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Exam 2 Help Session

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Exam 2 Help Session

Prepared by

Stephen M. Thebaut, Ph.D.

University of Florida

Software Testing and Verification

- A student writes:
I would like to request you to provide some tips on hypothesizing functions for given programs. I refer in particular to Example 2 of Lecture Notes #24 and Question 1 of the self check quiz in lesson plan for Lectures Notes #’s 24 and 25.

Although I followed the concept of synthesizing limited invariants, I found it difficult to come up with a function to represent the given program when I attempted these on my own.

- General Rule of Thumb for hypothesizing functions of compound programs:
- Work top-down, and
- Use the Axiom of Replacement

- Good example (nested if_then’s + sequencing): problem 4 of Problem Set 7
- For while loops, see examples 1 and 2 from Lecture Notes #21.

- Consider the assertion:
{n≥0}

p := 1

k := 0

while k<>n do

p := p*2

k := k+1

end_while

{p=2n}

What function, f, is computed by the while loop?

P = while k<>n do p,k := 2p,k+1

When will P terminate?

What measure would you use to prove this using the method of Well-Founded Sets?

Use the measure in one or more conditional rules describing the function.

For this case, the initial relationship between k and n determine three different loop “behaviors.” (What are they?)

- P = while k<>n do p,k := 2p,k+1
k<n p,k := p2n−k,n

k=n p,k := p,k

:= p2n−k,n

k>n undefined

Therefore,

[P] = (k≤n p,k := p2n−k,n)

Consider the assertion:

y := 0

t := x

while t<>k do

t := t–1

y := y+1

end_while

What function, f, is computed by the while loop?

P = while t<>k do t,y := t–1,y+1

t>k t,y := k,y+1*(t-k)

:= k,y+t-k

t=k t,y := t,y

:= k,y+t-k

t<k undefined

Therefore,

[P] = (t≥k t,y := k,y+t-k)

- Another student writes:
I have some questions about exam 2 for fall 07, problem No 6. And I do not know how to make up counterexample.

6. (4 pts.) It was noted in class that wp(while b do s, Q) is the weakest (while)loop invariant which guarantees termination. Is it also the case that the wp(Repeat s until b) is the weakest (Repeat_until) loop invariant which guarantees termination? Carefully justify your answer. (Hint: recall that in Problem Set 6, you were asked to prove “finalization” from the while loop ROI using the weakest pre-condition as an invariant. Does “finalization” from the Repeat_until ROI hold using the weakest pre-condition as an invariant?)

Answer: No. In general, the wp(Repeat s until b, Q) cannot be used as an invariant with the Repeat_until ROI. In particular, (wp(Repeat s until b) Лb ≠> Q in general). (Note that the ROI –- i.e., via the “initialization” antecedent {P} s {I} -- does not require “I” to hold until after s executes.

P I, {IЛ b} S {I}, (IЛb) Q

{P} while b do S {Q}

{P} S {I}, {IЛ b} S {I}, (IЛ b) Q

{P} repeat S until b {Q}

Note that for the repeat_until loop, "I" need not hold UNTIL AFTER S executes.

wp(repeat S until b, Q) = H1 V H2 V H3 V...

where:

H1 = wp(S, b ЛQ)

H2 = wp(S, ~b ЛH1)

H3 = wp(S, ~b ЛH2)

Hk = wp(S, ~b Л Hk-1)

Note that b Л (H1 V H2 V H3 V...) Q

in general.

- Suppose you wish to prove (A => B) is FALSE.
- This can be done by finding just one case for which A is true and B is false. This case is referred to as a "counter-example".
- So, to prove that the hypothesized ROI:
A, B, C

{P} while b do S {Q}

is FALSE, find one case for which A, B, and C are each true, but {P} while b do S {Q} is FALSE.

?

- How do you identify such a case? By exploiting the fallacy in the (FALSE) ROI.
- For example, what's the fallacy in the following ROI?
P I, (IЛb) Q

{P} while b do S {Q}

Answer: The two antecedents do not require that "I" holds after S executes! So, choose P, b, S, Q, and I such that the two antecedents hold, but neither I nor Q will hold after S executes when b becomes false.

?

