Synthesis of Reactive systems

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Synthesis of Reactive systems. Orna Kupferman Hebrew University. Moshe Vardi Rice University. Is the system correct?. The system has the required behavior. M satisfies . Formal Verification:. It Works!. System  A mathematical model M

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Synthesis of Reactive systems

Orna Kupferman Hebrew University

Moshe Vardi Rice University

The system has the required behavior

M satisfies 

Formal Verification:

It Works!

System A mathematical model M

Desired behavior A formal specification 

But…

Model checking

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!

truth assignment

for pq.

synthesis

satisfiability

Synthesis:

Input:a specification .

Output:a system satisfying .

Input:pq.

Output:p,q

A state of the system:   2AP

p,q

p,q

p

q

p,q

A computations of the system: 2)AP ( 

Satisfiability

Synthesis

of temporal logic specifications:

IsGpFpsatisfiable?

A specification:L 2)AP ( 

specifications  languages

An LTL specification .

LTL  nondeterministic Büchi word automata

 is satisfiable

A is nonempty

[VW86]

An automaton A.

L(A)=  :  satisfies  

 =G (req XF grant)

A:

req

req

req grant

req  grant

The automata-theoretic approach:

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

To: ornab@cs.technion.ac.il (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.

user 1

user 2

An example:

• Whenever user i sends a job, the job is eventually printed.
• The printer does not serve the two users simultaneously.
• G(j1 F p1)  G(j2  F p2)
• G((p1)  (p2))

Let’s synthesize a scheduler that satisfies the specification …

j1j2p1p2

Satisfiability of 

such a scheduler exists?

NO!

A model for 

help in constructing a scheduler?

NO!

A model for :a scheduler that is guaranteed to satisfy  forsomeinput sequence.

Wanted:a scheduler that is guaranteed to satisfy  forall input sequences.

synthesis

satisfiability

Closed vs. open systems

Closed system:no input!

o0, o1,o2,…,oi

o0

o0, o1

o0, o1,o2

all input sequences=some input sequence

Closed vs. open systems

Open system:interacts with an environment!

o0

o1=f(i0)

i0

o2=f(i0,i1)

i1

o3=f(i0,i1,i2)

i2

AP=IO

f:(2I)*  2O

An open system:labeled state-transition graph

synthesis

satisfiability

Closed vs. open systems

Open system:f:(2I)* 2O

In the printer example:I={j1,j2}, O={p1,p2}

f:({{},{j1},{j2},{j1,j2}})* {{},{p1},{p2},{p1,p2}}

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:

(2IO)

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

2IO-labeled 2I-tree I-exhaustive

f()

00 01 10 11

f(00)

f(01)

f(10)

f(11)

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:

(2IO)

The specification  is realizable if there is f:(2I)*2O such that all the computations of f satisfy .

 is satisfiable   is realizable ?

 is satisfiable   is realizable ?

Yes! (for all  exists)

NO!

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

From: ornab@cs.technion.ac.il (Orna Bernholtz)

To: vardi@cs.rice.edu

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

women

men

proofs

bugs

R

R

love(x,y)

in(x,y)

16 4

y2=x

Suppose that we have…

f: women  men

love(x,f(x))

f: proofs  bug

in(x,f(x))

f: R  R

f2(x)=x

16 4

Can we find such f?

X

Y

RX  Y

Can we find f: X  Y such that

R(x,f(x)) for every x  X?

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

Church’s problem1963

We will search for a “constructable” f.

X

(2I)

Y

(2O)

constructable

Synthesis:

R (2I)(2O)

R (2IO)

An LTL formulaover I  O

Can we find f: (2I) (2O) such that

R(x,f(x)) for every x (2I)?

Can we find f: (2I)* 2O such that

all the computations of f satisfy ?

X

(2I)

Y

(2O)

Synthesis:

Linear appraoch:

Branching appraoch:

An LTL formulaover I  O

CTL* formula

Can we find f: (2I) (2O) such that

R(x,f(x)) for every x (2I)?

Can we find f: (2I)* 2O such that

all the computations of f satisfy ?

Can we find f: (2I)* 2O such that

the computation tree of f satisfies ?

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

From: Moshe Vardi vardi@cs.rice.edu

To: ok@research.att.com

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…

For linear specifications

We easily extend to branching specifications

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

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

• Given a CTL* specification  over IO:
• Construct an automaton A on 2IO-labeled 2I-trees such that A accepts exactly all the trees that satisfy .
• Construct an automaton AI-exh on 2IO-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)*  2Owhose computation tree satisfies !

