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1. Verify the validity of the following correctness statements by adding all the intermediate assertions and so producing the proof tableau. State all the mathematical facts used. All variables are of type int. ASSERT (x == x0) int sign -1; if (x >= 0) sign = 13 x = x * sign; ASSERT (abs (x) == abs(x0) && x >= 0) abs(x) refers to the absolute value of x. This is surprisingly tricky for such a small and obviously correct code fragment, but still it is very satisfying to try it out on your own, so stop here and attempt it. If on the other hand you need a few hints, then read on. Hints: The tricky part is the missing else branch, which will force you to keep strengthening the post-condition until you find something that will match what you get from the then branch. The problem with this is that sign is free floating there, so you will need to strengthen its range. sign >= 0 is sufficient to infer the negated loop condition but is not enough. Do not be afraid to further restrict the range as much as you want; if it works to the end then you are all set, and if it does not work then you can always come back and modify the range.

1. Verify the validity of the following correctness statements by adding all the intermediate assertions and so producing the proof tableau. State all the mathematical facts used. All variables are of type int. ASSERT (x == x0) int sign -1; if (x >= 0) sign = 13 x = x * sign; ASSERT (abs (x) == abs(x0) && x >= 0) abs(x) refers to the absolute value of x. This is surprisingly tricky for such a small and obviously correct code fragment, but still it is very satisfying to try it out on your own, so stop here and attempt it. If on the other hand you need a few hints, then read on. Hints: The tricky part is the missing else branch, which will force you to keep strengthening the post-condition until you find something that will match what you get from the then branch. The problem with this is that sign is free floating there, so you will need to strengthen its range. sign >= 0 is sufficient to infer the negated loop condition but is not enough. Do not be afraid to further restrict the range as much as you want; if it works to the end then you are all set, and if it does not work then you can always come back and modify the range.

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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1. Verify the validity of the following correctness statements by adding all the intermediate assertions and so producing the proof tableau. State all the mathematical facts used. All variables are of type int. ASSERT (x == x0) int sign -1; if (x >= 0) sign = 13 x = x * sign; ASSERT (abs (x) == abs(x0) && x >= 0) abs(x) refers to the absolute value of x. This is surprisingly tricky for such a small and obviously correct code fragment, but still it is very satisfying to try it out on your own, so stop here and attempt it. If on the other hand you need a few hints, then read on. Hints: The tricky part is the missing else branch, which will force you to keep strengthening the post-condition until you find something that will match what you get from the then branch. The problem with this is that sign is free floating there, so you will need to strengthen its range. sign >= 0 is sufficient to infer the negated loop condition but is not enough. Do not be afraid to further restrict the range as much as you want; if it works to the end then you are all set, and if it does not work then you can always come back and modify the range.
Transcribed Image Text:1. Verify the validity of the following correctness statements by adding all the intermediate assertions and so producing the proof tableau. State all the mathematical facts used. All variables are of type int. ASSERT (x == x0) int sign -1; if (x >= 0) sign = 13 x = x * sign; ASSERT (abs (x) == abs(x0) && x >= 0) abs(x) refers to the absolute value of x. This is surprisingly tricky for such a small and obviously correct code fragment, but still it is very satisfying to try it out on your own, so stop here and attempt it. If on the other hand you need a few hints, then read on. Hints: The tricky part is the missing else branch, which will force you to keep strengthening the post-condition until you find something that will match what you get from the then branch. The problem with this is that sign is free floating there, so you will need to strengthen its range. sign >= 0 is sufficient to infer the negated loop condition but is not enough. Do not be afraid to further restrict the range as much as you want; if it works to the end then you are all set, and if it does not work then you can always come back and modify the range.
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