Formal methods: Model Checking and Testing Prof. Doron A. Peled University of Warwick, UK and Bar Ilan University, Israel.

Formal methods: Model Checking and Testing Prof. Doron A. Peled University of Warwick, UK and Bar Ilan University, Israel

Some related books: Also:Mainly:

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.,  0  i  n A[i] integer. Have some final assertion, e.g.,  0  i  n-1 A[i]  A[i+1]. (What is the problem with this spec?) Are supposed to terminate.

Concurrent Systems 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={v 0,v 1,v 2, …} - a set of variables, over some domain. p(v 0, v 1, …, v n ) - a parametrized assertion, e.g., v 0 =v 1 +v 2 /\ v 3 >v 4. A state is an assignment of values to the program variables. For example: s= For predicate (first order assertion) p: p(s) is p under the assignment s. Example: p is x>y /\ y>z. s=. 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:,,, …

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: P 1 :a=a+1 P 2 :a=a+1 Result: a=2. Is this always the case? Consider the actual translation: P 1 :load R1,a inc R1 store R1,a P 2 :load R2,a inc R2 store R2,a a may be also 1.

Scenario P 1 :load R1,a inc R1 store R1,a P 2 :load R2,a inc R2 store R2,a a=0 R1=0 R2=0 R1=1 R2=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 is switching 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 finite or infinite sequence of states s 0, s 1, s 2, … The initial state satisfies the initial condition, I.e., I (s 0 ). Moving from one state s i to s i+1 is by executing a transition e  t: e (s i ), I.e., s i satisfies e. s i+1 is obtained by applying t to s i.

Example: s0= s1= s2= s3= T={c>0  (c,e):=(c-1,e+1), d>0  (d,e):=(d-1,e+1)} I: c=a /\ d=b /\ e=0

L0:While True do NC0:wait(Turn=0); CR0:Turn=1 endwhile || L1:While True do NC1:wait(Turn=1); CR1:Turn=0 endwhile T0:PC0=L0  PC0:=NC0 T1:PC0=NC0/\Turn=0  PC0:=CR0 T2:PC0=CR0  (PC0,Turn):=(L0,1) T3:PC1=L1  PC1=NC1 T4:PC1=NC1/\Turn=1  PC1:=CR1 T5:PC1=CR1  (PC1,Turn):=(L1,0) Initially: PC0=L0/\PC1=L1 The transitions

The state graph:Successor relation between states. Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,CR1 Turn=1 L0,NC1 Turn=1 NC0,NC1 Turn=1 NC0,L1 Turn=1 L0,L1

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

Always ¬(PC0=CR0/\PC1=CR1) (Mutual exclusion) Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,CR1 Turn=1 L0,NC1 Turn=1 NC0,NC1 Turn=1 NC0,L1 Turn=1 L0,L1

Always if Turn=0 the at some point Turn=1 Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,CR1 Turn=1 L0,NC1 Turn=1 NC0,NC1 Turn=1 NC0,L1 Turn=1 L0,L1

Interleaving semantics: Execute one transition at a time. Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=1 L0,CR1 Turn=1 L0,NC1 Need to check the property for every possible interleaving!

Interleaving semantics Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=1 L0,CR1 Turn=1 L0,NC1 Turn=0 L0,L1 Turn=0 L0,NC1

L0:While True do NC0:wait(Turn=0); CR0:Turn=1 endwhile || L1:While True do NC1:wait(Turn=1); CR1:Turn=0 endwhile T0:PC0=L0  PC0:=NC0 T1:PC0=NC0/\Turn=0  PC0:=CR0 T1’:PC0=NC0/\Turn=1  PC0:=NC0 T2:PC0=CR0  (PC0,Turn):=(L0,1) T3:PC1==L1  PC1=NC1 T4:PC1=NC1/\Turn=1  PC1:=CR1 T4’:PC1=NC1/\Turn=0  PC1:=N1 T5:PC1=CR1  (PC1,Turn):=(L1,0) Initially: PC0=L0/\PC1=L1 Busy waiting

Always when Turn=0 then sometimes Turn=1 Now it does not hold! (Red subgraph generates a counterexample execution.) Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,CR1 Turn=1 L0,NC1 Turn=1 NC0,NC1 Turn=1 NC0,L1 Turn=1 L0,L1

Specification Formalisms (Book: Chapter 5)

Properties of formalisms Formal. Unique interpretation. Intuitive. Simple to understand (visual). Succinct. Spec. of reasonable size. Effective. Check that there are no contradictions. Check that the spec. is implementable. Check that the implementation satisfies spec. Expressive. May be used to generate initial code. Specifying the implementation or its properties?

