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Suppose you are given recurrences of the form \(T(n)=a T(n / b)+g(n)\), with \(T(1)=d>0\) and \(g(n)>0\) for all \(n\), and \(S(n)=a S(n / b)+g(n)\), with \(S(1)=0\) (and the same \(a, b\), and \(g(n))\). Is there any difference in the big \(\Theta\) behavior of the solutions to the two recurrences? What does this say about the influence of the initial condition on the big \(\Theta\) behavior of such recurrences?

Short Answer

Expert verified
The big \( \Theta \) behavior is the same for \( T(n) \) and \( S(n) \); initial values do not impact it.

Step by step solution

01

Understand the Recurrences

We have two recurrence relations: \( T(n) = a T(n/b) + g(n) \) with \( T(1) = d > 0 \) and \( S(n) = a S(n/b) + g(n) \) with \( S(1) = 0 \). The goal is to determine if the big \( \Theta \) behavior differs between these two recurrences.
02

Apply the Master Theorem

The Master Theorem provides a way to determine the asymptotic behavior of recurrences of the form \( T(n) = aT(n/b) + g(n) \). It states that for a recurrence \( T(n) = aT(n/b) + g(n) \), if \( g(n) = \Theta(n^c) \), compare \( c \) with \( \log_b a \):- If \( c < \log_b a \), then \( T(n) = \Theta(n^{\log_b a}) \).- If \( c = \log_b a \), then \( T(n) = \Theta(n^{\log_b a} \log n) \).- If \( c > \log_b a \), then \( T(n) = \Theta(g(n)) \).
03

Assess \( T(n) \) Using Master Theorem

For \( T(n) = aT(n/b) + g(n) \), apply the Master Theorem. Regardless of \( T(1) = d \), the asymptotic behavior \( \Theta(n^{\log_b a}) \), \( \Theta(n^{\log_b a} \log n) \), or \( \Theta(g(n)) \) is determined provided the conditions on \( g(n) \) hold.
04

Assess \( S(n) \) Using Master Theorem

Similarly, apply the Master Theorem to \( S(n) = a S(n/b) + g(n) \). The analysis of \( S(n) \) relies solely on the same \( a, b, \) and \( g(n) \) as \( T(n) \), so it will also have the same asymptotic behavior as \( T(n) \).
05

Conclusion on Initial Condition Influence

For both recurrences, \( T(n) \) and \( S(n) \), the initial conditions \( T(1) = d \) and \( S(1) = 0 \) do not affect the overall big \( \Theta \) behavior. Thus, the initial values do not influence the asymptotic behavior; it is defined entirely by \( a, b, \) and \( g(n) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Recurrence Relations
Recurrence relations are equations that recursively define sequences or multi-dimensional arrays. In simpler terms, they express how each term in the sequence relates to its predecessors. In computer science, these often arise when analyzing recursive algorithms, helping us understand their efficiency. For the given exercise, our recurrence relations are defined as
  • \( T(n) = aT(n/b) + g(n) \) with an initial condition \( T(1) = d > 0 \)
  • \( S(n) = aS(n/b) + g(n) \) with an initial condition \( S(1) = 0 \)
These equations model similar recursive processes, but they begin with different starting values. Understanding these expressions is crucial because solving them can give us insight into the overall runtime and behavior of an algorithm.
Asymptotic Behavior
Asymptotic behavior is crucial when it comes to analyzing algorithms. It helps us understand how functions behave as inputs grow towards infinity, which is critical for predicting the efficiency and limitations of an algorithm. This idea is captured using big \( \Theta \) notation, which provides bounds from above and below by constant factors.
In the context of the problem, we use the Master Theorem, a powerful tool to find asymptotic behavior in recurrence relations of the form \( T(n)=aT(n/b)+g(n) \). It allows us to categorize the growth of
  • \( \Theta(n^{\log_b a}) \) if \( c < \log_b a \)
  • \( \Theta(n^{\log_b a}\log n) \) if \( c = \log_b a \)
  • \( \Theta(g(n)) \) if \( c > \log_b a \)
Where \( g(n) = \Theta(n^c) \), the value of \( c \) compared to \( \log_b a \) determines which asymptotic behavior applies. Thus, it is the function \( g(n) \) and constants \( a \) and \( b \) that set the stage for the result, revealing that the initial conditions don't influence the asymptotic behavior.
Initial Conditions
Initial conditions can seem like a starting point; however, for recurrence relations analyzed with the Master Theorem, they surprisingly don't affect the asymptotic behavior. Initial conditions are the prescribed starting values, like \( T(1) = d \, (> 0) \) and \( S(1) = 0 \), which initially define the base case in recurrences.
While they might influence the first few terms of the sequence, their effect diminishes as \( n \) grows larger due to scaling and the repetitive nature of recurrrences. The overall growth rate, captured by the big \( \Theta \) notation, remains consistent irrespective of these starting values.
The concluding analysis in our problem showed that both \( T(n) \) and \( S(n) \) have identical asymptotic behavior. This clearly demonstrates that while initial conditions formalize the gateway to recursion, they are irrelevant to the long-term growth patterns described by asymptotic analysis.

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Most popular questions from this chapter

Draw a recursion tree diagram for $$ T(n)= \begin{cases}2 T(n / 4)+n & \text { if } n \geq 2 \\ 1 & \text { if } n=1\end{cases} $$ Use it to find a big \(\Theta\) bound on the solution to the recurrence. Assume \(n\) is a power of 4 .

Use the master theorem to give 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. \(T(n)=3 T(n / 3)+n^{2}\)

This problem explores ways to prove that $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{n}}=1-\left(\frac{1}{3}\right)^{n} $$ for all positive integers \(n\). a. First, we explore how to prove the formula by contradiction. In other words, assume that there is some integer \(n\) that makes the formula false. In this case, there must be some smallest \(n\) that makes the formula false. i. Can this smallest \(n\) be \(1 ?\) ii. What do you know about $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{i}} $$ when \(i\) is a positive integer smaller than this smallest \(n\) ? iii. Is \(n-1\) a positive integer for this smallest \(n\) ? iv. What do you know about $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{n-1}} $$ for this smallest \(n\) ? v. Write the answer to part iv as an equation, add \(2 / 3^{n}\) to both sides, and simplify the right side. vi. What does the equation that results from part \(\mathrm{v}\) say about your assumption that the formula is false? vii. What can you conclude about the truth of the formula? viii. If \(p(k)\) is the statement $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{k}}=1-\left(\frac{1}{3}\right)^{k} $$ what implication did you prove in the process of deriving your contradiction? b. i. What is the base case in a proof by mathematical induction that $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{n}}=1-\left(\frac{1}{3}\right)^{n} $$ for all positive integers \(n\) ? ii. What would you assume as an inductive hypothesis? iii. What would you prove in the inductive step of a proof of this formula by induction? iv. Prove it. v. What does the principle of mathematical induction allow you to conclude? vi. If \(p(k)\) is the statement $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{k}}=1-\left(\frac{1}{3}\right)^{k} $$ what implication did you prove in the process of doing your proof by induction?

Find the error in the following "proof" that all positive integers \(n\) are equal: Let \(p(n)\) be the statement that all numbers in an \(n\)-element set of positive integers are equal. Then \(p(1)\) is true. Now assume \(p(n-1)\) is true, and let \(N\) be the set of the first \(n\) integers. Let \(N^{\prime}\) be the set of the first \(n-1\) integers, and let \(N^{\prime \prime}\) be the set of the last \(n-1\) integers. By \(p(n-1)\), all members of \(N^{\prime}\) are equal, and all members of \(N^{\prime \prime}\) are equal. Thus, the first \(n-1\) elements of \(N\) are equal and the last \(n-1\) elements of \(N\) are equal, and so all elements of \(N\) are equal. Therefore, all positive integers are equal.

Prove that the weak principle of mathematical induction implies the strong principle of mathematical induction.

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