Model Checking

Doron A. PeledUniversity of WarwickCoventry, UKdoron@dcs.warwick.ac.uk

Modelling and specification for verification and validation How to specify what the software

is supposed to do? How to model it in a way that

allows us to check it?

Sequential systems. 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.

Concurrent SystemsInvolve 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).

Problems in modeling systems 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.

Modeling.The states based model.

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.

State space

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>…

Atomic Transitions

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 inc command.

Non atomicity

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

Consider the actual translation:

P1:load R1,a inc R1 store R1,aP2:load R2,a inc R2 store R2,a a may be also 1.

Scenario

P1:load R1,a

inc R1

store R1,a

P2:load R2,a

inc R2 store R2,a

Representing transitions 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.

Initial condition

A predicate I. The program can

start from states s such that I (s ) holds.

For example:I (s )=a>b /\ b>c.

A transition system A (finite) set of variables V over

some domain. A set of states . A (finite) set of transitions T, each

transition et has an enabling condition e, and a transformation t.

An initial condition I.

Example V={a, b, c, d, e}. : 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?

The interleaving model 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.

Example:

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=1endwhile ||L1:While True do NC1:wait(Turn=1); CR1:Turn=0endwhile

T0:PC0=L0PC0:=NC0T1:PC0=NC0/\Turn=0 PC0:=CR0T2:PC0=CR0 (PC0,Turn):=(L0,1)T3:PC1=L1PC1=NC1T4:PC1=NC1/\Turn=1 PC1:=CR1T5:PC1=CR1 (PC1,Turn):=(L1,0)

Initially: PC0=L0/\PC1=L1

The transitions

The state graph:Successor relation between states.Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

Some observations 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.

Always ¬(PC0=CR0/\PC1=CR1)(Mutual exclusion)

Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

Always if Turn=0 the at some point Turn=1Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

Always if Turn=0 the at some point Turn=1Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

Interleaving semantics:Execute one transition at a time.

Turn=0L0,L1

Turn=0L0,NC1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=1L0,CR1

Turn=1L0,NC1

Need to check the propertyfor every possible interleaving!

Interleaving semanticsTurn=0L0,L1

Turn=0L0,NC1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=1L0,CR1

Turn=1L0,NC1

Turn=0L0,L1

Turn=0L0,NC1

T0:PC0=L0PC0:=NC0T1:PC0=NC0/\Turn=0PC0:=CR0T1’:PC0=NC0/\

Turn=1PC0:=NC0T2:PC0=CR0(PC0,Turn):=(L0,1)

T3:PC1==L1PC1=NC1T4:PC1=NC1/\Turn=1PC1:=CR1T4’:PC1=NC1/\Turn=0PC1:=N1T5:PC1=CR1(PC1,Turn):=(L1,0) Initially: PC0=L0/\PC1=L1

Busy waiting

Always when Turn=0 then sometimes Turn=1Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

Now it does not hold!(Red subgraph generates a counterexample execution.)

How can we check the model?

The model is a graph. The specification should refer the

the graph representation. Apply graph theory algorithms.

What properties can we check?

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.

How to perform the checking?

Apply a search strategy (Depth first search, Breadth first search).

Check states/transitions during the search.

If property does not hold, report counter example!

If it is so good, why learn deductive verification methods?

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.

The state space explosion 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.

If it is so constrained, is it of any use?

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.

Depth First Search

Program DFSFor 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.

How can we check properties with 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.

¬(PC0=CR0/\PC1=CR1) is an invariant!Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

Want to do more!

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 --> <>Turn=1)

Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

Turn=0L0,L1

Turn=1L0,L1

•Add an additional initial node.

•Propositions are attached to incoming nodes.

•All nodes are accepting.

Turn=1L0,L1

Turn=0L0,L1

Correctness condition

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).

Correctness

All sequences

Sequences satisfying Spec

Program executions

How to prove correctness?

Show that L(Model) L(Spec). Equivalently: ______

Show that L(Model) L(Spec) = Ø. Also: can obtain L(Spec) by

translating from LTL!

