Exercise 3: (Time efficiency analysis and algorithm improvement) Consider the following algorithm for finding the distance between the two closest elements in an array of numbers. Algorithm MinDistance (A[0..n-1]) //Input: Array A[0..n-1] of numbers //Output: Minimum distance between two of its elements dmin x for i 0 to n - 1 do for j 0 to n - 1 do if i ‡ j and |A[i] – A[j]| < dmin dmin A[i] A[j]| return dmin - (a) How many comparisons does the above algorithm has? Answer: (b) Make as many improvements as you can in this algorithm to improve the time efficiency. If you need to, you can change the whole algorithm; if not, improve the implementation given. Then, explain why your algorithm is more efficient than the above algorithm.

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
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Exercise 3: (Time efficiency analysis and algorithm improvement)
Consider the following algorithm for finding the distance between the two
closest elements in an array of numbers.
Algorithm MinDistance (A[0..n - 1])
//Input: Array A[0..n-1] of numbers
//Output: Minimum distance between two of its elements
dmin – x
for i0 to n - 1 do
for j - 0 to n - 1 do
if i ‡ j and |A[i] – A[j]| < dmin
dmin A[i] — A[j]|
return dmin
-
(a) How many comparisons does the above algorithm has?
Answer:
(b) Make as many improvements as you can in this algorithm to improve the time
efficiency. If you need to, you can change the whole algorithm; if not,
improve the implementation given. Then, explain why your algorithm is more
efficient than the above algorithm.
Transcribed Image Text:Exercise 3: (Time efficiency analysis and algorithm improvement) Consider the following algorithm for finding the distance between the two closest elements in an array of numbers. Algorithm MinDistance (A[0..n - 1]) //Input: Array A[0..n-1] of numbers //Output: Minimum distance between two of its elements dmin – x for i0 to n - 1 do for j - 0 to n - 1 do if i ‡ j and |A[i] – A[j]| < dmin dmin A[i] — A[j]| return dmin - (a) How many comparisons does the above algorithm has? Answer: (b) Make as many improvements as you can in this algorithm to improve the time efficiency. If you need to, you can change the whole algorithm; if not, improve the implementation given. Then, explain why your algorithm is more efficient than the above algorithm.
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