1. Introduction to Infinite Games
Chapter 2 in “Automata, Logic and infinite games”
Jonathan Cederbaum

2. Infinite Games - Intro 2 Players
Game board is a (possibly infinite) directed Graph
We Define:
How the games is played
Who wins
Determinacy, Winning Strategies, Forgetful Strategies, Memoryless Strategies
Fundamental Results

3. Games Game: Arena + Winning Conditions
Arena: (V0, V1, E) (V = V0 U V1, E contained in VxV)
Successors of v in V:
vE := { v’ ∈ V | (v, v’) ∈ E }
May be infinite
2 Players: Player 0, Player 1
Notation: Player σ, Player σ̅ where σ ∈ {0, 1}
Note: (V, E) not necessarily bipartite

4. A Play – Informal Explanation
The token is placed on some Initial vertex v.
If v is a 0-Vertex, its Player 0’s turn.
Otherwise, it’s a 1-Vertex, and Player 1’s turn.
On each turn, the player moves the token to some successor v in vE.
Note: a player may play mutiple turns in concession.
The game goes on infinitely or until reaching a vertex without successors, a Dead End.
vE = ∅.

5. A Play – Formal Definition
A Prefix of a Play defined as expected

6. Arena - Example
Player 0
Player 1

7. Games & Winning Sets A Game G is a pair: (A, Win)
A is the arena
Win ⊆ Vω is the Winning Set.
Player 0 is the winner of a play π in the game G iff:
π is a finite play: π = v0v1…vr and vr is a 1-vertex where Player 1 can’t move anymore (Dead end).
π is an infinite play and π ∈ Win
Note: Player 1 wins if Player 0 doesn’t.
In an Initialized Game (G, vI) all plays start in vI

8. Winning Conditions How do we define the winning set (“Win”) ?
Use acceptance conditions of w-automata, to define a winning play.
Problem: the graph may be infinite, might not even have recurring states..
Solution: color each vertex (Finite set of colors),
Apply acceptance conditions on the color sequence induced by the play.

9. Winning Conditions - Cont
A Coloring Function: χ: V C
C is some finite set of “colors”
Given a play: π = v0v1v2…
Define the play’s coloring: χ(π)= χ (v0)χ(v1)χ(v2)…
C (the color set) is viewed as the state set of a finite w-automaton.
Acc is the acceptance condition for this automaton. (As defined in the previous lecture)
buchi, muller, Rabin, Street, etc.
Wχ (Acc) : Defines the winning set to be all plays π where χ(π) is accepted according to Acc.
note dependence on χ

10. Acceptance Conditions
Buchi: (acc=F⊂P(C))
π ∈ Wχ(Acc) iff Inf(χ(π)) ∩ F ≠ ∅
Muller: (acc=F⊂2C).
π ∈ Wχ(Acc) iff Inf(χ(π)) ∈ F
Max-Parity: C is some set of integers.
π ∈ Wχ(Acc) iff Max(Inf(χ(π))) is Even
Additional Acceptance Conditions:
Rabin, Rabin Chain Condition, Street etc.
A game is a regular game if its winning set equals Wχ(Acc) for some χ, and some Acc from above.
Notation: we write (A, χ, Acc) instead of (A,Wχ(Acc))
reminder: Inf

11. Arena - Example A = (V0, V1,E, χ) (colored arena) C = {1, 2, 3, 4}
- No dead ends
Choose Mullner accepting conditions given by:
F = {{1, 2}, {1, 2, 3, 4}}.
Example play 1:
π = z6z3z2z4z2z4z6z5(z2z4)ω
χ(π) = (12)ω
Inf(χ(π)) = {1, 2} ∈ F
So π is a winning play for player 0
π' = (z2z4z6z3)ω
χ(π) = (1223)ω
Inf(χ(π)) = {1, 2, 3} ∉ F
So π is a winning play for player 1
Player 0
Player 1
אין צמתים ללא מוצא
תנאי קבלה של מולנר

12. Fundamental Questions I
Given an initialized game can one player win regardless of what the other does?
In this case we say the game is determined.
We will formalize this question by introducing the notions of a strategy, and a winning strategy.
The answer is yes, every (regular) game is determined (this result will be proved in chapter 6).

