aa r X i v : . [ c s . L O ] J u l Reachability in succinct one-counter games ✩ Paul Hunter a a Université Libre de Bruxelles
Abstract
We consider the reachability problem on transition systems corresponding to succinctone-counter machines, that is, machines where the counter is incremented or decre-mented by a value given in binary.
1. Preliminaries
We are interested in reachability problems on transition graphs defined by one-counter machines where: • The counter may take any integer value (including negative values); • The counter is incremented or decremented by binary weights; • Additional transitions are available to the machine if the counter does or doesnot have value .Previous work [1, 2] has considered other variations on these initial assumptions.Formally, a transition graph of a one-counter machine (or one-counter graph) isgiven by a tuple ( V, V ∃ , E, E , E =0 , q , w ) where V is a finite set of states ; V ∃ ⊆ V are the states of Eve ( V \ V ∃ are the states of Adam ); E, E , E =0 ⊆ V × V [ E ( E =0 )is the set of edges (de)activated at ); q ∈ V is the initial state ; and w : E → Z is the weight function . The (infinite) unweighted arena defined by such a tuple has: • Vertex set: V × Z , • Eve’s vertices: V ∃ × Z , • Initial vertex: ( q , , • For every e = ( v, v ′ ) ∈ E and c ∈ Z an edge from ( v, c ) to ( v ′ , c + w ( e )) , • For every e = ( v, v ′ ) ∈ E an edge from ( v, to ( v ′ , , and • For every e = ( v, v ′ ) ∈ E =0 an edge from ( v, c ) to ( v ′ , c ) for c = 0 . We are interested in the following reachability problems listed in increasing orderof difficulty. They are all known to be in
EXPSPACE and
EXPTIME -hard. Finitememory strategies suffice for all but parity games. ✩ Work supported by ERC inVEST project
Preprint submitted to Elsevier October 9, 2018 lobal reachability.
Given t ∈ Z does Eve have a strategy to get the counter to value t (in any state)? That is, can she force the play to ( v, t ) for some v ∈ V ? EXPTIME -hardvia straightforward reduction from countdown games.
Reachability.
Given t ∈ Z and F ⊆ V does Eve have a strategy to get the counter tovalue t in a state of F ? Büchi (repeated reachability).
Given F ⊆ V does Eve have a strategy to infinitelyoften have the counter with value whilst in a state of F ? Parity.
Given a priority function
Ω : V → N which defines a priority function on theinfinite arena in the obvious way, does Eve win the (infinite) parity game? Known to bein EXPSPACE by the result in [3] which gave a
PSPACE algorithm for parity gameson unary-encoded one-counter graphs. EXPSPACE -completeness of succinct one-counter games
We will prove the following:
Theorem 1.
Determining if Eve has a winning strategy in any of these games is
EXPSPACE -complete.
From the above results it suffices to show
EXPSPACE -hardness.
It might seem that including edges that are (de)activatedwhen the counter is might yield a more powerful model, but we can use the antag-onistic nature of the game to simulate (de)activating edges. That is, we activate alltransitions but give the other player the ability to punish the (non-)zeroness of thecounter. Figures 1 and 2 show the gadgets that simulate an activating edge ( v, v ′ ) , andFigures 3 and 4 show gadgets that simulate a deactivating edge ( v, v ′ ) . In all figuresunlabelled edges have weight , square nodes represent states owned by Eve, circlenodes represent states owned by Adam, and all sinks are included in the target set. Target set.
