Skip to main content

Game of Two

This puzzle is taken from website of Krishnamurthy Iyer (Stanford)

Two immensely intelligent players, A & B, engage in a game, the rules of which are as follows. For some natural number N, the board consists of numbers from 1 to N. Each player takes turns to strike off a number from the board, with the added condition that if a number is struck off, then all its divisors should also be struck off. The player to strike off the last number on the board wins. A plays first. Is this game fair? If not, who has a winning strategy for each N and what is it? Else find the best strategy for each player.

Update (11/12/09): Solution in comments!!

Comments

  1. I bet player 1 will always win.Think about it.The answer lies on the smallest natural number.Player 1 can always choose to strike of number 1 or not to strike of number 1 in the first turn.

    ReplyDelete
  2. "The game can be reduced to an automata"

    "If you can reduce a game to an automaton, then at least one player has a winning strategy"

    "If a player has a winning strategy, it has got to be Player 1"

    Solution mailed by Jaadu. Does it make sense?

    ReplyDelete
  3. "All 2-player games where both players have to reach the same state and the game always finishes in finite time, there exists a winning strategy" -- Proved by Church and people of Automata theory in 40s

    So, in this game, at least one player has a strategy.

    Since there always exists a strategy, we can prove that a strategy for one player exists. If its of the first player, we know that a strategy exists for the first player. If its for the second player, we will prove that a strategy always exists for the first player.

    Let player II has a strategy. So, player II can win from any state {1,2,3,..n} - {k and Divisors of k}
    Our player I starts with canceling 1. Whatever player 2 cancels, the state is {1,2,3..n} - {k and Divisors of k}. So, now player I can use player II strategy to win the game.
    :)
    So, Player I always wins :D :D

    ReplyDelete
  4. I came across this problem even before and even then I just gave a "existential" proof. Can someone brave of heart try to get the constructive strategy ?

    ReplyDelete

Post a Comment

Popular posts from this blog

Asking a girl out

This is not a puzzle. So, for those of you who follow this puzzle blog, please bear with me for just one post. Interesting Math in this article though :P

Most of my friends already read an article that I wrote more than an year back - "Speak Up"


Here, inspired by the movie, The Beautiful Mind, I give a mathematical analysis of asking a girl out. Nice time it is. Feb 10. No plans for Feb 14 and I am sure this article makes me look even more geekier and all the more reason for me to believe that I will be alone, yet again. But what the hell, lets do it!

Note: This is not an independent analysis. There are many "mathematics sites" which does "similar" analysis.

@Consultants, correct me if I am wrong in my estimates. :P

Why is there a need to be selective?

From the age of 15, I guess there are approximately 3,600 girls I have liked (On average days, I don't see new girls. But going outside, I like about 30 girls. Saying that I go out once every week right …

Consecutive Heads

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.

1) Calculate P{X>Y}
2) What's the expected value of X
3) What's the expected value of Y

This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment.

Solution:
1)
(Solved by me finally after 13 months :))

Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently.

So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0"

f(State) = P(X>Y | "State" configuration initially)

f(0,0) = 1/4[f(…

Fraction Brainteaser

Source:
Sent to me by Gaurav Sinha

Problem:
Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20 out of 28 Geometry questions. In total, Siddhant scores 25 out of 34. 

Vaibhav writes another Maths test and correctly answers 20 out of 25 Arithmetic questions and 6 out of 9 Geometry questions. in total, Vaibhav scores 26 out of 34.

Note that
a) Vaibhav scores more than Siddhant
b) Siddhant score better than Vaibhav in both individual topics - 5/6 > 20/25 and 20/28 > 6/9

How is it possible?