1 Softrare Reliability Methods Prof. Doron A. Peled Bar Ilan University, Israel And Univeristy of...

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Softrare Reliability Methods

Prof. Doron A. PeledBar Ilan University,IsraelAndUniveristy of warwick,UK

Version 2008

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Some related books:

Also:Mainly:

http://www.dcs.warwick.ac.uk/~doron

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Goal: software reliability

Use software engineering methodologies to develop the code.

Use formal methods during code development

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What are formal methods?

Techniques for analyzing systems, based on some mathematics.

This does not mean that the user must be a mathematician.

Some of the work is done in an informal way, due to complexity.

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Examples for FM

Deductive verification:Using some logical formalism, prove formally that the software satisfies its specification.

Model checking:Use some software to automatically check that the software satisfies its specification.

Testing:Check executions of the software according to some coverage scheme.

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Typical situation:

Boss: Mark, I want that the new internet marketing software will be flawless. OK?

Mark: Hmmm. Well, ..., Aham, Oh! Ah??? Where do I start?

Bob: I have just the solution for you. It would solve everything.

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Some concerns

Which technique? Which tool? Which experts? What limitations? What methodology? At which points? How expensive? How many people?

Needed expertise. Kind of training. Size limitations. Exhaustiveness. Reliability. Expressiveness. Support.

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Badmouth

Formal methods can only be used by mathematicians.

The verification process is itself prone to errors, so why bother?

Using formal methods will slow down the project.

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Some answers...

Formal methods can only be used by mathematicians.

Wrong. They are based on some math but the user should not care.

The verification process is itself prone to errors, so why bother?

We opt to reduce the errors, not eliminate them.Using formal methods will slow down the

project.Maybe it will speed it up, once errors are found

earlier.

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Some exaggerations

Automatic verification can always find errors.

Deductive verification can show that the software is completely safe.

Testing is the only industrial practical method.

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Our approach

Learn several methods (deductive verification, model checking, testing process algebra).

Learn advantages and limitations, in order to choose the right methods and tools.

Learn how to combine existing methods.

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Emphasis

The process:Selecting the tools, Modeling,Verification, Locating errors.

Use of tools:Hands on. PVS, SPIN

Visual notation:Statecharts, MSCs, UML.

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Some emphasis

The process of selecting and using formal methods.

The appropriate notation. In particular, visual notation.

Hands-on experience with tools.

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The unbearable easiness of grading

During class, choose some small project in groups, e.g.,Explore some examples using tools.Implementing a simple algorithm.Dealing with new material.

or covering advanced subject.- Office presentation (1 hour).- Written description (2-3 pages +

computer output or 6-10 pages).- Class presentation (0.5-1.5 hours per

group).

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Example topics

Project:Verify some example

using some tools.Communication

protocols.Mutual exclusion.

Advanced topics:Abstractions.Reductions.Partitions.Static analysis.Verifying pushdown

automata.Verifying security

protocols.

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Where do we start?

Boss: Mark, can you verify this for me?

Mark: OK, first I have to ...

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Things to do

Check the kind ofsoftware to analyze.

Choose methods and tools.

Express system properties.

Model the software.

Apply methods.Obtain verification

results.Analyze results.Identify errors.Suggest correction.

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Different types of software

Sequential.Concurrent.Distributed.Reactive.Protocols.Abstract algorithms.Finite state.

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Specification:Informal, textual, visual

The value of x will be between 1 and 5, until some point where it will become 7. In any case it will never be negative.

(1<=x<=5 U (x=7/\ [] x>=0))

1<=x<=5 X=7

X>=0

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Verification methods

Finite state machines. Apply model checking.

Apply deductive verification (theorem proving).

Program too big, too complicated.Apply testing techniques.

Apply a combination of the above!

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Modeling

Use the program text.Translate to a programming language

embedded in some proof system.Translate to some notation (transition

system).Translate to finite automata.Use visual notation.Special case: black box system.

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Modeling Software Systems for Analysis

(Book: Chapter 4)

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Modelling and specification for verification and validation

How to specify what the software is supposed to do?

Can we use the UML model or parts of it?

How to model it in a way that allows us to check it?

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Systems of interest Sequential systems. Concurrent systems (multi-threaded).

1. Distributive systems.2. Reactive systems.3. Embedded systems

(software + hardware).

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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.

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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).

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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.

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Modeling.The states based model.

V={v0,v1,v2, …} - a set of variables, over some domain.

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,v1=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.

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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>…

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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.

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Non atomicity

Execute the following when a=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,aP2:load R2,a inc R2 store R2,a a may be also 1.

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Scenario

P1:load R1,a

inc R1

store R1,a

P2:load R2,a inc R2 store R2,a

a=0

R1=0

R2=0

R1=1

R2=1

a=1

a=1

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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.

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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.

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A transition system

A (finite) set of variables V over some domain.

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.

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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?

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The interleaving model

An execution is a maximal finite or infinite sequence of states s0, s1, s2, …That is: finite if nothing is enabled from the last state.

The first state s0 satisfies the initial condition, I.e., I (s0).

Moving from one state si to its successor si+1 is by executing a transition et: e (si), i.e., si satisfies e. si+1 is obtained by applying t to si.

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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

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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

Is this the only reasonable way to model this program?

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The state graph:Successor relation between reachable 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,L1T0 T0T3 T3

T1 T4T3

T0 T3

T0

T0 T4T1 T3

T2

T2

T5

T5

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Some important points

Reachable states: obtained from an initial state through a sequence of enabled transitions.

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.

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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

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Always if Turn=0 then at some point 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

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Always if Turn=0 then at some point 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

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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 property

for every possible interleaving!

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Interleaving semantics

Turn=0L0,L1

Turn=0L0,NC1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=1L0,CR1

Turn=1L0,NC1

Turn=0L0,L1

Turn=0L0,NC1

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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:=NC1T5:PC1=CR1(PC1,Turn):=(L1,0)

Initially: PC0=L0/\PC1=L1

Busy waiting

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Always when Turn=0 then at some point Turn=1

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

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,L1T4’ T1’

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Combinatorial explosion

V1:=1

V1:=3

V1:=2

Vn:=1

Vn:=3

Vn:=2…

How many states?

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Global states

3n states

v1=1,v2=1…vn=1

v1=2,v2=1…vn=1 v1=1,v2=1…vn=2…

v1=3,v2=1…vn=1 …

…

v1=1,v2=1…vn=3

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Specification Formalisms

(Book: Chapter 5)

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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?

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A transition system

A (finite) set of variables V. 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. Denote by R(s, s’) the fact that s’ is a

successor of s.

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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.

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

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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.

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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.)

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Semantics over suffixes of execution

O

U

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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.

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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”.

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

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

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What about

([]<>)/\([]<>)=[]<>(/\)?

([]<>)\/([]<>)=[]<>(\/)?

(<>[])/\(<>[])=<>[](/\)?

(<>[])\/(<>[])=<>[](\/)?

No, just

Yes!!!Yes!!!

No, just

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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 it is not the case that 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 |=.

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Then we interpret: For a state:

s|=p as in propositional logic. For an execution:|= is interpreted over a sequence, as in previous slide.

For a system/program:P|=holds if |= for every sequence of P.

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Spring Example

s1 s3s2

pull

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 …

…

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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.

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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.

+ proof rule: _p_ []p

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Traffic light example

Green Yellow Red Always has exactly one light:

[](¬(gr/\ye)/\¬(ye/\re)/\¬(re/\gr)/\(gr\/ye\/re))Correct change of color:

[]((grU ye)\/(yeU re)\/(reU gr))

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Another kind of traffic light

GreenYellowRedYellowFirst 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

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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/\[]

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Automata over finite words A=<, S, , I, F> (finite) - the alphabet. S (finite) - the states. S x x S - the transition relation. I S - the starting states. F S - the accepting states.

a

a

b

bs0 s1

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The transition relation

(s0, a, s0) (s0, b, s1) (s1, a, s0) (s1, b, s1)

a

a

b

bs0 s1

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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. Follows the transition relation (si, ci , si+1). Accepting if ends at accepting state.

a

a

b

bs0 s1

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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?

a

a

b

bs0 s1

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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

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Equivalent deterministic automaton

b

a

as0 s1

b

a,b a as0 s1

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

occurs infinitely often in a run.

a

a

b

bs0 s1

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Automata over infinite words Consider the word abababab… There is a run s0s0s1s0s1s0s1 … This run in accepting, since s0

appears infinitely many times.

a

a

b

bs0 s1

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Other runs For the word bbbbb… the run is

s0 s1 s1 s1 s1… and is not accepting. For the word aaabbbbb …, the

run is s0 s0 s0 s0 s1 s1 s1 s1 … What is the run for ababbabbb …?

a

a

b

bs0 s1

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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?

a,b a as0 s1

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No deterministic automaton for (a+b)*aω

In a deterministic automaton there is one run for each word.

After some sequence of a’s, i.e., aaa…a must reach some accepting state.

Now add b, obtaining aaa…ab. After some more a’s, i.e., aaa…abaaa…a must

reach some accepting state. Now add b, obtaining aaa…abaaa…ab. Continuing this way, one obtains a run that

has infinitely many b’s but reaches an accepting state(in a finite automaton, at least one would repeat) infinitely often.

83

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

a

a

b

bs0 s1

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Mutual Exclusion

a -- PC0=CR0/\PC1=CR1 b -- ¬(PC0=CR0/\PC1=CR1) c -- true []¬(PC0=CR0/\PC1=CR1)

b a cs0 s1

85


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

Apply now to our program:

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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

87

[]¬(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

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[](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

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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 property

for every possible interleaving!

90

[](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

91

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).

92

Correctness

All sequences

Sequences satisfying Spec

Program executions

93

Incorrectness

All sequences


Program executions

Counter

examples

94

Automatic Verification

(Book: Chapter 6)

95

How can we check the model?

The model is a graph. The specification should refer the

the graph representation. Apply graph theory algorithms.

96

What properties can we check?

Invariant: a property that needs 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.

97

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!

98

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.

99

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.

100

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.

101

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.

102

Start from an initial state

q3

q4

q2

q1

q5

q1

q1

Stack:

Hash table:

103

Continue with a successor

q3

q4

q2

q1

q5

q1 q2

q1

q2

Stack:

Hash table:

104

One successor of q2.

q3

q4

q2

q1

q5

q1 q2 q4

q1

q2

q4

Stack:

Hash table:

105

Backtrack to q2 (no new successors for q4).

q3

q4

q2

q1

q5

q1 q2 q4

q1

q2

Stack:

Hash table:

106

Backtracked to q1

q3

q4

q2

q1

q5

q1 q2 q4

q1

Stack:

Hash table:

107

Second successor to q1.

q3

q4

q2

q1

q5

q1 q2 q4 q3

q1

q3

Stack:

Hash table:

108

Backtrack again to q1.

q3

q4

q2

q1

q5

q1 q2 q4 q3

q1

Stack:

Hash table:

109

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.

110

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

111

¬(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

112

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.

113

[](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

114

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

init

New initial stateConvert graph into Buchi automaton

115

Turn=0L0,L1

Turn=1L0,L1

init

•Propositions are attached to incoming nodes.

•All nodes are accepting.

Turn=1L0,L1

Turn=0L0,L1

116

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).

117

Correctness

All sequences


Program executions

118

How to prove correctness?

