Puzzle 2 : Take the last chip
Turkey sandwich was worried about an upcoming test in discrete maths and was finding it hard to get to sleep. Turkey awoke early in the morning, aroused by a devilish laughter, only to see an impish looking homunclus sitting at the bottom of the bed next to a seemingly infinte pile of chips. Hello turkey, it said, would you like to play a little game ? This pile contains 43546758343209876 chips and the bottom chip represents your immortal soul. The rules are quite simple. The first player takes some chips, but not all of them. After that we take it in turns to take some chips.
The only rule now is that a player cant take more in their turn than the previous player took. The winner is the player who takes the last chip. If I win I get to keep your soul and if you win, you get an A in the test. Would you like to go first or second ? Can you give turkey a strategy for playing no matter how many chips are there ?
Was that easy ? What if a player can take up to twice the number of chips that the previous player took. What is the best strategy now ?
(I would love you to first attempt before seeing my attempt on the problem).
My attempt :
For the first part of the problem, I took three cases :
1) no of chips is odd : in that case, the player who plays first will win. I pick first, you pick second, that way when the odd number is reached, the first player will be the one to pick it.
2) no of chips is even but some odd multiple of 2 : In that case (take 2*3), I claim that first player wins. Divide the whole pair of 6 (2*3) chips as follows :
_ _ | _ _ | _ _
First player aims to win each division of chips.
Suppose player 1 chooses the first 2 chips. Player 2 has two choices, choose 1 chip or 2 of'em.
If he chooses 1, the problem reduces to 1) where no of chips becomes odd.
If he chooses 2, the problem is a reduced version of the same problem.
3) no of chips is even but even multiple of 2 : This is a tricky case. I claim that in such a case, the person who goes second will win. Take the no 4.
For 4, if each takes 1 chip, second person will win.
If first takes 2 chips, second will take will take 2 and win
If first takes 3 chips, then also second will win.
In general, for 2* even number, I will divide the whole set of chips like the following :
_ _ _ _ | _ _ _ _ | _ _ _ _ |..... and so on
And I will play second. I willl win no matter what.
For the second part, I didn't try much but my intuition says this :
Find out what range of 2^k-1 and 2^k the number lies in.
If k is even, I play first. Otherwise second.
Turkey sandwich was worried about an upcoming test in discrete maths and was finding it hard to get to sleep. Turkey awoke early in the morning, aroused by a devilish laughter, only to see an impish looking homunclus sitting at the bottom of the bed next to a seemingly infinte pile of chips. Hello turkey, it said, would you like to play a little game ? This pile contains 43546758343209876 chips and the bottom chip represents your immortal soul. The rules are quite simple. The first player takes some chips, but not all of them. After that we take it in turns to take some chips.
The only rule now is that a player cant take more in their turn than the previous player took. The winner is the player who takes the last chip. If I win I get to keep your soul and if you win, you get an A in the test. Would you like to go first or second ? Can you give turkey a strategy for playing no matter how many chips are there ?
Was that easy ? What if a player can take up to twice the number of chips that the previous player took. What is the best strategy now ?
(I would love you to first attempt before seeing my attempt on the problem).
My attempt :
For the first part of the problem, I took three cases :
1) no of chips is odd : in that case, the player who plays first will win. I pick first, you pick second, that way when the odd number is reached, the first player will be the one to pick it.
2) no of chips is even but some odd multiple of 2 : In that case (take 2*3), I claim that first player wins. Divide the whole pair of 6 (2*3) chips as follows :
_ _ | _ _ | _ _
First player aims to win each division of chips.
Suppose player 1 chooses the first 2 chips. Player 2 has two choices, choose 1 chip or 2 of'em.
If he chooses 1, the problem reduces to 1) where no of chips becomes odd.
If he chooses 2, the problem is a reduced version of the same problem.
3) no of chips is even but even multiple of 2 : This is a tricky case. I claim that in such a case, the person who goes second will win. Take the no 4.
For 4, if each takes 1 chip, second person will win.
If first takes 2 chips, second will take will take 2 and win
If first takes 3 chips, then also second will win.
In general, for 2* even number, I will divide the whole set of chips like the following :
_ _ _ _ | _ _ _ _ | _ _ _ _ |..... and so on
And I will play second. I willl win no matter what.
For the second part, I didn't try much but my intuition says this :
Find out what range of 2^k-1 and 2^k the number lies in.
If k is even, I play first. Otherwise second.
