Using standard Dynamic Programing logic:
http://codepad.org/IwBTor4F
On Thu, Oct 7, 2010 at 8:43 PM, Gene <gene.ress...@gmail.com> wrote:
> Nice problem. Let N_1, N_2, ... N_n be the values of the N notes.
> Let F(i,j) be the maximum possible score the current player can get
> using only notes i through j. Then
>
> F(i,j) = sum_{k = i to j}N_k - min( F(i+1, j), F(i, j-1) )
>
> This is saying that the current player always makes the choice that
> minimizes B the best score that the other player can achieve. When we
> know that choice, the best _we_ can do is the sum of all the available
> notes minus B.
>
> The base case is F(k,k) = N_k, and we are unconcerned with F(i,j)
> where i > j.
>
> For example, suppose we have notes 3 7 2 1 . The answer we want is
> F(1,4).
>
> Initially we have
> F(1,1) = 3
> F(2,2) = 7
> F(3,3) = 2
> F(4,4) = 1
>
> Now we can compute
> F(1,2) = 10 - min(F(2,2), F(1,1)) = 10 - min(7,3) = 7 (pick N_1=7).
> F(2,3) = 9 - min(F(3,3), F(2,2)) = 9 - min(2,7) = 7 (pick N_2=7).
> F(3,4) = 3 - min(F(4,4), F(3,3)) = 3 - min(1,2) = 2 (pick N_3=2).
> F(1,3) = 12 - min(F(2,3), F(1,2)) = 12 - min(7,7) = 5 (don't care).
> F(2,4) = 10 - min(F(3,4), F(2,3)) = 10 - min(2,7) = 8 (pick N_2=7).
> F(1,4) = 13 - min(F(2,4), F(1,3)) = 13 - min(8,5) = 8 (pick N_4=1).
>
>
> On Oct 7, 8:14 pm, Anand <anandut2...@gmail.com> wrote:
> > Given a row of notes (with specified values), two players play a game. At
> > each turn, any player can pick a note from any of the two ends. How will
> the
> > first player maximize his score? Both players will play optimally.
>
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