Model Checking and Testing combined Doron Peled, University of Warwick
Mar 28, 2015
Model Checking and Testing
combinedDoron Peled,
University of Warwick
Why “model checking”?
Want to verify hardware and code.
Want to perform verification automatically. Manual methods are time consuming, difficult.
Restricting to finite state systems.
Willing to give up exhaustiveness.
Checking a (mathematical) model of a system, not the system itself.
Want to obtain counterexamples.
A transition system
A (finite) set of variables V. A set of states . A (finite) set of transitions T, each transition
et has an enabling condition e and a transformation t.
An initial condition p. Denote that s’ is a successor of s by R(s,s’).
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., p(s0).
Moving from one state si to si+1 is by executing a transition et: e(si), i.e., si satisfies e. si+1 is obtained by applying t to si.
L0:While True do NC0:wait(Turn=0); CR0:Turn=1endwhile ||L1:While True do NC1:wait(Turn=1); CR1:Turn=0endwhile
T0:PC0=L0PC0=NC0T1:PC0=NC0/\Turn=0 PC0:=CR0T2:PC0=CR0 (PC0,Turn):=(L0,1)T3:PC1==L1PC1=NC1T4:PC1=NC1/\Turn=1 PC1:=CR1T5:PC1=CR1 (PC1,Turn):=(L1,0)
Initially: PC0=L0/\PC1=L1
The state space
Turn=0L0,L1
Turn=0L0,NC1
Turn=0NC0,L1
Turn=0CR0,NC1
Turn=0NC0,NC1
Turn=0CR0,L1
Turn=1L0,CR1
Turn=1NC0,CR1
Turn=1L0,NC1
Turn=1NC0,NC1
Turn=1NC0,L1
Turn=1L0,L1
Invariant: (PC0=CR0/\PC1=CR1)
Turn=0L0,L1
Turn=0L0,NC1
Turn=0NC0,L1
Turn=0CR0,NC1
Turn=0NC0,NC1
Turn=0CR0,L1
Turn=1L0,CR1
Turn=1NC0,CR1
Turn=1L0,NC1
Turn=1NC0,NC1
Turn=1NC0,L1
Turn=1L0,L1
How can we check the model?
The model is a graph. The specification should refer the
the graph representation. Apply graph theory algorithms.
What properties can we check without using temporal specification?
Invariants: a property that needs to hold in each state.
Deadlock detection: can we reach a state where the program is blocked?
Dead code: does the program have parts that are never executed.
How to perform the check?
Apply a search strategy (Depth first search, Breadth first search).
Check states/transitions during the search.
If property does not hold, report counterexample!
DFS – on-the-fly verification. BFS – finds the shortest
counterexample.
If it is so good, why learn deductive verification methods?
Model checking works for finite state systems. Would not work with Unconstrained integers. Unbounded message queues. General data structures:
queues trees stacks
parametric algorithms and systems. Some new algorithms for infinite
systems.
The state space explosion
Need to represent the state space of a program in the computer memory. Each state can be as big as the entire
memory! Many states:
Each integer variable has 2^32 possibilities. Two such variables have 2^64 possibilities.
In concurrent protocols, the number of states usually grows exponentially with the number of processes.
If it is so constrained, is it of any use?
Many protocols are finite state. Many programs or procedures 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 (BDDs, Partial Order Reduction).
Depth First Search
Program DFSFor each s such that
q(s) dfs(s)end DFS
Procedure dfs(s)for each s’ such
that R(s,s’) do
If new(s’) then dfs(s’)
end dfs.
Start from an initial state
q3
q4
q2
q1
q5
q1
q1
Stack:
Hash table:
Continue with a successor
q3
q4
q2
q1
q5
q1 q2
q1
q2
Stack:
Hash table:
One successor of q2.
q3
q4
q2
q1
q5
q1 q2 q4
q1
q2
q4
Stack:
Hash table:
Backtrack to q2(no new successors for q4).
q3
q4
q2
q1
q5
q1 q2 q4
q1
q2
Stack:
Hash table:
Backtracked to q1
q3
q4
q2
q1
q5
q1 q2 q4
q1
Stack:
Hash table:
Second successor to q1q4 has been already visited.
q3
q4
q2
q1
q5
q1 q2 q4 q3
q1
q3
Stack:
Hash table:
Backtrack again to q1.
q3
q4
q2
q1
q5
q1 q2 q4 q3
q1
Stack:
Hash table:
How can we check properties with DFS?
