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Model Checking

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Model Checking

Doron A. Peled

University of Warwick

Coventry, UK

doron@dcs.warwick.ac.uk

- How to specify what the software is supposed to do?
- How to model it in a way that allows us to check it?

- Perform some computational task.
- Have some initial condition, e.g.,0in A[i] integer.
- Have some final assertion, e.g.,0in-1 A[i]A[i+1].(What is the problem with this spec?)
- Are supposed to terminate.

Involve several computation agents.

Termination may indicate an abnormal event (interrupt, strike).

May exploit diverse computational power.

May involve remote components.

May interact with users (Reactive).

May involve hardware components (Embedded).

- Representing concurrency:- Allow one transition at a time, or- Allow coinciding transitions.
- Granularity of transitions.
- Assignments and checks?
- Application of methods?

- Global (all the system) or local (one thread at a time) states.

- V={v0,v1,v2, …} - set of variables.
- p(v0, v1, …, vn) - a parametrized assertion, e.g., v0=v1+v2 /\ v3>v4.
- A state is an assignment of values to the program variables. For example: s=<v0=1,v2=3,v3=7,…,v18=2>
- For predicate (first order assertion) p:p (s ) is p under the assignment s.Example: p is x>y /\ y>z. s=<x=4, y=3, z=5>.Then we have 4>3 /\ 3>5, which is false.

- The state space of a program is the set of all possible states for it.
- For example, if V={a, b, c} and the variables are over the naturals, then the state space includes: <a=0,b=0,c=0>,<a=1,b=0,c=0>,<a=1,b=1,c=0>,<a=932,b=5609,c=6658>…

- Each atomic transition represents a small piece of code such that no smaller piece of code is observable.
- Is a:=a+1 atomic?
- In some systems, e.g., when a is a register and the transition is executed using an inccommand.

Execute the following when x=0 in two concurrent processes:

P1:a=a+1

P2:a=a+1

Result: a=2.

Is this always the case?

Consider the actual translation:

P1:load R1,a

inc R1

store R1,a

P2:load R2,a

inc R2

store R2,a

a may be also 1.

P1:load R1,a

inc R1

store R1,a

P2:load R2,a

inc R2

store R2,a

a=0

R1=0

R2=0

R1=1

R2=1

a=1

a=1

- Each transition has two parts:
- The enabling condition: a predicate.
- The transformation: a multiple assignment.

- For example:a>b (c,d):=(d,c)This transition can be executed in states where a>b. The result of executing it isswitching the value of c with d.

- A predicate I.
- The program can start from states s such that I (s ) holds.
- For example:I (s )=a>b /\ b>c.

- A (finite) set of variables V over some domain.
- A set of states S.
- A (finite) set of transitions T, each transition et has
- an enabling condition e, and
- a transformation t.

- An initial condition I.

- V={a, b, c, d, e}.
- S: all assignments of natural numbers for variables in V.
- T={c>0(c,e):=(c-1,e+1), d>0(d,e):=(d-1,e+1)}
- I : c=a /\ d=b /\ e=0
- What does this transition system do?

- An execution is a finite or infinite sequence of states s0, s1, s2, …
- The initial state satisfies the initial condition, I.e., I (s0).
- Moving from one state si to si+1 is by executing a transition et:
- E (si ), I.e., sisatisfies e.
- si+1 is obtained by applying t to si.

- s0=<a=2, b=1, c=2, d=1, e=0>
- s1=<a=2, b=1, c=1, d=1, e=1>
- s2=<a=2, b=1, c=1, d=0, e=2>
- s3=<a=2, b=1 ,c=0, d=0, e=3>

T={c>0(c,e):=(c-1,e+1), d>0(d,e):=(d-1,e+1)}

I: c=a /\ d=b /\ e=0

L0:While True do

NC0:wait(Turn=0);

CR0:Turn=1

endwhile ||

L1:While True do

NC1:wait(Turn=1);

CR1:Turn=0

endwhile

T0:PC0=L0PC0:=NC0

T1:PC0=NC0/\Turn=0

PC0:=CR0

T2:PC0=CR0

(PC0,Turn):=(L0,1)

T3:PC1=L1PC1=NC1

T4:PC1=NC1/\Turn=1

PC1:=CR1

T5:PC1=CR1

(PC1,Turn):=(L1,0)

The transitions

Initially: PC0=L0/\PC1=L1

Turn=1

L0,L1

Turn=0

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

- Executions : the set of maximal paths (finite or terminating in a node where nothing is enabled).
- Nondeterministic choice : when more than a single transition is enabled at a given state. We have a nondeterministic choice when at least one node at the state graph has more than one successor.

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

Turn=0

L0,L1

Turn=0

L0,NC1

Turn=1

L0,NC1

Turn=0

NC0,NC1

Turn=1

L0,CR1

Turn=0

CR0,NC1

Need to check the property

for every possible interleaving!

Turn=0

L0,L1

Turn=0

L0,L1

Turn=0

L0,NC1

Turn=0

L0,NC1

Turn=1

L0,NC1

Turn=0

NC0,NC1

Turn=1

L0,CR1

Turn=0

CR0,NC1

L0:While True do

NC0:wait(Turn=0);

CR0:Turn=1

endwhile ||

L1:While True do

NC1:wait(Turn=1);

CR1:Turn=0

endwhile

T0:PC0=L0PC0:=NC0

T1:PC0=NC0/\Turn=0PC0:=CR0

T1’:PC0=NC0/\Turn=1PC0:=NC0

T2:PC0=CR0(PC0,Turn):=(L0,1)

T3:PC1==L1PC1=NC1

T4:PC1=NC1/\Turn=1PC1:=CR1

T4’:PC1=NC1/\Turn=0PC1:=N1

T5:PC1=CR1(PC1,Turn):=(L1,0)

Busy waiting

Initially: PC0=L0/\PC1=L1

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

Now it does not hold!

(Red subgraph generates a counterexample execution.)

- The model is a graph.
- The specification should refer the the graph representation.
- Apply graph theory algorithms.

- Invariants: a property that need to hold in each state.
- Deadlock detection: can we reach a state where the program is blocked?
- Dead code: does the program have parts that are never executed.

- Apply a search strategy (Depth first search, Breadth first search).
- Check states/transitions during the search.
- If property does not hold, report counter example!

- Model checking works only for finite state systems. Would not work with
- Unconstrained integers.
- Unbounded message queues.
- General data structures:
- queues
- trees
- stacks

- parametric algorithms and systems.

- Need to represent the state space of a program in the computer memory.
- Each state can be as big as the entire memory!
- Many states:
- Each integer variable has 2^32 possibilities. Two such variables have 2^64 possibilities.
- In concurrent protocols, the number of states usually grows exponentially with the number of processes.

- Many protocols are finite state.
- Many programs or procedure are finite state in nature. Can use abstraction techniques.
- Sometimes it is possible to decompose a program, and prove part of it by model checking and part by theorem proving.
- Many techniques to reduce the state space explosion.