P I, (IЛb) Q

{P} while b do S {Q}

For example, consider, for I: x=1

{x=1 Л y=-17}

while y<0 do

y := y+1

x := 2

end_while

{x=1}

?

- Suppose {P} while b do S {Q} for some P, Q, b, and S. Suppose, too, that K = wp(while b do S, Q). Circle “necessarily true” or “not necessarily true” for each of the following assertions.
b. {K Л b} S {K}true (See Lecture Notes #20.)

- In general, will loops terminate when
P wp ?

- For while loops, does {wp Л b} S {wp} ?
- Does (wp Л ¬b) Q ?

√

√

√

- Suppose {P} while b do S {Q} for some P, Q, b, and S. Suppose, too, that K = wp(while b do S, Q). Circle “necessarily true” or “not necessarily true” for each of the following assertions.
b. {K Л b} S {K}true (See Lecture Notes #20.)

e. {K Л b} repeat S until ¬b {Q} true

{K Лb}

{K Лb}

S

K (since {K Лb} S {K})

S

T

=

¬b

T

¬b

F

S

F

{Q} ?

{Q}

(since (K Л ¬b) Q)

3. Circle either “true” or “false” for each of the following assertions.

k. ({P} S {Q}) ({P} if b then S {(Q b)})

False

The assertion may seem plausible, but consider:

{z=1} y:=5 {z=1} {z=1} if x=0 then y:=5 {(z=1 x=0)} ?

2. Circle either “true” or “false” for each of the following assertions.

h. [{P Л b} S {Q}] [{P} while b do S {Q}]

False

Consider the counterexample:

{x=0} while x<5 do x:=x+1 {x=1}

- A student writes:
We've learned two ways of identifying loop invariant "I": a heuristic approach and a more systematic approach. My question is: since a systematic approach seems to be more effective, can we always use it to find I for all the problems?

- Unfortunately, no. The concept of an “invariant” as described in the context of axiomatic verification is directly related to a Rule of Inference (ROI), e.g.:
P I, {IЛ b} S {I}, (IЛb) Q

{P} while b do S {Q}

- The antecedents represent the necessary and sufficient requirementsfor I (in terms of P, b, S, and Q) in order to use the ROI to deduce {P} while b do S {Q}.
- The heuristics considered in class are motivated by these necessary and sufficient requirements, and are therefore dependent on the program’s specification (P and Q), as well as the program itself.
- In contrast, a (full) invariant as defined in Mill’s Invariant Status Theorem is a logical condition with properties:
q(X0), ( q(X)Лp(X) ) qog(X), and ( q(X)Л¬p(X) ) ( X=f(X0) )

where q(X)=( f(X)=f(X0) ).

- The function f = [while p do g],which is “characterized by q on termination,” need not be consistent with the pre- and post-condition used to specify the program by a user/designer.
- Thus, an invariant derived using the Invariant Status Theorem may or may not allow one to prove that a user/designer specified post-condition will hold on termination of a loop.
- In “reasonable” cases, however, q may be useful, at least as a starting point, in a trial-and-error process.
- Additional research is needed to fully explore this area.

- A student writes:
I still have trouble in providing counter examples...

- Consider the following assertion/ROI:
“People who wear red shirts do not smoke.”

=

Wears red shirts(X) => Does not smoke(X)

=

Wears red shirts(X)

Does not smoke(X)

- Is the assertion valid (true)?
- No. Proof by counterexample:
- This person satisfies the antecedent, but not the consequent!

Does [(P Л ¬b) Q] [{P} while b do S {Q}] ?

=

[(P Л ¬b) Q]

[{P} while b do S {Q}]

Counterexample:

{x=0} while y<>5 do x := x+1; y := y+1 {x=0 Л y=5}

?

True or False?

c. {x=5} while k <= 5 do k := k+3 {k-x≥0} strongly

e. {wp(S, Q)} x>0} x := 17; S {Q}

“I am confused about ‘undefined’ and ‘I’.

Suppose we have the program P like this:

if (x>0) x := 9 end_if

Is [P] = (x>0 -> x := 9|true ->I) or

[P] = (x>0 -> x := 9|true ->undefined)?

Exam 2 Help Session

Prepared by

Stephen M. Thebaut, Ph.D.

University of Florida

Software Testing and Verification