} //Postcondition: (index =1) ^ (2V < x) ^ (x<2v+1) Suppose we have already proved the first two parts of the postcondition, that is, (index 1) and (2Y < x). Now we are trying to prove the last part (x < 2Y+1). If (x < 2Y+1 index) is a loop invariant, can it be used to prove (x < 2Y*1)? Explain why or why not. Yes. Because the statement (x < 2Y+1 index) is true before and after loop execution. Yes. Because it is given that (r<2"+1\index) is an invariant, and (r<2+1*index) is equivalent to (r<2"+1). ONo. Since index = 1, the two sides of the proof %3D would be the same which means there is nothing to prove. O Yes. Because when the loop terminates we have (index=1), so x < 2Y+1index = 2y+1 .
} //Postcondition: (index =1) ^ (2V < x) ^ (x<2v+1) Suppose we have already proved the first two parts of the postcondition, that is, (index 1) and (2Y < x). Now we are trying to prove the last part (x < 2Y+1). If (x < 2Y+1 index) is a loop invariant, can it be used to prove (x < 2Y*1)? Explain why or why not. Yes. Because the statement (x < 2Y+1 index) is true before and after loop execution. Yes. Because it is given that (r<2"+1\index) is an invariant, and (r<2+1*index) is equivalent to (r<2"+1). ONo. Since index = 1, the two sides of the proof %3D would be the same which means there is nothing to prove. O Yes. Because when the loop terminates we have (index=1), so x < 2Y+1index = 2y+1 .
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
Related questions
Question
![Consider the following loop:
//Precondition: (index=x) ^ (x 2 1) ^ (y=0)
while (index >1)
{
index = index/2;
++y;
}
//Postcondition: (index =1) ^ (2° < x) ^ (x<2♥+1)
Suppose we have already proved the first
two parts of the postcondition, that is, (index
= 1) and (2Y < x). Now we are trying to prove
%D
the last part (x < 2y+1).
If (x < 2Y+1 index) is a loop invariant, can it
be used to prove (x < 2Y+1)? Explain why or
why not.
Yes. Because the statement (x < 2y+1 index) is
true before and after loop execution.
O Yes. Because it is given that (r<2+1*index) is
an invariant, and (x<2+1xindex) is equivalent
to (r<2+1).
O No. Since index = 1, the two sides of the proof
would be the same which means there is
nothing to prove.
Yes. Because when the loop terminates we
have (index=1), so
x < 2y+1index = 2y+1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F888f77e8-cf9b-4b32-b8a2-bb6248642d2d%2Fc259b1c8-46fc-47bc-b8aa-d621b19e666d%2Fn9kzqz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the following loop:
//Precondition: (index=x) ^ (x 2 1) ^ (y=0)
while (index >1)
{
index = index/2;
++y;
}
//Postcondition: (index =1) ^ (2° < x) ^ (x<2♥+1)
Suppose we have already proved the first
two parts of the postcondition, that is, (index
= 1) and (2Y < x). Now we are trying to prove
%D
the last part (x < 2y+1).
If (x < 2Y+1 index) is a loop invariant, can it
be used to prove (x < 2Y+1)? Explain why or
why not.
Yes. Because the statement (x < 2y+1 index) is
true before and after loop execution.
O Yes. Because it is given that (r<2+1*index) is
an invariant, and (x<2+1xindex) is equivalent
to (r<2+1).
O No. Since index = 1, the two sides of the proof
would be the same which means there is
nothing to prove.
Yes. Because when the loop terminates we
have (index=1), so
x < 2y+1index = 2y+1
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