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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