What do we need to know?

How to intersect two automata? How to complement an

automaton? How to translate from LTL to an

automaton?

Specification Formalisms

Properties of 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 transition system 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.

The interleaving model 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.

Temporal logic 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.

LTL: Syntax ::= () | ¬ | /\ \/ 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.)

Semantics over suffixes of execution

Combinations

[]<>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”.

Some relations: [](a/\b)=([]a)/\([]b) But <>(a/\b)(<>a)/\(<>b)

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

aa abb b bab

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

No, just <--Yes!!!Yes!!!No, just -->

Can discard some operators 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.

Formal semantic definition 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 |=.

Then we interpret: s|=p as in propositional logic. |= is interpreted over a

sequence, as in previous slide. P|=holds if |= for every

sequence of P.

Spring Example

s1 s3s2pull

release

releaseextended

malfunction

extended

r0 = s1 s2 s1 s2 s1 s2 s1 …r1 = s1 s2 s3 s3 s3 s3 s3 …r2 = s1 s2 s1 s2 s3 s3 s3 ……

LTL satisfaction by a single sequence

malfunction

s1 s3s2pull

release

releaseextende

dextended

r2 = s1 s2 s1 s2 s3 s3 s3 …

r2 |= <>[] extended ??r2 |= ¬ <>[] extended ??r2 |= (¬extended) U malfunction ??r2 |= [](¬extended->O extended) ??

LTL satisfaction by a system

malfunction

s1 s3s2pull

release

releaseextende

dextended

P |= <>[] extended ??P |= ¬ <>[] extended ??P |= (¬extended) U malfunction ??P |= [](¬extended->O extended) ??

The state space

Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

[] ¬(PC0=CR0/\PC1=CR1)(Mutual exclusion)

Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

[](Turn=0 --> <>Turn=1)

Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

Interleaving semantics:Execute one transition at a time.

Turn=0L0,L1

Turn=0L0,NC1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=1L0,CR1

Turn=1L0,NC1

Need to check the propertyfor every possible interleaving!

More specifications [](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.

Proof system ¬<>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

Traffic light example

Green --> Yellow --> Red --> GreenAlways has exactly one light:[](¬(gr/\ye)/\¬(ye/\re)/\¬(re/\gr)/\(gr\/ye\/re))Correct change of color:[]((grUye)\/(yeUre)\/(reUgr))

Another kind of traffic lightGreen-->Yellow-->Red-->Yellow-->GreenFirst attempt:[](((gr\/re) U ye)\/(ye U (gr\/re)))

Correct specification:[]( (gr-->(gr U (ye /\ ( ye U re )))) /\(re-->(re U (ye /\ ( ye U gr )))) /\(ye-->(ye U (gr \/ re))))

Needed only when wecan start with yellow

Properties of sequential programs 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/\[]

Automata over finite words 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.

bbS0 S1

The transition relation

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

bbS0 S1

A run over a word 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.

bbS0 S1

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

bbS0 S1

Nondeterministic automaton

Transitions: (S0,a,S0), (S0,b,S0), (S0,a,S1),(S1,a,S1).

What is the language of this automaton?

A,b aaS0 S1

Equivalent deterministic automaton

a,b aaS0 S1

aaS0 S1

Automata over infinite words Similar definition. Runs on infinite words over . Accepts when an accepting state

occurs infinitely often in a run.

bbS0 S1

Automata over infinite words

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.

BBS0 S1

Other runs 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 …, the

run is S0 S0 S0 S0 S1 S1 S1 S1 … What is the run for a b a b b a b b b …?

bbS0 S1

Nondeterministic automaton What is the language of this

automaton? What is the LTL specification if

b -- PC0=CR0, a=¬b?

a,b aaS0 S1

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

Specification using Automata Let each letter correspond to some

propositional property. Example: a -- P0 enters critical section,

b -- P0 does not enter section. []<>PC0=CR0

bbS0 S1

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

b acS0 S1

T0:PC0=L0==>PC0=NC0T1:PC0=NC0/\Turn=0==> PC0:=CR0T2:PC0=CR0==> (PC0,Turn):=(L0,1)T3:PC1==L1==>PC1=NC1T4:PC1=NC1/\Turn=1==> PC1:=CR1T5:PC1=CR1==> (PC1,Turn):=(L1,0)

Initially: PC0=L0/\PC1=L1

The state spaceTurn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

[]¬(PC0=CR0/\PC1=CR1)Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

[](Turn=0 --> <>Turn=1)

Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

Correctness condition

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).

