Open In App

Different types of recurrence relations and their solutions

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Share
Report issue
Report

In this article, we will see how we can solve different types of recurrence relations using different approaches. Before understanding this article, you should have idea about recurrence relations and different method to solve them (See : Worst, Average and Best Cases, Asymptotic Notations, Analysis of Loops).

Type 1: Divide and Conquer Recurrence Relations:

Following are some of the examples of recurrence relations based on divide and conquer.

T(n) = 2T(n/2) + cn
T(n) = 2T(n/2) + √n

These types of recurrence relations can be easily solved using Master Method

For recurrence relation T(n) = 2T(n/2) + cn, the values of a = 2, b = 2 and k =1. Here logb(a) = log2(2) = 1 = k. Therefore, the complexity will be Θ(nlog2(n)). Similarly for recurrence relation T(n) = 2T(n/2) + √n, the values of a = 2, b = 2 and k =1/2. Here logb(a) = log2(2) = 1 > k. Therefore, the complexity will be Θ(n).

Type 2: Linear Recurrence Relations:

Following are some of the examples of recurrence relations based on linear recurrence relation.

T(n) = T(n-1) + n for n>0 and T(0) = 1

These types of recurrence relations can be easily solved using substitution method

For example,

T(n) = T(n-1) + n
= T(n-2) + (n-1) + n
= T(n-k) + (n-(k-1))….. (n-1) + n

Substituting k = n, we get

T(n) = T(0) + 1 + 2+….. +n = n(n+1)/2 = O(n^2)

Type 3: Value Substitution Before Solving:

Sometimes, recurrence relations can’t be directly solved using techniques like substitution, recurrence tree or master method. Therefore, we need to convert the recurrence relation into appropriate form before solving. For example,

T(n) = T(√n) + 1

To solve this type of recurrence, substitute n = 2^m as:

T(2^m) = T(2^m /2) + 1
Let T(2^m) = S(m),
S(m) = S(m/2) + 1

Solving by master method, we get

S(m) = Θ(logm)
As n = 2^m or m = log2(n),
T(n) = T(2^m) = S(m) = Θ(logm) = Θ(loglogn)

Practice Problems on Recurrence Relations:

Question 1: What is the time complexity of Tower of Hanoi problem?

(A) T(n) = O(sqrt(n))
(B) T(n) = O(n^2)
(C) T(n) = O(2^n)
(D) None

Solution: For Tower of Hanoi, T(n) = 2T(n-1) + c for n>1 and T(1) = 1. Solving this,

T(n) = 2T(n-1) + c

= 2(2T(n-2)+ c) + c = 2^2*T(n-2) + (c + 2c)

= 2^k*T(n-k) + (c + 2c + .. kc)

Substituting k = (n-1), we get

T(n) = 2^(n-1)*T(1) + (c + 2c + (n-1)c) = O(2^n)

Question 2: Consider the following recurrence:

T(n) = 2 * T(ceil (sqrt(n) ) ) + 1, T(1) = 1

Which one of the following is true?

(A) T(n) = (loglogn)
(B) T(n) = (logn)
(C) T(n) = (sqrt(n))
(D) T(n) = (n)

Solution: To solve this type of recurrence, substitute n = 2^m as:

T(2^m) = 2T(2^m /2) + 1
Let T(2^m) = S(m),
S(m) = 2S(m/2) + 1
Solving by master method, we get
S(m) = Θ(m)
As n = 2^m or m = log2n,
T(n) = T(2^m) = S(m) = Θ(m) = Θ(logn)


Last Updated : 20 Dec, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads