Completeness and Complexity of Bounded Model Checking

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Completeness and Complexity of Bounded Model Checking

Ed Clarke

Daniel Kroening

Joel Ouaknine

Carnegie Mellon University, Pittsburgh, USA

Ofer Strichman

Technion, Haifa, Israel

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Overview

Bounded Model Checking of LTL: the (traditional) syntactic translation scheme

The semantic translation scheme The Completeness Threshold problem A solution to the Completeness Threshold problem The complexity of Bounded Model Checking (2exp) The complexity gap and how it can be closed

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Bounded Model Checking (Biere, Cimatti, Clarke, Zhu, 1999)

Model checking: is M a model of (M ² )?

Bounded Model Checking (BMC): is there a counterexample to M ² up to a given depth k ?

BMC is widely accepted as a complementary to Model-Checking.

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Bounded Model Checking (Biere, Cimatti, Clarke, Zhu, 1999)

BMC can be performed with SAT (no need to detect fixpoints).

SAT formulation of BMC: Keep k copies of each variable

Check if [M]k Æ[:]k is satisfiable, where:

[M]k represents all traces of M up to length k

[:]k represents all traces of length up to k that satisfy :

[:]k = (… formulation in next few slides)

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Generating []k is based on expansion formulas for LTL (Manna & Pnueli):

BMC (syntactic) translation(Biere, Cimatti, Clarke, Zhu, 1999)

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The no-loop case (finite traces)

Expansion rule BMC translation

Base case:

k

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The loop case (infinite traces)

Expansion rule BMC translation

Base case:

l

s(i) = i + 1 if i < k, and l otherwise

k=

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LTL model checking(Vardi-Wolper)

Given M,, construct a Buchi automaton B

LTL model checking: is : M £ B empty?

Emptiness checking: is there a path to a loop with an accepting state ?

s0

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“Unroll” k times Find a witness to Gtrue with the fairness constraint

s0

A semantic BMC translation(Based on Vardi-Wolper)

(Was mentioned by [De-Moura, Rushby, Sorea, 2002] in the context of infinite systems)

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Advantages of the semantic translation

Syntactic Translation Semantic Translation

Size of formula O(k ¢ |M| + k2 ¢||) O (k ¢ |M| + k ¢||)

Optimizations w.r.t. LTL formulas

None Efficient Buchi construction from LTL

Computing CT Only for Gp and Fp Full LTL

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Bounded Model Checking

k = 0

BMC(M,,k)

yes

k++

k ¸ ?no

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How big should k be?

For every model M and LTL property there exists k s.t.

We call the minimal such k the Completeness Threshold (CT)

Clearly if M ² then CT = 0

Conclusion: computing CT is at least as hard as model checking

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The Completeness Threshold

Computing CT is as hard as model checking

The value of CT depends on the model M, the property and the translation scheme.

Strategy: find over-approximations to CT based on graph theoretic properties of M

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Diameter d(M) = longest shortest path between any two reachable states.

Recurrence Diameter rd(M) = longest loop-free path between any two reachable states.

d(M) = 2

rd(M) = 3

Initialized Diameter dI(M) Initialized Recurrence Diameter rdI(M)

Basic notions…

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The Completeness Threshold

Theorem: for Gp properties CT = dI(M)

(Biere, Cimatti, Clarke, Zhu, 1999) s0

p

Arbitrary path

Theorem: for Fp properties CT= rdI(M)+1

(Kroening, Strichman, 2003)s0

ppppp

Theorem: for an LTL property CT = ?

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Completeness threshold for LTL

It cannot be longer than rdI()+1 It cannot be longer than dI() + d() Result: min(rdI()+1, dI() + d())

s0

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CT: examples

dI() + d() = 6rdI() + 1= 4

dI() + d() = 2rdI() + 1= 4 s0

s0

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Complexity of BMC

CT · (min(rdI()+1, dI() + d()))

The value of CT can be exponential in the # of state variables.

BMC SAT formula grows linearly with k

Conclusion: standard SAT based BMC is worst-case 2-exp

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The complexity GAP

SAT based BMC is 2-exp in the # state variables. LTL model checking is 1-exp in the # state variables.

So why use BMC ?

Finding bugs when k is small

In many cases rd(y) and d(y) are not exponential and are even rather

small.

SAT, in practice, is very efficient.

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Closing the complexity gap

Why is there a complexity gap ? LTL-MC with 2-dfs :

dfs1

dfs2

Every state is visited not more than twice

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2-dfs Each state is visited not more than twice

SAT Each state can potentially be visited an exponential no. of times,

because all paths are explored.

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Closing the complexity gap (for Gp)

Force a static order, following a forward traversal

Each time a state i is fully evaluated (assigned):

Prevent the search from revisiting it through deeper paths (by adding conflict clauses)

When backtracking from state i, prevent the search from revisiting it in step i

If :pi holds stop and return “Counterexample found”

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Is ‘1-exp SAT’ better or worse than BMC ? Bad news:

We gave up the main power of SAT: dynamic splitting heuristics. We may generate an exponential no. of added constraints

Good news Single exp. instead of double exp. No need to compute CT. (Instead of pre-computing CT we can

maintain a list of states and add their negation ‘when needed’).

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Is restricted SAT better or worse than explicit LTL-MC ? Not clear !

Unlike dfs, SAT has heuristics for progressing. SAT has pruning ability of sets of states

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Comparing the algorithms…

2-dfs LTL MC Restricted-SAT BMC

SAT - BMC

Time EXP EXP2 2-EXP

Memory* EXP EXP2 EXP

Guidance None Restricted Full

Pruning States Sets of states Sets of states

* Assuming the SAT solver restricts the size of its added clauses

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Completeness and Complexity of Bounded Model Checking