Laurent Doyen LSV, ENS Cachan & CNRS

Laurent DoyenLSV, ENS Cachan & CNRS

Quantitative Languages: Weighted Automata and

Beyond

Credits

• K. Chatterjee, H. Edelsbrunner, T. A. Henzinger (IST Austria)• A. Degorre, J.-F. Raskin (ULB Brussels)• R. Gentilini (Uni. Perugia)• S. Torunczyk (Uni. Warsaw) • P. Rannou (ENS Cachan)

(CSL’08, LICS’09, FCT’09, CSL’10, Concur’10)

This talk is based on joint work with…

Model-Checking

Property/ Specification

Yes / No

Satisfaction Relation

Program/ System

-perhaps a proof -perhaps some counterexamples

Model-Checking


Yes / No

Trace inclusion

Program/ System

Formula

Every request is followed by a grant

Finite-state machine

Model-Checking


Yes / No

Trace inclusion

Program/ System

Model-checking is boolean

Formula


- a trace is either good or bad

- a system is either good or bad


Quantitative Analysis


Value (R)


Program/ System

Formula


-Measure of “fit” between system and spec

-e.g. average number of requests immediately granted


Distance (R)



Program/ System #1

- Comparing two implementations

e.g. cost or quality measure

Program/ System #2



Quantitative Model-checking

Is there a Quantitative Framework with

- an appealing mathematical formulation, - useful expressive power, robustness and - good algorithmic properties ?

(Like the boolean theory of -regularity.)

Note: “Quantitative” is more than “timed” and “probabilistic”

A language is a boolean function:

Quantitative languages

A language is a boolean function:

Quantitative languages

A quantitative language is a function:

L(w) can be interpreted as:• the amount of some resource needed by the system to produce w (power, energy, time consumption),• a reliability measure (the average number of “faults” in w).

Is LA(w) LB(w) for all words w ?

Quantitative languagesQuantitative language

inclusion

LA(w)peak resource consumptionLong-run average response time

LB(w)resource bound a Average response-time requirement


Quantitative languagesQuantitative language

inclusion

LA(w)peak resource consumptionLong-run average response time

LB(w)resource bound a Average response-time requirement

Distance:

Outline

• Motivation• Weighted automata

• Decision problems• Expressive power• Closure properties

• Beyond weighted automata

AutomataBoolean languages are generated by finite automata.

Nondeterministic Büchi automaton



Value of a run r: Val(r)= 1 if an accepting state occurs -ly often in r 0 otherwise



Value of a word w: sup of {values of the runs r over w}

Value of a run r: Val(r)= 1 if an accepting state occurs -ly often in r



LA(w) = sup of {Val(r) | r is a run of A over w}

Weighted automataQuantitative languages are generated by weighted automata.

Weight function

Weighted automataQuantitative languages are generated by weighted automata.

Weight functionValue of a word w: sup of {values of the runs r over w}Value of a run r: Val(r)

where is a value function

Some value functions

(reachability)(Büchi)

(coBüchi)

(vi {0,1})

Some value functions

(reachability)(Büchi)

(coBüchi)

(vi {0,1})

Example: a motorSpecification

Deterministic LimAvg automaton

SpecificationImplementation(refined spec.)

Example: a motor


Example: a motor


Example: a motor


Example: a motor


Example: a motor


Example: a motor

In the long-run, implementation has average cost cheaper than specification (for all traces).

Outline




Emptiness

Given , is LA(w) ≥ for some word w ?

• solved by finding the maximal value of an infinite path in the graph of A,

• memoryless strategies exist in the corresponding quantitative 1-player games,• decidable in polynomial time for

Language Inclusion


• PSPACE-complete for

equivalent to

Language Inclusion


• PSPACE-complete for

• Solvable in polynomial-time when B is deterministic for and , • undecidable for nondeterministic automata. • open question for nondeterministic automata.

Language Inclusion

Language-inclusion gameLanguage

inclusion as a game

Discounted-sum automata, λ=3/4Is LA(w) LB(w) for all words w ?


inclusion as a game

Discounted-sum automata, λ=3/4Is LA(w) LB(w) for all words w ? Yes !


inclusion as a game


• turn-based, two players;• Challenger wins iff


inclusion as a game


• turn-based, two players;• Challenger wins iff

winning strategy

Language-inclusion game

Tokens on the initial states

Language inclusion as a

game


Challenger:

Simulator:

Challenger constructs a run r1 of A,Simulator constructs a run r2 of B.Challenger wins if Val(r1) > Val(r2).


game


Challenger:

Simulator:


game


Challenger:

Simulator:


game


Challenger:

Simulator:


game


Challenger:

Simulator:


game


Challenger:

Simulator:


game


Challenger:

Simulator:


game


Challenger:

Simulator:


game


Challenger:

Simulator:

Challenger wins since 4>2.


game

However, LA(w) LB(w) for all w.