Program DFS

For each s such that Init(s)

dfs(s)

end DFS

Procedure dfs(s)

for each s’ such that R(s,s’) do

If new(s’) then dfs(s’)

end dfs.

- Invariants: check that all reachable statessatisfy the invariant property. If not, showa path from an initial state to a bad state.
- Deadlocks: check whether a state where noprocess can continue is reached.
- Dead code: as you progress with the DFS, mark all the transitions that are executed at least once.

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

- Want to check more properties.
- Want to have a unique algorithm to deal with all kinds of properties.
- This is done by writing specification in more complicated formalisms.
- We will see that in the next lecture.

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

init

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

init

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,L1

Turn=1

L0,L1

- Add an additional initial node.
- Propositions are attached to incoming nodes.
- All nodes are accepting.

- We want to find a correctness condition for a model to satisfy a specification.
- Language of a model: L(Model)
- Language of a specification: L(Spec).
- We need: L(Model) L(Spec).

Sequencessatisfying Spec

Program executions

All sequences

- Show that L(Model) L(Spec).
- Equivalently: ______Show that L(Model) L(Spec) = Ø.
- Also: can obtain L(Spec) by translating from LTL!

- How to intersect two automata?
- How to complement an automaton?
- How to translate from LTL to an automaton?

Specification Formalisms

- Formal. Unique interpretation.
- Intuitive. Simple to understand (visual).
- Succinct. Spec. of reasonable size.
- Effective.
- Check that there are no contradictions.
- Check that the spec. is implementable.
- Check that the implementation satisfies spec.

- Expressive.
- May be used to generate initial code.
Specifying the implementation or its properties?

- A (finite) set of variables V.
- A set of states .
- A (finite) set of transitions T, each transition e==>t has
- an enabling condition e and a transformation t.

- An initial condition I.
- Denote by R(s, s’) the fact that s’ is a successor of s.

- An execution is a finite or infinite sequence of states s0, s1, s2, …
- The initial state satisfies the initial condition, I.e., I (s0).
- Moving from one state si to si+1 is by executing a transition e==>t:
- e(si), I.e., si satisfies e.
- si+1 is obtained by applying t to si.

- Lets assume all sequences are infinite by extending finite ones by “stuttering” the last state.

- Dynamic, speaks about several “worlds” and the relation between them.
- Our “worlds” are the states in an execution.
- There is a linear relation between them, each two sequences in our execution are ordered.
- Interpretation: over an execution, later over all executions.

::= () | ¬ | /\ \/ U |O | p

“box”, “always”, “forever”

“diamond”, “eventually”, “sometimes”

O “nexttime”

U“until”

Propositions p, q, r, … Each represents some state property (x>y+1, z=t, at-CR, etc.)

O

U

- []<>p “p will happen infinitely often”
- <>[]p “p will happen from some point forever”.
- ([]<>p) --> ([]<>q) “If p happens infinitely often, then q also happens infinitely often”.

b

a

a

a

b

b

a

b

a

b

b

- [](a/\b)=([]a)/\([]b)
- But <>(a/\b)(<>a)/\(<>b)
- <>(a\/b)=(<>a)\/(<>b)
- But [](a\/b)([]a)\/([]b)

- ([]<>A)/\([]<>B)=[]<>(A/\B)?
- ([]<>A)\/([]<>B)=[]<>(A\/B)?
- (<>[]A)/\(<>[]B)=<>[](A/\B)?
- (<>[]A)\/(<>[]B)=<>[](A\/B)?

No, just <--

Yes!!!

Yes!!!

No, just -->

- Instead of <>p, write true U p.
- Instead of []p, we can write ¬<>¬p,or ¬(true U ¬p).Because []p=¬¬[]p.¬[]p means it is not true that p holds forever, or at some point ¬p holds or <>¬p.

- Let be a sequence s0 s1 s2 …
- Let i be a suffix of : si si+1 si+2 … (0 = )
- i |= p, where p a proposition, if si|=p.
- i |= /\ if i |= and i |= .
- i |= \/ if i |= or i |= .
- i |= <> if for some ji, j |= .
- i |= [] if for each ji, j |= .
- i |= U if for some ji, j|=. and for each ik<j, k |=.

- s|=p as in propositional logic.
- |= is interpreted over a sequence, as in previous slide.
- P|=holds if |= for every sequence of P.

release

s1

s2

s3

pull

release

extended

extended

malfunction

r0 = s1 s2 s1 s2 s1 s2 s1 …

r1 = s1 s2 s3 s3 s3 s3 s3 …

r2 = s1 s2 s1 s2 s3 s3 s3 …

…

release

s1

s2

s3

pull

release

extended

extended

r2 = s1 s2 s1 s2 s3 s3 s3 …

malfunction

r2 |= extended ??

r2 |= O extended ??

r2 |= O O extended ??

r2 |= <> extended ??

r2 |= [] extended ??

r2 |= <>[] extended ??

r2 |= ¬ <>[] extended ??

r2 |= (¬extended) U malfunction ??

r2 |= [](¬extended->O extended) ??

release

s1

s2

s3

pull

release

extended

extended

malfunction

P |= extended ??

P |= O extended ??

P |= O O extended ??

P |= <> extended ??

P|= [] extended ??

P |= <>[] extended ??

P |= ¬ <>[] extended ??

P |= (¬extended) U malfunction ??

P |= [](¬extended->O extended) ??

Turn=1

L0,L1

Turn=0

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

Turn=0

L0,L1

Turn=0

L0,NC1

Turn=1

L0,NC1

Turn=0

NC0,NC1

Turn=1

L0,CR1

Turn=0

CR0,NC1

Need to check the property

for every possible interleaving!

- [](PC0=NC0 <> PC0=CR0)
- [](PC0=NC0 U Turn=0)
- Try at home:- The processes alternate in entering their critical sections.- Each process enters its critical section infinitely often.

¬<>p<-->[]¬p

[](pq)([]p[]q)

[]p(p/\O[]p)

O¬p<-->¬Op

[](pOp)(p[]p)

(pUq)<-->(q\/(p/\O(pUq)))

(pUq)<>q

+ propositional logic axiomatization.

+ axiom:p []p

Green --> Yellow --> Red --> Green

Always has exactly one light:

[](¬(gr/\ye)/\¬(ye/\re)/\¬(re/\gr)/\(gr\/ye\/re))

Correct change of color:

[]((grUye)\/(yeUre)\/(reUgr))

Needed only when we

can start with yellow

Green-->Yellow-->Red-->Yellow-->Green

First attempt:

[](((gr\/re) U ye)\/(ye U (gr\/re)))

Correct specification:

[]( (gr-->(grU (ye /\ ( yeUre ))))

/\(re-->(reU (ye /\ ( yeUgr ))))

/\(ye-->(yeU (gr \/ re))))

- init-when the program starts and satisfies the initial condition.
- finish-when the program terminates and nothing is enabled.
- Partial correctness: init/\[](finish)
- Termination: init/\<>finish
- Total correctness:init/\<>(finish/\ )
- Invariant:init/\[]

a

S0

b

S1

a

b

- A=<, S, , I, F>
- (finite) the alphabet, S (finite) the states.
- : S x x S is the transition relation
- I : S are the starting states
- F: S are the accepting states.

a

S0

b

S1

a

b

- (S0, a, S0)
- (S0, b, S1)
- (S1, a, S0)
- (S1, b, S1)

a

S0

b

S1

a

b

- A word over , e.g., abaab.
- A sequence of states, e.g. S0 S0 S1 S0 S0 S1.
- Starts with an initial state.
- Accepting if ends at accepting state.

a

S0

b

S1

a

b

- The words that are accepted by the automaton.
- Includes aabbba, abbbba.
- Does not include abab, abbb.
- What is the language?

