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

Some related books: Mainly: Also: http://www.dcs.warwick.ac.uk/~doron

Goal: software reliability Use software engineering methodologies to develop the code. Use formal methods during code development 2

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: Advanced topics: 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 of software 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)) X=7 1<=x<=5 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). Distributive systems. Reactive systems. 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={v0,v1,v2, …} - a set of variables, over some domain. p(v0, v1, …, vn) - a parametrized assertion, e.g., v0=v1+v2 /\ v3>v4. A state is an assignment of values to the program variables. For example: s=<v0=1,v1=3,v3=7,…,v18=2> For predicate (first order assertion) p: p(s) is p under the assignment s. Example: p is x>y /\ y>z. s=<x=4, y=3, z=5>. Then we have 4>3 /\ 3>5, which is false. 2

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>… 6

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. 7

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,a P2:load R2,a inc R2 store R2,a a may be also 1. 8

Scenario P1:load R1,a a=0 R1=0 P2:load R2,a R2=0 inc R1 R1=1 inc R2 store R2,a P1:load R1,a inc R1 store R1,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. 9

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. 10

A transition system A (finite) set of variables V over some domain. A set of states S. A (finite) set of transitions T, each transition e t has an enabling condition e, and a transformation t. An initial condition I. 11

Example V={a, b, c, d, e}. S: all assignments of natural numbers for variables in V. T={c >0(c,e):=(c -1,e +1), d >0(d,e):=(d -1,e +1)} I: c =a /\ d =b /\ e =0 What does this transition system do? 12

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 et: e (si), i.e., si satisfies e. si+1 is obtained by applying t to si. 13

Example: T={c>0(c,e):=(c -1,e +1), d>0(d,e):=(d-1,e+1)} I: c=a /\ d=b /\ e=0 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> 14

The transitions 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) L0:While True do NC0:wait(Turn=0); CR0:Turn=1 endwhile || L1:While True do NC1:wait(Turn=1); CR1:Turn=0 endwhile Initially: PC0=L0/\PC1=L1 Is this the only reasonable way to model this program? 17

The state graph:Successor relation between reachable states. Turn=0 L0,L1 Turn=1 L0,L1 T3 T0 T3 T0 T5 T2 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=1 L0,NC1 Turn=1 NC0,L1 T4 T1 T3 T0 T0 T3 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,NC1 T1 T3 T0 T4 Turn=0 CR0,NC1 Turn=1 NC0,CR1 T2 T5 18

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=0 L0,L1 Turn=1 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=1 L0,NC1 Turn=1 NC0,L1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,NC1 Turn=0 CR0,NC1 Turn=1 NC0,CR1 19

Always if Turn=0 then at some point Turn=1 L0,L1 Turn=1 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=1 L0,NC1 Turn=1 NC0,L1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,NC1 Turn=0 CR0,NC1 Turn=1 NC0,CR1 20

Always if Turn=0 then at some point Turn=1 L0,L1 Turn=1 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=1 L0,NC1 Turn=1 NC0,L1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,NC1 Turn=0 CR0,NC1 Turn=1 NC0,CR1 20

Interleaving semantics: Execute one transition at a time. Turn=0 L0,L1 Turn=0 L0,NC1 Turn=1 L0,NC1 Turn=0 NC0,NC1 Turn=1 L0,CR1 Turn=0 CR0,NC1 Need to check the property for every possible interleaving! 20

Interleaving semantics Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,NC1 Turn=0 CR0,NC1 Turn=1 L0,NC1 Turn=1 L0,CR1 Turn=0 L0,L1 Turn=0 L0,NC1 20

Busy waiting Initially: PC0=L0/\PC1=L1 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:=NC1 T5:PC1=CR1(PC1,Turn):=(L1,0) L0:While True do NC0:wait(Turn=0); CR0:Turn=1 endwhile || L1:While True do NC1:wait(Turn=1); CR1:Turn=0 endwhile Initially: PC0=L0/\PC1=L1 17

Always when Turn=0 then at some point Turn=1 L0,L1 Turn=1 L0,L1 T1’ T4’ Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=1 L0,NC1 Turn=1 NC0,L1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,NC1 Turn=0 CR0,NC1 Turn=1 NC0,CR1 Now it does not hold! (Red subgraph generates a counterexample execution.) 20

Combinatorial explosion V1:=1 Vn:=1 V1:=3 Vn:=3 … V1:=2 Vn:=2 How many states?

Global states … … … 3n states v1=1,v2=1…vn=1 v1=2,v2=1…vn=1

Specification Formalisms (Book: Chapter 5) 1

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. 11 2

The interleaving model An execution is a finite or infinite sequence of states s0, s1, s2, … The initial state satisfies the initial condition, I.e., I (s0). Moving from one state si to si+1 is by executing a transition et: e(si), I.e., si satisfies e. si+1 is obtained by applying t to si. Lets assume all sequences are infinite by extending finite ones by “stuttering” the last state. 13 3

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.) 2

Semantics over suffixes of execution     O  U 3

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.

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”. 4

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

What about ([]<>)/\([]<>)=[]<>(/\)? No, just  ([]<>)\/([]<>)=[]<>(\/)? (<>[])/\(<>[])=<>[](/\)? (<>[])\/(<>[])=<>[](\/)? No, just  Yes!!! Yes!!! No, just 

Formal semantic definition Let  be a sequence s0 s1 s2 … Let i be a suffix of : si si+1 si+2 … (0 = ) i |= p, where p a proposition, if si|=p. i |= /\ if i |=  and i |= . i |= \/ if i |=  or i |= . i |= ¬ if it is not the case that i |= . i |= <> if for some 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 s1 s2 s3 extended extended malfunction release pull r0 = s1 s2 s1 s2 s1 s2 s1 … r1 = s1 s2 s3 s3 s3 s3 s3 … r2 = s1 s2 s1 s2 s3 s3 s3 … …

LTL satisfaction by a single sequence r2 = s1 s2 s1 s2 s3 s3 s3 … s1 s3 s2 pull release extended malfunction r2 |= extended ?? r2 |= O extended ?? r2 |= O O extended ?? r2 |= <> extended ?? r2 |= [] extended ?? r2 |= <>[] extended ?? r2 |= ¬ <>[] extended ?? r2 |= (¬extended) U malfunction ?? r2 |= [](¬extended->O extended) ??

LTL satisfaction by a system pull release extended malfunction P |= extended ?? P |= O extended ?? P |= O O extended ?? P |= <> extended ?? P|= [] extended ?? P |= <>[] extended ?? P |= ¬ <>[] extended ?? P |= (¬extended) U malfunction ?? P |= [](¬extended->O extended) ??

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. + proof rule: _p_ []p

Traffic light example Green  Yellow  Red Always has exactly one light: [](¬(gr/\ye)/\¬(ye/\re)/\¬(re/\gr)/\(gr\/ye\/re)) Correct change of color: []((grU ye)\/(yeU re)\/(reU gr))

Another kind of traffic light GreenYellowRedYellow 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=<, S, , I, F>  (finite) - the alphabet. S (finite) - the states.   S x  x S - the transition relation. I  S - the starting states. F  S - the accepting states. a b s0 s1 5

The transition relation (s0, a, s0) (s0, b, s1) (s1, a, s0) (s1, b, s1) a b s0 s1 6

A run over a word A word over , e.g., abaab. A sequence of states, e.g. s0 s0 s1 s0 s0 s1. Starts with an initial state. Follows the transition relation (si, ci , si+1). Accepting if ends at accepting state. a b s0 s1 7

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 b s0 s1 8

Nondeterministic automaton Transitions: (s0,a ,s0), (s0,b ,s0), (s0,a ,s1),(s1,a ,s1). What is the language of this automaton? s0 s1 a a a,b 9

Equivalent deterministic automaton a,b a s0 s1 a b b 10

Automata over infinite words Similar definition. Runs on infinite words over . Accepts when an accepting state occurs infinitely often in a run. a b s0 s1 12

Automata over infinite words Consider the word abababab… There is a run s0s0s1s0s1s0s1 … This run in accepting, since s0 appears infinitely many times. a b s0 s1 13

Other runs For the word bbbbb… the run is s0 s1 s1 s1 s1… and is not accepting. For the word aaabbbbb …, the run is s0 s0 s0 s0 s1 s1 s1 s1 … What is the run for ababbabbb …? a b s0 s1 14

Nondeterministic automaton What is the language of this automaton? What is the LTL specification if b -- PC0=CR0, a =¬b? s0 s1 a a a,b Can you find a deterministic automaton with same language? Can you prove there is no such deterministic automaton? 15

No deterministic automaton for (a+b)*aω In a deterministic automaton there is one run for each word. After some sequence of a’s, i.e., aaa…a must reach some accepting state. Now add b, obtaining aaa…ab. After some more a’s, i.e., aaa…abaaa…a must reach some accepting state. Now add b, obtaining aaa…abaaa…ab. Continuing this way, one obtains a run that has infinitely many b’s but reaches an accepting state (in a finite automaton, at least one would repeat) infinitely often.

