问题

I am having some issues on how to solve recurrence relations.

T(n) = T(n/2) + log2(n), T(1) = 1, where n is a power of 2

This is a homework problem, so don't just give me the answer. I was just wondering how to start the problem.

In class we went over the Master theorem. But I don't think that would be the best way to solve this particular relation.

I don't really know how to start the problem... should I just be going

T(n) = T(n/2) + log_base2(n)
T(n/2) = [T(n/4)+log_base2(n/2)]
  T(n) = [T(n/4)+log_base2(n/2)] + log_base2(n)

And just keep working my way down to get something I can see makes a basic equation?

回答1:

This recurrence solves to Θ((log n)²). Here are two ways to see this.

Some Substitutions

If you know that n is a perfect power of two (that is, n = 2^k), you can rewrite the recurrence as

T(2^k) = T(2^k-1) + k

Let's define a new recurrence S(k) = T(2^k). Then we get that

S(k) = S(k - 1) + k

If we expand out this recurrence, we get that

S(k) = S(k - 1) + k

= S(k - 2) + (k - 1) + k

= S(k - 3) + (k - 2) + (k - 1) + k

= S(k - 4) + (k - 3) + (k - 2) + (k - 1) + k

...

= S(0) + 1 + 2 + 3 + ... + k

= S(0) + Θ(k²)

Assuming S(0) = 1, then this recurrence solves to Θ(k²).

Since S(k) = T(2^k) = T(n), we get that T(n) = Θ(k²) = Θ(log² n).

Iterating the Recurrence

Another option here is to expand out a few terms of the recurrence and to see if any nice patterns emerge. Here’s what we get:

T(n) = T(n / 2) + lg n

= T(n / 4) + lg (n / 2) + lg n

= T(n / 8) + lg (n / 4) + lg (n / 2) + lg n

...

Eventually, after lg n layers, this recurrence bottoms out and we’re left with this expression:

lg n + lg (n / 2) + lg (n / 4) + ... + lg (n / 2^{lg n)}

Using properties of logarithms, we can rewrite this as

lg n + (lg n - 1) + (lg n - 2) + (lg n - 3) + ... + (lg n - lg n)

Or, written in reverse, this is the sum

0 + 1 + 2 + 3 + ... + lg n

That sum is Gauss’s sum up to lg n, which evaluates to (lg n)(lg n + 1) / 2 = Θ((log n)2).

Hope this helps!

回答2:

If n is a power of 2 then you can just expand out the recurrence and solve exactly, using that lg(a/b) = lg(a) - lg(b).

T(n) = lg(n) + lg(n/2) + lg(n/4) + ... + lg(1) + 1
     = (lg(n) - 0) + (lg(n) - 1) .... + (lg(n) - lg(n)) + 1
     = lg(n)*lg(n) - lg(n)*(lg(n)+1)/2 + 1
     = lg(n)*lg(n)/2 - lg(n)/2 + 1

回答3:

This can be done with the Akra-Bazzi theorem. See the third example in http://people.mpi-inf.mpg.de/~mehlhorn/DatAlg2008/NewMasterTheorem.pdf.

回答4:

This can be solved with Master theorem. Your a=1 and b=2 and f(n) = log(n). Then c = log2(1) = 0. Because of your c and f(n) you fall into the second case (where k=1).

So the solution is Θ(log² n)

来源：https://stackoverflow.com/questions/14679790/solving-the-recurrence-tn-tn-2-lg-n

标签

algorithm

math

big-o

recurrence

master-theorem