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Model Checking. Doron A. Peled University of Warwick Coventry, UK [email protected] 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.

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model checking

Model Checking

Doron A. Peled

University of Warwick

Coventry, UK

[email protected]

modelling and specification for verification and validation
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
Sequential systems.
  • Perform some computational task.
  • Have some initial condition, e.g.,0in A[i] integer.
  • Have some final assertion, e.g.,0in-1 A[i]A[i+1].(What is the problem with this spec?)
  • Are supposed to terminate.
concurrent systems
Concurrent Systems

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).

problems in modeling systems
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
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
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
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 inccommand.
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 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.

Non atomicity
scenario
P1:load R1,a

inc R1

store R1,a

Scenario

P2:load R2,a

inc R2

store R2,a

a=0

R1=0

R2=0

R1=1

R2=1

a=1

a=1

representing transitions
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
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 transition system
  • A (finite) set of variables V over some domain.
  • A set of states S.
  • A (finite) set of transitions T, each transition et has
    • an enabling condition e, and
    • a transformation t.
  • An initial condition I.
example
Example
  • 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?
the interleaving model
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., sisatisfies e.
    • si+1 is obtained by applying t to si.
example1
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

slide17
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)

The transitions

Initially: PC0=L0/\PC1=L1

the state graph successor relation between states

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

The state graph:Successor relation between states.
some observations
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=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

Always ¬(PC0=CR0/\PC1=CR1)(Mutual exclusion)
always if turn 0 the at some point turn 1

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

Always if Turn=0 the at some point Turn=1
always if turn 0 the at some point turn 11

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

Always if Turn=0 the at some point Turn=1
interleaving semantics execute one transition at a time

Turn=0

L0,L1

Turn=0

L0,NC1

Turn=1

L0,NC1

Turn=0

NC0,NC1

Turn=1

L0,CR1

Turn=0

CR0,NC1

Interleaving semantics:Execute one transition at a time.

Need to check the property

for every possible interleaving!

interleaving semantics

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

Interleaving semantics
slide25
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

T1’:PC0=NC0/\Turn=1PC0:=NC0

T2:PC0=CR0(PC0,Turn):=(L0,1)

T3:PC1==L1PC1=NC1

T4:PC1=NC1/\Turn=1PC1:=CR1

T4’:PC1=NC1/\Turn=0PC1:=N1

T5:PC1=CR1(PC1,Turn):=(L1,0)

Busy waiting

Initially: PC0=L0/\PC1=L1

always when turn 0 then sometimes turn 1

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

Always when Turn=0 then sometimes Turn=1

Now it does not hold!

(Red subgraph generates a counterexample execution.)

how can we check the model
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
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
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
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
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
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 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.

Depth First Search
how can we check properties with 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=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

¬(PC0=CR0/\PC1=CR1) is an invariant!
want to do more
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=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 --> <>Turn=1)
slide38

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

slide39

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.
correctness condition
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
Correctness

Sequencessatisfying Spec

Program executions

All sequences

how to prove correctness
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
What do we need to know?
  • How to intersect two automata?
  • How to complement an automaton?
  • How to translate from LTL to an automaton?
properties of 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 system1
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 model1
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
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
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

Semanticsover suffixes of execution





O 

U

combinations
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

b

a

a

a

b

b

a

b

a

b

b

Some relations:
  • [](a/\b)=([]a)/\([]b)
  • But <>(a/\b)(<>a)/\(<>b)
  • <>(a\/b)=(<>a)\/(<>b)
  • But [](a\/b)([]a)\/([]b)
what about
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
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
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 ji, j |= .
  • i |= [] if for each ji, j |= .
  • i |= U  if for some ji, j|=. and for each ik<j, k |=.
then we interpret
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
Spring Example

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 …

ltl satisfaction by a single sequence

release

s1

s2

s3

pull

release

extended

extended

LTL satisfaction by a single sequence

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) ??

ltl satisfaction by a system

release

s1

s2

s3

pull

release

extended

extended

LTL satisfaction by a system

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) ??

the state space

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

The state space
pc0 cr0 pc1 cr1 mutual exclusion

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

[] ¬(PC0=CR0/\PC1=CR1)(Mutual exclusion)
turn 0 turn 11

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 --> <>Turn=1)
interleaving semantics execute one transition at a time1
Interleaving semantics:Execute one transition at a time.

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!

more specifications
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

[](pq)([]p[]q)

[]p(p/\O[]p)

O¬p<-->¬Op

[](pOp)(p[]p)

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

(pUq)<>q

+ propositional logic axiomatization.

+ axiom:p []p

Proof system
traffic light example
Traffic light example

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))

another kind of traffic light

Needed only when we

can start with yellow

Another kind of traffic light

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))))

properties of sequential programs
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

S0

b

S1

a

b

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.
the transition relation

a

S0

b

S1

a

b

The transition relation
  • (S0, a, S0)
  • (S0, b, S1)
  • (S1, a, S0)
  • (S1, b, S1)
a run over a word

a

S0

b

S1

a

b

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.
the language of an automaton

a

S0

b

S1

a

b

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?
nondeterministic automaton

S0

a

S1

A,b

a

Nondeterministic automaton
  • Transitions: (S0,a,S0), (S0,b,S0), (S0,a,S1),(S1,a,S1).
  • What is the language of this automaton?
automata over infinite words

a

S0

b

S1

a

b

Automata over infinite words
  • Similar definition.
  • Runs on infinite words over .
  • Accepts when an accepting state occurs infinitely often in a run.
automata over infinite words1

A

S0

B

S1

A

B

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.
other runs

a

S0

b

S1

a

b

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 …, therun is S0 S0 S0 S0 S1 S1 S1 S1 …
  • What is the run for a b a b b a b b b …?
nondeterministic automaton1

S0

a

S1

a,b

a

Nondeterministic automaton
  • 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?
specification using automata

a

S0

b

S1

a

b

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
mutual exclusion

S0

c

S1

b

a

Mutual Exclusion
  • A -- PC0=CR0/\PC1=CR1
  • B -- ¬(PC0=CR0/\PC1=CR1)
  • C -- TRUE
  • []¬(PC0=CR0/\PC1=CR1)
slide81
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

the state space1

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

The state space
pc0 cr0 pc1 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

[]¬(PC0=CR0/\PC1=CR1)
turn 0 turn 12

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 --> <>Turn=1)
correctness condition1
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).
correctness1
Correctness

Sequencessatisfying Spec

Program executions

All sequences

incorrectness
Incorrectness

Counter

examples

Sequencessatisfying Spec

Program executions

All sequences

intersecting m 1 s 1 t 1 i 1 a 1 and m 2 s 2 t 2 i 2 s 2
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
Example (all states of second automaton accepting!)

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).

slide90

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

more complicated when a 2 s 2
More complicated when A2S2

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.

more complicated when a 2 s 21
More complicated when A2S2

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)

intersection general case also propositions on nodes

A/\¬B

q0,q3

¬A/\B

q1,q2

¬A/\¬B

q1,q3

Intersection - general caseAlso: propositions on nodes

q0

q2

A/\¬B

¬A/\B

q0 and q2 give false.

q1

q3

¬A

A\/¬B

version 0 to catch q 0 version 1 to catch q 2
Version 0: to catch q0Version 1: to catch q2

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

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

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

how to check for emptiness
How to check for emptiness?

a

s0,t0

s0,t1

b

a

c

s1,t1

b

c

emptiness
Emptiness...

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

finding accepting runs
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
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
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 fairness
Model Checking under Fairness

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

Counter

example

Fair (φ)

Bad (¬ψ)

Program

model checking under fairness1
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 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.

Dekker’s algorithm

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.

dekker s algorithm1
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.

Dekker’s algorithm

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

dekker s algorithm2
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.

Dekker’s algorithm

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

dekker s algorithm3
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.

Dekker’s algorithm

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

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

What went wrong?
recall the interleaving model
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 et:
    • 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

some fairness definitions
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
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.
example2
P1::x=1

P2: do

:: y==0  if

:: true :: x==1  y=1

fi od

Example

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=r1pc2=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

fi od

Weak transition fairness

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.

strong transition fairness
P1::x=1

P2: do

:: y==0  if

:: true :: x==1  y=1

fi od

Strong transition fairness

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.

some fairness definitions1
Some fairness definitions
  • Weak transition fairness:/\T (<>[]en []<>exec).Equivalently: /\aT ¬<>[](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

weaker fairness condition
“Weaker fairness condition”

Strongtransitionfair execs

  • A is weaker than B if BA.
  • 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

model checking under fairness2
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 fairness3
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.
problems with software analysis
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
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
How to alleviate the complexity?
  • Abstraction
  • Compositionality
  • Partial Order Reduction
  • Symmetry
abstraction
Abstraction
  • Represent the program using a smaller model.
  • Pay attention to preserving the checked properties.
  • Do not affect the flow of control.
main idea
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
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
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)
slide126

go

Traffic light example

stop

stop

slide127

go

go

stop

stop

stop

what do we preserve
What do we preserve?

