Countdown games, and simulation on (succinct) one-counter nets

We answer an open complexity question by Hofman, Lasota, Mayr, Totzke (LMCS 2016) for simulation preorder on the class of succinct one-counter nets (i.e., one-counter automata with no zero tests where counter increments and decrements are integers written in binary); the problem was known to be PSPACE-hard and in EXPSPACE. We show that all relations between bisimulation equivalence and simulation preorder are EXPSPACE-hard for these nets; simulation preorder is thus EXPSPACE-complete. The result is proven by a reduction from reachability games whose EXPSPACE-completeness in the case of succinct one-counter nets was shown by Hunter (RP 2015), by using other results. We also provide a direct self-contained EXPSPACE-completeness proof for a special case of such reachability games, namely for a modification of countdown games that were shown EXPTIME-complete by Jurdzinski, Sproston, Laroussinie (LMCS 2008); in our modification the initial counter value is not given but is freely chosen by the first player. We also present an alternative proof for the upper bound by Hofman et al. In particular, we give a new simplified proof of the belt theorem that yields a simple graphic presentation of simulation preorder on (non-succinct) one-counter nets and leads to a polynomial-space algorithm (which is trivially extended to an exponential-space algorithm for succinct one-counter nets).


Introduction
One-counter automata (OCA), i.e., finite automata equipped with a nonnegative counter, are studied as one of the simplest models of infinite-state systems.They can be viewed as a special case of Minsky counter machines, or as a special case of pushdown automata.In general, OCA can test the value of the counter for zero, i.e., some transitions could be enabled only if the value of the counter is zero.One-counter nets (OCN) are a "monotonic" (cf., e.g., [JS08]) to reduce reachability games to any relation between simulation preorder and bisimulation equivalence.Further contributions are explained below in more detail.
As already mentioned, the EXPSPACE-hardness of reachability games on succinct OCNs was shown in [Hun15] by using [GHOW10].Here we present a direct proof of EXPSPACEhardness (and completeness) even for a special case of reachability games, which we call the "existential countdown games".It is a mild relaxation of the countdown games from [JSL08] (or their variant from [Kie13]), which is an interesting EXPTIME-complete problem.We thus provide a simple EXPSPACE-hardness proof (in fact, by a master reduction via a natural intermediate problem dealing with ultimately periodic words) that is independent of [Hun15] (and of the involved technique from [GHOW10] used by [Hun15]).
We now give an informal sketch of the results for countdown games.The left-hand part of Figure 1 shows an example of a very simple countdown game.It is, in fact, a special finite automaton, with Eve's states, in our case just p 1 , and Adam's states, in our case just p 2 .The game assumes a nonnegative counter whose value is modified by transitions; in the countdown games the counter is only decreased by transitions.E.g., in the configuration p 1 (185), i.e. in the situation where the current state is p 1 and the current counter value is 185, Eve can choose the transition p 1 −24 −−→ p 1 , which changes the current configuration to p 1 (185 − 24), i.e. to p 1 (161), or the transition p 1 −1 − − → p 2 , which changes the current configuration to p 2 (184).In p 2 (184) Adam has two choices, either going to p 2 (159) or to p 2 (181).Since the counter cannot become negative, in p 2 (17) Adam has one choice only, necessarily reaching p 2 (2) by several steps, where the respective play finishes.Eve's goal is that the play finishes in p win (0) for a distinguished state p win ; in our example we put p win = p 2 .
It is obvious that an initial configuration p 2 (n) is winning for Eve iff n ∈ {0, 3, 6, 9, 12, 15, 18, 21, 24} as partly depicted in the right-hand part of Figure 1 by the white points.(The black points thus correspond to the configurations where Adam has a winning strategy; e.g., p 2 (7) is winning for Adam.)An initial configurations p 1 (n) is winning for Eve (i.e., Eve has a strategy guaranteeing reaching p 2 (0)), iff n mod 3 = 1.
Our concrete example is simple, but also in the general case, with states p 1 , p 2 , . . ., p k (distributed between Eve and Adam), it is straightforward to stepwise fill the respective "black-white-points table" in the bottom-up fashion that we now describe.The "points" p 1 (0), p 2 (0), . . ., p k (0) are black except of p i (0) where p i = p win since p win (0) is white.If we have filled the rows 0, 1, . . ., (hence each p i (j) where i ∈ {1, 2, . . ., k} and j ∈ {0, 1, . . ., } has been determined to be black or white), it is trivial to fill the row +1: p i ( +1) becomes white iff p i belongs to Eve and there is a move from p i ( +1) to an already established white point (in the rows 0, 1, . . ., ), or p i belongs to Adam, there is a move from p i ( +1), and each such move leads to an already established white point.
It is thus clear that the question if p(n) is Eve's win is in EXPTIME: we fill the "rows" for 0, 1, 2, . . ., n, using exponential time and space in the input size, since n and the countdown values are given in binary or in decimal notation.We also note that if m is the maximum countdown value (which is 25 in Figure 1), the row ≥ m is determined by the rows −1, −2, . . ., −m.It is thus clear that deciding the "existential version" of the countdown games, i.e. the question, given a state p, if there is n such that p(n) is winning for Eve, can be decided in exponential space.(When filling the table in the bottom-up fashion, we can always keep just last m rows in memory.)We can also observe that the black-white table is thus (ultimately) periodic, with an at most double-exponential period; the period is indeed double-exponential in concrete cases, as we also show in this paper.
The lower bounds might look more surprising: deciding if p(n) is Eve's win is EXPTIMEcomplete [JSL08], while deciding if there is n such that p(n) is Eve's win is EXPSPACEcomplete, as we show in this paper.In fact, these lower bounds (EXPTIME-hardness and EXPSPACE-hardness) can be also relatively easily established, when looking at the "bottomup" computation-table of an exponential-space Turing machine in a convenient way, as is depicted in and discussed around Figure 5.
Regarding the simulation problem, we give an example of a succinct one-counter net in Figure 2; it is also a finite automaton equipped with a nonnegative counter.Now there is no Eve or Adam, and the counter changes associated with transitions can be also nonnegative; moreover, the transitions have action-labels (a, b in our example).On the set of all configurations p(n), the simulation preorder is the maximal relation such that for each pair p(m) q(n) and each move p(m) a − → p (m ) there is a move q(n) a − → q (n ) (with the same label a) such that p (m ) q (n ).In our example we can note that p 2 (3) p 1 (58), since the move p 2 (3) b − → p 2 (1) cannot be answered by any b-transition from p 1 (58); on the other hand, we have p 2 (1) p 1 (58), since no transition is enabled in p 2 (1).Slightly more subtle is to note that p 1 (0) p 2 (6): the move p 1 (0) a − → p 2 (5) is answered by p 2 (6) a − → p 2 (4), and we check that p 2 (5) p 2 (4).
We can also think in terms of games here.In the simulation game a position is not just one configuration, but a pair (p(m), q(n)) of configurations.The first player, called Attacker 11:5 (or Spoiler), chooses a move p(m) a − → p (m ) (Attacker loses if there is no such move), and the other player, called Defender (or Duplicator), answers by some q(n) a − → q (n ) (with the same label a); Defender loses if there is no such answer.The play then continues with the next round, with the current pair (p (m ), q (n )).An infinite play is deemed to be winning for Defender.It is standard to observe that p(m) q(n) iff Attacker has a winning strategy from (p(m), q(n)) (and p(m) q(n) iff Defender has a winning strategy from (p(m), q(n))).
Given a succinct one-counter net, we can represent the relation by the respective "black-white colourings" C p,q of the integer points in the first quadrant of the plane, for each ordered pair of states p, q ; white points correspond to Attacker's wins, black points correspond to Defender's wins.In our example we have four ordered pairs of states ((p 1 , p 1 ),(p 1 , p 2 ),(p 2 , p 1 ),(p 2 , p 2 )), and the respective four colourings are depicted in Figure 3. E.g., p 2 (1) p 1 (n) and p 2 (2) p 1 (n) for all n, p 1 (0) p 2 (5) and p 1 (0) p 2 (6), etc.We can easily observe the general monotonicity: if p(m) q(n) then p(m ) q(n ) for all m ≤ m and n ≥ n.Our example also suggests that the black-white frontier in each colouring is contained in a "linear belt", with a rational (or infinite) slope and a certain width.Such a belt (two parallel lines) in C p 1 ,p 2 in Figure 3 can be described as follows: the frontier points, which we can define as the rightmost black points in each row, have coordinates (0, 6), (0, 7), (1, 8), (1, 9), (2, 10), (2, 11), • • • and are contained in a belt with the slope 2 1 and the vertical width 1. (A more general form is depicted in Figure 10.) Contemplating a bit, it is not difficult to get an intuition captured by a "belt theorem", claiming that also in any general case of (succinct) one-counter nets the frontier in each plane is contained in a linear belt; moreover, it is intuitively obvious that each frontier is periodic, from some row onwards.(In this paper, we also show that the periods can be, and are at most, double-exponential.)By the previous figures and discussions, one can also easily get an intuition that deciding the countdown games, even in their (EXPSPACE-complete) existential form, could be reduced to deciding the simulation preorder on (succinct) one-counter nets; finding a solution like the one depicted in Figure 1 seems intuitively simpler than finding a solution like the one depicted in Figure 3.This is confirmed in this paper, by a reduction based on a "defender-choice technique" that in particular enables to mimic Adam's choices in a countdown game by Defender's choices in the corresponding simulation game.
The above mentioned belt theorem is important for the upper bounds, namely for showing that deciding simulation preorder is in PSPACE for one-counter nets where the counter changes are presented in unary, and in EXPSPACE for succinct one-counter nets.
Proving the belt theorem had turned out surprisingly difficult; we also contribute to this topic here, as we explain below.The (non)succinctness of the one-counter nets plays no important role in this discussion, so we restrict ourselves to the cases in which the counter can change by at most 1 in one move.The qualitative form of the belt theorem (claiming just the existence of belts, not caring about their slopes, widths, and positions) followed from [Av98] where an involved mechanism of two-player games was used; this qualitative result was shown in [JMS99] by another technique, based rather on "geometric" ideas.The quantitative form, stating that the linear belts, assumed to start in the origin (0, 0), have the slopes and widths that can be presented as (fractions of) numbers with polynomially bounded values, was shown in [HLMT16], by enhancing the technique of games from [Av98]; this is the crux of the PSPACE-membership of simulation preorder for OCN [HLMT16] (which also yields the EXPSPACE-membership for succinct OCN).
In this paper we give a new self-contained proof for both the qualitative and quantitative versions of the belt theorem.One important new ingredient is a simple observation that we call a black-white vector travel.We describe it here, using Figure 4, but the description can be safely skipped if the reader finds it too technical; it is later captured by Proposition 5.6 and Corollary 5.16.For any vector v 0 with a positive slope and with a black start and  a white end in some colouring C p 0 ,q 0 , like the vector v 0 in Figure 4, there are vectors v 1 , v 2 , . . ., v k of the same size and slope as v 0 , each v i being a black-white vector in some colouring C p i ,q i , such that • for all i ∈ {1, 2, . . ., k}, v i is a neighbour of v i−1 , i.e., the x-coordinates of the starts of v i and v i−1 differ by at most 1, and the same constraint holds for their y-coordinates; and • the start of v k is on the vertical axis.This can be easily verified: in the pair of configurations corresponding to the white end of v 0 Attacker has an optimal transition p 0 a,z − − → p 1 , for which each Defender's response q 0 a,z − − → q establishes a "white" pair with a smaller rank, i.e., closer to Attacker's final win.Attacker can perform the transition p 0 a,z − − → p 1 also in the pair of configurations corresponding to the black start of v 0 , if this start is not on the vertical axis (which would entail that Attacker's counter is zero); there is at least one Defender's response q 0 a,z 1 −−→ q 1 establishing another "black" pair.Hence by changing the x-coordinate of v 0 by z and its y-coordinate by z 1 we get the vector v 1 that is black-white in the colouring C p 1 ,q 1 .Etc.This black-white vector travel, together with other enhancements, allowed us to substantially simplify the proof of the qualitative belt theorem from [JMS99].Moreover, the quantitative version can be now derived from the qualitative version by a few simple observations.
Organization of the paper.Section 2 gives the basic definitions.In Section 3 we show that the "existential" countdown games are EXPSPACE-complete (which also yields an alternative proof for the known EXPTIME-completeness of countdown games).Section 4 describes the reductions from reachability games to (bi)simulation relations, in a general framework and then in the framework of succinct OCN.Section 5 contains new proofs for both the qualitative and quantitative versions of the belt theorem (leading to the PSPACE membership for OCN).Section 6 shows that the period of the belts in the succinct case can be double-exponential.We finish with some additional remarks in Section 7.

