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} //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
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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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
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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