Correctness

All sequences

Program executions

Incorrectness

All sequences

Program executions

Counterexamples

Intersecting M1=(S1,,T1,I1,A1) and M2=(S2,,T2,I2,S2)

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.

Example (all states of second automaton accepting!)

bct0 t1

b,cb,cs0 s1

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

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

bct0 t1

b,cb,cs0 s1

s1,t0b

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

More complicated when A2S2

bct0 t1

b,cb,cs0 s1

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

No, consider (ba)

It should be accepted, but never passes that state.

s0,t0s0,t1

More complicated when A2S2

bct0 t1

b,cb,cs0 s1

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)

s0,t0s0,t1

Intersection - general caseAlso: propositions on nodes

A/\¬Bq0,q3

¬A/\¬Bq1,q3

¬A/\Bq1,q2

A/\¬B

¬A/\B

A\/¬B

q0 and q2 give false.

Version 0: to catch q0Version 1: to catch q2

A/\¬Bq0,q3

¬A/\¬Bq1,q3

¬A/\Bq1,q2

A/\¬Bq0,q3

¬A/\¬Bq1,q3

¬A/\Bq1,q2

Move when see accepting of left (q0)

Move when see accepting of right (q2)

Version 0

Version 1

Make an accepting state in one of the version according to a component accepting state

A/\¬Bq0,q3,0

¬A/\¬Bq1,q3,0

¬A/\Bq1,q2,0

A/\¬Bq0,q3,1

¬A/\¬Bq1,q3 ,1

¬A/\Bq1,q2 ,1

Version 1

Version 0

How to check for emptiness?

Emptiness...

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

Finding accepting runs

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.

Equivalently... 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.

How to complement? 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!

Model Checking under FairnessExpress the fairness as a property φ.

To prove a property ψ under fairness,model check φψ.

Fair (φ)

Bad (¬ψ) Program

Counterexample

Model Checking under Fairness

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.

Dekker’s algorithm

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;

c1=c2=0,turn=1

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.

c1=c2=0,turn=1

What went wrong? 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

Recall:The interleaving model 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.

Fair executions

SequencesExecutions

Some fairness definitions 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.

How to use fairness constraints? 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.

ExampleP1::x=1 P2: do

:: y==0 if

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

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 */

Weak transition fairness

P1::x=1 P2: do :: y==0

if :: true

:: x==1 y=1

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.

Strong transition fairness

P1::x=1 P2: do :: y==0

if :: true

:: x==1 y=1 fi

Under strong transition fairness, 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.

Some fairness definitions 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 = \/ T en

execPi = \/ T exec

“Weaker fairness condition”

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.

Strongtransitionfair execs

Weakprocess

fair execs

Weaktransitionfair execs

Strongprocess

fair execs

Model Checking under fairness 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.

Model Checking under Fairness

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.

Weak transition fairness: similar. Strong fairness: much more difficult

algorithm.

Abstractions

Problems with software analysis Many possible outcomes and

interactions. Not manageable by an algorithm

(undecideable, complex). Requires a lot of practice and

ingenuity (e.g., finding invariants).

More problems

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.

How to alleviate the complexity? Abstraction Compositionality Partial Order Reduction Symmetry

Abstraction

Represent the program using a smaller model.

Pay attention to preserving the checked properties.

Do not affect the flow of control.

Main idea Use smaller data objects.

x:= f(m)y:=g(n)if x*y>0 then … else …x, y never used again.

How to abstract?

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).

Abstraction mapping 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)

Traffic light example

What do we preserve?go

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.

