What is the approach to solving a recurrence relation when I have more than one recurrent function calls on the right-hand side of the equation?

Question

I am trying to analyze the complexity of the function call to PEL(A[1..n]) where n is a certain power of 3 and PEL is defined by the following algorithm:

function PEL(A[m..n]){
  if(n - m <= 1) return 1;
  else { 
    p := [(n - m + 1)/3];
    MAKE(A[m..n]);
    PEL(A[m..n + p - 1]); PEL(A[m + p .. m + 3p - 1]);
  }
}

The complexity of MAKE(A[m..n]) is theta( (n-m)log(n-m) ).

From what I have gathered so far, we are dealing with the following recurrence relation:

C(N) = C(N/3) + C(2*N/3) + theta( (n-m)log(n-m) )

Where

C(1) = C(2) = 1

I understand that we need to apply the master theorem here, but in the master theorem we have recurrence relations of the form:

C(N) = a * C(N/b) + f(n)

And I have no idea how to get rid of the second recurrent call to C() in my recurrence relation, so how do I do it? I don't know how to derive the values of a and b.

You need Akra--Bazzi: http://en.wikipedia.org/wiki/Akra%E2%80%93Bazzi_method — David Eisenstat, Jun 03 '15 at 15:05
See my answer here ---> http://stackoverflow.com/a/30402465/3482140 — Am_I_Helpful, Jun 03 '15 at 15:07
@DavidEisenstat wow, cool, never hear of that method before. — Pavel, Jun 03 '15 at 15:07
@shekharsuman your answer doesn't provide anything more than just a reference, but that I already got from David. — Pavel, Jun 03 '15 at 15:08
@paulpaul1076- I mentioned the reason why master-theorem can't be used in my answer too, please refer to the 2 points. And, *`please write an answer to your question`*. Good luck implementing Akra-Bazzi Method. :) — Am_I_Helpful, Jun 03 '15 at 15:10
I wrote the solution, feel free to correct if anything is wrong, thanks for help. — Pavel, Jun 03 '15 at 16:46

Pavel · Accepted Answer · 2015-06-03T16:50:05.173

As all the commenters said, I need to use the Akra-Bazzi theorem.

C(1) = 1
C(2) = 1

For N > 2 we need to first find 'p' from the following equation : 
(1/3)^p + (2/3)^p = 1. It is obvious that p = 1.

Next we need to solve N^p * ( 1 + I[1,N](log(u)/u) ) with p = 1

I[1,N](x) denotes the integral of x, from 1 to N.
also I wrote log(u)/u instead of (u - 1)log(u-1)/u^2
since I((u-1)log(u-1)/u^2) looks like a monster.

I[1,N](log(u)/u) gives log^2(N)/2 so the end result is N + N*(log^2(N)/2).

All in all, the running time = theta( N + N*(log^2(N)/2) ).

What is the approach to solving a recurrence relation when I have more than one recurrent function calls on the right-hand side of the equation?

1 Answers1