Efficient Algorithms for Asymptotic Bounds on Termination Time in - - PowerPoint PPT Presentation

Efficient Algorithms for Asymptotic Bounds

• n Termination Time in VASS

Tomáš Brázdil 1 Krishnendu Chatterjee 2 Antonín Kučera 1 Petr Novotný 3 Dominik Velan 1 Florian Zuleger 3

1Masaryk University 2IST Austria 3Forsyte, TU Wien

July 2018

Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

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Motivation

Vector addition systems with states as a model of

concurrent processes, parametrized systems, programs in resource-bound analysis.

Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

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Motivation

Vector addition systems with states as a model of

concurrent processes, parametrized systems, programs in resource-bound analysis.

Termination of VASS is in P-time.

Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

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Motivation

Vector addition systems with states as a model of

concurrent processes, parametrized systems, programs in resource-bound analysis.

Termination of VASS is in P-time. Efficient programs terminate quickly.

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VASS

q1 q2 (0,0) (-1,1) (-1,0) (1,-1)

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VASS

q1 q2 (0,0) (-1,1) (-1,0) (1,-1) Definition Let d ∈ N. A d-dimensional vector addition system with states (VASS) is a pair A = (Q,T), where Q = ∅ is a finite set of states and T ⊆ Q ×Zd ×Q is a set of transitions such that for every q ∈ Q there exists p ∈ Q and u ∈ Zd such that (q,u,p) ∈ T.

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VASS - example

q1 q2 q3 q4 (0,0,0) (-1,1,0) (-1,0,1) (1,-1,0) (0,0,0) (0,0,0) (-1,1,0) (-1,0,0) (1,0,-1) (1,-1,0)

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Overview

1

The problem

2

Linear termination

3

Polynomial termination

4

The algorithm

Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

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Overview

1

The problem

2

Linear termination

3

Polynomial termination

4

The algorithm

Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

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Definitions and notation

Definition A finite path π of length n: p0,u1,p1,u2,p2,...,un,pn where (pi,ui+1,pi+1) ∈ T for all 0 ≤ i < n. If p0 = pn, then π is a cycle. A cycle is simple if all p1,...,pn−1 are pairwise different.

Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

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Definitions and notation

Definition A finite path π of length n: p0,u1,p1,u2,p2,...,un,pn where (pi,ui+1,pi+1) ∈ T for all 0 ≤ i < n. If p0 = pn, then π is a cycle. A cycle is simple if all p1,...,pn−1 are pairwise different. A configuration of A is a pair pv, where p ∈ Q and v ∈ Nd. The size of a configuration pv is defined as ||pv|| = ||v|| = max{v(i) | 1 ≤ i ≤ d}.

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Intuition

p q t1 = (0,0) t2 = (−1,1) t3 = (−2,0) t4 = (1,−2) p(2,2)

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Intuition

p q t1 = (0,0) t2 = (−1,1) t3 = (−2,0) t4 = (1,−2) p(2,2)

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Intuition

p q t1 = (0,0) t2 = (−1,1) t3 = (−2,0) t4 = (1,−2) p(2,2)

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Intuition

p q t1 = (0,0) t2 = (−1,1) t3 = (−2,0) t4 = (1,−2) p(2,2), q(2,2)

Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

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Intuition

p q t1 = (0,0) t2 = (−1,1) t3 = (−2,0) t4 = (1,−2) p(2,2), q(2,2)

Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

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Intuition

p q t1 = (0,0) t2 = (−1,1) t3 = (−2,0) t4 = (1,−2) p(2,2), q(2,2), q(3,0)

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Intuition

p q t1 = (0,0) t2 = (−1,1) t3 = (−2,0) t4 = (1,−2) p(2,2), q(2,2), q(3,0)

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Intuition

p q t1 = (0,0) t2 = (−1,1) t3 = (−2,0) t4 = (1,−2) p(2,2), q(2,2), q(3,0), p(1,0)

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Intuition

p q t1 = (0,0) t2 = (−1,1) t3 = (−2,0) t4 = (1,−2) p(2,2), q(2,2), q(3,0), p(1,0)

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Intuition

p q t1 = (0,0) t2 = (−1,1) t3 = (−2,0) t4 = (1,−2) p(2,2), q(2,2), q(3,0), p(1,0), q(1,0)

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Definitions and notation

Definition Let L(pv) be the least ℓ ∈ N∞ such that the length of every finite computation initiated in pv is bounded by ℓ. The termination complexity of A is a function L: N → N∞ defined by L(n) = sup

• L(pv) | pv ∈ Q ×Nd where ||pv|| = n
• .

If L(n) = ∞ for some n ∈ N, we say that A is non-terminating, otherwise it is terminating.

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The problem

Decide the asymptotic complexity of L.

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The problem

Decide the asymptotic complexity of L.

