1. Memoryless Determinacy of Parity Games
Itay Harel

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

3. 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

4. The main event

5. 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

6. Complexity Result

7. 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

8. 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?

9. 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

10. 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

11. 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

12. 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)!

13. Definitions and Lemmas

14. 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

15. 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

16. 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

17. 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

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

19. 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

20. 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…

21. 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

22. 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

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

24. 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

25. 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

26. 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 

27. 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!

28. 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

29. 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

30. 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

31. 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

32. Determinacy

33. 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 )

34. 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

35. 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)

36. 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

37. 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 and 1 paradises in 𝐺 𝑍

38. 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

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

40. 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…?

41. 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!

42. 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