Memoryless Determinacy of Infinite Parity Games: Another Simple Proof

03/01/2018 ∙ by Stephane Le Roux, et al. ∙ 0

The memoryless determinacy of infinite parity games was proven independently by Emerson and Jutla, and Mostowski, with various applications in computer science. Then Zielonka provided an elegant and simple argument. Several simpler proofs can be found in the literature in the case where the underlying graph is finite. These proofs usually proceed by induction on the number of relevant vertices. Recently, Haddad provided an even simpler argument by precisely defining what a relevant vertex is: one that has incoming edges and proper outgoing edges. The proof splits one relevant vertex into two non-relevant ones for the induction step, and concludes after a case disjunction. This note adapts Haddad's technique to the infinite case. A priority is relevant if it labels relevant vertices, and the proof proceeds by induction on the number of relevant priorities (as Zielonka). Haddad's vertex split is replaced with the splitting of many vertices, and the case disjunction with a transfinite induction over the winning regions (as Zielonka). The main difference with Zielonka is that here the notions of traps, attractors, etc., are not used. Memoryless determinacy was generalized by Grädel and Walukiewicz to infinite sets of priorities, but it is unclear whether Haddad's technique can be adapted to such a setting.

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1 Introduction

The memoryless determinacy of infinite parity games with finitely many priorities was proved independently by Emerson and Jutla [1], and Mostowski [4], with various applications in computer science. Then Zielonka [5] provided an elegant and simple argument for the same theorem.

Several simpler proofs can be found in the literature in the case where the underlying graph is finite. These proofs usually proceed by induction on the number of relevant vertices. Recently, Haddad [3] found a very simple argument by defining accurately what a relevant vertex is: one that has incoming edges and proper outgoing edges. His proof splits one relevant vertex into two non-relevant ones for the induction step, and concludes after a case disjunction.

Since Haddad [3] also proceeds by induction on the number of relevant vertices, it cannot be used verbatim to prove the determinacy of infinite games with finitely many priorities. However, this article adapts Haddad’s technique for these games by performing an induction on the number of relevant priorities, i.e. the priorities that label relevant vertices. For this purpose, Haddad’s single vertex split is replaced with the splitting of (in)finitely many vertices at once.

Zielonka [5] provided two proofs for the same theorem, a more constructive one and a shorter one, both highly relying on the notions of trap, attractor, etc. This article also provides a more constructive one and a shorter one. Both new proofs adapt Haddad’s technique for infinite games, in two fairly different ways but without traps or attractors. Both more constructive proofs, in [5] and in this article, nest a transfinite induction on the vertices within the induction on the finite number of priorities, and in turn the finite induction hypothesis is invoked within the transfinite induction. Unlike these, the shorter proof of this article avoids the second nesting and factors out a unique transfinite induction into Lemma 1 on prefix-independent winning conditions.

The same theorem was generalised by Grädel and Walukiewicz [2] for infinite sets of priorities, but the proofs from this article could not be used verbatim to prove this generalisation, since they proceed by induction on the number of relevant priorities.

Section 2 includes general definitions on games and Lemmas 1 and 2

about prefix-independent winning conditions. (These lemmas are probably folklore one way or another.) Section 

3 includes the definition of parity games and a shorter proof of memoryless determinacy of infinite parity games with finitely many priorities. Section 4 provides a more constructive proof.

2 Two-player win/lose games

A two-player win/lose game is a tuple , where (disjoint union), and satisfies , and , and , and .

The players Informally, there are two players in such a game: Player 0 controlling the vertices in , and Player 1 controlling the vertices in .

Runs A run in a game is an infinite path in the graph , i.e. a sequence such that for all . The run is winning for Player 0 (Player 1) if (), where . Also, finite paths in are called finite runs.

Strategies A Player 0 (Player 1) memoryless strategy is a function () such that () for all (). Only memoryless strategies are considered in this article, where they are often simply called strategies.

Compatible runs A run is compatible with a strategy () if whenever (). This notion is naturally extended to finite runs.

Winning strategies and regions A Player 0 strategy is said to win from some if all compatible runs such that are winning for Player 0. Let be the vertices from where wins, and let be the vertices from where some Player 0 strategy wins. The in stands for memoryless. A Player 0 strategy is said to be memoryless optimal (optimal for short) if . Likewise and are defined for Player 1. (It is straightforward to prove that .)

Prefix-independent winning condition is called prefix-independent if removing or adding prefixes to runs preserves membership to , i.e. .