Is L ∈ Θ(n)? Is L polynomial? What is the degree of the polynomial?

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Overview

1

The problem

2

Linear termination

3

Polynomial termination

4

The algorithm

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Ranking functions

Definition Let A = (Q,T) be a VASS. A linear map f is a function f : Q ×Nd → Q f (pv) = cf ·v+wf (p) where c ∈ Qd and wf : Q → Q is some function. A transition (p,u,q) is called f -ranked if cf ·u+wf (q) ≤ wf (p)−1 and f -neutral if cf ·u+wf (q) = wf (p).

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Ranking functions

Definition A linear map f is called a quasi-ranking function (QRF) if cf ≥ 0 and all transitions are either f -ranked or f -neutral. A QRF f is called ranking function (RF) if all transitions are f -ranked. A QRF f is called positive if cf > 0.

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Example

p q t1 = (0,0) t2 = (−1,1) t3 = (−2,0) t4 = (1,−2) f (pv) = (3,2)·v+1 f (qv) = (3,2)·v

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Example

p q t1 = (0,0) t2 = (−1,1) t3 = (−2,0) t4 = (1,−2) f (pv) = (3,2)·v+1 f (qv) = (3,2)·v configuration p(2,2) q(2,2) q(3,0) p(1,0) q(1,0) rank 11 10 9 4 3

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Increments

Definition The effect of π = p0,u1,p1,u2,p2,...,un,pn: eff (π) = u1 +···+un. Let Inc = {eff (π) | π is a simple cycle of A}. The elements of Inc are called increments.

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Linear termination - RF

q1 q2 (0,0) (-1,1) (-1,0) (1,-2)

Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

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Linear termination - RF

q1 q2 (0,0) (-1,1) (-1,0) (1,-2) x y

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Example - RF

q1 q2 (0,0) (-1,1) (-1,0) (1,-2) x y

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Example - positive QRF

q1 q2 (0,0) (-1,1) (-1,0) (1,-1)

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Example - positive QRF

q1 q2 (0,0) (-1,1) (-1,0) (1,-1) This is quadratic VASS.

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Example - positive QRF

q1 q2 (0,0) (-1,1) (-1,0) (1,-1) This is quadratic VASS. x y

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Example - no QRF

q1 q2 (0,-1) (-1,2) (1,0) This is non-terminating VASS. x y

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Example - no QRF

q1 q2 (0,-1) (-1,2) (-1,0) (1,-1) This is non-terminating VASS. x y

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Linear termination - the characterization

Let n ∈ Rd, n = 0, then we define the sets Hn = {u ∈ Rd | u·n < 0}, ˆ Hn = {u ∈ Rd | u·n ≤ 0}. Theorem Let A = (Q,T) be a d-dimensional VASS. If there is an open half-space Hn of Rd such that n > and Inc ⊆ Hn, then L(n) ∈ O(n). If there is no open half-space Hn of Rd such that n > and Inc ⊆ Hn, then L(n) ∈ Ω(n2). VASS A is linear if and only if there exists a RF f with cf = k ·n for some k > 0.

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Linear termination - idea of the proof

Lemma If there is no open half-space Hn of Rd such that n > 0 and Inc ⊆ Hn, then there exist v1,...,vk ∈ Inc and b1,...,bk ∈ N+ such that k ≥ 1 and k

i=1 bivi ≥

0. Proof. We distinguish two possibilities. There exists a closed half-space ˆ Hn of Rd such that n > 0 and Inc ⊆ ˆ Hn. There is no closed half-space ˆ Hn of Rd such that n > and Inc ⊆ ˆ Hn.

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Linear termination - idea of the proof

Proof. There exists a closed half-space ˆ Hn of Rd such that n > 0 and Inc ⊆ ˆ Hn. In this case for some u ∈ Inc we have −u ∈ cone(Inc),

• therwise we could tilt the normal n.

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Linear termination - idea of the proof

Proof. There exists a closed half-space ˆ Hn of Rd such that n > 0 and Inc ⊆ ˆ Hn. In this case for some u ∈ Inc we have −u ∈ cone(Inc),

• therwise we could tilt the normal n.

There is no closed half-space ˆ Hn of Rd such that n > and Inc ⊆ ˆ Hn. In this case the intersection of cone(Inc) with the first

• rthant is non-trivial.

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Overview

1

The problem

2

Linear termination

3

Polynomial termination

4

The algorithm

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Classification

There is no n ≥ 0, n = 0 such that Inc ⊆ ˆ Hn i.e., there is no QRF. ⇒ Non-terminating VASS.

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Classification

There is no n ≥ 0, n = 0 such that Inc ⊆ ˆ Hn i.e., there is no QRF. ⇒ Non-terminating VASS. There exists n > 0 such that Inc ⊆ ˆ Hn i.e., there exists positive QRF. ⇒ Polynomial or non-terminating VASS. In this case, we have complete classification.