Basic Definitions
By Z, N, N >0 we denote the sets of integers, of nonnegative integers, and of positive integers, respectively.We use [i, j], where i, j ∈ Z, for denoting the set {i, i+1, . . ., j}.
Labelled transition systems and (bi)simulations.A labelled transition system, an LTS for short, is a tuple L = (S, Act, ( where S is the set of states, Act is the set of actions, and a − → ⊆ S × S is the set of a-transitions (transitions labelled with a), for each a ∈ Act.We write s a − → t instead of (s, t) ∈ a − →.By s a − → we denote that a is enabled in s, i.e., s a − → t for some t.
The union of all simulations (on S) is the maximal simulation, denoted ; it is a preorder, called simulation preorder.The union of all bisimulations is the maximal bisimulation, denoted ∼; it is an equivalence, called bisimulation equivalence (or bisimilarity).We obviously have ∼ ⊆ .
We can write s 1 s 2 or s 1 ∼ s 2 also for states s 1 , s 2 from different LTSs L 1 , L 2 , in which case the LTS arising by the disjoint union of L 1 and L 2 is (implicitly) referred to.
It is useful to think in terms of two-player turn-based games, played by Attacker and Defender (or Spoiler and Duplicator).A round of the simulation game from a (current) pair (s, s ) proceeds as follows: Attacker chooses a transition s a − → t, and Defender responds with some s a − → t (for the action a chosen by Attacker); the play then continues with another round, now from the current pair (t, t ).If a player has no legal move in a round, then the other player wins; infinite plays are deemed to be Defender's wins.It is standard that s s iff Attacker has a winning strategy from (s, s ).
The bisimulation game is analogous, but in any round starting from (s, s ) Attacker can choose to play s a − → t or s a − → t , and Defender has to respond with some s a − → t or s a − → t, respectively.Here we have s ∼ s iff Attacker has a winning strategy from (s, s ).
Stratified simulation, and ranks of pairs of states.Given L = (S, Act, ( a − →) a∈Act ), we use (transfinite) induction to define the relations λ where λ ranges over the class Ord of ordinals.We put 0 = S × S. For λ > 0 we have s λ s if for each transition s a − → t and each λ < λ there is a transition s a − → t where t λ t .We note that λ ⊇ λ when λ ≤ λ, and that = λ∈Ord λ .For each pair (s, s ) ∈ , we define its rank rank(s, s ) as the least ordinal λ such that s λ s .We note in particular that rank(s, s ) = 1 iff s enables an action a (i.e., s a − →) that is not enabled in s (i.e., s a − →).
Remark 2.1.We use such a general definition for the purpose of the general reduction presented in Section 4. Otherwise we consider just (special cases of) LTSs that are imagefinite (i.e., in which the sets {t | s a − → t} are finite for all s ∈ S, a ∈ Act); in such systems we have = i∈N i and rank(s, s ) ∈ N for each (s, s ) ∈ .We could define the analogous concepts for bisimulation equivalence as well.
One-counter nets (OCNs and SOCNs), and their associated LTSs.A labelled onecounter net, or just a one-counter net or even just an OCN for short, is a triple where Q is the finite set of control states, Act the finite set of actions, and δ ⊆ Q × Act × {−1, 0, +1} × Q is the set of (labelled transition) rules.By allowing δ to be a finite subset of Q × Act × Z × Q, and presenting z ∈ Z in the rules (q, a, z, q ) in binary, we get a succinct one-counter net, or a SOCN for short.A rule (q, a, z, q ) is usually presented as q a,z − − → q .Each OCN or SOCN N = (Q, Act, δ) has the associated LTS where (q, m) We often write a state (q, m), which is also called a configuration, in the form q(m), and we view m as a value of a nonnegative counter.A rule q a,z − − → q thus induces transitions q(m) a − → q (m+z) for all m ≥ max{0, −z}.We remark that one-counter automata extend one-counter nets by the ability to test zero, i.e., by transitions that are enabled only if the counter value is zero.
Reachability games (r-games), winning areas, ranks of states.We are interested in reachability games played in LTSs associated with (succinct) one-counter nets, but we first define the respective notions generally.
By a reachability game, or an r-game for short, we mean a tuple where V is the set of states (or vertices), V ∃ ⊆ V is the set of Eve's states, − → ⊆ V × V is the transition relation (or the set of transitions), and T ⊆ V is the set of target states.By Adam's states we mean the elements of Eve's winning area is Win ∃ = λ∈Ord W λ , for Ord being the class of ordinals, where the sets W λ ⊆ V are defined inductively as follows.
We put W 0 = T ; for λ > 0 we put W <λ = λ <λ W λ , and we stipulate: a) if s ∈ W <λ , s ∈ V ∃ , and s − → s for some s ∈ W <λ , then (If (a) applies, then λ is surely a successor ordinal.)For each s ∈ Win ∃ , by rank(s) we denote (the unique) λ such that s ∈ W λ .A transition s − → s is rank-reducing if rank(s) > rank(s).We note that for any s ∈ Win ∃ with rank(s) > 0 we have: if s ∈ V ∃ , then there is at least one rank-reducing transition s − → s (in fact, rank(s) = rank(s)+1 in this case); if s ∈ V ∀ , then there is at least one transition s − → s and all such transitions are rank-reducing.This entails that Win ∃ is the set of states from which Eve has a strategy that guarantees reaching (some state in) T when Eve is choosing a next transition in Eve's states and Adam is choosing a next transition in Adam's states.
Remark 2.2.We are primarily interested in the games that have (at most) countably many states and are finitely branching (the sets {s | s − → s} are finite for all s).In such cases we have rank(s) ∈ N for each s ∈ Win ∃ .We have again introduced the general definition for the purpose of the reduction in Section 4.
Reachability games on succinct one-counter nets.We now define specific r-games, presented by SOCNs with partitioned control-state sets; these succinct one-counter nets are unlabelled, which means that the set of actions can be always deemed to be a singleton.
By a succinct one-counter net reachability game, a socn-r-game for short, we mean a tuple N = (Q, Q ∃ , δ, p win ) where Q is the finite set of (control) states, Q ∃ ⊆ Q is the set of Eve's (control) states, p win ∈ Q is the target (control) state, and δ ⊆ Q × Z × Q is the finite set of (transition) rules.We often present a rule (q, z, q ) ∈ δ as q z − → q .By Adam's (control) states we mean the elements of Q ∀ = Q Q ∃ .A socn-r-game N = (Q, Q ∃ , δ, p win ) has the associated r-game where (q, m) − → (q , n) iff q n−m − −− → q is a rule (in δ).We often write q(m) instead of (q, m) for states of G N .
We define the problem Socn-Rg (to decide succinct one-counter net r-games) as follows: Name: Socn-Rg Instance: a socn-r-game N (with integers z in rules q z − → q written in binary), and a control state p 0 .
Remark 2.3.We have defined the target states (in G N ) to be the singleton set {p win (0)}.There are other natural variants (e.g., one in [Hun15] defines the target set {p(0) | p = p 0 }) that can be easily shown to be essentially equivalent.
The EXPSPACE-hardness of Socn-Rg was announced in [Hun15], where an idea of a proof is sketched, also using a reference to an involved result [GHOW10] (which is further discussed in Section 7).In Section 3 we give a direct self-contained proof that does not rely on [Hun15] or involved techniques from [GHOW10], and that even shows that Socn-Rg is EXPSPACE-hard already in the special case that slightly generalizes the countdown games from [JSL08].(The EXPSPACE-membership follows from [Hun15], but we add a short proof to be self-contained.) Countdown games.We define a countdown game as a socn-r-game N = (Q, Q ∃ , δ, p win ), where in every rule q z − → q in δ we have z < 0. The problem Cg is defined as follows: Name: Cg (countdown games) Instance: a countdown game N (with integers in rules written in binary), and an initial configuration p 0 (n 0 ) where n 0 ∈ N (n 0 in binary).
The problem Cg (in an equivalent form) was shown EXPTIME-complete in [JSL08].Here we define an existential version, i.e.the problem Ecg: Name: Ecg (existential countdown games) Instance: a countdown game N and a control state p 0 .Question: is there some n ∈ N such that p 0 (n) ∈ Win ∃ ?We note that Ecg can be viewed as a subproblem of Socn-Rg: given an instance of Ecg, it suffices to add a fresh Eve's state p 0 and rules p 0

