Decidability and complexity issues for subclasses of counter systems - - PowerPoint PPT Presentation

Decidability and complexity issues for subclasses of counter systems Lecture 1 Vector Addition Systems with States

St´ ephane Demri demri@lsv.ens-cachan.fr

LSV, ENS Cachan, CNRS, INRIA

Course 2.9 – MPRI – 2010/2011 “Verification of parametrized and dynamic systems”

Decidability and complexity issues for subclasses

• f counter systems
• Lecture 1 (10/12/2010): Vector Addition Systems with

States.

• Lecture 2 (17/12/2010): Reversal-bounded counter

automata.

• Lecture 3 (07/01/2011): Counter systems with finite

monoid and flatness.

• Lecture 4 (14/01/2011): Linear-time temporal logics for

counter systems.

• Lecture 5 (21/01/2011): Exercises on data logics and

counter systems,improving Rackoff’s proof and model-checking (if time permits).

2

Organizational matters

• Slides available on-line on the 2.9 course web page:

http://mpri.master.univ-paris7.fr/C-2-9.html

• Structure of each lecture:
• Course 1h15-1h30.
• 10-min break.
• 30-45 min course
• 20 min exercises.

3

Internship proposals at LSV, ENS Cachan

• “Counter Systems with Presburger-definable Reachability

Sets: Decidability and Complexity” with Arnaud Sangnier (LIAFA, Paris VII).

• “D´

ecidabilit´ e et complexit´ e de la reconnaissance de langages alg´ ebriques” with Alain Finkel (LSV, Cachan).

• “Compl´

etude de la logique de s´ eparation” with Etienne Lozes (LSV, RWTH Aachen).

• Other proposals can be found at the course web page.

4

Plan of the lecture

• Recall on vector addition systems with states.
• Coverability graphs.
• EXPSPACE upper bound for the covering problem.
• Other properties that can be checked in EXPSPACE.
• If time permits, we start developments about the

EXPSPACE-hardness of problems on VASS.

• Exercise on VASS weakly computing multiplication.

5

Recapitulation about VASS

6

Recapitulation about VASS

q0 q1

B B @ −1 1 C C A B B @ 1 C C A B B @ 1 −1 1 1 C C A B B @ −1 1 1 C C A

• VASS is a counter system with transitions of the form

q

• b

− → q′ with b ∈ Zn, which is a shortcut for

• i∈[1,n]

x′

i = xi +

b(i)

• VAS = VASS with a unique control state.

7

Presburger arithmetic

• Terms: t ::= 0 | 1 | x | t + t.
• Presburger formulae (k ≥ 2)

ϕ ::= t ≡k t | t < t | ¬ϕ | ϕ ∧ ϕ | ∃x ϕ | ∀x ϕ

• Valuation v : VAR → N + extension to all terms with

v(0) = 0 v(1) = 1 v(t + t′) = v(t) + v(t′)

• Formula ϕ(x1, . . . , xn) with n free variables:

REL(ϕ(x1, . . . , xn))

def

= {(v(x1), . . . , v(xn)) ∈ Nn : v | = ϕ}.

8

Counter systems

q0 q1 q2 ϕ( x, x′) ϕ′( x, x′) x′

1 = x′ 2 = x′ 3 = 0

x′

1 = x1 + 1

x′

2 = x2 + 1

x′

3 = x3 + 1

• Counter system S = (Q, n, δ) of dimension n ≥ 1:
• Q is a nonempty finite set of control states.
• δ: finite set of transitions of the form t = (q, ϕ, q′) where

q, q′ ∈ Q and ϕ is a Presburger formula with free variables x1, . . . , xn, x′

1, . . . , x′ n.

• Configuration (q,

a) ∈ Q × Nn.

• (q,

a) t − → (q′, a′)

def

⇔ v[ x ← a, x′ ← a′] | = ϕ.

• Runs as nonempty (possibly infinite) sequences

ρ = (q0, a0) − → (q1, a1) · · · (qk, ak) · · ·

9

Subclasses of counter systems

• Standard counter automaton (Q, n, δ): transitions are of

the form either q

inc(i)

− − → q′ or q

dec(i)

− − → q′ or q

zero(i)

− − − → q′.