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

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

translating from LTL!

119

What do we need to know?

How to intersect two automata? How to complement an

automaton? How to translate from LTL to an

automaton?

120

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.

121

Example (all states of second automaton accepting!)

a

bct0 t1

a

a

b,c

b,cs0 s1

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

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

122

a

bct0 t1

a

a

b,c

b,cs0 s1

s0,t0

s0,t1

s1,t1

s1,t0b

b

a

c

a

c

123

More complicated when A2S2

a

b

ct0 t1

a

a

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,t0

s0,t1

s1,t1

b

a

c

a

c

124

More complicated when A2S2

a

b

ct0 t1

a

a

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,t0

s0,t1

s1,t1

b

c

a

c

a

125

Intersection - general case

q0 q2

q3q1

q0,q3 q1,q3q1,q2

a a, c

c

c, bb

c

c

b

a

126

Version 0: to catch q0


q0,q3 q1,q3q1,q2

q0,q3 q1,q3q1,q2

Move when see accepting of left (q0)

Move when see accepting of right (q2)

Version 0

Version 1

c

c

c

c

b

a

b

a

127



q0,q3 q1,q3q1,q2

q0,q3 q1,q3q1,q2

Move when see accepting of left (q0)

Move when see accepting of right (q2)

Version 0

Version 1

c

c

c

c

b

a

b

a

128

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

q0,q3,0 q1,q3,0q1,q2,0

q0,q3,1 q1,q3 ,1q1,q2 ,1

Version 1

Version 0

c

c

c

c

b

ab

a

129

How to check for emptiness?

s0,t0

s0,t1

s1,t1

b

a

c

a

c

130

Emptiness...

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

131

Strongly Connected Component (SCC)

A set of states with a path between each pair of them.

Can use Tarjan’s DFS algorithm for finding maximal SCC’s.

132

Finding accepting runs

If there is an accepting run, then at least one accepting state repeats on it forever.

Look at a suffix of this run where all the states appear infinitely often.

These states form a strongly connected component on the automaton graph, including an accepting state.

Find a component like that and form an accepting cycle including the accepting state.

133

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.

134

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!

135

Model Checking under Fairness

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

Fair (φ)

Bad (¬ψ) Program

Counter

example

136


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.

137

Translating from logic to automata

(Book: Chapter 6)

138

Why translating?

Want to write the specification in some logic.

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

139

Generalized Büchi automata

Acceptance condition F is a setF={f1 , f2 , … , fn } where each fi is a set of states.

To accept, a run needs to pass infinitely often through a state from every set fi .

140

Translating into simple Büchi automaton

q0 q2q1

q0 q2q1

Version 0

Version 1

c

c

c

c

b

a

b

a

141


q0 q2q1

q0 q2q1

Version 0

Version 1

c

c

c

c

b

b

a

142


q0 q2q1

q0 q2q1

Version 0

Version 1

c

c

c

c

b

a

b

a

143

Preprocessing

Convert into normal form, where negation only applies to propositional variables.

¬[] becomes <>¬. ¬<> becomes [] ¬. What about ¬ ( U )? Define operator R such that

¬ ( U ) = (¬) R (¬), ¬ ( R ) = (¬) U (¬).

144

Semantics of pR q

p

qqq qq qq

q

qq

q qqq

¬p¬p¬p

¬p ¬p ¬p ¬p ¬p ¬p ¬p ¬p¬p

¬p

145

Replace ¬true by false, and ¬false by true.

Replace ¬ ( \/ ) by (¬) /\ (¬) and ¬ ( /\ ) by (¬) \/ (¬)

146

Eliminate implications, <>, []

Replace -> by (¬ ) \/ . Replace <> by (true U ). Replace [] by (false R ).

147

Example

Translate ( []<>P ) ( []<>Q ) Eliminate implication ¬( []<>P ) \/ ( []<>Q ) Eliminate [], <>:

¬( false R ( true U P ) ) \/ ( false R ( true U Q ) )

Push negation inwards:(true U (false R ¬ P ) ) \/ ( false R ( true U Q ) )

148

The data structure

Incoming

New Old

NextName

149

The main idea

U = \/ ( /\ O ( U ) ) R = /\ ( \/ O ( R ) ) This separates the formulas to two

parts:one holds in the current state, and the otherin the next state.

150

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.

151

Splitting

Incoming

New Old

Next

Incoming

New Old

Next

Incoming

New Old

Next

Copy incoming edges, update other field.

152

Evolving

Incoming

New Old

Next

Incoming

New Old

Next

Copy incoming edges, update other field.

153

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 )).

154

More cases:

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

/\ , evolve: Add to New.

O , evolve: Add to Next.

155

How to start?

Incoming

New Old

Next

init

aU(bUc)

156

Incoming

init

aU(bUc)

Incoming Incoming

aU(bUc)aU(bUc) bUc

aU(bUc)

a

init init

157

Incoming

aU(bUc)bUc

init initIncoming Incoming

aU(bUc)aU(bUc) c

(bUc)

bbUc bUc

158

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.

159

Incoming

a,aU(bUc)

aU(bUc)

init

Incoming

aU(bUc)

Creating a successor node.

When we enter to Nodes a new node (with different Old or Next than any other node), we start a new node by copying Next to New, and making an edge to the new successor.

160

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 node is init.

Incoming

New Old

Next

X

Node Y

a, b, ¬c

161

The resulted nodes.

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

b, bUc c, bUc

162


b, bUc c, bUc

All nodes with incoming edge from “init”.

Initial nodes

163

Include only atomic propositions

Init

a

b

c

cb

164

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 .

165

Accepting w.r.t. bU c


b, bUc c, bUc

All nodes with c, or without bUc.

166

Acceptance w.r.t. aU (bU c)


b, bUc c, bUc

All nodes with bUc or without aU(bUc).

167

The SPIN System

168

What is SPIN?

Model-checker. Based on automata theory. Allows LTL or automata

specification Efficient (on-the-fly model

checking, partial order reduction). Developed in Bell Laboratories.

169

Documentation

Paper: The model checker SPIN,G.J. Holzmann, IEEE Transactions on Software Engineering, Vol 23, 279-295.

Web: http://www.spinroot.com

170

The language of SPIN

The expressions are from C. The communication is from CSP. The constructs are from Guarded

Command.

171

Expressions

Arithmetic: +, -, *, /, % Comparison: >, >=, <, <=, ==,

!= Boolean: &&, ||, ! Assignment: = Increment/decrement: ++, --

172

Declaration

byte name1, name2=4, name3; bit b1,b2,b3; short s1,s2; int arr1[5];

173

Message types and channels

mtype = {OK, READY, ACK} mtype Mvar = ACK

chan Ng=[2] of {byte, byte, mtype}, Next=[0] of {byte}Ng has a buffer of 2, each message consists of two bytes and an enumerable type (mtype).Next is used with handshake message passing.

174

Sending and receiving a message

Channel declaration:

chan qname=[3] of {mtype, byte, byte}

In sender:

qname!tag3(expr1, expr2)or equivalently:qname!tag3, expr1, expr2

In Receiver:

qname?tag3(var1,var2)

175

Defining an array of channels

Channel declaration: chan qname=[3] of {mtype, byte,

byte}defines a channel with buffer size 3.

chan comm[5]=[0] of {byte, byte}defines an array of channels (indexed 0 to 4. Communication is synchronous (handshaking), meaning that the sender waits for the receiver.

176

Condition

if:: x%2==1 -> z=z*y; x--:: x%2==0 -> y=y*y; x=x/2fi

If more than one guard is enabled: a nondeterministic choice.

If no guard is enabled: the process waits (until a guard becomes enabled).

177

Looping

do:: x>y -> x=x-y:: y>x -> y=y-x:: else breakod;

Normal way to terminate a loop: with break. (or goto).

As in condition, we may have a nondeterministic loop or have to wait.

178

Processes

Definition of a process: proctype prname (byte Id; chan Comm){ statements}

Activation of a process:run prname (7, Con[1]);

179

init process is the root of activating all others

init { statements }init {byte I=0; atomic{do ::I<10 -> run prname(I, chan[I]);

I=I+1 ::I=10 -> break; od}}atomic allows performing several actions as

one atomic step.

180

Exmaples of Mutual exclusion

Reference:A. Ben-Ari, Principles of Concurrent

and Distributed Programs, Prentice-Hall 1990.

181

General structure

loop

Non_Critical_Section;

TR:Pre_Protocol; CR:Critical_Section; Post_protocol;end loop;

Propositions:inCRi, inTRi.

182

Properties

loop

Non_Critical_Section;

TR:Pre_Protocol; CR:Critical_Section; Post_protocol;end loop;

Assumption:~<>[]inCRiRequirements:[]~(inCR0/\inCR1)[](inTRi<>inCRi)Not assuming:[]<>inTRi

183

Turn:bit:=1;

task P0 is

begin

loop

Non_Critical_Sec;

Wait Turn=0;

Critical_Sec;

Turn:=1;

end loop

end P0.

task P1 is

begin

loop

Non_Critical_Sec;

Wait Turn=1;

Critical_Sec;

Turn:=0;

end loop

end P1.

184

Translating into SPIN

#define critical (incrit[0] ||incrit[1])

byte turn=0, incrit[2]=0;proctype P (bool id){ do :: 1 -> do :: 1 -> skip :: 1 -> break od;

try:do ::turn==id -> break od; cr:incrit[id]=1; incrit[id]=0; turn=1-turn od}init { atomic{ run P(0); run P(1) } }

185

The leader election algorithm

A directed ring of computers. Each has a unique value. Communication is from left to right.

Find out which value is the greatest.

186

Example

7

2

312

9

4

187

Informal description:

Initially, all the processes are active.

A process that finds out it does not represent a value that can be maximal turns to be passive.

A passive process just transfers values from left to right.

188

More description

The algorithm executes in phases. In each phase, each process first sends

itscurrent value to the right.

Each process, when receiving the first value from its left compares it to its current value. If same: this is the maximum. Tell others. Not same: send current value again to left.

189

Continued

When receiving the second value: compare the three values received. These are values of the process itself. of the left active process. of the second active process on the left.

If the left active process has greatest value among three, then keep this value. Otherwise, become passive.

190

7

2

312

9

4

3

2

9

7

4

12

191

7

2

312

9

4

3, 7

2, 9

9, 4

7, 2

4, 12

12, 3

192

7

2

312

9

4

3, 7

2, 9

9, 4

7, 2

4, 12

12, 3

193

9

7

12

194

9

7

12

12, 7

7, 9

9, 12

195

12

196

send(1, my_number);state:=active;when

received(1,number) do

if state=active then if number!=max then send(2, number); neighbor:=number; else (max is

greatest, send to all processes); end if; else send(1,number); end if;end do;

when received(2,number) do

if state=active then if neighbor>number

and neighbor>max then

max:=neighbor; send(1, neighbor); else state:=passive; end if; else send(2, number); end if;end do;

197

Now, translate into SPIN (Promela) code

198

Running SPIN

Can download and implement (for free) using www.spinroot.com

Available in our system. Graphical interface: xspin

199

Homework: check properties

There is never more than one maximal value found.

A maximal value is eventually found.

From the time a maximal value is found, we continue to have one maximal value.

There is no maximal value until a moment where there is one such value, and from there, there is exactly one value until the end.

The maximal value is always 5.