A transition system A (finite) set of variables V. A set of states . A (finite) set of transitions T, each transition e  t has an enabling condition e and a transformation t. An initial condition I. Denote by R(s, s’) the fact that s’ is a successor of s.

The interleaving model An execution is a finite or infinite sequence of states s 0, s 1, s 2, … The initial state satisfies the initial condition, I.e., I (s 0 ). Moving from one state s i to s i+1 is by executing a transition e  t: e(s i ), I.e., s i satisfies e. s i+1 is obtained by applying t to s i. Lets assume all sequences are infinite by extending finite ones by “stuttering” the last state.

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

LTL: Syntax  ::= (  ) | ¬  |  /\   \/  U   |O  | p  “box”, “always”, “forever”  “diamond”, “eventually”, “sometimes” O  “nexttime”  U  “until” Propositions p, q, r, … Each represents some state property (x>y+1, z=t, at_CR, etc.)

Semantics over suffixes of execution   O   U        

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

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

What about ([]<>A)/\([]<>B)=[]<>(A/\B)? ([]<>A)\/([]<>B)=[]<>(A\/B)? (<>[]A)/\(<>[]B)=<>[](A/\B)? (<>[]A)\/(<>[]B)=<>[](A\/B)? No, just <-- Yes!!! No, just -->

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

Formal semantic definition Let  be a sequence s 0 s 1 s 2 … Let  i be a suffix of  : s i s i+1 s i+2 … (  0 =  )  i |= p, where p a proposition, if s i |=p.  i |=  /\  if  i |=  and  i |= .  i |=  \/  if  i |=  or  i |= .  i |= ¬  if it is not the case that  i |= .  i |= <>  if for some j  i,  j |= .  i |= []  if for each j  i,  j |= .  i |=  U  if for some j  i,  j |=  and for each i  k<j,  k |= .

Then we interpret: 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.

Spring Example s1s1 s3s3 s2s2 pull release extended malfunction extended r 0 = s 1 s 2 s 1 s 2 s 1 s 2 s 1 … r 1 = s 1 s 2 s 3 s 3 s 3 s 3 s 3 … r 2 = s 1 s 2 s 1 s 2 s 3 s 3 s 3 … …

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

Proof system ¬<>p []¬p [](p  q)  ([]p  []q) []p  (p/\O[]p) O¬p ¬Op [](p  Op)  (p  []p) (pUq) (q\/(p/\O(pUq))) (pUq)  <>q + propositional logic axiomatization. + axiom: _p_ []p

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

Another kind of traffic light Green-->Yellow-->Red-->Yellow-->Green First 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 we can start with yellow

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

Automata over finite words A=  (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 bs0s0 s1s1

The transition relation (s 0, a, s 0 ) (s 0, b, s 1 ) (s 1, a, s 0 ) (s 1, b, s 1 ) a a b bs0s0 s1s1

A run over a word A word over , e.g., abaab. A sequence of states, e.g. s 0 s 0 s 1 s 0 s 0 s 1. Starts with an initial state. Accepting if ends at accepting state. a a b bs0s0 s1s1

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 bs0s0 s1s1

Nondeterministic automaton Transitions: (s 0,a,s 0 ), (s 0,b,s 0 ), (s 0,a,s 1 ),(s 1,a,s 1 ). What is the language of this automaton? a,b a as0s0 s1s1

Equivalent deterministic automaton b a as0s0 s1s1 b a,b a a s0s0 s1s1

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 bs0s0 s1s1

Automata over infinite words Consider the word abababab… There is a run s 0 s 0 s 1 s 0 s 1 s 0 s 1 … This run in accepting, since s 0 appears infinitely many times. a a b bs0s0 s1s1

Other runs For the word bbbbb… the run is s 0 s 1 s 1 s 1 s 1 … and is not accepting. For the word aaabbbbb …, the run is s 0 s 0 s 0 s 0 s 1 s 1 s 1 s 1 … What is the run for ababbabbb …? a a b bs0s0 s1s1

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 a s0s0 s1s1

No deterministic automaton for (a+b)*a ω In a deterministic automaton there is one path for each run. 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.