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

preserved go o stop
Preserved: [](go->O stop)

go

Not preserved:

[]<>go

Counterexamples need to be checked.

go

stop

stop

stop

symmetry
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->[email protected]: 1->3, 2->2, 3->1
  • A set of permutations with @ is called a symmetry group.
using symmetry in analysis
Using symmetry in analysis
  • 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.
slide132

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

slide133

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

slide135

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

why translating
Why translating?
  • Want to write the specification in some logic.
  • Want model-checking tools to be able to check the specification automatically.
preprocessing
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 p r q the release operator dual to until
Semantics of pR q(the Release operator, dual to Until)

¬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

slide139
Replace ¬T by F, and ¬F by T.
  • Replace ¬ ( \/ ) by (¬) /\ (¬) and¬ ( /\ ) by (¬) \/ (¬)
eliminate implications
Eliminate implications, <>, []
  • Replace  ->  by (¬) \/ .
  • Replace <> by (T U).
  • Replace [] by (F R).
example3
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 main idea
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
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.
splitting

Incoming

New

Old

Next

Incoming

Incoming

New

Old

New

Old

Next

Next

Splitting

Copy incoming edges, update other field.

evolving

Incoming

New

Old

Next

Incoming

New

Old

Next

Evolving

Copy incoming edges, update other field.

possible cases
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
More cases:
  •  \/ , split:
    • Add  to New.
    • Add  to New.
  •  /\ , evolve:
    • Add  to New.
  • O , evolve:
    • Add  to Next.
how to start
How to start?

init

Incoming

New

Old

aU(bUc)

Next

slide150

Incoming

Incoming

a

aU(bUc)

bUc

aU(bUc)

aU(bUc)

init

Incoming

aU(bUc)

init

init

slide151

init

Incoming

bUc

aU(bUc)

init

init

Incoming

Incoming

b

aU(bUc)

c

aU(bUc)

(bUc)

when to stop splitting
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.
slide153

init

Incoming

a,aU(bUc)

Creating a successor node.

aU(bUc)

Incoming

aU(bUc)

how to obtain the automaton

Incoming

New

Old

Next

How to obtain the automaton?

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

the resulted nodes

a, aU(bUc)

b, bUc, aU(bUc)

c, bUc, aU(bUc)

b, bUc

c, bUc

The resulted nodes.
slide156

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
Acceptance conditions
  • 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).
accepting w r t buc green

a, aU(bUc)

b, bUc, aU(bUc)

c, bUc, aU(bUc)

b, bUc

c, bUc

All nodes with c, or without bUc.

Accepting w.r.t. bUc(green)
acceptance w r t au buc blue

a, aU(bUc)

b, bUc, aU(bUc)

c, bUc, aU(bUc)

b, bUc

c, bUc

All nodes with bUc or without aU(bUc).

Acceptance w.r.t. aU(bUc)(blue)
translating multiple buchi into buchi automata
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
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.
why testing
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
Part 1: Testing of black box finite state machine

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
finite automata mealy machines
Finite automata (Mealy machines)

S - finite set of states. (size n)

S– set of inputs. (size d)

O– set of outputs, for each transition.

(s0S - initial state).

  • S  S S - transition relation.

 S  S O – output on edge.

why deterministic machines
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).
determ i nism

a/1

b/1

a/1

Determinism

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

preliminaries separating sequences
Preliminaries: separating sequences

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}.

a separate to blocks of states with different output
A: separate to blocks of states with different output.

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}.

repeat b separate blocks based on moving to different blocks
Repeat B: Separate blocks based on moving to different blocks.

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.

want to know the state of the machine at end homing 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

b/1

s1

s2

a/0

b/1

b/0

a/0

s3

a/0

Example (homing sequence)

{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.

synchronizing sequence
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
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/1

a/1

b/1

s1

s2

a/1

a/1

s3

b/0

Sometimes cannot identify initial state

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.

c onformance testing
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

b/1

a/1

b/1

s1

s2

?=

a/1

a/1

s3

b/0

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

a/1

a/1

b/1

b/1

Conformance testing [Ch,V]

a/1

b/1

b/1

a/1

a/1

b/1

Cannot distinguish if reduced or not.

conformance testing cont

a

Conformance testing (cont.)

b

b

a

a

a

a

b

b

a

a

b

Need: bound on number of states of B.

preparation construct a spanning tree

b/1

a/1

b/1

s1

s1

s2

a/1

b/1

a/1

s3

s2

a/1

s3

b/0

Preparation:Construct a spanning tree
how the algorithm works
How the algorithm works?

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

comments
Comments
  • 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.
combination lock automaton
Combination lock automaton
  • 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

when only a bound on size of black box is known
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.

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

conformance testing algorithm vc
Conformance testing algorithm [VC]

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

model checking1
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 w automata
Buchi automata (w-automata)

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

S0S - initial states. (P has m states)

S - finite alphabet. (contains p letters)

d S  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

a

<>a

a

a, a

example check a1

a

a

a

a

Example: check <>a

<>a

example check a2
Example: check  <>a

a, a

<>a

a

a

Use automatic translation algorithms, e.g.,

[Gerth,Peled,Vardi,Wolper 95]

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

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

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.