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 b s0 s1 16

Mutual Exclusion a -- PC0=CR0/\PC1=CR1 b -- ¬(PC0=CR0/\PC1=CR1) c -- true []¬(PC0=CR0/\PC1=CR1) b a c s0 s1 17

Apply now to our program: 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) L0:While True do NC0:wait(Turn=0); CR0:Turn=1 endwhile || L1:While True do NC1:wait(Turn=1); CR1:Turn=0 endwhile Initially: PC0=L0/\PC1=L1 17 5

The state space Turn=0 L0,L1 Turn=1 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,NC1 Turn=0 CR0,NC1 Turn=1 NC0,CR1 18

[]¬(PC0=CR0/\PC1=CR1) (Mutual exclusion) Turn=0 L0,L1 Turn=1 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=1 L0,NC1 Turn=1 NC0,L1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,NC1 Turn=0 CR0,NC1 Turn=1 NC0,CR1 19

[](Turn=0 <>Turn=1) L0,L1 Turn=1 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=1 L0,NC1 Turn=1 NC0,L1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,NC1 Turn=0 CR0,NC1 Turn=1 NC0,CR1 20

Interleaving semantics: Execute one transition at a time. Turn=0 L0,L1 Turn=0 L0,NC1 Turn=1 L0,NC1 Turn=0 NC0,NC1 Turn=1 L0,CR1 Turn=0 CR0,NC1 Need to check the property for every possible interleaving! 20

[](Turn=0  <>Turn=1) L0,L1 Turn=1 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=1 L0,NC1 Turn=1 NC0,L1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,NC1 Turn=0 CR0,NC1 Turn=1 NC0,CR1 20 8

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). 21

Correctness Sequences satisfying Spec Program executions All sequences 22

Incorrectness Sequences satisfying Spec Program executions Counter examples Sequences satisfying Spec Program executions All sequences 22

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. 9

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. 10

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! 11

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. 12

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. 13

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. 14

Depth First Search Procedure dfs(s) Program DFS for each s’ such that R(s,s’) do If new(s’) then dfs(s’) end dfs. Program DFS For each s such that Init(s) dfs(s) end DFS 15

Start from an initial state Hash table: q1 q1 q2 q3 Stack: q1 q4 q5 16

Continue with a successor Hash table: q1 q1 q2 q2 q3 Stack: q1 q2 q4 q5 17

One successor of q2. Hash table: q1 q1 q2 q4 q2 q3 Stack: q1 q2 q4 q4 18

Backtrack to q2 (no new successors for q4). Hash table: q1 q1 q2 q4 q2 q3 Stack: q1 q2 q4 q5 19

Backtracked to q1 Hash table: q1 q1 q2 q4 q2 q3 Stack: q1 q4 q5 20

Second successor to q1. Hash table: q1 q1 q2 q4 q3 q2 q3 Stack: q1 q3 21

Backtrack again to q1. Hash table: q1 q1 q2 q4 q3 q2 q3 Stack: q1 q4 22

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. 23

The state graph:Successor relation between states. Turn=0 L0,L1 L0,NC1 NC0,L1 CR0,NC1 NC0,NC1 CR0,L1 Turn=1 L0,CR1 NC0,CR1 18

¬(PC0=CR0/\PC1=CR1) is an invariant! Turn=0 L0,L1 L0,NC1 NC0,L1 CR0,NC1 NC0,NC1 CR0,L1 Turn=1 L0,CR1 NC0,CR1 19 24

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. 25

[](Turn=0  <>Turn=1) L0,L1 Turn=1 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=1 L0,NC1 Turn=1 NC0,L1 Turn=0 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,NC1 Turn=0 CR0,NC1 Turn=1 NC0,CR1 20 8

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

Propositions are attached to incoming nodes. All nodes are accepting. init Turn=0 L0,L1 Turn=1 L0,L1 Turn=0 L0,L1 Turn=1 L0,L1 Propositions are attached to incoming nodes. All nodes are accepting. 19 20

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). 21

Correctness Sequences satisfying Spec Program executions All sequences 22

How to prove correctness? Show that L(Model)  L(Spec). Equivalently: ______ Show that L(Model)  L(Spec) = Ø. Also: can obtain Spec by translating from LTL! 23

What do we need to know? How to intersect two automata? How to complement an automaton? How to translate from LTL to an automaton? 24

Intersecting M1=(S1,,T1,I1,A1) and M2=(S2,,T2,I2,S2) Run the two automata in parallel. Each state is a pair of states: S1 x S2 Initial states are pairs of initials: I1 x I2 Acceptance depends on first component: A1 x S2 Conforms with transition relation: (x1,y1)-a->(x2,y2) when x1-a->x2 and y1-a->y2.

Example (all states of second automaton accepting!) b,c s1 a b,c a t0 c t1 b States: (s0,t0), (s0,t1), (s1,t0), (s1,t1). Accepting: (s0,t0), (s0,t1). Initial: (s0,t0). 26

a s0 s1 b,c a b,c a t0 t1 c b a s0,t0 s0,t1 b a s1,t0 c s1,t1 b c 27

More complicated when A2S2 b,c s1 a a s0,t0 b,c s0,t1 b a a c s1,t1 t0 c t1 b c Should we have acceptance when both components accepting? I.e., {(s0,t1)}? No, consider (ba) It should be accepted, but never passes that state. 26

More complicated when A2S2 b,c s1 a a b,c s0,t0 s0,t1 b a a c s1,t1 t0 c t1 c b Should we have acceptance when at least one components is accepting? I.e., {(s0,t0),(s0,t1),(s1,t1)}? No, consider b c It should not be accepted, but here will loop through (s1,t1) 26

Intersection - general case q0 q2 b a, c a c, b q1 q3 c b c q0,q3 q1,q3 q1,q2 a c

Version 0: to catch q0 Version 1: to catch q2 b c q0,q3 q1,q2 q1,q3 a c Move when see accepting of left (q0) Move when see accepting of right (q2) c b q0,q3 q1,q2 q1,q3 a c Version 1

Version 0: to catch q0 Version 1: to catch q2 b c q0,q3 q1,q2 q1,q3 a c Move when see accepting of left (q0) Move when see accepting of right (q2) c b q0,q3 q1,q2 q1,q3 a c Version 1

Make an accepting state in one of the version according to a component accepting state q0,q3,0 q1,q2,0 q1,q3,0 a c a b c b q0,q3,1 q1,q2 ,1 q1,q3 ,1 c Version 1

How to check for emptiness? a s0,t0 s0,t1 b a c s1,t1 c 28

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! 29

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

Model Checking under 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={f1 , f2 , … , fn } where each fi is a set of states. To accept, a run needs to pass infinitely often through a state from every set fi .

Translating into simple Büchi automaton Version 0 b c q0 q1 q2 a c c b q0 q1 q2 a c Version 1

Translating into simple Büchi automaton Version 0 c q0 q1 q2 a c b c b q0 q1 q2 c Version 1

Translating into simple Büchi automaton Version 0 c q0 q1 q2 a c b c b q0 q1 q2 a c Version 1

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

Semantics of pR q q q q q q q q q q p q q q q q ¬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 R ¬ P ) ) \/ ( false R ( true U Q ) )

The data structure Incoming New Old Next Name

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

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

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

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

Possible cases:  U  , split:  R  , 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:  /\ , evolve: O , evolve: Add  to New. Add  to Next.

How to start? init Incoming New Old aU(bUc) Next

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

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

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.

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

How to obtain the automaton? X Incoming New Old Next 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. a, b, ¬c Node Y

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

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

Include only atomic propositions Init b b c

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 c, bUc b, bUc, aU(bUc) c, bUc, aU(bUc) b, bUc c, 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, bUc c, bUc All nodes with bUc or without aU(bUc).

The SPIN System

What is SPIN? Model-checker. Based on automata theory. Allows LTL or automata specification Efficient (on-the-fly model checking, partial order reduction). Developed in Bell Laboratories.

Documentation Paper: The model checker SPIN, G.J. Holzmann, IEEE Transactions on Software Engineering, Vol 23, 279-295. Web: http://www.spinroot.com

The language of SPIN The expressions are from C. The communication is from CSP. The constructs are from Guarded Command.

Expressions Arithmetic: +, -, *, /, % Comparison: >, >=, <, <=, ==, != Boolean: &&, ||, ! Assignment: = Increment/decrement: ++, --

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

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

Sending and receiving a message Channel declaration: chan qname=[3] of {mtype, byte, byte} In sender: qname!tag3(expr1, expr2) or equivalently: qname!tag3, expr1, expr2 In Receiver: qname?tag3(var1,var2)

Defining an array of channels Channel declaration: chan qname=[3] of {mtype, byte, byte} defines a channel with buffer size 3. chan comm[5]=[0] of {byte, byte} defines an array of channels (indexed 0 to 4. Communication is synchronous (handshaking), meaning that the sender waits for the receiver.

Condition if :: x%2==1 -> z=z*y; x-- :: x%2==0 -> y=y*y; x=x/2 fi If more than one guard is enabled: a nondeterministic choice. If no guard is enabled: the process waits (until a guard becomes enabled).

Looping do :: x>y -> x=x-y :: y>x -> y=y-x :: else break od; Normal way to terminate a loop: with break. (or goto). As in condition, we may have a nondeterministic loop or have to wait.

Processes Definition of a process: proctype prname (byte Id; chan Comm) { statements } Activation of a process: run prname (7, Con[1]);

init process is the root of activating all others init { statements } init {byte I=0; atomic{do ::I<10 -> run prname(I, chan[I]); I=I+1 ::I=10 -> break; od}} atomic allows performing several actions as one atomic step.

Exmaples of Mutual exclusion Reference: A. Ben-Ari, Principles of Concurrent and Distributed Programs, Prentice-Hall 1990.

General structure loop Non_Critical_Section; TR:Pre_Protocol; CR:Critical_Section; Post_protocol; end loop; Propositions: inCRi, inTRi.

Properties loop Non_Critical_Section; TR:Pre_Protocol; CR:Critical_Section; Post_protocol; end loop; Assumption: ~<>[]inCRi Requirements: []~(inCR0/\inCR1) [](inTRi<>inCRi) Not assuming: []<>inTRi

Turn:bit:=1; task P1 is task P0 is begin begin loop loop Non_Critical_Sec; Wait Turn=1; Critical_Sec; Turn:=0; end loop end P1. task P0 is begin loop Non_Critical_Sec; Wait Turn=0; Critical_Sec; Turn:=1; end loop end P0.