• 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

Independent of H…

Synthesis with incomplete information:

“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

The system’s computation tree:

10

0

11

1

00

01

A tree with a binary branching degree

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 branching degree four

The system’s computation tree:

11

10

0

0

0

01

0

0

10

00

1

11

1

00

01

For someone that has complete information:

I={job}, H={bug}

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

2I-tree

What the system sees

The fat tree:

2IH-tree

10

11

00

01

0000

0001

0100

0101

0010

0011

0110

0111

1000

1001

1100

1101

1010

1011

1110

What reality is; the thing that should satisfy .

The system’s computation tree:

The thin tree:

0

1

10

11

00

01

1111

The system’s computation tree:

The thin tree:

0

1

10

11

00

01

The fat tree:

10

11

00

01

0000

0001

0100

0101

0010

0011

0110

0111

1000

1001

1100

1101

1010

1011

1110

1111

A consistent tree: indistinguishable nodes agree on their label.

indistinguishable by the system

Solving the synthesis problem:

• Given a CTL* specification  over IOH :
• Construct an automaton A on 2IOH -labeled 2IH -trees such that A accepts exactly all the trees that satisfy .
• Construct an automaton Aexh on 2IOH -labeled 2IH -trees such that Aexh accepts exactly all the consistent (IH)-exhaustive trees.

A tree accepted by both A and Aexh:

f: (2I)*  2Owhose fat computation tree satisfies !

• Check A  Aexh for emptiness.
• (with respect to regular trees)

Solving the synthesis problem:

• Given a CTL* specification  over IOH :
• Construct an automaton A on 2IOH-labeled 2IH-trees such that A accepts exactly all the trees that satisfy .
• Construct an automaton Aexh on 2IOH-labeled 2IH-trees such that Aexh accepts exactly all the consistent (IH)-exhaustive trees.

A tree accepted by both A and Aexh:

f: (2I)*  2Owhose fat computation tree satisfies !

• Check A  Aexh for emptiness.
• (with respect to regular trees)

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?

Unusual effectiveness of alternating automata!

Solving the synthesis problem:

Given a CTL* specification  over IOH :

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

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

A tree accepted by A:

f: (2I)*  2Owhose fat computation tree satisfies !

Check A for emptiness.

(with respect to regular trees)

A is a Büchi automaton with linearly many states

Complexity:

• Satisfiability:
• LTL: PSPACE-complete.
• CTL: EXPTIME-complete.
• CTL*: 2EXPTIME-complete.
• Synthesis with complete information:
• LTL: 2EXPTIME-complete.
• CTL: EXPTIME-complete.
• CTL*: 2EXPTIME-complete.
• Synthesis with incomplete information:
• LTL: 2EXPTIME-complete.
• CTL: EXPTIME-complete.
• CTL*: 2EXPTIME-complete.

Let’s synthesis five dining philosophers.

So far…

O I

...systems with a single component.

HMMMM…

P0

P2

P1

• An architecture:
• I0Oenv
• I1Oenv
• I2O0
• I3O1 O2

Synthesis of distributed systems:

P3

Each process Pi has Ii, Oi, and Hi

composition??

Synthesis of distributed systems:

• Input:
• A specification  overIOH.
• An architecture A.

Output:

Strategiesfi: (2Ii)* 2IiHisuch that their composition satisfies  (if exist).

P0

P1

Two independent input streams

Two player games with incomplete information

[Peterson Reif 79]

Solving synthesis of distributed systems:

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

can simulate a Turing machine.

Pn

Solving synthesis of distributed systems:

[PR90]:hierarchical architectures are decidable.

P0

P1

P2

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

From: Orna Kupferman <orna@cs.huji.ac.il>

To: vardi@cs.rice.edu

Subject: Re: hierarchies

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

Orna

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

From: Moshe Vardi <vardi@cs.rice.edu>

To: ornak@cs.huji.ac.il

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

P0

P1

P2

Pn

P0

P1

P2

Pn

Solving synthesis of distributed systems:

[PR90]:hierarchical architectures are decidable.

[KV00]:using alternating automata:

One/two-way chains are decidable.

P0

P1

P2

Pn

One/two-way rings are decidable.

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

From: Moshe Vardi <vardi@cs.rice.edu>

To: ornak@cs.huji.ac.il

Subject: Re: hierarchies

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:

• A specification   an alternating automaton A.
• 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)

From: Moshe Vardi <vardi@cs.rice.edu>

To: ornak@cs.huji.ac.il

Subject: Re: hierarchies

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

Synthesis is not harder than verification!

How come? Verification is linear in the system and at most exponential in the specification.

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)

From: Moshe Vardi <vardi@cs.rice.edu>

To: orna@eecs.berkeley.edu

I think we are done.

Moshe