200

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

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

boolean c1 initially 1;


integer (1..2) turn initially 1;

201

Project

Model in Spin Specify properties Do model checking Can this work without fairness? What to do with fairness?

202

Modeling issues

Book: chapters 4.12, 5.4, 8.4, 10.1

203

Fairness

(Book: Chapter 4.12, 8.3, 8.4)

204


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.






205









c1=c2=0,turn=1

206









c1=c2=0,turn=1

207






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

208

0:START P10:START P1

11:c1:=111:c1:=1

12:true12:true

13:end13:end2:c1:=02:c1:=0

8:c2=0?8:c2=0?

7:turn=2?7:turn=2?

6:c1:=06:c1:=0

3:c1:=13:c1:=1

11:turn:=211:turn:=2

10:c1:=110:c1:=1

9:critical-19:critical-1

4:no-op4:no-op

5:turn=2?5:turn=2?

nono

nono

nono

nonoyesyes

yesyes

yesyes

yesyes

0:START P20:START P2

11:c2:=111:c2:=1

12:true12:true

13:end13:end2:c2:=02:c2:=0

8:c1=0?8:c1=0?

7:turn=1?7:turn=1?

6:c2:=06:c2:=0

3:c2:=13:c2:=1

11:turn:=111:turn:=1

10:c2:=110:c2:=1

9:critical-29:critical-2

4:no-op4:no-op

5:turn=1?5:turn=1?

nono

nono

nono

nonoyesyes

yesyes

yesyes

yesyes

Initially:

turn=1

209

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

210

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

Sequences Executions

211

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.

212

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.

213

Example

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

if :: true

:: x==1 -> y=1 fi

od

Initially: x=0; y=0;

In order for the loop to terminate (in a deadlock !) 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=0pc2=r1 /* y==0*/

pc2=r1pc2=r0 /* true */

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

214

Weak transition fairness

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

if :: true

:: x==1 -> y=1

fi od


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 P1 to execute, but P2 can still choose the true guard.Strong process fairness: same.

215

Strong transition fairness

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

if :: true

:: x==1 -> y=1 fi

od


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.

216

Specifying fairness conditions

Express properties over an alternating sequence of states and transitions:s0 1 s1 1 s2 …

Use transition predicates exec.

217

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 = \/ Pi en

execPi = \/ Pi exec

218

“Weaker fairness condition”

A is weaker than B if B A.(Means A has more executions than B.)

Consider the executions L(A) and L(B). Then L(B) L(A).

If an execution is strong {process/transition} fair, then 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

219

Fairness is an abstraction; no scheduler can guarantee exactly all fair executions!

Initially: x=0, y=0P1::x=1

||P2::do

:: x==0 -> y=y+1 :: x==1 -> break od

x=0,y=0

x=0,y=1x=1,y=0

x=1,y=1x=0,y=2

x=1,y=2Under fairness assumption (any of the four defined),P1 will execute the assignment, and consequently, P2 will terminate. All executions are finite and there are infinitely many of them, and infinitely many states. Thus, an execution tree (the state space) will potentially look like the one on the right, but with infinitely many states, finite branching and only finite sequences. But according to König’s Lemma there is no such tree!

220

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.

221


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.

222

Abstractions

(Book: Chapter 10.1)

223

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).

224

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.

225

How to alleviate the complexity?

Abstraction Compositionality Partial Order Reduction Symmetry

226

Abstraction

Represent the program using a smaller model.

Pay attention to preserving the checked properties.

Do not affect the flow of control.

227

Main idea

Use smaller data objects.

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

228

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).

229

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)

230

Traffic light example

go

stop

stop

231

go

stop

stop

go

stop

232

What do we preserve?

go

stop

stop

go

stop

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

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

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

233

Preserved: [](go->O stop)

go

stop

stop

go

stop

Not preserved:

[]<>go

Counterexamples need to be checked.

234

Symmetry

A permutation is a one-one and onto function p:AA.For example, 13, 24, 31, 45, 52.

One can combine permutations, e.g.,p1: 13, 21, 32p2: 12, 21, 33p1@p2: 13, 22, 31

A set of permutations with @ is called a symmetry group.

235

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.

236

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

init

237

Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

init

The quotient model

238

Homework: what is preserved in the following buffer abstraction? What is not preserved?

e

empty

quasi

full

q

q

q

f

239

BDD representation

240

Computation Tree Logic. . .

. . .

. . .

. . .

p p

p

. . .

. . .

. . .

. . .

EG p

p p p p

p

p p

AF p

241

Computation Tree Logic

q q

q

p

. . .

. . .

. . .

. . .

p

q

p

. . .

. . .

. . .

. . .

E pUq

p

A pUq

242

Example formulas

CTL formulas: mutual exclusion: AG ( cs1

cs2) non starvation: AG (request

AF grant) “sanity” check: EF request

243

Model Checking M |= f[Clarke, Emerson, Sistla 83]

The Model Checking algorithm works iterativelyon subformulas of f , from simpler subformulas to more complex ones

When checking subformula g of f we assume that all subformulas of g have already been checked

For subformula g, the algorithm returns the set of states that satisfy g ( Sg ) The algorithm has time complexity: O(|M||f|)

244

Model checking f = EF g

Given a model M= < S, I, R, L >and Sg the sets of states satisfying g in M

procedure CheckEF (Sg )

Q := emptyset; Q’ := Sg ;while Q Q’ do Q := Q’; Q’ := Q { s | s' [ R(s,s’)

Q(s’) ] }end whileSf := Q ; return(Sf )

245

g

g

g

f

f

f

f

f

f

f

Example: f = EF g

246

Model checking f = EG g

CheckEG gets M= < S, I, R, L > and Sg

and returns Sf

procedure CheckEG (Sg)

Q := S ; Q’ := Sg ;

while Q Q’ do Q := Q’; Q’ := Q { s | s' [ R(s,s’) Q(s’) ] }end whileSf := Q ; return(Sf )

247

g

g

g

g

g

g

Example: f = EG g

248

Symbolic model checking[Burch, Clarke, McMillan, Dill 1990]

If the model is given explicitly (e.g. by adjacent

matrix) then only systems with about ten Boolean

variables (~1000 states) can be handledSymbolic model checking uses Binary Decision Diagrams ( BDDs )to represent the model and sets of states. It

can handle systems with hundreds of Boolean variables.

249

Binary decision diagrams (BDDs) [Bryant 86]

Data structure for representing Boolean functions

Often concise in memory Canonical representation Boolean operations on BDDs can

be done in polynomial time in the BDD size

250

Assume that states in model M are encoded by {0,1}n and described by Boolean variables v1...vn

Sf can be represented by a BDD over v1...vn

R (a set of pairs of states (s,s’) ) can be represented by a BDD over v1...vn v1’...vn’

BDDs in model checking

251

BDD definition A tree representation of a Boolean

formula. Each leaf represents 0 (false) or 1 (true). Each internal leaf represents a node. If we follow a path in the tree and go

from a node left (low) on 0 and right (high) on 1, we obtain a leaf that corresponds to the value of the formula under this truth assignment.

252

Examplea

b c

c c b b

)a/\(b\/¬c)¬)/\(a/\(b/\c)(

0 0 0 11 1 0 1

253

OBDD: there is some fixed appearance order between variables, e.g., a<b<c

)a/\(b\/¬c)¬)/\(a/\(b/\c)(

a

b b

c c c c

0 0 0 11 0 1 1

254

In reduced form: combine all leafs with same values, all isomorphic subgraph.

a

b b

c c c c

)a/\(b\/¬c)¬)/\(a/\(b/\c)(

0 0 0 11 0 1 1

In addition, remove nodes with identical children (low(x)=high(x)).

255


a

b b

c c c c

)a/\(b\/¬c)¬)/\(a/\(b/\c)(

0 1

In addition, remove (shortcut) nodes with identical children (low(x)=high(x)).Apply bottom up until not possible.

256


a

b b

c c c c

)a/\(b\/¬c)¬)/\(a/\(b/\c)(

0 1

In addition, remove (shortcut) nodes with identical children (low(x)=high(x)).

257


a

b b

c c

)a/\(b\/¬c)¬)/\(a/\(b/\c)(

0 1

In addition, remove (shortcut) nodes with identical children (low(x)=high(x)).

258

Example, even parity, 3 bits

a

b b

c c c c

1 0 0 01 1 1 0

259

Apply reducea

b b

c c c c

01

260

Apply reducea

b b

c c c c

01

261

Apply reducea

b b

c c c

01

262

Apply reducea

b b

c c c

01

263

Apply reducea

b b

c c

01

264

f[0/x], f[1/x] (“restrict” algorithm)

Obtain the replacement of a variable x by 0 or 1, in formula f, respectively.

For f[0/x], incoming edges to node x are redirected to low(x), and x is removed.

For f[1/x], incoming edges to node x are redirected to high(x), and x is removed.

Then we reduce the OBBD.

265

Calculate x x= [0/x]\/[1/x] Thus, we apply “restrict” twice to

and then “apply” the disjunction.

266

Shannon expansion of Boolean expression f. f=(¬x/\f[0/x])\/(x/\f[1/x]) Thus, f#g, for some logical operator #

is f#g=(¬x/\f#g [0/x])\/(x/\f#g [1/x])= (¬x/\f [0/x]#g [0/x])\/(x/\f [1/x]#g[1/x])

267

Now compute f#g recursively:

Let rf be the root of the OBDD for f, and rg be the root of the OBDD for g.

If rf and rg are terminals, then apply rf#rg and put the result.

If both roots are same node (say x), then create a low edge to low(rf)#low(rg), and a high edge to high(rf)#high(rg).

If rf is x and rg is y, and x<y, there is no x node in g, so g=g[0/x]=g[1/x]. So we create a low edge to low(rf)#g and a high edge to high(rf)#g. The symmetric case is handled similarly.

We reduce.

268

Same subgraphs are not needed to be explored again (use memoising, i.e., dynamic programming, complexity: exponential 2xmultiplications of sizes.

R1,S1

R6,S5

R2,S3 R3,S2

R3,S3 R4,S3

R5,S4

R6,S3

R4,S3

R4,S3

R5,S4 R6,S5

R6,S5

R6,S4

R6,S5

R5,S4 R6,S5

x

z

t

z

y

x

t

0 1 0 1

R1

R3

R2

R5

R6

R4

S1

S3

S2

S4 S5

x

z

z tt

tt

y

269

Same subgraphs are not needed to be explored again (use memoising,

i.e., dynamic programming, complexity: exponential 2xmultiplications of sizes.

R1,S1

R6,S5

R2,S3 R3,S2

R3,S3 R4,S3

R5,S4

R6,S3

R4,S3

R4,S3

R5,S4 R6,S5

R6,S5

R6,S4

R6,S5

R5,S4 R6,S5

x

z

t

z

y

x

y

0 1 0 1

R1

R3

R2

R5

R6

R4

S1

S3

S2

S4 S5

x

z

z tt

tt

y

270

Symbolic Model Checking Characterize CTL formulas using fixpoints. AF= \/AX AF EF= EX EF Z.EX Z AG= /\AX A G EG= EX EG Z.EX Z AU = \/(/\AXU) EU = (EXU) Z.(EXZ) AR = /\(\/AXR) ER = (EXR) Z.(EXZ)

271

Representing the successor relation formula R A relation between the current state and the

next state can be represented as a BDD with prime variables representing the variables at next states.