Language-inclusion gameThe game is blind if the Challenger cannot observe the state of the Simulator.

Challenger has no winning strategy in the blind game

if and only if LA(w) LB(w) for all words w.

Language-inclusion gameThe game is blind if the Challenger cannot observe the state of the Simulator.

Challenger has no winning strategy in the blind game

if and only if LA(w) LB(w) for all words w.

When the game is not blind, we say that B simulates A if the Challenger has no winning strategy.

Simulation implies language inclusion.

Simulation is decidable

Universality and Equivalence

Is LA(w) = LB(w) for all words w ?

Language equivalence problem:

Universality problem:Given , is LA(w) ≥ for all words w ?

Complexity/decidability: same situation as Language inclusion.

Undecidability proof

Quantitative language universality is undecidable for LimAvg automata.

Theorem

Proof:Using a reduction from the halting problem of 2-counter machines.

2-counter machines

q1: inc c1 goto q2

q2: inc c1 goto q3

q3: if c1 == 0 goto q6 else dec c1 goto q4

q4: inc c2 goto q5

q5: inc c2 goto q3

q6: halt

• 2 counters c1, c2

• increment, decrement, zero test

2-counter machines

q1: inc c1 goto q2

q2: inc c1 goto q3


q4: inc c2 goto q5

q5: inc c2 goto q3

q6: halt

• 2 counters c1, c2

• increment, decrement, zero test

Reduction

q1: inc c1 goto q2

q2: inc c1 goto q3


q4: inc c2 goto q5

q5: inc c2 goto q3

q6: halt

Reduction:Given M, construct LimAvg-automaton AM such that M halts iff AM is not universal (for threshold 0).

! • Deterministic machine• Nonnegative counters

Given M and state qhalt, decide if qhalt is reachable (i.e., M halts).

Halting problem:

Reductionq1: inc c1 goto q2

q2: inc c1 goto q3


q4: inc c2 goto q5

q5: inc c2 goto q3

q6: halt

• Initial nondeterministic jump to several gadgets• Witness word (with negative value) = (#AcceptingRun)ω

Gadgets

Reminder: Witness word = #AcceptingRun#AcceptingRun#...

Gadget 1:« First symbol is # »

Gadgets

Gadget 2:« Every σ1 is followed by σ2 »


Gadgets

Gadget 3:« Infinitely many # »

Guess: this is the last #


Gadgets

Gadget 4:« Counter correctness »

Check zero tests on c

# # # #...

-1 -1 -1 -1

Gadgets



# # # #...

-1 -1 -1 -1cheat? ≥1

cheat? ≥1

cheat? ≥1

Gadgets



Check non-zero test on c

Gadgets



Check non-zero test on c

Correctnessq1: inc c1 goto q2

q2: inc c1 goto q3


q4: inc c2 goto q5

q5: inc c2 goto q3

q6: halt

• If M halts, then (#AcceptingRun)ω is a witness word (with LimAvg value -1/Length(AcceptingRun).• If there exists a witness word, then # occurs infinitely often, and finitely many cheats occur. Hence, M has an accepting run.

Outline

Outline




ReducibilityA class C of weighted automata can be reduced to a class C’ of weighted automata iffor all A C, there is A’ C’ such that LA = LA’.


E.g. for boolean languages:• Nondet. coBüchi can be reduced to nondet. Büchi• Nondet. Büchi cannot be reduced to det. Büchi (nondet. Büchi cannot be determinized)


NBW

DBW DCW=NCW

Some easy facts

and cannot be reduced to and

and can define quantitative languages with infinite range, and cannot.

Some easy factsFor discounted-sum, prefixes provide good approximations of the value.

For LimSup, LimInf and LimAvg, suffixes determine the value.

and cannot be reduced to

cannot be reduced to and

Büchi does not reduce to LimAvg

“infinitely many ’’Deterministic Büchi automaton

Assume that L1 is definable by a LimAvg automaton A.Then, all -cycles in A have average weight 0.


Hence, the maximal average weight of a run over any word in tends to (at most) 0 when


All -cycles in A have average weight 0.


Let We have where for sufficienly large


where for sufficienly large Hence,


Let We have

and A cannot exist !


