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**azair****Member**- Registered: 2013-01-08
- Posts: 3

Hello Everyone,

I was trying to understand the calculation regarding the Time-Complexity of an Algorithm, but got stuck @line no :04 and 06

Here is the problem :

01: T(n)=T(n-1)+T(n-2)+4 and T(0)=T(1)=1

02: Lets assume that T(n-1)~T(n-2)

03: then, T(n)=2T(n-2)+c where c=4

04: 2{2T(n-4)+c}+c <-- HOW???

05: 4T(n-4)+3c

06: 2{4T(n-4)+3c}+c <-- HOW??? and HOW "...3c}+c"???

07: 16T(n-8)+15c

and therefore,

08: T(n)=2^k T(n-2k)+(2^k -1)c

09: T(n)=2^n/2 T(0) + (2^n/2 -1)c since: n-2k=0; k=n/2.

10: T(n)=(1+c)2^k - c.

=>I just don't understand as to what is happening @line 04 and 06.

Can anyone help me?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,738

hi azair

01: T(n)=T(n-1)+T(n-2)+4 and T(0)=T(1)=1

02: Lets assume that T(n-1)~T(n-2)

03: then, T(n)=2T(n-2)+c where c=4

There's a line missing here that should make it clearer.

replace n by n-2 in line 03

03.5: T(n-2) = 2T(n-4) + c

Now replace T(n-2) in line 03 using line 03.5

04: T(n) = 2{2T(n-4)+c}+c

and simplfy

05: =4T(n-4)+3c

Another missing line here

replace n by n-2 again

05.5: T(n-2) = 4(n-6) + 3c

but, from line 03,

T(n) = 2T(n-2) + c

and replace T(n-2) using line 05.5

And then I think there's a typo; my correction in red

06: T(n) = 2{4T(n-6)+3c}+c

and simplify

06.5: = 8T(n-6) + 7c

and with n replaced by n-6 in line 03

06.75: T(n-6)=2T(n-8)+c

and replace T(n-6) in line 06.5

06.8: T(n) = 8(2T(n-8) + c) + 7c

and finally simplify to get

07: T(n) = 16T(n-8)+15c

By now you can, hopefully, spot a pattern in the two numbers:

2,4,8,16,.......powers of 2

and 1,3,7,15,......powers of 2 minus 1

which is where the general formula comes from.

Hope that helps,

Bob

*Last edited by bob bundy (2014-11-18 21:50:45)*

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**azair****Member**- Registered: 2013-01-08
- Posts: 3

Ohhh!! yes!!! I got it...Hurray

Thank You very much Bob.

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❄ Music in the Sky ❅

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