13. Fundamental Questions II
Given a final graph, can we compute who the winner is? Can compute it efficiently?
This and more – in chapter 7, stay tuned.
Complexity of determining the winner depends on the type of the game (The winning condition).

14. Fundamental Questions III
What does the winning strategy look like?
A general strategy tells the player the next move to make given all previous moves .
When might we have a winning strategy that requires remembering a bounded number of moves - a Forgetful Strategy.
When can we decide the next move given just the current vertex - a Memoryless Strategy.
Answer:
regular games have Forgetful winning strategies.
Parity games have Memoryless winning strategies.

15. Strategies
A is an arena.
σ ∈ {0, 1}.
fσ : V*Vσ → V - a partial function.
A prefix of a play π = v0v1…vr conforms with fσ if for every i, where 0 ≤ i ≤ r-1 and vi ∈ Vσ , the function fσ is defined at v0…vi and: vi+1 = fσ(v0 …vi).
A Play conforms with fσ if all its prefixes conform with fσ .
fσ is a strategy for Player σ on U ⊆ V if it is defined for every prefix of a play which starts in U, conform with it, and doesn’t end in a Dead End of Player σ.
When U is a singleton {v} we say fσ is a strategy for Player σ on v.
fσ is a Winning Startegy for Player σ on U if all plays which conform to fσ and start in U are wins for Player σ.
- רישא של משחק פאי, היא קונפורמית עם (או עוקבת אחר ) f..
- f היא אסט' אם היא מוגדרת לכל משחק שקונפורמי איתה

16. Strategies - Cont
Player σ wins a game G on U ⊆ V if he has a winning strategy on U.
When (G, vI ) is an initialized game, we say Player σ wins it if he wins G on the singleton set {v}.
Given a game G, we define the winning region for Player σ, denoted Wσ(G), or Wσ for short, to be the set of all vertices v such that Player 0 wins (G, v).
Remark: For any game G, Player σ wins G on Wσ(G).
Since if Player σ wins on each group A ∈ T, then Player σ wins on U A.
לגבי הערה בסוף:
אפשר לזכור את כל המסלול מתחילת המשחק, ולפי הוק' ההתחלתי נדע איזה אסט' לבחור (נתון לנו אסט' לכל קוד')

17. Strategy - Example A = (V0, V1,E, χ) (colored arena) C = {1, 2, 3, 4}
Mullner accepting conditions given by:
F = {{1, 2}, {1, 2, 3, 4}}.
f1(yz0) = z0
f1(yz4) = z6 (or = z1)
f1(yz3) = z2 (or = z1)
is a winning strategy for
Player 1 on W1 = {z0, z1}.
-Player 0 must move z2 to z4
-so z2, z4 appear infinite times.
-player 1 should not always choose z4->z2
-Player 1 then should choose z4->z6 occasionally.
-Player 0 wins iff on z6 he alternates between z3 and z5.
Player 0
Player 1
במקרה השני, שחקן 0 חייב להזיז מz2 לz4
שחקן 1 חייב אינסוף לפעמיים להזיז מ z4 לz6
שחקן אפס חיייב לבחור איןסוף פעמיים גם z3 וגם z5 מz6

18. Strategy - Example z4 if π ∈ V∗ z2 z3 if π ∈ V∗ z5z2z4(z2z4)∗ z6
A = (V0, V1,E, χ) (colored arena)
C = {1, 2, 3, 4}
Mullner accepting conditions given by:
F = {{1, 2}, {1, 2, 3, 4}}.
To summarize, the following is a winning strategy for Player 0, on W0 = {z2, z3, z4, z5, z6}.
z4 if π ∈ V∗ z2
z3 if π ∈ V∗ z5z2z4(z2z4)∗ z6
z5 if π ∈ V ∗z3z2z4(z2z4)∗ z6
z3 if π ∈ (V \ {z3, z5})∗z6
Player 0
Player 1
Z3 אם הוא ביקר בz5 לפני Z6
Z5 הוא ביקר בZ3 לפני Z6