EXPSPACE-Completeness of Existential Countdown Games
In this section we prove the following new theorem.
Our EXPSPACE-hardness proof of ECG is, in fact, a particular instance of a simple general method.A slight modification also yields a proof of EXPTIME-hardness of countdown games that is an alternative to the proof in [JSL08] and in [Kie13].
In the rest of this section we present the proofs of Theorems 3.1 and 3.2 together, since they only differ by a small detail.We start with the upper bounds since these are obvious.

ECG is in EXPSPACE (and CG in EXPTIME). Let us consider an
δ, p win ), p 0 .By m we denote the maximum value by which the counter can be decremented in one step (i.e., m = max { |z| ; there is some q z − → q in δ }); the value m is at most exponential in the size of the instance N , p 0 .
We can stepwise construct for some j.We have W (0) = {p win }, and for determining W (n) (n ≥ 1) it suffices to know the segment W (n−m'), W (n−m'+1), . . ., W (n−1) where m' = min{m, n}.Hence, during the construction of W (j), j = 0, 1, 2, . . ., it suffices to remember just the segment W (j−m'), W (j−m'+1), . . ., W (j−1) (where m' ≤ m).Obviously, if for some j, j , where m ≤ j < j , the segments W (j−m), . . ., W (j−1) and W (j −m), . . ., W (j −1) are the same (i.e., if W (j−k) = W (j −k) for each k = 1, 2, . . ., m), then also W (j+k) = W (j +k) for each k ∈ N, so the sequence repeats itself with a period j − j.By the pigeonhole principle, this surely happens for some j, j such that j < j ≤ m+2 |Q|•m because there are at most 2 |Q|•m segments of length m.This means that in the algorithm that checks whether p 0 ∈ W (j) for some j ∈ N, the computation can be stopped after constructing W (m + 2 |Q|•m ).The size of a binary counter serving to count till the (at most double-exponential) value m + 2 |Q|•m is at most exponential.Therefore Ecg belongs to EXPSPACE.

ECG is EXPSPACE-hard (and CG EXPTIME-hard).
In principle, we use a "master" reduction.We fix an arbitrary language L in EXPSPACE, in an alphabet Σ (hence L ⊆ Σ * ), decided by a (deterministic) Turing machine M in space 2 p(n) for a fixed polynomial p.For any word w ∈ Σ * , |w| = n, there is the respective computation of M using at most m = 2 p(n) tape cells, which is accepting iff w ∈ L. Our aim is to show a construction of a countdown game N M w,m , with a specified control state p 0 , such that there is k ∈ N for which p 0 (k) ∈ Win ∃ if, and only if, M accepts w.The construction of N M w,m will be polynomial, in the size n = |w|; this will establish that Ecg is EXPSPACE-hard.(In fact, this polynomial construction easily yields a logspace-reduction, but this detail is unimportant at our level of discussion.) In fact, the same construction of N M w,m will also show EXPTIME-hardness of Cg.In this case we assume that L is decided by a Turing machine M in time (and thus also space) 2 p(n) , and we construct a concrete exponential value n 0 guaranteeing that p 0 (n 0 ) ∈ Win ∃ iff M accepts w.

Construction informally. The construction of the countdown game N M
w,m elaborates an idea that is already present in [CKS81] (in Theorem 3.4) and that was also used, e.g., in [JS07].We first present the construction informally.
Figure 5 presents an accepting computation of M, on a word w = a 1 a 2 . . .a n ; it starts in the initial control state q 0 with the head scanning a 1 .The computation is a sequence of configurations C w 0 , C w 1 , . . ., C w t , where C w t is accepting (since the control state is q acc ).We assume that M never leaves cells 0 . . .m−1 of its tape during the computation.Hence each C w i can be presented as a word of length m over the alphabet ∆ = (Q × Γ) ∪ Γ where Q and Γ are the set of control states and the tape alphabet of M, respectively; by ∈ Γ we denote the special blank tape symbol.We refer to the (bottom-up) presentation of C w 0 , C w 1 , . . ., C w t depicted in Figure 5 as to a computation table.
Given m, each number k ∈ N determines the cell j in the "row" i (i.e., in the potential C w i ) where i = k ÷ m (÷ being integer division) and j = k mod m; we refer by cell(k) to this cell j in the row i. 5), then the symbol β in cell(k) in the table is surely determined by the symbols β 1 , β 2 , β 3 in the cells cell(k−m−1), cell(k−m), cell(k−m+1) (and by the transition function of the respective Turing machine M); see Figure 5 for illustration (where also the cases β and β on the "border" are depicted).The transition function of M allows us to define which triples (β 1 , β 2 , β 3 ) are eligible for β, i.e., those that can be in the cells cell(k−m−1), cell(k−m), cell(k−m+1) when β is in cell(k) (these triples are independent of k, assuming k > m).
Let us now imagine a game between Eve and Adam where Eve, given w, claims that w is accepted by M, in space m (in our case m = 2 p(|w|) for a respective polynomial p).Eve does not present a respective accepting computation table but she starts a play by producing a tape-symbol x and a number k 0 ∈ N, i.e. sets a counter to k 0 , claiming that cell(k 0 ) in the computation table contains (q acc , x) (in Figure 5 the correct values are k 0 = t • m and x = ).Then the play proceeds as follows.If Eve claims that β is the symbol in cell(k) for the current counter value k, while also claiming that k > m, then she decreases the counter by m−2 and produces a triple ( Adam then decreases the counter either by 3, asking to verify the claim that cell(k−m−1) contains β 1 , or by 2, asking to verify that cell(k−m) contains β 2 , or by 1, asking to verify that cell(k−m+1) contains β 3 .Eve then again produces an eligible triple for the current symbol, etc., until claiming that the counter value is k ≤ m (which can be contradicted by Adam when he is able to decrease the counter by m+1).The last phase just implements checking if the symbol claimed for cell(k) corresponds to the initial configuration.
It is clear that Eve has a winning strategy in this game if w is accepted by M (in space m).If w is not accepted, then Eve's first claim does not correspond to the real computation table.Moreover, if a claim by Eve is incorrect, then at least one claim for any respective eligible triple is also incorrect (as can be easily checked), hence Adam can be always asking to verify incorrect claims, which is revealed when the consistency with the initial configuration is verified in the end.Hence Eve has no winning strategy if w is not accepted by M.
In the formal construction presented below we proceed by introducing an intermediate auxiliary problem (in two versions) that allows us to avoid some technicalities in the construction of countdown games N M w,m .Roughly speaking, instead of the computation of M on w we consider the computation of M w on the empty word, which first writes w on the tape and then invokes M; checking the consistency with the (empty) initial configuration is then technically easier to handle in the constructed countdown game.
Construction formally.Now we formalize the above idea, using the announced intermediate problem.
By a sequence description we mean a tuple D = (∆, D, m), where ∆ is its finite alphabet, always containing two special symbols # and (and other symbols), D : ∆ 3 → ∆ is its description function, and m ≥ 3 is its initial length.The sequence description D = (∆, D, m) defines the infinite sequence S D in ∆ ω , i.e. the function S D : N → ∆, that is defined inductively as follows: Proof.To show EXPSPACE-hardness of Eseq, we assume an arbitrary fixed language L ⊆ Σ * in EXPSPACE, decided by a Turing machine M in space 2 p(n) for a polynomial p.
Using a standard notation, let M = (Q, Σ, Γ, δ, q 0 , {q acc , q rej }) where Σ ⊆ Γ, ∈ Γ Σ, # ∈ Γ, and δ : Q × Γ → Q × Γ × {−1, 0, +1} is the transition function of M, satisfying δ(q acc , x) = (q acc , x, 0) and δ(q rej , x) = (q rej , x, 0) (hence an accepting or rejecting configuration is formally viewed as repeated forever).Moreover, w.l.o.g.we assume that the computation of M on w = a 1 a 2 • • • a n ∈ Σ * starts with w written in tape cells 1, 2, . . ., n with the head scanning the cell 1 (the control state being q 0 ), the computation never leaves the cells 0, 1, . . ., m−1 for m = 2 p(n) , never rewrites in the cell 0, and the state q acc , q rej can only be entered when the head is scanning the cell 0 (as is also depicted in Figure 5); we also assume that p is such that m = 2 p(n) satisfies m > n and m ≥ 3.
Given w ∈ Σ * , we now aim to show a polynomial construction of D w = (∆ w , D w , m), where m = 2 p(|w|) , such that w ∈ L iff there is i ∈ N such that S Dw (i) = (q acc , ). (Hence w is reduced to the Eseq-instance D w , β 0 where β 0 = (q acc , ).) As already suggested in the previous discussion, for w = a 1 a 2 • • • a n we first construct a Turing machine M w that starts with the empty tape while scanning the cell 0 in its initial state q 0 , then by moving to the right it writes w = a 1 a 2 • • • a n in the cells 1, 2, . . ., n, after which it moves the head to the cell 1 and enters q 0 , thus invoking the computation of M on w.
When we consider the computation of M w on the empty word as the sequence S = C 0 C 1 C 2 . . . of configurations of length m, where m = 2 p(|w|) , and we view the symbol (q 0 , ) as #, then we observe that S(0 A polynomial reduction from L to Eseq is therefore clear, yielding EXPSPACE-hardness of Eseq. In the case of Seq, we assume that M deciding L works in time (and thus also space) 2 p(n) ; to the Eseq-instance S Dw , β 0 = (q acc , ) constructed to w as above we simply add n 0 = m 2 (for m = 2 p(|w|) ), to get a Seq-instance.Here it is clear that w ∈ L iff S Dw (n 0 ) = β 0 .This yields EXPTIME-hardness of Seq.
We now show polynomial (in fact, logspace) reductions from Eseq to Ecg and from Seq to Cg.Again, we present both reductions together.
Given a sequence description D = (∆, D, m) (where #, ∈ ∆), we construct the countdown game where , and the set δ consists of the rules in Figure 6 (for all β, β 1 , β 2 , β 3 ∈ ∆).(Note that the rules for states s and s # also include the rules of the form (5).) The idea is that the configuration s β (k) of N D should "claim" that S D (k) = β (and Eve should have a winning strategy from s β (k) iff this claim is correct).For technical reasons we add 2 to the counter, hence it is s β (k+2) that "claims" S D (k) = β.Lemma 3.5.For each β ∈ ∆ we have s β (0) ∈ Win ∃ , s β (1) ∈ Win ∃ , and for each k ∈ N we have

