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In the following article, we have presented the Iteration method for finding the Time complexity of an algorithm in detail.

**Table of Contents**:

- Introduction to Recurrence relations
- Iteration Method
- MCQs on Recurrence relations

Pre-requisites:

# Introduction to Recurrence relations

Recurrence relation is way of determining the running time of a recursive algorithm or program. It's a equation or a inequality that describes a functions in terms of its values and smaller inputs. Now before jumping on to various methods of solving recurrence relation, let's first take a look at the example of recurrence relation.

Let us consider **T(n)** to be the running time of a given problems on size **n**, the problem in our case will be finding the *nth fibonacci number*. Let *F(n)* denote the *nth fibonacci number*, therefore *F(n) = F(n-1) + F(n-2)* and *T(0) = a , a is a constant*.

The recurrence relation will be :

**T(n) = c + T(n-1)**, where c is a constant

There are mainly four methods of solving recurrence relation :

- Iteration method
- Substitution method
- Master method
- Recursion tree method

In this article our primary focus is *Solving recurrence relation via Iteration method*, hence we will deep dive into the process through examples and explanations.

# Iteration Method

Iteration method can be also be called a *Brute force* method because we have to substitute the recurrent part value until a pattern is observed, thereafter we use mathematical summation technique is used to find the recurrence.

Now let us take an example :

**Recurrence relation :** T(1) = theta(1) and T(n) = n^3 + 2T(n/2)

**Solution :**

T(n) n^3 + 2(T(n/2)) ------------ (i)

*On putting value of T(n/2) in (i) we get :*

T(n) = n^3 + 2((n/2)^3 + 2(T(n/4))) ------------ (ii)

*Putting value of T(n/4) in (ii) we get :*

T(n) = n^3 + 2((n/2)^3 + 2((n/4)^3 + 2T(n/8))))

*Using similar subtitutions further we get :*

T(n) = n^3 + 2((n/2)^3 + 2((n/4)^3 + 2((n/8)^3 + 2(T(n/16)))))

*We notice that n^3 term is replaced by (n/2)^3, then by (n/4)^3 and so on ............*

*Now let's simply it further :*

T(n) = n^3 + 2(n/2)^3 + 4((n/4)^3 + 2((n/8)^3 + 2((T(n/16))))

T(n) = n^3 + 2(n/2)^3 + 4(n/4)^3 + 8((n/8)^3 + 2(T(n/16)))

T(n) = n^3 + ((n^3)/4) + ((n^3)/4^2) + ((n^3)/4^3) + 16(T(n/16))

*Now form the original problem we know that :*

T(1) = theta(1)

*hence,*

T(n/2^i) = 1 when i = log n (base 2)

*therefore*

T(n/2^4) = 1 when 4 = log n (base 2)

T(n) = n^3(1 + 1/4^1 + ------------ + 1/4^i-1) + 2^i(T(n/2^i))

*Now we know that i = log n (base 2)*

T(n) = n^3(1/4^0 + 1/4^1 + --------- + 1/4^i-1) + 2^i(T(n/2^i))

*On simplifying further we get:*

T(n) = n^3(1/4^0 + 1/4^1 + ----------- + 1/4^i-1) + n*T(1)

*Now n(T(1)) becomes theta(n)*, *we also have finite G.P. with first term equal to 1 and common ratio equal to 1/4, lets apply the sum formula for finding the sum of the terms of finite G.P.*

**S(n) = a(1 - r^n)/(1 - r)** *is the sum of first n terms of G.P.*

T(n) = theta(n) + n^3(1(1 - n^i-1)/(1-1/4))

*Since its a finite G.P. hence finite time will be taken to execute*

T(n) = theta(n) + n^3(theta(1))

T(n) = theta(n) + theta(n^3)

T(n) = theta(n^3) **Ans**

# MCQs on Recurrence relations

**(i) Find the solution of given recurrence relation F(n) = 20F(n-1) - 25F(n-2), where F(0) = 4 and F(1) = 14**

*Ans : b*

**Explanation :** First find the characteristic polynomial i.e. x^2-20x+36=0 in this case, having found the roots find the form of recurrence equation i.e. an=a(2^n)+b(18^n), find the value of a and b using equations F(0) = 4 and F(1) = 14 i.e. a=7/2 and b=-1/2, hence the answer is option b.

**(ii) Determine the recurrence relation for the following series 1,7,31,127,499**

*Ans : c*

**Explanation :** On observing carefully, we see that the difference between the consecutive terms is a factor of 4, therefore 1.4=4,7.4=28,31.4=124 so on, ending term is 3 less than the next term therefore the answer will be option c.

**(iii) Find the solution of given recurrence relation F(n) = 6T(n-1) - 8T(n-2), given that T(0) = 3 and T(1) = 5**

*Ans : b*

**Explanation :** Follow similar steps as mentioned in the first problem.

**(iv) Find the solution of the recurrence relation T(n) = 5T(n-1) + 6T(n-2)**

*Ans : b*

**Explanation :** Just check on the left side of the given equation for all the options individually, we find that 6n satisfies the equation hence answer is option b.

**(v) Find the value of T(2) for the recurrence relation T(n) = 17T(n-1) + 30n, given that T(0) = 3**

*Ans : d*

**Explanation :** First calculate the value for T(1) i.e. T(1) = 17T(0) + 30 = 81, now calculate value of T(2) by putting the value of T(1) in it i.e. T(2) = 17T(1) + 60 = 1437, that is our required answer.

**Thank You!**

With this article at OpenGenus, you must have a strong idea of Iteration Method to find Time Complexity of different algorithms.