19. Reduction of Muller Games to Parity Games
Note: For every regular game (A, χ, Acc) there exists a Muller winning condition Acc’ such that (A, χ, Acc) and (A, χ, Acc’) have the same winning regions.
e.g for Buchi games define: F ‘:= {G ∈ 2C | G ∩ F = ∅}.
Result: For every Muller game G=(A, χ, F) there exists a (max) parity game G’=(A’, χ’, Acc’) and an injection r : V → V’ such that every (unique) play π in G starting in v ∈ V induces a (unique) play π’ in G’ starting in r(v), and Player σ wins π in G iff he wins π’ in G’.
this result explains why we focus on Parity Games in chapter 6 & 7
Remainder:
Muller: (acc=F⊂2C).
π ∈ Wχ(Acc) iff Inf(χ(π)) ∈ F
Max-Parity: C is some set of integers.
π ∈ Wχ(Acc) iff Max(Inf(χ(π))) is Even

20. The Construction - Vertices
V ‘ = Vx
בצד שמאל הגרף המקורי
בצד ימין הבניה
הצבעים בגרף המקורי: {a,b}

21. The Construction - Edges
Connect vertex (v, q) to vertex (v’, q’) if:
v’∈ vE
q’ = ϕ(v', q)
q’ is the “updated” q after seeing color c = χ(v’)
ϕ is the memory update function.
3 cases for updating q:
Case #1: (color c appears before ♮):
q = xcy ♮ z
q’ = x ♮ yzc
רעיון: כשאנחנו עוברים לקוד (v’, q’) אנחנו רוצים לזכור שראינו את הצבע c=X(v’) נעדכן את q’ בהתאם, זאת אומרת כשעוברים מ-(v, q) ל v’ אז q’ נקבע באופן יחיד בהתאם.
להדגים על הדוגמא
Case #2: (c appears after ♮):
q = x ♮ ycz
q’ = xy ♮ zc
Case #3: (c not in q):
(q = q)
q’ = qc

22. The Construction – Initial Vertex
r(v) := (v, ♮χ(v))

23. The Construction – Coloring Function
Given F, that defines the Muller condition in A:
If “ y ∈ F “
otherwise
F = {{b}}
y ∈ F - ז.א קב' הצבעים ב-y הם איבר ב-F.
נשים לב שלמחרוזת y הארוכה ביותר יהיה צבע גדול ביותר ללא תלות במקרים.
המקרים קובעים את הזוגיות.

24. Muller to Parity – Full Example
For every play π in A, there is a corresponding play π’ in A’ , for example:
π = z1z2z0z1z2ω
χ(π) = abaabω (winning for Player 0)
π’ = (z1, ♮a)(z2, ♮ab)(z0, ♮ba)(z1, b♮a)(z2, ♮ab)(z2, a♮b)ω
χ’(π’) = ω (winning for Player 0)
Player 0
Player 1
לכל משחק π ב-A מתאים משחק יחיד 'π ב-'A. נבחן אותו:
F = {{b}}

25. Correctness of Construction
Theorem: For every Muller game (A, χ, F) there exists a (max) parity game (A’, χ’, Acc’) and a function r : V → V’ such that for every v ∈ V, Player σ wins ((A, χ, F), v) if and only if Player σ wins ((A’, χ’, Acc’), r(v)).
π = v0v1… ∈ Vω – infinite play in A.
π’ – the corresponding play in A’
p1(π’ ) = π
p2(π’ ) = q0q ∈ ω where qi = xi ♮ yi is defined by: q0 := ♮ χ(v0) and qi+1 := ϕ (vi+1, qi) for i ∈ ω.
F := Inf(χ(π))
Argument 1: from some point only c ∈ F appear on the right of the ♮.
From some i ∈ ω colors not in F do not appear in π.
at some j > i we have seen again all colors in F.
at that point colors in F appear on the right side of the string.
from step j+1 only colors in F appear right of the ♮.
conclusion: |yi| ≤ |F| from some i ∈ ω
Argument 2: There are infinite times where all colors in F are right of the ♮.
since the left most color in F must appear at some point
conclusion: there are infinite times k where | yk | = |F| and
Thus the largest length of y that appears infinitely often is |F| and “y=F” –> the highest color that appears infinite times is:
2*|F| if y ∈ F (even)
2*|F| if not (odd)
Summary: π is winning for player 0 in A iff π’ is winning for him in A’. So Player 0 Wins (A, v) iff he wins (A’, r(v))
טענה 1:
אם צבע c2 מופיע מימין לc1 ב-q, זה אומר שראינו פעם אחרונה שראינו את c1 הייתה לפני הפעם האחרונה שראינו את c2.
משלב מזמן מסוים לא יופיעו יותר צבעים שאינם ב-F.
מצד שני כל הצבעים ב-F מופיעים איןסוף פעמים.

