Practice Set for Recurrence Relations

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(n^2.8)
(b) O(n³)
(c) θ(n^2.8)
(d) θ(n³)

Correct answers: (a), (b), and (c)

Explanation:

• a = 7, b = 2, f(n) = 3n² + 2
• c = 2 for f(n) = O(n^c)
• log⁡_ba = log⁡₂7 ≈ 2.81 > 2
• By Master Theorem case 1, T(n) = θ(n^2.81)
• This implies O(n^2.8) 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⋅2^n/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 = 2^n/2⋅2^n/2 and f₄ = n¹⁰⋅2^n/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 log⁡f₁ = √nlog⁡n

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(n^3/2)
(c) O(nlog⁡n)
(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)
• log⁡₂3 ≈ 1.58 < 2, so case 3 applies.
• T(n) = O(n²)

Read in detail about Recurrence Relation here.