S0

a

S1

A,b

a

- Transitions: (S0,a,S0), (S0,b,S0), (S0,a,S1),(S1,a,S1).
- What is the language of this automaton?

S0

a

S1

a,b

a

a

S0

a

S1

b

b

a

S0

b

S1

a

b

- Similar definition.
- Runs on infinite words over .
- Accepts when an accepting state occurs infinitely often in a run.

A

S0

B

S1

A

B

- Consider the word a b a b a b a b …
- There is a run S0 S0 S1 S0 S1 S0 S1 …
- This run in accepting, since S0 appears infinitely many times.

a

S0

b

S1

a

b

- For the word b b b b b … the run is S0 S1 S1 S1 S1… and is not accepting.
- For the word a a a b b b b b …, therun is S0 S0 S0 S0 S1 S1 S1 S1 …
- What is the run for a b a b b a b b b …?

S0

a

S1

a,b

a

- What is the language of this automaton?
- What is the LTL specification if b -- PC0=CR0, a=¬b?

- Can you find a deterministic automaton with same language?
- Can you prove there is no such deterministic automaton?

a

S0

b

S1

a

b

- Let each letter correspond to some propositional property.
- Example: a -- P0 enters critical section, b -- P0 does not enter section.
- []<>PC0=CR0

S0

c

S1

b

a

- A -- PC0=CR0/\PC1=CR1
- B -- ¬(PC0=CR0/\PC1=CR1)
- C -- TRUE
- []¬(PC0=CR0/\PC1=CR1)

L0:While True do

NC0:wait(Turn=0);

CR0:Turn=1

endwhile ||

L1:While True do

NC1:wait(Turn=1);

CR1:Turn=0

endwhile

T0:PC0=L0==>PC0=NC0

T1:PC0=NC0/\Turn=0==>

PC0:=CR0

T2:PC0=CR0==>

(PC0,Turn):=(L0,1)

T3:PC1==L1==>PC1=NC1

T4:PC1=NC1/\Turn=1==>

PC1:=CR1

T5:PC1=CR1==>

(PC1,Turn):=(L1,0)

Initially: PC0=L0/\PC1=L1

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

- We want to find a correctness condition for a model to satisfy a specification.
- Language of a model: L(Model)
- Language of a specification: L(Spec).
- We need: L(Model) L(Spec).

Sequencessatisfying Spec

Program executions

All sequences

Counter

examples

Sequencessatisfying Spec

Program executions

All sequences

- Run the two automata in parallel.
- Each state is a pair of states: S1 x S2
- Initial states are pairs of initials: I1 x I2
- Acceptance depends on first component: A1 x S2
- Conforms with transition relation:(x1,y1)-a->(x2,y2) whenx1-a->x2 and y1-a->y2.

a

s0

b,c

s1

a

b,c

a

t0

c

t1

b

States: (s0,t0), (s0,t1), (s1,t0), (s1,t1).

Accepting: (s0,t0), (s0,t1). Initial: (s0,t0).

a

s0

b,c

s1

a

b,c

a

t0

c

t1

b

a

s0,t0

s0,t1

s1,t0

b

a

c

Also state (s0,t1) is unreachable but we will keep it for thetime being!

s1,t1

b

c

a

s0

b,c

s1

a

a

b,c

s0,t0

s0,t1

b

a

a

c

s1,t1

t0

c

t1

b

c

Should we have acceptance when both components accepting? I.e., {(s0,t1)}?

No, consider (ba) It should be accepted, but never passes that state.

a

s0

b,c

s1

a

A

b,c

s0,t0

s0,t1

b

a

a

c

s1,t1

t0

c

t1

c

b

Should we have acceptance when at least one components is accepting? I.e., {(s0,t0),(s0,t1),(s1,t1)}?No, consider b c It should not be accepted, but here will loop through (s1,t1)

A/\¬B

q0,q3

¬A/\B

q1,q2

¬A/\¬B

q1,q3

q0

q2

A/\¬B

¬A/\B

q0 and q2 give false.

q1

q3

¬A

A\/¬B

Version 0

A/\¬B

q0,q3

¬A/\B

q1,q2

¬A/\¬B

q1,q3

Move when see accepting of left (q0)

Move when see accepting of right (q2)

A/\¬B

q0,q3

¬A/\B

q1,q2

¬A/\¬B

q1,q3

Version 1

Version 0

A/\¬B

q0,q3,0

¬A/\B

q1,q2,0

¬A/\¬B

q1,q3,0

A/\¬B

q0,q3,1

¬A/\B

q1,q2 ,1

¬A/\¬B

q1,q3 ,1

Version 1

a

s0,t0

s0,t1

b

a

c

s1,t1

b

c

Need to check if there exists an accepting run (passes through an accepting state infinitely often).

If there is an accepting run, then at least one accepting state repeats on it forever. This state appears on a cycle. So, find a reachable accepting state on a cycle.

- A strongly connected component: a set of nodes where each node is reachable by a path from each other node. Find a reachable strongly connected component with an accepting node.

- Complementation is hard!
- Can ask for the negated property (the sequences that should never occur).
- Can translate from LTL formula to automaton A, and complement A. But:can translate ¬ into an automaton directly!

Express the fairness as a property φ.To prove a property ψ under fairness,model check φψ.

Counter

example

Fair (φ)

Bad (¬ψ)

Program

Specialize model checking. For weak process fairness: search for a reachable strongly connected component, where for each process P either

- it contains on occurrence of a transition from P, or
- it contains a state where P is disabled.

P1::while true dobeginnon-critical section 1c1:=0; while c2=0 do begin if turn=2 then begin c1:=1; wait until turn=1;

c1:=0; end end critical section 1 c1:=1; turn:=2 end.

boolean c1 initially 1;

boolean c2 initially 1;

integer (1..2) turn initially 1;

P2::while true dobeginnon-critical section 2c2:=0; while c1=0 do begin if turn=1 then begin c2:=1; wait until turn=2;

c2:=0; end end critical section 2 c2:=1; turn:=1 end.

