更好地猜测上限

Question

It's a question from 《introduction to algorithms》whose number is 4.4-5 and is described like this:

Use a recursion tree to determine a good asymptotic upper bound on the recurrence T(n) = T(n-1) + T(n/2) + n.Use the substitution method to verify your answer.

I found it is difficult to me to calculate the recursion tree's recurrence. The answer I gave

Math.pow(2,n)

seems too loose.Maybe there is some better guess existed.Thanks for any help.

Answer 1

Hope I did no mistakes :)

let's A(n)=T(n/2)+n

0. T(n)=T(n-1)+A(n)=T(n-2)+A(n-1)+A(n)=...=A(1)+A(2)+...+A(n)
   T(n)=sum[1..n]A(n)
   T(n)=sum[i=1..n]T(i/2)+sum[i=1..n]i

assuming n/2 is integer division, T(n/2)=T((n+1)/2) for even n , so the first sum consists of two equal halves: T(1)+T(1)+T(2)+T(2)+...

1. T(n)=2*sum[1..n/2]T(i)+n*(n-1)/2

since T(n)<=T(m) for every n<=m

2. T(n)<=n*T(n/2)+n*(n-1)/2

since T(n/2)>=n/2>=(n-1)/2

3. T(n)<=n*T(n/2)+n*T(n/2)=2*n*T(n/2)

let's consider this for only n=2^k , since T is monotonous: n=2^k and U(k)=T(2^k)

4. U(k)<=2*(2^k)*U(k-1)=2^(k+1)*U(k-1)

let L(k)=log2 U(k)

5. L(k)<=k+1+L(k-1)

just like we did between step0 and step1

6. L(k)<=k*(k-1)/2+k=k*k/2-k/2+k<=k*k

7. U(k)=2^L(k)<=2^squared(k)

8. T(n)=U(log2 n)<=2^squared(log2 n)

Answer 2

The recursion relation seems to give rise to a sub-exponential and super-linear computation-time, which means that any chosen base would work as an upper bound given a large enough n .

Your choice of 2^n is a good answer and possibly the one they were looking for in the book. It is a simple solution that is valid even for quite small values of n . (Still, I understand why you are asking the question because it does grow much faster than T(n) even for moderately large n .)

Given T(1) = 1 (or some other constant) the recursion equation gives us the running time as follows for the first few values of n .

T(1) = 1          n^1 = 2
T(2) = 4          n^2 = 4
T(3) = 11         n^3 = 8
T(4) = 19         n^4 = 16
T(5) = 35         n^5 = 32
T(6) = 52         n^6 = 64
T(7) = 78         n^7 = 128
T(8) = 105        n^8 = 256
T(9) = 149        n^9 = 512

We can see that the choice of 2^n as an upper limit is valid for all values T(6) and above.

If you want a lower bound than 2^n you could choose a lower base (with the trade-off that it will only be valid for higher numbers of n ). But I must add that it will still be basically the same solution as the one you already have.

Any base larger than one would do but to be a bit more specific we could for example see that the recursion equation T(n) = T(n-1) + T(n/2) + n is bounded by the equation T(n) = T(n-1) + T(n-2) for n>5 .

This is the same recursion relation as for the Fibonacci sequence and following the steps in the answers to this question it has a computational complexity matching the golden ratio (1+sqrt(5))/2 = 1,618 to the power of n .

Plotting the actual values we can see for which n the value of T(n) is bounded by ((1+sqrt(5))/2)^n . From the figure it seems to be values n=13 and above.

All this said, I have thought a bit about approximating the running time with some sub-exponential function. It doesn't seem like it can be easily done and as I said I believe you have found the expected answer.