Preserved: [](go->O stop)

Not preserved:

[]<>go

Counterexamples need to be checked.

Symmetry 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.

Using symmetry in analysis Want to find some symmetry group such

that 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.

Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

The quotient model

What is preserved in the following buffer abstraction? What is not preserved?

eempty

Translating from logic to automata

Comment: 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

Why translating?

Want to write the specification in some logic.

Want model-checking tools to be able to check the specification automatically.

Preprocessing 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 (¬).

Semantics of pR q(the Release operator, dual to Until)

qqq qq qq

¬p¬p¬p

¬p ¬p ¬p ¬p ¬p ¬p ¬p ¬p¬p

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

and ¬ ( /\ ) by (¬) \/ (¬)

Eliminate implications, <>, []

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

Example

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 ) )

The data structureIncoming

New Old

NextName

The main idea

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.

How to translate?

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.

SplittingIncoming

New Old

Incoming

New Old

Incoming

New Old

Copy incoming edges, update other field.

EvolvingIncoming

New Old

Incoming

New Old

Copy incoming edges, update other field.

Possible cases: 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 )).

More cases:

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

/\ , evolve: Add to New.

O , evolve: Add to Next.

How to start?Incoming

New Old

aU(bUc)

Incoming

aU(bUc)

Incoming Incoming

aU(bUc)aU(bUc) bUc

aU(bUc)

init init

Incoming

aU(bUc)bUc

init initIncoming Incoming

aU(bUc)aU(bUc) c

When to stop splitting? 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.

Incoming

a,aU(bUc)

aU(bUc)

Incoming

aU(bUc)

Creating a successor node.

How to obtain the automaton?

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.

Incoming

New Old

Node Y

a, b, ¬c

The resulted nodes.

a, aU(bUc) b, bUc, aU(bUc) c, bUc, aU(bUc)

b, bUc c, bUc

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

Acceptance conditions Use “generalized Buchi automata”, where

there 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).

Accepting w.r.t. bUc (green)

b, bUc c, bUc

All nodes with c, or without bUc.

Acceptance w.r.t. aU(bUc)(blue)

b, bUc c, bUc

All nodes with bUc or without aU(bUc).

Translating Multiple Buchi into Buchi automata

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…).

Conclusions 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

Why testing? 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.

Part 1: Testing of black box finite state machine

Know:Transition relationSize or bound on size

Wants to know:In what state we started?In what state we are?Transition relationConformanceSatisfaction of a temporal property

Finite automata (Mealy machines)

S - finite set of states. (size n)– set of inputs. (size d)O – set of outputs, for each transition.(s0 S - initial state). S S - transition relation. S O – output on edge.

Why deterministic machines?

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).

Determinism

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

b/1a/1

Preliminaries: separating sequences

a/0b/1 b/0

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

A: separate to blocks of states with different output.

a/0b/1 b/0

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

Repeat B: Separate blocks based on moving to different blocks.

a/0b/1 b/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.

Want to know the state of the machine (at end). Homing 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.)

Example (homing sequence)

a/0b/1 b/0

a/0{s1, s2, s3}

{s1, s2} {s3}{s1} {s2} {s3}

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

Synchronizing sequence 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.

State identification: 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].

Sometimes cannot identify initial state

b/1a/1 s1

Start with a:in case of being in s1 or s3 we’ll move to s1 and cannot distinguish.Start with b:In case of being in s1 or s2 we’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.

Conformance testing 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.

Check conformance with a given state machine

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).

Conformance testing [Ch,V]

Cannot distinguish if reduced or not.

b/1a/1

Conformance testing (cont.)

Need: bound on number of states of B.

Preparation:Construct a spanning tree

b/1a/1 s1

b/1a/1

How the algorithm works?According to the spanning

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

1. From each state, perform the distinguishing sequences.

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

b/1a/1

Reset or hom

Distinguishing sequences

Comments1. Checking the different distinguishing

sequences (m-1 of them) means each time resetting and returning to the state under experiment.

2. 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.

3. Since there are no more than m states, and according to the experiment, no less than m states, there are m states exactly.

