如何解决“博弈论”中属于“发现者”类别的问题?
注意-Finderskeepers守门员游戏在“博弈论”中属于“公平游戏”类别。
什么是“公平游戏”?
让游戏在两个玩家“ A”和“ B”之间进行。如果两个玩家都具有相同的动作,则他们之间的游戏被称为“不公平”。在游戏的特定时刻选择哪个动作取决于游戏的状态。
发现者保持游戏
Let two players A and B playing a game on a pile of coins. Each player has to remove a minimum of ‘a’ coins or maximum of ‘b’ coins from the pile in his turn until there are less than ‘a’ coins left.
Types:
- Finder-Winner -> In this format the player to play the last move wins.
- Keeper-Loser -> In this format the player to play the last move loses.
策略
Note- If there are less than ‘a’ coins available then last player to move must pick all coins.
- FINDER-WINNER: Let ‘A’ be the player to start. For a game of FINDER-WINNER the main strategy is to reduce the present number coins to a multiple (a+b) for ‘A’ to win the game. Else he loses and ‘B’ wins.
- KEEPER-LOSER: Let ‘A’ starts the game. For ‘A’ to win the game the main condition for winning it is to reduce the pile of coins to – (a+b)k + [1, a], i.e. the number of coins can be reduced to more then a multiple of (a+b) by [1 upto a]. If he can do so he wins else ‘B’ wins.
例子:
- 让A和B在20个硬币上玩FINDER-WINNER游戏。一个玩家的举动最少可以取出5个硬币,最多可以取出7个硬币。预测游戏的获胜者:如果A首先开始并且两个玩家都玩得最佳。
Given number of coins = 20.
Minimum coins that can be removed = 5 – Let it be ‘a’.
Maximum coins that can be removed = 7 – Let it be ‘b’.根据针对FINDER-WINNER的既定策略:
为了使A获胜: A必须将硬币数目去除为(a + b)的倍数。 20的(A + B)最接近的倍数为12。因此, A可以从集合中扣除5或6或7个硬币,因此无论如何他都无法达到12。因此,为了让B采取行动,他将剩下15或14或13个硬币。现在,对于A的下一个动作,相应的状态是(10,9,8) , (9,8,7)和(8,7,6) 。由于B的比赛也处于最佳状态,因此他将使A的举动的硬币数保持大于7,以使A不能获胜。因此,他将获得至少1个硬币,该硬币将在最后一步中获胜。
- 让A和B以20个硬币为一组玩Keeper-Loser游戏。一个棋手可以取出最少2个硬币,最多5个硬币。预测游戏的获胜者:如果A首先开始,并且两个玩家都玩得最佳。
Given number of coins = 20.
Minimum coins that can be removed = 2 – Let it be ‘a’.
Maximum coins that can be removed = 5 – Let it be ‘b’.根据给定的KEEPER-LOSER策略:
为了使A获胜: A必须将硬币数目去除为(a + b)k + [1,a]的倍数。这里a + b = 7 。
现在,为了发挥最佳状态,让我们看看A在接下来的回合中会做什么。如果A将一组硬币减少到15个,则形式为(2 + 5)* 2 +1,因为从20个硬币中删除5个硬币是可以达到的。 B从此处可以取[2,5]中的任意数量的硬币,将其命名为c 。现在,A的主要策略是删除他的举动中的7 – c。为什么会这样呢?
假设15点之后, B通过拿走5个硬币将集合减少到10个。现在, A的最佳策略是删除2个硬币。如果他这样做,则减少到8个硬币。现在, B可以通过分别删除(2、3、4、5)个硬币,将集合减少到(6、5、4、3)。再次从这里开始, A的主要想法是拿走7 – c个硬币,这将为B保留1个硬币。因此, A获胜。