(Weiss 7.40): The following divide-and-conquer algorithm is proposed for finding the simultaneous maximum and minimum: Base Case// If there is one item, it is the maximum and minimum If there are two items, then compare them, and in one comparison you can find the maximum and minimum. Inductive Step// Otherwise, split the input into two halves, divided as evenly as possibly (if N is odd, one of the two halves will have one more element than the other). Recursively find the maximum and minimum of each half, and then in two additional comparisons produce the maximum and minimum for the entire problem. Find a function that will take in a vector and solve the problem, producing a vector of two elements, the min and max.
(Weiss 7.40): The following divide-and-conquer algorithm is proposed for finding the simultaneous maximum and minimum: Base Case// If there is one item, it is the maximum and minimum If there are two items, then compare them, and in one comparison you can find the maximum and minimum. Inductive Step// Otherwise, split the input into two halves, divided as evenly as possibly (if N is odd, one of the two halves will have one more element than the other). Recursively find the maximum and minimum of each half, and then in two additional comparisons produce the maximum and minimum for the entire problem. Find a function that will take in a vector and solve the problem, producing a vector of two elements, the min and max.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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(Weiss 7.40): The following divide-and-conquer
Base Case//
- If there is one item, it is the maximum and minimum
- If there are two items, then compare them, and in one comparison you can find the maximum and minimum.
Inductive Step//
- Otherwise, split the input into two halves, divided as evenly as possibly (if N is odd, one of the two halves will have one more element than the other).
Recursively find the maximum and minimum of each half, and then in two additional comparisons produce the maximum and minimum for the entire problem.
Find a function that will take in a vector and solve the problem, producing a vector of two elements, the min and max.
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