Friday, July 2, 2010
Prisoners again
Tuesday, July 1, 2008
Cards puzzle
I give you 4 cards from a deck of 52 playing cards. You intelligently select 4 out of those 5 and give those to your friend. He analyzes the 4 cards and tells you the fifth one. You have to pre-decide a strategy to do so. The puzzle is – What will the strategy be?
Sunday, May 25, 2008
Encryption- decryption !
You are Alice and you want to send a coded message to Bob. The way you will send the message is with up and down arrows or diagonal arrows, shown below.
The way you can generate a message is with zeros and ones. When zeros and ones are passed though these + and X, arrows are generated. The only way the message can be measured is with + and X.
The compatibility of + and X with arrows is such that plus aligns the arrows along its direction and crosses align the arrows along itself, for example as shown in the figure.

If someone intercepts these messages in between and sends them to Bob, the situation will change and number of matchings will differ between Alice and Bob. The question is – How many such arrows need to be send in order for probability of matching exceeding 0.99999 even if someone intercepts in between.
Monday, February 18, 2008
Hats !
A group of n people meet, and agree to play as a team in a game whose rules are explained to them. They are then allowed to discuss strategy. After the strategy session, an adversary puts red or blue hats on everyone's head; from this point on, no communication is allowed between the players. Each person can see the color of all hats but their own. After exactly one minute, the players simultaneously predict their hat color by sending email saying either "my hat is red" or "my hat is blue." The team wins if the n emailed statements are either all true or all false.
Devise a strategy that guarantees that the team wins, no matter how the adversary chooses to place the hats.
(extra credit) Consider the same situation, except that the goal of the group is to maximize the number of correct answers, assuming that the adversary knows exactly what strategy every member of the group is using. If each member of the group guesses red or blue with equal probability (1/2), then the expected number of correct answers will be n/2, no matter how the adversary arranges the hats. Is there a strategy (either deterministic or randomized) that does better than n/2?
Tuesday, February 12, 2008
Knight's win
Call them A and B. When A makes a move in a particular cell of the chess board, that cell is marked with A. Same is true with B. A can not move to a cell already marked by B and B can not move to a cell already marked by A. The one marking the most cells wins - Lets say - A wins - He can often trap B into a situation where he can not move to any of the cells which are already marked by A.
Two questions - Is that possible ?
If possible what is the strategy for A to win this ?
Wednesday, February 6, 2008
Who is the last man standing ?
1) Who is the last man standing ?
2) Generalize this with a formula to find the last man standing given a finite number of men starting the killing cycles.
Saturday, February 2, 2008
Break the jars
Now with three jars you can reduce your attempts easily sacrificing all three of them. The question is - How many attempts are required ?