Question

Complete the proof of Theorem \(8.8 .\) That is, show that a deterministic algorithm that finds the smallest and largest of \(n\) keys only by comparisons of keys must in the worst case do at least \((3 n-3) / 2\) comparisons if \(n\) is odd.

Short answer

The theorem is proven valid by demonstrating the minimum number of comparisons, \((3n-3) / 2\), required for the worst-case scenario of a deterministic algorithm finding the smallest and largest of \(n\) keys through comparison of the keys, with \(n\) being an odd number.

Step-by-step solution

Establishing the Initial Case

If we have just one key which is minimum \(n = 1\), no comparisons are needed. Thus, in this case, the algorithm obviously makes 0 comparisons which is \(\geq (3 \cdot 1 - 3) /2\). So, the theorem holds for \(n = 1\).

Consider \(n+2\) keys

Assume we have \(n+2\) keys and we want to find the smallest and largest among them. We first do \((n+2) / 2\) comparisons to pair off the keys and possibly one key left over if \(n+2\) is odd. The smaller of each pair advances to the next round.

Second Round of Comparisons

In the second round, we do \(n/2\) or \((n+1)/2\) comparisons depending on whether \(n+2\) is even or odd respectively. At this point, we have found the smallest key with at most \((3n+3)/2\) comparisons.

Finding the Largest Key

Now we want to find the largest key. This key must be one of the keys in the first round that beat some other key or a leftover key. There are a total of \(n/2 + 2\) or \((n+1)/2 +2\) keys to compare, for even and odd \(n+2\) respectively. We have hence ideally required \((3n+3)/2 = n+2 + n/2 + 2\) or \((n+3+(n+1)/2 + 2)\) comparisons.

Prove by Inductive Step

This number of comparisons is more than \((3n+3-3)/2\) comparisons for both odd and even \(n\). Hence, by the principle of mathematical induction, the theorem is correct for all odd \(n\).

Key concepts

Deterministic Algorithms

A deterministic algorithm is a type of algorithm that follows a pre-defined set of rules and procedures for its execution. This means that given a particular input, it will always produce the same output, and the path it takes to reach this output is predictable. In simpler terms, deterministic algorithms operate in a fixed manner without involving any randomness.

In the context of the given exercise, we are looking at a deterministic algorithm designed to find the smallest and largest keys in a set of numbers by merely comparing them. The algorithm systematically pairs and compares the keys, ensuring that every decision point, such as which key advances to the next round, is determined by established rules. No element of chance will influence the outcome.

This predictability is critical for mathematical proofs, like the one in the original exercise, as it allows us to accurately calculate the number of steps or comparisons the algorithm will make.

Comparative Analysis

Comparative analysis is a method used in algorithm design to evaluate different solutions or steps taken by an algorithm. This involves comparing the performance of these steps in terms of time, space, or in this case, the number of comparisons made. The goal is to identify the most efficient approach.

The exercise focuses on comparative analysis by estimating the number of comparisons needed in the worst case. Different scenarios are considered, such as whether the number of keys, `n`, is odd or even. By systematically performing comparisons, the algorithm evaluates each pair of keys, gradually narrowing down potential candidates for the minimum and maximum.

Understanding how comparisons reduce the number of possible options in finding the minimum and maximum helps in finely optimizing the algorithm. This is vital because minimizing comparisons leads to improved efficiency in terms of execution time.

Mathematical Induction

Mathematical induction is a powerful tool in proving statements that are true for all integers, generally starting from a base case and proving that if it holds for an integer `n`, then it must also pose true for `n+1`. This helps establish the correctness of a conjecture or formula.

In our exercise, induction is used to prove that the formula for the minimum number of comparisons holds for all odd values of `n`. The process starts by verifying the case when `n=1`, then assuming it’s true for some odd integer `n`, and eventually proving it holds for `n+2`, since we only need the proof to be complete when increasing by even increments.

By successfully completing these steps, we demonstrate a comprehensive proof, showing that no fewer than `(3n-3)/2` comparisons suffice, ensuring the theorem holds for all odd `n`.

Worst-case Scenario

The worst-case scenario in algorithm analysis refers to the maximum amount of time or resources required by an algorithm for any input of size `n`. This is particularly useful in understanding an algorithm’s efficiency and reliability under the most demanding conditions.

For the given problem, the worst-case scenario is calculated by finding both the smallest and largest keys among `n` keys using comparisons. The question investigates the largest number of comparisons needed, emphasizing that no matter how the keys are organized, at least `(3n-3)/2` comparisons are inevitable for an odd `n`.

Understanding the worst-case helps set upper bounds on algorithm performance, allowing developers and users to anticipate the maximum time complexity and resources required, ensuring that even under suboptimal conditions, the algorithm's performance remains within acceptable limits.