P1::while true do begin non-critical section 1 c1:=0;while c2=0 do begin if turn=2 then begin c1:=1; wait until turn=1;

c1:=0; end end critical section 1 c1:=1; turn:=2 end.

boolean c1 initially 1;

boolean c2 initially 1;

integer (1..2) turn initially 1;

P2::while true do begin non-critical section 2 c2:=0; while c1=0 do begin if turn=1 then begin c2:=1; wait until turn=2;

c2:=0; end end critical section 2 c2:=1; turn:=1 end.

c1=c2=0,turn=1

P1::while true do begin non-critical section 1 c1:=0; while c2=0 dobegin if turn=2 then begin c1:=1; wait until turn=1;

c1:=0; end end critical section 1 c1:=1; turn:=2 end.

boolean c1 initially 1;

boolean c2 initially 1;

integer (1..2) turn initially 1;

P2::while true do begin non-critical section 2 c2:=0; while c1=0 do begin if turn=1 then begin c2:=1; wait until turn=2;

c2:=0; end end critical section 2 c2:=1; turn:=1 end.

c1=c2=0,turn=1

P1::while true do begin non-critical section 1 c1:=0;while c2=0 do begin if turn=2 then begin c1:=1; wait until turn=1;

c1:=0; end end critical section 1 c1:=1; turn:=2 end.

P1 waits for P2 to set c2 to 1 again.Since turn=1 (priority for P1), P2 is ready to do that. But never gets the chance, since P1 is constantly active checking c2 in its while loop.

P2::while true do begin non-critical section 2 c2:=0; while c1=0 do begin if turn=1 then begin c2:=1; wait until turn=2;

c2:=0; end end critical section 2 c2:=1; turn:=1 end.

c1=c2=0,turn=1

The execution is unfair to P2. It is not allowed a chance to execute.

Such an execution is due to the interleaving model (just picking an enabled transition.

If it did, it would continue and set c2 to 0, which would allow P1 to progress.

Fairness = excluding some of the executions in the interleaving model, which do not correspond to actual behavior of the system.

while c1=0 do begin if turn=1 then begin c2:=1; wait until turn=2;

c2:=0; end end

- An execution is a finite or infinite sequence of states s0, s1, s2, …
- The initial state satisfies the initial condition, I.e., I (s0).
- Moving from one state si to si+1 is by executing a transition et:
- e(si), I.e., si satisfies e.
- si+1 is obtained by applying t to si.

Now: consider only “fair” executions. Fairness constrains sequences that are considered to be executions.

Sequences

Executions

Fair executions

- Weak transition fairness:It cannot happen that a transition is enabled indefinitely, but is never executed.
- Weak process fairness: It cannot happen that a process is enabled indefinitely, but non of its transitions is ever executed
- Strong transition fairness:If a transition is infinitely often enabled, it will get executed.
- Strong process fairness:If at least one transition of a process is infinitely often enabled, a transition of this process will be executed.

- Assume in the negative that some property (e.g., termination) does not hold, because of some transitions or processes prevented from execution.
- Show that such executions are impossible, as they contradict fairness assumption.
- We sometimes do not know in reality which fairness constraint is guaranteed.

P1::x=1

P2: do

:: y==0 if

:: true :: x==1 y=1

fi od

Initially: x=0; y=0;

In order for the loop to terminate we need P1 to execute the assignment. But P1 may never execute, since P2 is in a loop executing true. Consequently, x==1 never holds, and y is never assigned a 1.

pc1=l0-->(pc1,x):=(l1,1) /* x=1 */

pc2=r0/\y=0-->pc2=r1 /* y==0*/

pc2=r1pc2=r0 /* true */

pc2=r1/\x=1(pc2,y):=(r0,1) /* x==1 --> y:=1 */

P1::x=1

P2: do

:: y==0 if

:: true :: x==1 y=1

fi od

Initially: x=0; y=0;

Under weak transition fairness, P1 would assign 1 to x, but this does not guarantee that 1 is assigned to y and thus the P2 loop will terminates, since the transition for checking x==1 is not continuously enabled (program counter not always there).

Weak process fairness only guarantees P2 to execute, but it can still choose to do the true.

Strong process fairness: same.

P1::x=1

P2: do

:: y==0 if

:: true :: x==1 y=1

fi od

Initially: x=0; y=0;

Under strongtransitionfairness, P1 would assign 1 to x. If the execution was infinite, the transition checking x==1 was infinitely often enabled. Hence it would be eventually selected. Then assigning y=1, the main loop is not enabled anymore.

- Weak transition fairness:/\T (<>[]en []<>exec).Equivalently: /\aT ¬<>[](en /\¬exec)
- Weak process fairness: /\Pi (<>[]enPi []<>execPi )
- Strong transition fairness:/\T ([]<>en []<>exec)
- Strong process fairness:/\Pi ([]<>enPi []<>execPi )

exec is executed.

execPi some transition of Pi is executed.

en is enabled.

enPi some transition of process Pi is enabled.

enPi = \/Ten

execPi = \/Texec

Strongtransitionfair execs

- A is weaker than B if BA.
- Consider the executions L(A) and L(B).Then L(B)L(A).
- An execution is strong {process,transition} fair implies that it is also weak {process,transition} fair.
- There are fewer strong {process,transition} fair executions.

Strongprocessfair execs

Weaktransitionfair execs

Weakprocessfair execs

- Instead of verifying that the program satisfies , verify it satisfies fair
- Problem: may be inefficient. Also fairness formula may involves special arrangement for specifying what exec means.
- May specialize model checking algorithm instead.

Specialize model checking. For weak process fairness: search for a reachable strongly connected component, where for each process P either

- it contains on occurrence of a transition from P, or
- it contains a state where P is disabled.

- Many possible outcomes and interactions.
- Not manageable by an algorithm (undecideable, complex).
- Requires a lot of practice and ingenuity (e.g., finding invariants).

- Testing methods fail to cover potential errors.
- Deductive verification techniques require
- too much time,
- mathematical expertise,
- ingenuity.

- Model checking requires a lot of time/space and may introduce modeling errors.

- Abstraction
- Compositionality
- Partial Order Reduction
- Symmetry

- Represent the program using a smaller model.
- Pay attention to preserving the checked properties.
- Do not affect the flow of control.

- Use smaller data objects.
x:= f(m)

y:=g(n)

if x*y>0 then …

else …

x, y never used again.

- Assign values {-1, 0, 1} to x and y.
- Based on the following connection:sgn(x) = 1 if x>0, 0 if x=0, and -1 if x<0.sgn(x)*sgn(y)=sgn(x*y).

- S - states, I - initial states. L(s) - labeling.
- R(S,S) - transition relation.
- h(s) maps s into its abstract image.Full model --h--> Abstract model I(s) --> I(h(s)) R(s, t) --> R(h(s),h(t)) L(h(s))=L(s)

go

Traffic light example

stop

stop

go

go

stop

stop

stop

Every execution of the full model can be simulated by an execution of the reduced one.

Every LTL property that holds in the reduced model hold in the full one.

But there can be properties holding for the original model but not the abstract one.

go

go

stop

stop

stop

go

Not preserved:

[]<>go

Counterexamples need to be checked.

go

stop

stop

stop

- A permutation is a one-one and onto function p:A->A.For example, 1->3, 2->4, 3->1, 4->5, 5->2.
- One can combine permutations, e.g.,p1: 1->3, 2->1, 3->2p2: 1->2, 2->1, 3->3p1@p2: 1->3, 2->2, 3->1
- A set of permutations with @ is called a symmetry group.

