Softrare Reliability Methods Prof. Doron A. Peled Bar Ilan University, Israel And Univeristy of warwick, UK Version 2008.

Softrare Reliability Methods

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

Version 2008

Some related books:

Also:Mainly:

Goal: software reliability

Use software engineering methodologies to develop the code.

Use formal methods during code development

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.

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.

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.

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.

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.

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.

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.

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.

Emphasis

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

Use of tools:Hands on. PVS, SPIN

Visual notation:Statecharts, MSCs, UML.

Some emphasis

The process of selecting and using formal methods.

The appropriate notation. In particular, visual notation.

Hands-on experience with tools.

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).

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.

Where do we start?

Boss: Mark, can you verify this for me?

Mark: OK, first I have to ...

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.

Different types of software

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

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

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!

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.

Modeling Software Systems for Analysis

(Book: Chapter 4)

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?

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

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

(software + hardware).

Sequential systems.

Perform some computational task. Have some initial condition, e.g.,

0in A[i] integer. Have some final assertion, e.g.,

0in-1 A[i]A[i+1].(What is the problem with this spec?)

Are supposed to terminate.

Concurrent Systems

Involve several computation agents.Termination may indicate an abnormal

event (interrupt, strike).May exploit diverse computational

power.May involve remote components.May interact with users (Reactive).May involve hardware components

(Embedded).

Problems in modeling systems Representing concurrency:

- Allow one transition at a time, or- Allow coinciding transitions.

Granularity of transitions. Assignments and checks? Application of methods?

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

Modeling.The states based model.

V={v0,v1,v2, …} - 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,v2=3,v3=7,…,v18=2>

For predicate (first order assertion) p:p(s) is p under the assignment s.Example: p is x>y /\ y>z. s=<x=4, y=3, z=5>.Then we have 4>3 /\ 3>5, which is false.

State space

The state space of a program is the set of all possible states for it.

For example, if V={a, b, c} and the variables are over the naturals, then the state space includes: <a=0,b=0,c=0>,<a=1,b=0,c=0>, <a=1,b=1,c=0>,<a=932,b=5609,c=6658>…

Atomic Transitions

Each atomic transition represents a small piece of code such that no smaller piece of code is observable.

Is a:=a+1 atomic? In some systems, e.g., when a is a

register and the transition is executed using an inc command.

Non atomicity

Execute the following when 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.

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

Representing transitions

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

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

Initial condition

A predicate I. The program can

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

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

A transition system

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

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

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

An initial condition I.

Example

V={a, b, c, d, e}. : all assignments of natural

numbers for variables in V. T={c >0(c,e):=(c -1,e +1),

d >0(d,e):=(d -1,e +1)} I: c =a /\ d =b /\ e =0 What does this transition system

do?

The interleaving model

An execution is a 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.

Example:

s0=<a=2, b=1, c=2, d=1, e=0>

s1=<a=2, b=1, c=1, d=1, e=1>

s2=<a=2, b=1, c=1, d=0, e=2>

s3=<a=2, b=1 ,c=0, d=0, e=3>

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

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

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

L0:While True do NC0:wait(Turn=0); CR0:Turn=1endwhile ||L1:While True do NC1:wait(Turn=1); CR1:Turn=0endwhile

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

The transitions

Is this the only reasonable way to model this program?

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

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.

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

Turn=0L0,L1

Turn=0L0,NC1

Turn=0NC0,L1

Turn=0CR0,NC1

Turn=0NC0,NC1

Turn=0CR0,L1

Turn=1L0,CR1

Turn=1NC0,CR1

Turn=1L0,NC1

Turn=1NC0,NC1

Turn=1NC0,L1

Turn=1L0,L1

Always if Turn=0 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

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

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!

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

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=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

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’

Combinatorial explosion

V1:=1

V1:=3

V1:=2

Vn:=1

Vn:=3

Vn:=2…

How many states?

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

Program verification: flowchart programs

(Book: chapter 7)

History

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

Program Verification

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

termination) Hoare’s logic

Predicate (first order logic)

Variables, functions, predicates

Terms

Formulas (assertions)

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).

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)

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)

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) :

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)

Assignments and tests

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

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

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

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.

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

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

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).

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

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)

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)

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)

(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

(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

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

Verification conditions: tests

y2>=x2

B

C

D

B

C

Dt(X,Y)

FT

FT

t(X,Y)¬t(X,Y)

B)

C)

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

BC

D

EF

G

H

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

BC

D

EF

G

H

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

(B)’rel

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

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

BC

D

EF

G

H

(B)/\¬(y1=y2) (C)

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

BC

D

EF

G

H

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

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

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

BC

D

EF

G

H

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

(D)/\¬(y1>y2) (E)

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

BC

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

Part 5

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

BC

D

EF

G

H

(H)’rel= y1=gcd(x1,x2)

(G) (H)’rel2

Proving termination

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>…

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].

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

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).

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

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

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.

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

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

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

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

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

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

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

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

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.

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

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)

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

Why deterministic machines?

Otherwise no amount of experiments would guarantee anything.

If dependent on some parameter (e.g., temperature), we can determinize, by taking parameter as additional input.

We still can model concurrent system. It means just that the transitions are deterministic.

All kinds of equivalences are unified into language equivalence.

Also: connected machine (otherwise we may never get to the completely separate parts).

Determinism

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

b/1a/1

a/1

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 )

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.

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.

