对于2人游戏的最佳策略 [英] Optimal strategy for a 2 player game
问题描述
- 2玩家A和B的玩游戏，其中一些的 N 的
- 玩家A，使第一招：放;双方球员发挥交替进行。
- 在每个移动玩家需要数n，选择一个数字的我的这样的 2 ^ I＆LT; ñ的和替代的 N 的用的 K = N - 2 ^我的当且仅当1秒中的 K 大于或等于1秒中的二进制重新presentation的 N 的 数
- 在游戏的时候没有球员可以使一招，即不存在结束这样的的我的
- 2 players A & B are playing a game involving a number n
- Player A makes the first move & both players play alternately.
- In each move the player takes the number n,chooses a number i such that 2^i < n and replaces n with k = n - 2^i iff the number of 1s in the binary representation of k is greater than or equal to the number of 1s in the binary representation of n
- Game ends when no player can make a move, ie there does not exist such an i
例如：
n = 13 = b1101
只能I = 1
k = n - 2^i = 11 = b1011
此外，唯一可能的I = 2
Again,only possible i = 2
k = n - 2^i = 7 = b111
由于玩家A不能做任何更多的动作，玩家B赢
Since Player A cant make any more moves, Player B wins
我推断，在任何步骤，我们只能选择一个我，这样有一个0在n的二进制再presentation相应的位置。照片
例如：
如果n = 1010010，那我也只能是{0,2,3,5}。
但我不能将任何further.A极大极小的算法是不完全打击我，我会AP preciate提前任何help.Thanks
I've deduced that at any step,we can only choose an i,such that there is a 0 at the corresponding position in the binary representation of n.
For Example:
if n=1010010,then i can only be {0,2,3,5}.
But I cant move any further.A minimax algorithm isnt exactly striking me.I would appreciate any help.Thanks in advance
推荐答案
假定n是不是太大了，我们可以用动态规划来解决这个问题。 定义一个数组A [1：n]，其中A [1]再presents是否球员，我会赢得输入我。 让我们用跨pretation：
Assuming n is not too big, we can use dynamic programming to solve this problem. Define an array A[1:n], where A[i] represents whether player i will win on input i. Let's use the interpretation:
A[i] = 1, if A wins on input i,
A[i] = 0, if A loses on input i.
现在可以计算自下而上如下：
Now A can be computed bottom-up as follows:
A[1] = 0, A[2] = 1.
For j=3:n {
Assign A[j] = 1 if there exists a number i such that (A[j-2^i] = 0) AND
(number of 1's in i >= number of 1's in j)
Otherwise Assign A[j] = 0
}
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