States Rules
p win (∃) - Proof.We start by noting the following facts that are easy to check (recall that Eve wins iff the configuration p win (0 ).The statement of the lemma for s β (0) and s β (1) follows from the fact (c).To prove that s β (k+2) ∈ Win ∃ iff S D (k) = β, we proceed by induction on k: • Base case k ∈ [0, m]: First we note that Eve cannot win in s β (k+2) by playing any rule of the form (5) because either this cannot be played at all, or where k ≤ 4, and Adam either cannot continue in t (β 1 ,β 2 ,β 3 ) (k ), or he can play to s β (k ) for some ∈ {1, 2, 3} with k < 2, which is losing for Eve (fact (c)).Now it easily follows from the facts (d) and (e) that Eve wins in s β (k+2) exactly in those cases where either β = # and k = 0 (fact (d)), or β = and k ∈ [1, m] (fact (e)).• Induction step for k > m: Eve cannot win in s β (k+2) by playing a rule of the form (3) or (4), so she is forced to use a rule from (5), i.e., to play a transition of the form where where k = k−m+ ≥ 2. Either β is correct (i.e., S D (k) = β), and then Eve can choose correct β 1 , β 2 , β 3 , or β is incorrect (i.e., S D (k) = β), and then at least one of β 1 , β 2 , β 3 must be also incorrect.Since, by the induction hypothesis, Adam can win in t (β 1 ,β 2 ,β 3 ) (k ) iff one of β 1 , β 2 , β 3 is incorrect (by choosing the corresponding move for this incorrect β ), Eve can win in s β (k+2) iff β is correct.From this we derive that s β (k+2 Hence, for a given Eseq-instance D, β 0 (where Recalling Proposition 3.4, we have thus established the lower bounds in Theorems 3.1 and 3.2.

Reachability Game Reduces to (Bi)simulation Game
We show a reduction for general r-games, and then apply it to the case of socn-r-games.This yields a logspace reduction of Socn-Rg to behavioural relations between bisimulation equivalence and simulation preorder.
Recalling the EXPSPACE-hardness of Socn-Rg (from [Hun15], or from the stronger statement of Theorem 3.1), the respective lemmas (Lemma 4.2 and 4.3) will yield the following theorem (which also answers the respective open question from [HLMT16]): Theorem 4.1.For succinct labelled one-counter net (SOCNs), deciding membership problem in any relation containing bisimulation equivalence and contained in simulation preorder (of the associated LTSs) is EXPSPACE-hard.
4.1.Reduction in a General Framework.We start with an informal introduction to the reduction, which is an application of the technique called "Defender's forcing" in [JS08].
Any r-game G = (V, V ∃ , − →, T ) gives rise to the LTS L = (V, Act, ( a − →) a∈Act ) where A technical problem is how to achieve (4.1) in the case V ∃ = V ; in this case also Adam's choices in the r-game have to be faithfully mimicked in the (bi)simulation game, where it is now Defender who should force the outcome of the relevant game rounds.This is accomplished by adding "intermediate" states and transitions, and the "choice action" a c , as depicted in Figure 8 (and discussed later in detail); Attacker must let Defender to really choose since otherwise Defender wins by reaching a pair (s, s) with the equal sides (where s ∼ s).11:17 ; it is thus Attacker who chooses (s 2 , s 2 ) or (s 3 , s 3 ) as the next current pair.
Now we formalize the above sketch.We assume an r-game G, and we define a "mimicking" LTS L(G) (the enhanced union of the above LTSs L and L ).In illustrating Figures 7 and 8 we now ignore the bracketed parts of transition-labels; hence, e.g., in Figure 7 we can see the transition s 1 − → s 2 in G on the left and the (corresponding) transitions s 1 Proof.a) For the sake of contradiction suppose that there is s ∈ Win ∃ such that s s ; we consider such s ∈ Win ∃ with the least rank.We note that rank(s) > 0, since s ∈ T entails s s due to the transition s → s be a rank-reducing transition.Attacker's move s a s,s − −− → s, from the pair (s, s ), must be responded with s a s,s − −− → s ; but we have s s by the "least-rank" assumption, which contradicts the assumption s s .If s ∈ V ∀ , then X = {s | s − → s} is nonempty (since s ∈ Win ∃ ) and rank(s) < rank(s) for all s ∈ X.For the pair (s, s ) we now consider Attacker's move s ac − → s, X .Defender can choose s ac − → s, s for any s ∈ X (recall that rank(s) < rank(s)).In the current pair 11:18 A Figure 8.In (s 1 , s 1 ) it is, in fact, Defender who chooses (s 2 , s 2 ) or (s 3 , s 3 ) (when Attacker avoids pairs with equal states); to take the counter-changes into account correctly, we put x = min {x, 0}, x = max {x, 0}, and y = min {y, 0}, y = max {y, 0} (hence x = x +x and y = y +y ).
( s, X , s, s ) Attacker can play s, X a s,s − −− → s, and this must be responded by s, s a s,s − −− → s .But we again have s s by the "least-rank" assumption, which contradicts s s .
b) It is easy to verify that the following set is a bisimulation in L(G): We note that the transitions s 1 ac − → s 1 2 and s 1 ac − → s 1 3 in Figure 8 could be omitted if we only wanted to show that s ∈ Win ∃ iff s s .4.2.Socn-Rg Reduces to Behavioural Relations on SOCNs.We now note that the LTS L(G N ) "mimicking" the r-game G N associated with a socn-r-game N (recall (2.2)) can be presented as L N for a SOCN N (recall (2.1)) that is efficiently constructible from N : Lemma 4.3.There is a logspace algorithm that, given a socn-r-game N , constructs a SOCN N such that the LTSs L(G N ) and L N are isomorphic.
Proof.We again use Figures 7 and 8 for illustration; now s i are viewed as control states and the bracketed parts of edge-labels are counter-changes (in binary).
Given a socn-r-game N = (Q, Q ∃ , δ, p win ), we first consider the r-game ("the control-state game of N ") arising from N by forgetting the counter-changes; hence q − → q iff there is a rule q z − → q.In fact, we will assume that there is at most one rule q z − → q in δ (of N ) for any pair (q, q) ∈ Q × Q; this can be achieved by harmless modifications.
We construct the (finite) LTS L(N csg ) ("mimicking" N ).Hence each q ∈ Q has the copies q, q in L(N csg ), and other states are added (as also depicted in Figure 8 where s i are now in the role of control states); there are also the respective labelled transitions in L(N csg ), with labels a q,q , a c , a win .
It remains to add the counter changes (integer increments and decrements in binary), to create the required SOCN N .For q ∈ Q ∃ this adding is simple, as depicted in Figure 7: if q z − → q (in N ), then we simply extend the label a q,q in L(N csg ) with z; for q a q,q − −− → q and q a q,q − −− → q in L(N csg ) we get q a q,q ,z −−−−→ q and q a q,q ,z −−−−→ q in N .For q ∈ Q ∀ (where is tempting to the same, i.e. to extend the label a q,q with z when q z − → q, and extend a c with 0. But this might allow cheating for Defender: she could thus mimic choosing a transition q(k) x − → q(k+x) even if k+x < 0. This is avoided by the modification that is demonstrated in Figure 8 (by x = x +x , etc.); put simply: Defender must immediately prove that the transition she is choosing to mimic is indeed performable.Formally, if X = {q | q − → q} = ∅ (in L(N csg )), then in N we put q ac,0 −−→ q, X and q, X a q,q ,z −−−−→ q for each q z − → q (in N ); for each q z − → q we also define z = min{z, 0}, z = max{z, 0} and put q ac,z − −− → q, q , q, q a q,q ,z − −−−− → q .Then for any pair q z − → q, q z − → q where q = q we put q, q a q,q , z−z Finally, p win We have thus finished the proof of Theorem 4.1.

