T(n)=n+4n2+8+16n4+32+⋯+Θ(nlg4)=lg(n−1)∑i=0 2in+2lg(n−1)∑i=0 4i+Θ(n2)=2lgn−12−1n+2⋅4lgn−14−1+Θ(n2)=(2lgn−1)n+23(4lgn−1)+Θ(n2)=(n−1)n+23(n2−1)+Θ(n2)=Θ(n2)
Question
T(n)=n+4n²+8+16n⁴+32+⋯+Θ(nlg4)=lg(n−1)∑ᵢ=0 2ⁱn+2lg(n−1)∑ᵢ=0 4ⁱ+Θ(n²)=2lg(n)−12⁻¹n+2⋅4lg(n)−14⁻¹+Θ(n²)=(2lg(n)−1)n+23(4lg(n)−1)+Θ(n²)=(n−1)n+23(n²−1)+Θ(n²)=Θ(n²)
Solution
Break Down the Problem
- Identify the expression for .
- Analyze the individual components of the expression.
- Combine the terms to simplify .
Relevant Concepts
- Big Theta notation indicates asymptotic tight bounds.
- Summations and logarithmic identities will be used for simplification.
- The behavior of polynomial and logarithmic terms as grows.
Analysis and Detail
-
The given expression is:
-
Simplify the summations:
-
The first summation: Therefore,
-
The second summation: Therefore,
-
-
Combining terms together:
Verify and Summarize
- Through simplification and collecting the terms, we observe that as increases:
- The leading term comes from .
- Thus, dominates the logarithmic terms.
Final Answer
Similar Questions
T(n)=n+4n2+8+16n4+32+⋯+Θ(nlg4)=lg(n−1)∑i=0 2in+2lg(n−1)∑i=0 4i+Θ(n2)=2lgn−12−1n+2⋅4lgn−14−1+Θ(n2)=(2lgn−1)n+23(4lgn−1)+Θ(n2)=(n−1)n+23(n2−1)+Θ(n2)=Θ(n2)
Consider this recurrence relation: T(1) = 1 T(2) = 1 T(n) = 4 T(n-2) + 2n2 for n>2 The Master Theorem tells us Group of answer choices
Select the correct recurrence relation for Tower of Hanoi?ans.T(n)= 2T(n-1)+1T(n)= 2T(n-1)+2T(n)= 2T(n)+1T(n)= 2T(n-2)+2 Previous Marked for Review Next
Select the correct recurrence relation for Tower of Hanoi?ans.T(n)= 2T(n-1)+2T(n)= 2T(n-2)+2T(n)= 2T(n)+1T(n)= 2T(n-1)+1 Previous Marked for Review Next
Consider the following recurrence relation: T(n)=3T(n/3)+n2/3,T(1)=1,T(0)=0 What is the complexity of T(n)? Options : Θ(n) Θ(nlogn) Θ(n^2/3) Θ(n^1.5)
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.