Question

Show by induction that any solution to a recurrence of the form $$ T(n) \leq 2 T\left(\frac{n}{3}\right)+c \log _{3} n $$ is \(O\left(n \log _{3} n\right)\). What happens if you replace 2 with 3 ? Explain why. Would it make a difference if you used a different base for the logarithm (only an intuitive explanation is needed here)?

Short answer

The recurrence is \(O(n \log_{3} n)\). Replacing 2 with 3 changes the solution to \(O(n)\). Different log base only changes constants.

Step-by-step solution

Understand the Problem

We need to show that the recurrence relation \( T(n) \leq 2 T\left(\frac{n}{3}\right)+c \log_{3} n \) has a solution in the form of \( O(n \log_{3} n) \). Additionally, we must consider how this changes if the 2 is replaced with a 3 and the impact of using a different logarithmic base.

Basis Step for Induction

Assume the initial condition \( T(n) = O(n \log_{3} n) \) holds for a small value of \( n \), such as \( n = 1 \). In this trivial case, \( T(1) \leq c \log_{3} 1 = 0 \), which holds trivially for \( O(n \log_{3} n) \), as \( T(1) \) can be bounded by a constant.

Inductive Hypothesis

Assume that the induction hypothesis is true for \( n = k \), meaning \( T(k) \leq c k \log_{3} k \). We need to show that it holds for \( n = 3k \), i.e., \( T(3k) \leq c (3k) \log_{3} (3k) \).

Inductive Step

Using the recurrence relation, we have:\[ T(3k) \leq 2T(k) + c \log_{3}(3k) \]Substitute \( T(k) \) from the hypothesis:\[ \leq 2(c k \log_{3} k) + c (\log_{3} 3 + \log_{3} k) \]Combine terms:\[ = 2c k \log_{3} k + c \log_{3} 3 + c \log_{3} k \]This can be rearranged as:\[ = c (3k) \log_{3} k \]thus proving the statement for \( n = 3k \).

Conclusion of Induction

By the principle of mathematical induction, the solution \( T(n) = O(n \log_{3} n) \) holds for all natural numbers \( n \). Replacing 2 with 3 changes the recurrence to one with larger growth rate, \( O(n^{\log_3{3}}) = O(n) \), thus escaping \( O(n \log_{3} n) \).

Effect of Logarithm Base

The base of the logarithm (3 in this case) only affects the constant factor of the \( O \) bound and does not affect the asymptotic growth rate due to logarithmic base change rule.

Key concepts

Mathematical Induction

Mathematical induction is a powerful technique used to prove statements or formulas that are true for all natural numbers. It consists of two main parts:

• Base Case: You prove the statement for the initial value, often the smallest number you can use, such as 1 or 0. This shows that the property holds for this first element.
• Inductive Step: Assume the statement holds for some arbitrary natural number, say \( n = k \), then prove it for \( n = k+1 \). This step is crucial as it links the initial case to all subsequent ones.

Mathematical induction is like dominos; confirm it works for the first one, ensure the subsequent ones fall when the previous does, and the entire sequence follows. For example, showing that a recurrence relation holds for \( n \) involves confirming it for a base case (like \( n = 1 \)), and proving if it holds for \( n = k \), it must also hold for \( n = 3k \) in the context of this problem. This creates a chain proving the relationship for all \( n \).

This method is often used in problems involving sequences, series, or structures that are defined recursively - a key application is in asymptotic analysis, which studies how functions behave as inputs become very large.

Recurrence Relations

Recurrence relations are equations that define sequences or multidimensional arrays recursively. In simpler terms, they provide a way to compute the sequence by relating current terms to previous ones. For instance, a common form is \( T(n) = aT(n/b) + f(n) \). This form is widely recognized in computational sciences for analyzing algorithms, especially divide-and-conquer strategies.

The goal in working with recurrence relations is often to find a closed-form or asymptotic expression that describes the sequence's long-term behavior. In our specific problem, the recurrence \( T(n) \leq 2 T\left(\frac{n}{3}\right) + c \log_{3} n \) describes an algorithm whose execution time for a problem of size \( n \) is a combination of execution time for smaller subproblems and an additional logarithmic factor.

By solving such relations, we can determine the efficiency of an algorithm. Replacing 2 with 3 in our example, changes this relation to \( T(n) \leq 3 T\left(\frac{n}{3}\right) + c \log_{3} n \), greatly impacting the growth rate and implying a more significant increase as \( n \) grows, moving from logarithmic to linear complexity in the asymptotic notation.

Logarithmic Functions

Logarithms are functions that help us solve equations involving exponents by "undoing" the exponentiation process. Essential in disciplines like computer science and mathematics, these functions are especially useful for simplifying problems and expressions, particularly those involving multiplicative processes over large ranges.

The logarithmic function used in our problem, \( \log_{3} n \), relates to the power needed for the base 3 to reach \( n \). Changing the logarithmic base impacts scaling but not the functional form in asymptotic terms. According to the change of base formula, any logarithm base can be converted into another with a constant multiplicative factor, preserving order of growth for the big O notation.

• For example, \( \log_{a} b = \frac{\log_{c} b}{\log_{c} a} \).
• Different bases affect only constant factors, making them interchangeable for asymptotic analysis.

Logarithmic functions thus provide a rigorous way of comparing asymptotic behaviors, enabling us to describe complex algorithms succinctly.