- Want to find some symmetry group suchthat for each permutation p in it,R(s,t) if and only if R(p(s), p(t))and L(p(s))=L(s).
- Let K(s) be all the states that can be permuted to s. This is a set of states such that each one can be permuted to the other.

init

Turn=0

L0,L1

Turn=1

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=1

L0,NC1

Turn=1

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=1

L0,CR1

Turn=1

NC0,NC1

Turn=0

CR0,NC1

Turn=1

NC0,CR1

init

Turn=0

L0,L1

Turn=0

L0,NC1

Turn=0

NC0,L1

Turn=0

NC0,NC1

Turn=0

CR0,L1

Turn=0

CR0,NC1

The quotient model

e

empty

q

quasi

q

q

full

f

Translating from logic to automataComment: the language of LTL is a proper subset of the language of Buchi Automata. For example,one cannot express in LTL the following automaton [Wolper]:

a

a ,¬a

- Want to write the specification in some logic.
- Want model-checking tools to be able to check the specification automatically.

- Convert into normal form, where negation only applies to propositional variables.
- ¬[] becomes <>¬.
- ¬<> becomes [] ¬.
- What about ¬ ( U )?
- Define operator V such that¬ ( U) = (¬) R (¬),
¬ ( R) = (¬) U (¬).

¬p

¬p

¬p

¬p

¬p

¬p

¬p

¬p

¬p

q

q

q

q

q

q

q

q

q

¬p

¬p

¬p

¬p

p

q

q

q

q

q

- Replace ¬T by F, and ¬F by T.
- Replace ¬ ( \/ ) by (¬) /\ (¬) and¬ ( /\ ) by (¬) \/ (¬)

- Replace -> by (¬) \/ .
- Replace <> by (T U).
- Replace [] by (F R).

- Translate ( []<>P ) ( []<>Q )
- Eliminate implication ¬( []<>P ) \/ ( []<>Q )
- Eliminate [], <>:¬( F R ( T U P ) ) \/ ( F R ( T U Q ) )
- Push negation inwards:(T U (F R ¬P ) ) \/ ( F R ( T U Q ) )

Incoming

New

Old

Next

Name

- U = \/ ( /\ O ( U ) )
- R = /\ ( \/ O ( R ) )
This separates the formulas to two parts:one holds in the current state, and the other in the next state.

- Take one formula from “New” and add it to “Old”.
- According to the formula, either
- Split the current node into two, or
- Evolve the node into a new version.

Incoming

New

Old

Next

Incoming

Incoming

New

Old

New

Old

Next

Next

Copy incoming edges, update other field.

Incoming

New

Old

Next

Incoming

New

Old

Next

Copy incoming edges, update other field.

- U , split:
- Add to New, add U to Next.
- Add to New.
Because U = \/ ( /\ O (U )).

- R , split:
- Add to New.
- Add to New, R to Next.
Because R = /\ ( \/ O (R )).

- \/ , split:
- Add to New.
- Add to New.

- /\ , evolve:
- Add to New.

- O , evolve:
- Add to Next.

init

Incoming

New

Old

aU(bUc)

Next

Incoming

Incoming

a

aU(bUc)

bUc

aU(bUc)

aU(bUc)

init

Incoming

aU(bUc)

init

init

init

Incoming

bUc

aU(bUc)

init

init

Incoming

Incoming

b

aU(bUc)

c

aU(bUc)

(bUc)

- When “New” is empty.
- Then compare against a list of existing nodes “Nodes”:
- If such a with same “Old”, “Next” exists,just add the incoming edges of the new versionto the old one.
- Otherwise, add the node to “Nodes”. Generate a successor with “New” set to “Next” of father.

init

Incoming

a,aU(bUc)

Creating a successor node.

aU(bUc)

Incoming

aU(bUc)

Incoming

New

Old

Next

X

There is an edge from node X to Y labeled with propositions P (negated or non negated), if X is in the incoming list of Y, and Y has propositions P in field “Old”.

Initial nodes are those marked as init.

a, b, ¬c

Node Y

a, aU(bUc)

b, bUc, aU(bUc)

c, bUc, aU(bUc)

b, bUc

c, bUc

a, aU(bUc)

b, bUc, aU(bUc)

c, bUc, aU(bUc)

b, bUc

c, bUc

Initial nodes: All nodes with incoming edge from “init”.

- Use “generalized Buchi automata”, wherethere are several acceptance sets F1, F2, …, Fn, and each accepted infinite sequence must include at least one state from each set infinitely often.
- Each set corresponds to a subformula of form U . Guarantees that it is never the case that U holds forever, without .
- Translate to Buchi acceptance condition (will be shown later).

a, aU(bUc)

b, bUc, aU(bUc)

c, bUc, aU(bUc)

b, bUc

c, bUc

All nodes with c, or without bUc.

a, aU(bUc)

b, bUc, aU(bUc)

c, bUc, aU(bUc)

b, bUc

c, bUc

All nodes with bUc or without aU(bUc).

- For acceptance sets F1, F2, …, Fn, generate n copies of the automaton.
- Move from the ith copy to the i+1th copy when passing a state of Fi.
- In the first copy, the states of F1 are accepting.
- Same intuition as the unrestricted intersection:Need to see each accepting set infinitely often, and it does not matter that we miss some of them (there must be an infinite supply anyway…).

- Model checking: automatic verification of finite state systems.
- Modeling: affects how we can view and check the system’s properties.
- Specification formalisms, e.g., LTL or Buchi automata
- Apply graph algorithms (or BDD, or BMD).
- Methods to alleviate state space explosion.
- Translate LTL into Buchi automata.

Algorithmic Testing and Black Box Checking

- Reduce design/programming errors.
- Can be done during development, beforeproduction/marketing.
- Practical, simple to do.
- Check the real thing, not a model.
- Scales up reasonably.
- Being state of the practice for decades.

Wants to know:

- In what state we started?
- In what state we are?
- Transition relation
- Conformance
- Satisfaction of a temporal property

Know:

- Transition relation
- Size or bound on size

S - finite set of states. (size n)

S– set of inputs. (size d)

O– set of outputs, for each transition.

(s0S - initial state).

- S S S - transition relation.
S S O – output on edge.

- Otherwise no amount of experiments would guarantee anything.
- If dependent on some parameter (e.g., temperature), we can determinize, by taking parameter as additional input.
- We still can model concurrent system. It means just that the transitions are deterministic.
- All kinds of equivalences are unified into language equivalence.
- Also: connected machine (otherwise we may never get to the completely separate parts).

a/1

b/1

a/1

When the black box is nondeterministic, we might never test some choices.

b/1

s1

s2

a/0

b/1

b/0

a/0

s3

a/0

Start with one block containing all states {s1, s2, s3}.

b/1

s1

s2

a/0

b/1

b/0

a/0

s3

a/0

Two sets, separated using the string b {s1, s3}, {s2}.

b/1

s1

s2

a/0

b/1

b/0

a/0

s3

a/0

Separate first block using b to three singleton blocks.Separating sequences: b, bb.Max rounds: n-1, sequences: n-1, length: n-1.For each pair of states there is a separating sequence.