Structure of simulation preorder on one-counter nets
In this section we give a new self-contained proof clarifying the structure of simulation preorder on the LTS L N associated with a given one-counter net N ; this will also yield a polynomial-space algorithm generating a description of on L N that can be used to decide if p(m) q(n).
We first show a natural graphic presentation of the relation , and in Section 5.1 we show its linear-belt form; the result is captured by the belt theorem.The proof is inspired by [JMS99] but is substantially different.A main new ingredient is the notion of so called down-black and up-white lines, and their limits in which we do not a priori exclude lines with irrational slopes; this allows us to avoid many technicalities used in previous proofs, while the presented "geometric" ideas should be straightforward, and transparent due to the respective figures.
In Section 5.2 we prove that the slopes and the widths of belts are presented/bounded by small integers.Though this quantitative belt theorem is in principle equivalent to the respective theorem proved in [HLMT16], our proof is again conceptually different; using the novel notions, the quantitative characteristics are derived from the (qualitative) belt theorem easily.
For completeness, in Section 5.3 we briefly recall the idea from the previous papers that shows how the achieved structural results yield a polynomial-space algorithm generating a description of for a given one-counter net.
In the sequel we assume a fixed OCN N = (Q, Act, δ) if not said otherwise.By R, R ≥0 , R >0 we denote the sets of reals, of nonnegative reals, and of positive reals, respectively.We also use ∞ for an infinite amount (in particular for the slope of a vertical line); hence α < ∞ for all α ∈ R. By γ, γ , where γ ∈ R ≥0 and γ ∈ R ≥0 ∪ {∞}, we denote the set Monotonic black-white presentation of the simulation preorder .For each pair (p, q) ∈ Q × Q we define the (black-white) colouring C p,q of the integer points in the first quadrant of the plane R × R; we put C p,q : N × N → {black, white} where We recall the definition of rank(s, s ) in the paragraph "Stratified simulation, and ranks of pairs of states" in Section 2; this yields the definition of rank(p(m), q(n)) for our fixed OCN N = (Q, Act, δ).Hence for each (p, q) ∈ Q × Q we have that each white point (m, n) in C p,q has an associated finite rank, namely rank(p(m), q(n)) ∈ N.
The next proposition captures the trivial fact that "black is upwards-and leftwardsclosed " and "white is downwards-and rightwards-closed " (as is also depicted in Figure 3 in Introduction, which also shows the "linear-belt form" of the colourings).
Proposition 5.1 (Black and white monotonicity).If C p,q (m, n) = black, then C p,q (m , n ) = black for all m ≤ m and n ≥ n.Hence if C p,q (m, n) = white, then C p,q (m , n ) = white for all m ≥ m and n ≤ n.
Proof.Since OCNs are monotonic in the sense that p(m) a − → q(n) implies p(m+i) a − → q(n+i) for all i ∈ N, we will easily verify that the relation ⊆ R, we will thus get R = , and the proof will be finished.To verify that R is indeed a simulation, we consider p(m ), q(n ) ∈ R and fix m ≥ m and n ≤ n so that p(m) q(n).If p(m ) a − → p (m +i) (for i ∈ {−1, 0, +1}), then p(m) a − → p (m+i).Since p(m) q(n), we have q(n) a − → q (n+j) where j ∈ {−1, 0, +1} and p (m+i) q (n+j).Hence q(n ) a − → q (n +j), and p (m +i), q (n +j) ∈ R.
5.1.Belt theorem.Below we state the (qualitative) belt theorem, illustrated in Figure 10, after introducing the needed notions.We can remark that the validity of the belt theorem can be easily intuitively anticipated (by a quick thought about the problem) but it has turned out surprisingly hard to be rigorously proven.(It can be also intuitively anticipated that the colouring inside the belt determined by the lines L , R in Figure 10 is ultimately periodic, but this is only used in Section 5.3.)Lines, slopes, axes, points, points above or below lines, points on the left of or on the right of vertical lines.(See Figure 9.) A vertical line (in the plane R × R) is the set = {(x, y) | y ∈ R} for a fixed x ∈ R; we put slope( ) = ∞.A non-vertical line is the set = {(x, y) | y = γ • x + c} for some fixed γ ∈ R and c ∈ R; here slope( ) = γ.A line is either a vertical line or a non-vertical line.By a γ-line, γ ∈ R ∪ {∞}, we mean a line whose slope is γ.As expected, by the horizontal axis, h-axis, we mean the 0-line that goes through the origin (0, 0); the vertical axis, v-axis, is the ∞-line that goes through (0, 0).
By points we mean the pairs (m, n) ∈ N × N (hence only the integer points in the first quadrant of the plane R × R), unless explicitly stated that we consider all integer points (elements of Z × Z).Given a non-vertical line = {(x, y) below , or also on the right of , if m ≥ x.A point p is strictly above if p is above but not on ; similarly, p is strictly below if p is below but not on .
Theorem 5.2 (Belt theorem).For each (p, q) ∈ Q × Q, and its respective colouring C = C p,q , there are (a slope) γ ∈ 0, ∞ that is rational or ∞, and two (parallel) γ-lines L and R such that all points above L are black in C and all points below R are white in C. (See Figure 10.) We recall that we do not discuss how the mapping C p,q looks inside the belt depicted in Figure 10; only in Section 5.3 we refer to the fact that it is ultimately periodic there.Now we introduce further notions useful in the proof of the belt theorem.
Down-black and up-white lines and slopes, down-limits α C and up-limits β C .A line , with slope( ) ∈ 0, ∞ , is down-black in a colouring C = C p,q if there are infinitely many points below that are black in C. A (nonnegative) slope γ ∈ 0, ∞ is down-black in C if there is a γ-line that is down-black in C. By the down-limit of C we mean the value We say that α is a down-limit if α = α C for some C = C p,q .For a down-limit α, by α-colourings we mean the colourings C such that α C = α.
A line is up-white in a colouring C = C p,q if there are infinitely many points above that are white in C. A (nonnegative) slope γ ∈ 0, ∞ is up-white in C if there is a γ-line that is up-white in C. By the up-limit of C we mean the value We highlight the following trivial fact.
We note that we cannot a priori exclude that some α C is irrational and/or not down-black; similarly we cannot exclude that β C differs from α C , and that it is irrational and/or not up-white.But we immediately note a simple fact: Proof.Let α C > β C for C = C p,q , and let α C > γ > β C ; hence γ is not down-black and not up-white.Thus for each γ-line we have that almost all points (i.e., all but finitely many) below are white and almost all points above are black; by an obvious shift of we deduce that γ is up-white and/or down-black after all, which is a contradiction.
We now state a lemma that trivially entails Theorem 5.2: the slope γ claimed for C in Theorem 5.2 is equal to α C (which is claimed to be equal to β C by the lemma).
Lemma 5.5.For each (p, q) ∈ Q × Q and the respective colouring C = C p,q the following conditions hold:

and
(3) there are α C -lines L and R such that all points above L are black in C, and all points below R are white in C.
Before proving the lemma we introduce a main technical ingredient of the proof, namely Proposition 5.7 and Corollary 5.9, preceded by the needed notions and by a simple, yet very useful, observation (Proposition 5.6).We can remark that these technical ingredients can hardly be "intuitively anticipated"; they have been "distilled" from the overall nontrivial proof, as useful technical claims.Some intuition about a possible proof strategy might be perhaps got by looking at the simulation games described in the papers [Av98] and [HLMT16], but our proof is, in fact, not so tightly 11:23  Black-white vectors.A vector v is black-white in a colouring C if start(v) is black and end(v) is white in C.

Neighbour points and vectors
and v, v have the same slopes and sizes (hence v is a "small shift" of v).
The next proposition is the announced simple observation.It states that for any vector v where start(v) is not on the vertical axis, end(v) is not on the horizontal axis, and v is black-white in some C p,q there is its neighbour vector v that is black-white in some C p ,q and the rank of the white end-point of v in C p ,q is smaller than the rank of the white end-point of v in C p,q .(Recall the vectors v 0 , v 1 in Figure 4a in Introduction, and the discussion of ranks rank(p(m), q(n)) between the definition of C p,q (m, n) and Proposition 5.1.)Proposition 5.6 (Neighbour black-white vector with smaller rank).Let where m > 0 and n > 0. Then there are p , q ∈ Q and i, j ∈ {−1, 0, 1} such that C p ,q (m+i, n+j) = black, C p ,q (m +i, n +j) = white, and rank(p (m +i), q (n +j)) < rank(p(m ), q(n )).
Proof.Let the assumptions hold; we put rank(p(m ), q(n )) = r ∈ N. We can thus fix a transition p(m ) a − → p (m +i) (related to a rule p a,i −→ p ) such that for each q(n ) a − → q (n +j) we have rank(p (m +i), q (n +j)) < r.Since m > 0, we have p(m) a − → p (m+i), and since p(m) q(n), there is a transition q(n) a − → q (n+j) such that p (m+i) q (n+j); since n > 0, we also have q(n ) a − → q (n +j).Hence C p ,q (m+i, n+j) = black, and C p ,q (m +i, n +j) = white, and rank(p (m +i), q (n +j)) < r.Now we aim to formulate a crucial ingredient of the proof of Lemma 5.5, namely Proposition 5.7 and its corollary, for which we need further notions.Areas, border and interior points of areas.By an area we mean just a set A ⊆ N × N of points.A point p ∈ A is a border point of A if it has a neighbour point outside A (hence in (N × N) A); we put border(A) = {p | p is a border point of A}.Each point p ∈ A border(A) is an interior point of A. (E.g., (0, 0) is an interior point of A = {(0, 0), (0, 1), (1, 0), (1, 1)}.)Line-level-line areas area(( , b), ) and area( , (b, )).(See Figure 12b.)Given two lines , where ∞ ≥ slope( ) ≥ slope( ) ≥ 0, and a bottom-level b ∈ R ≥0 , we define area(( , b), ) as the set of points (in N × N) that are below , above , and also above the line ⊥ b ; an exception is the case with slope( ) = 0 (hence slope( ) = 0 as well) where the points in area(( , b), ) are to the right of (the vertical line) ⊥ b .By area( , (b, )) we mean the set of points that are below , above , and also above the line ( ) ⊥ b ; again, an exception is the case with slope( ) = 0, where the points in area( , (b, )) are to the right of (the vertical line) ( ) ⊥ b .In fact, we will not encounter the "pathological" case where slope( ) = slope( ) and is strictly above (to the left of) , in which case the sets area(( , b), ) and area( , (b, )) are empty.In the special case where slope( ) = ∞ and slope( ) = 0, we only