For example:p/\¬q/\r/\¬p’/\q’/\r’ says that the current state satisfies p/\¬q/\r and the next state satisfies ¬p/\q/\r. (typically, for one transition, represented as a Boolean relation).

If ti represents this relation for transition i, we can

write for the entire code R=\/i ti.

272

Calculating (Z) for (Z)=EX Z Z is a BDD. Rename variables in Z by their primed

version to obtain BDD Z’. Calculate the BDD R/\Z’. Let y1’…yn’ be the primed variables,

Then calculate the BDD T=y1’… yn’ R/\Z’ to remove primed variables.

Calculate the BDD T.

273

Model checking Z (least fixed point)For example, = EX ZFor formulas with main operator .

procedure Check LFP ()Q :=False; Q’ := (Q) ;while Q Q’ do Q := Q’; Q’ := (Q) ;end whilereturn(Q)

274

Model checking Z (Greatest fixed point)For example, = (EX Z)For formulas with main operator .

procedure Check GFP ()Q :=True; Q’ := (Q) ;while Q Q’ do Q := Q’; Q’ := (Q) ;end whilereturn(Q)

275

Program verification: flowchart programs

(Book: chapter 7)

276

History

Verification of flowchart programs: Floyd, 1967 Hoare’s logic: Hoare, 1969 Linear Temporal Logic: Pnueli, Krueger, 1977 Model Checking: Clarke & Emerson, 1981

277

Program Verification

Predicate (first order) logic. Partial correctness, Total correctness Flowchart programs Invariants, annotated programs Well founded ordering (for

termination) Hoare’s logic

278

Predicate (first order logic)

Variables, functions, predicates

Terms

Formulas (assertions)

279

Signature

Variables: v1, x, y18Each variable represents a value of some given

domain (int, real, string, …). Function symbols: f(_,_), g2(_), h(_,_,_).Each function has an arity (number of

paramenters), a domain for each parameter, and a range.

f:int*int->int (e.g., addition), g:real->real (e.g., square root)

A constant is a predicate with arity 0. Relation symbols: R(_,_), Q(_).Each relation has an arity, and a domain for each

parameter.R : real*real (e.g., greater than).Q : int (e.g., is a prime).

280

Terms

Terms are objects that have values. Each variable is a term. Applying a function with arity n to n

terms results in a new term.Examples: v1, 5.0, f(v1,5.0),

g2(f(v1,5.0))

More familiar notation: sqr(v1+5.0)

281

Formulas

Applying predicates to terms results in a formula.

R(v1,5.0), Q(x)More familiar notation: v1>5.0 One can combine formulas with the

boolean operators (and, or, not, implies).

R(v1,5.0)->Q(x)x>1 -> x*x>x One can apply existentail and universal

quantification to formulas.x Q(X) x1 R(x1,5.0) x y R(x,y)

282

A model, A proofs

A model gives a meaning (semantics) to a first order formula: A relation for each relation symbol. A function for each function symbol. A value for each variable.

An important concept in first order logic is that of a proof. We assume the ability to prove that a formula holds for a given model.

Example proof rule (MP) :

283

Flowchart programs

Input variables: X=x1,x2,…,xlProgram variables: Y=y1,y2,…,ymOutput variables: Z=z1,z2,…,zn

start

haltY=f(X)

Z=h(X,Y)

284

Assignments and tests

Y=g(X,Y) t(X,Y)FT

285

start

halt

(y1,y2)=(0,x1)

y2>=x2

(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)

Initial condition

Initial condition: the values for the input variables for which the program must work.

x1>=0 /\ x2>0

FT

286

start

halt

(y1,y2)=(0,x1)

y2>=x2

(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)

The input-output claim

The relation between the values of the input and the output variables at termination.

x1=z1*x2+z2 /\ 0<=z2<x2

FT

287

Partial correctness, Termination, Total correctness

Partial correctness: if the initial condition holds and the program terminates then the input-output claim holds.

Termination: if the initial condition holds, the program terminates.

Total correctness: if the initial condition holds, the program terminates and the input-output claim holds.

288

Subtle point:

The program ispartially correct

withrespect tox1>=0/\x2>=0and totally correctwith respect tox1>=0/\x2>0

start

halt

(y1,y2)=(0,x1)

y2>=x2

(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)

T F

289

start

halt

(y1,y2)=(0,x1)

y2>=x2

(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)

Annotating a scheme

Assign an assertion for each pair of nodes. The assertion expresses the relation between the variable when the program counter is located between these nodes.

A

B

C D

E

FT

290

Invariants Invariants are assertions that hold at each state

throughout the execution of the program. One can attach an assertion to a particular

location in the code:e.g., at(B) (B).This is also an invariant; in other locations, at(B) does not hold hence the implication holds.

If there is an assertion attached to each location, (A), (B), (C), (D), (E), then their disjunction is also an invariant: (A)\/(B)\/(C)\/(D)\/(E)(since location is always at one of these locations).

291

Annotating a scheme with invariants

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

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

y2>=0 /\ y2>=x2D):x1=y1*x2+y2 /\

y2>=0 /\ y2<x2E):x1=z1*x2+z2 /\ 0<=z2<x2Notice: (A) is the initial

condition, Eis the input-output condition.

start

halt

(y1,y2)=(0,x1)

y2>=x2

(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)

A

B

C D

E

FT

A) Is the precondition of (y1,y2)=(0,x1) and B) is its postcondition

292

Preliminary:Relativizing assertions

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

obtaining B) [Y\g(X,Y)]e(B) expressed w.r.t. variables at

A.) (B)A =x1=0 * x2 + x1 /\ x1>=0Think 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 by substitution.

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

(y1,y2)=(0,x1)

A

B

A

B

(y1,y2)=(0,x1)

Y=g(X,Y)

293

Preliminary:Relativizing assertions

(y1,y2)=(0,x1)

A

B

A

B

(y1,y2)=(0,x1)

A):

(B)A

(B)

Y=g(X,Y)

Y=g(X,Y)

294

Verification conditions: assignment

A) B)A

where B)A = B)[Y\g(X,Y)]

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

y2>=0

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

x1>=0

(y1,y2)=(0,x1)

A

B

A

B

(y1,y2)=(0,x1)

Y=g(X,Y)

295

(y1,y2)=(y1+1,y2-x2)

Second assignment

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

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

B)C: x1=(y1+1)*x2+y2-x2 /\ y2-x2>=0

C

B

296

(z1,z2)=(y1,y2)

Third assignment

D):x1=y1*x2+y2 /\ y2>=0 /\ y2<x2

E):x1=z1*x2+z2 /\ 0<=z2<x2

E)D:

x1=y1*x2+y2 /\ 0<=y2<x2

E

D

297

Verification conditions: tests

B) /\ t(X,Y) C)B) /\¬t(X,Y) D)

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

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

D):x1=y1*x2+y2 /\ y2>=0 /\ y2<x2

y2>=x2

B

C

D

B

C

Dt(X,Y)

FT

FT

298


y2>=x2

B

C

D

B

C

Dt(X,Y)

FT

FT

t(X,Y)¬t(X,Y)

B)

C)

299

Partial correctness proof:An induction on length of execution

B)

B)

D)

C)

Initially, states satisfy the initial conditions.

Then, passing from one set of states to another, we preserve the invariants at the appropriate location.

We prove: starting with a state satisfying the initial conditions, if are at a point in the execution, the invariant there holds.

Not a proof of termination!

start

halt

(y1,y2)=(0,x1)

y2>=x2

(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)

A

B

C D

E

A)

no

no

yes

yes

T F

300

Exercise: prove partial correctness

Initial condition: x>=0

Input-output claim:

z=x!

start

halt

(y1,y2)=(0,1)

y1=x

(y1,y2)=(y1+1,(y1+1)*y2) z=y2

TF

301

What have we achieved?

For each statement S that appears between points X and Y we showed that if the control is in X when (X) holds (the precondition of S) and S is executed, then (Y) (the postcondition of S) holds.

Initially, we know that (A) holds. The above two conditions can be combined into

an induction on the number of statements that were executed: If after n steps we are at point X, then (X)

holds.

302

Another example

(A) : x>=0

(F) : z^2<=x<(z+1)^2

z is the biggest numberthat is not greaterthan sqrt x.

start

(y1,y2,y3)=(0,0,1)

A

halt

y2>x

(y1,y3)=(y1+1,y3+2) z=y1

B

C

D

F

truefalse

E

y2=y2+y3

303

Some insight

1+3+5+…+(2n+1)=(n+1)^2

y2 accumulates theabove sum, untilit is bigger than x.

y3 ranges over oddnumbers 1,3,5,…

y1 is n-1.

start

(y1,y2,y3)=(0,0,1)

A

halt

y2>x

(y1,y3)=(y1+1,y3+2) z=y1

B

C

D

F

truefalse

E

y2=y2+y3

304

Invariants

It is sufficient to have one invariant for every loop(cycle in the program’sgraph).

We will have(C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1

start

(y1,y2,y3)=(0,0,1)

A

halt

y2>x

(y1,y3)=(y1+1,y3+2) z=y1

B

C

D

F

truefalse

E

y2=y2+y3

305

Obtaining (B)

By backwards substitution in (C).

(C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1

(B)=y1^2<=x /\ y2+y3=(y1+1)^2 /\ y3=2*y1+1

start

(y1,y2,y3)=(0,0,1)

A

halt

y2>x

(y1,y3)=(y1+1,y3+2) z=y1

B

C

D

F

truefalse

E

y2=y2+y3

306

Check assignment condition

(A)=x>=0(B)=y1^2<=x /\ y2+y3=(y1+1)^2 /\ y3=2*y1+1(B) relativized is 0^2<=x /\ 0+1=(0+1)^2 /\ 1=2*0+1Simplified: x>=0

start

(y1,y2,y3)=(0,0,1)

A

halt

y2>x

(y1,y3)=(y1+1,y3+2) z=y1

B

C

D

F

truefalse

E

y2=y2+y3

307

Obtaining (D)

By backwards substitution in

(B).

(B)=y1^2<=x /\ y2+y3=(y1+1)^2 /\ y3=2*y1+1

(D)=(y1+1)^2<=x /\ y2+y3+2=(y1+2)^2 /\ y3+2=2*(y1+1)+1

start

(y1,y2,y3)=(0,0,1)

A

halt

y2>x

(y1,y3)=(y1+1,y3+2) z=y1

B

C

D

F

truefalse

E

y2=y2+y3

308

Checking

(C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1

(C)/\y2<=x) (D)

(D)=(y1+1)^2<=x /\ y2+y3+2=(y1+2)^2 /\ y3+2=2*(y1+1)+1

start

(y1,y2,y3)=(0,0,1)

A

halt

y2>x

(y1,y3)=(y1+1,y3+2) z=y1

B

C

D

F

truefalse

E

y2=y2+y3

309

y1^2<=x /\

y2=(y1+1)^2 /\ y3=2*y1+1 /\y2<=x (y1+1)^2<=x /\

y2+y3+2=(y1+2)^2 /\

y3+2=2*(y1+1)+1y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 /\y2<=x (y1+1)^2<=x /\ y2+y3+2=(y1+2)^2

/\ y3+2=2*(y1+1)+1

y1^2<=x /\

y2=(y1+1)^2 /\

y3=2*y1+1 /\y2<=x (y1+1)^2<=x /\

y2+y3+2=(y1+2)^2 /\

y3+2=2*(y1+1)+1

310

Not finished!

