Math Problem Statement

Show that T(n) = T(floor(n/2)) + 1 is O(log n).

Solution

The image presents an exercise where you are asked to show that the recurrence relation T(n)=T(n2)+1T(n) = T\left(\left\lfloor \frac{n}{2} \right\rfloor\right) + 1 is O(logn)O(\log n), meaning it grows logarithmically. The attempts in the image try to solve this using two different approaches:

Key Points:

  1. Attempt 1:

    • The goal is to prove that T(n)clognT(n) \leq c \log n for some constant cc.
    • The steps try to bound the recurrence relation by substituting T(n)T(n) with a logarithmic expression.
    • The step where log(n+1)\log(n+1) appears doesn't successfully reduce the problem, and hence, the attempt fails (as marked with the red face).
  2. Attempt 2:

    • In this attempt, the approach is to subtract a constant α\alpha, assuming αR\alpha \in \mathbb{R}, so that T(n)clog(nα)T(n) \leq c \log(n - \alpha).
    • The key idea is to use a stronger assumption: T(n)clog(nα)T(n) \leq c \log(n - \alpha), for some constant α\alpha, where α1\alpha \geq 1.
    • After simplification, the expression bounds the recurrence as desired, leading to a logarithmic bound.
    • A base case check is suggested to finalize the proof.

Summary of Approaches:

  • First Attempt: Unsuccessful due to an incorrect reduction step when trying to simplify the expression involving log(n+1)\log(n+1).
  • Second Attempt: Successful, provided a constant α\alpha is subtracted, and the base case holds.

Detailed Steps to Show T(n)=O(logn)T(n) = O(\log n):

The recurrence T(n)=T(n2)+1T(n) = T\left(\left\lfloor \frac{n}{2} \right\rfloor\right) + 1 suggests that for each step, we reduce the size of nn by half. This is characteristic of logarithmic time behavior.

  1. Base Case: T(1)=1T(1) = 1, which is constant and fits within O(logn)O(\log n).
  2. Recurrence Analysis: T(n)T(n2)+1.T(n) \leq T\left(\frac{n}{2}\right) + 1. Unfolding the recurrence: T(n)=T(n2)+1=T(n4)+1+1==T(1)+log2n=log2n.T(n) = T\left(\frac{n}{2}\right) + 1 = T\left(\frac{n}{4}\right) + 1 + 1 = \dots = T(1) + \log_2 n = \log_2 n. Therefore, T(n)T(n) is bounded by O(logn)O(\log n).

Would you like further details on any specific step or concept here?

Related Questions:

  1. What is the general method for solving recurrence relations using the master theorem?
  2. How can we prove that a recurrence is O(logn)O(\log n) using substitution methods?
  3. What are some common pitfalls when trying to bound recurrences?
  4. How can we verify the correctness of a base case in recursive algorithms?
  5. Can the recurrence T(n)=T(n3)+1T(n) = T\left(\frac{n}{3}\right) + 1 also be O(logn)O(\log n)?

Tip:

When solving recurrence relations, it's important to carefully check the base case to ensure that the derived asymptotic behavior holds across all nn.

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Math Problem Analysis

Mathematical Concepts

Recurrence Relations
Logarithmic Growth
Complexity Analysis

Formulas

T(n) = T(floor(n/2)) + 1
O(log n)

Theorems

Recurrence Relation Analysis

Suitable Grade Level

University Level (Algorithm Design, Computer Science)