Point
The following crucial technical proposition is illustrated in Figure 13.It will be used several times later to bound the up-limits under the described circumstances.
Proposition 5.7 (Bounding the up-limits β C for a class C of colourings).We assume a nonempty set C of colourings (from the set {C p,q | p, q ∈ Q}), two lines , and a level b ∈ R ≥0 such that the following conditions are satisfied (cf. Figure 13a): (1) ∞ ≥ slope( ) ≥ slope( ) ≥ 0, and area(( , b), ) has no point on the vertical axis; (2) each colouring C ∈ C is monochromatic (all-black or all-white) inside area(( , b), ); (3) for each C ∈ C:  Proof.To prove Proposition 5.7, we let its assumptions hold, and we put ρ = slope( ).If ρ = ∞, then the claim is trivial: there are no points above, i.e. left of, any vertical line ¯ that does not intersect the first quadrant; such ¯ trivially satisfies the claim.Hence we further assume We also note that for almost all m ∈ N there is n such that (m, n) ∈ area(( , b), ); this is obvious if slope( ) > slope( ), and in the case ρ = slope( ) = slope( ) it is guaranteed by the existence of interior points of area(( , b), ) (they exist due to the points P C ): if p = (m, n) is interior in area(( , b), ), then we have (m, n+1) ∈ area(( , b), ) and for any m ≥ m there is n such that (m , n ) lies below and above the parallel ρ-line going through p.
We call a vector v eligible if start(v) ∈ area(( , b), ), v is black-white in some colouring, and both the ρ-d-size of v and the co-ρ-d-size of v are nonnegative.(We note that end(v) can be outside area(( , b), ), but then it necessarily lies strictly above .)E.g., v in Figure 13b is eligible.

