Synthesis of Reactive systems Orna Kupferman Hebrew University Moshe Vardi Rice University.

Synthesis of Reactive systems

Orna Kupferman Hebrew University

Moshe Vardi Rice University

Is the system correct?

Formal Verification:

System A mathematical model M

Desired behavior A formal specification

The system has the required behavior

M satisfies

Model checking

It Works!

But…

It’s hard to design systems:It’s even harder to design correct systems:

Synthesis:

Input: a specification .

Output: a system satisfying .

WOW!!!An unusual effectiveness of logic in

computer science!

Synthesis:

Input: a specification .

Output: a system satisfying .

Input: pq.

Output: p,q

truth assignment

for pq.

synthesis satisfiability

of temporal logic specifications:

Synthesis

Satisfiability

A specification: L 2)AP (

A state of the system: 2AP

p,q

A computations of the system: 2)AP (

p,q p q p,q

specifications languages

Is GpFp satisfiable?

An LTL specification .

An automaton A.

[VW86]

L(A)= : satisfies

LTL nondeterministic Büchi word automata

req

req grant

req grant

req

=G (req XF grant)

A:

The automata-theoretic approach:

is satisfiable

A is nonempty

Date: Mon, 28 Dec 92 18:12:25 PST

From: Moshe Vardi [email protected]

To: [email protected] (Orna Bernholtz)

Yes, the VW86 algorithm can be easily extended to give you a finite representation of an accepting run. Thus, it can be used as a synthesis algorithm .

You can view this as the automata-theoretic prespective on the Clarke&Emerson-style synthesis. For further elaboration on this perspective, see the paper by P. Wolper: On the relations of programs and computations to models of temporal logic, LNCS 398, 1989.

Moshe

P.S. Let me know if you’d like me to mail you the paper.

mailto:[email protected]


user 1

user 2

1. Whenever user i sends a job, the job is eventually printed.

2. The printer does not serve the two users simultaneously.

1. G(j1 F p1) G(j2 F p2)

2. G((p1) (p2))

Let’s synthesize a scheduler that satisfies the specification …

An example:

Satisfiability of such a scheduler exists?

NO!

A model for help in constructing a scheduler?

NO!

j1 j2 p1 p2

A model for : a scheduler that is guaranteed to satisfy for some input sequence.

Wanted: a scheduler that is guaranteed to satisfy for all input sequences.

Closed vs. open systems

Closed system: no input!

o0o0, o1o0, o1,o2o0, o1,o2,…,oi

all input sequences=some input sequence



Open system: interacts with an environment! o0

o1=f(i0)

o2=f(i0,i1)

o3=f(i0,i1,i2)

i2

i1

i0

An open system: labeled state-transition graph

AP=IO

f:(2I)* 2O


Open system: f:(2I)* 2O

In the printer example: I={j1,j2}, O={p1,p2}f:({{},{j1},{j2},{j1,j2}})* {{},{p1},{p2},{p1,p2}}


f()

f(01)f(00) f(10) f(11)

00 01 10 11

The computation tree of f (|I|=2):

2IO-labeled 2I-tree I-exhaustive

A computation of f:

)f() ( (i0,f(i0)) (i1,f(i0,i1)) (i2,f(i0,i1,i2)) …

A path in the computation tree, which embodies all computations:(2IO)

The specification is realizable if there is f:(2I)*2O such that all the

computations of f satisfy .

NO!Yes! (for all exists)

is satisfiable is realizable? is satisfiable is realizable?

A computation of f:

)f() ( (i0,f(i0)) (i1,f(i0,i1)) (i2,f(i0,i1,i2)) …

A path in the computation tree, which embodies all computations:(2IO)

Date: Thu, 27 Jan 94 13:46:43 IST

From: [email protected] (Orna Bernholtz)

To :[email protected]

Subject: Church’s problem

We mentioned it in the summer. You referred me to Pnueli and Rozner work about “synthesis as a game between the environment and the

system.”

Orna


love(x,y)

in(x,y)

y2=x

women

men

proofs bugs

R R

16 4

f: women men

love(x,f(x))

f: proofs bug

in(x,f(x))

f: R R

f2(x)=x16 4

Suppose that we have…

Can we find such f?

