Take into account the MAXMIN methodology below. It uses how many comparisons? Is it more likely to operate quicker or slower in practise than the divide-and-conquer algorithm? procedure The remark maxmin2(S) calculates the maximum and minimum of S[1..n] in max and min, respectively. 1. if n is odd then max:=S[n]; min:=S[n] 2. else max:=−∞; min:=∞ 3. for i := 1 to n/2 do 4. if S[2i − 1] ≤ S[2i] 5. then small:=S[2i − 1]; large:=S[2i] 6. else small:=S[2i]; large:=S[2i − 1] 7. if small < min then min:=small 8. if large > max then min:=small
Take into account the MAXMIN methodology below. It uses how many comparisons? Is it more likely to operate quicker or slower in practise than the divide-and-conquer algorithm? procedure The remark maxmin2(S) calculates the maximum and minimum of S[1..n] in max and min, respectively. 1. if n is odd then max:=S[n]; min:=S[n] 2. else max:=−∞; min:=∞ 3. for i := 1 to n/2 do 4. if S[2i − 1] ≤ S[2i] 5. then small:=S[2i − 1]; large:=S[2i] 6. else small:=S[2i]; large:=S[2i − 1] 7. if small < min then min:=small 8. if large > max then min:=small
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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Take into account the MAXMIN methodology below. It uses how many comparisons? Is it more likely to operate quicker or slower in practise than the divide-and-conquer
1. if n is odd then max:=S[n]; min:=S[n]
2. else max:=−∞; min:=∞
3. for i := 1 to n/2 do
4. if S[2i − 1] ≤ S[2i]
5. then small:=S[2i − 1]; large:=S[2i]
6. else small:=S[2i]; large:=S[2i − 1]
7. if small < min then min:=small
8. if large > max then min:=small
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