Figure 14
To finish the proof, it suffices to show that the co-ρ-d-sizes of eligible vectors are bounded by some B 0 ∈ R ≥0 ; the claimed line ¯ then surely exists (as is illustrated in Figure 13b): we can take ¯ above in the distance (to ) greater than B 0 so that, moreover, ¯ intersects v-axis above the vertical coordinates of all P C , C ∈ C.
For the sake of contradiction, we assume that the co-ρ-d-sizes of eligible vectors are not bounded; hence for every B ∈ R ≥0 the set is eligible and the co-ρ-d-size of v is greater than B } is nonempty.We first note that there is some B ∈ R ≥0 such that for each B ≥ B and each v ∈ E B we have that v is not black-white in any C ∈ C (and thus must be black-white in some C ∈ C).
We verify this claim by considering a fixed colouring C ∈ C, which is monochromatic in area(( , b), ), and by noting that for each B ∈ R ≥0 and each v ∈ E B we have: This entails that for this m there is no n such that (m, n ) ∈ area(( , b), ).As discussed above, this can be the case only for finitely many m, which entails that there are only finitely many eligible vectors that are black-white in some C ∈ C. Hence the existence of B is clear.Now for each B ≥ B we fix a vector v B ∈ E B with the least possible rank of its white end; thus v B is black-white in some C p,q ∈ C, end(v B ) = (m, n), and rank(p(m), q(n)) is the least possible, when considering all v ∈ E B and all C ∈ C. Since start(v B ) is in area(( , b), ), it is not on the vertical axis, and thus Proposition 5.6 entails that start(v B ) ∈ border(area(( , b), )) (otherwise v B has a neighbour vector that is also in E B and its white end has a lesser rank).
For any fixed B ≥ B , by 3(b) we thus have dist(start(v B ), ) ≤ dist(p C , ) for the respective colouring C = C p,q ∈ C (in which end(v B ) has the least possible rank); the vector (p C , end(v B )), which is also black-white in C, thus has the co-ρ-d-size also greater than B (since it is not smaller than the co-ρ-d-size of v B ).If the vector (p C , end(v B )) were eligible (as is depicted in Figure 14a), thus belonging to E B , Proposition 5.6 would yield a contradiction (since p C is an interior point of area(( , b), ), the vector (p C , end(v B )) would have a neighbour vector that is also in E B and its white end has a lesser rank).Hence the ρ-d-size of (p C , end(v B )) is negative (as depicted in Figure 14b); this entails that ρ = slope( ) > 0 (since in the case ρ = 0 the fact C(p C ) = black would entail C(end(v B )) = black, by Proposition 5.1, which contradicts the assumption C(end(v B )) = white).
The fact that ρ = slope( ) > 0 and the ρ-d-size of (p C , end(v B )) is negative entails that end(v B ) is strictly below the line that is perpendicular to and goes through p C ; moreover, end(v B ) is below the horizontal line going through p C since C(p C ) = black and C(end(v B )) = white (see Figure 14b).There are thus only finitely many points that can be end(v B ), independently of the chosen B ≥ B .In other words, the set {end(v B ) | B ≥ B } is finite, which also entails that the set {v B | B ≥ B } is finite (recall that v B is eligible, and thus the ρ-d-size of v B is nonnegative).This contradicts our choice of v B that entails that the co-ρ-d-size of v B is greater than B, for each B ∈ R ≥0 .
The co-ρ-d-sizes of eligible vectors are thus indeed bounded by some B 0 ∈ R ≥0 .
Besides Corollary 5.8, which is trivial, we also derive the next corollary, showing that if the assumptions of Proposition 5.7 are slightly strengthened, then we get a strict upper bound on the up-limits β C .Proof.Let the described conditions be satisfied (in fact, such a situation is depicted already in Figure 13a; Figure 15a makes clear that all neighbour points of p C are strictly below ).
Since ρ = slope( ) > slope( ) ≥ 0, we can very slightly rotate to the right around the intersection-point of and ⊥ b (see Figure 15b), by which we get with a slope ρ , ρ > ρ ≥ slope( ), so that the resulting area(( , b ), ) ( ⊥ b has rotated to ( ) ⊥ b ) is a subset of area(( , b), ) (in other words, the rotation of ⊥ b to ( ) ⊥ b is so small that there is no [integer] point in area(( , b ), ) area(( , b), ), i.e. in the grey triangle in Figure 15b) and so that the original points p C are interior points also in area(( , b ), ) while the conditions of Proposition 5.7 (in particular dist(p C , ) ≥ bd C ) are kept.(To be very precise, in the case where slope( ) = ∞ and slope( ) = 0 we recall our stipulation that ⊥ b coincides with ; the discussed slight rotation then trivially has the property that there is no point in area(( , b ), ) area(( , b), ).) Hence Proposition 5.7 entails that ρ ≥ β C , and thus ρ > β C (since ρ > ρ ).
Proof of Lemma 5.5.We denote the down-limits, i.e. the elements of the set (5.1) (a) Assumptions of Corollary 5.9 (the unit square shows that p C has no neighbour on ).
Claim 5.10.For each α i -colouring C we have β C = α i (hence the down-limit α C = α i and the up-limit β C coincide).
We fix a ρ-line and a ρ -line , both going through the origin (0, 0).(We can recall and from Figure 13a, and imagine that they go through (0, 0).)By C we denote the set of all α i -colourings, and note: • for each C ∈ C (where α C = α i ), there are infinitely many black points in area(( , 0), ), since the ρ-line must be down-black in C due to the fact ρ > α i = α C (recall Proposition 5.3), but almost all border points of area(( , 0), ) along (i.e.those having neighbours strictly below ) are white: if α i = α C > ρ , then no ρ -line can be down-black (in particular, no shift of to the left can be down-black), and if α i = ρ = 0, then is the horizontal axis and almost all points on are interior, having no neighbours (i.e.neighbour integer points in the first quadrant) outside area(( , 0), ), since ρ = slope( ) > 0; • for each C ∈ C (where α C = α i ) we have: if α C > α i , hence α C = α j for some j > i, then only finitely many points of area(( , 0), ) are black in C (since α C > ρ = slope( ), and thus cannot be down-black in C); if α C < α i , hence α C = α j for some j < i, then only finitely many points of area(( , 0), ) are white in C (here α i > ρ = slope( ) > α C = α j , and we use that β C = α C = α j by the induction hypothesis of our proof of Lemma 5.5, which entails that is not up-white in C).
We will use the notion of the rightmost down-black α i -lines C for α i -colourings C, after which we again apply Proposition 5.7.But, in fact, we can not a priori assume that the rightmost down-black α i -lines exist and that α i is rational, so we partition and analyse the class C of α i -colourings without any such assumptions.We define the line C as the "right-hand limit" for the respective down-black lines, if some α i -lines that are down-black in C exist, which does not exclude the case that C itself is not down-black.
Partition of α i -colourings by rightmost down-black lines C .Let C be the set of all α i -colourings.For technical convenience we use a fixed α i -line R guaranteed by Claim 5.11 (or Corollary 5.12), to define the partition of C into the following classes: • C 1 contains each C ∈ C for which there is an α i -line that is down-black in C; for each C ∈ C 1 we define the rightmost down-black line C as the α i -line whose distance from R is the infimum of distances of the α i -lines that are down-black in C. We further partition C 1 as follows: -C 11 contains all C ∈ C 1 for which C is down-black; -C 12 contains all C ∈ C 1 for which C is not down-black.• C 2 = C C 1 ; hence for each C ∈ C 2 and each α i -line only finitely many points below are black in C.
We now show that there is a desired "left-bound-line" for all C ∈ C 1 , by Claim 5.14 that is preceded by a simple auxiliary fact in Claim 5.13; it turns out that C 12 is empty, while we still do not exclude that C 11 is empty as well.Then we show that C 2 is empty (by Claim 5.15); hence the (nonempty) set C of α i -colourings is equal to C 11 , and Claim 5.14 thus finishes the proof of Lemma 5.5 (including the fact that α i is rational or ∞).
(1) If γ ∈ R ≥0 is rational, then for any γ-line we have: (a) either contains infinitely many integer points (in Z × Z), or no integer point; (b) there exists the least value µ ∈ R >0 such that µ = dist(p, ) for some integer point p that is not on .(2) If γ ∈ R >0 is irrational, then for each B ∈ R >0 there is a (tiny) real number µ > 0 such that for any two γ-lines , whose (Euclidean) distance is µ we have that the distance of any two different integer points lying between the lines and is larger than B.
Proof. 1.For γ = 0 both parts of the claim are obvious; so we assume that γ = ∆y ∆x for ∆ x , ∆ y ∈ N >0 , and consider a γ-line .We note that if an integer point (z 1 , z 2 ) ∈ Z × Z is on , then the integer points (z 1 + i • ∆ x , z 2 + i • ∆ y ), for all i ∈ Z, are on as well.Hence 1(a) is clear.Now we note that for any integer point p = (z 1 , z 2 ) ∈ Z × Z outside we have that dist(p, ) = dist(p , ) for all (integer) points p = (z which obviously exists (since it exists for each fixed z 2 ∈ {0, 1, . . ., ∆ y }).
2. We fix an irrational γ ∈ R >0 and some B ∈ R >0 .We note that for each vector v = (p, p ) where p, p are two different integer points and dist(p, p ) ≤ B (by dist(p, p ) we refer to the Euclidean distance of points) we have that the absolute value of the co-γ-d-size of v is greater than zero (since slope(v) is rational and thus differs from γ).Moreover, the set M γ,B = µ µ is the absolute value of the co-γ-d-size of some (p, p ) where p, p are two different integer points and dist(p, p ) ≤ B is obviously finite.We can thus choose µ satisfying 0 < µ < min M γ,B ; such µ satisfies the claim.Indeed: if there were two γ-lines , whose (Euclidean) distance is µ and two different integer points p, p lying between and and satisfying dist(p, p ) ≤ B, then the absolute value of the co-γ-d-size of the vector (p, p ) would belong to M γ,B and would be not bigger than µ -a contradiction.
Claim 5.14.If C 1 = ∅, then there is an α i -line L such that all points above L are black in all colourings from C 1 ; moreover, α i is rational and C 12 is empty.(We recall that ∞ > α i , as stipulated after Corollary 5.12.)Proof.We fix an α i -line R so that all points in border ( area(← R ) ) are white in each C ∈ C = C 1 ∪ C 2 (such R exists by Corollary 5.12).We assume that C 1 = ∅, and we fix an α i -line sufficiently above all rightmost down-black lines C , C ∈ C 1 (defined after Corollary 5.12), so that area(( , 0), C ) has infinitely many interior points for each C ∈ C 1 .(See Figure 16.) Finally we choose (a sufficiently large) b ∈ R ≥0 so that area(( , b), R ) does not intersect with the vertical axis and is monochromatic in each colouring C ∈ C 1 ; we can easily verify that such b exists since we have: • for each (α i -colouring) C ∈ C 2 and for each colouring C for which α C > α i there are only finitely many points below the α i -line that are black in C (and we choose b so that these colourings are all-white in area(( , b), R )); • for each colouring C for which α C < α i , hence α C = α j for some j ∈ {0, 1, . . ., i−1}, we have α C = β C by the induction hypothesis of our proof of Lemma 5.5, and there are thus only finitely many points above the α i -line R that are white in C (and we choose b so that these colourings are all-black in area(( , b), R )). Figure 16 depicts area(( , b), R ), and C just for one C ∈ C 1 ; let us ignore the additional dark-grey belt and the line C,µ for this moment.Let us now assume that α i (the slope of , C , R ) is rational.By using Claim 5.13(1) we easily deduce that for each C ∈ C 1 the rightmost down-black line C must be down-black; moreover, it contains infinitely many points that are black in C, while only finitely many points strictly below C can be black in C. (If C was not down-black then also a line arising by a tiny shift of C to the left, by less than µ in Claim 5.13(1b), would be not down-black; this would contradict the definition of C .On the other hand, if there were infinitely many points strictly below C that are black in C, then a tiny shift of C to the right would give us a down-black line, which also contradicts the definition of C .) Hence if α i is rational, then C 12 is empty, and thus C 1 = C 11 .Moreover, we can simply increase b so that for each C ∈ C 1 all the infinitely many points in area(( C , b), R ) that are black in C are on the line C .We can thus apply Proposition 5.7 (depicted in Figure 13) to C 1 , , R , b; for each C ∈ C 1 , the point p C can be chosen as one of the infinitely many points on C that are black in C and interior in area(( , b), R ) (hence not close to ⊥ b ).Proposition 5.7 gives us the claimed L .
To finish the proof, we now assume that α i is irrational (which entails that any α i -line contains at most one integer point), and we will lead this assumption to a contradiction with the assumption C 1 = ∅.Recalling Figure 16, we aim to increase b so that not only Proposition 5.7, but even Corollary 5.9, could be applied; this will entail that α C > β C for some C ∈ C 1 , which is the desired contradiction.rank of its white start-point start(v), where start(v) is necessarily an interior point of area( , (b, )), leads to a contradiction by Proposition 5.6 applied to the opposite black-white vector (which would yield a violating (B 0 , µ 0 )-vector with a smaller rank of its start-point).
This entails that the slope of (B 0 , µ 0 )-vectors, which is larger than α i , is up-white in each C ∈ C 2 (consider the line containing white-white vectors v 1 , v 2 , v 3 , . . . in Figure 18b); this is a contradiction since β C = α i for all C ∈ C 2 (by Claim 5.10).Hence C 2 = ∅ after all.5.2.Quantitative belt theorem.We have proven the belt theorem (Theorem 5.2), but we have not derived anything specific about the slopes, widths, and positions of the belts (one of which is depicted in Figure 10).Now we show that these parameters can be presented by polynomially bounded integers (in the size of the underlying one-counter net N = (Q, Act, δ) that we have fixed).
We start with the belt-slopes, which is another name for the down-limits α C (that coincide with the up-limits β C by Lemma 5.5); we recall that we have ordered the set {α C | C = C p,q , (p, q) ∈ Q × Q}, denoting its elements (in (5.1)) as (5.3) Hence when we refer to a belt-slope α, we understand that α = α i for some i ∈ [1, k].We can also recall that by an α-colouring we mean a colouring C (C = C p,q for some (p, q) ∈ Q × Q) for which α C = α.Before clarifying that the belt-slopes are fractions of small integers (or ∞), which is shown by Proposition 5.17, we formulate a useful corollary of Proposition 5.6, which captures the black-white vector travel discussed around Figure 4 in Introduction.
(The black-white vectors in Figure 4 are directed upwards but the corollary also handles the vectors directed downwards, like v i 1 in Figure 19.) Corollary 5.16 (of Proposition 5.6).Any vector v 0 that is black-white in some C 0 gives rise to a sequence v 0 , v 1 , . . ., v n where start(v n ) ∈ v-axis or end(v n ) ∈ h-axis, and for i = 0, 1, 2, . . ., n−1 we have that v i+1 is a neighbour vector of v i that is black-white in some C i+1 and the rank of the white end of v i+1 in C i+1 is smaller than the rank of the white end of v i in C i .
Proof.Let C be the set of α-colourings, for some fixed belt-slope α where ∞ > α > 0, and let L , R be some α-lines such that all points above L are black in all C ∈ C and all points below R are white in all C ∈ C; the belt theorem (Theorem 5.2) guarantees that such lines exist, and that α is rational.
By reasoning as in the proof of Claim 5.14 (when assuming that α i is rational) we derive that for each C ∈ C there is the rightmost α-line C that is down-black in C; we also recall that there are infinitely many points on C that are black in C (but only finitely many points strictly below C are black in C).Analogously we derive that there is the leftmost α-line C that is up-white in C; there are infinitely many points on C that are white in C. (The colourings depicted in Figure 3, related to the simple net in Figure 2, might suggest that C is below C , but Figure 19  (where ms refers to "Maximal Span").Now we choose v 0 so that v 0 is black-white in some C 0 ∈ C ms , start(v 0 ) ∈ C 0 , and end(v 0 ) ∈ C 0 ; moreover, we choose v 0 sufficiently high in the respective α-belt so that each v i in the prefix v 0 , v 1 , v 2 , . . ., v |Cms| of the sequence guaranteed by Corollary 5.16 must necessarily satisfy that start(v i ) ∈ C i and end(v i ) ∈ C i for some C i ∈ C ms .Figure 19 depicts this "sufficiently high", where b is chosen so that By the pigeonhole principle, in the above discussed sequence v 0 , v 1 , v 2 , . . ., v |Cms| we have some i 1 , i 2 such that 0 ≤ i 1 < i 2 ≤ |C ms | and both start(v i 1 ) and start(v i 2 ) are on C for the same C ∈ C ms (and end(v i 1 ) and end(v i 2 ) are on C ).Since slope( C ) = α, and start(v i 1 ), start(v i 2 ) are two different points on C , we deduce that α is the slope of the vector (start(v i 1 ), start(v i 2 )).This entails that α = ∆y ∆x where ∆ x , ∆ y ∈ {1, 2, • • • , i 2 −i 1 }; ∆ x is the absolute value of the 0-d-size [horizontal d-size] of the vector (start(v i 1 ), start(v i 2 )), and ∆ y is the absolute value of its co-0-d-size [vertical d-size] (see Figure 19).
To deal with the belt-widths and the belt positions, it is useful to define two quasi-orders on the set of colourings {C | C = C p,q , (p, q) ∈ Q × Q}, based on the belt-slopes and the lines C L , C R defined below (which can differ from the lines L , R in Theorem 5.2 and C , C used in the proof of Proposition 5.17).For technical reasons we also extend the notion of border points to almost-border points of particular areas.

