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Draw the tree structure that describes binary search on a list with 16 elements. What is the number of comparisons in the worst case?

Short Answer

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The worst-case number of comparisons is 5.

Step by step solution

01

Understand Binary Search

Binary search is a technique for finding a specific element in a sorted list by repeatedly dividing the search interval in half. Starting with the whole list, compare the target value to the middle element of the list. If the target value is equal to the middle element, the search is complete. If the target value is smaller than the middle element, continue the search with the left half of the list. If the target value is larger, continue with the right half.
02

Determine the Tree Depth

For a list with 16 elements, consider a full binary search process. Each level of the tree represents a new division of the search process. In binary search, the number of levels in the tree is equal to the logarithm base 2 of the number of elements, rounded up. So, for 16 elements, we calculate: \[ \log_2 (16) = 4 \]So, the tree will have 4 levels.
03

Draw and Analyze the Binary Tree

Create the binary tree by repeatedly splitting the list into halves: - Level 0: Start with the whole list (16 elements). - Level 1: Split into two parts (8 elements each). - Level 2: Each part from Level 1 is split again (4 elements each). - Level 3: Split each part from Level 2 (2 elements each). - Level 4: Each element from Level 3 split provides 1 element each. This tree will have a total of 31 nodes, including intermediates that guide towards a potential match.
04

Calculate Maximum Comparisons

In the worst-case scenario, at each level of the binary search tree, a comparison is made. The number of comparisons corresponds to the height of the tree, which is 4 levels in binary search for a list of 16 elements. Therefore, in the worst case, the number of comparisons is 4 + 1 (to determine the middle at each level), totaling 5 comparisons.

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Key Concepts

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

Binary Tree
A binary tree is a structure that organizes data hierarchically. Each node in this structure has at most two children, often referred to as the left child and the right child. This makes it a flexible method for organizing data and is widely used in computer science.
A special type of binary tree used in search algorithms is a Binary Search Tree (BST). In BSTs, each node follows the rule that any left child contains a value less than or equal to its parent node and any right child contains a value greater than its parent node. This property, combined with binary trees' hierarchical nature, allows efficient search operations.
In constructing a binary tree, especially for binary search operations, the tree becomes a powerful tool for organizing data levels such that searching, inserting, and deleting become efficient. Understanding the concept of a binary tree is crucial for mastering many search algorithms, including binary search.
Search Algorithm
Search algorithms are fundamental techniques in computer science that allow us to find a specific value or piece of data within a dataset. One of the most efficient search algorithms is the binary search algorithm.
The binary search algorithm finds a target in a sorted list by repeatedly dividing the search interval into halves. Each step involves:
  • Comparing the target value to the middle element of the list.
  • If they match, the search ends successfully.
  • If the target is smaller, consider only the left half for the next round.
  • If greater, focus on the right half instead.
This methodology, rooted in a divide-and-conquer approach, significantly reduces the number of required comparisons and makes binary search much faster than simple linear searches, especially for large datasets.
Algorithm Complexity
Algorithm complexity provides a way to analyze and compare the efficiency of algorithms, particularly in terms of time and space. In the context of binary search, the time complexity is of significant interest.
Binary search has a time complexity of \( O(\log n) \), where \( n \) is the number of elements in the dataset. This logarithmic complexity arises because each search step effectively halves the dataset size, allowing the search process to quickly hone in on the target value.
Such efficiency makes binary search exceptionally suited for large, sorted datasets where quick look-ups are essential. When calculating complexity, understanding the number of operations, like comparisons, gives insight into how quickly and effectively an algorithm performs.
Comparisons in Algorithms
In algorithms, comparisons are operations where two values are evaluated to determine their relationship. Understanding how many comparisons an algorithm requires can help gauge its efficiency.
For binary search, the worst-case scenario yields a number of comparisons equal to the height of the binary search tree created from dividing the list. For our specific example with 16 elements, the maximum number of comparisons needed would be 5. This includes checking each middle element of the binary tree until the target value is found or the search list is reduced too far.
Comparison efficiency is crucial because it directly impacts how quickly an algorithm can perform its task. In efficient algorithms like binary search, comparison count is minimized, which is a key factor in its speed and suitability for large datasets.

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Most popular questions from this chapter

a. Use Gauss's approach to find the following sum: $$ 2+4+6+\ldots+100 $$ b. Use Gauss's approach to find a formula for the sum of the even numbers from 2 to \(2 n\) : $$ 2+4+6+\ldots+2 n $$ Your formula will be an expression involving \(n\).

A tennis tournarnsnt has 342 players. A singlo match imolves 2 players. The winner of a match will play the winner of a match in the next round, wheress losers are eliminated from the toumament. The 2 players who have won all previous rounds play in the final game, and the wirner wins the tournament. What is the total number of matches needed to determine the winner? a. Here is one algorithm to answer this question. Compute \(342 / 2=171\) to get the number of pairs (matches) in the first round, which results in 171 winners to go on to the second round. Compute \(171 / 2=85\) with 1 left over, which results in 85 matches in the second round and 85 winners, plus the 1 left over, to go on to the third round. For the third round compute \(86 / 2=43,50\) the third round has 43 matches, and so on. The total number of matches is \(171+85+43+\ldots\) Finish this process to find the total number of matches. b. Here is another algorithm to solve this problem, Each match results in exactly one loser, so there must be the same number of matches as losers in the tournament. Compute the total number of losers in the entire tournament. (Hint: This isf't really a computation; it is a one-sentence argument.) c. What is your opinion on the relative clarity, elegance, and efficiency of the two algorithms?

Suppose selection sort and bubble sort are both performed on a list that is already sorted. Does bubble sort do fewer exchanges than seloction sort? Explain.

Bubble sort can be improved. Smart bubble sort keeps track of how many exchanges are done within any single poss through the unsorted section of the list. If no ewchanges occur, then the list is sorted and the algorithm should stop. a. Write a pseudocode version of the smart bubble sort algorithm. b. Perform a smart bubble sort on the following list. How many comparisons are required? \(7,4,12,9,11\) c. Describe the best-case scenario for smart bubble sort on an n-element list. How many comparisons are required? How many exchanges are required? d. Under what circumstances does smart bubble sort do the same number of comparisons as regular bubble sort?

If a list is already sorted in ascending order, a modified sequential search algorithm can be used that compares against each element in turn, stopping if a list element exceeds the target value. Write a pseudocode version of this short sequential search algorithm.

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