Apply algorithm 4.3.2 in P.155 with n=6 and A=(3,5,4,1,3,2). Draw only the right half of the corresponding walkthrough as shown in P.155, showing only the list A at the end of the outer “for” loop.
Apply algorithm 4.3.2 in P.155 with n=6 and A=(3,5,4,1,3,2). Draw only the right half of the corresponding walkthrough as shown in P.155, showing only the list A at the end of the outer “for” loop.
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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![We can sort the A-values by finding the minimum of all of them and
interchanging it with A[1], then finding the minimum of all the A-values from
A[2] to A[n] of them and interchanging it with A[2], and so on.
Algorithm 4.3.2: MinSort
Begin
For k + 1 To (n-1) Do
min – A[k];
// the kh pass selects a correct value for A[k]
// min will be the smallest in A[k]..A[n]
// & min occurs at entry A[index]
index + k;
For j + k+1 To n Do
If (A[j] < min) Then
min + A[j];
index + j;
// the if
// the for-j loop
A[index] E A[k];
End;
End;
A[k] + min;
End;
// the for-k loop
End.
// If A[1], A[2], ... A[n – 1] have been selected correctly, what about A[n]?
// Can we remove the variable min and just use A[index]?
Walkthrough with n = 5 and A = (3.1, 5.7, 4.3, 1.9, 3.1)
//
Al] %3 3.1, А2] — 5.7, А[3] — 4.3, A[4] —D 1.9, A|5] — 3.1
k
min
index
j
A[j]
A
3.1
2
5.7
3.1
5.7
4.3
1.9
3.1
3
4.3
4
1.9
1.9
4
3.1
1.9
5.7
4.3
3.1
3.1
5.7
3
4.3
4.3
3
4
3.1
3.1
4
3.1
1.9
3.1
4.3
5.7
3.1
3
4.3
4
5.7
3.1
3.1
1.9
3.1
3.1
5.7
4.3
4
5.7
4.3
4.3
5
1.9
3.1
3.1
4.3
5.7
We can prove that
"A[1] <= A[2] <= ... <= A[k] <= A[k + 1], A[k+2], ..., A[n]"
is a loop invariant of the for-k loop.
That will confirm that MinSort is correct.
// using MI on k
// But is it efficient?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f7775f3-edad-4f27-ba1b-f7886a7505bd%2F9051f5af-1cc1-4d4b-b4a9-32a59b5270cf%2Focxk6wg_processed.png&w=3840&q=75)
Transcribed Image Text:We can sort the A-values by finding the minimum of all of them and
interchanging it with A[1], then finding the minimum of all the A-values from
A[2] to A[n] of them and interchanging it with A[2], and so on.
Algorithm 4.3.2: MinSort
Begin
For k + 1 To (n-1) Do
min – A[k];
// the kh pass selects a correct value for A[k]
// min will be the smallest in A[k]..A[n]
// & min occurs at entry A[index]
index + k;
For j + k+1 To n Do
If (A[j] < min) Then
min + A[j];
index + j;
// the if
// the for-j loop
A[index] E A[k];
End;
End;
A[k] + min;
End;
// the for-k loop
End.
// If A[1], A[2], ... A[n – 1] have been selected correctly, what about A[n]?
// Can we remove the variable min and just use A[index]?
Walkthrough with n = 5 and A = (3.1, 5.7, 4.3, 1.9, 3.1)
//
Al] %3 3.1, А2] — 5.7, А[3] — 4.3, A[4] —D 1.9, A|5] — 3.1
k
min
index
j
A[j]
A
3.1
2
5.7
3.1
5.7
4.3
1.9
3.1
3
4.3
4
1.9
1.9
4
3.1
1.9
5.7
4.3
3.1
3.1
5.7
3
4.3
4.3
3
4
3.1
3.1
4
3.1
1.9
3.1
4.3
5.7
3.1
3
4.3
4
5.7
3.1
3.1
1.9
3.1
3.1
5.7
4.3
4
5.7
4.3
4.3
5
1.9
3.1
3.1
4.3
5.7
We can prove that
"A[1] <= A[2] <= ... <= A[k] <= A[k + 1], A[k+2], ..., A[n]"
is a loop invariant of the for-k loop.
That will confirm that MinSort is correct.
// using MI on k
// But is it efficient?
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