Still needs to:

Calculate (E) bysubstituting backwardsfrom (F).

Check that(C)/\y2>x(E)

start

(y1,y2,y3)=(0,0,1)

A

halt

y2>x

(y1,y3)=(y1+1,y3+2) z=y1

B

C

D

F

truefalse

E

y2=y2+y3

311

Exercise: prove partial correctness. Initially: x1>0/\x2>0. At termination: z1=gcd(x1,x2).

halthalt

startstart

(y1,y2)=(x1,x2)(y1,y2)=(x1,x2)

z1=y1z1=y1

y1=y2F T

y1>y2y1>y2

y2=y2-y1y2=y2-y1y1=y1-y2y1=y1-y2

TF

312

Annotation of program with invariants

halthalt

startstart

(y1,y2)=(x1,x2)(y1,y2)=(x1,x2)

z1=y1z1=y1

y1=y2F

T

y1>y2y1>y2

y2=y2-y1y2=y2-y1y1=y1-y2y1=y1-y2

TF

z1=gcd(x1,x2)

x1>0 /\ x2>0

gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0

gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1y2

gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1<y2

gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1>y2

y1=gcd(x1,x2)

A

B

D

EF

G

H

313

Part 1

halthalt

startstart

(y1,y2)=(x1,x2)(y1,y2)=(x1,x2)

z1=y1z1=y1

y1=y2F

T

y1>y2y1>y2

y2=y2-y1y2=y2-y1y1=y1-y2y1=y1-y2

TF

(A)= x1>0 /\ x2>0

(B)=gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0 A

B

D

EF

G

H

(B)’rel= gcd(x1,x2)=gcd(x1,x2)/\x1>0/\x2>0 (A)

(B)’rel

314

Part 2a

halthalt

startstart

(y1,y2)=(x1,x2)(y1,y2)=(x1,x2)

z1=y1z1=y1

y1=y2F

T

y1>y2y1>y2

y2=y2-y1y2=y2-y1y1=y1-y2y1=y1-y2

TF

(B)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0

(D)=gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1y2 A

B

D

EF

G

H

(B)/\¬(y1=y2) (D)

315

Part 2b

halthalt

startstart

(y1,y2)=(x1,x2)(y1,y2)=(x1,x2)

z1=y1z1=y1

y1=y2F

T

y1>y2y1>y2

y2=y2-y1y2=y2-y1y1=y1-y2y1=y1-y2

TF

(G)= y1=gcd(x1,x2)

A

B

D

EF

G

H

(B)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0

(B)/\(y1=y2) (G)

316

Part 3

halthalt

startstart

(y1,y2)=(x1,x2)(y1,y2)=(x1,x2)

z1=y1z1=y1

y1=y2F

T

y1>y2y1>y2

y2=y2-y1y2=y2-y1y1=y1-y2y1=y1-y2

TF

(D)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1y2

(E)=gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1<y2

(F)=(gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1>y2

A

B

D

EF

G

H

(D)/\(y1>y2) (F)

(D)/\¬(y1>y2) (E)

317

Part 4

halthalt

startstart

(y1,y2)=(x1,x2)(y1,y2)=(x1,x2)

z1=y1z1=y1

y1=y2F T

y1>y2y1>y2

y2=y2-y1y2=y2-y1y1=y1-y2y1=y1-y2

TF

x1>0 /\ x2>0

(B)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0

(E)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1<y2

(F)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1>y2

A

B

D

EF

G

H

(B)’rel1=gcd(y1,y2-y1)=gcd(x1,x2)/\y1>0/\y2-y1>0(B)’rel2=gcd(y1-y2,y2)=gcd(x1,x2)/\y1-y2>0/\y2>0

(E) (B)’rel1 (F) (B)’rel2

318

Annotation of program with invariants

halthalt

startstart

(y1,y2)=(x1,x2)(y1,y2)=(x1,x2)

z1=y1z1=y1

y1=y2F

T

y1>y2y1>y2

y2=y2-y1y2=y2-y1y1=y1-y2y1=y1-y2

TF

(H)= z1=gcd(x1,x2)

(G)= y1=gcd(x1,x2)

A

B

D

EF

G

H

(H)’rel= y1=gcd(x1,x2)

(G) (H)’rel2

319

Proving termination

320

Well-founded sets

Partially ordered set (W,<): If a<b and b<c then a<c (transitivity). If a<b then not b<a (asymmetry). Not a<a (irreflexivity).

Well-founded set (W,<): Partially ordered. No infinite decreasing chain a1>a2>a3>…

321

Examples for well founded sets Natural numbers with the bigger than

relation. Finite sets with the set inclusion relation. Strings with the substring relation. Tuples with alphabetic order:

(a1,b1)>(a2,b2) iff a1>a2 or [a1=a2 and b1>b2].

(a1,b1,c1)>(a2,b2,c2) iff a1>a2 or [a1=a2 and b1>b2] or [a1=a2 and b1=b2 and c1>c2].

322

Why does the program terminate

y2 starts as x1. Each time the loop is

executed, y2 is decremented.

y2 is natural number The loop cannot be

entered again when y2<x2.

start

halt

(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)

(y1,y2)=(0,x1)

A

B

D

E

falsey2>=x2

C

true

323

Proving termination

Choose a well-founded set (W,<). Attach a function u(N) to each

point N. Annotate the flowchart with

invariants, and prove their consistency conditions.

Prove that (N) (u(N) in W).

324

How not to stay in a loop?

Show that u(M)>=u(N)’rel.

At least once in each loop, show that u(M)>u(N).

S

M

N

TN

M

325

How not to stay in a loop?

For stmt: (M)(u(M)>=u(N)’rel)

Relativize since we need to compare values not syntactic expressions.

For test (true side):((M)/\test)(u(M)>=u(N))

For test (false side):((M)/\

¬test)(u(M)>=u(L))

stmt

M

N

test

N

M

true

L

false

326

What did we achieve?

There are finitely many control points. The value of the function u cannot

increase. If we return to the same control point,

the value of u must decrease (its a loop!).

The value of u can decrease only a finite number of times.

327

Why does the program terminate

u(A)=x1u(B)=y2u(C)=y2u(D)=y2u(E)=z2

W: naturals> : greater than

start

halt

(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)

(y1,y2)=(0,x1)

A

B

D

E

falsey2>=x2

C

true

328

Recall partial correctness annotation

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

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

y2>=0 /\ y2>=x2D):x1=y1*x2+y2 /\

y2>=0 /\ y2<x2E):x1=z1*x2+z2 /\ 0<=z2<x2

start

halt

(y1,y2)=(0,x1)

y2>=x2

(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)

A

B

C D

E

falsetrue

329

Strengthen for termination

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

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

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

y2>=0 /\ y2<x2/\x2>0E):x1=z1*x2+z2 /\ 0<=z2<x2

start

halt

(y1,y2)=(0,x1)

y2>=x2

(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)

A

B

C D

E

falsetrue

330

Strengthen for termination

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

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

/\y2>=x2/\x2>0u(c)>=0D):x1=y1*x2+y2 /\ y2>=0 /\

y2<x2/\x2>0u(D)>=0E):x1=z1*x2+z2 /\ 0<=z2<x2u(E)>=0This proves that u(M) is natural for

each point M.

u(A)=x1u(B)=y2u(C)=y2u(D)=y2u(E)=z2

331

We shall show:

u(A)=x1u(B)=y2u(C)=y2u(D)=y2u(E)=z2A)u(A)>=u(B)’relB)u(B)>=u(C)C)u(C)>u(B)’relB)u(B)>=u(D)D)u(D)>=u(E)’re

l

start

halt

(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)

(y1,y2)=(0,x1)

A

B

D

E

falsey2>=x2

C

true

332

Proving decrement

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

u(C)=y2u(B)=y2u(B)’rel=y2-x2

C) y2>y2-x2(notice that C) x2>0)

start

halt

(y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2)

(y1,y2)=(0,x1)

A

B

D

E

falsey2>=x2

C

true

333

Integer square prog.

(C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1

(B)=y1^2<=x /\ y2+y3=(y1+1)^2 /\y3=2*y1+1

start

(y1,y2,y3)=(0,0,1)

A

halt

y2>x

(y1,y3)=(y1+1,y3+2) z=y1

B

C

D

F

truefalse

E

y2=y2+y3

334

u(A)=x+1u(B)=x-y2+1u(C)=max(0,x-y2+1)u(D)=x-y2+1u(E)=u(F)=0u(A)>=u(B)’relu(B)>u(C)’relu(C)>=u(D)u(C)>=u(E)u(D)>=u(B)’relNeed some invariants,i.e., y2<=x/\y3>0at points B and D,and y3>0 at point C.

start

(y1,y2,y3)=(0,0,1)

A

halt

y2>x

(y1,y3)=(y1+1,y3+2) z=y1

B

C

D

F

truefalse

E

y2=y2+y3

335

Program VerificationUsing Hoare’s Logic

Hoare triple is of the form{Precondition} Prog-segment {Postcondition}

It expresses partial correctness: if the segment starts with a state satisfying the precondition and it terminates, the final state satisfies the postscondition.

The idea is that one can decompose the proof of the program into smaller and smaller segments, depending on its structure.

336

While programs

Assignments y:=e Composition S1; S2 If-then-else if t then S1 else S2 fi While while e do S od

337

Greatest common divisor

{x1>0/\x2>0}y1:=x1;y2:=x2;while ¬(y1=y2) do if y1>y2 then y1:=y1-y2 else y2:=y2-y1 fiod{y1=gcd(x1,x2)}

338

Why it works?

Suppose that y1,y2 are both positive integers. If y1>y2 then gcd(y1,y2)=gcd(y1-y2,y2) If y2>y1 then gcd(y1,y2)=gcd(y1,y2-y1) If y1=y2 then gcd(y1,y2)=y1=y2

339

Assignment axiom

{p[e/y] } y:=e {p}

For example:{y+5=10} y:=y+5 {y=10}{y+y<z} x:=y {x+y<z}{2*(y+5)>20} y:=2*(y+5) {y>20}Justification: write p with y’ instead of y,

and add the conjunct y’=e. Next, eliminate y’ by replacing y’ by e.

340

Why axiom works backwards?