(co)Büchi and LimAvg

cannot be reduced to By analogous arguments, cannot be reduced to

“finitely many ’’Deterministic coBüchi automaton

(co)Büchi and LimAvgDet. coBüchi automaton

L2 is defined by the following nondet. LimAvg automaton:

(co)Büchi and LimAvgDet. coBüchi automaton

L2 is defined by the following nondet. LimAvg automaton:

Hence, cannot be determinized.

Reducibility relations



is reducible to .

Expressive power of {0,1}-automata

is not reducible to .

is reducible to .


Store the value

AB

is reducible to .


AB

Key: can take finitely many different values.

in the example,

Store the value

is reducible to .


AB

is reducible to .


AB

is reducible to .


AB

is reducible to .


BA

Therefore for

is reducible to .


A B

for all


What about Discounted Sum ?

Last result

cannot be determinized.

λ=3/4

This automaton can be determinized for λ ≤ 1/2,

not for rational λ > 1/2.

Discλ cannot be determinized

This automaton can be determinized for λ ≤ 1/2,

not for rational λ > 1/2.


Discλ cannot be determinizedValue of a word :

disc. sum of ’s

disc. sum of ’s


Let


Let


Let


Let


Let


Let


Let


Let


Let


Let

If

then


Let

If

then




How many different values can take ?



infinitely many if for all



infinitely many if for all

By a careful analysis of the shape of the family of equations, it can be proven that no rational can be a solution.

Last result

cannot be determinized.

λ=3/4


Outline




Operations

Operations on quantitative languages:

• shift(L1,c)(w) = L1(w) + c• scale(L1,c)(w) = c·L1(w) (c>0)

Operations


• shift(L1,c)(w) = L1(w) + c• scale(L1,c)(w) = c·L1(w) (c>0)• max(L1,L2)(w) = max(L1(w),L2(w)) • min(L1,L2)(w) = min(L1(w),L2(w)) • complement(L1)(w) = 1-L1(w)

Operations


• shift(L1,c)(w) = L1(w) + c• scale(L1,c)(w) = c·L1(w) (c>0)• max(L1,L2)(w) = max(L1(w),L2(w)) • min(L1,L2)(w) = min(L1(w),L2(w)) • complement(L1)(w) = 1-L1(w)• sum(L1,L2)(w) = L1(w) + L2(w)

Closure propertiesAll classes of weighted automata are closed under shift and scale.

All classes of nondeterministic weighted automata are closed under max.

Closure properties

Closure properties

Analogous results for boolean languages.

Closure propertiesThere is no nondeterministic LimAvg automaton for the language Lm = min(La,Lb).


Assume that Lm = min(La,Lb) is definable by a LimAvg automaton C.


Assume that Lm = min(La,Lb) is definable by a LimAvg automaton C.Then, some a-cycle or b-cycle in C has average weight >0.(consider the word for large)

Closure properties

Then, some a-cycle (or b-cycle) in C has average weight >0.Then, some word gets value >0…

There is no nondeterministic LimAvg automaton for the language Lm = min(La,Lb).

Assume that Lm = min(La,Lb) is definable by a LimAvg automaton C.


There is no nondeterministic Discounted automaton for the language Lm = min(La,Lb).

Proof: analogous argument.

Closure properties

Closure properties

min(L1,L2) = 1-max(1-L1,1-L2)

Closure properties

By analogous arguments (analysis of cycles).

Outline




Alternating weighted automata

Weight function

Alternating transition relation

Value of word outcome of two-player game:

Maximizer choosesMinimizer chooses



Maximizer choosesMinimizer chooses


2



2

-1

LimAvg Automata

LimAvg Automata

LimAvg Automata

E ::= A | max(E,E) | min(E,E) | Sum(E,E)

LimAvg Automaton ExpressionsLimAvg-automaton expressions are defined by:

E.g.: max(A1 + A2, min (A3, A4))

where A is a deterministic Lim{sup/inf}Avg-automaton.



where A is a deterministic Lim{sup/inf}Avg-automaton.Closure properties:

• Closed under max, min, sum• Deterministic automata are closed under complement• -max(E1,E2) = min(-E1,-E2)• -min(E1,E2) = max(-E1,-E2)• -sum(E1,E2) = sum (-E1,-E2)



Closure properties:




Expressive power:




Decision problems: all questions reduce to quant. emptiness


Given , is LE(w) ≥ for some word w ?

Value setSolve decision problems using the value set:

E.g.: max(A1 + A2, min (A3, A4))

Value set = { (LA1(w),LA2(w),LA3(w),LA4(w)) | w Σω} R4

How to compute this set ?

Solve decision problems using the value set:

E.g.: max(A1 + A2, min (A3, A4))

A1 A2 A3 A4

Synchronized product is deterministicWeights are vectors in R4

Value set

Solve decision problems using the value set:

E.g.: max(A1 + A2, min (A3, A4))

A1 A2 A3 A4

Synchronized product is deterministicWeights are vectors in R4

Alur/Degorre/Maler/Weiss/09. On Omega-Languages Defined by Mean-Payoff Conditions.

Value set

where A1 and A2 are LimInfAvg-automata.

Value set

Value set

Consider a ‘crazy’ word:

Value set

Accumulation points

sum is 1 !

Value set

Accumulation points

sum is 1 !

Value set

Accumulation points

According to E = A1+A2, value is 0

sum is 1 !

sum is 0 !

Value set

• [ADMW09] considers accumulation points• we need individual limits

Value set

• [ADMW09] considers accumulation points• we need individual limitsThe two notions coincide for lasso-

words !

Value set

Accumulation set

Corners of accumulation set correspond to simple cycles in A1 A2

E = A1+A2

Value set

≠Accumulation set Value

set

E = A1+A2


Value set


set

E = A1+A2

Points in value set are fmin(points in accumulation set) where fmin(x,y,z,…) = u if ui = min(xi,yi,zi,…) 1 ≤ i ≤ n


Value set


set

E = A1+A2

Points in value set are fmin(points in accumulation set) where fmin(x,y,z,…) = u if ui = min(xI,yi,zi,…) 1 ≤ i ≤ n


Conjecture: Corners in value set are fmin(corners in accumulation set).

Value set

Lemma: Fmin(conv(S)) is a convex set.

Points in value set are fmin(points in accumulation set)

Conjecture: Corners in value set are fmin(corners in accumulation set)

ValueSet = Fmin(AccSet) where

Fmin(A) = {fmin(x1,…,xn) | x1,…,xn A} for A Rn

Conjecture: Fmin(conv(S)) = conv(Fmin(S))

Value set

Fmin(conv(S))Conjecture: Fmin(conv(S)) = conv(Fmin(S))This holds in two dimensions (R2),

conv(S)


conv(S)


conv(S)


conv(S)

Fmin(conv(S))Conjecture: Fmin(conv(S)) = conv(Fmin(S))This holds in two dimensions (R2), but it is false in R3 !

Counter-example:S = {q,r,s} with q=(0,1,0) r=(-1,-1,1) s=(1,1,1)fmin(r,s) = rfmin(q,r) = fmin(q,r,s) = t = (-1,-1,0)fmin(q,s) = qFmin(S) = {q,r,s,t}

Fmin(conv(S))Conjecture: Fmin(conv(S)) = conv(Fmin(S))This holds in two dimensions (R2), but it is false in R3 !

Counter-example:S = {q,r,s} with q=(0,1,0) r=(-1,-1,1) s=(1,1,1)Fmin(S) = {q,r,s,t}Let p = (r+s)/2 = (0,0,1)fmin(p,q) = (0,0,0) Fmin(conv(S)) but (0,0,0) conv(Fmin(S))

Fmin(conv(S))Conjecture: Fmin(conv(S)) = conv(Fmin(S))

How to compute Fmin(conv(S)) ? (given finite set S)

conv(S)



conv(S)



conv(S)

Set of points that agree with some p conv(S) on one coordinate, and have other coordinates smaller

pu



conv(S)

Fmin(conv(S))

conv(S)

Fmin(conv(S))

conv(S)

Fmin(conv(S))

conv(S)

Fmin(conv(S))

conv(S)

Fmin(conv(S))

conv(S)

Fmin(conv(S))

conv(S)

Fmin(conv(S))

conv(S)

Fmin(conv(S))

conv(S)

Fmin(conv(S))

conv(S)

Fmin(conv(S))

conv(S)

EmptinessSolve decision problems using the value set:

E.g.: E =max(A1 + A2, min (A3, A4))

Value set = { (LA1(w),LA2(w),LA3(w),LA4(w)) | w Σω} R4

LE(Σω) = { max(x+y, min(z,t)) | (x,y,z,t) Value Set}is a union of intervals.Find maximum of LE to solve emptiness

Distance

LEF(Σω) = { (LE(w),LF(w)) | w Σω}

Distance

and maximize |x-y| in this set.

LimAvg Automaton Expressions

Conclusion• Quantitative generalization of languages to model programs/systems more accurately.

• Decision problems, expressive power, closure properties.• Mean-payoff automaton expressions: robust and decidable. Distance computable.• Other results (not shown): probabilistic extension, cut-point languages, stability/robustness.

• Outlook: open questions for discounted sum synthesis question quantitative logic

Thank you !

Questions ?

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Laurent Doyen LSV, ENS Cachan & CNRS