Memoryless Determinacy of Parity Games Itay Harel

Table of Contents Quick recap Complexity results Definitions and lemmas sub-games 𝜎-traps Attractors 𝜎-paradise Determinacy 3 important lemmas non-constructive proof

Quick recap – parity games A – arena : 𝑉 0 , 𝑉 1 ,𝐸 𝐸⊆𝑉 𝑥 𝑉 𝜒 – coloring function: 𝜒:𝑉 ⟶𝐶, 𝐶 ⊆ ℕ Acc – winning condition for inf. In parity games: 0-player wins if max⁡(𝐼𝑛𝑓 𝜒 𝜋 is even Strategy: 𝑓 𝜎 : 𝑉 ∗ 𝑉 𝜎 ⟶𝑉, a partial function

The main event

Determinacy 𝜎 𝑤𝑖𝑛𝑠 𝜎 𝑤𝑖𝑛𝑠 M.W.S M.W.S Theorem: The set of vertices of a parity game is partitioned into a 0-paradise and a 1-paradise 𝜎 𝑤𝑖𝑛𝑠 M.W.S 𝜎 𝑤𝑖𝑛𝑠 M.W.S

Complexity Result

Complexity class of finite parity games 𝑊𝑖𝑛𝑠 = 𝐺,𝑣 | 𝐺 𝑖𝑠 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑝𝑎𝑟𝑖𝑡𝑦 𝑔𝑎𝑚𝑒 𝑎𝑛𝑑 𝑣 𝑖𝑠 𝑎 𝑤𝑖𝑛𝑛𝑖𝑛𝑔 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑝𝑙𝑎𝑦𝑒𝑟 0 Theorem: 𝑊𝑖𝑛𝑠 ∈𝑁𝑃 𝐶𝑂−𝑁𝑃

𝑊𝑖𝑛𝑠 ∈𝑁𝑃 A non-deterministic algorithm: Given G and v, guess a memoryless strategy w Check whether w is a M.W.S Why is it O.K. to only guess memoryless? How can we check a strategy quickly?

Creating 𝐺 𝑤 A memoryless strategy w can be represented using a sub-graph of G: 𝑍 3 𝑍 0 𝑍 1 𝑍 2 𝑍 4

Checking a strategy (1) Check whether there exists a vertex 𝑣′ such that: 𝑣′ is reachable from 𝑣 𝜒 𝑣 ′ is odd 𝑣′ lies on a cycle in 𝐺 𝑤 containing only vertices of priority less or equal 𝜒 𝑣 ′ Claim: w is a winning strategy iff 𝑣 ′ such a vertex doesn’t exist

Checking a strategy (2) Let’s assume such a vertex 𝑣′ exists Observation: In 𝐺 𝑤 , if a vertex is reachable from v, player-1 can force the token into it (formal proof with induction) Observation: In 𝐺 𝑤 , if the token is inside a cycle of vertices, player-1 can force the token to go over the entire cycle A winning play for player-1: Force the token into 𝑣′ Go over the cycle forever If 𝑣′ exists, w is not a M.W.S Other direction – very similar

𝑊𝑖𝑛𝑠 ∈𝑁𝑃 𝐶𝑂−𝑁𝑃 We saw that: 𝑊𝑖𝑛𝑠 ∈𝑁𝑃 𝑊𝑖𝑛𝑠 ∈𝑁𝑃 𝐶𝑂−𝑁𝑃 We saw that: 𝑊𝑖𝑛𝑠 ∈𝑁𝑃 For 𝑊𝑖𝑛𝑠 ∈𝐶𝑂−𝑁𝑃, we need to show that for: 𝑊𝑖𝑛𝑠 = 𝐺,𝑣 | 𝐺 𝑖𝑠 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑝𝑎𝑟𝑖𝑡𝑦 𝑔𝑎𝑚𝑒 𝑎𝑛𝑑 𝑣 𝑖𝑠 𝑎 𝑤𝑖𝑛𝑛𝑖𝑛𝑔 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑝𝑙𝑎𝑦𝑒𝑟 1 𝑊𝑖𝑛𝑠 ∈𝑁𝑃 Exactly the same (switch odd with even)!