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 bs0s0 s1s1

Mutual Exclusion a -- PC0=CR0/\PC1=CR1 b -- ¬(PC0=CR0/\PC1=CR1) c -- true []¬(PC0=CR0/\PC1=CR1) b a c s0s0 s1s1

L0:While True do NC0:wait(Turn=0); CR0:Turn=1 endwhile || L1:While True do NC1:wait(Turn=1); CR1:Turn=0 endwhile T0:PC0=L0  PC0=NC0 T1:PC0=NC0/\Turn=0  PC0:=CR0 T2:PC0=CR0  (PC0,Turn):=(L0,1) T3:PC1==L1  PC1=NC1 T4:PC1=NC1/\Turn=1  PC1:=CR1 T5:PC1=CR1  (PC1,Turn):=(L1,0) Initially: PC0=L0/\PC1=L1 Apply now to our program:

The state space Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,CR1 Turn=1 L0,NC1 Turn=1 NC0,NC1 Turn=1 NC0,L1 Turn=1 L0,L1

[]¬(PC0=CR0/\PC1=CR1) (Mutual exclusion) Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,CR1 Turn=1 L0,NC1 Turn=1 NC0,NC1 Turn=1 NC0,L1 Turn=1 L0,L1

[](Turn=0 --> <>Turn=1) Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,CR1 Turn=1 L0,NC1 Turn=1 NC0,NC1 Turn=1 NC0,L1 Turn=1 L0,L1

Interleaving semantics: Execute one transition at a time. Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=1 L0,CR1 Turn=1 L0,NC1 Need to check the property for every possible interleaving!

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

Correctness All sequences Sequences satisfying Spec Program executions

Incorrectness All sequences Sequences satisfying Spec Program executions Counter examples

Automatic Verification (Book: Chapter 6)

How can we check the model? The model is a graph. The specification should refer the the graph representation. Apply graph theory algorithms.

What properties can we check? 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.

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

If it is so good, why learn deductive verification methods? Model checking works only for finite state systems. Would not work with Unconstrained integers. Unbounded message queues. General data structures: queues trees stacks parametric algorithms and systems.

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

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

Depth First Search Program DFS For each s such that Init(s) dfs(s) end DFS Procedure dfs(s) for each s’ such that R(s,s’) do If new(s’) then dfs(s’) end dfs.

Start from an initial state q3q3 q4q4 q2q2 q1q1 q5q5 q1q1 q1q1 Stack: Hash table:

Continue with a successor q3q3 q4q4 q2q2 q1q1 q5q5 q 1 q 2 q1q2q1q2 Stack: Hash table:

One successor of q 2. q3q3 q4q4 q2q2 q1q1 q5q5 q 1 q 2 q 4 q1q2q4q1q2q4 Stack: Hash table:

Backtrack to q 2 (no new successors for q 4 ). q3q3 q4q4 q2q2 q1q1 q5q5 q 1 q 2 q 4 q1q2q1q2 Stack: Hash table:

Backtracked to q 1 q3q3 q4q4 q2q2 q1q1 q5q5 q 1 q 2 q 4 q1q1 Stack: Hash table:

Second successor to q1. q3q3 q4q4 q2q2 q1q1 q5q5 q 1 q 2 q 4 q 3 q1q3q1q3 Stack: Hash table:

Backtrack again to q 1. q3q3 q4q4 q2q2 q1q1 q5q5 q 1 q 2 q 4 q 3 q1q1 Stack: Hash table:

How can we check properties with DFS? Invariants: check that all reachable states satisfy the invariant property. If not, show a path from an initial state to a bad state. Deadlocks: check whether a state where no process can continue is reached. Dead code: as you progress with the DFS, mark all the transitions that are executed at least once.

The state graph:Successor relation between states. Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,CR1 Turn=1 L0,NC1 Turn=1 NC0,NC1 Turn=1 NC0,L1 Turn=1 L0,L1

¬ (PC0=CR0/\PC1=CR1) is an invariant! Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,CR1 Turn=1 L0,NC1 Turn=1 NC0,NC1 Turn=1 NC0,L1 Turn=1 L0,L1

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

Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=0 CR0,NC1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,CR1 Turn=1 L0,NC1 Turn=1 NC0,NC1 Turn=1 NC0,L1 Turn=1 L0,L1 init New initial state Convert graph into Buchi automaton

Turn=0 L0,L1 Turn=1 L0,L1 init Propositions are attached to incoming nodes. All nodes are accepting. Turn=1 L0,L1 Turn=0 L0,L1

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

Correctness All sequences Sequences satisfying Spec Program executions

How to prove correctness? Show that L(Model)  L(Spec). Equivalently: ______ Show that L(Model)  L(Spec) = Ø. Also: can obtain L(Spec) by translating from LTL!

What do we need to know? How to intersect two automata? How to complement an automaton? How to translate from LTL to an automaton?