Lemma 1 below is only used in Section 3. It states that for all games with prefix-independent winning condition, there exists a memoryless strategy that wins from all vertices from where the player can win without memory. It shares similarities with [5, Second proof, p150].

Lemma 1

Let be a prefix-independent winning condition. For all games using colors and winning condition , there exists a Player 0 memoryless strategy such that . (And likewise for Player 1.) Let be some ordinal of cardinality , and let be an enumeration of . For all let be a Player 0 strategy that is winning from . Let be such that is the least with , which is well-defined since every non-empty subset of an ordinal has a least element. Let be a Player 0 strategy satisfying for all .

Let a run start in and be compatible with . If is also compatible with , it makes Player 0 win, by definition of . Otherwise let be the least such that and , so by definition of . To apply the argument recursively, note that since is compatible with and by prefix independence. By property of the ordinals this situation can occur only finitely many times, so the tail of is eventually compatible with some from some on. So by prefix-independence, makes Player 0 win.

Uniform memoryless determinacy A game is uniformly memoryless determined if there exist Player 0 and Player 1 memoryless strategies and , respectively, such that .

The remainder of Section 2 is only used in Section 4.

Unfair-win vertex For all games a vertex (resp. ) is called an unfair win if it has a self-loop and proper outgoing edges, and (resp. ).

Lemma 2 below states that when trying to prove memoryless determinacy of prefix-independent winning conditions, it suffices to consider games void of unfair-win vertices.

Lemma 2

Fix and a prefix-independent . If the games void of unfair-win vertices are uniformly memoryless determined, so are all games, i.e. without the unfair-win restriction. Let us transform an arbitrary game into by removing the proper outgoing edges of the unfair-win vertices, thus making them absorbing, as in Figure 1. Let be a Player 0 memoryless strategy in , so is also a strategy in . Let a run start in and be compatible with in . Let us show that is also compatible with in . Towards a contradiction, let be the first edge of that is not present in , so is a unfair-win vertex in , and (as Player 0 just follows ), so

is odd. It implies that the prefix

, which is compatible with in by choice of , leads in to an absorbing state with odd priority, contradicting (by prefix independence) the assumption that . Therefore is also compatible with in and thus makes Player 0 win, which shows that for all . By symmetry for all Player 1 strategy . By assumption let and be Player 0 and Player 1 memoryless strategies, respectively, such that . So by the above inclusions.

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Figure 1: From (left-hand side), removing edges from unfair-win vertices if makes Player win (middle) or useless self-loops if (right-hand side).

3 A shorter proof

Parity games A parity game with finitely many priorities is a two-player win/lose game where , the function is bounded, and the prefix-independent is defined as follows. For all runs let , which is well-defined since is bounded. The run is winning for Player 0 (Player 1) if is even (odd). So a parity game amounts to a tuple where is bounded.

Absorbing, vanishing, and relevant vertices [3] A vertex is absorbing if it has no proper outgoing edge, i.e. ; it is vanishing if , i.e. it has no incoming edge; it is relevant if it is neither absorbing nor vanishing.

Relevant priorities Let be the set of the relevant vertices of a parity game. Then is the set of the relevant priorities.

Theorem 3 below is proved without interleaving the inductions.

Theorem 3 ([5])

The parity games are uniformly memoryless determined. Let us prove the claim by induction on the number of relevant priorities. Base case: if there are no relevant priorities in a game, there is no relevant vertices either, so all vertices are either absorbing or vanishing. In this case every run visits at most two vertices, and it is straightforward to prove the claim by backward induction.

For the inductive case, let us assume that the claim holds for all games with relevant priorities less than some , and let us proceed in two steps. The first step will show that for all games with relevant priorities at most , and the second step will invoke Lemma 1 to complete the proof.

First step Let be the relevant vertices from with priority . Wlog let us assume that is even, i.e. up to adding to every priority and swapping and .

Let us modify to make the priority irrelevant and to be able to invoke the induction hypothesis. More specifically, let us derive from by splitting each vertex into one vanishing split vertex that keeps the proper outgoing edges and the same controller, and one absorbing split vertex that keeps the incoming edges and receives a self-loop. (The controller of and the priority of are irrelevant, so, e.g., both split vertices preserve them.) This is depicted in Figure 2.

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Figure 2: Splitting a vertex controlled by Player and with priority : from (left-hand side) to (right-hand side). The dashed arrow means that a self-loop may or may not be present.

Formally, for all let , where and . Also let and . Let be defined by and for all , and let .

The new game has one less relevant priority than , namely , so by induction hypothesis there exist Player 0 and Player 1 memoryless strategies and , respectively, such that . Either or will induce a winning strategy in , which will complete the first step.