Translating into SPIN #define critical (incrit[0] ||incrit[1]) byte turn=0, incrit[2]=0; proctype P (bool id) { do :: 1 -> do :: 1 -> skip :: 1 -> break od; try:do ::turn==id -> break od; cr:incrit[id]=1; incrit[id]=0; turn=1-turn od} init { atomic{ run P(0); run P(1) } }

The leader election algorithm A directed ring of computers. Each has a unique value. Communication is from left to right. Find out which value is the greatest. 2

Example 7 2 3 12 9 4 3

Informal description: Initially, all the processes are active. A process that finds out it does not represent a value that can be maximal turns to be passive. A passive process just transfers values from left to right. 4

More description The algorithm executes in phases. In each phase, each process first sends its current value to the right. Each process, when receiving the first value from its left compares it to its current value. If same: this is the maximum. Tell others. Not same: send current value again to left. 5

Continued When receiving the second value: compare the three values received. These are values of the process itself. of the left active process. of the second active process on the left. If the left active process has greatest value among three, then keep this value. Otherwise, become passive. 6

7 2 3 12 9 4 7

9, 4 4, 12 2 9 2, 9 12, 3 7 4 7, 2 3 12 3, 7 8

9, 4 4, 12 2 9 2, 9 12, 3 7 4 7, 2 3 12 3, 7 9

9 12 7 11

9 7 12 12, 7 7, 9 9, 12 11

12 12

when received(1,number) do if state=active then if number!=max then send(1, my_number); state:=active; when received(1,number) do if state=active then if number!=max then send(2, number); neighbor:=number; else (max is greatest, send to all processes); end if; else send(1,number); end do; when received(2,number) do if state=active then if neighbor>number and neighbor>max then max:=neighbor; send(1, neighbor); else state:=passive; end if; else send(2, number); end do; 13

Now, translate into SPIN (Promela) code 14

Running SPIN Can download and implement (for free) using www.spinroot.com Available in our system. Graphical interface: xspin

Homework: check properties There is no maximal value until a moment where there is one such value, and from there, there is exactly one value until the end. The maximal value is always 5. There is never more than one maximal value found. A maximal value is eventually found. From the time a maximal value is found, we continue to have one maximal value. 15

Dekker’s algorithm boolean c1 initially 1; boolean c2 initially 1; integer (1..2) turn initially 1; P2::while true do begin non-critical section 2 c2:=0; while c1=0 do begin if turn=1 then begin c2:=1; wait until turn=2; end end critical section 2 c2:=1; turn:=1 end. P1::while true do begin non-critical section 1 c1:=0; while c2=0 do begin if turn=2 then begin c1:=1; wait until turn=1; end end critical section 1 c1:=1; turn:=2 end.

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

Modeling issues Book: chapters 4.12, 5.4, 8.4, 10.1

Fairness (Book: Chapter 4.12, 8.3, 8.4)

Dekker’s algorithm boolean c1 initially 1; boolean c2 initially 1; integer (1..2) turn initially 1; P2::while true do begin non-critical section 2 c2:=0; while c1=0 do begin if turn=1 then begin c2:=1; wait until turn=2; c2:=0; end end critical section 2 c2:=1; turn:=1 end. P1::while true do begin non-critical section 1 c1:=0; while c2=0 do begin if turn=2 then begin c1:=1; wait until turn=1; c1:=0; end end critical section 1 c1:=1; turn:=2 end.

Dekker’s algorithm c1=c2=0, turn=1 boolean c1 initially 1; integer (1..2) turn initially 1; P2::while true do begin non-critical section 2 c2:=0; while c1=0 do begin if turn=1 then begin c2:=1; wait until turn=2; c2:=0; end end critical section 2 c2:=1; turn:=1 end. P1::while true do begin non-critical section 1 c1:=0; while c2=0 do begin if turn=2 then begin c1:=1; wait until turn=1; c1:=0; end end critical section 1 c1:=1; turn:=2 end. c1=c2=0, turn=1

Dekker’s algorithm c1=c2=0, turn=1 boolean c1 initially 1; integer (1..2) turn initially 1; P2::while true do begin non-critical section 2 c2:=0; while c1=0 do begin if turn=1 then begin c2:=1; wait until turn=2; c2:=0; end end critical section 2 c2:=1; turn:=1 end. P1::while true do begin non-critical section 1 c1:=0; while c2=0 do begin if turn=2 then begin c1:=1; wait until turn=1; c1:=0; end end critical section 1 c1:=1; turn:=2 end. c1=c2=0, turn=1

Dekker’s algorithm c1=c2=0, turn=1 P1 waits for P2 to set c2 to 1 again. Since turn=1 (priority for P1), P2 is ready to do that. But never gets the chance, since P1 is constantly active checking c2 in its while loop. P2::while true do begin non-critical section 2 c2:=0; while c1=0 do begin if turn=1 then begin c2:=1; wait until turn=2; c2:=0; end end critical section 2 c2:=1; turn:=1 end. P1::while true do begin non-critical section 1 c1:=0; while c2=0 do begin if turn=2 then begin c1:=1; wait until turn=1; c1:=0; end end critical section 1 c1:=1; turn:=2 end. c1=c2=0, turn=1

Initially: turn=1 11:c1:=1 11:c2:=1 12:true 12:true yes no yes no 0:START P1 Initially: turn=1 0:START P2 11:c1:=1 11:c2:=1 12:true 12:true yes no yes no 2:c1:=0 13:end 2:c2:=0 13:end 8:c2=0? 8:c1=0? yes no yes no 7:turn=2? no 7:turn=1? no 9:critical-1 9:critical-2 yes yes 3:c1:=1 10:c1:=1 3:c2:=1 10:c2:=1 5:turn=2? 5:turn=1? 11:turn:=2 yes 11:turn:=1 no yes no 4:no-op 6:c1:=0 4:no-op 6:c2:=0

What went wrong? c2:=0; end end The execution is unfair to P2. It is not allowed a chance to execute. Such an execution is due to the interleaving model (just picking an enabled transition). If it did, it would continue and set c2 to 0, which would allow P1 to progress. Fairness = excluding some of the executions in the interleaving model, which do not correspond to actual behavior of the system. while c1=0 do begin if turn=1 then begin c2:=1; wait until turn=2; c2:=0; end end

Recall: The interleaving model An execution is a finite or infinite sequence of states s0, s1, s2, … The initial state satisfies the initial condition, I.e., I (s0). Moving from one state si to si+1 is by executing a transition et: e(si), I.e., si satisfies e. si+1 is obtained by applying t to si. Now: consider only “fair” executions. Fairness constrains sequences that are considered to be executions. Sequences Executions Fair executions 13

Some fairness definitions Weak transition fairness: It cannot happen that a transition is enabled indefinitely, but is never executed. Weak process fairness: It cannot happen that a process is enabled indefinitely, but non of its transitions is ever executed Strong transition fairness: If a transition is infinitely often enabled, it will get executed. Strong process fairness: If at least one transition of a process is infinitely often enabled, a transition of this process will be executed.

How to use fairness constraints? Assume in the negative that some property (e.g., termination) does not hold, because of some transitions or processes prevented from execution. Show that such executions are impossible, as they contradict fairness assumption. We sometimes do not know in reality which fairness constraint is guaranteed.

Example Initially: x=0; y=0; P1::x=1 P2: do :: y==0 -> if :: true :: x==1 -> y=1 fi od In order for the loop to terminate (in a deadlock !) we need P1 to execute the assignment. But P1 may never execute, since P2 is in a loop executing true. Consequently, x==1 never holds, and y is never assigned a 1. pc1=l0(pc1,x):=(l1,1) /* x=1 */ pc2=r0/\y=0pc2=r1 /* y==0*/ pc2=r1pc2=r0 /* true */ pc2=r1/\x=1(pc2,y):=(r0,1) /* x==1  y:=1 */

Weak transition fairness Initially: x=0; y=0; P2: do :: y==0 -> if :: true :: x==1 -> y=1 fi od P1::x=1 Under weak transition fairness, P1 would assign 1 to x, but this does not guarantee that 1 is assigned to y and thus the P2 loop will terminates, since the transition for checking x==1 is not continuously enabled (program counter not always there). Weak process fairness only guarantees P1 to execute, but P2 can still choose the true guard. Strong process fairness: same.

Strong transition fairness Initially: x=0; y=0; P1::x=1 P2: do :: y==0 -> if :: true :: x==1 -> y=1 fi od Under strong transition fairness, P1 would assign 1 to x. If the execution was infinite, the transition checking x==1 was infinitely often enabled. Hence it would be eventually selected. Then assigning y=1, the main loop is not enabled anymore.

Specifying fairness conditions Express properties over an alternating sequence of states and transitions: s0 1 s1 1 s2 … Use transition predicates exec.

Some fairness definitions exec  is executed. execPi some transition of Pi is executed. en  is enabled. enPi some transition of process Pi is enabled. enPi = \/ Pi en execPi = \/ Pi exec Weak transition fairness: /\ T (<>[]en []<>exec). Equivalently: /\aT ¬<>[](en /\¬exec) Weak process fairness: /\Pi (<>[]enPi []<>execPi ) Strong transition fairness: /\ T ([]<>en []<>exec ) Strong process fairness: /\Pi ([]<>enPi []<>execPi )

“Weaker fairness condition” A is weaker than B if B A. (Means A has more executions than B.) Consider the executions L(A) and L(B). Then L(B)  L(A). If an execution is strong {process/transition} fair, then it is also weak {process/transition} fair. There are fewer strong {process,transition} fair executions. Strong transition fair execs Strong process fair execs Weak transition fair execs Weak process fair execs

Fairness is an abstraction; no scheduler can guarantee exactly all fair executions! Initially: x=0, y=0 P1::x=1 || P2::do :: x==0 -> y=y+1 :: x==1 -> break od x=0,y=0 x=1,y=0 x=0,y=1 x=1,y=1 x=0,y=2 x=1,y=2 Under fairness assumption (any of the four defined), P1 will execute the assignment, and consequently, P2 will terminate. All executions are finite and there are infinitely many of them, and infinitely many states. Thus, an execution tree (the state space) will potentially look like the one on the right, but with infinitely many states, finite branching and only finite sequences. But according to König’s Lemma there is no such tree!

Model Checking under fairness Instead of verifying that the program satisfies , verify it satisfies fair Problem: may be inefficient. Also fairness formula may involves special arrangement for specifying what exec means. May specialize model checking algorithm instead.

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

Abstractions (Book: Chapter 10.1) 1

Problems with software analysis Many possible outcomes and interactions. Not manageable by an algorithm (undecideable, complex). Requires a lot of practice and ingenuity (e.g., finding invariants). 2

More problems Testing methods fail to cover potential errors. Deductive verification techniques require too much time, mathematical expertise, ingenuity. Model checking requires a lot of time/space and may introduce modeling errors. 3

How to alleviate the complexity? Abstraction Compositionality Partial Order Reduction Symmetry 4

Abstraction Represent the program using a smaller model. Pay attention to preserving the checked properties. Do not affect the flow of control. 5

Main idea Use smaller data objects. x:= f(m) y:=g(n) if x*y>0 then … else … x, y never used again. 6

How to abstract? Assign values {-1, 0, 1} to x and y. Based on the following connection: sgn(x) = 1 if x>0, 0 if x=0, and -1 if x<0. sgn(x)*sgn(y)=sgn(x*y). 7

Abstraction mapping S - states, I - initial states. L(s) - labeling. R(S,S) - transition relation. h(s) maps s into its abstract image. Full model -h Abstract model I(s)  I(h(s)) R(s, t)  R(h(s),h(t)) L(h(s))=L(s) 8

go Traffic light example stop stop 9

go go stop stop stop 10

What do we preserve? Every execution of the full model can be simulated by an execution of the reduced one. Every LTL property that holds in the reduced model hold in the full one. But there can be properties holding for the original model but not the abstract one. go go stop stop stop 11

Preserved: [](go->O stop) Not preserved: []<>go Counterexamples need to be checked. go stop stop stop 12

Symmetry A permutation is a one-one and onto function p:AA. For example, 13, 24, 31, 45, 52. One can combine permutations, e.g., p1: 13, 21, 32 p2: 12, 21, 33 p1@p2: 13, 22, 31 A set of permutations with @ is called a symmetry group. 13

Using symmetry in analysis Want to find some symmetry group such that for each permutation p in it, R(s,t) if and only if R(p(s), p(t)) and L(p(s))=L(s). Let K(s) be all the states that can be permuted to s. This is a set of states such that each one can be permuted to the other. 14

init Turn=0 L0,L1 Turn=1 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 Turn=1 NC0,NC1 Turn=0 CR0,L1 Turn=1 L0,CR1 Turn=1 NC0,NC1 Turn=0 CR0,NC1 Turn=1 NC0,CR1

The quotient model init Turn=0 L0,L1 Turn=0 L0,NC1 Turn=0 NC0,L1 NC0,NC1 Turn=0 CR0,L1 Turn=0 CR0,NC1

Homework: what is preserved in the following buffer abstraction Homework: what is preserved in the following buffer abstraction? What is not preserved? e empty q quasi q q full f

BDD representation

Computation Tree Logic EG p AF p p p p p p p p p p p . . . . . . . . . . . . . . . . . . . . . . . .

Computation Tree Logic E pUq A pUq p p p p q q q q . . . . . . . . . . . . . . . . . . . . . . . .

Example formulas CTL formulas: mutual exclusion: AG ( cs1  cs2) non starvation: AG (request  AF grant) “sanity” check: EF request in 4 cycles the result of the addition will get into z.

Model Checking M |= f [Clarke, Emerson, Sistla 83] The Model Checking algorithm works iteratively on subformulas of f , from simpler subformulas to more complex ones When checking subformula g of f we assume that all subformulas of g have already been checked For subformula g, the algorithm returns the set of states that satisfy g ( Sg ) The algorithm has time complexity: O(|M||f|)

Model checking f = EF g Given a model M= < S, I, R, L > and Sg the sets of states satisfying g in M procedure CheckEF (Sg ) Q := emptyset; Q’ := Sg ; while Q  Q’ do Q := Q’; Q’ := Q  { s | s' [ R(s,s’)  Q(s’) ] } end while Sf := Q ; return(Sf ) CheckEF gets as parameter the set Sg (and implicitly the model M) and returns the set Sf

Example: f = EF g g f f f

Model checking f = EG g CheckEG gets M= < S, I, R, L > and Sg and returns Sf procedure CheckEG (Sg) Q := S ; Q’ := Sg ; while Q  Q’ do Q := Q’; Q’ := Q { s | s' [ R(s,s’)  Q(s’) ] } end while Sf := Q ; return(Sf )

Example: f = EG g g

Symbolic model checking [Burch, Clarke, McMillan, Dill 1990] If the model is given explicitly (e.g. by adjacent matrix) then only systems with about ten Boolean variables (~1000 states) can be handled Symbolic model checking uses Binary Decision Diagrams ( BDDs ) to represent the model and sets of states. It can handle systems with hundreds of Boolean variables.

Binary decision diagrams (BDDs) [Bryant 86] Data structure for representing Boolean functions Often concise in memory Canonical representation Boolean operations on BDDs can be done in polynomial time in the BDD size

BDDs in model checking Assume that states in model M are encoded by {0,1}n and described by Boolean variables v1...vn Sf can be represented by a BDD over v1...vn R (a set of pairs of states (s,s’) ) can be represented by a BDD over v1...vn v1’...vn’

BDD definition A tree representation of a Boolean formula. Each leaf represents 0 (false) or 1 (true). Each internal leaf represents a node. If we follow a path in the tree and go from a node left (low) on 0 and right (high) on 1, we obtain a leaf that corresponds to the value of the formula under this truth assignment.

Example a b c (a/\(b\/¬c))\/(¬a/\(b/\c)) 1

OBDD: there is some fixed appearance order between variables, e. g OBDD: there is some fixed appearance order between variables, e.g., a<b<c a b c 1 (a/\(b\/¬c))\/(¬a/\(b/\c))

In reduced form: combine all leafs with same values, all isomorphic subgraph. 1 1 1 1 (a/\(b\/¬c))\/(¬a/\(b/\c)) In addition, remove nodes with identical children (low(x)=high(x)).

In reduced form: combine all leafs with same values, all isomorphic subgraph. 1 (a/\(b\/¬c))\/(¬a/\(b/\c)) In addition, remove (shortcut) nodes with identical children (low(x)=high(x)). Apply bottom up until not possible.

In reduced form: combine all leafs with same values, all isomorphic subgraph. 1 (a/\(b\/¬c))\/(¬a/\(b/\c)) In addition, remove (shortcut) nodes with identical children (low(x)=high(x)).

In reduced form: combine all leafs with same values, all isomorphic subgraph. 1 (a/\(b\/¬c))\/(¬a/\(b/\c)) In addition, remove (shortcut) nodes with identical children (low(x)=high(x)).

Example, even parity, 3 bits c c c c 1 1 1 1

Apply reduce a b b c c c c 1

Apply reduce a b b c c c c 1

Apply reduce a b b c c c 1

Apply reduce a b b c c c 1

Apply reduce a b b c c 1

f[0/x], f[1/x] (“restrict” algorithm) Obtain the replacement of a variable x by 0 or 1, in formula f, respectively. For f[0/x], incoming edges to node x are redirected to low(x), and x is removed. For f[1/x], incoming edges to node x are redirected to high(x), and x is removed. Then we reduce the OBBD.

Calculate x x= [0/x]\/[1/x] Thus, we apply “restrict” twice to  and then “apply” the disjunction.

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

Now compute f#g recursively: Let rf be the root of the OBDD for f, and rg be the root of the OBDD for g. If rf and rg are terminals, then apply rf#rg and put the result. If both roots are same node (say x), then create a low edge to low(rf)#low(rg), and a high edge to high(rf)#high(rg). If rf is x and rg is y, and x<y, there is no x node in g, so g=g[0/x]=g[1/x]. So we create a low edge to low(rf)#g and a high edge to high(rf)#g. The symmetric case is handled similarly. We reduce.

Same subgraphs are not needed to be explored again (use memoising, i.e., dynamic programming, complexity: exponential 2xmultiplications of sizes. R1,S1 x R2,S3 R3,S2 y z R4,S3 R3,S3 R4,S3 R6,S5 z t x z t y 1 R1 R3 R2 R5 R6 R4 S1 S3 S2 S4 S5 t R4,S3 R6,S3 t R5,S4 R6,S5 R5,S4 R6,S5 t R6,S5 R6,S4 R5,S4 R6,S5

Same subgraphs are not needed to be explored again (use memoising, i.e., dynamic programming, complexity: exponential 2xmultiplications of sizes. R1,S1 x R2,S3 R3,S2 y z R4,S3 R3,S3 R4,S3 R6,S5 z t x z t y 1 R1 R3 R2 R5 R6 R4 S1 S3 S2 S4 S5 t R4,S3 R6,S3 t R5,S4 R6,S5 R5,S4 R6,S5 t R6,S5 R6,S4 R5,S4 R6,S5

Symbolic Model Checking Characterize CTL formulas using fixpoints. AF= \/AX AF EF= EX EF  Z.EX Z AG= /\AX A G EG= EX EG  Z.EX Z AU = \/(/\AXU) EU =  (EXU) Z.(EXZ) AR = /\(\/AXR) ER = (EXR)  Z.(EXZ)

Representing the successor relation formula R A relation between the current state and the next state can be represented as a BDD with prime variables representing the variables at next states. For example: p/\¬q/\r/\¬p’/\q’/\r’ says that the current state satisfies p/\¬q/\r and the next state satisfies ¬p/\q/\r. (typically, for one transition, represented as a Boolean relation). If ti represents this relation for transition i, we can write for the entire code R=\/i ti.

Calculating (Z) for (Z)=EX Z Z is a BDD. Rename variables in Z by their primed version to obtain BDD Z’. Calculate the BDD R/\Z’. Let y1’…yn’ be the primed variables, Then calculate the BDD T=y1’… yn’ R/\Z’ to remove primed variables. Calculate the BDD T.

procedure Check LFP () Q :=False; Q’ := (Q) ; while Q  Q’ do Model checking Z  (least fixed point) For example, = EX Z For formulas with main operator . procedure Check LFP () Q :=False; Q’ := (Q) ; while Q  Q’ do Q := Q’; Q’ := (Q) ; end while return(Q) CheckEF gets as parameter the set Sg (and implicitly the model M) and returns the set Sf

procedure Check GFP () Q :=True; Q’ := (Q) ; while Q  Q’ do Model checking Z  (Greatest fixed point) For example, = (EX Z) For formulas with main operator . procedure Check GFP () Q :=True; Q’ := (Q) ; while Q  Q’ do Q := Q’; Q’ := (Q) ; end while return(Q) CheckEF gets as parameter the set Sg (and implicitly the model M) and returns the set Sf

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, y18 Each 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,…,xl Program variables: Y=y1,y2,…,ym Output variables: Z=z1,z2,…,zn start Z=h(X,Y) Y=f(X) halt

Assignments and tests T F Y=g(X,Y) t(X,Y)

Initial condition start Initial condition: the values for the input variables for which the program must work. x1>=0 /\ x2>0 (y1,y2)=(0,x1) y2>=x2 F T (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) halt

The input-output claim start The relation between the values of the input and the output variables at termination. x1=z1*x2+z2 /\ 0<=z2<x2 (y1,y2)=(0,x1) y2>=x2 T F (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) halt

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 is partially correct with respect to start halt (y1,y2)=(0,x1) y2>=x2 (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) T F The program is partially correct with respect to x1>=0/\x2>=0 and totally correct with respect to x1>=0/\x2>0

Annotating a scheme start A 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. (y1,y2)=(0,x1) B T F y2>=x2 C D (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) E halt

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 start A A): x1>=0 /\ x2>=0 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 E):x1=z1*x2+z2 /\ 0<=z2<x2 Notice: (A) is the initial condition,  Eis the input-output condition. (y1,y2)=(0,x1) B T F y2>=x2 C D (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) E A) Is the precondition of (y1,y2)=(0,x1) and B) is its postcondition halt

Preliminary: Relativizing assertions (B) : x1= y1 * x2 + y2 /\ y2 >= 0 Relativize B) w.r.t. the assignment, obtaining B) [Y\g(X,Y)] (I.e., (B) expressed w.r.t. variables at A.)  (B)A =x1=0 * x2 + x1 /\ x1>=0 Think about two sets of variables, before={x, y, z, …} after={x’,y’,z’…}. Rewrite (B) using after, and the assignment as a relation between the set of variables. Then eliminate after by substitution. Here: x1’=y1’ * x2’ + y2’ /\ y2’>=0 /\ x1’=x1 /\ x2’=x2 /\ y1’=0 /\ y2’=x1 now eliminate x1’, x2’, y1’, y2’. A (y1,y2)=(0,x1) Y=g(X,Y) B A (y1,y2)=(0,x1) B

Preliminary: Relativizing assertions (B)A (y1,y2)=(0,x1) Y=g(X,Y) A): B Y=g(X,Y) A (B) (y1,y2)=(0,x1) B

Verification conditions: assignment A)  B)A where B)A = B)[Y\g(X,Y)] A): x1>=0 /\ x2>=0 B): x1=y1*x2+y2 /\ y2>=0 B)A= x1=0*x2+x1 /\ x1>=0 A (y1,y2)=(0,x1) Y=g(X,Y) B A (y1,y2)=(0,x1) B

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

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 D (z1,z2)=(y1,y2) E

Verification conditions: tests B T F t(X,Y) 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 C D B T F y2>=x2 C D

Verification conditions: tests B T F t(X,Y) C C) D t(X,Y) ¬t(X,Y) B) B T F y2>=x2 C D

Partial correctness proof: An induction on length of execution 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! A) no B) yes start A C) (y1,y2)=(0,x1) B T F no B) y2>=x2 C D yes (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) E D) halt

Exercise: prove partial correctness start (y1,y2)=(0,1) Initial condition: x>=0 Input-output claim: z=x! F T y1=x (y1,y2)=(y1+1,(y1+1)*y2) z=y2 halt

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. 15

Another example start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2) z=y1 B C D F true false E y2=y2+y3 (A) : x>=0 (F) : z^2<=x<(z+1)^2 z is the biggest number that is not greater than sqrt x. 16

Some insight start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2) z=y1 B C D F true false E y2=y2+y3 1+3+5+…+(2n+1)=(n+1)^2 y2 accumulates the above sum, until it is bigger than x. y3 ranges over odd numbers 1,3,5,… y1 is n-1. 17

Invariants start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2) z=y1 B C D F true false E y2=y2+y3 It is sufficient to have one invariant for every loop (cycle in the program’s graph). We will have (C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 18

Obtaining (B) start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2) z=y1 B C D F true false E y2=y2+y3 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 /\ 19

Check assignment condition start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2) z=y1 B C D F true false E y2=y2+y3 (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+1 Simplified: x>=0 20

Obtaining (D) start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2) z=y1 B C D F true false E y2=y2+y3 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 21

Checking start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2) z=y1 B C D F true false E y2=y2+y3 (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 22

y1^2<=x /\ y1^2<=x /\ y2=(y1+1)^2 /\ 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 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 23

Not finished! start A (y1,y2,y3)=(0,0,1) Still needs to: Calculate (E) by substituting backwards from (F). Check that (C)/\y2>x(E) B y2=y2+y3 C false true y2>x D E (y1,y3)=(y1+1,y3+2) z=y1 F halt 24

Exercise: prove partial correctness. Initially: x1>0/\x2>0 Exercise: prove partial correctness. Initially: x1>0/\x2>0. At termination: z1=gcd(x1,x2). start (y1,y2)=(x1,x2) y1=y2 F T y1>y2 F T z1=y1 y2=y2-y1 y1=y1-y2 halt

Annotation of program with invariants gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1>y2 start gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1<y2 A x1>0 /\ x2>0 (y1,y2)=(x1,x2) gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0 B gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1y2 y1=y2 T F D G y1=gcd(x1,x2) F T y1>y2 z1=y1 E F y2=y2-y1 y1=y1-y2 H z1=gcd(x1,x2) halt

Part 1 y1=y2 T F F T start y1>y2 z1=y1 y2=y2-y1 y1=y1-y2 (A)= x1>0 /\ x2>0 start (B)=gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0 A (B)’rel= gcd(x1,x2)=gcd(x1,x2)/\x1>0/\x2>0 (y1,y2)=(x1,x2) (A) (B)’rel B y1=y2 T F D G F T y1>y2 z1=y1 E F y2=y2-y1 y1=y1-y2 H halt

Part 2a y1=y2 T F F T start y1>y2 z1=y1 y2=y2-y1 y1=y1-y2 (B)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0 start (D)=gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1y2 A (y1,y2)=(x1,x2) (B)/\¬(y1=y2) (D) B y1=y2 T F D G F T y1>y2 z1=y1 E F y2=y2-y1 y1=y1-y2 H halt

Part 2b y1=y2 T F F T start y1>y2 z1=y1 y2=y2-y1 y1=y1-y2 (B)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0 start (G)= y1=gcd(x1,x2) A (B)/\(y1=y2) (G) (y1,y2)=(x1,x2) B y1=y2 T F D G F T y1>y2 z1=y1 E F y2=y2-y1 y1=y1-y2 H halt

Part 3 y1=y2 T F F T start y1>y2 z1=y1 y2=y2-y1 y1=y1-y2 (F)=(gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0/\y1>y2 start (D)/\(y1>y2) (F) (E)=gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0/\y1<y2 A (D)/\¬(y1>y2) (E) (y1,y2)=(x1,x2) B (D)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1y2 y1=y2 T F G D F T y1>y2 z1=y1 E F y2=y2-y1 y1=y1-y2 H halt

Part 4 y1=y2 T F F T start y1>y2 z1=y1 y2=y2-y1 y1=y1-y2 (B)’rel1= gcd(y1,y2-y1)=gcd(x1,x2) /\y1>0/\y2-y1>0 (F)= gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0/\y1>y2 (B)’rel2= gcd(y1-y2,y2)=gcd(x1,x2) /\y1-y2>0/\y2>0 start (E)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1<y2 A x1>0 /\ x2>0 (y1,y2)=(x1,x2) (B)= gcd(y1,y2)=gcd(x1,x2) /\y1>0/\y2>0 B y1=y2 T F D G F T y1>y2 z1=y1 E F y2=y2-y1 y1=y1-y2 H halt (E) (B)’rel1 (F) (B)’rel2

Annotation of program with invariants start (H)’rel= y1=gcd(x1,x2) A (y1,y2)=(x1,x2) B (G)= y1=gcd(x1,x2) y1=y2 T F D G F T y1>y2 z1=y1 E F (H)= z1=gcd(x1,x2) y2=y2-y1 y1=y1-y2 H halt (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 start 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. A (y1,y2)=(0,x1) B y2>=x2 C true false D (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) E halt

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 j(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). M S N M T N

How not to stay in a loop? M For stmt: j(M)(u(M)>=u(N)’rel) Relativize since we need to compare values not syntactic expressions. For test (true side): (j(M)/\test)(u(M)>=u(N)) For test (false side): (j(M)/\¬test)(u(M)>=u(L)) stmt N M true false test N L

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 start u(A)=x1 u(B)=y2 u(C)=y2 u(D)=y2 u(E)=z2 W: naturals > : greater than A (y1,y2)=(0,x1) B y2>=x2 C true false D (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) E halt

Recall partial correctness annotation start A (y1,y2)=(0,x1) j(A): x1>=0 /\ x2>=0 j(B): x1=y1*x2+y2 /\ y2>=0 j(C): x1=y1*x2+y2 /\ y2>=0 /\ y2>=x2 j(D):x1=y1*x2+y2 /\ y2>=0 /\ y2<x2 j(E):x1=z1*x2+z2 /\ 0<=z2<x2 B true false y2>=x2 C D (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) E halt

Strengthen for termination start A j(A): x1>=0 /\ x2>0 j(B): x1=y1*x2+y2 /\ y2>=0/\x2>0 j(C): x1=y1*x2+y2 /\ y2>=0/\y2>=x2/\x2>0 j(D):x1=y1*x2+y2 /\ y2>=0 /\ y2<x2/\x2>0 j(E):x1=z1*x2+z2 /\ 0<=z2<x2 (y1,y2)=(0,x1) B true false y2>=x2 C D (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) E halt

Strengthen for termination j(A): x1>=0 /\ x2>0 u(A)>=0 j(B): x1=y1*x2+y2 /\ y2>=0/\x2>0u(B)>=0 j(C): x1=y1*x2+y2 /\y2>=0 /\y2>=x2/\x2>0u(c)>=0 j(D):x1=y1*x2+y2 /\ y2>=0 /\ y2<x2/\x2>0u(D)>=0 j(E):x1=z1*x2+z2 /\ 0<=z2<x2u(E)>=0 This proves that u(M) is natural for each point M. u(A)=x1 u(B)=y2 u(C)=y2 u(D)=y2 u(E)=z2

We shall show: start halt (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) B D E false y2>=x2 C true u(A)=x1 u(B)=y2 u(C)=y2 u(D)=y2 u(E)=z2 j(A)u(A)>=u(B)’rel j(B)u(B)>=u(C) j(C)u(C)>u(B)’rel j(B)u(B)>=u(D) j(D)u(D)>=u(E)’rel

Proving decrement start halt (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) B D E false y2>=x2 C true j(C): x1=y1*x2+y2 /\ y2>=0 /\ y2>=x2/\x2>0 u(C)=y2 u(B)=y2 u(B)’rel=y2-x2 j(C)  y2>y2-x2 (notice that j(C)  x2>0)

Integer square prog. start (y1,y2,y3)=(0,0,1) A halt y2>x (y1,y3)=(y1+1,y3+2) z=y1 B C D F true false E y2=y2+y3 j(C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 j(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 true false E y2=y2+y3 u(A)=x+1 u(B)=x-y2+1 u(C)=max(0,x-y2+1) u(D)=x-y2+1 u(E)=u(F)=0 u(A)>=u(B)’rel u(B)>u(C)’rel u(C)>=u(D) u(C)>=u(E) u(D)>=u(B)’rel Need some invariants, i.e., y2<=x/\y3>0 at points B and D, and y3>0 at point C.