Intersecting M 1 =(S 1, ,T 1,I 1,A 1 ) and M 2 =(S 2, ,T 2,I 2,S 2 ) Run the two automata in parallel. Each state is a pair of states: S 1 x S 2 Initial states are pairs of initials: I 1 x I 2 Acceptance depends on first component: A 1 x S 2 Conforms with transition relation: (x 1,y 1 )-a->(x 2,y 2 ) when x 1 -a->x 2 and y 1 -a->y 2.

Example (all states of second automaton accepting!) a b ct0t0 t1t1 a a b,c s0s0 s1s1 States: (s 0,t 0 ), (s 0,t 1 ), (s 1,t 0 ), (s 1,t 1 ). Accepting: (s 0,t 0 ), (s 0,t 1 ). Initial: (s 0,t 0 ).

a b c t0t0 t1t1 a a b,c s0s0 s1s1 s 0,t 0 s 0,t 1 s 1,t 1 s 1,t 0 b b a c a c

More complicated when A 2  S 2 a b ct0t0 t1t1 a a b,c s0s0 s1s1 Should we have acceptance when both components accepting? I.e., {(s 0,t 1 )}? No, consider (ba)  It should be accepted, but never passes that state. s 0,t 0 s 0,t 1 s 1,t 1 b a c a c

More complicated when A 2  S 2 a b ct0t0 t1t1 a a b,c s0s0 s1s1 Should we have acceptance when at least one components is accepting? I.e., {(s 0,t 0 ),(s 0,t 1 ),(s 1,t 1 )}? No, consider b c  It should not be accepted, but here will loop through (s 1,t 1 ) s 0,t 0 s 0,t 1 s 1,t 1 b c a c a

Intersection - general case q0q0 q2q2 q3q3 q1q1 q 0,q3q 1,q 3 q1,q2q1,q2 a a, c c c, b b c c b a

Version 0: to catch q 0 Version 1: to catch q 2 q 0,q 3 q 1,q 3 q1,q2q1,q2 q 0,q 3 q 1,q 3 q 1,q 2 Move when see accepting of left (q 0 )Move when see accepting of right (q 2 ) Version 0 Version 1 c c c c b a b a

Make an accepting state in one of the version according to a component accepting state q 0,q 3,0 q 1,q 3,0q 1,q 2,0 q 0,q 3,1q 1,q 3,1 q 1,q 2,1 Version 1 Version 0 c c c c b a b a

How to check for emptiness? s 0,t 0 s 0,t 1 s 1,t 1 b a c a c

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

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.

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.

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

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

Model Checking under Fairness Express the fairness as a property φ. To prove a property ψ under fairness, model check φ  ψ. Fair (φ) Bad (¬ψ)Program Counter example

Model Checking under Fairness Specialize model checking. For weak process fairness: search for a reachable strongly connected component, where for each process P either it contains on occurrence of a transition from P, or it contains a state where P is disabled.

Translating from logic to automata (Book: Chapter 6)

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

Generalized Büchi automata Acceptance condition F is a set F={f 1, f 2, …, f n } where each f i is a set of states. To accept, a run needs to pass infinitely often through a state from every set f i.

Translating into simple Büchi automaton q0q0 q2q2 q1q1 q0q0 q2q2 q1q1 Version 0 Version 1 c c c c b a b a

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

Semantics of pR q p qqqqqqq q qq qqqq ¬p¬p ¬p¬p ¬p¬p ¬p¬p ¬p¬p ¬p¬p ¬p¬p ¬p¬p ¬p¬p ¬p¬p ¬p¬p ¬p¬p ¬p¬p

Replace ¬ true by false, and ¬ false by true. Replace ¬ (  \/  ) by ( ¬  ) /\ ( ¬  ) and ¬ (  /\  ) by ( ¬  ) \/ ( ¬  )

Eliminate implications, <>, [] Replace  ->  by ( ¬  ) \/ . Replace <>  by (true U  ). Replace []  by (false R  ).

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

The data structure Incoming NewOld Next Name

The main idea  U  =  \/ (  /\ O (  U  ) )  V  =  /\ (  \/ O (  R  ) ) This separates the formulas to two parts: one holds in the current state, and the other in the next state.

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

Splitting Incoming New Old Next Incoming New Old Next Incoming New Old Next Copy incoming edges, update other field.

Evolving Incoming New Old Next Incoming New Old Next Copy incoming edges, update other field.