{p} y:=t {?}Strategy: write p and the conjunct y=t, where

y’ replaces y in both p and t. Eliminate y’.This y’ represents value of y before the

assignment.{y>5} y:=2*(y+5) {? } {p} y:=t { y’ (p[y’/y] /\ t[y’/y]=y) }y’>5 /\ y=2*(y’+5) y>20

341

Composition rule

{p} S1 {r }, {r} S2 {q }

{p} S1;S2 {q}For example: if the antecedents are1. {x+1=y+2} x:=x+1 {x=y+2}2. {x=y+2} y:=y+2 {x=y}Then the consequent is {x+1=y+2} x:=x+1; y:=y+2 {x=y}

342

More examples

{p} S1 {r}, {r} S2 {q} {p} S1;S2 {q}{x1>0/\x2>0} y1:=x1

{gcd(x1,x2)=gcd(y1,x2)/\y1>0/\x2>0}

{gcd(x1,x2)=gcd(y1,x2)/\y1>0/\x2>0} y2:=x2

___{gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0}____

{x1>0/\x2>0} y1:=x1 ; y2:=x2 {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0}

343

If-then-else rule

{p/\t} S1 {q}, {p/\¬t} S2 {q}

{p} if t then S1 else S2 fi {q}For example: p is gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0/\¬(y1=y2)t is y1>y2S1 is y1:=y1-y2S2 is y2:=y2-y1q is gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0

344

While rule

{p/\t} S {p} {p} while t do S od {p/\¬t}Example:p is {gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0}t is ¬ (y1=y2)S is if y1>y2 then y1:=y1-y2 else y2:=y2-y1 fi

345

Consequence rules

Strengthen a precondition rp, {p } S {q } {r } S {q } Weaken a postcondition {p } S {q }, qr {p } S {r }

346

Use of first consequence rule

Want to prove{x1>0/\x2>0} y1:=x1

{gcd(x1,x2)=gcd(y1,x2)/\y1>0/\x2>0}By assignment rule:{gcd(x1,x2)=gcd(x1,x2)/\x1>0/\x2>0}

y1:=x1 {gcd(x1,x2)=gcd(y1,x2)/\y1>0/\x2>0}

x1>0/\x2>0 gcd(x1,x2)=gcd(x1,x2)/\x1>0/\x2>0

347

Combining program

{x1>0 /\ x2>0} y1:=x1; y2:=x1;{gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0} while S do if e then S1 else S2 fi od{gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0/\

y1=y2}Combine the above using concatenation

rule!

348

Not completely finished

{x1>0/\x2>0} y1:=x1; y2:=x1; while ¬(y1=y2) do if e then S1 else S2 fi od{gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0/\

y1=y2}But we wanted to prove:{x1>0/\x1>0} Prog {y1=gcd(x1,x2)}

349

Use of second consequence rule

{x1>0/\x2>0} Prog{gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0/\

y1=y2}And the implicationgcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0/\y1=y2 y1=gcd(x1,x2)Thus,{x1>0/\x2>0} Prog {y1=gcd(x1,x2)}

350

Annotating a while program

{x1>0/\x2>0}y1:=x1; {gcd(x1,x2)=gcd(y1,x2

) /\y1>0/\x2>0}y2:=x2; {gcd(x1,x2)=gcd(y1,y2

) /\y1>0/\y2>0}

while ¬(y1=y2) do{gcd(x1,x2)=gcd(y1,y2)/\

y1>0/\y2>0/\¬(y1=y2)}

if y1>y2 then y1:=y1-y2 else y2:=y2-y1 fiod{y1=gcd(x1,x2)}

351

While rule

{p/\e} S {p} {p} while e do S od {p/\¬e}

352

Consequence rules

Strengthen a precondition rp, {p} S {q} {r} S {q} Weaken a postcondition {p} S {q}, qr {p} S {r}

353

Soundness

Hoare logic is sound in the sense thateverything that can be proved is correct!

This follows from the fact that each axiomand proof rule preserves soundness.

354

Completeness

A proof system is called complete if every

correct assertion can be proved.

Propositional logic is complete. No deductive system for the

standard arithmetic can be complete (Godel).

355

And for Hoare’s logic?

Let S be a program and p its precondition.

Then {p} S {false} means that S never terminates when started from p. This is undecideable. Thus, Hoare’s logic cannot be complete.

356

Weakest prendition, Strongest postcondition

For an assertion p and code S, let post(p,S) be the strongest assertion such that {p}S{post(p,S) }

That is, if {p}S{q} then post(p,S)q. For an assertion q and code S, let

pre(S,q) be the weakest assertion such that {pre(S,q)}S{q}

That is, if {p}S{q} then ppre(S,q).

357

Relative completeness

Suppose that either post(p,S) exists for each p, S, or pre(S,q) exists for each S, q.

Some oracle decides on pure implications.Then each correct Hoare triple can be proved.What does that mean? The weakness of theproof system stem from the weakness of the

(FO) logic, not of Hoare’s proof system.

358

Extensions

Many extensions for Hoare’s proof rules:

Total correctness Arrays Subroutines Concurrent programs Fairness

359

Proof rule for total correctnessSimilar idea to Floyd’s termination: Well foundedness

{p/\t/\f=z} S {p/\f<z}, p(f>=0) {p} while t do S od {p/\¬t}

wherez - an int. variable, not appearing in

p,t,e,S.f - an int. expression.

360

Verification with Array Variables

Book: Chapter 7.2

361

The problem

Using array variables can lead to complication:

{x[1]=1/\x[2]=3} x[x[1]]:=2

{x[x[1]]=2} No!!!Why?Because the assignment changes x[1] as

well. Now it is also 2, and x[x[1]], which is x[2] is 3 and not 2!

362

What went wrong?

Take the postcondition {x[x[1]]=2} and substitute 2 instead of x[x[1]].

We obtain {2=2} (which is equivalent to {true}).

Now, (x[1]=1/\x[2]=3) 2=2. So we may wrongly conclude that

the above Hoare triple is correct.

363

How to fix this?

`Backward substitution’ should be done with arrays as complete elements.

Define (x; e1: e2) : an array like x, with value at the index e1 changed to e2.

(x; e1: e2)[e3]=e2 if e1=e3 x[e3] otherwise

(x; e1: e2)[e3]=if (e1=e3, e2, x[e3])

364

Solved the problem?

How to deal with if(φ, e1, e2)?Suppose that formula ψ contains this

expression.Replace if(φ, e1, e2) by new variable v

in ψ.The original formula ψ is equivalent to: (φ/\ ψ[e1/v])\/(¬φ/\ ψ[e2/v])

365

Returning to our case

Our postcondition is {x[x[1]]=2}. The assignment x[x[1]]:=2 causes the

substitution in the postcondition ofthe (array) variable x by a new array, which is (x; x[1] : 2), resulting in

{x[x[1]]=2}

(x; x[1] : 2)[(x; x[1] : 2)[1]] = 2

366

Are we done?

Not yet. It remains to Convert the array form into an if form. Get rid of the if form.

Will not be done in class. All we say is that we obtain an

expression that is not implied by the precondition x[1]=1/\x[2]=3.

367

Proof checking with PVSBook: Chapter 3

368

A Theory

Name: THEORY BEGIN Definitions (types,

variables, constants)

Axioms Lemmas

(conjectures, theorems)

END Name

369

Group theory

(*, e), where * is the operator and e the unity element.

Associativity (G1): (x*y)*z=x*(y*z).

Unity (G2): (x*e)=x

Right complement (G3):

x y x*y=e.

Want to prove: x y y*x=e.

370

Informal proof

Choose x arbitrarily.By G3, there exists y

s.t.(1) x*y=e.By G3, we have z s.t.(2) y*z=e.y*x=(y*x)*e (by

G2) =(y*x)*(y*z) (by

(2))

=y*(x*(y*z)) (by G1)

=y*((x*y)*z) (by G1) =y*(e*z) (by (1)) =(y*e)*z (by G1) =y*z (by

(G2)) =e (by (2))

371

Example: groups

Group: THEORY BEGIN element: TYPE unit: element *: [element, element-

> element] < some axioms>

left:CONJECTURE FORALL (x: element): EXISTS (y: element): y*x=unitEND Group

372

Axioms

associativity: AXIOM FORALL (x, y, z:element): (x*y)*z=x*(y*z)

unity: AXIOM FORALL (x:element): x*unit=x

complement: AXIOM FORALL(x:element): EXISTS (y:element): x*y=unity

373

Skolemization

Corresponds to choosing some arbitrary constant and proving “without loss of generality”.

Want to prove (…/\…)(…\/x(x)\/…).

Choose a new constant x’. Prove (…/\…)(…\/(x’)\/…).

374

Skolemization

Corresponds to choosing some unconstrained arbitrary constant when one is known to exist.

Want to prove (…/\x(x)/\…)(…\/…).

Choose a new constant x’. Prove (…/\(x’)/\…)(…\/…).

375

Skolem in PVS

(skolem 2 (“a1” “b2” “c7”)) (skolem -3 (“a1” “_” “c7”)) (skolem! -3)

invents new constants, e.g., for x will invent x!1, x!2, … when applied repeatedly.

376

Instantiation

Corresponds to restricting the generality.

Want to prove (…/\x(x)/\…) (…\/…).

Choose a some term t. Prove (…/\(t)/\…)(…\/…).

377

Instantiation

Corresponds to proving the existence of an element by showing an evidence.

Want to prove (…/\…)(…\/x(x)\/…).

Choose some term t. Prove (…/\…)(…\/(t)\/…).

378

Instantiating in PVS

(inst -1 “x*y” “a” “b+c”) (inst 2 “a” “_” “x”)

379

Other useful rules

(replace -1 (-1 2 3))Formula -1 is of the form le=ri. Replace any occurrence of le by ri in lines -1, 2, 3.

(replace -1 (-1 2 3) RL)Similar, but replace ri by le instead.

(assert), (assert -) (assert +) (assert 7) Apply algebraic simplification.

(lemma “<axiom-name>”) - add axiom as additional antecedent.

381

Why testing?

Reduce design/programming errors. Can be done during development,

before production/marketing. Practical, simple to do. Check the real thing, not a model. Scales up reasonably. Being state of the practice for

decades.

382

Part 1: Testing of black box finite state machine

Know:

Transition relation

Size or bound on size

Wants to know:

In what state we started?

In what state we are?

Transition relation

Conformance

Satisfaction of a temporal property

383

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). δ S S - transition relation (deterministic but

sometimes not defined for each input per each state). S O - output on edges.

Notation: δ(s,a1a2..an)= δ(… (δ(δ(s,a1),a2) … ),an)(s,a1a2..an)= (s,a1)(δ(s,a1),a2)…(δ(… δ(δ(s,a1),a2) … ),an)

384

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

S={s1, s2, s3 }, {a, b }, O={0,1 }.

δ(s1,a)=s3 (also s1=a=>s3),δ(s1,b)=s2,(also s1=b=>s2)…(s1,a)=0 , (s1,b)=1,…δ(s1,ab)=s1, (s1,ab)=01

s1

s3

s2

a/0b/1 b/0

b/1

a/0

a/0

385

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).

386

Determinism

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

b/1a/1

a/1

387

Preliminaries: separating sequences

s1

s3

s2

a/0b/1 b/0

b/1

a/0

a/0

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

A set of sequences X such that if we execute them from different states, at least one of them will give a different output sequence.

s≠t µX (s, µ )≠(s, µ )

388

A: separate to blocks of states with different output.

s1

s3

s2

a/0b/1

b/0

b/1

a/0

a/0

The states are separated into {s1, s3}, {s2} using the string b.

But s1 and s3 have the same outputs to same inputs.

389

Repeat B : Separate blocks based on input that cause moving to different blocks.

s1

s3

s2

a/0b/1 b/0

b/1

a/0

a/0

Separate first block using b to three singleton blocks obtaining separation sequence bb.Separating sequences: b, bb.Max rounds: n-1, sequences: n-1, length: n-1.

If string x separated blocks B1, B2, and letter a splits part of block C to move to B1 and part to B2, then ax separates C into C1, C2, accordingly.

390

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

After firning homing sequence, we know in what state we are by observing outputs.

Find a sequence µ such thatδ(s, µ )≠δ(t, µ ) (s, µ )≠(s, µ )

So, given an input µ that is executed from state s, we look at a table of outputs and according to a table, know in which state r we are.