• Succinct counter automaton (Q, n, δ): transitions of the

form either q

add( b)

− − − → q′ with b ∈ Zn or q

zero( b′)

− − − → q′ with

• b′ ∈ {0, 1}n (simultaneous zero-tests).
• Vector addition systems with states (VASS): succinct

counter automata without zero-tests. A transition t is an element in Q × Zn × Q.

• VAS T ⊆ Zn (finite sets of tuples).

10

Reachability problems

• REACHABILITY PROBLEM:

Input: VASS V, (q, x) and (q′, x′). Question: is there a finite run with initial configuration (q, x) and final configuration (q′, x′)? (in symbols (q, x) ∗ − → (q′, x′)?)

• CONTROL STATE REACHABILITY PROBLEM:

Input: VASS V, (q, x) and q′. Question: is there a finite run with initial configuration (q, x) and whose final configuration has control state q′? (∃ x′ (q, x) ∗ − → (q′, x′)?)

• CONTROL STATE REPEATED REACHABILITY PROBLEM:

Input: VASS V, (q, x) and qf. Question: is there an infinite run with initial configuration (q, x) such that the control state qf is repeated infinitely often?

11

Variant problems

• COVERING PROBLEM:

Input: VASS V, (q, x) and (q′, x′). Question: is there a finite run with initial configuration (q, x) and whose final configuration is (q′, x′′) with x′ x′′? (control state reachability is an instance with x′ = 0)

• BOUNDEDNESS PROBLEM:

Input: VASS V and (q, x). Question: is the set {(q′, x′) ∈ Q × Nn : (q, x) ∗ − → (q′, x′)} finite?

• TERMINATION PROBLEM:

Input: VASS V and (q, x). Question: is there an infinite run with initial configuration (q, x)?

12

Witness run characterization for termination problem

• VAS T ⊆f Zn and initial configuration

x0 ∈ Nn.

• Propositions below are equivalent:

1 There is an infinite run from

x0.

2 There is a finite run

x0

∗

− → y

+

− → y′ such that y y′.

y y′

def

⇔ for i ∈ [1, n], we have y(i) ≤ y′(i).

• ∗

− →: reflexive and transitive closure of − →.

• +

− →: transitive closure of − →.

• Use of Dickson’s Lemma: for any infinite sequence
• z0,

z1, . . . of tuples in Nn, there are i < j such that zi zj.

13

From VASS to VAS (other direction is obvious)

A B C

+1 −1

(x1, A, B, C) (1, −1, 1, 0) (0, 1, −1, 0) (−1, 0, −1, 1) (0, 0, 1, −1) (A, 4) ≈ (4, 1, 0, 0) and (C, 2) ≈ (2, 0, 0, 1) Reduction is correct from VASS without self-loops

14

Reduction

• W.l.o.g., V has no transition of the form q
• b

− → q. Otherwise, replace q

• b

− → q by q

• −

→ qnew and qnew

• b

− → q.

• As an exercise, show that the reachability [resp. covering,

boundedness] problem for VASS can be reduced to the same problem for VASS without self-loops.

• VAS T built from VASS V = (Q, n, δ) has dimension

n + card(Q). Control states are encoded in the card(Q) last components.

• Alternative reduction from VASS of dimension n to VAS of

dimension n + 3 (instead of n + card(Q)). [Hopcroft & Pansiot, TCS 79]

15

Bijection between configurations

• VASS V = (Q, n, δ) without self-loop.
• Bijection h : Q → {n + 1, . . . , n + card(Q)} dedicated to

relate each control state of V with a unique component in the VAS we shall build.

• Bijection between configurations in V and elements from

the set X: X = { x ∈ Nn+card(Q) : x([n + 1, n + card(Q)]) = ei ∈ Ncard(Q) for some i ∈ [1, card(Q)]},

• ei ∈ Ncard(Q): unit element with 1 for the ith component and

zero otherwise.

x([n + 1, n + card(Q)]) is the tuple in Ncard(Q) restricted to the card(Q) last components of x.

16

• X = N × {(1, 0, 0), (0, 1, 0), (0, 0, 1)} for the VASS below:

A B C

+1 −1

17

Defining the VAS T

• VAS T such that for t = q
• b

− → q′ ∈ δ (q = q′), the transition t′ ∈ T is defined as follows:

• (t′)([1, n]) =

b,

• for q′′ ∈ Q \ {q, q′}, t′(h(q′′)) = 0,
• t′(h(q)) = −1 and t′(h(q′)) = 1.
• For each run (q0,

x0) . . . (qk, xk) of V, f((q0, x0)) . . . f((qk, xk)) is a run in T .