Possible cases:  U , split: Add  to New, add  U  to Next. Add  to New. Because  U  =  \/ (  /\ O (  U  )).  R , split: Add  to New. Add  to New,  R  to Next. Because  R  =  /\ (  \/ O (  R  )).

More cases:  \/ , split: Add  to New. Add  to New.  /\ , evolve: Add  to New. O , evolve: Add  to Next.

How to start? Incoming NewOld Next init aU(bUc)

Incoming init aU(bUc) Incoming aU(bUc) bUcbUc a init

Incoming aU(bUc)bUcbUc init Incoming aU(bUc) c (bUc) b

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 version to the old one. Otherwise, add the node to “Nodes”. Generate a successor with “New” set to “Next” of father.

Incoming a,aU(bUc) aU(bUc) init Incoming aU(bUc) Creating a successor node.

How to obtain the automaton? There is an edge from node X to Y labeled with propositions P (negated or non negated), if X is in the incoming list of Y, and Y has propositions P in field “Old”. Initial node is init. Incoming NewOld Next X Node Y a, b, ¬c

The resulted nodes. a, aU(bUc) b, bUc, aU(bUc)c, bUc, aU(bUc) b, bUcc, bUc

a, aU(bUc) b, bUc, aU(bUc)c, bUc, aU(bUc) b, bUcc, bUc All nodes with incoming edge from “init”. Initial nodes

Include only atomic propositions Init a b c cb

Acceptance conditions Use “generalized Buchi automata”, where there are several acceptance sets F1, F2, …, Fn, and each accepted infinite sequence must include at least one state from each set infinitely often. Each set corresponds to a subformula of form  U . Guarantees that it is never the case that  U  holds forever, without .

Accepting w.r.t. bU c a, aU(bUc) b, bUc, aU(bUc)c, bUc, aU(bUc) b, bUcc, bUc All nodes with c, or without bUc.

Acceptance w.r.t. aU (bU c) a, aU(bUc) b, bUc, aU(bUc)c, bUc, aU(bUc) b, bUcc, bUc All nodes with bUc or without aU(bUc).

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: lTransition relation lSize or bound on size Wants to know: lIn what state we started? lIn what state we are? lTransition relation lConformance lSatisfaction 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. (s 0  S - initial state). δ   S    S - transition relation.   S    O – output on edge. Notation: δ(s,a 1 a 2..a n )= δ(… (δ(δ(s,a 1 ),a 2 ) … ),a n ) (s,a 1 a 2..a n )= (s,a 1 ) (δ(s,a 1 ),a 2 )… (δ(… δ(δ(s,a 1 ),a 2 ) … ),a n )

Finite automata (Mealy machines) S - finite set of states. (size n)  – set of inputs. (size d) O – set of outputs, for each transition. (s 0  S - initial state).   S    S - transition relation.  S   O – output on edge. S={s 1, s 2, s 3 },  {a, b}, O={0,1}. δ(s 1,a)=s 3 (also s 1 =a=>s 3 ), δ(s 1,b)=s 2,(also s 1 =b=>s 2 )… (s 1,a)=0, δ(s 1,b)=1,… δ(s 1,ab)=s 1, (s 1,ab)=01 s1s1 s3s3 s2s2 a/0 b/1 b/ 0 b/1 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/1 a/1

Preliminaries: separating sequences s1s1 s3s3 s2s2 a/0 b/1 b/0 b/1 a/0 Start with one block containing all states {s 1, s 2, s 3 }.

A: separate to blocks of states with different output. s1s1 s3s3 s2s2 a/0 b/1 b/0 b/1 a/0 Two sets, separated using the string b {s 1, s 3 }, {s 2 }.

Repeat B: Separate blocks based on moving to different blocks. s1s1 s3s3 s2s2 a/0 b/1 b/0 b/1 a/0 Separate first block using b to three singleton blocks. Separating sequences: b, bb. Max rounds: n-1, sequences: n-1, length: n-1. For each pair of states there is a separating sequence.

Want to know the state of the machine (at end): Homing sequence. Depending on output, we would know in what state we are. Find a sequence µ such that δ(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 ended. For any other state, the output will be different.

Want to know the state of the machine (at end): Homing sequence. Algorithm: Put all the states in one block (initially we do not know what is the current state). Then repeatedly partitions blocks of states, as long as they are not singletons, as follows: Take a non singleton block, append a distinguishing sequence µ that separates at least two states in that block. Update each block to the states after executing µ. Max length: (n-1) 2 (Lower bound: n(n-1)/2.)

