Diversification
Some product line diversification to see if I can energize my reader base.
I found these on Anil Krishna's webpage. He has some good stuff there.
This one is quite do-able. I mean I could do it, so probably the difficulty rating is -1 :)
Anil and Kavita attend a party with 3 other couples,including Srini and Shyamala (a couple). During the party everyone shakes hands with a certain number of other people that doesn't include oneself and one's spouse. At the end of the party Anil asks each person (apart from himself) how many people they shook hands with. He finds that each person answered truthfully and each one gave a different number. Anil did shake hands with Srini. From this information find out
1: Did Kavita shake hands with Srini?
2: Did Kavita shake hands with Shyamala?
And another one...
There are 2 numbers X and Y (integers).
2<= X < Y <= 100
Person 1, P1, knows X * Y (product).
Person 2, P2, knows X + Y (sum).
P1 knows that P2 knows the sum, P2 knows that P1 knows the product.
They have this conversation:
P1: I do not know what X and Y are.
P2: I knew you would not know.
P1: Now I know what X and Y are.
P2: Now I too know what X and Y are.
What are X and Y?
P.S. Ah ha... the difficulty rating of the third one is tending to -1. :)
I found these on Anil Krishna's webpage. He has some good stuff there.
This one is quite do-able. I mean I could do it, so probably the difficulty rating is -1 :)
Anil and Kavita attend a party with 3 other couples,including Srini and Shyamala (a couple). During the party everyone shakes hands with a certain number of other people that doesn't include oneself and one's spouse. At the end of the party Anil asks each person (apart from himself) how many people they shook hands with. He finds that each person answered truthfully and each one gave a different number. Anil did shake hands with Srini. From this information find out
1: Did Kavita shake hands with Srini?
2: Did Kavita shake hands with Shyamala?
And another one...
There are 2 numbers X and Y (integers).
2<= X < Y <= 100
Person 1, P1, knows X * Y (product).
Person 2, P2, knows X + Y (sum).
P1 knows that P2 knows the sum, P2 knows that P1 knows the product.
They have this conversation:
P1: I do not know what X and Y are.
P2: I knew you would not know.
P1: Now I know what X and Y are.
P2: Now I too know what X and Y are.
What are X and Y?
P.S. Ah ha... the difficulty rating of the third one is tending to -1. :)
Comments
Ans - Yes, No.
2) Hmmm... Interestingly, the set to look for is [2, 100] i.e 1 is kept out. Does this point to something related to prime nos?... Also P2's 1st remark (P2: I knew you would not know), I feel, again hints to usage of prime nos. to eliminate certain numbers from the set [2, 100]. Not sure how to approach this problem by means of 'probability' though...
1 --> the numbers are not prime.
2 --> the sum of the numbers is not even, since all even numbers are sum of primes.
3,4 --> now write a program to find all possible odd numbers [4,200] which are not product of primes and find the number where you can be sure that there is no alternative.
Don't think probability is required in this one.
http://www.cs.utexas.edu/users/EWD/ewd06xx/EWD666.PDF
For the second one:
P1 doesn't know the numbers from the product, which means that the two numbers aren't primes, and aren't also of the form p and p**2 for prime p.
P2 already knew that P1 could not figure the numbers out, so the sum of the numbers isn't even. That is because all even numbers can be written as sums of two primes—at least we can assume that in this case even though this may not be true in general :-)
So, P1 knows that the one number is even and one is odd (otherwise the sum is even). Since this helps him arrive at a unique solution, it means that he could eliminate one potential pair by using this information. That is, there is a possible factorization of X*Y such that both the factors are even. Which means that the product is divisible by 4. Since one number is even and one is odd, the even number is divisible by 4.
So now we have a list of candidates: 4, 8, 12, ..., 100 for X and 3, 5, 7, ..., 99 for Y. In fact, we can do better - but that needs some programming or a lot of patience. :-)
If I am bored enough, I will write a program to solve the problem. Otherwise, I think (4,13) might be one potential solution.