• Each configuration f((qi,

xi)) belongs to X.

• Similarly, for each run

x0 · · · xk in T with x0 ∈ X, f −1( x0) · · · f −1( xk) is a run of V.

18

Reductions

• (q′,

x′) is reachable from (q, x) in V iff f((q′, x′)) is reachable from f((q, x)) in T .

• This can be easily shown by induction on the lenght of the

run.

• Consequently, the reachability problem for VASS can be

reduced to the reachability problem for VAS.

• Given configurations (q,

x) and (q′, x′), the propositions below are equivalent:

• in V, there is a run of the form (q,

x)

∗

− → (q′, x′′) with x′ x′′,

• in T , there is a run of the form f((q,

x))

∗

− → z with f((q′, x′)) z.

• Consequently, the covering problem for VASS can be

reduced to the covering problem for VAS.

19

Reduction for the boundedness problem

• Given a configuration (q,

x) for V, the propositions below are equivalent (one direction is obvious):

• (q,

x) is unbounded (in V),

• f((q,

x)) is unbounded (in T ).

• For any configuration reachable from f((q,

x)), its restriction to the card(Q) last components can take at most card(Q) distinct values.

• If f((q,

x)) is unbounded, then there is q′ ∈ Q such that there is an infinite amount of configurations reachable from f((q, x)) of the form f((q′, y)).

• Hence, f((q,

x)) is unbounded implies (q, x) unbounded.

• The boundedness problem for VASS can be reduced to the

boundedness problem for VAS.

20

Solving the covering problem for VAS

21

About the covering problem for VAS

• COVERING PROBLEM:

Input: a VAS T and two configurations x, x′ ∈ Nn, Question: is there some configuration x′′ reachable from

• x such that

x′ x′′?

• The covering problem for VAS is EXPSPACE-complete:
• Decidability with nonprimitive recursive complexity.

[Karp & Miller, TCS 69]

• EXPSPACE lower bound from [Lipton, TR 76].
• EXPSPACE upper bound from [Rackoff, TCS 78].
• The control state reachability problem for VASS can be

reduced to the covering problem for VAS: require that one specific component has at least value 1. (reaching the control state A in VASS is equivalent to cover (0, 1, 0, 0) in corresponding VAS)

22

Coverability Graph

23

Coverability graphs in a nutshell

• Finite graph whose set of nodes is a finite subset of

(N ∪ {∞})n that can be effectively computed.

• It approximates the set of reachable configurations.
• Simple properties on it allow to solve various problems:

boundedness, covering, termination, etc.

• . . . but in the worst-case, its size can be nonprimitive

recursive.

• First, we need to define relations and operations on

(N ∪ {∞})n.

24

A digression on a variant of Ackermann function

• A0(m) = 2m + 1, An+1(0) = 1.
• An+1(m + 1) = An(An+1(m)) ( =

m+1 times

• An(· · · (An(m)) · · · ) ).
• A(n) = An(2).
• The function A(n) majorizes the primitive recursive

functions.

• The size of the coverability graph can be in O(A(n)).

(n: size of T and x0), see e.g. [Jantzen, APN’97].

25

How to calculate with ∞?

• For k, k′ ∈ N ∪ {∞},

k ≤ k′

def

⇔ either k, k′ ∈ N and k ≤ k′ or k′ = ∞.

• k < k′ whenever k ≤ k′ and k = k′.
• (N ∪ {∞}, <) is isomorphic to the ordinal ω + 1.
• ≤ and < are extended component-wise to (N ∪ {∞})n.
• 2

3 1

• <
• 2

4 1

• → acc(
• 2

3 1

• ,
• 2

4 1

• )

def

=

• 2

∞ 1

• .
• For

x < x′, let us define acc( x, x′) ∈ (N ∪ {∞})n:

• acc(

x, x′)(i)

def

= x′(i) when x(i) = x′(i),

• acc(

x, x′)(i)

def

= ∞ when x(i) < x′(i). “The ith component can be as large as we wish.”