4. Isomorphism between the transition relation is found, hence from minimality the two automata recognize the same languages.

Combination lock automaton

Assume accepting states. Accepts only words with a specific suffix

(cdab in the example).

s1 s2 s3 s4 s5bdc a

Any other input

When only a bound on size of black box is known… Black box can “pretend” to behave

as a specification automaton for a long time, then upon using the right combination, make a mistake.

b/1a/1s1

Pretends to be S3

Pretends to be S1

Conformance testing algorithm [VC]

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. Distinguishing sequences

b/1a/1

Words of length n-m+1

Reset or hom

Model Checking 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.

Buchi automata (-automata) S - finite set of states. (B has l n states) S0 S - initial states. (P has m states) - finite alphabet. (contains p letters) S S - 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.

Example: check a

Example: check <>a

Use automatic translation algorithms, e.g., [Gerth,Peled,Vardi,Wolper 95]

a, a<>a

System

Every element in the product is a counter example for the checked property.

s1,q2 s3,q2

s2,q1a

Acceptance isdetermined byautomaton P.

Model Checking / Testing 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.

Black box checking [PVY]

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.

Experimentsaa

try b a

Simpler problem: deadlock?

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.

Deadlock complexity Nondeterministic algorithm:

Linear time, logarithmic space. Deterministic algorithm:

Exponential (p n-1) time, linear space. Lower bound: Exponential time (use

combination lock automata). How does this conform with what we

know about complexity theory?

Modeling black box checking

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.

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..

Modeling BBC as games Each configuration contains an automaton

and its current state (and more). Moves of the player are labeled with

try a, reset... Moves of the -player withsuccess, fail.

c1 c2 when the automata in c1 and c2 would respond in the same way to the experiments so far.

A naive strategy for BBC 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 new automata,

construct a distinguishing sequence. Only one of them survives.

Complexity: O((n+1)p (n+1)/n!)

On-the-fly strategy 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).

Learning an automaton 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 Mi

and asks for a counterexample. It then generates an automaton with

more states Mi+1 and so forth.

A strategy based on learning 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 v2

. We test B with v1 v2

n. If this succeeds: error, otherwise, this is a counterexample for Mi .

Complexity 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.

Some experiments 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.

Conclusions 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

Part 2: Software testing 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.

Some software testing stages

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.

Some drawbacks of testing

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

Black-Box (data-driven, input-output) testing

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.

testingIs 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!)

Some testing principles 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!

Inspections and Walkthroughs Manual testing methods. Done by a team of people. Performed at a meeting

(brainstorming). Takes 90-120 minutes. Can find 30%-70% of errors.

Code Inspection 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.

Checklist for inspections Data declarationAll variables

declared?Default values

understood?Arrays and strings

initialized?Variables with similar

names?Correct initialization?

Control flowEach loop terminates?DO/END statements

match?

Input/outputOPEN statements

correct?Format specification

correct?End-of-file case handled?

Walkthrough 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.

Selection of test cases (for white-box testing)The main problem is to select a good

coveragecriterion. 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.

Cover all the paths of the program

Infeasible.Consider the flow diagram

on the left.It corresponds to a loop.The loop body has 5 paths.If the loops executes 20times there are 5^20

different paths!May also be unbounded!

How to cover the executions?IF (A>1)&(B=0) THEN X=X/A;

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

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

Statements? Conditions?

Statement coverageExecute every statement at least onceBy choosingA=2,B=0,X=3each 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

Decision coverageEach decision has a true and false outcome at least once.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.

Decision coverage

A=3,B=0,X=3 IF (A>1)&(B=0) THEN X=X/A; END;

Now x=1

Decision coverage

A=2,B=1,X=1

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

Condition coverageEach condition has a true and false value at least once.

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

Condition coverage

A=1,B=0,X=3 IF (A>1) (A>1) & (B=0) THEN X=X/A; END;

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

Condition coverage

A=2,B=1,X=0

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

Can use condition+decision coverage.

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

Multiple Condition CoverageTest all combinations of all conditions in each test.

