Question

This is a simple game:
There is a set, A={a1,…,an}, the opponents can choose one of the first or last elements of set, and at the end the one who collect bigger numbers wins. Now say each participants dose his best, what I need to do is write a Dynamic algorithm to estimate their score.
any idea or clue is truly appreciated.

Answer 1

EXAMPLE CODE

Here is Python code to compute the optimal score for first and second players.

A=[3,1,1,3,1,1,3]

cache={}
def go(a,b):
    """Find greatest difference between player 1 coins and player 2 coins when choosing from A[a:b]"""
    if a==b: return 0 # no stacks left
    if a==b-1: return A[a] # only one stack left
    key=a,b
    if key in cache:
        return cache[key]

    v=A[a]-go(a+1,b) # taking first stack
    v=max(v,A[b-1]-go(a,b-1)) # taking last stack

    cache[key]=v
    return v

v = go(0,len(A))
n=sum(A)
print (n+v)/2,(n-v)/2

COUNTEREXAMPLE

Note that the code includes a counter example to one of the other answers to this question.

Consider the case [3,1,1,3,1,1,3].

By symmetry, the first players move always leaves the pattern [1,1,3,1,1,3].

For this the sum of even elements is 1+3+1=5, while the sum of odd is 1+1+3=5, so the argument is that from this position the second player will always win 5, and the first player will always win 5, so the first player will win (as he gets 5 in addition to the 3 from the first move).

However, this logic is flawed because the second player can actually get more.

First player takes 3, leaves [1,1,3,1,1,3] (only choice by symmetry)
Second player takes 3, leaves [1,1,3,1,1]
First player takes 1, leaves [1,3,1,1] (only choice by symmetry)
Second player takes 1, leaves [1,3,1]
First player takes 1, leaves [3,1] (only choice by symmetry)
Second player takes 3, leaves [1]
First player takes 1

So overall first player gets 3+1+1+1=6, while second gets 3+1+3=7 and second player wins.

The flaw is that although it is true that the second player can play such that they will win all even or all odd positions, this is not optimal play and they can actually do better than this in some cases.