26

How to calculate with ∞? (II)

• Given

x ∈ (N ∪ {∞})n and t ∈ Zn, let us define

• x + t ∈ (Z ∪ {∞})n:
• (

x + t)(i)

def

= x(i) + t(i) if x(i) ∈ N,

• (

x + t)(i)

def

= ∞ otherwise.

(i ∈ [1, n])

• 2

∞ 1

• +
• −3

−6 2

• =
• −1

∞ 3

• .
• The construction of the coverability graph CG(T ,

x0) uses these operations on N ∪ {∞}.

27

Example

A B C (t1) +1

0 (t2)

(t3) −1

0 (t4)

0, 1, 0, 0 ≈ (A, 0) 1, 0, 1, 0 ≈ (B, 1) 0, 0, 0, 1 ∞, 1, 0, 0 0, 0, 1, 0 ∞, 0, 1, 0 ∞, 0, 0, 1 t1 t3 t2 t4 t1 t2 t2 t3 t4

28

Properties of CG(T , x0)

• Given T and

x0, coverability graph CG(T , x0) is a structure (V, E) with V ⊆ (N ∪ {∞})n and E ⊆ V × T × V. (a) CG(T , x0) is a finite structure with “root” x0. (K¨

• nig’s Lemma and Dickson’s Lemma)

(b) Every configuration reachable from x0 can be covered in CG(T , x0), i.e.

• for

y reachable from x0, there is y′ in CG(T , x0) such that

• y

y′.

(c) For all y in CG(T , x0) and bounds B ∈ N, there is a configuration y′ reachable from x0 such that for i ∈ [1, n],

y(i) = ∞ implies y′(i) ≥ B,

y(i) = ∞ implies y′(i) = y(i).

29

A quick presentation of the construction

E := ∅; V := ∅; ToBeTreated := { x0}; while ToBeTreated = ∅ do

• Select an element

x from ToBeTreated;

• ToBeTreated := ToBeTreated \ {

x};

• for t ∈ T such that

x + t ∈ (N ∪ {∞})n do

x′ := x + t;

• if there is

y ∈ V such that y

∗

− → x in (V, E) and y < x′ then

• Let

y0 be the extended configuration the closest to x in (V, E) such that y0 < x′;

x′ := acc( y0, x′);

• if

x′ ∈ V then

• V := V ∪ {

x′};

• ToBeTreated := ToBeTreated ∪ {

x′};

• E := E ∪ {

x

t

− → x′};

30

Characterizations with coverability graph Just look at the finite graph CG(T , x0)!

• There is

x′′ reachable from x0 such that x′ x′′ iff there is

• y in CG(T ,

x0) such that x′ y.

• The set of configurations reachable from

x0 is infinite iff ∞ appears in CG(T , x0).

• Every run from

x0 terminates iff there is no cycle in CG(T , x0).

31

Finiteness of CG(T , x0)

• By K¨
• nig’s Lemma, CG(T ,

x0) is infinite iff there is an infinite branch (finite-branching is due to the finite set of transitions).

• In an infinite branch, after some node nd, no new ∞ is

inserted since the dimension n is finite.

• So after that node nd by Dickson’s Lemma, there are

nodes nd1 and nd2 labelled by y and y′ respectively, such that y y′.

• If

y ≺ y′, then this leads to a contradiction by construction

• f nd.
• If

y = y′, then this leads to a contradiction since the branch is infinite but in case y = y′, we are supposed to stop the construction.

32

Covering

There is x′′ reachable from x0 such that x′ x′′ iff there is

• y in CG(T ,

x0) s.t. x′ y.

• Suppose that

x′′ reachable from x0 and x′ x′′.

• By (b), there is

y in CG(T , x0) s.t. x′′ y.

• Since is transitive on (N ∪ {∞})n,

x′ y.

( x′ x′′ and x′′ y imply x′ y)

• Suppose that there is

y in CG(T , x0) such that x′ y.

• B: maximal value occurring in

x′.

• By (c), there is

y′ reachable from x0 such that for i ∈ [1, n], if y(i) = ∞ then y′(i) ≥ B otherwise y′(i) = y(i).

• Hence,

x′ y′.

33

Boundedness

The set of configurations reachable from x0 is infinite iff ∞ appears in CG(T , x0).

• Suppose the set of configurations reachable from

x0 is infinite.