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

A smaller number of cases: A=2,B=0,X=4 A=2,B=1,X=1 A=1,B=0,X=2 A=1,B=1,X=1Note the X=4 in the firstcase: it is due to the factthat X changes beforebeing used!

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).

Preliminary:Relativizing assertions

(B) : x1= y1 * x2 + y2 /\ y2 >= 0Relativize B) w.r.t. the assignment

becomes B) [Y\g(X,Y)]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’.

(y1,y2)=(0,x1)

Y=g(X,Y)

Verification conditions: tests

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

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

B)= D) /\ y2x2

y2>=x2

Dt(X,Y)

How to find values for coverage?

•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

A=2 | X>1

X=X/Ano

How to find values for coverage?

A>1 & B=0

A=2 | X>1

X=X/Ano

A≠2 /\ X>1

(A≠2 /\ X/A>1) /\ (A>1 & B=0)

A≠2 /\ X/A>1Need to find a

satisfying assignment:A=3, X=6, B=0Can also calculate path condition forwards.

How to cover a flow chart? 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 traversal

Find minimal number of times time to travel each edge using linear programming or dataflow algorithms.Duplicate edges and find an Euler path.

Test cases based on data-flow analysis 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.

z:=z+x

Test case design for black box testing Equivalence partition Boundary value analysis Cause-effect graphs

Equivalence partition 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.

Specificationcondition

Valid equivalenceclass

Invalid equivalenceclass

Example: A legal variable Begins with A-Z Contains [A-Z0-9] Has 1-6 characters.

Starting char

Length

Starts A-Z Starts other

[A-Z0-9] Has others

1-6 chars 0 chars, >6 chars

Equivalence partition (cont.)

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.

Example

AB36P (1,3,5) 1XY12 (2) A17#%X (4)

Starting char

Length

Starts A-Z Starts other

[A-Z0-9] Has others

1-6 chars 0 chars, >6 chars

(6) VERYLONG (7)

Boundary value analysis

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.

Test case generation based on LTL specification

Compiler ModelChecker

Path conditioncalculation

First orderinstantiator

Testmonitoring

Transitions

Path Flow

LTLAut

Goals 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.

Divide and Conquer Intersect property automatonproperty automaton with the

flow chartflow 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).

Spec:¬at l2U (at l2/\ ¬at l2/\(¬at l2U at l2))

¬at l2

l2:x:=x+z

l3:x<t

l1:…

l2:x:=x+z

l3:x<t

l2:x:=x+z

Spec: ¬at l2U (at l2/\ xy /\ (¬at l2/\(¬at l2U at l2 /\ x2y )))

¬at l2

at l2/\xy

¬at l2

at l2/\x2y

l2:x:=x+z

l3:x<t

l1:…

l2:x:=x+z

l3:x<t

l2:x:=x+z

Example: GCD

l1:x:=a

l5:y:=z

l4:x:=y

l3:z:=x rem y

l2:y:=b

l6:z=0? yesno

Example: GCD

l1:x:=a

l5:x:=y

l4:y:=z

l3:z:=x rem y

l2:y:=b

l6:z=0? yesno

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

Why use Temporal specification Temporal specification for sequential

software? Deadlock? Liveness? – No! Captures the tester’s intuitionintuition 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.”

Example: GCD

l1:x:=a

l5:x:=y

l4:y:=z

l3:z:=x rem y

l2:y:=b

l6:z=0? yesno

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

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

Example: GCD

l1:x:=a

l5:x:=y

l4:y:=z

l3:z:=x rem y

l2:y:=b

l6:z=0? yesno

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

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

Path 2: l0l1l2l3l4l5l6l3l4l5l6l7 a>0/\b>0/\a rem b0

Potential explosion

Bad point: potential explosionGood point: may be chopped on-the-fly

Drivers and Stubs 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.

l1:x:=a

l5:x:=y

l4:y:=z

l3:z’=x rem y/\x’=x/\y’=x

l2:y:=b

l6:z=0? yesno

Conclusions 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.

Model Checking

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