26. Determinacy
In all examples so far, the Winning regions for Player 0 and Player 1 partition V.
When a game has this property we say it is determined.
Theorem: Every parity game is determined.
proven in chapter 6.
Corollary: Every regular game is determined
(by the equivalence result proven earlier).

27. Forgetfull & Memoryless Strategies
Player 0
Player 1
A = (V0, V1,E, χ) (colored arena)
C = {1, 2, 3, 4}
Mullner accepting conditions given by:
F = {{1, 2}, {1, 2, 3, 4}}.
To win, if the token is placed in z6, Player 0 must alternate between choosing z3 and z5
Player 0 has to remember 1 bit. We saw that this is enough.
We showed a Forgetfull strategy for Player 0.
Player 1 just moves z0 to zo. This is a Memoryless strategy.
אסט' שכחניות וחסרות זיכרון

28. Forgetful Strategies - Definition
A Strategy fσ is said to be forgetful (finite memory) if there exists:
A finite set M
An element mI∈ M (the initial memory value)
A function δ: V × M → M (memory update function)
A function g: V × M → V (next move function)
such that when π = v0v1…vr is a prefix of a play, in the domain of fσ , and the sequence: m0,m1,...,mr is determined by m0=mI and mi+1= δ(vi+1,mi), then fσ (π)=g(vr , mr).
A forgetful strategy where |M|=1 is called Memoryless.
notation: fσ:Vσ → V

29. Memoryless & Forgetful Strategy Example
Player 1 memoryless Strategy:
M = {m}
g(z0,m) = z0
g(z3,m) = g(z4,m) = z2
is winning on {z0, z1}
A = (V0, V1,E, χ) (colored arena)
C = {1, 2, 3, 4}
Mullner accepting conditions given by:
F = {{1, 2}, {1, 2, 3, 4}}.
Player 0 forgetfull Strategy:
M = {3, 4}
mI = 3
Winning On {z2, z3, z4, z5, z6}.
Player 0
Player 1
האסט' שהגצנו למשחק הנ"ל הם למעשה forgetfull וmemoryless..

30. Forgetful Strategies – Cont.
Player σ wins a game G forgetful when he has a (maybe different) forgetful winning strategy for each point in his winning region.
winning a game G memoryless is defined accordingly.
Exercise: Give an example for a game G such that Player 0 wins forgetful on each {v} for v ∈ W0, but he has no (single) forgetful winning strategy on W0.
this is in contrast to the result on “regular” strategies.
The Lemma following shows that the statement is true for memoryless strategies (under certain conditions).

31. Lemma Let G=(A, Win) by any game such that:
V (vertex set) is countable
V*Win ⊆ Win (adding a finite prefix)
Win/V*⊆ Win (removing a finite prefix)
Win/V ∗ := { η ∈ Vω | ∃w ∈ V∗ with wη ∈ Win }
Let U be a set of vertices that Player σ has a memoryless winning strategy on each u ∈ U.
Then Player σ has a memoryless winning strategy on (all) U.
Condition 2 & 3 are satisfied for every regular game since Inf(X(π)) is unaffected.
conditions 2&3 state that winning a game doesn’t depend on a final prefix.

32. Lemma – Proof (construction)
goal: construct a memoryless winning strategy on U.
for every u∈U, Let fσu : Vσ → V be a memoryless winning strategy for Player σ on u.
wlog assume the domain of fσu , denoted Du is minimal.
Let < be a well ordering on U (V is countable)
Define
For each v ∈ D, let u(v) be the minimal vertex in U s.t the strategy for u is defined for v.
Define strategy: fσ(v) := fσ u(v) (v)
choose the “minimal strategy” defined for v
- u(v) : the minimal vertex in U that v appears in its strategy. (knows how to win from v)