Depending on output, would know in what state we are.

Algorithm: Put all the states in one block (initially we do not know what is the state).

Then repeatedly partitions blocks of states, as long as they are not singletons, as follows:

- Take a non singleton block, append a distinguishing sequence that separates at least two states.
- Update all blocks to the states after executing .
Max length: (n-1)2 (Lower bound: n(n-1)/2.)

b/1

s1

s2

a/0

b/1

b/0

a/0

s3

a/0

{s1, s2, s3}

b

1

1

0

{s1, s2} {s3}

b

1

1

0

{s1} {s2} {s3}

On input b and output 1, still don’t know if was in s1or s3, i.e., if currently in s2 or s1.So separate these cases with another b.

- One sequence takes the machine to the same final state, regardless of the initial state or the outputs.
- Not every machine has a synchronizing sequence.
- Can be checked whether exists and can be found in polynomial time.

- Want to know in which state the system has started (was reset).
- Can be a preset distinguishing sequence (fixed), or a tree (adaptive).
- May not exist (PSPACE complete to check if preset exists, polynomial for adaptive).
- Best known algorithm: exponential length for preset,polynomial for adaptive [LY].

b/1

a/1

b/1

s1

s2

a/1

a/1

s3

b/0

Start with a:in case of being in s1or s3we’ll move to s1and cannot distinguish.Start with b:In case of being in s1or s2we’ll move to s2 and cannot distinguish.

The kind of experiment we do affects what we can distinguish. Much like the Heisenberg principle in Physics.

- Unknown deterministic finite state system B.
- Known: n states and alphabet .
- An abstract model C of B. C satisfies all the properties we want from B.C has m states.
- Check conformance of B and C.
- Another version: only a bound n on the number of states l is known.

b/1

a/1

b/1

s1

s2

?=

a/1

a/1

s3

b/0

- Black box machine has no more states than specification machine (errors are mistakes in outputs, mistargeted edges).
- Specification machine is reduced, connected, deterministic.
- Machine resets reliably to a single initial state (or use homing sequence).

a/1

a/1

b/1

b/1

a/1

b/1

b/1

a/1

a/1

b/1

Cannot distinguish if reduced or not.

a

b

b

a

a

a

a

b

b

a

a

b

Need: bound on number of states of B.

b/1

a/1

b/1

s1

s1

s2

a/1

b/1

a/1

s3

s2

a/1

s3

b/0

Reset or homing

Reset or homing

According to the spanning tree, force a sequence of inputs to go to each state.

- From each state, perform the distinguishing sequences.
- From each state, make a single transition, check output, and use distinguishing sequences to check that in correct target state.

s1

a/1

b/1

s3

s2

Distinguishing sequences

- Checking the different distinguishing sequences (m-1 of them) means each time resetting and returning to the state under experiment.
- A reset can be performed to a distinguished state through a homing sequence. Then we can perform a sequence that brings us to the distinguished initial state.
- Since there are no more than m states, and according to the experiment, no less than m states, there are m states exactly.
- Isomorphism between the transition relation is found, hence from minimality the two automata recognize the same languages.

- Assume accepting states.
- Accepts only words with a specific suffix (cdab in the example).

c

d

a

b

s1

s2

s3

s4

s5

Any other input

- Black box can “pretend” to behave as a specification automaton for a long time, then upon using the right combination, make a mistake.

a/1

Pretends to be S1

b/1

a/1

b/1

b/1

s1

s2

a/1

a/1

s3

Pretends to be S3

b/0

Reset or homing

Reset or homing

- The worst that can happen is a combination lock automaton that behaves differently only in the last state. The length of it is the difference between the size n of the black box and the specification m.
- Reach every state on the spanning tree and check every word of length n-m+1 or less. Check that after the combination we are at the state we are supposed to be, using the distinguishing sequences.
- No need to check transitions: already included in above check.
- Complexity: m2 n dn-m+1 Probabilistic complexity: Polynomial.

s1

a/1

b/1

s3

s2

Words of length n-m+1

Distinguishing sequences

- Finite state description of a system B.
- LTL formula . Translate into an automaton P.
- Check whether L(B) L(P)=.
- If so, S satisfies . Otherwise, the intersection includes a counterexample.
- Repeat for different properties.

S - finite set of states. (B has l n states)

S0S - initial states. (P has m states)

S - finite alphabet. (contains p letters)

d S SS - transition relation.

F S - accepting states.

Accepting run: passes a state in F infinitely often.

System automata: F=S, deterministic, one initial state.

Property automaton: not necessarily deterministic.

a

<>a

a

a, a

a

a

a

a

<>a

a, a

<>a

a

a

Use automatic translation algorithms, e.g.,

[Gerth,Peled,Vardi,Wolper 95]

a

c

b

a

a

<>a

s1

s2

q1

a

c

b

a

a

s3

q2

a

s1,q1

s2,q1

Acceptance isdetermined byautomaton P.

b

a

s1,q2

s3,q2

c

Given Finite state system B.

Transition relation of B known.

Property represent by automaton P.

Check if L(B) L(P)=.

Graph theory or BDD techniques.

Complexity: polynomial.

Unknown Finite state system B.

Alphabet and number of states of B or upper bound known.

Specification given as an abstract system C.

Check if B C.

Complexity: polynomial if number states known. Exponential otherwise.

Property represent by automaton P.

Check if L(B) L(P)=.

Graph theory techniques.

Unknown Finite state system B.

Alphabet and Upper bound on Number of states of B known.

Complexity: exponential.

reset

a

a

b

b

c

c

try c

try b

a

a

b

b

c

c

a

a

b

c

b

c

fail

- Nondeterministic algorithm:guess a path of length n from the initial state to a deadlock state.Linear time, logarithmic space.
- Deterministic algorithm:systematically try paths of length n, one after the other (and use reset), until deadlock is reached.Exponential time, linear space.

- Nondeterministic algorithm:Linear time, logarithmic space.
- Deterministic algorithm:Exponential (p n-1) time, linear space.
- Lower bound: Exponential time (usecombination lock automata).
- How does this conform with what we know about complexity theory?

- Cannot model using Turing machines: not all the information about B is given. Only certain experiments are allowed.
- We learn the model as we make the experiments.
- Can use the model of games of incomplete information.

- Two players: $-player, -player (here, deterministic).
- Finitely many configurations C. Including:Initial Ci , Winning : W+ and W- .
- An equivalence relation @ on C (the $-player cannot distinguish between equivalent states).
- Labels L on moves (try a, reset, success, fail).
- The $-player has the moves labeled the same from configurations that are equivalent.
- Deterministic strategy for the $-player: will lead to a configuration in W+ W-. Cannot distinguish between equivalent configurations.
- Nondeterministic strategy: Can distinguish between equivalent configurations..

- Each configuration contains an automaton and its current state (and more).
- Moves of the $-player are labeled withtry a, reset... Moves of the -player withsuccess, fail.
- c1@ c2 when the automata in c1and c2 would respond in the same way to the experiments so far.