Before letting the players use or and play in , the strategies need modifying, as their domains and especially codomains are different in and (i.e. vs ). To prepare the modification, let be defined by for all , and for all . Now, the function is a Player 0 strategy in , and is a Player 1 strategy in . The effect of is depicted in Figure 3, where the double lines represent partial strategies towards and , respectively.

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Figure 3: Merging back vertices, from (left-hand side) to (right-hand side)

An important property of is that a (finite) run is compatible with (resp. ) in iff it is compatible with (resp. ) in . Indeed,

(1)

To prove the forthcoming Inequality (2) let us make a case disjunction. First case, . Let be a run compatible with in . If sees priority infinitely often, it makes Player 0 win. If sees priority finitely many times only, visits only finitely often, so from some point on, is compatible with , by Equivalence (1). By prefix-independence and since is winning from everywhere, makes Player 0 win. This shows that wins in (from everywhere), thus implying Inequality (2).

Second case, , so let , and let be a run that starts at and that is compatible with in . Towards a contradiction let us assume that enters for the first time at time . It implies that the prefix is compatible with by Equivalence (1), and that . Subsequently, it implies that is compatible with , whereas it makes Player 0 win, contradiction, so does not enter . So is compatible with by Equivalence (1), thus making Player 1 win. This shows that wins in from , thus implying Inequality (2). Therefore, for all games with relevant priorities at most ,

(2)

Second step Let be a game with relevant priorities at most . By Lemma 1 let and be Player 0 and Player 1 memoryless strategies, respectively, such that and . Towards a contradiction, which will prove the claim, let us assume that . Let and let us argue two useful facts about (thus using one Zielonka’s trap without defining the concept).

  1. For all there exists such that . Otherwise the player controlling would have no other choice than entering , which would imply , a contradiction.

  2. If , there is in no edge from to . Otherwise Player could reach from , thus implying , a contradiction since .

Let the game be the restriction of to , i.e. , where and and , and . By Fact 1 above, is indeed a parity game. By Fact 2 above, for all , so by definition of

(3)

However, the game uses priorities at most , so Inequality (2) contradicts Equality (3) and the assumption made at the beginning of the second step.

Note that the first step of the proof of Theorem 3, which uses Haddad’s technique, could be replaced with something similar to the end of [5, First proof, p149] from Equation onwards.

4 A more constructive proof

Useless self-loops For all parity games an edge with (resp. ) is called useless if proper outgoing edges also start from , and is odd (resp. even).

Lemma 4

If the parity games where only absorbing states have self-loops are uniformly memoryless determined, so are all parity games. By Lemma 2 it suffices to show that the parity games void of unfair-win vertices are uniformly memoryless determined, since the parity condition is prefix independent. Note that a self-loop on a non-absorbing vertex is either useless or it makes the vertex unfair-win.

Let us first transform an arbitrary parity game into by removing the useless self-loops, as in Figure 1 (righthand side). Let be a Player 0 memoryless strategy in , so is also a strategy in . Let be a run starting in and that is compatible with in , which is less constrained than being compatible with in : the run amounts to the interleaving of some run that is compatible with on , which makes Player 0 win, with finitely or infinitely many uses of useless self-loops on vertices controlled by Player 1. These vertices have even priorities, so makes Player 0 win just like does, which shows that for all . By symmetry for all Player 1 strategy .

Let us now assume that is void of unfair-win vertices. So in only absorbing states have self-loops, so is uniformly memoryless determined by assumption. Let and be Player 0 and Player 1 memoryless strategies, respectively, such that . So by the above inclusions.

Theorem 5 ([5])

The parity games are uniformly memoryless determined. By Lemma 4 it suffices to prove the claim for games where only absorbing states have self-loops. Let us proceed by induction on the number of relevant priorities in . The beginning of the argument is the same as the proof of Theorem 3 until Equivalence 1. (The only difference is that there should not be dashed self-loops in the new Figures 2 and 3.)

Unfortunately, composing the optimal strategies and from with does not always yield optimal strategies for . Figure 4 exemplifies this with : going from to , as suggested by the double arrow, is a Player 0 winning strategy in but going from to is not winning in .

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Figure 4: (right-hand side) is derived from (left-hand side).

To solve the above issue, is further modified as shown in Figure 5, where the parity of is irrelevant.

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Figure 5: with (left-hand side) to (right-hand side)

Yet, changing some priorities to may create new issues elsewhere, which in turn require to change more priorities to . Transfinitely many modifications may even be required, as suggested in Figure 6. Note that only the priorities of may be modified: its structure remains the same, so runs and strategies that are valid in one of the modifications are also valid in the other ones.