Example (homing sequence) s1s1 s3s3 s2s2 a/0 b/1 b/0 b/1 a/0 {s 1, s 2, s 3 } {s 1, s 2 } {s 3 } {s 1 } {s 2 } {s 3 } b b 1 0 011 1 On input b and output 1, we still don’t know if we were in s 1 or s 3, i.e., if we are currently in s 2 or s 1. 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 can be found in polynomial time. a/1 b/1 a/0 b/0b/1 a/0

Algorithm for synchronizing sequeneces s1s1 s2s2 s3s3 a/1 b/1 a/0 b/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 (s 2,s 2 ).) s 1,s 1 s 2,s 2 s 3,s 3 s 1,s 2 s 2,s 3 s 1,s 3 bb b b b b a a a

Algorithm continues (2) There is an input sequence from 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). s 1,s 1 s 2,s 2 s 3,s 3 s 1,s 2 s 2,s 3 s 1,s 3 b b b b b b a a a

Algorithm continues (3) Notation:δ(S,x)= {t|  s  S,δ(s,x)=t} 1. i=1; S i =S 2. Take some nodes s≠t  S i, and find a shortest path labeled x i to some (r,r). 3. i=i+1; S i :=δ(S i-1,x). If |S i |>1,goto 4., else S:=S 1, and goto 3. 4. Concatenate x 1 x 2 …x k. s 1,s 1 s 2,s 2 s 3,s 3 s 1,s 2 s 2,s 3 s 1,s 3 b b b b b b a a a Number of sequences ≤ n-1. Each of length ≤ n(n-1)/2. Overall O(n(n-1) 2 /2).

Example: (s 2,s 3 )=a=>(s 2,s 2 ) x 1 :=a δ({s 1,s 2,s 3 },a)={s 1,s 2 } (s 1,s 2 )=ba=>(s 2,s 2 ) x 2 :=ba δ({s 1,s 2 },ba)={s 2 } So x 1 x 2 =aba is a synchronization sequence, bringing every state into state s 2. s 1,s 1 s 2,s 2 s 3,s 3 s 1,s 2 s 2,s 3 s 1,s 3 b b b b b b a a a

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

Sometimes cannot identify initial state b/1 a/1 s1s1 s3s3 s2s2 b/0 b/1 a/1 Start with a: in case of being in s 1 or s 3 we’ll move to s 1 and cannot distinguish. Start with b: In case of being in s 1 or s 2 we’ll move to s 2 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). s1s1 s3s3 s2s2 a/1 b/0 b/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/1 a/1 b/1

Conformance testing (cont.) a bb a a a a b b b a Need: bound on number of states of B.   a

Preparation: Construct a spanning tree b/1 a/1 s1s1 s3s3 s2s2 b/0 b/1 a/1 s1s1 s2s2 s3s3 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. s1s1 s2s2 s3s3 b/1a/1 Reset Distinguishing sequences

Comments 1. 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). s1s1 s2s2 s3s3 s4s4 s5s5 bdca 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/1 a/1 s1s1 s3s3 s2s2 b/0 b/1 a/1 b/1 Pretends to be S 3 Pretends to be S 1 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: m 2 n d n-m+1 Probabilistic complexity: Polynomial. Distinguishing sequences s1s1 s2s2 s3s3 b/1a/1 Words of length  n-m+1 Reset

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

Buchi automata (  -automata) S - finite set of states. (B has l  n states) S 0  S - initial states. (P has m states)  - finite alphabet. (contains p letters)    S    S - transition relation. F  S - accepting states. Accepting run: passes a state in F infinitely often. System automata: F=S, deterministic, one initial state. Property automaton: not necessarily deterministic.

Example: check a  a, a aa a <>  a

Example: check <>  a aa aa a a  <>  a

Example: check  <>a Use automatic translation algorithms, e.g., [Gerth,Peled,Vardi,Wolper 95] aa aa  a, a <>   a

System cb a

Every element in the product is a counter example for the checked property. cb a aa aa a a s1s1 s2s2 s3s3 q2q2 q1q1 s 1,q 1 s 1,q 2 s 3,q 2 s 2,q 1 a b c a Acceptance is determined by automaton P.  <>  a

Model Checking / Testing Given Finite state system B. Transition relation of B known. Property represent by automaton P. Check if L(B)  L(P)= . Graph theory or BDD techniques. Complexity: polynomial. Unknown Finite state system B. Alphabet and number of states of B or upper bound known. Specification given as an abstract system C. Check if B  C. Complexity: polynomial if number states known. Exponential otherwise.