• Ad absurdum, assume that ∞ does not occur in CG(T ,

x0).

• By (b), there is

y in CG(T , x0) s.t. for an infinite amount of configurations x reachable from x0, we have x y.

• There are at most (1 + max(

y))n distinct configurations smaller than y ∈ Nn, contradiction.

• Suppose ∞ occurs in CG(T ,

x0).

• By (c) the set of configurations reachable from

x0 is infinite.

• For instance, consider bounds B greater and greater when

applying (c).

34

Exponential-space decision procedure

35

Small covering property

• VAS T with configurations

x, x′. Equivalence between

• there is a run from

x leading to y such that x′ y,

• there is a run from

x leading to y′ such that x′ y′ and its length is at most double-exponential in the size of the instance T , x and x′ (numbers in binary).

• A run of double-exponential length requires

double-exponential space to be fully encoded.

• In the worst-case, there is a triple-exponential amount of

such runs.

• Solution: guess nondeterministically the small run and

invoke Savitch’s theorem.

36

Example of small covering

A B C [t1]

„ 1 « „ « [t2]

[t3]

„ 1 −1 « „ « [t4]

How to cover (A, (1, K)) from (A, (0, 0))?

Long covering: (A, (0, 0))

(t1t2)2K t1

− − − − → (B, (0, 2K + 1))

t3t4t2

− − → (A, (1, 2K )) (A, (1, K)) Short covering: (A, (0, 0))

(t1t2)K t1

− − − − → (B, (0, K + 1))

t3t4t2

− − → (A, (1, K)) (A, (1, K))

(t1t2)K t1t3t4t2 subword of (t1t2)2K t1t3t4t2

37

How to be clever enough to guarantee a “short” covering?

38

A nondeterministic algorithm with lenght L

• Algorithm for T ,

x, x′ and L:

1 i := 0;

xc := x (current configuration);

2 While

x′ xc and i < L do

1

Guess a transition t ∈ T ; (nondeterministic step !)

2

If xc + t ∈ Nn then abort;

3

i := i + 1; xc := xc + t.

3 If

x′ xc then accept else abort (i = L).

• If the maximal absolute value in T ,

x, x′ is 2N and L = 22N3 , then the maximal absolute value appearing in the algorithm is 2N + 2N × 22N3 (can be encoded with exponential space in N).

• Determinism can be regained with recursive calls to a

function F(T , x, x′, L) since the number of transitions is finite.

39

When small covering property implies EXPSPACE

Design a decision procedure that nondeterministically guesses the small run and only requires exponential space:

• A counter with an exponential amount of bits can count

until a double-exponential value.

• Only two configurations need to be store thanks to

nondeterminism.

• 22N3

× 2N is still of double-exponential magnitude.

• Comparing or adding two natural numbers requires

logarithmic space only.

• [Savitch, JCSS 70]: a nondeterministic procedure for a

given problem using space f(N) ≥ log(N) can be turned into a deterministic procedure using f(N) × f(N) space.

• Exponential functions are closed under multiplication.

40

Remarks before the EXPSPACE proof

• The EXPSPACE proof for the covering problem is based on

an induction on the dimension.

• The proof for EXPSPACE upper bound for boundedness is

a bit more complex [Rackoff, TCS 78].

• EXPSPACE-hardness proof for covering, boundedness and

reachability problems is presented later. It is based on [Lipton, TR 76; Esparza, 98].

• The proof for decidability of reachability problem is much

more complex:

• The book [Reutenauer, 1990] presents the full proof.
• Nice hints about the proof can be found in [Haddad, 01].
• A simpler proof has been found by J. Leroux.

[Leroux, LICS’09; Leroux POPL ’11]

41

Special PAVAS event January 20th, 2011, Cachan

• MPRI students are welcome to attend the special PAVAS

event.

• This special event focuses on two recent major

claims/results:

1 a new proof, by a P

. Janˇ car, for the decidability of equivalence for deterministic pushdown automata, first established by G. S´ enizergues, and

2 a new proof, by J. Leroux, for the decidability of accessibility

in vector addition systems, or equivalently in Petri nets, first established by E. Mayr (and S. R. Kosaraju).

• Registration on

http://www.lsv.ens-cachan.fr/Events/Pavas/ See also Course 2.9 web page.

