Chapter 4: Problem 9
Draw recursion trees, and use them to find big \(\Theta\) bounds on the solutions to the following recurrences. For each, assume that \(T(1)=1\) and that \(n\) is a power of the appropriate integer. a. \(T(n)=8 T(n / 2)+n\) b. \(T(n)=8 T(n / 2)+n^{3}\) c. \(T(n)=3 T(n / 2)+n\) d. \(T(n)=T(n / 4)+1\) e. \(T(n)=3 T(n / 3)+n^{2}\)
Short Answer
Step by step solution
- Understanding Recurrence (a)
- Understanding Recurrence (b)
- Understanding Recurrence (c)
- Understanding Recurrence (d)
- Understanding Recurrence (e)
- Apply Master Theorem to Recurrence (a)
- Apply Master Theorem to Recurrence (b)
- Apply Master Theorem to Recurrence (c)
- Apply Recursion Tree to Recurrence (d)
- Apply Master Theorem to Recurrence (e)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Master Theorem
- **Case 1:** If \( f(n) \) is \( O(n^{c}) \) where \( c < \log_b a \), this means the function \( f(n) \) grows slower than the threshold function \( n^{\log_b a} \), and the solution is \( T(n) = \Theta(n^{\log_b a}) \).
- **Case 2:** If \( f(n) \) is \( \Theta(n^{c}) \) where \( c = \log_b a \), the function matches the threshold exactly, leading to an extra logarithmic factor: \( T(n) = \Theta(n^{\log_b a} \log n) \).
- **Case 3:** If \( f(n) \) is \( \Omega(n^{c}) \) with \( c > \log_b a \), then \( f(n) \) dominates growth, and \( T(n) = \Theta(f(n)) \).
An example is recurrence (a): \( T(n) = 8T(n/2) + n \), where \( n^{\log_2 8} = n^3 \). Since \( f(n) = n \) is \( O(n^1) \) and \( 1 < 3 \), we have \( T(n) = \Theta(n^3) \) according to Case 1.
Recursion Trees
The basic idea is to:
- Represent the initial problem at the root.
- Each child node represents a subproblem from the parent node's problem.
- Repeat until reaching a base case, typically a known value or simple computation.
- The depth of the tree, \( \log_b n = \log_4 n \) where each level sums to 1, tells us the complete height.
- The total work can then be calculated by multiplying the constant contribution by the number of levels, resulting in \( T(n) = \Theta(\log n) \).
Big Theta Notation
When a function \( f(n) = \Theta(g(n)) \), this indicates that:
- There exist positive constants \( c_1 \), \( c_2 \), and \( n_0 \) such that for all \( n \geq n_0 \),
- the function \( f(n) \) is sandwiched between \( c_1g(n) \) and \( c_2g(n) \).
Applying this to recurrence (b), \( T(n) = 8T(n/2) + n^3 \) uses the Master Theorem by directly comparing \( n^3 \) with the threshold \( n^{\log_2 8} = n^3 \), meaning both functions grow at the same rate. Hence, \( T(n) = \Theta(n^3 \log n) \), indicating precise bounds while capturing the logarithmic factor required in Case 2 of the theorem.