Black box checking [PVY] Property represent by automaton P. Check if L(B)  L(P)= . Graph theory techniques. Unknown Finite state system B. Alphabet and Upper bound on Number of states of B known. Complexity: exponential.  

Experiments aa bb cc reset a a b b c c try b a a b b c c try c fail

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

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

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

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

Modeling BBC as games Each configuration contains an automaton and its current state (and more). Moves of the  player are labeled with try a, reset... Moves of the  -player with success, fail. c 1  c 2 when the automata in c 1 and c 2 would respond in the same way to the experiments so far.

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

On-the-fly strategy Systematically (as in the deadlock case), find two sequences v 1 and v 2 of length <=m n. Applying v 1 to P brings us to a state t that is accepting. Applying v 2 to P brings us back to t. Apply v 1 v 2 n to B. If this succeeds, there is a cycle in the intersection labeled with v 2, with t as the P (accepting) component. Complexity: O(n 2 p 2mn m). v1v1 v2v2

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 M i and asks for a counterexample. It then generates an automaton with more states M i+1 and so forth.

A strategy based on learning Start the learning algorithm. Queries are just experiments to B. For a conjectured automaton M i, check if M i  P =  If so, we check conformance of M i with B ([VC] algorithm). If nonempty, it contains some v 1 v 2 . We test B with v 1 v 2 n. If this succeeds: error, otherwise, this is a counterexample for M i.

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 2 mn) if there is an error. O(l 3 d l + l 2 n d n-l+1 + l 2 mn) if there is no error. If n is not known, check while time allows. Probabilistic complexity: polynomial.

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

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 declaration All variables declared? Default values understood? Arrays and strings initialized? Variables with similar names? Correct initialization? Control flow Each loop terminates? DO/END statements match? Input/output OPEN statements correct? Format specification correct? End-of-file case handled?

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 20 times 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 coverage Execute every statement at least once By choosing A=2,B=0,X=3 each statement will be chosen. The case where the tests fail is not checked! IF (A>1) & (B=0) THEN X=X/A; END; IF (A=2) | (X>1) THEN X=X+1; END; Now x=1.5

Decision coverage Each decision has a true and false outcome at least once. Can be achieved using A=3,B=0,X=3 A=2,B=1,X=1 Problem: Does not test individual conditions. E.g., when X>1 is erroneous in second decision. IF (A>1) & (B=0) THEN X=X/A; END; IF (A=2) | (X>1) THEN X=X+1; END;

Decision 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 A  1 and the case where x>1 where not checked! IF (A>1) & (B=0) THEN X=X/A; END; IF (A=2)|(X>1) THEN X=X+1; END;

Condition coverage Each condition has a true and false value at least once. For example: A=1,B=0,X=3 A=2,B=1,X=0 lets each condition be true and false once. Problem:covers only the path where the first test fails and the second succeeds. (A>1) IF (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  (A>1) IF (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+decisio n coverage. (A>1) IF (A>1) & (B=0) THEN X=X/A; END; IF (A=2) | (X>1) THEN X=X+1; END;

Multiple Condition Coverage Test all combinations of all conditions in each test. A>1,B=0 A>1,B≠0 A  1,B=0 A  1,B≠0 A=2,X>1 A=2,X  1 A≠2,X>1 A≠2,X  1 IF (A>1) & (B=0) THEN X=X/A; END; IF (A=2) | (X>1) THEN X=X+1; END;

A smaller number of cases: A=2,B=0,X=4 A=2,B=1,X=1 A=1,B=0,X=2 A=1,B=1,X=1 Note the X=4 in the first case: it is due to the fact that X changes before being used! IF (A>1) & (B=0) THEN X=X/A; END; IF (A=2) | (X>1) THEN X=X+1; END; 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 >= 0 Relativize  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’=x1 now eliminate x1’, x2’, y1’, y2’. (y1,y2)=(0,x1) A B A B Y=g(X,Y)

Verification conditions: tests  C)   B)= t(X,Y) /\  C)  D)   B)=  t(X,Y) /\  D)  B)=  D) /\  y2  x2 y2>=x2 B C D B C D t(X,Y) F T 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/A no yes true

How to find values for coverage? A>1 & B=0 A=2 | X>1 X=X+1 X=X/A no yes true A≠2 /\ X>1 (A≠2 /\ X/A>1) /\ (A>1 & B=0) A≠2 /\ X/A>1 Need 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. Duplicate edges and find an Euler path.