Model Checking / Testing
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.

Black box checking [PVY]



experiments

reset

a

a

b

b

c

c

try c

try b

a

a

b

b

c

c

a

a

b

c

b

c

fail

Experiments
simpler problem deadlock
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
Deadlock complexity
  • 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?
modeling black box checking
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
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
Modeling BBC as games
  • 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.
  • [email protected] c2 when the automata in c1and c2 would respond in the same way to the experiments so far.
a naive strategy for bbc
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 newautomata, construct a distinguishing sequence. Only one of them survives.
  • Complexity: O((n+1)p (n+1)/n!)
on the fly strategy
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).

v1

v2

learning an automaton
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 Miand asks for a counterexample.
  • It then generates an automaton with more states Mi+1and so forth.
a strategy based on learning
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 v2w . We test B with v1 v2n. If this succeeds: error, otherwise, this is a counterexample for Mi .
complexity
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
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.
conclusions1
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.
part 2 software testing
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
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
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 task.
black box data driven input output testing
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.

testing
testing

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!)

some testing principles
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
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
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 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?

Checklist for inspections
walkthrough
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
Selection of test cases (for white-box testing)

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.
cover all the paths of the program
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 20

times there are 5^20 different paths!

May also be unbounded!

how to cover the executions
How to cover the executions?

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?
statement coverage execute every statement at least once
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;

Statement coverageExecute every statement at least once

Now x=1.5

decision coverage each 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.

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

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

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

Decision coverage

Now x=1

decision coverage1
A=2,B=1,X=1 

The case where A1 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;

Decision coverage
condition coverage each 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)&(B=0) THEN X=X/A; END;

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

Condition coverageEach condition has a trueand false value at least once.
condition coverage
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;

Condition coverage
condition coverage1
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;

Condition coverage
multiple condition coverage test all combinations of all conditions in each test
A>1,B=0

A>1,B≠0

A1,B=0

A1,B≠0

A=2,X>1

A=2,X1

A≠2,X>1

A≠2,X1

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

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

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

A smaller number of cases:

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
Preliminary:Relativizing assertions

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

verification conditions tests
Verification conditions: tests

B

T

F

C)  B)= t(X,Y) /\C)

D)  B)=t(X,Y) /\ D)

B)= D) /\ y2x2

t(X,Y)

C

D

B

T

F

y2>=x2

C

D

how to find values for coverage
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

no

yes

X=X/A

A=2 | X>1

true

no

yes

X=X+1

true

how to find values for coverage1
How to find values for coverage?

(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

how to cover a flow chart
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 traversalFind 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
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.

X:=3

t>y

x>y

z:=z+x

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

Valid equivalence

class

Invalid equivalence

class

Specification

condition

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.
example a legal variable
Example: A legal variable
  • 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

equivalence partition cont
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.

Valid equivalence

class

Invalid equivalence

class

Specification

condition

example4
(6)

VERYLONG (7)

AB36P (1,3,5)

1XY12 (2)

A17#%X (4)

Example

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

boundary value analysis
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

LTLAut

Model

Checker

Path

Path condition

calculation

Flowchart

Compiler

Transitions

First order

instantiator

Test

monitoring

Test case generation based on LTL specification
goals
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
Divide and Conquer
  • 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).
spec at l 2 u at l 2 at l 2 at l 2 u at l 2

l2:x:=x+z

l3:x<t

l1:…

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

l2:x:=x+z

¬at l2

X

=

at l2

l3:x<t

¬at l2

l2:x:=x+z

at l2

spec at l 2 u at l 2 x y at l 2 at l 2 u at l 2 x 2 y

l2:x:=x+z

l3:x<t

l1:…

Spec: ¬at l2U (at l2/\ xy /\ (¬at l2/\(¬at l2U at l2 /\ x2y )))

xy

l2:x:=x+z

¬at l2

X

=

at l2/\

xy

l3:x<t

x2y

¬at l2

l2:x:=x+z

at l2/\

x2y

example gcd

l0

Example: GCD

l1:x:=a

l2:y:=b

l3:z:=x rem y

l4:x:=y

l5:y:=z

no

l6:z=0?

yes

l7

example gcd1

l0

Example: GCD

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

why use temporal specification
Why use Temporal specification
  • 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.”
example gcd2

l0

Example: GCD

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

example gcd3

l0

Example: GCD

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 b0

l4:y:=z

l5:x:=y

no

l6:z=0?

yes

l7

potential explosion
Potential explosion

Bad point: potential explosion

Good point: may be chopped on-the-fly

drivers and stubs
Drivers and Stubs

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

conclusions2
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.
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