Topic 1: Big-O Analysis #
Question 1 #
For each of the following functions, state the tightest possible asymptotic bound (Θ) in terms of n:
- f(n) = 5n³ + 2n² + 100
- f(n) = log₂(n) + √n
- f(n) = 2^(n+1)
- f(n) = n! / 1000
Answer:
- f(n) = 5n³ + 2n² + 100 = Θ(n³) — The highest-degree term dominates.
- f(n) = log₂(n) + √n = Θ(√n) — √n grows faster than log n.
- f(n) = 2^(n+1) = Θ(2^n) — The constant factor 2 in the exponent doesn’t change the asymptotic class.
- f(n) = n! / 1000 = Θ(n!) — Division by constant doesn’t affect asymptotic growth.
Question 2 #
Prove that n² + 10n + 100 = Θ(n²).
Answer: To prove Θ(n²), we need to show both O(n²) and Ω(n²).
O(n²): Find c, n₀ such that n² + 10n + 100 ≤ c·n² for n ≥ n₀. n² + 10n + 100 ≤ n² + 10n² + 100n² = 111n² for n ≥ 1 So c = 111, n₀ = 1 works.
Ω(n²): Find c, n₀ such that n² + 10n + 100 ≥ c·n² for n ≥ n₀. For n ≥ 11: n² + 10n + 100 ≥ n² + 10n + n² = 2n² (since n ≥ 11 ⇒ 10n ≥ n²/11 + 10) Actually simpler: For n ≥ 10: n² + 10n + 100 ≥ n² So c = 1, n₀ = 10 works.
Since both hold, n² + 10n + 100 = Θ(n²). ∎
Question 10 #
Solve the recurrence T(n) = 2T(n-1) + 1 with T(0) = 0 using recursion trees.
Answer: Recursion Tree:
- Level 0: 1 node, work = 1
- Level 1: 2 nodes, each does 1, total = 2
- Level 2: 4 nodes, total = 4
- Level i: 2^i nodes, total = 2^i
Number of levels = n (from n down to 0)
Total = 1 + 2 + 4 + … + 2^{n-1} + 2^n - 1 (base case) = 2^n - 1 = Θ(2^n)
This is exponential - the recurrence describes algorithms like the naive recursive subset sum.
Question 6 #
Prove that for all n ≥ 1, the sum 1 + 2 + … + n = n(n+1)/2.
Answer: Base Case: n = 1: Left side = 1. Right side = 1(1+1)/2 = 1. ✓
Inductive Case: Assume the formula holds for n = k: 1 + 2 + … + k = k(k+1)/2
For n = k+1:
1 + 2 + ... + k + (k+1)
= [k(k+1)/2] + (k+1) (by induction hypothesis)
= [k(k+1) + 2(k+1)] / 2
= (k+1)(k+2) / 2
= (k+1)((k+1)+1) / 2
This is exactly the formula with n = k+1.
By induction, the formula holds for all n ≥ 1. ∎
Question 10 #
Prove that the following algorithm correctly determines whether a string is a palindrome:
def is_palindrome(s):
if len(s) <= 1:
return True
return s[0] == s[-1] and is_palindrome(s[1:-1])
Answer: Base Cases:
- Empty string (len = 0): returns True. Empty string is a palindrome. ✓
- Single character (len = 1): returns True. Single character is a palindrome. ✓
Inductive Case: Assume is_palindrome correctly determines if any string of length < k is a palindrome. Consider string s of length k ≥ 2.
The function checks if s[0] == s[-1]:
- If false: first and last characters differ, s cannot be palindrome. Returns False correctly.
- If true: The outer characters match. For s to be a palindrome, the inner substring s[1:-1] must also be a palindrome.
By the induction hypothesis, is_palindrome(s[1:-1]) correctly determines if the inner substring is a palindrome.
Therefore, the function correctly determines if s is a palindrome.
By induction, the algorithm is correct for all strings. ∎
Question 4 #
For the game of Tic-Tac-Toe on a 3×3 board, estimate the size of the complete game tree (upper bound). Explain your reasoning.
Answer: Upper Bound: 9! = 362,880 nodes
Reasoning:
- Player X goes first and can place in any of 9 positions
- Player O then has 8 remaining positions
- Player X then has 7 remaining, and so on
- Total orderings: 9 × 8 × 7 × … × 1 = 9!
Tighter bound:
- Game ends early when someone wins or board is full
- After 5 X’s and 4 O’s (9 total), the game definitely ends
- More accurate upper bound: 9! / 5!4! ≈ 12,654 (considering game ends)
The actual number of distinct games is around 255,168 (with optimal play and early termination).
The grocery store has priced apples at 5 for a dollar.
You want the most weight of apple.
Does the greedy algorithm work? Prove or disprove.