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Classification

There is no n ≥ 0, n = 0 such that Inc ⊆ ˆ Hn i.e., there is no QRF. ⇒ Non-terminating VASS. There exists n > 0 such that Inc ⊆ ˆ Hn i.e., there exists positive QRF. ⇒ Polynomial or non-terminating VASS. In this case, we have complete classification. There exists n ≥ 0 but not > 0 such that Inc ⊆ ˆ Hn i.e., there exists QRF but not necessarily positive one. ⇒ We do not know.

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Polynomial termination - example

q1 q2 (0,0) (-1,1) (-1,0) (1,-1) This is quadratic VASS. q1 q2 (0,0) (-1,1) (1,-1) (-1,0) This is non-terminating VASS.

x y

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Polynomial termination - good normal

Observation: Let n > 0 be such that Inc ⊆ ˆ

• Hn. We can do at most O(n)

transitions from the set {u ∈ Inc | u·n < 0}. We are looking for such n that maximizes this set.

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Polynomial termination - good normal

Observation: Let n > 0 be such that Inc ⊆ ˆ

• Hn. We can do at most O(n)

transitions from the set {u ∈ Inc | u·n < 0}. We are looking for such n that maximizes this set. Definition Let n > 0 such that Inc ⊆ ˆ

• Hn. We say that n is a good normal

if ∀u ∈ Inc we have u·n = 0 ⇔ −u ∈ cone(Inc). A good normal corresponds to the vector cf of a QRF f maximizing the set of f -ranked transitions.

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Polynomial termination

The idea now is to take a VASS A = (Q,T) and do the following: compute a QRF maximizing the set of f -ranked transitions, find a set T ′ = {t ∈ T | t is f -neutral}, compute all SCCs of (Q,T ′), repeat the same steps for every SCC of (Q,T ′). Let k ∈ N be the depth of the recursion tree (with linear component in the leaves). Then L ∈ Θ(nk+1).

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Polynomial termination - example

q1 q2 q3 q4 (0,0,0) (-1,1,0) (-1,0,1) (1,-1,0) (0,0,0) (0,0,0) (-1,1,0) (-1,0,0) (1,0,-1) (1,-1,0)

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Polynomial termination - example

q1 q2 q3 q4 (0,0,0) (-1,1,0) (-1,0,1) (1,-1,0) (0,0,0) (0,0,0) (-1,1,0) (-1,0,0) (1,0,-1) (1,-1,0)

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Polynomial termination - example

q1 q2 q3 q4 (0,0,0) (-1,1,0) (-1,0,1) (1,-1,0) (0,0,0) (-1,1,0) (1,0,-1) (1,-1,0)

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Polynomial termination - example

q1 q2 q3 q4 (0,0,0) (-1,1,0) (-1,0,1) (1,-1,0) (0,0,0) (-1,1,0) (1,0,-1) (1,-1,0)

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Polynomial termination - example

q1 q2 q3 q4 (-1,1,0) (1,-1,0) (-1,1,0) (1,-1,0)

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Polynomial termination - idea of the proof

The proof is by induction on the depth of recursion. The O-bound follows from the following:

the size of every reachable configuration is in O(n), we can move at most linear distance from any point in any direction.

In order to obtain the Ω-bound, we construct the path of required length. Here we use that fact that for a good normal n and u ∈ Inc we have u·n = 0 iff −u ∈ cone(Inc).

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Polynomial termination - idea of the proof

The proof is by induction on the depth of recursion. The O-bound follows from the following:

the size of every reachable configuration is in O(n), we can move at most linear distance from any point in any direction.

In order to obtain the Ω-bound, we construct the path of required length. Here we use that fact that for a good normal n and u ∈ Inc we have u·n = 0 iff −u ∈ cone(Inc). If we consider n ≥ 0 we can use the same idea and obtain a valid Ω-bound (but not valid O-bound in general).

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Overview

1

The problem

2

Linear termination

3

Polynomial termination

4

The algorithm

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The algorithm

Decompose(A)

1

f ← a QRF for A maximizing the number of ranked transitions,

2

Tf ← {f -neutral transitions of A}

3

if Tf = T then return ∞

4

if Tf = ∅ then return 1

5

let A1,...,Al be all SCCs of (Q,Tf ) with at least one transition

6

return 1+max{Decompose(A1),...,Decompose(Al)}

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Conclusion

VASS termination time complete characterization of linear VASS

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Conclusion

VASS termination time complete characterization of linear VASS partial characterization of polynomial VASS (including the degree of the polynomial)

no QRF ⇒ non-terminating VASS positive QRF ⇒ complete classification QRF ⇒ open problem

Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS