Checkers Problem

Source: Nikhil Garg (Sophomore, IITD) mailed them to me.

Problem 1:
A checker starts at point (1,1). You can move checker using following moves :

1) if it is at (x,y) take it to (2x ,y ) or (x,2y)
2) if it is at (x,y) & x>y take it to (x-y,y)
3) if it is at (x,y) & x<y take it to (x,y-x)

Characterise all lattice points which can be reached.

Problem 2:
You have a checker at (0,0) , (0,1) , (1,0), (2,0), (0,2), (1,1) each. You can make a move as follows:

if(x,y) is filled  & (x+1,y) and (x,y+1) both are empty, remove checker from (x,y) & put one at each of (x+1,y) and (x,y+1)

Prove that under this move , you can not remove checker from all the six initial points.

Solution:
Update (02/03/10): Solution posted by Nikhil Garg in comments!

Comments

  1. NG11:37 AM, March 02, 2010

    Since none has posted the solutions , I am posting them :

    1) Note that after steps 2 or 3 , gcd(x,y) does not change.
    2) After step 1 , gcd either does not change or if changes , get doubled.

    So this is the invariant on gcd. Initial gcd is 1. So finally it can either be 1 or some power of 2.
    It is also easy to prove ( possibly induction ) that all such points can infact be reached.


    2) This is indeed a tough one.
    Consider a weight function w(x,y) associated with every lattice point.
    w(x,y)= 1/( 2^(x+y) )
    So this ensures that every move leaves total weight of board unchanged.

    Now total weight of all the board ( all first quadrant ) is 4. & weight of given 6 squares is more then 2. So these 6 checkers are too heavy for the rest of board !

    ReplyDelete
  2. Pratik Poddar1:45 PM, March 02, 2010

    \m/
    Creativity at its best! Couldn't reach even close. Nice solution. Thanx.

    ReplyDelete

Post a Comment

Popular posts from this blog

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? 
Read more

Buying Dimsums

Source: Alok Goyal (Stellaris VP, Ex-Helion VC) puzzle blog Problem: A fast food restaurant sells dimsums in boxes of 7 and 3. What’s the greatest number of dimsums a person cannot buy. Generalize it for p and q where p and q are relatively prime. I loved the puzzle. Hope you enjoy it too.
Read more

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)
Read more