Anyone have any idea how to do #3 on:
http://www.bioalgorithms.info/moodle...ks/hw0/hw0.pdf
My professor said she solved it using a checkerboard.
Thanks.
Anyone have any idea how to do #3 on:
http://www.bioalgorithms.info/moodle...ks/hw0/hw0.pdf
My professor said she solved it using a checkerboard.
Thanks.
Shouldn't be. I certainly did not need a password.
http://www.bioalgorithms.info/moodle...ks/hw0/hw0.pdf
Hmm. Not sure why you're getting a password. I can type it out if need be.
Two players play the following game with two “chromosomes” of length
n and m nucleotides. At every turn a player can destroy one of the chromosomes and break another one into two nonempty parts. For example, the first player can destroy a chromosome of length n and break another chromosome into two chromosomes of length m/3 and m − m/3 . The player left with two single-nucleotide chromosomes loses.
Who will win? Describe the winning strategy for each n and m.
Solve the game backwards as in this thread and this one.
I spent some more time looking at the problem. I tried doing a 13 by 13 matrix.
I got the following:
...and the pattern just repeats.Code:LWLWLWLWLWLWL WWWWWWWWWWWWW LWLWLWLWLWLWL WWWWWWWWWWWWW LWLWLWLWLWLWL WWWWWWWWWWWWW LWLWLWLWLWLWL
Assuming this is right, I think the strategy would be to give your opponent a pile of odd # nucleotides and another pile of odd nucleotides; if you give your opponent a pile of even and a pile of odd, they will win (assuming they know the strategy). Player one will always win if he goes first (will win if at least one chromosome pile has an even number of nucleotides). Player 2 will always win if the start piles both have an odd # of nucleotides.
I think your solution is correct as is. The m/3 restriction is not needed.
A simple proof by induction is possible. Establish the base case that a 2-pile is a winning position and a 3-pile is a losing one. Then for n > 3, an n-pile with n even is winning because it breaks into two odd piles, both of which are losing for the opponent. An n-pile with n odd is losing because it must break into an even pile and an odd pile. The even pile is a winning one for the opponent.