Church’s problem 1963

Any f: does every x have y such that R(x,y)?

We will search for a “constructable” f.

X Y

RX Y

Can we find f: X Y such that

R(x,f(x)) for every x X?

X

)2I(

R (2I) (2O)

Can we find f: (2I) (2O) such that

R(x,f(x)) for every x (2I) ?

Y

)2O(

R (2IO)

An LTL formula over I Oconstructabl

e

Can we find f: (2I)* 2O such that

all the computations of f satisfy ?

Synthesis:

X

)2I(

Can we find f: (2I) (2O) such that

R(x,f(x)) for every x (2I) ?

Y

)2O(

An LTL formula over I O


all the computations of f satisfy ?

Synthesis:

Linear appraoch:

Branching appraoch:


the computation tree of f satisfies ?

CTL* formula

Date: Sat, 6 Jan 1996 10:28:16 CST

From: Moshe Vardi [email protected]

To: [email protected]

We need some motivation for the branching specs. I think Antioniotti looked at synthesis with CTL specs, but I am not sure that he fully solved it .

Didn’t I give you some of his papers?

Moshe

“Whenever user 1 sends a job, the printer may print it”

AG(j1 EFp1)Exists an input

sequence…

Solving the synthesis problem: [Rabin 70, Pnueli Rozner 88]

For linear specifications

We easily extend to branching specifications

Solving the synthesis problem: [Rabin 70, Pnueli Rozner 88]

Given a CTL* specification over IO:

1. Construct an automaton A on 2IO-labeled 2I-trees such that A accepts exactly all the trees that satisfy .

2. Construct an automaton AI-exh on 2IO-labeled 2I-trees such that AI-exh accepts exactly all the I-exhaustive trees.

A tree accepted by both A and AI-exh :

f: (2I)* 2O whose computation tree satisfies !

3. Check A AI-exh for emptiness.

(with respect to regular trees)

Synthesis with incomplete information:

“The printer should not print papers containing bugs”.

Hidden information, unknown to the system!

•Partial observability…

•Internal signals…

•Incomplete information…

The system does not see the full picture!

The system does not see the full picture!

Still has to be correct with respect to the most hostile environment


“The printer should not print papers containing bugs”.

Hidden information, unknown to the system!The setting:

•I: input signals

•O: output signals

•H: hidden signals.

A strategy for the system:

f:(2I)* 2O

Independent of H…

What about the computation tree?

00 01 10 11

0 1

The system’s computation tree:

For someone that has incomplete information:

I={job}

2I={{},{job}}

For someone that has complete information:

I={job}, H={bug}

2I x2H={{},{job}}x{{},{bug}}

A tree with a binary branching degree

A tree with branching degree four

00 01 10 11

0 1


0000 01 10010 11

For someone that has complete information:

I={job}, H={bug}

2I x2H={{},{job}}x{{},{bug}}


00 01 10 11

0 1The thin tree:

1111

The fat tree:

00 01 10 11

1000 1001 1100 11010010 0011 0110 01110000 0001 0100 0101 1010 1011 1110

What the system sees

What reality is; the thing that should satisfy .

2I-tree

2IH-tree

1111


00 01 10 11

0 1The thin tree:

The fat tree:

00 01 10 11

1000 1001 1100 11010010 0011 0110 01110000 0001 0100 0101 1010 1011 1110

indistinguishable by the system

A consistent tree: indistinguishable nodes agree on their label.

Solving the synthesis problem:

Given a CTL* specification over IOH:

1. Construct an automaton A on 2IOH -labeled 2IH -trees such that A accepts exactly all the trees that satisfy .

2. Construct an automaton Aexh on 2IOH -labeled 2IH -trees such that Aexh accepts exactly all the consistent (IH)-exhaustive trees.A tree accepted by both A and Aexh :

f: (2I)* 2O whose fat computation tree satisfies !

3. Check A Aexh for emptiness.

(with respect to regular trees)

consistent

Consistency is not a regular property!

The idea:Wanted: is there a fat tree that is both good and consistent?

We cannot check whether a tree is consistent.

