将此递归解决方案转换为DP

Question

Given a stack of integers, players take turns at removing either 1, 2, or 3 numbers from the top of the stack. Assuming that the opponent plays optimally and you select first, I came up with the following recursion:

int score(int n) {
    if (n <= 0) return 0;

    if (n <= 3) {
        return sum(v[0..n-1]);
    }

    // maximize over picking 1, 2, or 3 + value after opponent picks optimally
    return max(v[n-1] + min(score(n-2), score(n-3), score(n-4)),
               v[n-1] + v[n-2] + min(score(n-3), score(n-4), score(n-5)),
               v[n-1] + v[n-2] + v[n-3] + min(score(n-4), score(n-5), score(n-6)));
}

Basically, at each level comparing the outcomes of selecting 1, 2, or 3 and then your opponent selecting either 1, 2, or 3.

I was wondering how I could convert this to a DP solution as it is clearly exponential. I was struggling with the fact that there seem to be 3 dimensions to it: num of your pick, num of opponent's pick, and sub problem size, ie, it seems the best solution for table[p][o][n] would need to be maintained, where p is the number of values you choose, o is the number your opponent chooses and n is the size of the sub problem.

Do I actually need the 3 dimensions? I have seen this similar problem: http://www.geeksforgeeks.org/dynamic-programming-set-31-optimal-strategy-for-a-game/ , but couldn't seem to adapt it.

Answer 1

Here is way the problem can be converted into DP :-

score[i] = maximum{ sum[i] - score[i+1] , sum[i] - score[i+2] , sum[i] - score[i+3]  } 

Here score[i] means max score generated from game [i to n] where v[i] is top of stack . sum[i] is sum of all elements on the stack from i onwards . sum[i] can be evaluated using a separate DP in O(N) . The above DP can be solved using table in O(N)

Edit :- Following is a DP solution in JAVA :-

public class game {

    static boolean play_game(int[] stack) {
        if(stack.length<=3)
            return true;
        int[] score = new int[stack.length];
        int n = stack.length;
        score[n-1] = stack[n-1];
        score[n-2] = score[n-1]+stack[n-2];
        score[n-3] = score[n-2]+stack[n-3];
        int sum = score[n-3]; 
        for(int i=n-4;i>=0;i--) {
            sum = stack[i]+sum;
            int min = Math.min(Math.min(score[i+1],score[i+2]),score[i+3]);
            score[i] = sum-min;
        }
        if(sum-score[0]<score[0]) 
            return true;

        return false;
    }

    public static void main(String args[]) {
        int[] stack = {12,1,7,99,3};
        System.out.printf("I win => "+play_game(stack));
    }

EDIT:-

For getting a DP solution you need to visualize a problems solution in terms of the smaller instances of itself. For example in this case as both players are playing optimally , after the choice made by first one ,the second player also obtains an optimal score for remaining stack which the subproblem of the first one. The only problem here is that how represent it in a recurrence . To solve DP you must first define a recurrence relation in terms of subproblem which precedes the current problem in any way of computation. Now we know that whatever second player wins , first player loses so effectively first player gains total sum - score of second player. As second player as well plays optimally we can express the solution in terms of recursion.