Definitions and Lemmas

Sub-games Definition: Let U ⊆V be a subset of V. Denote: 𝐺 𝑈 = Α| 𝑈 , 𝜒| 𝑈 The graph G[U] is a sub-game of G if every dead-end in G[U] is also a dead-end in G

Sub-game example 𝑉= 𝑍 0 , 𝑍 1 , 𝑍 2 , 𝑍 3 , 𝑍 4 , 𝑍 5 𝑉= 𝑍 0 , 𝑍 1 , 𝑍 2 , 𝑍 3 , 𝑍 4 , 𝑍 5 𝑍 3 is the only dead-end Let’s look at 𝑈= 𝑍 0 , 𝑍 1 , 𝑍 2 , 𝑍 3 G[U] is a sub-game of G Let’s look at 𝑊= 𝑍 0 , 𝑍 1 , 𝑍 2 , 𝑍 4 G[W] not a sub-game 𝑍 0 𝑍 1 𝑍 4 𝑍 2 𝑍 5 𝑍 3

Sub-games lemma Claim: Let 𝑈 and 𝑈′ be subsets of V s.t. 𝑈 ′ ⊆𝑈⊆𝑉 If 𝐺[𝑈] is a sub-game of 𝐺 and 𝐺 𝑈 [ 𝑈 ′ ] is a sub-game of 𝐺[𝑈] then 𝐺[𝑈′] is a sub-game of G Proof: Notice that 𝐺 𝑈 𝑈 ′ =𝐺[ 𝑈 ′ ] If 𝑣∈ 𝑈 ′ is a dead-end in 𝐺 𝑈 ′ , it is a dead-end in 𝐺 𝑈 𝑈 ′ Since 𝐺 𝑈 𝑈 ′ is a sub-game of 𝐺 𝑈 , v is a dead-end in 𝐺 𝑈 Using the same argument, v is a dead-end in 𝐺

A set 𝑈⊆𝑉 will be called a 𝜎−𝑡𝑟𝑎𝑝 if: 𝜎−𝑡𝑟𝑎𝑝𝑠 In English: 𝜎 can’t force the token out of U, 𝜎 can always stay inside U. 𝜎 is ‘trapped’. Definition: A set 𝑈⊆𝑉 will be called a 𝜎−𝑡𝑟𝑎𝑝 if: ∀ 𝑣∈ 𝑉 𝜎 ⋂𝑈 : 𝑣𝐸 ⊆𝑈 ∀ 𝑣∈ 𝑉 𝜎 ⋂𝑈 : 𝑣𝐸⋂𝑈 ≠ 𝜙

𝜎−𝑡𝑟𝑎𝑝𝑠 example 1−𝑡𝑟𝑎𝑝: 𝑣 0 , 𝑣 7 0−𝑡𝑟𝑎𝑝 𝑣 0 , 𝑣 1 , 𝑣 2 , 𝑣 3 , 𝑣 7

𝜎−𝑡𝑟𝑎𝑝𝑠 lemmas (1) Claim 1: For every 𝜎 – trap U in G , 𝐺[𝑈] is a sub-game of 𝐺 Proof: Let 𝑣∈𝑈 be a dead-end in 𝐺 𝑈 . If 𝑣∈ 𝑉 𝜎 ⋂𝑈 then 𝑣𝐸 ⊆𝑈 , which means v was a dead-end in 𝐺. If 𝑣∈ 𝑉 𝜎 ⋂𝑈 then 𝑣𝐸⋂𝑈 ≠ 𝜙, which means it can’t be a dead-end in 𝐺 𝑈

𝜎−𝑡𝑟𝑎𝑝𝑠 lemmas (2) Claim 2: For every family 𝑈 𝑖 𝑖𝜖𝐼 of 𝜎 – traps the union 𝑈 𝑖 is a σ-trap as well. Proof: Trivial…

𝜎−𝑡𝑟𝑎𝑝𝑠 lemmas (3) Claim 3: Let X be a σ-trap in G and Y is a subset of X Y is a σ-trap in G iff Y is a σ-trap in G[X] Proof : Doodle time

