Math Problem Statement

Problem 1: Recurrence Relation with Division by 3 and Linear Term Solve the recurrence relation: T(N)=T(N/3)+N with the base case T(1)=1. Solution to Problem 1: T(N)=O(N) Hint: The summation of the geometric series will be ..which is eventually Final Answer: T(N)=O(N) Problem 2: Recurrence Relation with Multiplication by 2 and Linear Term Problem: Solve the recurrence relation: T(N)=2T(N/2)+N with the base case T(1)=1. Solution to Problem 2: Final Answer: T(N)=O(NlogN)

Solution

Solution to Problem 1: Recurrence Relation with Division by 3 and Linear Term

The recurrence relation given is:

T(N)=T(N3)+NT(N) = T\left(\frac{N}{3}\right) + N with the base case T(1)=1T(1) = 1.

Approach

  1. Recursive Expansion: Expand the recurrence repeatedly: T(N)=T(N3)+NT(N) = T\left(\frac{N}{3}\right) + N Substitute for T(N3)T\left(\frac{N}{3}\right): T(N)=T(N32)+N3+NT(N) = T\left(\frac{N}{3^2}\right) + \frac{N}{3} + N Repeat kk times: T(N)=T(N3k)+N(1+13+132++13k1)T(N) = T\left(\frac{N}{3^k}\right) + N \left(1 + \frac{1}{3} + \frac{1}{3^2} + \cdots + \frac{1}{3^{k-1}}\right)

  2. Geometric Series: The summation i=0k113i\sum_{i=0}^{k-1} \frac{1}{3^i} converges as kk \to \infty, and the sum becomes: Sum=1113=32.\text{Sum} = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}. Thus: T(N)=T(1)+N32.T(N) = T(1) + N \cdot \frac{3}{2}.

  3. Asymptotic Analysis: The dominant term is 32N\frac{3}{2}N, so: T(N)=O(N).T(N) = O(N).

Solution to Problem 2: Recurrence Relation with Multiplication by 2 and Linear Term

The recurrence relation is:

T(N)=2T(N2)+NT(N) = 2T\left(\frac{N}{2}\right) + N with the base case T(1)=1T(1) = 1.

Approach

  1. Recursive Expansion: Expand the recurrence repeatedly: T(N)=2T(N2)+NT(N) = 2T\left(\frac{N}{2}\right) + N Substitute for T(N2)T\left(\frac{N}{2}\right): T(N)=2(2T(N4)+N2)+NT(N) = 2\left(2T\left(\frac{N}{4}\right) + \frac{N}{2}\right) + N Simplify: T(N)=4T(N4)+2N.T(N) = 4T\left(\frac{N}{4}\right) + 2N. Repeat kk times: T(N)=2kT(N2k)+kN.T(N) = 2^k T\left(\frac{N}{2^k}\right) + kN.

  2. Stopping Condition: When N2k=1\frac{N}{2^k} = 1, we have k=log2(N)k = \log_2(N). At this point: T(1)=1.T(1) = 1.

  3. Final Expression: Substitute k=log2(N)k = \log_2(N): T(N)=2log2(N)T(1)+Nlog2(N).T(N) = 2^{\log_2(N)}T(1) + N \cdot \log_2(N). Simplify 2log2(N)=N2^{\log_2(N)} = N: T(N)=N+Nlog2(N).T(N) = N + N\log_2(N).

  4. Asymptotic Analysis: The dominant term is Nlog2(N)N\log_2(N), so: T(N)=O(NlogN).T(N) = O(N\log N).

Summary

  • Problem 1 Final Answer: T(N)=O(N)T(N) = O(N)
  • Problem 2 Final Answer: T(N)=O(NlogN)T(N) = O(N\log N)

Let me know if you need any clarifications or details about these solutions!


Questions for Deeper Understanding:

  1. What happens if the base case is not T(1)=1T(1) = 1 but some other value?
  2. How does changing the constant in T(N)=T(N/3)+cNT(N) = T(N/3) + cN affect the solution?
  3. Can the recurrence T(N)=T(N/3)+NT(N) = T(N/3) + N be solved using a Master Theorem?
  4. What are the real-world scenarios where these recurrences arise?
  5. How would the analysis change if NN is not a power of 3 or 2?

Tip:

When solving recurrences, always check if the Master Theorem applies first—it can save a lot of time in analysis!

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Math Problem Analysis

Mathematical Concepts

Recurrence Relations
Geometric Series
Asymptotic Analysis
Divide and Conquer

Formulas

Geometric series sum: S = a / (1 - r) for |r| < 1
Master Theorem for divide-and-conquer recurrences: T(N) = aT(N/b) + O(N^d)

Theorems

Master Theorem
Geometric Series Formula

Suitable Grade Level

Undergraduate Level (Computer Science, Discrete Mathematics)