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

Know: transition relation.Don’t know: which state we start.Want to know: which state we end up.After firing homing sequence, we know in

what state we are by observing the outputs. Find a sequence µ such that

δ(s, µ )≠δ(t, µ ) (s, µ )≠(t, 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.

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!)

Note: blocks do not have to be disjoint!

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

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.

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

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

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 twins 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

If original automaton is not strongly connected, there is no synchronizing sequence. Otherwise, all twins are reachable from each other in product graph. Thus, in this case, if there is not twins node (r,r) reachable from all pairs, there is some pair (Si,Sj) from which no twins node is reachable, thus no sequence will synchronize Si and Sj.

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 twins (r,r ).

3. i=i+1; Si:=δ(Si-1,x ). If |Si |>1,goto 2., else goto 4.

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).

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

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].

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.

So…

We’ll assume resets from now on!

Conformance testing Unknown deterministic finite state system B. Known: n states and alphabet . An abstract model C of B. C satisfies all the

properties we want from B. C has m states. Check conformance of B and C. Another version: only a bound n on the number of

states l is known.

Check conformance with a given state machine

Black box machine has no more states than specification machine (errors are mistakes in outputs, mistargeted edges).

Specification machine is reduced, connected, deterministic. Machine resets reliably to a single initial state (or use homing

sequence).

s1

s3

s2

a/1

b/0

b/1

a/1

?=

a/1

b/1

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

Conformance testing (cont.)

ab b

a

a

a

a b

b

b

a

Need: bound on number of states of B.

a

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.

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

Comments1. Checking the different distinguishing

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

2. A reset can be performed to a distinguished state through a homing sequence. Then we can perform a sequence that brings us to the distinguished initial state.

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

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

Combination lock automaton

Assume accepting states. Accepts only words with a specific suffix

(cdab in the example).

s1 s2 s3 s4 s5

bdc a

Any other input

When only a bound on size of black box is known…

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

b/1a/1s1

s3

s2

a/1

b/0

b/1

a/1

b/1

Pretends to be s3

Pretends to be s1

a/1

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

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.

Some software testing stages

Unit testing – the lowest level, testing some procedures.

Integration testing – different pieces of code. System testing – testing a system as a whole. Acceptance testing – performed by the

customer. Regression testing – performed after updates. Stress testing – checking the code under

extreme conditions. Mutation testing – testing the quality of the

test suite.

Some drawbacks of testing

There are never sufficiently many test cases.

Testing does not find all the errors. Testing is not trivial and requires

considerable time and effort. Testing is still a largely informal task.

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

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

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

The goal: to maximize the number of errors found.

testing

Is based on the internal structure of the program.

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

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

Some testing principles

A programmer should not test his/her own program. One should test not only that the program does

what it is supposed to do, but that it does not do what it is not supposed to.

The goal of testing is to find errors, not to show that the program is errorless.

No amount of testing can guarantee error-free program.

Parts of programs where a lot of errors have already been found are a good place to look for more errors.

The goal is not to humiliate the programmer!

Inspections and Walkthroughs

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

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

errors.

Code Inspection

Team of 3-5 people. One is the moderator.

He distributes materials and records the errors.

The programmer explains the program line by line.

Questions are raised. The program is

analyzed w.r.t. a checklist of errors.

Checklist for inspections

Data declarationAll variables

declared?Default values

understood?Arrays and strings

initialized?Variables with similar

names?Correct initialization?

Control flowEach loop terminates?DO/END statements

match?

Input/outputOPEN statements

correct?Format specification

correct?End-of-file case

handled?

Walkthrough

Team of 3-5 people. Moderator, as

before. Secretary, records

errors. Tester, play the role

of a computer on some test suits on paper and board.

Selection of test cases (for white-box testing)

The main problem is to select a good coverage

criterion. Some options are: Cover all paths of the program. Execute every statement at least once. Each decision has a true or false value at

least once. Each condition is taking each truth value at

least once. Check all possible combinations of conditions

in each decision.

Cover all the paths of the program

Infeasible.Consider the flow

diagram on the left.It corresponds to a

loop.The loop body has 5

paths.If the loops executes

20times there are 5^20

different paths!May also be

unbounded!

How to cover the executions?

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

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

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

Statements? Conditions?

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

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

Now x=1.5

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

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

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

Decision coverage

A=2,B=1,X=1

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

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

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

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;

Condition coverage

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

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

Condition coverage

A=2,B=1,X=0

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

Can use condition+decision coverage.

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

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

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

A>1,B=0 A>1,B≠0 A1,B=0 A1,B≠0 A=2,X>1 A=2,X1 A≠2,X>1 A≠2,X1

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

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

A smaller number of cases: 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>1) & (B=0) THEN X=X/A; END;

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

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

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)

Verification conditions: tests

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

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

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.

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.

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

Test case design for black box testing

Equivalence partition Boundary value

analysis Cause-effect graphs

Equivalence partition

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

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

Specificationcondition

Valid equivalenceclass

Invalid equivalenceclass

Example: A legal variable

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

Specificationcondition

Valid equivalenceclass

Invalid equivalenceclass

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

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.

Specificationcondition

Valid equivalenceclass

Invalid equivalenceclass

Example

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

Specificationcondition

Valid equivalenceclass

Invalid equivalenceclass

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)

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.

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

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++)//BUG should be 10 instead of 11 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++)//BUG needs to start with k=0 { if(la.get(i).containsKey(k)) System.out.print(la.get(i).get(k)+" "); } System.out.println("]"); }}

Bucket sort:With bugs!!

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!!

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"); }}

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"); }}