Practice Set for Recurrence Relations
Last Updated :
23 May, 2025
A recurrence relation is an equation that expresses each term of a sequence as a function of its preceding terms. In other words, it defines a sequence where each term depends on one or more of its previous terms. Recurrence relations commonly arise in divide-and-conquer algorithms, dynamic programming, and combinatorial problems.
Below are practice problems with solutions on recurrence relation.
Problem 1:
Solve the following recurrence relation?
7T(n/2)+3n2+2
Options:
(a) O(n2.8)
(b) O(n3)
(c) θ(n2.8)
(d) θ(n3)
Correct answers: (a), (b), and (c)
Explanation:
- a = 7, b = 2, f(n) = 3n2 + 2
- c = 2 for f(n) = O(nc)
- logba = log27 ≈ 2.81 > 2
- By Master Theorem case 1, T(n) = θ(n2.81)
- This implies O(n2.8) and O(n3) as well.
Problem 2:
Sort the following functions in decreasing order of their asymptotic (big-O) complexity:
- f1(n)=n√n
- f2(n)=2n
- f3(n)=(1.000001)n
- f4(n)=n10⋅2n/2
Options:
(a) f2 > f4 > f1 > f3
(b) f2 > f4 > f3 > f1
(c) f1 > f2 > f3 > f4
(d) f2 > f1 > f4 > f3
Correct answer: (b) f2 > f4 > f3 > f1
Explanation:
- f2 > f4 because 2n = 2n/2⋅2n/2 and f4 = n10⋅2n/2
- f4 > f3 because f4=n10⋅(1.414)n and 1.414n grows faster than 1.000001n
- f3 > f1 since log f3=nlog(1.000001) grows faster than logf1 = √nlogn
Problem 3:
Given: f(n)=22n, which of the following correctly represents f(n)?
Options:
(a) O(2n)
(b) Ω(2n)
(c) Θ(2n)
(d) None of these
Correct answer: (b) Ω(2n)
Explanation:
- 22n = (2n)2
- f(n) grows faster than 2n
- f(n) ≥ c⋅2nf(n) some c, so Ω(2n) holds.
- O(2n) and Θ(2n) do not hold because f(n) grows faster.
Problem 4:
Master’s theorem can be applied on which of the following recurrence relations?
Options:
(a) T(n) = 2T(n/2) + 2n
(b) T(n) = 2T(n/3) + sin(n)
(c) T(n) = T(n − 2) + 2n2 + 1
(d) None of these
Correct answer: (d) None of these
Explanation:
- Master theorem applies to recurrences of form T(n) = aT(n/b) + f(n) where f(n) is polynomial or monotonic.
- (a) is invalid because f(n) = 2n is exponential, not polynomial.
- (b) is invalid because sin(n) is not monotonic.
- (c) is a decreasing function recurrence (not dividing), so Master theorem does not apply here in the usual form.
Problem 5:
Given: T(n) = 3T(n/2 +47) + 2n2 + 10n − 1/2
What is the order of T(n)?
Options:
(a) O(n2)
(b) O(n3/2)
(c) O(nlogn)
(d) None of these
Correct answer: (a) O(n2)
Explanation:
- For large n, n/2 + 47 ≈ n/2
- So, T(n) ≈ 3T(n/2) + O(n2)
- By Master theorem: a = 3, b = 2, f(n) = O(n2)
- log23 ≈ 1.58 < 2, so case 3 applies.
- T(n) = O(n2)
Read in detail about Recurrence Relation here.
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