- Learn first the structure of the black box.
- Then apply the intersection.
- Enumerate automata with n states (without repeating isomorphic automata).
- For a current automata and newautomata, construct a distinguishing sequence. Only one of them survives.
- Complexity: O((n+1)p (n+1)/n!)

- Systematically (as in the deadlock case), find two sequences v1 and v2 of length <=m n.
- Applying v1 to P brings us to a state t that is accepting.
- Applying v2 to P brings us back to t.
- Apply v1 v2 n to B. If this succeeds,there is a cycle in the intersection labeled with v2, with t as the P (accepting) component.
- Complexity: O(n2p2mnm).

v1

v2

- Use Angluin’s algorithm for learning an automaton.
- The learning algorithm queries whether some strings are in the automaton B.
- It can also conjecture an automaton Miand asks for a counterexample.
- It then generates an automaton with more states Mi+1and so forth.

- Start the learning algorithm.
- Queries are just experiments to B.
- For a conjectured automaton Mi , check if Mi P =
- If so, we check conformance of Mi with B ([VC] algorithm).
- If nonempty, it contains some v1 v2w . We test B with v1 v2n. If this succeeds: error, otherwise, this is a counterexample for Mi .

- l - actual size of B.
- n - an upper bound of size of B.
- d - size of alphabet.
- Lower bound: reachability is similar to deadlock.
- O(l 3 d l + l 2mn) if there is an error.
- O(l 3 d l + l 2 n dn-l+1+ l 2mn) if there is no error.
If n is not known, check while time allows.

- Probabilistic complexity: polynomial.

- Basic system written in SML (by Alex Groce, CMU).
- Experiment with black box using Unix I/O.
- Allows model-free model checking of C code with inter-process communication.
- Compiling tested code in SML with BBC program as one process.

- Black Box Checking: automatic verification of unspecified system.
- A hard problem, exponential in number of states, but polynomial on average.
- Implemented, tested.
- Another use: when the model is given, but is not exact.

The rest is not part of this lecture

- Testing is not about showing that there are no errors in the program.
- Testing cannot show that the program performs its intended goal correctly.
So, what is software testing?

Testing is the process of executing the program in order to find errors.

A successful test is one that finds an error.

- Unit testing – the lowest level, testing some procedures.
- Integration testing – different pieces of code.
- System testing – testing a system as a whole.
- Acceptance testing – performed by the customer.
- Regression testing – performed after updates.
- Stress testing – checking the code under extreme conditions.
- Mutation testing – testing the quality of the test suite.

- There are never sufficiently many test cases.
- Testing does not find all the errors.
- Testing is not trivial and requires considerable time and effort.
- Testing is still a largely informal task.

The testing is not based on the structure of the program (which is unknown).

In order to ensure correctness, every possible input needs to be tested - this is impossible!

The goal: to maximize the number of errors found.

White Box

Is based on the internal structure of the program.

There are several alternative criterions for checking “enough” paths in the program.

Even checking all paths (highly impractical) does not guarantee finding all errors (e.g., missing paths!)

- A programmer should not test his/her own program.
- One should test not only that the program does what it is supposed to do, but that it does not do what it is not supposed to.
- The goal of testing is to find errors, not to show that the program is errorless.
- No amount of testing can guarantee error-free program.
- Parts of programs where a lot of errors have already been found are a good place to look for more errors.
- The goal is not to humiliate the programmer!

- Manual testing methods.
- Done by a team of people.
- Performed at a meeting (brainstorming).
- Takes 90-120 minutes.
- Can find 30%-70% of errors.

- Team of 3-5 people.
- One is the moderator. He distributes materials and records the errors.
- The programmer explains the program line by line.
- Questions are raised.
- The program is analyzed w.r.t. a checklist of errors.

Data declaration

All variables declared?

Default values understood?

Arrays and strings initialized?

Variables with similar names?

Correct initialization?

Control flow

Each loop terminates?

DO/END statements match?

Input/output

OPEN statements correct?

Format specification correct?

End-of-file case handled?

- Team of 3-5 people.
- Moderator, as before.
- Secretary, records errors.
- Tester, play the role of a computer on some test suits on paper and board.

The main problem is to select a good coverage

criterion. Some options are:

- Cover all paths of the program.
- Execute every statement at least once.
- Each decision has a true or false value at least once.
- Each condition is taking each truth value at least once.
- Check all possible combinations of conditions in each decision.

Infeasible.

Consider the flow diagram on the left.

It corresponds to a loop.

The loop body has 5 paths.

If the loops executes 20

times there are 5^20 different paths!

May also be unbounded!

IF (A>1)&(B=0) THEN X=X/A; END;

IF (A=2)|(X>1) THEN X=X+1; END;

- Choose values for A,B,X.
- Value of X may change, depending on A,B.
- What do we want to cover? Paths? Statements? Conditions?

By choosing

A=2,B=0,X=3

each statement will be chosen.

The case where the tests fail is not checked!

IF (A>1)&(B=0) THEN X=X/A; END;

IF (A=2)|(X>1) THEN X=X+1; END;

Now x=1.5

Can be achieved using

A=3,B=0,X=3

A=2,B=1,X=1

Problem: Does not test individual conditions. E.g., when X>1 is erroneous in second decision.

IF (A>1)&(B=0) THEN X=X/A; END;

IF (A=2)|(X>1) THEN X=X+1; END;

A=3,B=0,X=3

IF (A>1)&(B=0) THEN X=X/A; END;

IF (A=2)|(X>1) THEN X=X+1; END;

Now x=1

A=2,B=1,X=1

The case where A1 and the case where x>1 where not checked!

IF (A>1)&(B=0) THEN X=X/A; END;

IF (A=2)|(X>1) THEN X=X+1; END;

For example:

A=1,B=0,X=3

A=2,B=1,X=0

lets each condition be true and false once.

Problem:covers only the path where the first test fails and the second succeeds.

IF (A>1)&(B=0) THEN X=X/A; END;

IF (A=2)|(X>1) THEN X=X+1; END;

A=1,B=0,X=3

IF (A>1) & (B=0) THEN X=X/A; END;

IF (A=2) | (X>1) THEN X=X+1; END;

A=2,B=1,X=0

Did not check the first THEN part at all!!!

Can use condition+decision coverage.

IF (A>1)&(B=0) THEN X=X/A; END;

IF (A=2)|(X>1) THEN X=X+1; END;

A>1,B=0

A>1,B≠0

A1,B=0

A1,B≠0

A=2,X>1

A=2,X1

A≠2,X>1

A≠2,X1

IF (A>1)&(B=0) THEN X=X/A; END;

IF (A=2)|(X>1) THEN X=X+1; END;

A=2,B=0,X=4

A=2,B=1,X=1

A=1,B=0,X=2

A=1,B=1,X=1

Note the X=4 in the first

case: it is due to the fact

that X changes before

being used!

IF (A>1)&(B=0) THEN X=X/A; END;

IF (A=2)|(X>1) THEN X=X+1; END;

Further optimization: not all combinations.For C /\ D, check (C, D), (C, D), (C, D).For C \/ D, check (C, D), (C, D), (C, D).

