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Algorithms/Phase 1 · Asymptotic Analysis & Recurrences

Chapter review LI-1.2-1.3 · Item 20 of 79

Chapters 1.2–1.3 Lock-in — Recurrences & Cases and space

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ACompressed recap

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Recurrences from code: Σ calls + outside work. Unrolling: T(n−1)+n → n², 2T(n−1)+1 → 2ⁿ−1, 2T(n/2)+n → n log n (+ n). Recursion tree: level i = aⁱ f(n/bⁱ), leaves n^(log_b a); shrinking → f, equal → f log n, growing → leaves. Master theorem with w = n^(log_b a); for f = nᵈ compare a with bᵈ; case 2 with logᵏ adds one log; gaps (n/log n), T(n−1) not covered. Change of variables m = log n for roots; subtract-and-conquer sums; uneven splits by fractions. Cases: worst is default; average needs a distribution; space includes recursion depth; amortised = total/ops.

BMerged rule card

1.2.1 Recurrences from recursive code; solving by unrolling (substitution)

  • T(n) = Σ recursive calls + outside work; base case.
  • T(n−1)+c → n; T(n−1)+n → n²; T(n/2)+c → log n; 2T(n/2)+n → n log n; 2T(n−1)+1 → 2ⁿ − 1.
  • Unroll, find the pattern, sum; for exact values compute bottom-up.

1.2.2 The recursion-tree method

  • Level i: aⁱ nodes of size n/bⁱ; cost aⁱ·f(n/bⁱ); depth log_b n; leaves n^(log_b a).
  • Shrinking levels → Θ(f(n)); equal → Θ(f(n) log n); growing → Θ(n^(log_b a)).
  • Unequal splits: fractions sum to 1 → n log n; less than 1 → n (for f = n).

1.2.3 The master theorem

  • T(n) = aT(n/b) + f(n); w = n^(log_b a).
  • f = O(w/n^ε) → Θ(w); f = Θ(w logᵏ n), k ≥ 0 → Θ(w logᵏ⁺¹ n); f = Ω(w·n^ε) + regularity → Θ(f).
  • f = nᵈ: compare a with bᵈ.
  • Not for T(n−1), a < 1, gaps (n/log n), non-constant a.

1.2.4 Harder recurrences: change of variables, T(n−1) forms, unequal splits

  • T(n^(1/k)) → set m = log n; T(√n) + 1 → log log n; 2T(√n) + 1 → log n; 2T(√n) + log n → log n log log n.
  • T(n−1) + f → Σ f; a·T(n−1) with a > 1 → exponential; Fibonacci → φⁿ.
  • Uneven splits: fractions sum 1 → n log n; < 1 → n (for f = n).
  • Substitution: guess, then prove by induction.

1.3.1 Best, worst and average case; space complexity and recursion depth; amortised cost

  • Worst (default), best, average (needs a distribution).
  • Linear search: 1 / n / (n+1)/2.
  • Space: auxiliary memory + recursion depth × frame; in-place = O(1).
  • Amortised = total ÷ operations; doubling array O(1) amortised (copies < 2n).

CInterleaved GATE-style questions

Q1NAT2 marksExact value

T(1) = 2, T(n) = T(n/2) + 3 (n a power of 2). T(64) = ?

Numerical answer: ________

Q2MCQ1 markTree

T(n) = 9T(n/3) + n is:

  • (A) Θ(n log n)
  • (B) Θ(n)
  • (C) Θ(n³)
  • (D) Θ(n²)

Q3MCQ1 markMaster

T(n) = 3T(n/3) + n log n is:

  • (A) Θ(n²)
  • (B) Θ(n log² n)
  • (C) Θ(n log log n)
  • (D) Θ(n log n)

Q4NAT2 marksMaster exponent

T(n) = 64T(n/4) + n² log n. T(n) = Θ(nᵏ). k = ?

Numerical answer: ________

Q5MCQ1 markRoot recurrence

T(n) = 2T(√n) + log n is:

  • (A) Θ(log n log log n)
  • (B) Θ(√n)
  • (C) Θ(log² n)
  • (D) Θ(log n)

Q6NAT2 marksExact

T(1) = 1, T(n) = T(n − 1) + 2n. T(10) = ?

Numerical answer: ________

Q7NAT1 markAverage

Linear search in 99 distinct elements, x present uniformly. Average comparisons?

Numerical answer: ________

Q8MCQ1 markSpace

A recursive function f(n) calls f(n − 1) once, with constant work and frame. Its time and space are:

  • (A) Θ(n) time, Θ(n) space
  • (B) Θ(log n) time, Θ(log n) space
  • (C) Θ(n²) time, Θ(n) space
  • (D) Θ(n) time, Θ(1) space

DAnswer Vault

Answer Vault

Try the questions first. Open when you are ready to check.

Q1 Answer: 202 marks

log 64 = 6 steps of +3: 2 + 18.

Re-read: Lesson 1.2.1 — Recurrences from recursive code; solving by unrolling (substitution)

Q2 Answer: (D)1 mark

Leaves n² dominate.

  • (A) is wrong: Levels are not equal.
  • (B) is wrong: Root only.
  • (C) is wrong: Too large.

Re-read: Lesson 1.2.2 — The recursion-tree method

Q3 Answer: (B)1 mark

w = n, f = n log n → case 2 with k = 1.

  • (A) is wrong: Too large.
  • (C) is wrong: No.
  • (D) is wrong: Missing a log.

Re-read: Lesson 1.2.3 — The master theorem

Q4 Answer: 32 marks

w = n^(log₄ 64) = n³, which beats n² log n.

Re-read: Lesson 1.2.3 — The master theorem

Q5 Answer: (A)1 mark

S(m) = 2S(m/2) + m.

  • (B) is wrong: Too large.
  • (C) is wrong: Too large.
  • (D) is wrong: That is 2T(√n) + 1.

Re-read: Lesson 1.2.4 — Harder recurrences: change of variables, T(n−1) forms, unequal splits

Q6 Answer: 1092 marks

1 + 2(2 + 3 + … + 10) = 1 + 2 × 54.

Re-read: Lesson 1.2.4 — Harder recurrences: change of variables, T(n−1) forms, unequal splits

Q7 Answer: 501 mark

(99 + 1)/2.

Re-read: Lesson 1.3.1 — Best, worst and average case; space complexity and recursion depth; amortised cost

Q8 Answer: (A)1 mark

n calls, all on the stack at the deepest point.

  • (B) is wrong: No halving.
  • (C) is wrong: Constant work per call.
  • (D) is wrong: Recursion depth n uses Θ(n) stack.

Re-read: Lesson 1.3.1 — Best, worst and average case; space complexity and recursion depth; amortised cost

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