Lines C
L (leftmost with a white point) and C R (rightmost with a black point).For each colouring C we define C L as the leftmost α C -line that contains a point that is white in C, and C R as the rightmost α C -line that contains a point that is black in C; we put C L = C R = h-axis if C is all-black (in which case α C = 0), and C L = C R = v-axis if C is all-white (in which case α C = ∞).We note that Theorem 5.2 (depicted in Figure 10), which includes the fact that the belt-slopes α C are rational or ∞, entails that the lines C L and C R are well-defined.We again note that the simulation preorder depicted in Figure 3, related to our simple example in Figure 2, can give an impression that C L is typically below (to the right of) C R ; in this case (whenever C L is below C R ), the distance between C L and C R is clearly at most 1 (and the distance 1 is achieved in the colouring C p 2 ,p 1 in Figure 3).In more complicated examples, C L can be surely above (to the left of) C R ; in Figure 19, C L would be between L and C , and C R would be between C and R .In the case when C L is above C R the distance between C L and C R can be surely larger than 1; nevertheless it is polynomially bounded (in the size of Q of our fixed OCN N = (Q, Act, δ)), as we discuss below.
A counterclockwise quasi-order L and a clockwise quasi-order R on the set of colourings.The quasi-orders L and R on the set { C | C = C p,q , (p, q) ∈ Q × Q } are defined as follows: We can recall the order α 1 < α 2 • • • < α k of belt-slopes (5.3).Hence C L C iff α C = α i and α C = α j where i < j, or i = j and C L is below (to the right of) C L ; analogously we can express C R C .
Almost-border points of area(← ) and area( →).We say that p ∈ area(← ) is an almost-border point of area(← ) if p has a neighbour point in area( →) (hence p is a border point of area(← ) or has a neighbour point on ).Similarly, p ∈ area( →) is an almost-border point of area( →) if p has a neighbour point in area(← ).
The next proposition (Proposition 5.18) and its corollary (Corollary 5.19) show that the distances of lines C L and C R to the origin (0, 0) are "small", bounded by a polynomial in the number of colourings (hence in |Q × Q| for the underlying OCN N = (Q, Act, δ)); this also yields the announced polynomial bound on the distance between C L and C R (for any colouring C = C p,q ).
Figure 20 (the notation on which will be explained below) aims to illustrate that C R cannot intersect h-axis in a large distance to the right from the origin (0, 0).This follows by Proposition 5.18(1) that entails that if the intersection of some C R with h-axis is to the right of (0, 0), then the distance of this intersection to (0, 0) is bounded by the product of a small number with the cardinality of the set {C | C R C} (as follows by a repeated use of Proposition 5.18(1)).
Proposition 5.18.For each colouring C, and its respective lines C L and C R , we have: (1) If (0, 0) is strictly above C R , then there is C R C such that area(← C R ) contains a point that is an almost-border point of area(← C R ).

Figure 1 .
Figure1.A simple countdown game (with p win = p 2 ), and its solution.

Figure 2 .
Figure 2.An example of a succinct one-counter net.

Figure 3 .
Figure 3.A depiction of the simulation preorder related to the net from Figure 2.
A neighbour smaller-rank black-white vector.A vector travel.
and β 3 = S D (i−m+1).The two versions of the announced intermediate problem are defined as follows: Name: Seq (sequence problem) Instance: A sequence description D = (∆, D, m), n 0 ∈ N, and β 0 ∈ ∆ (with m and n 0 written in binary).Question: Is S D (n 0 ) = β 0 ?Name: Eseq (existential sequence problem) Instance: A sequence description D = (∆, D, m), and β 0 ∈ ∆ (with m written in binary).Question: Is there i ∈ N such that S D (i) = β 0 ?Our informal discussion around Figure 5 (including the remark on M w starting with the empty tape) makes (almost) clear that • Seq is EXPSPACE-complete, and • Eseq is EXPTIME-complete.Remark 3.3.In fact, we would get the same complexity results even if we restricted D in D = (∆, D, m) to D : ∆ 2 → ∆ and defined S D (i) = D(S D (i−m−1, S D (i−m)) for i > m. (We would simulate M by a Turing machine M that only moves to the right in each step, while working on a circular tape of length m.)But this technical enhancement would not simplify our construction of the countdown games N M w,m , in fact.Formally it suffices for us to claim just the lower bounds: Proposition 3.4.Eseq is EXPSPACE-hard and Seq is EXPTIME-hard.
and for i > m the symbol S(i) is determined by S(i−m−1), S(i−m), S(i−m+1) and the transition function of M w , independently of the actual value of i.The symbols in ∆ w and the function D w , guaranteeing S Dw = S, are thus obvious.

a
s,s − −− →= {(s, s)}, and a win − −− →= {(s, s) | s ∈ T }; hence each transition gets its unique action-label, and each target state gets a loop labelled by Eve's winning action a win .Let L be a copy of L with the state set V = {s | s ∈ V } but without the action a win .We thus have s a s,s − −− → s in L iff s a s,s − −− → s in L , and for s ∈ T we have s s in the disjoint union of L and L (since a win is enabled in s but not in s ).If V ∃ = V (in each state it is Eve who chooses the next transition), then we easily observe that s ∈ Win ∃ entails s s and s ∈ Win ∃ entails s ∼ s .(4.1) Figure 8 we write, e.g., s 1 3 instead of s 1 , s 3 , and s 1 23 instead of s 1 , {s 2 , s 3 } .)We put Act = {a c , a win } ∪ {a s,s | s − → s} and define a − → for a ∈ Act as follows.If s ∈ V ∃ and s − → s, then s a s,s − −− → s and s a s,s − −− → s (in Figure 7 we write, e.g., a 1 3 instead of a s 1 ,s 3 ).If s ∈ V ∀ and X = {s | s − → s} = ∅, then: a) s ac − → s, X , and s ac − → s, s , s ac − → s, s for all s ∈ X (cf. Figure 8 where s = s 1 and X = {s 2 , s 3 }; a c is a "choice-action"); b) for each s ∈ X we have s, X a s,s − −− → s and s, s a s,s − −− → s ; moreover, for each s ∈ X {s} we have s, s a s,s − −− → s (e.g., in Figure 8 we thus have s s 3 ).For each s ∈ T we have s a win − −− → s (for special a win that is not enabled in s ).Lemma 4.2.For an r-game G = (V, V ∃ , − →, T ), and the LTS L(G) = (S, Act, ( a − →) a∈Act ), the following conditions hold for every s ∈ V and every relation ρ satisfying ∼ ⊆ ρ ⊆ : a) if s ∈ Win ∃ (in G), then s s (in L(G)) and thus (s, s ) ∈ ρ; b) if s ∈ Win ∃ , then s ∼ s and thus (s, s ) ∈ ρ.
a) The d-size of a vector.The d-sizes related to γ ∈ 0, ∞ .

Figure 12
Figure12 -to-line distance, perpendicular lines ⊥ b .For an integer point p (in Z × Z) and a line , by dist(p, ) we mean the standard (Euclidean) distance of p to .Given a line where slope( ) ∈ 0, ∞ and a level (which might be also called a bottom-level ) b ∈ R ≥0 , by ⊥ b we denote the line that is perpendicular to , intersects the horizontal and/or vertical axis in the first quadrant, and dist((0, 0), ⊥ b ) = b.(Hence ( ) ⊥ b = ⊥ b for any that is parallel with .See Figure 12a.) consider the sets area(( , b), ) where ( ) ⊥ b coincides with , and the sets area( , (b, )) where ( ) ⊥ b coincides with .Special cases of line-level-line areas: area(← ) and area( →).Given a line , with slope( ) ∈ 0, ∞ , we put area(← ) = { p ∈ N × N | p is above }, and area( →) = { p ∈ N × N | p is below }. (Hence area(← ) = area((v-axis, 0), ) and area( →) = area( , (0, h-axis)).)Border points of area(( , b), ) along , along , and along ⊥ b .(See the dotted boundaries in Figure 12b.)Given a point p ∈ border(area(( , b), )), we say that p lies along if p ∈ border(area( →)), hence if p has a neighbour point (integer point in the first quadrant) strictly above ; p lies along if p ∈ border(area(← )).If p ∈ border(area(( , b), )) does not lie along nor along , then we say that p lies along ⊥ b .D-sizes of vectors related to γ ∈ 0, ∞ , γ-d-size and co-γ-d-size.(See Figure 11b.)For γ ∈ 0, ∞ , the γ-d-size of v and the co-γ-d-size of v are the following real numbers.If start(v) = end(v), then both γ-d-size and co-γ-d-size of v are 0. If start(v) = end(v), then we consider the γ-line going through start(v), and the line that is perpendicular to and goes through end (a)  the points in border(area(( , b), )) that lie along are white in C, B 0 in the proof yields ¯ .

Corollary 5. 9 (
Bounding the up-limits β C strictly).If C, , , b satisfy the conditions of Proposition 5.7 and, moreover, slope( ) > slope( ) and for each C ∈ C we can choose p C (which is an interior point of area(( , b), )) so that it has no neighbour point on and the inequality in 3(b) is strict (dist(p C , ) > bd C ), then slope( ) > β C for each C ∈ C.
Rotation is so slight that the grey area contains no integer points, except possibly on ⊥ b .

Figure 16 .
Figure16.The situation considered in the proof of Claim 5.14.

Figure 17 .
Figure 17.A closer look at C when α C is irrational.
Now we cannot assume that C 12 = ∅, but the colourings C ∈ C 12 (where C are not down-black) create no problem for our aim: We can increase b so that for each C ∈ C 12 there are no points in area(( C , b), R ) that are black in C. By definition of C , now for each point p in area(( , b), R ) that is black in some C ∈ C 12 (hence p is strictly above C ) there are infinitely many interior points p in area(( , b), R ) that are black in C and for which dist(p , C ) < dist(p, C ), hence dist(p, ) < dist(p , ). (Figure 17b aims to illustrate this, by depicting black points strictly above C that "converge" to C .) Hence for each C ∈ C 12 and each sufficiently large b there surely is some p C , C(p C ) = black, such that dist(p, ) < dist(p C , ) for all points p ∈ border(area(( , b), R )) that are black in C.

•
for each C ∈ C the points strictly above C in area(v-axis, (b, C )) are black in C and the points strictly below C in area(( C , b), h-axis) are white in C, • and each colouring C ∈ C (for which α C = α) is monochromatic in area(( L , b), R ).
C and C L is below (to the right of) C L ; C R C if either α C > α C , or α C = α C and CR is above (to the left of) C R .By C L C we denote that C L C and C L C; the case C R C is analogous.