33. Lemma – Proof (correctness)
Need to show that fσ(v) := fσ u(v) (v) is a winning strategy on U.
Let π = v0v1… be a play, starting in U and conforms with fσ .
In each σ-vertex vj in π, Player σ chooses fσ u(vj)
Let
Then from i on, fσ = fσu(vi)
vivi+1… is a suffix of a play π’ that starts in u(vi), visits vi , and conforms to fσu(vi)
by minimality of Du(vi)
Then π ∈ V∗(Win/V ∗)
And so, by condition 2 & 3 π ∈ Win

34. Example Max Parity Game
Player 0 Wins (memoryless) on each v in {z1, z2}, eg:
f0z1 wins on z1: f0z1 (z1) = z2 f0z1(z2) = z3
f0z2 wins on z2: f0z2 (z2) = z1 f0z2(z1) = z0
Strategy f0(z)= (f0z1 if z=z1 or f0z2 if z=z2) is not winning.
But if z1<z2 we get according to the construction: f0 = f0z1 which is indeed a winning strategy on {z1, z2} for Player 0.
Player 0
Player 1

35. Forgetful Determinacy of Regular Games
We prove: In every Regular Game both Players win forgetful.
“Forgetfull determinacy of Regular games”
We use the following theorem: In every Parity game, both players win memoryless. (proven in chapter 6)
“memoryless determinacy of parity games”

36. Forgetful Determinacy of Regular Games – Proof
Let (A, χ,F) be a Muller game
remainder: a regular game (A, χ, acc) has the same winning regions as some muller game (A, χ,F)
Let A’ be the “equivalent” Max-Parity game
V‘ = V × …
Let f’0, f’1 be memoryless winning strategies on W0’, W1’
by “memoryless determinacy of parity games”
idea: use finite memory to simulate playing on A’ when playing on A.
Winning regions are: Wσ = { v ∈ V | (v, r(v)) ∈ W’σ}
uses ideas of proof 2.7

37. Forgetful Determinacy of Regular Games – Proof
set memory M=
set mI=r(v)= ♮ χ(v)
set δ=ϕ
update memory according to the update rule of 2.7
gσ(v, q) := f’σ( (v, q) )0 ≡ (v’, q’)0 = v’
f’σ generates a winning play in A’, then by previous theorem, the sequence of the first components is a winning play win A.
Thus gσ is a winning strategy for Player σ.

38. Ilustration of Forgetfull Strategy (same example from before)
Player 0
Player 1
F = {{b}}
לכל משחק π ב-A מתאים משחק יחיד 'π ב-'A. נבחן אותו:
- π = z1z2z0z1z2ω is winning for Player 0
- Winning regions: W0 = {z2} and W1 = {z0, z1}.
π’ = (z1, ♮a)(z2, ♮ab)(z0, ♮ba)(z1, b♮a)(z2, ♮ab)(z2, a♮b)ω
χ’(π’) = ω (winning for Player 0)

39. Some (unproven) Results regarding memoryless determinacy
Parity games enjoy memoryless determinacy
In some Muller games both player need memory to win.
Rabin and Street are in between: one of the two players always has a memoryless winning strategy, specifically:
In Rabin games Player 0 has a memoryless strat.
In Streett games Player 1 has a memoryless strat.

40. Reachability Games & Attractors
A new kind of games, given:
Arena A = (V0, V1, E)
X ⊆ V
Reachability Game R(A, X):
π is a winning play for Player 0 if:
some v∈X occurs in π, or:
A dead end of Player 1 occurs in π.
Different then former games: dead end of player 0 isn’t necessarily a loss for him
strategies for player 0 aren’t required to be defined on plays ending in some v∈X.

41. Ordinals and Transfinite Induction – A Short Digression
Motivation: we want to prove some property using induction
Problem: dealing with non countable sets
Well Ordered sets:
A Totally ordered set, s.t each (non emtpy) subset has a least element.
equivalently (AOIC) every strictly descending series is finite
canonical example: ℕ
Ordinals: we think of them as the position of an element in an ordered set.
An ordinal that is no a successor (other then 0) is called a limit ordrinal.
Example: 0,1,…, ω, ω+1, ω+2…, ω*2,… ,ω*ω…
Theorem: every set can be well ordered (using AOC)

42. Transfinte induction (and recursion)
prove P(0) (base case)
successor step: assuming P(a) prove P(a+1)
Limit step: for every limit ordinal λ assume P(a) for all ordinals λ<a to prove P(λ)
Recursion:
Define S0
for every ordinal a define Sa+1 using definition Sa
Limit step: for every limit ordinal λ define S λ using the definition of sa for all ordinals a<λ.