A

(B) : x1= y1 * x2 + y2 /\ y2 >= 0

Relativize B) w.r.t. the assignment becomes B) [Y\g(X,Y)]

(I.e., (B) expressed w.r.t. variables at A.)

(B)A =x1=0 * x2 + x1 /\ x1>=0

Think about two sets of variables,before={x, y, z, …} after={x’,y’,z’…}.

Rewrite (B) using after, and the assignment as a relation between the set of variables. Then eliminate after.

Here: x1’=y1’ * x2’ + y2’ /\ y2’>=0 /\x1=x1’ /\ x2=x2’ /\ y1’=0 /\ y2’=x1now eliminate x1’, x2’, y1’, y2’.

Y=g(X,Y)

(y1,y2)=(0,x1)

B

A

(y1,y2)=(0,x1)

B

B

T

F

C) B)= t(X,Y) /\C)

D) B)=t(X,Y) /\ D)

B)= D) /\ y2x2

t(X,Y)

C

D

B

T

F

y2>=x2

C

D

- Put true at end of path.
- Propagate path backwards.
- On assignment, relativize expression.
- On “yes” edge of decision, add decision as conjunction.
- On “no” edge, add negation of decision as conjunction.
- Can be more specific when calculating condition with multiple condition coverage.

A>1 & B=0

no

yes

X=X/A

A=2 | X>1

true

no

yes

X=X+1

true

(A≠2 /\ X/A>1) /\ (A>1 & B=0)

A>1 & B=0

A≠2 /\ X/A>1

no

yes

Need to find a satisfying assignment:

A=3, X=6, B=0

Can also calculate path condition forwards.

X=X/A

A≠2 /\ X>1

A=2 | X>1

true

no

yes

X=X+1

true

- Cover all nodes, e.g., using search strategies: DFS, BFS.
- Cover all paths (usually impractical).
- Cover each adjacent sequence of N nodes.
- Probabilistic testing. Using random number generator simulation. Based on typical use.
- Chinese Postman: minimize edge traversalFind minimal number of times time to travel each edge using linear programming or dataflow algorithms.Duplicate edges and find an Euler path.

- Partition the program into pieces of code with a single entry/exit point.
- For each piece find which variables are set/used/tested.
- Various covering criteria:
- from each set to each use/test
- From each set to some use/test.

X:=3

t>y

x>y

z:=z+x

- Equivalence partition
- Boundary value analysis
- Cause-effect graphs

Valid equivalence

class

Invalid equivalence

class

Specification

condition

- Goals:
- Find a small number of test cases.
- Cover as much possibilities as you can.

- Try to group together inputs for which the program is likely to behave the same.

- Begins with A-Z
- Contains [A-Z0-9]
- Has 1-6 characters.

Valid equivalence

class

Specification

condition

Invalid equivalence

class

Starting char

Starts A-Z

Starts other

1

2

Chars

[A-Z0-9]

Has others

3

4

1-6 chars

0 chars, >6 chars

Length

5

6

7

- Add a new test case until all valid equivalence classes have been covered. A test case can cover multiple such classes.
- Add a new test case until all invalid equivalence class have been covered. Each test case can cover only one such class.

Valid equivalence

class

Invalid equivalence

class

Specification

condition

(6)

VERYLONG (7)

AB36P (1,3,5)

1XY12 (2)

A17#%X (4)

Valid equivalence

class

Specification

condition

Invalid equivalence

class

Starting char

Starts A-Z

Starts other

1

2

Chars

[A-Z0-9]

Has others

3

4

1-6 chars

0 chars, >6 chars

Length

5

6

7

- In every element class, select values that are closed to the boundary.
- If input is within range -1.0 to +1.0, select values -1.001, -1.0, -0.999, 0.999, 1.0, 1.001.
- If needs to read N data elements, check with N-1, N, N+1. Also, check with N=0.

LTLAut

Model

Checker

Path

Path condition

calculation

Flowchart

Compiler

Transitions

First order

instantiator

Test

monitoring

- Verification of software.
- Compositional verification. Only a unit of code.
- Parametrized verification.
- Generating test cases.
A path found with some truth assignment satisfying the path condition. In deterministic code, this assignment guarantees to derive the execution of the path.

In nondeterministic code, this is one of the possibilities.Can transform the code to force replying the path.

- Intersect property automaton with theflow chart, regardless of the statements and program variables expressions.
- Add assertions from the property automaton to further restrict the path condition.
- Calculate path conditions for sequences found in the intersection.
- Calculate path conditions on-the-fly. Backtrack when condition is false.Thus, advantage to forward calculation of path conditions (incrementally).

l2:x:=x+z

l3:x<t

l1:…

l2:x:=x+z

¬at l2

X

=

at l2

l3:x<t

¬at l2

l2:x:=x+z

at l2

l2:x:=x+z

l3:x<t

l1:…

xy

l2:x:=x+z

¬at l2

X

=

at l2/\

xy

l3:x<t

x2y

¬at l2

l2:x:=x+z

at l2/\

x2y

l0

l1:x:=a

l2:y:=b

l3:z:=x rem y

l4:x:=y

l5:y:=z

no

l6:z=0?

yes

l7

l0

l1:x:=a

l2:y:=b

Oops…with an error (l4 and l5 were switched).

l3:z:=x rem y

l4:y:=z

l5:x:=y

no

l6:z=0?

yes

l7

- Temporal specification for sequential software?
- Deadlock? Liveness? – No!
- Captures the tester’s intuition about the location of an error:“I think a problem may occur when the program runs through the main while loop twice, then the if condition holds, while t>17.”

l0

l1:x:=a

l2:y:=b

a>0/\b>0/\at l0 /\at l7

l3:z:=x rem y

at l0/\a>0/\b>0

l4:y:=z

l5:x:=y

no

l6:z=0?

yes

at l7

l7

l0

l1:x:=a

l2:y:=b

a>0/\b>0/\at l0/\at l7

l3:z:=x rem y

Path 1: l0l1l2l3l4l5l6l7a>0/\b>0/\a rem b=0

Path 2: l0l1l2l3l4l5l6l3l4l5l6l7a>0/\b>0/\a rem b0

l4:y:=z

l5:x:=y

no

l6:z=0?

yes

l7

Bad point: potential explosion

Good point: may be chopped on-the-fly

l0

l1:x:=a

l2:y:=b

- Driver: represents the program or procedure that called our checked unit.
- Stub: represents a procedure called by our checked unit.
- In our approach: replace both of them with a formula representing the effect the missing code has on the program variables.
- Integrate the driver and stub specification into the calculation of the path condition.

l3:z’=x rem y

/\x’=x/\y’=x

l4:y:=z

l5:x:=y

no

l6:z=0?

yes

l7

- Black box testing: Know transition relation,or bound on number of states, want to find initialstate, structure, conformance, temporal property.
- Software testing:Unit testing, code inspection, coverage, test case generation.
- Model checking and testing have a lot in common.