391

Homing sequence

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

Then repeatedly separate blocks, as long as they are not singletons, as follows:

Take a non singleton block, append a distinguishing sequence µ that separates at least two states in that block.

Update each block to the states after executing µ. (Blocks may not be disjoint, its not a partition!)

Max length of sequence: (n-1)2 (Lower bound: n(n-1)/2.)

392

Example (homing sequence)

s1

s3

s2

a/0b/1 b/0

b/1

a/0

a/0

{s1, s2, s3}

{s1, s2} {s3}

{s1} {s2} {s3}

b

b1 0

011

1

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

393

Synchronizing sequence

One sequence takes the machine to the same final state, regardless of the initial state or the outputs.That is: find a sequence µ such that for each states s, t,

δ(s, µ )=δ(t, µ ) Not every machine has a

synchronizing sequence. Can be checked whether

exists and if so, can be found in polynomial time.

a/1

b/1

a/0b/0b/1

a/0

394

Algorithm for synchronizing sequeneces

s1

s2 s3

a/1

b/1

a/0b/0b/1

a/0

Construct a graph with ordered pairs of nodes (s,t) such that (s,t )=a=>(s’,t’ ) when s=a=>s’, t=a=>t’.(Ignore self loops, e.g., on (s2,s2).)s1,s1

s2,s2

s3,s3

s1,s2

s2,s3 s1,s3

b b

b

b

b

baa

a

395

Algorithm continues (2)There is an input

sequence from both s≠t to some r iff there is a path in this graph from (s,t ) to (r,r ).

There is a synchronization sequence iff some node (r,r ) is reachable from every pair of distinct nodes.

In this case it is (s2,s2 ).

s1,s1

s2,s2

s3,s3

s1,s2

s2,s3 s1,s3

b b

b

b

b

baa

a

396

Algorithm continues (3)Notation:δ(S,x)=

{t |sS,δ(s,x)=t }1. i=1; S1=S

2. Take some nodes s≠tSi, and find a shortest path labeled xi to some (r,r ).

3. i=i+1; Si:=δ(Si-1,x ). If |Si |>1,goto 2., else goto 3.

4. Concatenate x1x2…xk.

s1,s1

s2,s2

s3,s3

s1,s2

s2,s3 s1,s3

b b

b

b

b

baa

a

Number of sequences ≤ n-1.Each of length ≤ n(n-1)/2.Overall O(n(n-1)2/2).

397

Example:

(s2,s3)=a=>(s2,s2)

x1:=a

δ({s1,s2,s3},a)={s1,s2}

(s1,s2)=ba=>(s2,s2)

x2:=ba

δ({s1,s2},ba)={s2}

So x1x2=aba is a synchronization sequence, bringing every state into state s2.

s1,s1

s2,s2

s3,s3

s1,s2

s2,s3 s1,s3

b b

b

b

b

baa

a

398

State identification: Want to know in which state the

system has started (was reset to).

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].

399

Sometimes cannot identify initial state

b/1a/1 s1

s3

s2

a/1

b/0

b/1

a/1

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.

400

So…

We’ll assume resets from now on!

401

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.

402

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).

s1

s3

s2

a/1

b/0

b/1

a/1

?=

a/1

b/1

403

Conformance testing [Ch,V]

a/1

b/1

Cannot distinguish if reduced or not.

a/1

b/1

a/1

b/1

a/1

b/1a/1

b/1

404

Conformance testing (cont.)

ab b

a

a

a

a b

b

b

a

Need: bound on number of states of B.

a

405

Preparation: Construct a spanning tree, and separating sequences

b/1a/1 s1

s3

s2

a/1

b/0

b/1

a/1

s1

s2s3

b/1a/1

Given an initial state, we can reach any state of the automaton.

406

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.

s1

s2s3

b/1a/1

Rese

t

Rese

t

Distinguishing sequences

407

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.

408

Combination lock automaton

Assume accepting states. Accepts only words with a specific suffix

(cdab in the example).

s1 s2 s3 s4 s5

bdc a

Any other input

409

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

s3

s2

a/1

b/0

b/1

a/1

b/1

Pretends to be s3

Pretends to be s1

a/1

410

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

s1

s2s3

b/1a/1

Words of length n-m+1

Rese

t

Rese

t

411

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.

412

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.

413

Example: check a

a, aa

a<>a

414

Example: check <>a

a

a

a

a

<>a

415

Example: check <>a

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

a

a

a, a<>a

416

System

c b

a

417

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

c b

a

a

a

a

a

s1 s2

s3 q2

q1

s1,q1

s1,q2 s3,q2

s2,q1a

b

c

aAcceptance isdetermined byautomaton P.

<>a

418

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.

419

Black box checking [PVY]

Outputs: {0,1} (for enabled/disabled)

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.

420

Experiments

aa

bb cc

reset

a

a

b

b

c

c

try ba

a

b

b

c

c

try c

(Output 1+ progress)

(Output 0=stay)

421

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.

422

Deadlock complexity

Nondeterministic algorithm:Linear time, logarithmic space.

Deterministic algorithm:Exponential (pn-1) time, linear space.

Lower bound: Exponential time (usecombination lock automata).

How does this conform with what we know about complexity theory?

423

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.

424

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..

425

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.

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

426

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!)

427

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

428

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.

429

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 v2n. If this succeeds:

error, otherwise, this is a counterexample for Mi .

430

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 d n-l+1+ l 2mn) if there is no error.If n is not known, check while time allows. Probabilistic complexity: polynomial.

431

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.

432

Part 2: Software testing(Book: chapter 9)

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.

433

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.

434

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.

435

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.

436

testing

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!)

437

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!

438

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.

439

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.

440

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?

441

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.

442

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.

443

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!

444

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?

445

Statement coverageExecute every statement at least once

By choosingA=2,B=0,X=3each statement will

be chosen.The case where the

tests fail is not checked!


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

Now x=1.5

446

Decision coverage (if, while checks)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=2) | (X>1) THEN X=X+1; END;

447

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;

Now x=1

448

Decision coverage

A=2,B=1,X=1

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


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

449

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;

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

450

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;

451

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;

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

452

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


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

453

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!


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

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

454

Preliminary:Relativizing assertions(Book: Chapter 7)

(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)

A

B

A

B

(y1,y2)=(0,x1)

Y=g(X,Y)

455


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

B)= D) /\ y2x2y2>=x2

B

C

D

B

C

Dt(X,Y)

FT

FT

456

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+1

X=X/Ano

no

yes

yes

true

true

457

How to find values for coverage?

A>1 & B=0

A=2 | X>1

X=X+1

X=X/Ano

no

yes

yes

true

true

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=0

Can also calculate path condition forwards.

458

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.

459

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

z:=z+x

x>y

t>y

460

Test case design for black box testing: how to find a good test suite? Equivalence partition Boundary value

analysis

461

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

462

Example: A legal variable

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




Starting char

Chars

Length

Starts A-Z Starts other

[A-Z0-9] Has others

1-6 chars 0 chars, >6 chars

1 2

3 4

56 7

463

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.




464

Example

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




Starting char

Chars

Length

Starts A-Z Starts other

[A-Z0-9] Has others

1-6 chars 0 chars, >6 chars

1 2

3 4

56 7

(6) VERYLONG (7)

465

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.

466

For Unit Testing

Write “drivers”, which replaces procedures calling the code.

Write “stubs”, which replaces procedures called by the code.

Tested unit

Driver

StubStubStub

Driver

467

package bucket;import java.util.*;import java.math.*;public class BugBucketSort{ public static void main(String[] args) { Random r = new Random(100); r.nextInt(100); LinkedList<Integer> ls = new LinkedList<Integer>(); for (int j = 0; j < 10; j++) ls.push(r.nextInt(100)); System.out.println(ls); bucketSort(ls); } public static void bucketSort(LinkedList<Integer> ls) { LinkedList<HashMap<Integer,Integer>> la = new LinkedList<HashMap<Integer,Integer>>(); for (int i = 0 ; i<11; i++) la.add(new HashMap<Integer,Integer>()); System.out.println("\nValues in the array: "+la.size()); for (int j = 0; j < ls.size(); j++) { la.get((int)ls.get(j)/10).put((int)ls.get(j)%10, ls.get(j)); } System.out.println("\nThe ordered array is:"); System.out.print("["); for (int i = 0; i < la.size(); i++) for (int k = 1; k < 10; k++) { if(la.get(i).containsKey(k)) System.out.print(la.get(i).get(k)+" "); } System.out.println("]"); }}

Bucket sort:With bugs!!

468

package bucket;import java.util.*;import java.math.*;public class BucketSort{ public static void main(String[] args) { Random r = new Random(100); r.nextInt(100); LinkedList<Integer> ls = new LinkedList<Integer>(); for (int j = 0; j < 10; j++) ls.push(r.nextInt(100)); System.out.println(ls);// System.out.println((int) 17 / 5); bucketSort(ls); } public static void bucketSort(LinkedList<Integer> ls) { LinkedList<HashMap<Integer,Integer>> la = new LinkedList<HashMap<Integer,Integer>>(); for (int i = 0 ; i<10; i++) la.add(new HashMap<Integer,Integer>()); System.out.println("\nValues in the array: "+la.size()); for (int j = 0; j < ls.size(); j++) { la.get((int)ls.get(j)/10).put((int)ls.get(j)%10, ls.get(j)); } System.out.println("\nThe ordered array is:"); System.out.print("["); for (int i = 0; i < la.size(); i++) for (int k = 0; k < 10; k++) { if(la.get(i).containsKey(k)) System.out.print(la.get(i).get(k)+" "); } System.out.println("]"); }}

Bucket sort:Without bugs!!

469

class MergeSortAlgorithm{ public static int [] mergeSort(int array[])

{ int loop;int rightIndex = 0;int [] leftArr = new int[array.length / 2];int [] rightArr = new int[(array.length - array.length) / 2];if (array.length <= 1)

return array;for (loop = 0 ; loop < leftArr.length ; ++loop)

leftArr[loop] = array[loop];for (; rightIndex < rightArr.length ; ++loop)

rightArr[rightIndex++] = array[loop];mergeSort(leftArr);

rightArr = mergeSort(rightArr);array = merge(leftArr,rightArr);return array; }

public static int [] merge(int [] leftArr , int [] rightArr){ int [] newArray = new int[leftArr.length + rightArr.length];

int leftIndex = 0; int rightIndex = 0; int newArrayIndex = 0;while (leftIndex < leftArr.length && rightIndex < rightArr.length){ if (leftArr[leftIndex] < rightArr[rightIndex])

newArray[newArrayIndex] = leftArr[leftIndex++];newArray[newArrayIndex] = rightArr[rightIndex++];}

while (rightIndex < rightArr.length) newArray[newArrayIndex++] = rightArr[rightIndex++];

while (leftIndex < leftArr.length) newArray[newArrayIndex++] = leftArr[leftIndex++];

return newArray;}public static void print(int [] arr , String caller){ int loop;

for (loop = 0 ; loop < arr.length ; ++loop) System.out.println("caller " + caller + " : " + arr[loop]);}public static void main(String [] args){ int []originalArray = {1,5,4,3,7};

print(originalArray,"before"); originalArray = mergeSort(originalArray);print(originalArray,"after"); }}

470

class MergeSortAlgorithm{ public static int [] mergeSort(int array[])