Test cases based on data-flow analysis Partition the program into pieces of code with a single entry/exit point. For each piece find which variables are set/used/tested. Various covering criteria: from each set to each use/test From each set to some use/test. 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. Specification condition Valid equivalence class Invalid equivalence class

Example: A legal variable Begins with A-Z Contains [A-Z0-9] Has 1-6 characters. Specification condition Valid equivalence class Invalid equivalence class Starting char Chars Length Starts A-ZStarts other [A-Z0-9]Has others 1-6 chars0 chars, >6 chars 12 34 5 67

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. Specification condition Valid equivalence class Invalid equivalence class

Example AB36P (1,3,5) 1XY12 (2) A17#%X (4) Specification condition Valid equivalence class Invalid equivalence class Starting char Chars Length Starts A-ZStarts other [A-Z0-9]Has others 1-6 chars0 chars, >6 chars 12 34 5 67 (6) VERYLONG (7)

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

Test case generation based on LTL specification Compiler Model Checker Path condition calculation First order instantiator Test monitoring Transitions Path Flow chart LTL  Aut

Goals Verification of software. Compositional verification. Only a unit of code. Parametrized verification. Generating test cases. A path found with some truth assignment satisfying the path condition. In deterministic code, this assignment guarantees to derive the execution of the path. In nondeterministic code, this is one of the possibilities. Can transform the code to force replying the path.

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

Spec: ¬at l 2 U (at l 2 /\  ¬at l 2 /\(¬at l 2 U at l 2 )) ¬ at l 2 at l 2 ¬ at l 2 at l 2 l 2 :x:=x+z l 3 :x<t l1:…l1:… l 2 :x:=x+z l 3 :x<t l 2 :x:=x+z X =

Spec: ¬at l 2 U (at l 2 /\ x  y /\  (¬at l 2 /\(¬at l 2 U at l 2 /\ x  2  y ))) ¬ at l 2 at l 2 /\ x  y ¬ at l 2 at l 2 /\ x  2  y l 2 :x:=x+z l 3 :x<t l1:…l1:… l 2 :x:=x+z l 3 :x<t l 2 :x:=x+z X = xyxy x2yx2y

Example: GCD l 1 :x:=a l 5 :y:=z l 4 :x:=y l 3 :z:=x rem y l 2 :y:=b l 6 :z=0? yesno l0l0 l7l7

Example: GCD l 1 :x:=a l 5 :x:=y l 4 :y:=z l 3 :z:=x rem y l 2 :y:=b l 6 :z=0? yesno Oops … with an error (l4 and l5 were switched). l0l0 l7l7

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

Example: GCD l 1 :x:=a l 5 :x:=y l 4 :y:=z l 3 :z:=x rem y l 2 :y:=b l 6 :z=0? yesno l0l0 l7l7 a>0/\b>0/\at l 0 /\  at l 7 at l 0 /\ a>0/\ b>0 at l 7

Example: GCD l 1 :x:=a l 5 :x:=y l 4 :y:=z l 3 :z:=x rem y l 2 :y:=b l 6 :z=0? yesno l0l0 l7l7 a>0/\b>0/\at l 0 /\  at l 7 Path 1: l 0 l 1 l 2 l 3 l 4 l 5 l 6 l 7 a>0/\b>0/\a rem b=0 Path 2: l 0 l 1 l 2 l 3 l 4 l 5 l 6 l 3 l 4 l 5 l 6 l 7 a>0/\b>0/\a rem b  0

Potential explosion Bad point: potential explosion Good point: may be chopped on-the-fly

Drivers and Stubs Driver: represents the program or procedure that called our checked unit. Stub: represents a procedure called by our checked unit. In our approach: replace both of them with a formula representing the effect the missing code has on the program variables. Integrate the driver and stub specification into the calculation of the path condition. l 1 :x:=a l 5 :x:=y l 4 :y:=z l 3 :z ’ =x rem y /\x ’ =x/\y ’ =x l 2 :y:=b l 6 :z=0? yesno l0l0 l7l7

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

Formal methods: Model Checking and Testing Prof. Doron A. Peled University of Warwick, UK and Bar Ilan University, Israel.

Similar presentations

Presentation on theme: "Formal methods: Model Checking and Testing Prof. Doron A. Peled University of Warwick, UK and Bar Ilan University, Israel."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Formal methods: Model Checking and Testing Prof. Doron A. Peled University of Warwick, UK and Bar Ilan University, Israel.

Similar presentations

Presentation on theme: "Formal methods: Model Checking and Testing Prof. Doron A. Peled University of Warwick, UK and Bar Ilan University, Israel."— Presentation transcript:

Similar presentations

About project

Feedback