43. Theorem: Reachability Games Enjoy Memoryless Determinacy
A is an arena, X ⊆ V
The winning region and the winning strategy for player 0 in R(A, X) are defined inductively:
X0 = X
For ordinals v: Xν+1 = Xv ∪ pre(Xv)
pre(Y ) = { v ∈ V0 | vE ∩ Y = ∅ } ∪ { v ∈ V1 | vE ⊆ Y }
vertices in V0 who have a neighbour in Y
verticesi in V1 where all neighbours are in Y.
for each limit ordinal ξ:
Idea: A ⊂ W0 ⟹ A ∪ pre(A) ⊂ W0
Let ξ be the smallest ordinal such that Xξ = Xξ+1
Claim: W:= Xξ is Player 0’s winning region (W=W0).

44. Illustration X ω =Ui Xi Xω+1 =Xω U Pre(Xω) X3=X2UPre(X2) X2=X1UPre(X1)

45. Reachability games enjoy memoryless determinacy – First Direction.
First direction, show W⊆W0:
For every v ∈ W\X there exists a unique ordinal ξv < ξ such that v ∈ Xξv+1 \ Xξv
Further, if v∈V0 (v∈W∩V0\X) then by Pre there is a neigbour of v, v’ s.t v’∈ Xξv
Choose (memoryless) strategy: f0(v) = v’
Claim: f0 is winning on W.
Show by transfinite induction: f0 is winning on Xν for every ν ≤ ξ.

46. Illustration Xξv+1 Xξv v∈ V1 v∈ V0 f0 V’ =XξvU Pre(Xξv)
להסביר את רעיון האינדוקציה
Xξv

47. Reachability games enjoy memoryless determinacy – Second direction.
Define W’ = V\W . Second direction: show W’ ⊆ W1
Let v∈ W’
v ∉ X
If v is a dead end:
It is a dead end of Player 0
Since all dead ends of player 1 are in X1 (by Pre)
Then player 1 win on v
If v is not a dead end and v ∈ V0:
then for every v’ ∈ vE v’ ∉ W (v’ ∈ W’)
If v is not a dead end v ∈ V1:
there exists v’ ∈ vE such that v’ ∉ W (v’ ∈ W’)
Set f1(v) = v’ (memoryless strategy for Player 1)
Every play conform with f1 and starting in W’ is contained in W’
Since W’ doesn’t contain vertices from X or dead ends of Player 1, every such play is winning for Player 1.
Finally: W=W0 and W’=W1

48. Reachability Games-definitions
The winning region of Player 0 in the reachability game R(A, X) is denoted Attr0(A, X) and called attractor of the set X in the arena A.
The winning strategy (denoted f0 in the proof) is called the a corresponding attractor strategy for Player 0.
1-attractor and strategy are symmetrically defined by exchanging V0 and V1 in the arena.

49. σ-trap - Definition
A σ-trap is a subset Y ⊆ V such that vE ⊆ Y for every v ∈ Y∩Vσ and vE∩Y ≠ ∅ for every v∈Y∩Vσ̅
A function which picks for every v∈Y∩Vσ̅ a vertex v’∈ vE ∩ Y is called a trapping strategy for Player σ̅.
The complement of a σ-attractor is a σ-trap
in last proof W’=V\W is a 0-trap.
מלכודת סיגמא
-לוכדת את שחקן סיגמא בתת קבוצה
-השחקן השני משתמש באסט' לוכדת

50. Conclusion – WLOG arenas have no dead ends (no proof)
Let (A, Acc) be an arbitrary game
A = (V0, V1,E)
set Uσ = Attrσ(A, ∅)
Player σ wins on Uσ Memoryless (by forcing the player σ̅ to a dead end)
Arena A’: remove vertices from which you can force the other player to a dead end:
V0’ := V0 \ (U0∪ U1)
V1’ := V1 \ (U0∪ U1)
V’ = V0’ ∪ V1’
E’ = E∩(V’xV’)
A’ = (v0’, v1’, E’)
Claim: for every for every v∈ V’, Player σ wins (A’, Acc, v) iff he wins (A, Acc, v)
Winning strategies for (A’, Acc) can be used in (A, Acc).
use: “the complement of a σ-attractor is a σ-trap”
hw..