42

Definitions about sizes

• For

x ∈ Zn,

• maxneg(

x)

def

= max({max(0, − x(i)) : i ∈ [1, n]}): maximal absolute negative value.

• For example, maxneg(
• −1

−2 −8 7

• ) = max(0, −(−8)) = 8.
• max(

x)

def

= max({ x(i) : i ∈ [1, n]}): maximal value.

• For instance, max(
• −1

−2 −8 7

• ) = 7.
• scale(T )

def

= max({|t(i)| : t ∈ T , i ∈ [1, n]}): maximal absolute value.

• For instance, scale(
• −1

−2 −8 7

• ) = | − 8| = 8.
• Note that maxneg(T ) ≤ scale(T ).

43

Definitions about sizes (II) (or reasonably succinct encodings)

• 2 + ⌈log2(1 + K)⌉ is a sufficient number of bits to encode

integers in [−K, K] for K > 0.

• Size |T |

def

= n × card(T ) × (2 + ⌈log2(1 + scale(T ))⌉).

• If N = |T | + |{

x}| + |{ x′}|, then maxneg(T ), card(T ), max( x′), max( x) ≤ 2N

44

Paths and pseudo-runs

• Path π: finite sequence of transitions (below in blue).

B B @ 1 1 C C A

B B B @

−1 1 1

1 C C C A

− − − − − − → B B @ 1 1 1 C C A

B B B @

1 −1

1 C C C A

− − − − − − → B B @ 1 1 1 C C A

B B B @

−1 1 1

1 C C C A

− − − − − − → B B @ 1 2 1 C C A

B B B @

1 −1

1 C C C A

− − − − − − → B B @ 1 2 1 C C A

• π′ is a subpath of π = t1 . . . tk

def

⇔ there are 1 ≤ j1 < j2 · · · < jk′ ≤ k s.t. π′ = tj1 . . . tjk′.

• Pseudo-configuration

x ∈ Zn (a configuration is in Nn).

• Given π = t1 . . . tk and

x ∈ Zn, pseudo-run (π, x) = x0 · · · xk s.t. x0 = x and for i ∈ [1, k], xi = xi−1 + ti.

• The length of (π,

x) [resp. π] is k + 1 [resp. k].

• A pseudo-run

x0 · · · xk is a covering of x′ when x′ xk.

45

Defining length of shortest coverings

• Assume that (π,

x) is a run covering x′, m(T , x, x′, π)

def

= the length of the shortest subpath π′ of π such that (π′, x) is a run covering x′.

• m(T ,

x, x′, π) ≤ length of π.

• MB(n) (B, n ≥ 1): be the supremum of the set below
• m(T ,

y, y′, π) : (π, y) is a run covering y′ T is a VAS of dimension n and maxneg(T ) + max( y′) ≤ B

• 46

Upper bound for MB(n)

• Let us show that MB(n) ≤ gB(n) for all n, B ≥ 1 with

gB(n) =

• B

if n = 1,

• B · gB(n − 1)

n + gB(n − 1) if n ≥ 2.

• For n ≥ 1 and B ≥ 2, gB(n) ≤ B3n! (gB(1) = B):

gB(n) ≤

• B·gB(n−1)

n+gB(n−1) ≤

• B·gB(n−1)

n+1 ≤ . . . . . . ≤

• B1+(3(n−1))!n+1 ≤ B(3n)!

47

Towards the rough bound 22N3

• Existence of a covering of

x′ from x in T is equivalent to the existence of a covering of length at most α = (maxneg(T ) + max( x′) + 2)(3n)!

• With N = |T | + |{

x}| + |{ x′}|, α ≤ (2N + 2N + 2)2N log2(N) ≤ (2N+2)2N2 ≤ 22N3 .

• The covering problem for VAS, VASS (and Petri nets) can

be solved in exponential space.

48

Back to inequalities (main proof)

• We show for n, B ≥ 1:

MB(n) ≤

• B = gB(1)

if n = 1,

• B · gB(n − 1)

n + gB(n − 1) if n ≥ 2.

• Base case n = 1.
• T of dim. 1,

x, x′ ∈ N and maxneg(T ) + max( x′) ≤ B.

x′ x implies empty path produces a run covering x′.

• Otherwise, no need to use negative values from T and

m(T , x, x′, π) is bounded by max( x′).

