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DAA Recurrence Relation

DAA Recurrence Relation with daa tutorial, introduction, Algorithm, Asymptotic Analysis, Control Structure, Recurrence, Master Method, Recursion Tree Method, Sorting Algorithm, Bubble Sort, Selection Sort, Insertion Sort, Binary Search, Merge Sort, Counting Sort, etc.

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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
 2TDAA Recurrence Relation + 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) = TDAA Recurrence Relation + 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 

   T (n) ≤c logn.

Put this in given Recurrence Equation.

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

Example2 Consider the Recurrence

T (n) = 2TDAA Recurrence Relation+ 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) ≤2cDAA Recurrence Relationlog DAA Recurrence Relation+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

T (n) = 1  if n=1
      = 2T (n-1) if n>1

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

T (n) = T (n-1) +1 and T (1) = 	θ (1).

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).
Next TopicRecursion Tree Method




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