Invariants: check that all reachable statessatisfy the invariant property. If not, showa path from an initial state to a bad state.
Deadlocks: check whether a state where noprocess can continue is reached.
Dead code: as we progress with the DFS, mark all the transitions that are executed at least once.
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
is temporal logics. Temporal logic specification can be
translated into graphs (finite automata).
Temporal Logic First order logic or propositional
assertions describe a state. Modalities:
<>p means p will happen eventually.
[]p means p will happen always.
p
p pppppp
More temporal logic p U q – p has to hold until q holds.
p p qpp
• []<>p --- its always the case that there is a later p, i.e., p happens infinitely often.
• <>[]p --- At some point p will hold forever, i.e., p is stable.
• <>p/\<>q both p and q would happen eventually.Note, this is not the same as <>(p/\q).
[](Turn=0<>Turn=1)
Turn=0L0,L1
Turn=0L0,NC1
Turn=0NC0,L1
Turn=0CR0,NC1
Turn=0NC0,NC1
Turn=0CR0,L1
Turn=1L0,CR1
Turn=1NC0,CR1
Turn=1L0,NC1
Turn=1NC0,NC1
Turn=1NC0,L1
Turn=1L0,L1
Correctness condition
We want to find a correctness condition for a model to satisfy a specification.
Language of a model: L(Model) Language of a specification:
L(Spec).
We need: L(Model) L(Spec).
Correctness
All sequences
Sequences satisfying Spec
Program executions
Incorrectness
All sequences
Sequences satisfying Spec
Program executions
CounterexamplesCounterexamples
are sometimes more interesting and useful than finding that the program is “correct” due to:
•Underspecification
•Modeling errors
•Algorithm and tool limitation
How to check correctness?
Show that L(Model) L(Spec). Equivalently: ______
Show that L(Model) L(Spec) = Ø. Also: can obtain L(Spec) by
translating from LTL!
What do we need to know?
How to intersect two automata? How to complement an
automaton?Well, not really, if the specification is given in LTL, we can negate the specification and then translate.
How to translate from LTL to an automaton?
Büchi automata (-automata)
S - finite set of states. S0 S - initial states.
- finite alphabet. S S - transition relation. F S - accepting states.
Accepting run: passes a state in F infinitely often.
System automata: F=S.
Example: check a
a, aa
a<>a
Example: check <>a
a
a
a
a
<>a
Example: check <>a
Use automatic translation algorithms, e.g., [Gerth,Peled,Vardi,Wolper 95]
a
a
a, a<>a
Turn=0L0,L1
Turn=1L0,L1
init
•Add an initial node.
•Propositions are attached to incoming nodes.
•All nodes are accepting.
Turn=1L0,L1
Turn=0L0,L1
Technicality…
System
s1
s3
s2
c b
a
All states are accepting! = no fairness conditions
Every element in the product is a counter example for the checked property.
q1
q2a
a
a
a
Acceptance isdetermined byautomaton P.
s1,q1 s2,q1
s1,q2
a
b
c
as3,q2
s1
s3
s2
c b
a
How to check for (non)emptiness?
s1,q1 s2,q1
s1,q2
a
b
c
as3,q2
Nonemptiness...
Need to check if there exists an accepting run, i.e., infinite sequence that passes through an accepting state infinitely often.
Finding accepting runs
If there is an accepting run, then at least one accepting state repeats on it forever. This state must appear on a cycle. So, find a reachable accepting state on a cycle.
Equivalently...
A strongly connected component: a maximal set of nodes where each node is reachable by a path from each other node. Find a reachable strongly connected component with an accepting node.
How to complement?
Complementation is hard! Can ask for the negated property (the
sequences that should never occur). Can translate from LTL formula to
automaton A, and complement A. But: can translate ¬ into an automaton directly!
Model Checking under Fairness
Express the fairness as a property φ.To prove a property ψ under fairness,model check φψ.
Fair (φ)
Bad (¬ψ) Program
Counter
example
Model Checking under Fairness
Specialize model checking. For weak process fairness: search for a reachable strongly connected component, where for each process P either
it contains on occurrence of a transition from P, or
it contains a state where P is disabled.
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.
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
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.