• MB(1) ≤ B since max(

x′) ≤ B.

49

Induction step

• Suppose the property holds true for n − 1 ≥ 1.
• It is sufficient to show:

m(T , x, x′, π) ≤

= gB(n−1)

• B · gB(n − 1)

n + gB(n − 1) whenever maxneg(T ) + max( x′) ≤ B and T of dimension n.

• After gB(n − 1) steps, a component greater than

gB(n − 1) maxneg(T ) + max( x′), has value at least max( x′). (useful when using induction hypothesis)

• B′ = gB(n − 1)maxneg(T ) + max(

x′) ≤ B gB(n − 1).

• Pseudo-run

x0 · · · xk is r-bounded (r > 0)

def

⇔ for i ∈ [0, k], we have xi ∈ [0, r − 1]n.

50

Two cases are distinguished

• (π,

x): run covering x′ for the VAS T .

• π = t1 · · · tK , (π,

x) = x0 · · · xK and x′ xK .

• We distinguish the case when (π,

x) is B′-bounded or not.

• If (π,

x) is B′-bounded, then subpaths between identical configurations can be removed (pigeonhole principle).

• Otherwise, the path is divided in two:
• the first part is B′-bounded and can be shortened,
• shortening the second part can be done by using induction

hypothesis. 51

Case 1: (π, x) is B′-bounded

• In the run

x0 · · · xK , each xl ∈ [0, B′ − 1]n.

• Subruns between two identical configurations can be

eliminated: if xi = xj with 0 ≤ i < j ≤ K then

• (π′,

x) is also a run covering of x′,

• π′ = t1 · · · titj+1 · · · tK ,
• π′ is a strict subpath of π.
• This situation occurs as soon as K ≥ (B′)n.
• This transformation can be repeated until K < (B′)n.

(pigeonhole principle)

• Conclusion: there is a subpath π′ s.t. (π′,

x) is also a run covering x′ of length bounded by (B′)n ≤ (B gB(n − 1))n.

52

Case 2: (π, x) is not B′-bounded

• Unique decomposition π = π1π2 s.t.
• π1 is of length k1,
• all values in

x0 · · · xk1−1 are < B′.

• (π1,

x) is not B′-bounded (“faulty” configuration xk1).

• x0

t1

− → · · ·

tk1−1

− − →

• xk1−1
• B′−bounded

tk1

− → xk1 ∈ [0, B′ − 1]n tk1+1 − − →

• xk1+1 · · ·

tK

− → xK π1 = t1 · · · tk1 π2 = tk1+1 · · · tK

• There is π′

1, subpath of π1, such that

• its length is bounded by (B gB(n − 1))n + 1,

(again pigeonhole principle!)

• (π1,

x) and (π′

1,

x) have the same final configuration xk1.

• (π′

1π2,

x) and (π2, xk1) are both runs covering x′.

53

Case 2: (π, x) is not B′-bounded (II)

• Let i ∈ [1, n] such that

xk1(i) ≥ B′.

• xk1 =

    . . . · ≥ B′ . . .    

tk1+1

− − → · · ·

tK

− → xK =     . . . · . . .     x′

• T −, π−

2 ,

xk1

−,

x′−: restrictions of T , π2, xk1, x′ to the components in [1, n] \ {i}.

• xk1

− =

  . . . . . .  

t−

k1+1

− − → · · ·

t−

K

− → xK =   . . . . . .   x′−

• (t−

k1+1 · · · t− K ,

xk1

−) is a run covering

x′− in T −.

• maxneg(T −)+max(

x′−) ≤ B and T − is of dimension n −1.

• m(T −,

xk1

−,

x′−, t−

k1+1 · · · t− K ) ≤ MB(n − 1).

54

Case 2: (π, x) is not B′-bounded (III)

• By induction hypothesis, there is π′

2, subpath of

t−

k1+1 · · · t− K such that

• (π′

2,

xk1

−) is a run covering

x′−,

• its length is bounded by MB(n − 1) ≤ gB(n − 1).
• xk1

− =

  . . . . . .   t′

1

− → · · ·

t′

K′

− →

• x′

k1+K ′ =

  . . . . . .   x′−

• K ′≤gB(n−1)
• Path π′′

2 obtained from π′ 2 by adding component i.

