimGame(N, A, B));
 
  }
}
 
  // This code is contributed by Potta Lokesh

Python3




# Python3 program of the above approach
 
# Function to find the winner of the game
def NimGame(N, A, B):
     
    # Stores sum of A and B
    sum = A + B
     
    # If N is of the form
    # m*(A+B)+y
    if (N % sum <= A - 1):
        return "Bob"
         
    # Otherwise,
    else:
        return "Alice"
 
# Driver code
 
# Input
N = 3
A = 1
B = 2
 
# Function call
print(NimGame(N, A, B))
 
# This code is contributed by amreshkumar3

C#




// C# program of the above approach
using System;
 
class GFG{
 
// Function to find the winner of the
// game
public static String NimGame(int N, int A, int B)
{
     
    // Stores sum of A and B
    int sum = A + B;
     
    // If N is of the form
    // m*(A+B)+y
    if (N % sum <= A - 1)
        return "Bob";
     
    // Otherwise,
    else
        return "Alice";
}
 
// Driver code
static void Main()
{
     
    // Input
    int N = 3, A = 1, B = 2;
     
    // Function call
    Console.Write(NimGame(N, A, B));
}
}
 
// This code is contributed by SoumikMondal

Javascript




<script>
        // JavaScript program for the above approach
 
        // Function to find the winner of the
        // game
        function NimGame(N, A, B) {
            // Stores sum of A and B
            let sum = A + B;
 
            // If N is of the form
            // m*(A+B)+y
            if (N % sum <= A - 1)
                return "Bob";
            // Otherwise,
            else
                return "Alice";
        }
 
        // Driver code
 
        // Input
        let N = 3, A = 1, B = 2;
        // Function call
        document.write(NimGame(N, A, B) + "<br>");
 
  // This code is contributed by Potta Lokesh
    </script>
Output
Bob

Time Complexity: O(1)
Auxiliary Space: O(1)




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