Re: [algogeeks] Maths

[Abhishek Sharma]

This is simple..

Find the values for f(n) for n=1,2,3,4,... n-1 which are 0, 1, 2, 3, ... n-2
respectively. (Solve the equation for n=2,3, etc to get the values).

From the pattern you can easily find out that f(n+1)= n.

On Wed, Feb 16, 2011 at 9:15 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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[algogeeks] Maths

[Akshata Sharma]

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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Re: [algogeeks] Maths

[Vikas Kumar]

f(n)=n-1.

On Wed, Feb 16, 2011 at 7:39 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:

 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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