Said strategy can be memoryless Attractor Sets Mathematical definition was given in chapter 2 For a game G and a set 𝑋⊆𝑉, denote 𝑨𝒕𝒕 𝒓 𝝈 (𝑮,𝑿) as: The set of vertices from which Player σ has a strategy to attract the token to X or a dead-end in 𝑽 𝝈 in a finite (possibly 0) number of steps Claim: Said strategy can be memoryless

Attractor Sets example 𝑍 0 𝑍 1 𝑍 5 𝑍 2 𝑍 4 𝑍 3 𝑍 6 𝑍 7

Attractor sets lemmas (1) Claim 1: The set 𝑉\ Attr 𝜎 (𝐺,𝑋) is a σ-trap in G. Proof: Let us look at 𝑣∈𝑉\ Attr 𝜎 (𝐺,𝑋) If 𝑣∈ 𝑉 𝜎 and 𝑣𝐸 ∩Attr 𝜎 𝐺,𝑋 ≠𝜙 then: There is a move σ can do to take the token to some 𝑢∈𝐴𝑡𝑡𝑟 𝐺,𝑋 From 𝑢, there is a σ-strategy that reaches a member of X in a finite set of moves This means that there is a σ-strategy that reaches X from 𝑣 in a finite set of moves 𝑣∈𝐴𝑡𝑡𝑟 𝐺,𝑋 in contradiction If 𝒗∈ 𝑽 𝝈 then 𝒗𝑬 ⊆𝑽\ 𝐀𝐭𝐭𝐫 𝝈 𝑮,𝑿

If 𝒗∈ 𝑽 𝝈 then 𝒗𝑬 ∩(𝐕\𝑨𝒕𝒕𝒓 𝝈 𝑮,𝑿 )≠𝝓 Attractor sets lemmas (2) Claim 1: The set 𝑉\ Attr 𝜎 (𝐺,𝑋) is a σ-trap in G. Proof (cont.): If 𝑣∈ 𝑉 𝜎 and 𝑣𝐸 ∩(V\Attr 𝜎 𝐺,𝑋 )=𝜙 then: Notice that v can’t be a dead-end. 𝑣𝐸 ∩Attr 𝜎 𝐺,𝑋 ≠𝜙 𝜎 must move the token to some 𝑢∈𝐴𝑡𝑡𝑟 𝐺,𝑋 From 𝑢, there is a σ-strategy that reaches a member of X in a finite set of moves This means that there is a σ-strategy that reaches X from 𝑣 in a finite set of moves 𝑣∈𝐴𝑡𝑡𝑟 𝐺,𝑋 in contradiction If 𝒗∈ 𝑽 𝝈 then 𝒗𝑬 ∩(𝐕\𝑨𝒕𝒕𝒓 𝝈 𝑮,𝑿 )≠𝝓

Attractor sets lemmas (3) Claim 2: If X is a σ-trap in G, then so is 𝐴𝑡𝑡 𝑟 𝜎 𝐺,𝑋 Proof: Trivial… Do a doodle proof! Claim 3: X is a σ-trap in G iff 𝐴𝑡𝑡 𝑟 𝜎 𝐺,𝑉\X =𝑉\X Another doodle proof 

𝜙 is a σ-trap so there is at least one σ-trap. Attractor sets lemmas (4) Claim 4: 𝐴𝑡𝑡 𝑟 𝜎 𝐺,𝑋 =𝑉 \ U where U is the greatest (w.r.t. set inclusion) σ-trap contained in 𝑉 \ 𝑋 Proof: Is U well defined? 𝜙 is a σ-trap so there is at least one σ-trap. Further more, union of σ-traps is a σ-trap, which means U is well defined! Now… doodle away!