We can assume that F ⊆ V ∃ as follows: for every v ∈ F \ V ∃ we add anew vertex v ′ ∈ V ∃ ∩ F and edge ( v ′ , v ) [with weight 0] and replace all edges ( u, v ) with ( u, v ′ ) [with the same weight]. EXPSPACE -hardness of Büchi games
It is well known that CTL model checking (on a transition system) reduces to a two-player game with a Büchi winning condition [4]. The same reduction shows that CTLmodel checking on succinct one-counter automata reduces to Büchi games on one-counter graphs. In [5], CTL model checking on succinct one-counter automata wasshown to be
EXPSPACE -complete, hence one-counter games with a Büchi winningcondition are also
EXPSPACE -hard. 2 v ′ < > -1, Figure 1: Simulating an activating edge from anEve state v v ′ < > - -1, Figure 2: Simulating an activating edge from anAdam state v =0 < > v ′ - v =0 v ′ Figure 4: Simulating a deactivating edge from anAdam state .3. From Büchi games to Reachability The following lemma, which is readily established using pumping techniques willbe useful.Given a Büchi game G = ( V, E, E , E =0 , q , F ) with target set F ⊆ V ∃ , weconstruct a new one-counter reachability game as follows: • The game graph consists of | F | + 1 copies of G with a -activated edge from ( v, i ) to ( v, i + 1) for all v ∈ F and ≤ i ≤ | F | , • The initial state is ( q , , • The target set is { ( v, | F | + 1) : v ∈ F } , and • The target value is .Clearly Eve wins this game if and only if in the original game she can reach F withcounter value | F | + 1 times. Hence if she wins the Büchi game she has a winning inthe reachability game. We now show the converse, that is if she can reach F | F | + 1 times then she can reach some vertex in F with counter value infinitely often. Moreprecisely we will show how to defeat any positional (w.r.t. the current state and countervalue) strategy for Adam in the original Büchi game. It is well known [6] that suchstrategies are sufficient for winning strategies, thus this is sufficient for our result. Sucha strategy has a natural interpretation in the reachability game, so Eve has a counter-strategy to ensure F is visited with counter value | F | + 1 times against this strategy.By the pigeon-hole principle there is some vertex v ∈ F visited at least twice in theplay. Hence Eve has a strategy (against Adam’s strategy) to reach v with counter value from both q and v . Hence Eve can visit v with counter value infinitely often in theorignal game. Given a reachability game G with target set F ⊆ V ∃ and E = E =0 = ∅ , weconstruct a new arena as follows: • Double the weights of the edges in G ; • Add a new (initial) vertex v and a new sink (with -weighted loops) v f ; • Add an edge of weight +1 from v to the original initial vertex; • Add edges of weight − from F to v f .Due to parity arguments the counter can only have value at v f . Clearly v f can bereached with value if and only if the target set F can be reached with value in theoriginal game. Thus Eve wins the Global Reachability game on this new arena if andonly if she has a winning strategy in the original Reachability game. Thus this givesa reduction from Reachability to Global Reachability. Note that this does not work inthe unary case as we utilize the “long-reach” ability of doubled weight values to avoidcounter values of in the original game. 4 . Super-exponential counter values For one-counter machines without alternation (i.e. one player games) it is known [7,Lemma 42] that the reachability problem can be solved without the counter value ex-ceeding an exponential bound. Our results show that such a bound in the case of al-ternating machines is unlikely – it would yield an alternating
PSPACE (i.e.
EXPTIME )algorithm, thereby implying
EXPTIME = EXPSPACE . We now give a concrete ex-ample that shows super-exponential counter values are in fact necessary in succinctone-counter games.
Game summary.
The game G n proceeds as follows:1. Eve increments the counter to a multiple of n , M · n ,2. Adam chooses some odd m ∈ (0 , n ) and adds it to the counter,3. Eve removes a multiple (at least one) of m · n from the counter, and4. Eve removes some m ′ ∈ (0 , n ) from the counter. Implementation.
Step 1 is implemented by a single Eve vertex with a loop of weight n . Steps 2 and 4 can be implemented by a sequence of n nodes (belonging to therelevant player) where two edges from the i -th to ( i + 1) -th vertex allow the relevantplayer to choose the i -th bit. Step 3 is implemented by a sub-game of n rounds repeatedas often as Eve chooses (but at least once). In the i -th round of the subgame Adamchooses the i -th bit b of m . If he chooses correctly then b · i + n is subtracted fromthe counter and the subgame continues, if he chooses incorrectly then the i -th bit iscleared, Eve exits the subgame and clears all but the i -th bit. Note that if Eve tries toexit when Adam chose correctly then the i -th bit is never cleared so Eve is unable toreach a counter value of . Correctness.