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

Efficient Algorithms for Asymptotic Bounds on Termination Time in VASS Tomáš Brázdil 1 Krishnendu Chatterjee 2 Antonín Kučera 1 Petr Novotný 3 Dominik Velan 1 Florian Zuleger 3 1 Masaryk University 2 IST Austria 3 Forsyte, TU Wien July 2018 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Motivation Vector addition systems with states as a model of concurrent processes, parametrized systems, programs in resource-bound analysis. 2/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

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. 2/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

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. 2/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

VASS (0,0) q 1 q 2 (-1,1) (1,-1) (-1,0) 3/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

VASS (0,0) q 1 q 2 (-1,1) (1,-1) (-1,0) 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 × Z d × Q is a set of transitions such that for every q ∈ Q there exists p ∈ Q and u ∈ Z d such that ( q , u , p ) ∈ T . 3/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

VASS - example (0,0,0) q 1 q 2 (-1,1,0) (1,-1,0) (-1,0,1) (-1,0,0) (0,0,0) (0,0,0) q 3 q 4 (-1,1,0) (1,-1,0) (1,0,-1) 4/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Overview The problem 1 Linear termination 2 Polynomial termination 3 The algorithm 4 5/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Overview The problem 1 Linear termination 2 Polynomial termination 3 The algorithm 4 5/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Definitions and notation Definition A finite path π of length n : p 0 , u 1 , p 1 , u 2 , p 2 ,..., u n , p n where ( p i , u i +1 , p i +1 ) ∈ T for all 0 ≤ i < n . If p 0 = p n , then π is a cycle . A cycle is simple if all p 1 ,..., p n − 1 are pairwise different. 6/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Definitions and notation Definition A finite path π of length n : p 0 , u 1 , p 1 , u 2 , p 2 ,..., u n , p n where ( p i , u i +1 , p i +1 ) ∈ T for all 0 ≤ i < n . If p 0 = p n , then π is a cycle . A cycle is simple if all p 1 ,..., p n − 1 are pairwise different. A configuration of A is a pair p v , where p ∈ Q and v ∈ N d . The size of a configuration p v is defined as || p v || = || v || = max { v ( i ) | 1 ≤ i ≤ d } . 6/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Intuition t 1 = (0 , 0) p q t 2 = ( − 1 , 1) t 4 = (1 , − 2) t 3 = ( − 2 , 0) p(2,2) 7/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Intuition t 1 = (0 , 0) p q t 2 = ( − 1 , 1) t 4 = (1 , − 2) t 3 = ( − 2 , 0) p(2,2) 7/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Intuition t 1 = (0 , 0) p q t 2 = ( − 1 , 1) t 4 = (1 , − 2) t 3 = ( − 2 , 0) p(2,2) 7/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Intuition t 1 = (0 , 0) p q t 2 = ( − 1 , 1) t 4 = (1 , − 2) t 3 = ( − 2 , 0) p(2,2), q(2,2) 7/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Intuition t 1 = (0 , 0) p q t 2 = ( − 1 , 1) t 4 = (1 , − 2) t 3 = ( − 2 , 0) p(2,2), q(2,2) 7/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Intuition t 1 = (0 , 0) p q t 2 = ( − 1 , 1) t 4 = (1 , − 2) t 3 = ( − 2 , 0) p(2,2), q(2,2), q(3,0) 7/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Intuition t 1 = (0 , 0) p q t 2 = ( − 1 , 1) t 4 = (1 , − 2) t 3 = ( − 2 , 0) p(2,2), q(2,2), q(3,0) 7/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Intuition t 1 = (0 , 0) p q t 2 = ( − 1 , 1) t 4 = (1 , − 2) t 3 = ( − 2 , 0) p(2,2), q(2,2), q(3,0), p(1,0) 7/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Intuition t 1 = (0 , 0) p q t 2 = ( − 1 , 1) t 4 = (1 , − 2) t 3 = ( − 2 , 0) p(2,2), q(2,2), q(3,0), p(1,0) 7/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Intuition t 1 = (0 , 0) p q t 2 = ( − 1 , 1) t 4 = (1 , − 2) t 3 = ( − 2 , 0) p(2,2), q(2,2), q(3,0), p(1,0), q(1,0) 7/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Definitions and notation Definition Let L( p v ) be the least ℓ ∈ N ∞ such that the length of every finite computation initiated in p v is bounded by ℓ . The termination complexity of A is a function L : N → N ∞ defined by L( p v ) | p v ∈ Q × N d where || p v || = n � � L ( n ) = sup . If L ( n ) = ∞ for some n ∈ N , we say that A is non-terminating , otherwise it is terminating . 8/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

The problem Decide the asymptotic complexity of L . 9/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

The problem Decide the asymptotic complexity of L . Is L ∈ Θ( n )? Is L polynomial? What is the degree of the polynomial? 9/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Overview The problem 1 Linear termination 2 Polynomial termination 3 The algorithm 4 10/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Ranking functions Definition Let A = ( Q , T ) be a VASS. A linear map f is a function f : Q × N d → Q f ( p v ) = c f · v + w f ( p ) where c ∈ Q d and w f : Q → Q is some function. A transition ( p , u , q ) is called f -ranked if c f · u + w f ( q ) ≤ w f ( p ) − 1 and f -neutral if c f · u + w f ( q ) = w f ( p ) . 11/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Ranking functions Definition A linear map f is called a quasi-ranking function (QRF) if c f ≥ � 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 c f >� 0. 12/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Example t 1 = (0 , 0) p q t 2 = ( − 1 , 1) t 4 = (1 , − 2) t 3 = ( − 2 , 0) f ( p v ) = (3 , 2) · v +1 f ( q v ) = (3 , 2) · v 13/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Example t 1 = (0 , 0) p q t 2 = ( − 1 , 1) t 4 = (1 , − 2) t 3 = ( − 2 , 0) f ( p v ) = (3 , 2) · v +1 f ( q v ) = (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 13/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Increments Definition The effect of π = p 0 , u 1 , p 1 , u 2 , p 2 ,..., u n , p n : eff ( π ) = u 1 + ··· + u n . Let Inc = { eff ( π ) | π is a simple cycle of A } . The elements of Inc are called increments . 14/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Linear termination - RF (0,0) q 1 q 2 (-1,1) (1,-2) (-1,0) 15/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Linear termination - RF (0,0) q 1 q 2 (-1,1) (1,-2) (-1,0) y x 15/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Example - RF (0,0) q 1 q 2 (-1,1) (1,-2) (-1,0) y x 15/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Example - positive QRF (0,0) q 1 q 2 (-1,1) (1,-1) (-1,0) 16/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Example - positive QRF (0,0) q 1 q 2 (-1,1) (1,-1) (-1,0) This is quadratic VASS. 16/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Example - positive QRF (0,0) q 1 q 2 (-1,1) (1,-1) (-1,0) This is quadratic VASS. y x 16/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Example - no QRF (0,-1) q 1 q 2 (-1,2) (1,0) This is non-terminating VASS. y x 17/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS

Example - no QRF (0,-1) q 1 q 2 (-1,2) (1,-1) (-1,0) This is non-terminating VASS. y x 18/30 Brázdil, Chatterjee, Kučera, Novotný, Velan, Zuleger Asymptotic termination in VASS