Combination lock automaton
Accepts only words with a specific suffix (cdab in the example).
s1 s2 s3 s4 s5
bdc a
Any other input
Conformance testing
aba
a
b
b
Cannot distinguish if reduced or not.
ab ab
Conformance testing (cont.)
When the black box is nondeterministic, we might never test some choices.
b a
a
Conformance testing (cont.)
ab b
a
a
a
a b
b
b
a
Need: bound on number of states of B.
a
Need reliable RESET
s1
s3
s2
a
a
a
bb
Start with a:in case of being in s1 or s3 we’ll move to s1 and cannot distinguish.Start with b:In case of being in s1 or s2 we’ll move to s2 and cannot distinguish.
The kind of experiment we do affects what we can distinguish. Much like the Heisenberg principle in Physics.
[VC] algorithm
Known automaton A has l states.
Black box automaton has up to n states.
Check each transition. Check that there are no "combination lock" errors.
Complexity: O(l2 n p n-l+1). When n=l: O(l3p).
s1
s2s3
b/1a/1
Words of length n-m+1
Rese
t or
hom
ing
Rese
t or
hom
ing
Distinguishing sequences
Experiments
aa
bb cc
reset
a
a
b
b
c
c
try ba
a
b
b
c
c
try c
fail
Simpler problem: deadlock?
Nondeterministic algorithm:guess a path of length n from the initial state to a deadlock state.Linear time, logarithmic space.
Deterministic algorithm:systematically try paths of length n, one after the other (and use reset), until deadlock is reached.Exponential time, linear space.
Deadlock complexity Nondeterministic algorithm:
Linear time, logarithmic space. Deterministic algorithm:
Exponential (pn-1) time, linear space.
Lower bound: Exponential time (usecombination lock automata).
How does this conform with what we know about complexity theory?
Modeling black box checking
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 equivalent conference.
Nondeterministic strategy: Can distinguish.
Modeling BBC as games
Each configuration contains an automaton and its current state (and more).
Moves of the player are labeled withtry a, reset... Moves of the -player withsuccess, fail.
c1 c2 when the automata in c1 and c2 would respond in the same way to the experiments so far.
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 [PVY]
Start the learning algorithm. Queries are just experiments to B. For a conjectured automaton Mi ,
check if Mi P =
If so, we check conformance of Mi with B ([VC] algorithm).
If nonempty, it contains some v1 (v2) . We test B with v1 v2
n+1. If this succeeds: error, otherwise, this is a counterexample for Mi .
Complexity
l - real size of B. n - an upper bound of size of B. p - size of alphabet. Lower bound: reachability is similar to
deadlock. O(l 3 p l + l 2mn) if there is an error. O(l 3 p l + l 2 n p n-l+1+ l 2mn) if there is no
error.If n is not known, check while time allows.Average 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. Another application: Adaptive Model
Checking when the model may not be accurate [GPY].
Unit checking [GP93]
Check a unit of code, e.g., a bunch of interacting procedures, a-la unit testing.
No initial states are given, not finite state, parametric, compositional.
Use temporal properties to describe suspicious paths in the execution.
Guide the path search with property. Use flow chart instead of “state space”. Cannot check whether a state occurred,
use DFS and iterative deepening.
Unit testing of code:Calculating path condition
A>1 & B=0
A=2 | X>1
X=X+1
X=X/Ano
no
yes
yes
true
true
A≠2 /\ X>1
(A≠2 /\ X/A>1) /\ (A>1 & B=0)
A≠2 /\ X/A>1
Need to find a satisfying assignment:
A=3, X=6, B=0
If deterministic code, starting with such initial values will enforce executing this path
Spec: ¬at l2U (at l2/\ xy /\
(¬at l2/\(¬at l2U at l2 /\ x2y )))
¬at l2
at l2/\xy
¬at l2
at l2/\x2y
l2:x:=x+z
l3:x<t
l1:…
l2:x:=x+z
l3:x<t
l2:x:=x+z
XX==
xy
x2y
Now simplify condition using theorem proving.
Test case generation based on LTL specification
CompilerModel
CheckerPath condition
calculation
First orderinstantiator
Testmonitoring
Transitions
Path
Flowchart
LTLAut
Conclusions
Model checking is useful for automatically finding errors in hardware/software design.
Testing is nonexhaustive yet practical. Combining model checking and testing
methods enhances capabilities and alleviates limitations.
Black Box Checking allows model checking a system directly, without first modeling it.
Unit Checking allows systematic testing of temporal properties of systems.