Recurrence Relation

A recurrence is an equation or inequality that describes a function in terms of its values on smaller inputs. To solve a Recurrence Relation means to obtain a function defined on the natural numbers that satisfy the recurrence.

For Example, the Worst Case Running Time T(n) of the MERGE SORT Procedures is described by the recurrence.

T (n) = θ (1) if n=1  2T + θ (n) if n>1 

There are four methods for solving Recurrence:

1. Substitution Method
2. Iteration Method
3. Recursion Tree Method
4. Master Method

1. Substitution Method:

The Substitution Method Consists of two main steps:

1. Guess the Solution.
2. Use the mathematical induction to find the boundary condition and shows that the guess is correct.

For Example1 Solve the equation by Substitution Method.

T (n) = T + n 

We have to show that it is asymptotically bound by O (log n).

Solution:

For T (n) = O (log n) 

We have to show that for some constant c

Put this in given Recurrence Equation.

T (n) ≤c log+ 1 ≤c log+ 1 = c logn-clog2 2+1 ≤c logn for c≥1 Thus T (n) =O logn. 

Example2 Consider the Recurrence

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

Find an Asymptotic bound on T.

Solution:

We guess the solution is O (n (logn)).Thus for constant 'c'.  T (n) ≤c n logn Put this in given Recurrence Equation. Now,   T (n) ≤2clog +n       ≤cnlogn-cnlog2+n       =cn logn-n (clog2-1)       ≤cn logn for (c≥1) Thus T (n) = O (n logn). 

2. Iteration Methods

It means to expand the recurrence and express it as a summation of terms of n and initial condition.

Example1: Consider the Recurrence

Solution:

   T (n) = 2T (n-1)       = 2[2T (n-2)] = 22T (n-2)       = 4[2T (n-3)] = 23T (n-3)       = 8[2T (n-4)] = 24T (n-4)   (Eq.1)  Repeat the procedure for i times  T (n) = 2i T (n-i) Put n-i=1 or i= n-1 in    (Eq.1) T (n) = 2n-1 T (1)       = 2n-1 .1    {T (1) =1 .....given}       = 2n-1  

Example2: Consider the Recurrence

Solution:

 T (n) = T (n-1) +1        = (T (n-2) +1) +1 = (T (n-3) +1) +1+1        = T (n-4) +4 = T (n-5) +1+4        = T (n-5) +5= T (n-k) + k Where k = n-1    T (n-k) = T (1) = θ (1)    T (n) = θ (1) + (n-1) = 1+n-1=n= θ (n).