What is the time complexity of the recurrence T(n) = 2T(n-1) + 4

问题

What is the time complexity of the recurrence T(n) = 2T(n-1) + 4 ?

I'm having serious problems with this. I tried:

T(n) = 2T(n-1)+4 = 2(2T(n-2)+4)+4 = 4T(n-2)+12= 4(2T(n-3)+4)+4 = 8T(n-3)+20 = 8(2T(n-4)+4)+4 = 16T(n-4)+36 =…

T(n) = 2^kT(n-k) + (4+2^(k+1))

so it looks like T(n) = 2^n + (4+2^(n+1)) but it doesn't seem right... please help :(

回答1:

Your computation are wrong. I'm assuming here that T(0)=0

T(n) =                       2T(n-1)+  4 
     =   2(2T(n-2)+4)+  4 =  4T(n-2)+ 12
     =   4(2T(n-3)+4)+ 12 =  8T(n-3)+ 28 
     =   8(2T(n-4)+4)+ 28 = 16T(n-4)+ 60
     =  16(2T(n-5)+4)+ 60 = 32T(n-5)+124
     =  32(2T(n-6)+4)+124 = 64T(n-6)+252

Then now: look at the sequence

0,4,12,28,60,124,252,508,1020,2044,...

It is very tempting to add 4 to all these numbers:

4,8,16,32,64,128,256,512,1024,2048,...

Do you recognize it ? So the guess is clearly

T(n) = 2^(n+2) - 4

Now, you can easily prove it by induction.

By the way if T(0) is not equal to 0 the formula is

T(n) = 2^(n+2) - 4 + T(0)*2^n

回答2:

Solving the recurrence relation, I found this:

回答3:

T(n) = 2T(n-1) + 1

Let n = log m then, we get T(log m)=2T(log m - log 2) + 1

Assume S(a) = 2S(a/2) + 1

Applying Master's theorem we get: n^logb^a = n^log2^2 = n.

n^log2^2-1 = 1, here € = 1

By first case we get: S(a) = O(a)

therefore T(n) = O(2^n)

来源：https://stackoverflow.com/questions/22741590/what-is-the-time-complexity-of-the-recurrence-tn-2tn-1-4