Suppose that, even unrealistically, we are to search a list of 700 million items using Binary Search, Recursion (Algorithm 2.1). What is the maximum number of comparisons that this algorithm must perform before finding a given item or concluding that it is not in the list?

Short Answer

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The maximum number of comparisons that the recursive binary search algorithm must perform before finding a given item or concluding that it's not in the given list of 700 million items is 30.

Step by step solution

01

Understanding Binary Search

Binary search works by repeatedly dividing the search space in half. If the search key is equal to the middle item, the search ends. If the search key is less than the middle item, the algorithm repeats the process on the left sublist; if the search key is greater, the algorithm repeats on the right sublist.
02

Applying Logarithm Formula

Since each step of the binary search algorithm splits the search space in half, the problem of finding the maximum comparisons is equivalent to finding how many times the list of 700 million items can be split in half until only one item (or less) remains. This can be computed by taking the base 2 logarithm of the total number of items. Thus, the maximum number of comparisons is given by the expression \(\lceil \log_2(700000000) \rceil\). The ceiling function \(\lceil a \rceil\) is used, as the number of comparisons cannot be a partial value.
03

Computing the Logarithm Value

Use a calculator to find the logarithm value. The base 2 logarithm of 700 million is approximately 29.897352853986263. We round up this number to the nearest whole number using the ceiling function, because the number of operations, in this case, comparisons, must be an integer.

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Most popular questions from this chapter

Write a divide-and-conquer algorithm for the Towers of Hanoi problem. The Towers of Hanoi problem consists of three pegs and \(n\) disks of different sizes. The object is to move the disks that are stacked, in decreasing order of their size, on one of the three pegs to a new peg using the third one as a temporary peg. The problem should be solved according to the following rules: (1) when a disk is moved, it must be placed on one of the three pegs; (2) only one disk may be moved at a time, and it must be the top disk on one of the pegs; and (3) a larger disk may never be placed on top of a smaller disk. a. Show for your algorithm that \(S(n)=2^{n}-1\). (Here \(S(n)\) denotes the number of steps (moves), given an input of \(n\) disks.) b. Prove that any other algorithm takes at least as many moves as given in part (a).

Write an algorithm that searches a sorted list of \(n\) items by dividing it into three sublists of almost \(n / 3\) items. This algorithm finds the sublist that might contain the given item and divides it into three smaller sublists of almost equal size. The algorithm repeats this process until it finds the item or concludes that the item is not in the list. Analyze your algorithm and give the results using order notation.

When a divide-and-conquer algorithm divides an instance of size \(n\) of a problem into subinstances each of size \(n / c,\) the recurrence relation is typically given by \\[\begin{array}{l}T(n)=a T\left(\frac{n}{c}\right)+g(n) \quad \text { for } n>1 \\ T(1)=d\end{array}\\] where \(g(n)\) is the cost of the dividing and combining processes, and \(d\) is a constant. Let \(n=c^{k}\) a. Show that \\[T\left(c^{k}\right)=d \times a^{k}+\sum_{j=1}^{k}\left[a^{k-j} \times g\left(c^{j}\right)\right]\\] b. Solve the recurrence relation given that \(g(n) \in \Theta(n)\)

Suppose that, in a divide-and-conquer algorithm, we always divide an instance of size \(n\) of a problem into \(n\) subinstances of size \(n / 3\), and the dividing and combining steps take linear time. Write a recurrence equation for the running time \(T(n),\) and solve this recurrence equation for \(T(n)\) Show your solution in order notation.

Suppose that, in a divide-and-conquer algorithm, we always divide an instance of size \(n\) of a problem into 10 subinstances of size \(n / 3\), and the dividing and combining steps take a time in \(\Theta\left(n^{2}\right)\). Write a recurrence equation for the running time \(T(n),\) and solve the equation for \(T(n)\).

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