Program Verification Using Hoare’s Logic Hoare triple is of the form {Precondition} Prog-segment {Postcondition} It expresses partial correctness: if the segment starts with a state satisfying the precondition and it terminates, the final state satisfies the postscondition. The idea is that one can decompose the proof of the program into smaller and smaller segments, depending on its structure.

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

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

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

Assignment axiom {p[e/y] } y:=e {p} For example: {y+5=10} y:=y+5 {y=10} {y+y<z} x:=y {x+y<z} {2*(y+5)>20} y:=2*(y+5) {y>20} Justification: write p with y’ instead of y, and add the conjunct y’=e. Next, eliminate y’ by replacing y’ by e.

Why axiom works backwards? {p} y:=t {?} Strategy: write p and the conjunct y=t, where y’ replaces y in both p and t. Eliminate y’. This y’ represents value of y before the assignment. {y>5} y:=2*(y+5) {? } {p} y:=t { $y’ (p[y’/y] /\ t[y’/y]=y) } y’>5 /\ y=2*(y’+5)  y>20

Composition rule {p} S1 {r }, {r} S2 {q } {p} S1;S2 {q} For example: if the antecedents are 1. {x+1=y+2} x:=x+1 {x=y+2} 2. {x=y+2} y:=y+2 {x=y} Then the consequent is {x+1=y+2} x:=x+1; y:=y+2 {x=y}

More examples {p} S1 {r}, {r} S2 {q} {p} S1;S2 {q} {x1>0/\x2>0} y1:=x1 {gcd(x1,x2)=gcd(y1,x2)/\y1>0/\x2>0} {gcd(x1,x2)=gcd(y1,x2)/\y1>0/\x2>0} y2:=x2 ___{gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0}____ {x1>0/\x2>0} y1:=x1 ; y2:=x2 {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0}

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

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

Consequence rules Strengthen a precondition rp, {p } S {q } {r } S {q } Weaken a postcondition {p } S {q }, qr {p } S {r }

Use of first consequence rule Want to prove {x1>0/\x2>0} y1:=x1 {gcd(x1,x2)=gcd(y1,x2)/\y1>0/\x2>0} By assignment rule: {gcd(x1,x2)=gcd(x1,x2)/\x1>0/\x2>0} y1:=x1 {gcd(x1,x2)=gcd(y1,x2)/\y1>0/\x2>0} x1>0/\x2>0 gcd(x1,x2)=gcd(x1,x2)/\x1>0/\x2>0

Combining program {x1>0 /\ x2>0} y1:=x1; y2:=x1; {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0} while S do if e then S1 else S2 fi od {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0/\y1=y2} Combine the above using concatenation rule!

Not completely finished {x1>0/\x2>0} y1:=x1; y2:=x1; while ¬(y1=y2) do if e then S1 else S2 fi od {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0/\y1=y2} But we wanted to prove: {x1>0/\x1>0} Prog {y1=gcd(x1,x2)}

Use of second consequence rule {x1>0/\x2>0} Prog {gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0/\y1=y2} And the implication gcd(x1,x2)=gcd(y1,y2)/\y1>0/\y2>0/\y1=y2 y1=gcd(x1,x2) Thus, {x1>0/\x2>0} Prog {y1=gcd(x1,x2)}

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

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

Consequence rules Strengthen a precondition rp, {p} S {q} {r} S {q} Weaken a postcondition {p} S {q}, qr {p} S {r}

Soundness Hoare logic is sound in the sense that everything that can be proved is correct! This follows from the fact that each axiom and proof rule preserves soundness.

Completeness A proof system is called complete if every correct assertion can be proved. Propositional logic is complete. No deductive system for the standard arithmetic can be complete (Godel).

And for Hoare’s logic? Let S be a program and p its precondition. Then {p} S {false} means that S never terminates when started from p. This is undecideable. Thus, Hoare’s logic cannot be complete.

Weakest prendition, Strongest postcondition For an assertion p and code S, let post(p,S) be the strongest assertion such that {p}S{post(p,S) } That is, if {p}S{q} then post(p,S)q. For an assertion q and code S, let pre(S,q) be the weakest assertion such that {pre(S,q)}S{q} That is, if {p}S{q} then ppre(S,q).

Relative completeness Suppose that either post(p,S) exists for each p, S, or pre(S,q) exists for each S, q. Some oracle decides on pure implications. Then each correct Hoare triple can be proved. What does that mean? The weakness of the proof system stem from the weakness of the (FO) logic, not of Hoare’s proof system.

Extensions Many extensions for Hoare’s proof rules: Total correctness Arrays Subroutines Concurrent programs Fairness

Proof rule for total correctness Similar idea to Floyd’s termination: Well foundedness {p/\t/\f=z} S {p/\f<z}, p(f>=0) {p} while t do S od {p/\¬t} where z - an int. variable, not appearing in p,t,e,S. f - an int. expression.

Verification with Array Variables Book: Chapter 7.2

The problem Using array variables can lead to complication: {x[1]=1/\x[2]=3} x[x[1]]:=2 {x[x[1]]=2} No!!! Why? Because the assignment changes x[1] as well. Now it is also 2, and x[x[1]], which is x[2] is 3 and not 2!

What went wrong? Take the postcondition {x[x[1]]=2} and substitute 2 instead of x[x[1]]. We obtain {2=2} (which is equivalent to {true}). Now, (x[1]=1/\x[2]=3) 2=2. So we may wrongly conclude that the above Hoare triple is correct.

How to fix this? `Backward substitution’ should be done with arrays as complete elements. Define (x; e1: e2) : an array like x, with value at the index e1 changed to e2. (x; e1: e2)[e3]=e2 if e1=e3 x[e3] otherwise (x; e1: e2)[e3]=if (e1=e3, e2, x[e3])

Solved the problem? How to deal with if(φ, e1, e2)? Suppose that formula ψ contains this expression. Replace if(φ, e1, e2) by new variable v in ψ. The original formula ψ is equivalent to: (φ/\ ψ[e1/v])\/(¬φ/\ ψ[e2/v])

Returning to our case Our postcondition is {x[x[1]]=2}. The assignment x[x[1]]:=2 causes the substitution in the postcondition of the (array) variable x by a new array, which is (x; x[1] : 2), resulting in {x[x[1]]=2} (x; x[1] : 2)[(x; x[1] : 2)[1]] = 2

Are we done? Not yet. It remains to Will not be done in class. Convert the array form into an if form. Get rid of the if form. Will not be done in class. All we say is that we obtain an expression that is not implied by the precondition x[1]=1/\x[2]=3.

Proof checking with PVS Book: Chapter 3 1

A Theory Name: THEORY BEGIN Definitions (types, variables, constants) Axioms Lemmas (conjectures, theorems) END Name 2

Group theory (*, e), where * is the operator and e the unity element. Associativity (G1): (x*y)*z=x*(y*z). Unity (G2): (x*e)=x Right complement (G3): x y x*y=e. Want to prove: x y y*x=e. 3

Informal proof =y*(x*(y*z)) (by G1) Choose x arbitrarily. By G3, there exists y s.t. (1) x*y=e. By G3, we have z s.t. (2) y*z=e. y*x=(y*x)*e (by G2) =(y*x)*(y*z) (by (2)) =y*(x*(y*z)) (by G1) =y*((x*y)*z) (by G1) =y*(e*z) (by (1)) =(y*e)*z (by G1) =y*z (by (G2)) =e (by (2)) 4

Example: groups Group: THEORY BEGIN element: TYPE unit: element *: [element, element-> element] < some axioms> left:CONJECTURE FORALL (x: element): EXISTS (y: element): y*x=unit END Group 5

Axioms associativity: AXIOM FORALL (x, y, z:element): (x*y)*z=x*(y*z) unity: AXIOM FORALL (x:element): x*unit=x complement: AXIOM FORALL(x:element): EXISTS (y:element): x*y=unity 6

Skolemization Corresponds to choosing some arbitrary constant and proving “without loss of generality”. Want to prove (…/\…)(…\/x(x)\/…). Choose a new constant x’. Prove (…/\…)(…\/(x’)\/…). 7

Skolemization Corresponds to choosing some unconstrained arbitrary constant when one is known to exist. Want to prove (…/\x(x)/\…)(…\/…). Choose a new constant x’. Prove (…/\(x’)/\…)(…\/…). 8

Skolem in PVS (skolem 2 (“a1” “b2” “c7”)) (skolem -3 (“a1” “_” “c7”)) (skolem! -3) invents new constants, e.g., for x will invent x!1, x!2, … when applied repeatedly. 9

Instantiation Corresponds to restricting the generality. Want to prove (…/\x(x)/\…) (…\/…). Choose a some term t. Prove (…/\(t)/\…)(…\/…). 10

Instantiation Corresponds to proving the existence of an element by showing an evidence. Want to prove (…/\…)(…\/x(x)\/…). Choose some term t. Prove (…/\…)(…\/(t)\/…). 11

Instantiating in PVS (inst -1 “x*y” “a” “b+c”) (inst 2 “a” “_” “x”) 12

Other useful rules (replace -1 (-1 2 3)) Formula -1 is of the form le=ri. Replace any occurrence of le by ri in lines -1, 2, 3. (replace -1 (-1 2 3) RL) Similar, but replace ri by le instead. (assert), (assert -) (assert +) (assert 7) Apply algebraic simplification. (lemma “<axiom-name>”) - add axiom as additional antecedent. 13

Algorithmic Testing

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 Wants to know: In what state we started? In what state we are? Transition relation Conformance Satisfaction of a temporal property Know: Transition relation Size or bound on size

Finite automata (Mealy machines) S - finite set of states. (size n) S - set of inputs. (size d) O - set of outputs, for each transition. (s0  S - initial state). δ  S  S  S - transition relation (deterministic but sometimes not defined for each input per each state).   S  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) S– set of inputs. (size d) O – set of outputs, for each transition. (s0  S - initial state). δ  S  S  S - transition relation.   S  S O – output on edge. S={s1, s2, s3 }, S={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/0 b/1 b/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 b/1 a/1 When the black box is nondeterministic, we might never test some choices.