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Figure 6: (lefthand side) induces which may need to be adjusted a transfinite number of times.

The transfinite modification of

is now formally defined by mutual induction with the number of relevant properties. Let us assume that for some ordinal number

there exist sequences (priority functions) and (Player 1 strategies in ) satisfying the following, where . For all

  1. A run that is winning for Player 1 in is also winning in , and .

  2. is optimal for Player 1 in .

  3. makes Player 1 win from in .

The intermediate goal is to define (and thus ) and , a Player 1 strategies in , and to show that the above four properties also hold for the extended sequences and .

Let , let for all , and let for all . Note that and . Let be a Player 1 strategy such that for all , where is the least such that . By I.H. on the relevant priorities, let be a Player 1 optimal strategy in , and let be a Player 1 strategy, also in , that coincides with on and with on . Let us now show that the above four properties also hold for the extended sequences and .

  1. For all and , if then and , where is even by assumption. It is then straightforward to show that a run that is winning for Player 1 in is also winning in , and subsequently that .

  2. Let and . Since , by definition of and we have , where is the least such that . This implies since by choice of , so by I.H (item  2). Evaluating this equation at yields , i.e. . This shows that .

  3. Let a run in start from and be compatible with , and let us make a case disjunction to show that is optimal in .

    • First case, . Let be any ordinal such that , and let us prove that is compatible with . If , then is compatible with , and ; if , the definition of implies that , where is the least such that . So since by I.H (item 2), so on the one hand is compatible with , and on the other hand : the membership holds since is optimal for Player 1 in by I.H. (item 3), and the inclusion holds by I.H. (item 1). Invoking these two facts recursively shows that is compatible with . So is winning for Player 1 in by I.H. (item 3), and also in by the above item 1.

    • Second case, enters at some point, so the first case applies to its tail and prefix independence implies that is winning for Player 1.

    • Third case, avoids , so it stays in and is compatible with , so it is also winning for Player 1.

  4. Let a run in start from and be compatible with , and let us make a case disjunction. If never visits (but possibly at the start), it is also compatible with by the equivalence (1), and it is winning for Player 1 since is optimal in by the above item 3. If visits for the first time at some time , then and coincide along until , also by the equivalence (1). Since is compatible with , also is compatible with . So (instead of ) since by assumption, and since is optimal for Player 1 in by the above item 3. By construction of this implies that for some . Since and coincide on by the above item 2, so do and (especially on ). Since wins from in by I.H. (item 4), since , and since only absorbing states have self-loops by assumption, a straightforward induction shows that follows (in addition to ) and remains in from on. So the corresponding tail of is winning for Player 1 by I.H. (item 4). By prefix independence is also winning.

By definition implies , so for cardinality reasons there exists a (least) ordinal such that . This implies , so

(4)

First witness strategy By item 4 above, makes Player 1 win from in .

Second witness strategy By I.H. (on the relevant priorities) let be such that . Let us argue that wins from in , which will prove the claim since .

Let a run start from and be compatible with , and let us make a case disjunction: if never sees the priority , it avoids and is also compatible with by the equivalence (1), and it is winning for Player 0 since ; if sees the priority infinitely often, it is winning for Player 0.

The remaining case is that sees the priority first at some time , and then only finitely many times. In particular, is not an absorbing vertex. Since it is not a vanishing vertex either, . By Equivalence (1), and coincide along since priority (i.e on a vertex in ) is first seen at . Since and is optimal for Player 0 in , (as opposed to ), so by Equivalence (4). By applying the argument iteratively, one shows that stays in , and after the last time sees , its tail is winning for Player 0 by the first case above.

References

  • [1] A. Emerson and C. Jutla. Tree automata, mu-calculus and determinacy. In 32nd IEEE Symp. on Foundations of Computer Science, pages 368–377, 1991.
  • [2] Erich Grädel and Igor Walukiewicz. Positional determinacy of games with infinitely many priorities. Logical Methods in Computer Science, 2006.
  • [3] Serge Haddad. Memoryless determinacy of finite parity games: Another simple proof. Information Processing Letters, 132:19 – 21, 2018.
  • [4] A. Mostowski. Games with forbidden positions. Research report 78, University of Gdansk, 1991.
  • [5] Wieslaw Zielonka. Infinite games on finitely coloured graphs with applications to automata on infinite trees. Theoretical Computer Science, 200(1):135 – 183, 1998.