𝜎−𝑃𝑎𝑟𝑎𝑑𝑖𝑠𝑒 In English: A region from which: 𝜎 cannot escape σ wins from all vertices of this region using a memoryless strategy Definition: A set 𝑈⊆𝑉 will be called a 𝜎−𝑃𝑎𝑟𝑎𝑑𝑖𝑠𝑒 if: U is a 𝜎 −𝑡𝑟𝑎𝑝 𝑓 𝜎 𝑖𝑠 𝑎 𝑀.𝑊.𝑆 𝑓𝑜𝑟 𝜎 𝑜𝑛 𝑈

𝜎−𝑃𝑎𝑟𝑎𝑑𝑖𝑠𝑒 𝑙𝑒𝑚𝑚𝑎𝑠 (1) Claim 1: If U is a σ-paradise, then so is 𝐴𝑡𝑡 𝑟 𝜎 𝐺, 𝑈 Proof: U is a 𝜎 −𝑡𝑟𝑎𝑝, which means that 𝐴𝑡𝑡 𝑟 𝜎 𝐺,𝑈 is also a a 𝜎 −𝑡𝑟𝑎𝑝 Also, we know that player 𝜎 has a memoryless strategy from 𝐴𝑡𝑡 𝑟 𝜎 𝐺,𝑈 that either: Brings the token to U Brings the token to a dead-end for 𝜎 Combine said strategy with 𝒇 𝑼 and you get a M.W.S for 𝐴𝑡𝑡 𝑟 𝜎 𝐺,𝑈

𝜎−𝑃𝑎𝑟𝑎𝑑𝑖𝑠𝑒 𝑙𝑒𝑚𝑚𝑎𝑠 (2) Claim 2: For every family 𝑈 𝑖 𝑖𝜖𝐼 of 𝜎 – paradises the union 𝑈 𝑖 is a σ-paradise as well Proof: A union of 𝜎 −𝑡𝑟𝑎𝑝𝑠 is a 𝜎 −𝑡𝑟𝑎𝑝, which means that U is a 𝜎 −𝑡𝑟𝑎𝑝 We will find a M.W.S using Rom’s trick! Denote 𝑤 𝑖 ≔ the M.W.S on 𝑈 𝑖 for player 𝜎 Define a well-ordering relation < on 𝐼. Define the strategy: ∀ 𝑣∈𝑈 ∩ 𝑉 𝜎 𝑠𝑒𝑡 𝑤 𝑣 = 𝑤 𝑖 (𝑣) where i is the least element of I s.t. 𝑣∈ 𝑈 𝑖 It’s easy to see that using said construction, a play p is either finite and ends with a dead-end to 𝝈 , or infinite and its suffix conforms with some 𝒘 𝒊

Another quick recap! Sub-game: a sub graph with no new dead-ends 𝜎−𝑡𝑟𝑎𝑝: a sub set of the arena from which: 𝜎 can’t leave 𝜎 can always stay 𝑉 0 , 𝑉 1 ,𝐸 𝐸⊆𝑉 𝑥 𝑉 𝐴𝑡𝑡 𝑟 𝜎 (𝐺,𝑋) – all vertices from which 𝜎 can: Force 𝜎 to a dead-end Force the game into X 𝜎− paradise: 𝜎 can’t leave 𝜎 has a M.W.S

Determinacy

Determinacy Theorem: The set of vertices of a parity game is partitioned into a 0-paradise and a 1-paradise We will prove this using an induction over 𝑛≡max⁡(𝐼𝑚 𝜒 )

Base case: n = 0 Lemma 0: If the maximum parity of G is 0, then V is partitioned into a 0 and a 1-paradise Proof: Observation: player 1 can only win by taking player 0 to a dead-end The winning region of player 1 is 𝐴𝑡𝑡 𝑟 1 (𝐺,𝜙) From above lemmas, 𝐴𝑡𝑡 𝑟 1 (𝐺,𝜙) is a 1-paradise (why?) Let’s look at 𝑉\ Attr 1 𝐺,𝜙 From above lemmas, we know that it is a 1-trap. Also, since the maximum parity is 0, and there are no dead-ends for 0 in 𝑉\ Attr 1 𝐺,𝜙 , 0 always has a winning strategy! 𝑉\ Attr 1 𝐺,𝜙 is a 0-paradise and 𝐴𝑡𝑡 𝑟 1 (𝐺,𝜙) is a 1-paradise

The construction (1) Induction step: assume the theorem holds for every parity game with maximum parity less than n Let us mark 𝜎 ≡𝑛 𝑚𝑜𝑑 2 Let 𝑋 𝜎 be a 𝜎 −𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒 s.t. 𝑋 𝜎 ≔𝑉\ 𝑋 𝜎 𝑖𝑠 𝑎 𝜎 −𝑡𝑟𝑎𝑝 Define: 𝑁= 𝑣 ∈ 𝑋 𝜎 𝜒 𝑣 =𝑛} Define 𝑍= 𝑋 𝜎 \Attr(𝐺 𝑋 𝜎 ,𝑁)

