Optimal Strategy for a Game

Situatie

Problem statement: Consider a row of n coins of values v1 . . . vn, where n is even. We play a game against an opponent by alternating turns. In each turn, a player selects either the first or last coin from the row, removes it from the row permanently, and receives the value of the coin. Determine the maximum possible amount of money we can definitely win if we move first.

Solutie

Pasi de urmat

1. 5, 3, 7, 10 : The user collects maximum value as 15(10 + 5)

2. 8, 15, 3, 7 : The user collects maximum value as 22(7 + 15)

Does choosing the best at each move give an optimal solution?

No. In the second example, this is how the game can finish:

1.
…….User chooses 8.
…….Opponent chooses 15.
…….User chooses 7.
…….Opponent chooses 3.
Total value collected by user is 15(8 + 7)
2.
…….User chooses 7.
…….Opponent chooses 8.
…….User chooses 15.
…….Opponent chooses 3.
Total value collected by user is 22(7 + 15)
So if the user follows the second game state, maximum value can be collected although the first move is not the best.

// C++ program to find out maximum value from a 
// given sequence of coins 
#include <bits/stdc++.h> 
using namespace std; 
 
// Returns optimal value possible that a player can 
// collect from an array of coins of size n. Note 
// than n must be even 
int optimalStrategyOfGame(int* arr, int n) 
{ 
    // Create a table to store solutions of subproblems 
    int table[n][n]; 
 
    // Fill table using above recursive formula. Note  
    // that the table is filled in diagonal fashion (similar 
    // to http://goo.gl/PQqoS), from diagonal elements to 
    // table[0][n-1] which is the result. 
    for (int gap = 0; gap < n; ++gap) { 
        for (int i = 0, j = gap; j < n; ++i, ++j) { 
 
            // Here x is value of F(i+2, j), y is F(i+1, j-1) and 
            // z is F(i, j-2) in above recursive formula 
            int x = ((i + 2) <= j) ? table[i + 2][j] : 0; 
            int y = ((i + 1) <= (j - 1)) ? table[i + 1][j - 1] : 0; 
            int z = (i <= (j - 2)) ? table[i][j - 2] : 0; 
 
            table[i][j] = max(arr[i] + min(x, y), arr[j] + min(y, z)); 
        } 
    } 
 
    return table[0][n - 1]; 
} 
 
// Driver program to test above function 
int main() 
{ 
    int arr1[] = { 8, 15, 3, 7 }; 
    int n = sizeof(arr1) / sizeof(arr1[0]); 
    printf("%d\n", optimalStrategyOfGame(arr1, n)); 
 
    int arr2[] = { 2, 2, 2, 2 }; 
    n = sizeof(arr2) / sizeof(arr2[0]); 
    printf("%d\n", optimalStrategyOfGame(arr2, n)); 
 
    int arr3[] = { 20, 30, 2, 2, 2, 10 }; 
    n = sizeof(arr3) / sizeof(arr3[0]); 
    printf("%d\n", optimalStrategyOfGame(arr3, n)); 
 
    return 0; 
} 

Tip solutie

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