he first few f(i):
f(1)=0
f(2)=max(f(1)+f(1-1+1)+1) = 1
f(3)=max{f(1)+f(2)+1, f(2)+f(1)+1} = 2
f(4)=max{f(1)+f(3)+1, f(2)+f(2)+1, f(3)+f(1)+1} =3
f(5)=max{f(1)+f(4)+1, f(2)+f(3)+1, ....,f(4)+f(1)+1} = 4
f(6)=max{.....} =5
what I can see here is in each f(i), all comma separated expressions
evaluate to the same value. For exampe in f(5), all f(1)+f(4)+1,
f(2)+f(3)+1....evaluate to same value 4.
so by inspection it is easy to see that the answer is: f(n+1)
= n, for all n. This can be generalized on the base case to: f(n+1) =
(n+1)*f(1) + n. But, how to go to prove such a thing formally?
On Feb 16, 8:45 pm, Vikas Kumar <dev.vika...@gmail.com> wrote:
> f(n)=n-1.
>
> On Wed, Feb 16, 2011 at 7:39 PM, Akshata Sharma
> <akshatasharm...@gmail.com>wrote:
>
> > please help..
>
> > if f(n+1) = max{ f(k) + f(n-k+1) + 1} for 1 <= k <= n; f(1) = 0.
> > Find f(n+1) in terms of n.
> > Eg: f(4) = ? n = 3; 1<= k <= 3; f(4) = max{f(1) + f(3) + 1, f(2) +
> > f(2)+1, f(3) + f(1) +1}
>
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