There is a transformation g:thin trees fat trees that generates only consistent fat trees.

So we check: is there a thin tree t such that g(t) is good?

The automaton reads t, but pretends to read g(t).

Unusual effectiveness of alternating automata!

Solving the synthesis problem:

Construct an alternating automaton A on 2IO -labeled 2I -trees such that A accepts an I-exhaustive (thin) tree iff its fat version satisfies .

A tree accepted by A:

f: (2I)* 2O whose fat computation tree satisfies !

Check A for emptiness. (with respect to regular

trees)

Given a CTL* specification over IOH:

Construct an alternating automaton A on 2IO -labeled 2I -trees such that A accepts an I-exhaustive (thin) tree iff its fat version satisfies .

Synthesis with complete information:

•LTL: 2EXPTIME-complete.

•CTL: EXPTIME-complete.

•CTL*: 2EXPTIME-complete.

Satisfiability:

•LTL: PSPACE-complete.



Complexity:


•LTL: 2EXPTIME-complete.



A is a Rabin automaton with exponentially many states and a linear index

A is a Büchi automaton with linearly many states

So far…

O I

...systems with a single component.

Let’s synthesis five dining philosophers.

HMMMM…

Synthesis of distributed systems:

P0

P2P1

P3

Each process Pi has Ii, Oi, and HiAn architecture:

•I0Oenv

•I1Oenv

•I2O0

•I3O1 O2

Synthesis of distributed systems:

Input:

• A specification over IOH.

• An architecture A.

Output:

Strategies fi: (2Ii)* 2IiHi

such that their composition satisfies (if exist).

composition??

Solving synthesis of distributed systems:

Pnueli Rozner 90: distributed systems are hard to synthesize; undecidable in the general case.

can simulate a Turing machine.

Two independent input streams

Two player games with incomplete information

[Peterson Reif 79]

P0 P1


[PR90:]hierarchical architectures are decidable.

P2 PnP0 P1

Date: Sat, 6 Feb 1999 10:34:25 –0600 (CST)

From: Moshe Vardi< [email protected]>


Subject: Re: hierarchies

In fact, I think we might be able to handle even a more general case, where I_j \subset O_{j_1} \cup O_{j+1}, which allows information to flow up and down the chain.

Moshe

Date: Sun, 7 Feb 1999 17:07:19 +0200

From: Orna Kupferman >[email protected]>



We should be able to generalize even more… …the dependencies induce a flow that alternating automata can handle.

Orna



[PR90:]hierarchical architectures are decidable.

P2 PnP0 P1

[KV00:]using alternating automata:

One/two-way chains are decidable.

One/two-way rings are decidable.

P2 PnP0 P1

P2 PnP0 P1

Date: Sun, 7 Feb 1999 22:17:29 –0600 (CST)




This is nice because these architectures are actually quite realistic. In communication protocol architecture, we typically have layers, where the upper layer is the application layer and the lower level is the physical layer, and information flows between the layers.

Moshe


The solution:

1. A specification an alternating automaton A.

2. Reapet:

A and an architecture with n components.

A’ (of size exponential in A) and an architecture with n-1 components.

Complexity:

nonelementary.

Date: Mon, 8 Feb 1999 14:18:13 –0600 (CST)




BTW, regarding the nonelementary complexity, we can cite the MONA experience that shows that nonelementary algorithms can nevertheless be practical, since the worst-case complexity does not always arise.

Moshe


More about the nonelementary complexity:

Synthesis is not harder than verification!

How come? Verification is linear in the system and at

most exponential in the specification.

More about the nonelementary complexity:

Input to verification: M and .

Input to synthesis: and A.

[Rozner92 :]a specification such that the smallest system satisfying has a nonelementary size.

Other related work:

Synthesis against a non-maximal environment.

The computatin tree may not be I-exhaustive; makes a difference for existential requirements [joint work with P. Madhusudan and P.S. Thiagaragan].

-calculus synthesis.

Many technical problems…

Date: Thu, 27 Aug 1998 12:08:42 –0500 (CST)



I think we are done.Moshe


Synthesis of Reactive systems Orna Kupferman Hebrew University Moshe Vardi Rice University.

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