Preliminaries: separating sequences b/1 b/0 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 µX (s, µ )≠(s, µ ) Start with one block containing all states {s1, s2, s3}.

A: separate to blocks of states with different output. 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. 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. s1 s2 a/0 b/1 b/0 a/0 s3 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.

Want to know the state of the machine (at end): Homing sequence. After firning homing sequence, we know in what state we are by observing outputs. Find a sequence µ such 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 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!) Max length of sequence: (n-1)2 (Lower bound: n(n-1)/2.)

Example (homing sequence) b/1 b/0 {s1, s2, s3} b 1 1 {s1, s2} {s3} b 1 1 {s1} {s2} {s3} 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/0 b/0

Algorithm for synchronizing sequeneces 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).) a/0 s1 b/1 b/0 s1,s1 a/0 b s2 s3 b/1 s2,s2 s1,s2 b a/1 a a b b b a s3,s3 s2,s3 s1,s3 b

Algorithm continues (2) There is an input sequence from both s≠t to some r iff there is a path in this graph from (s,t ) to (r,r ). There is a synchronization sequence iff some node (r,r ) is reachable from every pair of distinct nodes. In this case it is (s2,s2 ). s1,s1 b s2,s2 s1,s2 b a a b b b a s3,s3 s2,s3 s1,s3 b

Algorithm continues (3) Notation:δ(S,x)= {t |sS,δ(s,x)=t } i=1; S1=S Take some nodes s≠tSi, and find a shortest path labeled xi to some (r,r ). i=i+1; Si:=δ(Si-1,x ). If |Si |>1,goto 2., else goto 3. Concatenate x1x2…xk. s1,s1 b s2,s2 s1,s2 b a a b b b a s3,s3 s2,s3 s1,s3 b Number of sequences ≤ n-1. Each of length ≤ n(n-1)/2. Overall O(n(n-1)2/2).

Example: s1,s1 s2,s2 s1,s2 s3,s3 s2,s3 s1,s3 (s2,s3)=a=>(s2,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 b s2,s2 s1,s2 b a a b b b a s3,s3 s2,s3 s1,s3 b

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 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. b/1 a/1 s1 s3 s2 b/0 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 b/0 b/1 ? = Black box machine has no more states than specification machine (errors are mistakes in outputs, mistargeted edges). Specification machine is reduced, connected, deterministic. Machine resets reliably to a single initial state (or use homing sequence).

Conformance testing [Ch,V] b/1  b/1 b/1 a/1 a/1 b/1  a/1 b/1 Cannot distinguish if reduced or not.

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

Preparation: Construct a spanning tree, and separating sequences b/1 a/1 s1 s3 s2 b/0 s1 s2 s3 b/1 a/1 Given an initial state, we can reach any state of the automaton.

How the algorithm works? Reset According to the spanning tree, force a sequence of inputs to go to each state. From each state, perform the distinguishing sequences. From each state, make a single transition, check output, and use distinguishing sequences to check that in correct target state. Reset s1 a/1 b/1 s3 s2 Distinguishing sequences

Comments Checking the different distinguishing sequences (m-1 of them) means each time resetting and returning to the state under experiment. A reset can be performed to a distinguished state through a homing sequence. Then we can perform a sequence that brings us to the distinguished initial state. Since there are no more than m states, and according to the experiment, no less than m states, there are m states exactly. Isomorphism between the transition relation is found, hence from minimality the two automata recognize the same languages.

Combination lock automaton Assume accepting states. Accepts only words with a specific suffix (cdab in the example). c d a b s1 s2 s3 s4 s5 Any other input

When only a bound on size of black box is known… Black box can “pretend” to behave as a specification automaton for a long time, then upon using the right combination, make a mistake. a/1 Pretends to be s1 a/1 b/1 b/1 s1 s2 b/1 a/1 a/1 s3 Pretends to be s3 b/0

Conformance testing algorithm [VC] Reset Reset The worst that can happen is a combination lock automaton that behaves differently only in the last state. The length of it is the difference between the size n of the black box and the specification m. Reach every state on the spanning tree and check every word of length n-m+1 or less. Check that after the combination we are at the state we are supposed to be, using the distinguishing sequences. No need to check transitions: already included in above check. Complexity: m2 n dn-m+1 Probabilistic complexity: Polynomial. s1 a/1 b/1 s3 s2 Words of length n-m+1 Distinguishing sequences

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

Buchi automata (w-automata) S - finite set of states. (B has l  n states) S0  S - initial states. (P has m states) S - finite alphabet. (contains p letters) d  S  S  S - transition relation. F  S - accepting states. Accepting run: passes a state in F infinitely often. System automata: F=S, deterministic, one initial state. Property automaton: not necessarily deterministic.

Example: check a a <>a a a, a

Example: check <>a

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

System c b a

Acceptance is determined by automaton P. Every element in the product is a counter example for the checked property. a a <>a s1 s2 q1 a c b a a s3 q2 a s1,q1 s2,q1 Acceptance is determined by automaton P. b a s1,q2 s3,q2 c

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

Black box checking [PVY] Outputs: {0,1} (for enabled/disabled) Property represent by automaton P. Check if L(B )  L(P )=. Graph theory techniques. Unknown Finite state system B. Alphabet and Upper bound on Number of states of B known. Complexity: exponential.  

Experiments reset try c try b (Output 1+ progress) (Output 0 =stay) a

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

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

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

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

Modeling BBC as games Each configuration contains an automaton and its current state (and more). Moves of the $-player are labeled with try a, reset... Moves of the -player with success, fail. c1 @ c2 when the automata in c1 and c2 would respond in the same way to the experiments so far.

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

On-the-fly strategy Systematically (as in the deadlock case), find two sequences v1 and v2 of length <=m n. Applying v1 to P brings us to a state t that is accepting. Applying v2 to P brings us back to t. Apply v1 v2 n to B. If this succeeds, there is a cycle in the intersection labeled with v2, with t as the P (accepting) component. Complexity: O(n2p2mnm). v1 v2

Learning an automaton Use Angluin’s algorithm for learning an automaton. The learning algorithm queries whether some strings are in the automaton B. It can also conjecture an automaton Mi and asks for a counterexample. It then generates an automaton with more states Mi+1 and so forth.

A strategy based on learning Start the learning algorithm. Queries are just experiments to B. For a conjectured automaton Mi , check if Mi  P =  If so, we check conformance of Mi with B ([VC] algorithm). If nonempty, it contains some v1 v2w . We test B with v1 v2n. If this succeeds: error, otherwise, this is a counterexample for Mi .

Complexity l - actual size of B. n - an upper bound of size of B. d - size of alphabet. Lower bound: reachability is similar to deadlock. O(l 3 d l + l 2mn) if there is an error. O(l 3 d l + l 2 n d n-l+1+ l 2mn) if there is no error. If n is not known, check while time allows. Probabilistic complexity: polynomial.

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

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 White Box Is based on the internal structure of the program. There are several alternative criterions for checking “enough” paths in the program. Even checking all paths (highly impractical) does not guarantee finding all errors (e.g., missing paths!)

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

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 (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 IF (A>1) & (B=0) THEN X=X/A; END; 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. 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  IF (A>1) & (B=0) THEN X=X/A; END; IF (A=2) | (X>1) THEN X=X+1; END; A=1,B=0,X=3 

Condition coverage A=2,B=1,X=0  IF (A>1) & (B=0) THEN X=X/A; END; IF (A=2) | (X>1) THEN X=X+1; END; A=2,B=1,X=0  Did not check the first THEN part at all!!! Can use condition+decision coverage.

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

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)] (I.e., ( B) expressed w.r.t. variables at A.)  (B)A =x1=0 * x2 + x1 /\ x1>=0 Think about two sets of variables, before={x, y, z, …} after={x’,y’,z’…}. Rewrite (B) using after, and the assignment as a relation between the set of variables. Then eliminate after. Here: x1’=y1’ * x2’ + y2’ /\ y2’>=0 /\ x1=x1’ /\ x2=x2’ /\ y1’=0 /\ y2’=x1 now eliminate x1’, x2’, y1’, y2’. A (y1,y2)=(0,x1) Y=g(X,Y) B A (y1,y2)=(0,x1) B

Verification conditions: tests B T F t(X,Y) C)  B)= t(X,Y) /\C) D)  B)=t(X,Y) /\D) B)= D) /\ y2x2 C D B T F y2>=x2 C D

How to find values for coverage? Put true at end of path. Propagate path backwards. On assignment, relativize expression. On “yes” edge of decision, add decision as conjunction. On “no” edge, add negation of decision as conjunction. Can be more specific when calculating condition with multiple condition coverage. A>1 & B=0 no yes X=X/A A=2 | X>1 true no yes X=X+1 true

How to find values for coverage? (A≠2 /\ X/A>1) /\ (A>1 & B=0) A>1 & B=0 A≠2 /\ X/A>1 no yes Need to find a satisfying assignment: A=3, X=6, B=0 Can also calculate path condition forwards. X=X/A A≠2 /\ X>1 A=2 | X>1 true no yes X=X+1 true

How to cover a flow chart? 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 t>y x>y z:=z+x