{ int loop;int rightIndex = 0;int [] leftArr = new int[array.length / 2];int [] rightArr = new int[(array.length - array.length) / 2]; // BUGif (array.length <= 1)

return array;for (loop = 0 ; loop < leftArr.length ; ++loop)

leftArr[loop] = array[loop];for (; rightIndex < rightArr.length ; ++loop)

rightArr[rightIndex++] = array[loop];mergeSort(leftArr); // BUGrightArr = mergeSort(rightArr);array = merge(leftArr,rightArr);return array; }

public static int [] merge(int [] leftArr , int [] rightArr){ int [] newArray = new int[leftArr.length + rightArr.length];

int leftIndex = 0; int rightIndex = 0; int newArrayIndex = 0;while (leftIndex < leftArr.length && rightIndex < rightArr.length){ if (leftArr[leftIndex] < rightArr[rightIndex])

newArray[newArrayIndex] = leftArr[leftIndex++];newArray[newArrayIndex] = rightArr[rightIndex++]; //

TWO BUGS}while (rightIndex < rightArr.length) newArray[newArrayIndex++] =

rightArr[rightIndex++];while (leftIndex < leftArr.length) newArray[newArrayIndex++] =

leftArr[leftIndex++];return newArray;}

public static void print(int [] arr , String caller){ int loop;

for (loop = 0 ; loop < arr.length ; ++loop) System.out.println("caller " + caller + " : " + arr[loop]);}public static void main(String [] args){ int []originalArray = {1,5,4,3,7};

print(originalArray,"before"); originalArray = mergeSort(originalArray);print(originalArray,"after"); }}

471

Test case generation based on LTL specification

CompilerModel

CheckerPath condition

calculation

First orderinstantiator

Testmonitoring

Transitions

Path

Flowchart

LTLAut

472

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.

473

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).

474

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

¬at l2

at l2

¬at l2

at l2

l2:x:=x+z

l3:x<t

l1:…

l2:x:=x+z

l3:x<t

l2:x:=x+z

XX==

475

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

XX==

xy

x2y

476

Example: GCD

l1:x:=a

l5:y:=z

l4:x:=y

l3:z:=x rem y

l2:y:=b

l6:z=0?yesn

o

l0

l7

477

Example: GCD

l1:x:=a

l5:x:=y

l4:y:=z

l3:z:=x rem y

l2:y:=b

l6:z=0?yesn

o

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

l0

l7

478

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.”

479

Example: GCD

l1:x:=a

l5:x:=y

l4:y:=z

l3:z:=x rem y

l2:y:=b

l6:z=0?yesn

o

l0

l7

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

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

at l7

480

Example: GCD

l1:x:=a

l5:x:=y

l4:y:=z

l3:z:=x rem y

l2:y:=b

l6:z=0?yesn

o

l0

l7

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

481

Potential explosion

Bad point: potential explosion

Good point: may be chopped on-the-fly

496

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?yesn

o

l0

l7

497

Process Algebra

(Book: Chapter 8)

498

The Main Issue

Q: When are two models equivalent?

A: When they satisfy different properties.

Q: Does this mean that the models have different executions?

499

What is process algebra? An abstract description for

nondeterministic and concurrent systems.

Focuses on the transitions observed rather than on the states reached.

Main correctness criterion: conformance between two models.

Uses: system refinement, model checking, testing.

500

Different models may have the same set of executions!

aa a

b b cc

a-insert coin, b-press pepsi, c-press pepsi-light

d-obtain pepsi, e-obtain pepsi-light

dd ee

501

Actions: Act={a,b,c,d}{}.Agents: E, E’, F, F1, F2, G1, G2, …

E

E’

G2G1

F1 F2

F

aa a

b b cc

Agent E may evolve into agent E’.

Agent F may evolve into F1 or F2.

dd ee

502

Events.

E

E’

G2G1

F1 F2

F

aa a

b b cc

E—aE’, F—aF1, F—aF2, F1—aG1,

F2—aG2. G1—F, G1—F.

503

Actions and co-actions

For each action a, except for , there is a co-action a. a and a interact (a input, a output).The coaction of a is a.

G2G1

F1 F2

Fa a

b c

E

E’

a

b c

504

Notation

a.E – execute a, then continue according to E.E+F – execute according to E or to F.E||F – execute E and F in parallel.

E

G H

F

a

b c

a.(b+c)(actually, a.((b.0)+(c.0))

E—aFF—bGF—cH

0 – deadlock/termination.

505

Conventions

“.” has higher priority than “+”. “.0” or “.(0||0||…||0)” is omitted.

506

CCS - calculus of concurrent systems [Milner]. Syntax

a,b,c, … actions, A, B, C - agents. a,b,c, coactions of a,b,c. -silent action. nil - terminate. a.E - execute a, then behave like E. + - nondeterministic choice. || - parallel composition. \L - restriction: cannot use letters of L. [f] - apply mapping function f between

between letters.

507

Semantics (proof rule and axioms).Structural Operational Semantics SOS

a.p –a p p—ap’ |-- p+q –a p’ q—aq’ |-- p+q –a q’ p—ap’ |-- p|q –a p’|q q—aq’ |-- p|q –a p|q’ p—ap’, q—aq’ |-- p|q – p’|q’ p—ap’ , a R |-- p\L –a p’\R p—ap’ |-- p[m]—m(a)p’[m]

508

Action Prefixing

a.E—aE (Axiom)

Thus, a.(b.(c||c)+d)—a(b.(c||c)+d).

509

Choice

E—aE’ F—aF’ (E+F)—aE’ (E+F)—aF’

b.(c||c)—b(c||c).Thus,

(b.(c||c)+e)—b(c||c).

If E—aE’ and F—aF’, then E+F has a nondeterministic choice.

510

Concurrent Composition

E—aE’ F—aF’ E||F—aE’||F E||F—aE||F’

E—aE’, F—aF’ ———————— E||F—E’||F’

c—c0, c—c0, c||c—0||0, c||c—c0||c, c||c—cc||0.

511

Restriction

E—aE’, a, a R —————————

E\R –aE’\RIn this case: allows only

internal interaction of c.c||c—0||0 c||c—c0||c c||c—cc||0(c||c) \ {c}—(0||0) \{c}

512

Relabeling

E—aE’ ———— E[m] –m(a)E’[m]

No axioms/rules for agent 0.

513

Examples

a.E||b.F

a.E||FE||b.F

E||F

b

b

a

a

514

Derivations

(0||0)

a.(b.(c||c)+d)

b.(c||c)+d

(c||c) 0

(0||c) (c||0)

a

b d

c

cc

c

515

Modeling binary variable

C0=is_0? . C0 + set_1 . C1 + set_0 . C0

C1=is_1? . C1 + set_0 . C0 + set_1 . C1

C0 C1

set_1

set_0

set_0

is_0?

set_1

is_1?

516

Equational Definition

E=a.(b..E+c..E) E—aE’, A=EF=a.b..F+a.c..F A—aE’

G2G1

F1 F2

Fa a

b c

E

E’

a

b c

517

Trace equivalence:Systems have same finite sequences.

Same traces

Fa a

b b

E

a

b c c

E=a.(b+c) F=(a.b)+a.(b+c)

518

Failures: comparing also what wecannot do after a finite sequence.

Fa a

b b

E

a

b c c

Failure of agent E: (σ, X), where after executing σ from E, none of the events in X is enabled.Agent F has failure (a, {c}), which is not a failure of E.

519

Simulation equivalence

Relation over set of agents S. RSS. E R F If E’ R F’ and E’—aE’’, then there

exists F’’, F’—aF’’, and E’’ R F’’.

E

c d

b b

aaF

c d

b b

a

520

Simulation equivalence

Relation over set of agents S. RSS. E R F If E’ R F’ and E’—aE’’, then there exists F’’,

F’—aF’’, and E’’ R F’’.

E

c d

b b

aaF

c d

b b

a

521

Here, simulation works only in one direction. No equivalence!

Relation over set of agents S. RSS. E R F If E’ R F’ and E’—aE’’, then there exists F’’,

F’—aF’’, and E’’ R F’’.

E

c d

b b

aaF

c d

b b

a

want to establish

symmetricallynecessarily

problem!!!

522

Simulation equivalentbut not failure equivalent

Left agent a.b+a has a failure (a,{b}).

E

b

aaF

b

a

523

Bisimulation: same relation simulates in both directions

Not in this case: different simulation relations.

E

b

aaF

b

a

524

Hierarchy of equivalences

Bisimulation

Trace

FailureSimulation

525

Example:

A=a.((b.nil)+(c.d.A))

B=(a.(b.nil))+(a.c.d.B)

a b

c

ds0

s1 s2

s3 a

d

b

ac

t0

t1

t4

t2

t3

526

Bisimulation between G1 and G2

Let N= N1 U N2

A relation R : N1 x N2 is a bisumulation ifIf (m,n) in R then1. If m—am’ then n’:n—an’ and (m’,n’) in R2. If n—an’ then m’:m—am’ and (m’,n’) in R.

Other simulation relations are possible, I.e., m=a=> m’ when m—…—a—m’.

527

Algorithm for bisimulation:

Partition N into blocks B1B2…Bn=N. Initially: one block, containing all of N. Repeat until no change:

Choose a block Bi and a letter a. If some of the transitions of Bi move to some block Bj and some not, partition Bi accordingly.

At the end: Structures bisimilar if initial states of two structures are in same blocks.

528

Correctness of algorithm

Invariant: if (m,n) in R then m and n remain in the same block throughout the algorithm.

Termination: can split only a finite number of times.

529

Example:

a b

c

ds0

s1 s2

s3 a

d

b

ac

t0

t1

t4

t2

t3{s0,t0},{s1,s2,s3,t1,t2,t3,t4} split on b

{s0,t0},{s1,t1},{s0,s2,s3,t2,t3,t4}

530

Example:a b

c

ds0

s1 s2

s3 a

d

b

ac

t0

t1

t4

t2

t3

{s0,t0},{s1,t1},{s2,s3,t2,t3,t4} split on c

{s0,t0},{s1},{t1},{s2,s3,t2,t3,t4}

531

Example:

{s0,t0},{s1},{t1},{s2,s3,t2,t3,t4} split on c

{s0,t0},{s1},{t1},{t4},{s2,s3,t2,t3}

a b

c

ds0

s1 s2

s3 a

d

b

ac

t0

t1

t4

t2

t3

532

Example:

{s0,t0},{s1},{t1},{t4},{s2,s3,t2,t3} split on d

{s0,t0},{s1},{t1},{t4},{s3, t3},{s2,t2}

a b

c

ds0

s1 s2

s3 a

d

b

ac

t0

t1

t4

t2

t3

533

Example:

{s0,t0},{s1},{t1},{t4},{s2,t2},{s3,t3} split on a

{s0},{t0},{s1},{t1},{t4},{s3, t3},{s2,t2}

a b

c

ds0

s1 s2

s3 a

d

b

ac

t0

t1

t4

t2

t3

534

Example:

{s0},{t0},{s1},{t1},{t4},{s2,s3,t2,t3} split on d

{s0},{t0},{s1},{t1},{t4},{s3},{t3},{s2,t2}

a b

c

d

s0

s1 s2

s3a

d

b

ac

t0

t1

t4

t2

t3

1 Softrare Reliability Methods Prof. Doron A. Peled Bar Ilan University, Israel And Univeristy of...

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