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Chapter 8: Problem 8

Show that the average-case time complexity of Interpolation Search is in \(\Theta(\lg (\lg n)),\) assuming the keys are uniformly distributed and that search key \(x\) is equally probable to be in cach of the array slots.

Short Answer

Expert verified
The Interpolation Search operates by estimating the position of the key and continuously adjusting the search range. In the case that the keys are uniformly distributed and equally likely to be in any array slot, the algorithm would partition the array into two equally small or large parts at each recursive step on average. Therefore, the average-case time complexity for this algorithm ends up being \(\Theta(\lg(\lg n))\). The proof requires understanding of the Interpolation Search algorithm and some calculations based on the probabilities and properties of logarithmic functions.

Step by step solution

01

Understanding Interpolation Search

Interpolation search is an algorithm for searching for a given key value within an array. It estimates the position of the key based on the first element, the last element, and the size of the array. The order of the elements in the array must be sorted for the algorithm to work. Assuming array indices start at 0 and end at \(n-1\), the interpolation formula is \(\text{Position} = \text{low} + \frac{((\text{high}-\text{low}) / (\text{array}[\text{high}] - \text{array}[\text{low}])) * (\text{key} - \text{array}[\text{low}])\). We keep adjusting low and high based on this estimated position until the element is found or the subarray has size 0.
02

Deriving Average Case Complexity

For average case, we assume that every position in the array is equally likely to be searched. Therefore, every recursive step of the algorithm divides the array into two equally likely subarrays. Since the partitioning is uniform and we discard one partition at each step, an average-case analysis reveals that the number of steps is proportional to the binary logarithm of the binary logarithm of the size of the array, that is \(n\). Hence, we conclude the average-case time complexity in this scenario is \(\Theta(\lg(\lg n))\)

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Most popular questions from this chapter

Discuss the advantages of using each of the following instead of the other: (a) A binary search tree with a balancing mechanism (b) \(A 3-2\) tree

Write an algorithm that lists all the keys in a \(3-2\) tree in their natural order. Analyze your algorithm, and show the results using order notation.

Implement the Binary Scarch, Interpolation Search, and Robust Interpolation Search algorithms on your system, and study their best-case, average-case. and worst-case performances using several problem instances.

Write an algorithm that creates a \(3-2\) tree from a list of keys. Analyze your algorithm, and show the results using order notation.

Suppose a very large sorted list is stored in external storage. Assuming that this list cannot be brought into internal memory, develop a searching algorithm that looks for a key in this list. What major factor(s) should be considered when an external search algorithm is developed? Define the major fac\(\operatorname{tor}(s),\) analyze your algorithm, and show the results using order notation.
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