• (π′′

2,

xk1) is a pseudo-run with final pseudo-configuration z such that for j ∈ ([1, n] \ {i}), z(j) ≥ x′(j).

• Since

xk1(i) ≥ B′ = gB(n − 1)maxneg(T ) + max( x′), after gB(n − 1) steps, the ith component is greater or equal to max( x′) and the ith component is never negative.

55

Conclusion

• (π′′

2,

xk1) is a run covering x′.

• Length of π′

1π′′ 2 is at most (B × gB(n − 1))n + gB(n − 1).

• π′

1π′′ 2 is a subpath of π.

56

Other properties in EXPSPACE

• Termination problem.
• Boundedness problem.
• LTL model-checking problem for VASS is

EXPSPACE-complete [Habermehl, ICATPN 97].

• Given (T ,

x) and i ∈ [1, n], checking whether { y(i) : x

∗

− → y} is finite.

• Generalization:

∃ x1, . . . , xn, π1, . . . , πn x0

π1

− → x1

π2

− → x2 · · · πn − → xn ∧C( x1, . . . , xn)∧ x1 xn

• Satisfiability problem is in EXPSPACE.

[Atig & Habermehl, RP’09]

57

Concluding remarks

• Today’s lecture:
• Coverability graphs.
• Covering problem for VAS in EXPSPACE by induction on the

dimension (Rackoff’s proof).

• Other properties in EXPSPACE.
• Next lecture: EXPSPACE-hardness of covering problem for

VAS and basic definitions on reversal-bounded counter automata.

58

Exercises

59

• Exo. 1
• By the characterization of unboundedness from CG(T ,

x0), (T , x0) is unbounded iff there is x0

∗

− → y

∗

− → y′ with y ≺ y′.

• Assuming that T has dimension n, (T ,

x0) is i-unbounded

def

⇔ { y(i) : x0

∗

− → y} is infinite.

• (T ,

x0) is unbounded iff there is i ∈ [1, n] such that (T , x0) is i-unbounded.

• Are the propositions below equivalent?

1 (T ,

x0) is i-unbounded.

2 There is a run of the form

x0

∗

− → y

∗

− → y′ with y ≺ y′ and

• y(i) <

y′(i).

60

Exos 2–4

• 2. Show that the covering problem restricted to VAS of

dimension 1 can be solved in linear time.

• 3. Show that the reachability [resp. covering, boundedness]

problem for VASS can be reduced to the same problem for VASS without self-loops.

• 4. Show that the covering problem for VAS augmented the

multiplication by 2 can be solved in exponential space.

61

VASS weakly computing multiplication

A B

B B @ −1 1 C C A B B @ 1 C C A B B @ 1 −1 1 1 C C A B B @ −1 1 1 C C A

• 1. Compute

{

• a

b c d

• ∈ N4 : (A,
• 2

1

• ) ∗

− → (A,

• a

b c d

• )}
• 2. Show that

{ @ a b d 1 A ∈ N3 : d ≤ a × b} = { @ a b d 1 A ∈ N3 : ∃ @ a′ b′ c′ 1 A ∈ N3, (A, B B @ a b 1 C C A) ∗ − → (A, B B @ a′ b′ c′ d 1 C C A)}

62

A B

B B @ −1 1 C C A B B @ 1 C C A B B @ 1 −1 1 1 C C A B B @ −1 1 1 C C A

• 3. Is there
• a

b c d

• ∈ N4 such that

{

• a′

b′ c′ d′

• ∈ N4 : (A,
• a

b c d

• ) ∗

− → (A,

• a′

b′ c′ d′

• )}

is infinite?

63

C A B Compute the map f such that

{ @ a b e 1 A ∈ N3 : e ≤ f(a, b)} = { @ a b e 1 A ∈ N3 : ∃ B B @ a′ b′ c′ d′ 1 C C A ∈ N4, (C, B B B @ a b 1 C C C A) ∗ − → (A, B B B B @ a′ b′ c′ d′ e 1 C C C C A )}

[t5]

B B B @ −1 1 1 1 C C C A B B B @ 1 C C C A [t6] B B B @ −1 1 C C C A [t2]

[t3]

B B B @ 1 C C C A B B B @ 1 −1 1 1 C C C A [t4]

[t1]

B B B @ −1 1 1 C C C A

64