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

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

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 Starts A-Z Starts other 1 2 Chars [A-Z0-9] Has others 3 4 1-6 chars 0 chars, >6 chars Length 5 6 7

Equivalence partition (cont.) Add a new test case until all valid equivalence classes have been covered. A test case can cover multiple such classes. Add a new test case until all invalid equivalence class have been covered. Each test case can cover only one such class. Valid equivalence class Invalid equivalence class Specification condition

Example AB36P (1,3,5) 1XY12 (2) A17#%X (4) (6) VERYLONG (7) Specification condition Valid equivalence class Invalid equivalence class Starting char Starts A-Z Starts other 1 2 Chars [A-Z0-9] Has others 3 4 1-6 chars 0 chars, >6 chars Length 5 6 7

Boundary value analysis 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 Driver Driver Write “drivers”, which replaces procedures calling the code. Write “stubs”, which replaces procedures called by the code. Tested unit Stub Stub Stub

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

Bucket sort: Without 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]; // BUG 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); // BUG 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++]; // 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"); }}

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 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 l2U (at l2/\ ¬at l2/\(¬at l2U at l2)) l2:x:=x+z l3:x<t l1:… l2:x:=x+z ¬at l2 X = at l2 l3:x<t ¬at l2 l2:x:=x+z at l2

Spec: ¬at l2U (at l2/\ xy /\ (¬at l2/\(¬at l2U at l2 /\ x2y ))) l2:x:=x+z l3:x<t l1:… l2:x:=x+z ¬at l2 X = at l2/\ xy l3:x<t x2y ¬at l2 l2:x:=x+z at l2/\ x2y

Example: GCD l1:x:=a l2:y:=b l3:z:=x rem y l4:x:=y l5:y:=z no l6:z=0? yes l7

Example: GCD Oops…with an error (l4 and l5 were switched). l1:x:=a l2:y:=b Oops…with an error (l4 and l5 were switched). l3:z:=x rem y l4:y:=z l5:x:=y no l6:z=0? yes l7

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

Example: GCD l1:x:=a l2:y:=b a>0/\b>0/\at l0 /\at l7 l3:z:=x rem y at l0/\ a>0/\ b>0 l4:y:=z l5:x:=y no l6:z=0? yes at l7 l7

Example: GCD l1:x:=a l2:y:=b a>0/\b>0/\at l0/\at l7 l3:z:=x rem y Path 1: l0l1l2l3l4l5l6l7 a>0/\b>0/\a rem b=0 Path 2: l0l1l2l3l4l5l6l3l4l5l6l7 a>0/\b>0/\a rem b0 l4:y:=z l5:x:=y no l6:z=0? yes l7

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

Drivers and Stubs l1:x:=a l2:y:=b Driver: represents the program or procedure that called our checked unit. Stub: represents a procedure called by our checked unit. In our approach: replace both of them with a formula representing the effect the missing code has on the program variables. Integrate the driver and stub specification into the calculation of the path condition. l3:z’=x rem y /\x’=x/\y’=x l4:y:=z l5:x:=y no l6:z=0? yes l7

Process Algebra (Book: Chapter 8)

The Main Issue Q: When are two models equivalent? A: When they satisfy different properties. Q: Does this mean that the models have different executions?

What is process algebra? An abstract description for nondeterministic and concurrent systems. Focuses on the transitions observed rather than on the states reached. Main correctness criterion: conformance between two models. Uses: system refinement, model checking, testing.

Different models may have the same set of executions! b c b c a-insert coin, b-press pepsi, c-press pepsi-light d-obtain pepsi, e-obtain pepsi-light

Actions: Act={a,b,c,d}{}. Agents: E, E’, F, F1, F2, G1, G2, … Agent E may evolve into agent E’. Agent F may evolve into F1 or F2.

Events. E E’ G2 G1 F1 F2 F a b c E—aE’, F—aF1, F—aF2, F1—aG1, F2—aG2. G1—F, G1—F. 

Actions and co-actions E E’ a b c  G2 G1 F1 F2 F a b c   For each action a, except for , there is a co-action a. a and a interact (a input, a output). The coaction of a is a.

Notation a.(b+c) (actually, a.((b.0)+(c.0)) E—aF F—bG F—cH 0 – deadlock/termination. a F b c G H a.E – execute a, then continue according to E. E+F – execute according to E or to F. E||F – execute E and F in parallel.

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

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

Semantics (proof rule and axioms). Structural Operational Semantics SOS a.p –a p p—ap’ |-- p+q –a p’ q—aq’ |-- p+q –a q’ p—ap’ |-- p|q –a p’|q q—aq’ |-- p|q –a p|q’ p—ap’, q—aq’ |-- p|q –t p’|q’ p—ap’ , a  R |-- p\L –a p’\R p—ap’ |-- p[m]—m(a)p’[m]

Action Prefixing a.E—aE (Axiom) Thus, a.(b.(c||c)+d)—a(b.(c||c)+d).

Choice E—aE’ F—aF’ (E+F)—aE’ (E+F)—aF’ b.(c||c)—b(c||c). Thus, (b.(c||c)+e)—b(c||c). If E—aE’ and F—aF’, then E+F has a nondeterministic choice.

Concurrent Composition E—aE’ F—aF’ E||F—aE’||F E||F—aE||F’ E—aE’, F—aF’ ———————— E||F—E’||F’ c—c0, c—c0, c||c—0||0, c||c—c0||c, c||c—cc||0.

Restriction E—aE’, a, a R ————————— E\R –aE’\R In this case: allows only internal interaction of c. c||c—0||0 c||c—c0||c c||c—cc||0 (c||c) \ {c}—(0||0) \{c}

Relabeling E—aE’ ———— E[m] –m(a)E’[m] No axioms/rules for agent 0.

Examples a a.E||b.F b E||b.F a.E||F a b E||F

Derivations a.(b.(c||c)+d) a b.(c||c)+d b d (c||c) c c  (0||c) (c||0) c c  (0||c) (c||0) c c (0||0)

Modeling binary variable set_1 C0 C1 set_0 is_0? set_1 is_1? set_0 C0=is_0? . C0 + set_1 . C1 + set_0 . C0 C1=is_1? . C1 + set_0 . C0 + set_1 . C1

Equational Definition b c  G2 G1 F1 F2 F a b c   E=a.(b..E+c..E) E—aE’, A=E F=a.b..F+a.c..F A—aE’

Trace equivalence: Systems have same finite sequences. b b b E=a.(b+c) F=(a.b)+a.(b+c) Same traces

Failures: comparing also what we cannot do after a finite sequence. b c b b Failure of agent E: (σ, X), where after executing σ from E, none of the events in X is enabled. Agent F has failure (a, {c}), which is not a failure of E.

Simulation equivalence F E a a a b b b b c d c d Relation over set of agents S. RSS. E R F If E’ R F’ and E’—aE’’, then there exists F’’, F’—aF’’, and E’’ R F’’.

Simulation equivalence F E a a a b b b b c d c d Relation over set of agents S. RSS. E R F If E’ R F’ and E’—aE’’, then there exists F’’, F’—aF’’, and E’’ R F’’.

Here, simulation works only in one direction. No equivalence! want to establish F E a a a symmetrically b b b b necessarily c d c d problem!!! Relation over set of agents S. RSS. E R F If E’ R F’ and E’—aE’’, then there exists F’’, F’—aF’’, and E’’ R F’’.

Simulation equivalent but not failure equivalent Left agent a.b+a has a failure (a,{b}).

Bisimulation: same relation simulates in both directions F E a a a b b Not in this case: different simulation relations.

Hierarchy of equivalences Bisimulation Simulation Failure Trace

Example: A=a.((b.nil)+(c.d.A)) B=(a.(b.nil))+(a.c.d.B) a b c d s0 s1 t0 t1 t4 t2 t3 A=a.((b.nil)+(c.d.A)) B=(a.(b.nil))+(a.c.d.B)

Bisimulation between G1 and G2 Let N= N1 U N2 A relation R : N1 x N2 is a bisumulation if If (m,n) in R then 1. If m—am’ then $n’:n—an’ and (m’,n’) in R 2. If n—an’ then $m’:m—am’ and (m’,n’) in R. Other simulation relations are possible, I.e., m=a=> m’ when m—t…—a... —tm’.

Algorithm for bisimulation: Partition N into blocks B1B2…Bn=N. Initially: one block, containing all of N. Repeat until no change: Choose a block Bi and a letter a. If some of the transitions of Bi move to some block Bj and some not, partition Bi accordingly. At the end: Structures bisimilar if initial states of two structures are in same blocks.

Correctness of algorithm Invariant: if (m,n) in R then m and n remain in the same block throughout the algorithm. Termination: can split only a finite number of times.

Example: {s0,t0},{s1,s2,s3,t1,t2,t3,t4} split on b a b c d s0 s1 s2 s3 {s0,t0},{s1,t1},{s0,s2,s3,t2,t3,t4}

Example: {s0,t0},{s1,t1},{s2,s3,t2,t3,t4} split on c b c d s0 s1 s2 s3 t0 t1 t4 t2 t3 {s0,t0},{s1,t1},{s2,s3,t2,t3,t4} split on c {s0,t0},{s1},{t1},{s2,s3,t2,t3,t4}

Example: {s0,t0},{s1},{t1},{s2,s3,t2,t3,t4} split on c b c d s0 s1 s2 s3 t0 t1 t4 t2 t3 {s0,t0},{s1},{t1},{s2,s3,t2,t3,t4} split on c {s0,t0},{s1},{t1},{t4},{s2,s3,t2,t3}

Example: {s0,t0},{s1},{t1},{t4},{s2,s3,t2,t3} split on d b c d s0 s1 s2 s3 t0 t1 t4 t2 t3 {s0,t0},{s1},{t1},{t4},{s2,s3,t2,t3} split on d {s0,t0},{s1},{t1},{t4},{s3, t3},{s2,t2}

Example: {s0,t0},{s1},{t1},{t4},{s2,t2},{s3,t3} split on a a b c d s0

Example: {s0},{t0},{s1},{t1},{t4},{s2,s3,t2,t3} split on d a b c d s0 {s0},{t0},{s1},{t1},{t4},{s3},{t3},{s2,t2}