The construction (2) A few observations: 𝑋 𝜎 is a trap ⟹ 𝐺 𝑋 𝜎 is a sub-game of 𝐺 Z is a σ-trap in 𝐺 𝑋 𝜎 since it is the complement of an Attr set From the above 𝐺 𝑋 𝜎 [𝑍] is a sub-game of 𝐺 𝑋 𝜎 From a previous lemma: 𝐺 𝑍 is a sub-game of 𝐺

The construction (3) A few more observations: 𝐼𝑚𝑎𝑔𝑒 𝜒 ​ 𝑍 ⊆ 0,1,2,…,𝑛−1 The induction hypothesis applies to 𝐺 𝑍 ! We can partition Z to 𝑍 0 and 𝑍 1 - 0 and 1 paradises in 𝐺 𝑍

Important lemma Lemma 1: 𝑇ℎ𝑒 𝑢𝑛𝑖𝑜𝑛 𝑋 𝜎 ∪ 𝑍 𝜎 𝑖𝑠 𝑎 𝜎 −𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒 Proof: 𝑇ℎ𝑒 𝑢𝑛𝑖𝑜𝑛 𝑋 𝜎 ∪ 𝑍 𝜎 𝑖𝑠 𝑎 𝜎 −𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒 Proof: Let’s look at the sketch

Last lemma before we win! 𝐼𝑓 𝑍 𝜎 =𝜙, 𝑡ℎ𝑒𝑛 𝑋 𝜎 𝑖𝑠 𝑎 𝜎−𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒 Proof: Let’s look at the sketch

So… what’s left? We need to find 𝑋 𝜎 , 𝑋 𝜎 such that: 𝑋 𝜎 is a 𝜎 −𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒 𝑋 𝜎 ( ≔𝑉\ 𝑋 𝜎 ) is a 𝜎 −𝑡𝑟𝑎𝑝 𝑍 𝜎 is the empty set Reminder: we saw that 𝑋 𝜎 ∪ 𝑍 𝜎 𝑖𝑠 𝑎 𝜎 −𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒 Maybe, if 𝑋 𝜎 is the maximal 𝜎 −𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒, we can finish…?

Creating 𝑋 𝜎 𝐷𝑒𝑓𝑖𝑛𝑒 𝑊 𝜎 𝑡𝑜 𝑏𝑒 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑎𝑙𝑙 𝜎 −𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒s 𝑋 𝜎 = 𝑊 ∈ 𝑊 𝜎 𝑊 : since paradises are close under union, 𝑋 𝜎 is the largest 𝜎 −𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒 We have seen before that the attractor set of a paradise is still a paradise 𝐴𝑡𝑡 𝑟 𝜎 𝐺, 𝑋 𝜎 𝑖𝑠 𝑎 𝜎 −𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒 Since 𝑋 𝜎 is the largest paradise: 𝑋 𝜎 = 𝐴𝑡𝑡 𝑟 𝜎 𝐺, 𝑋 𝜎 We have seen before that the complement of an attractor set is a trap: 𝑋 𝜎 =𝑉\ 𝑋 𝜎 =𝑉\𝐴𝑡𝑡 𝑟 𝜎 𝐺, 𝑋 𝜎 𝑖𝑠 𝑎 𝜎 −𝑡𝑟𝑎𝑝 Just like we wanted!

Proving determinacy Theorem: The set of vertices of a parity game is partitioned into a 0-paradise and a 1-paradise. Outline: Do an induction over 𝑛≡max⁡(𝐼𝑚 𝜒 ) We have proved the base case We shall define 𝑋 𝜎 as the union of all 𝜎 −𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒𝑠 and 𝑋 𝜎 as its complement Use lemma 1 to show that 𝑋 𝜎 is a 𝜎 −𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒 Use lemma 2 to show that 𝑋 𝜎 is a 𝜎−𝑝𝑎𝑟𝑎𝑑𝑖𝑠𝑒