# Some Tricks to solve problems on Impartial games

How to solve problems that fall in the Finders keepers category in ‘Game Theory’?

Note- Finders keepers game fall into the category of ‘Impartial Games’ in ‘Game Theory’.

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**What are ‘Impartial Games’?**

Let a game is being played between two players ‘A’ and ‘B’. A game between them is said to be ‘Impartial’ if both the players have same set of moves. Which move to choose at a particular moment of the game depends on the state of the game.

**FINDERS KEEPERS GAME**

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.

**STRATEGIES**

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.

**Examples:**

- Let
**A**and**B**play a game of FINDER-WINNER on a set of**20 coins**. A player in his move can remove a**minimum of 5 coins**and a**maximum of 7 coins**. predict the Winner of the Game If**A**starts first and if both players play optimally.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’.**According to given strategy for FINDER-WINNER:**__For__**A to win****:****A**must remove the number of coins to a**multiple of (a + b)**. Nearest multiple of**(a + b)**for**20**is**12**. So**A**can deduct either**5**or**6**or**7**coins from the set so he cant reach**12**anyhow. So for**B**to make a move, he will be left with**15**or**14**or**13**coins. Now for**A’s**next move corresponding states are**(10, 9, 8)**,**(9, 8, 7)**and**(8, 7, 6)**. Since**B**is also playing optimally he will keep the number of coins for**A’s**move greater than**7**so that**A**cant win. Hence, he will get at least 1 coin to be removed in the last move and WIN. - Let
**A**and**B**play a game of Keeper-Loser on a set of**20 coins**. A player in his move can remove a**minimum of 2 coins**and a**maximum of 5 coins**. Predict the Winner of the Game If**A**starts first and if both players play optimally.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’.**According to given strategy for KEEPER-LOSER :**__For__**A to win****:****A**must remove the number of coins to a multiple of**(a + b)k + [1, a]**. Here**a + b = 7**.

Now, for playing optimally lets see what**A**will do in further rounds. If**A**reduces the set of coins to lets say**15**which is of the form**(2 + 5) * 2 + 1**as it is reachable by removing**5**coins from a set of**20**coins. From here**B**can take any number of coins from**[2, 5]**lets name it as**c**. The main strategy of**A**now will be to remove**7 – c**in his moves.**Why is it so?**Lets say after

**15**,**B**reduces the set to**10**by taking away**5**coins. Now optimal strategy for**A**is to remove**2**coins. If he does so, the set reduces to**8**coins. Now from here**B**can reduce the set to**(6, 5, 4, 3)**by removing**(2, 3, 4, 5)**coins respectively. From here again, the main idea for**A**is to take away**7 – c**coins which will keep**1**coin for**B**to remove. Hence**A**wins.