Loop invariants 2000 Consider the following code, assuming that i, x, y, and n are integers, with n ≥ 0. • State a non-trivial loop invariant for variable . • Prove the loop invariant. Be sure that the proof includes the final conditions after the loop has ended. • Note: a correct proof of an incorrect invariant will still receive partial credit. • Give the final value of a in terms of n. Hint: is the pattern close to (e.g. off by one) from some other pattern you might recognize? Hint: you might need to include a condition for y somewhere within your proof.

Database System Concepts
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Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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Loop invariants
Consider the following code, assuming that i, x, y, and n are integers, with n > 0.
• State a non-trivial loop invariant for variable x.
• Prove the loop invariant. Be sure that the proof includes the final conditions after the loop has ended.
Note: a correct proof of an incorrect invariant will still receive partial credit.
• Give the final value of a in terms of n.
Hint: is the pattern close to (e.g. off by one) from some other pattern you might recognize?
Hint: you might need to include a condition for y somewhere within your proof.
i = 0;
x = 1;
while in do
y = x + 2;
x=x*y;
i = 1 + 1;
After the loop, x =
(the invariant goes here)
Transcribed Image Text:Loop invariants Consider the following code, assuming that i, x, y, and n are integers, with n > 0. • State a non-trivial loop invariant for variable x. • Prove the loop invariant. Be sure that the proof includes the final conditions after the loop has ended. Note: a correct proof of an incorrect invariant will still receive partial credit. • Give the final value of a in terms of n. Hint: is the pattern close to (e.g. off by one) from some other pattern you might recognize? Hint: you might need to include a condition for y somewhere within your proof. i = 0; x = 1; while in do y = x + 2; x=x*y; i = 1 + 1; After the loop, x = (the invariant goes here)
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