51. Memoryless determinacy in Buchi games
As mentioned, all regular games enjoy memoryless determinacy (chapter 6).
We show a simple proof for Buchi games.
Remainder - Buchi acceptance condition:
(Acc = F ⊆ C): π ∈ Wχ(Acc) iff Inf(χ(π)) ∩ F = ∅.
מוכרעים ללא זכרון

52. Memoryless Determinacy in Buchi Games - Construction
Inductive construction of a winning region for Player 0 in the Buchi game (A, χ, F):
set Y := χ−1(F)
Z0 =V (all vertices)
Xξ = Attr0(A, Zξ)
Yξ = Pre(Xξ)
Zξ+1 = Yξ ∩Y
Zξ = (for limit ordinals ξ)
Let ξ be the least ordinal ≥ 1 s.t Zξ = Zξ+1
Zξ ⊂Xξ ⟹ Xξ+1 =Attr0(Zξ) ⊂ Attr0(Xξ)=Xξ (by monotonicity of Attr0)
Xξ+1 ⊂ Xξ
Y ξ+1 ⊂ Yξ (by monotonicity of pre)
Zξ+1 ⊂ Zξ
claim: W := Attr0(A, Zξ) is the winning region of player 0. (i.e W=W0)
our goal – reach some vertex in Y

53. Illustration of the Induction Step
Xξ =Attr0(A, Zξ)
Zξ+1= Yξ ∩Y Zξ
Yξ =Pre(Xξ )

54. Illustration - Zξ = Zξ+1 W=Attr0(A, Zξ) ⊂ W0
Xξ =Attr0(A, Zξ)=W
Zξ=Zξ+1= Yξ ∩Y
אם אנחנו ב-W קיימת אסט להגיע ל Zξ
אם אנחנו ב Zξ קיים קוד קוד שכן ב-W
אז אין סוף פעמיים נבקר ב Zξ ולכן ננצח
Yξ =Pre(Xξ )

55. Correctness - W⊆ W0 for other v∈ V0∩W\ Zξ, v∈ W=Attr0(A, Zξ):
Define a memoryless winning strategy for Player 0 on W:
For every v∈ V0∩Zξ, there exists v’∈ vE∩Attr0(A, Zξ)
since Zξ = Zξ+1 = Yξ ∩Y= Pre(Xξ) ∩ Y = Pre(Attr0(A, Zξ)) ∩ Y
set f0(v) = v’
for other v∈ V0∩W\ Zξ, v∈ W=Attr0(A, Zξ):
set f0(v) according to the respective attractor strategy
For a play starting in W, that conform with f0:
If finite: must end in dead end of Player 1.
If infinite: must reach Zξ at some point, from that point it will reach Zξ infinite more times, Since Zξ⊆ Y (ξ ≥1) then it is a winning game for Player 0.

56. illustration - V\W ⊆ W1 Xk Xk+1 v0 Xξ v1 Y∩Xk\Xk+1 V\Xk+1 is a 0-trap
v ∈ Xk \ Xk+1
שחקן 0 כלוא ב V\XK+1
אם v ∈ Y שחקן 0 חייב להוציא אותו מחוץ ל Xk, ולשחקן אחד תמיד יש אפשרות להוציא אותו מחוץ, לכן באינדקוציה..
v ∉ Pre(Xk) v1

57. Correctness – V\W ⊆ W1 Xξ = Attr0(A, Zξ) – a zero attractor
We show a memoryless winning strategy on W’ :=V\W
For each v ∈ W’ there’s a least k s.t v ∈ Xk \ Xk+1
Since X0=V
and Xi is decreasing in i ( Xa ⊆ Xb for every b < a)
(notice: W=Xξ )
Xk+1 is a 0-attractor (by def) so the complement V\Xk+1 is a 0-trap.
If v ∉ Y:
If v ∈ V1: Set f1(v) according to the trapping strategy
If v ∈ V0: Player 0 is trapped in V\Xk+1
If v ∈ Y: then v ∉ Pre(Xk) ≡ Yk (v∈ Yk and v∈ Y  v∈ Zk+1  v∈ Xk+1 )
If v ∈ V1: there exists v’∈ vE and v’ ∉ Xk
Set f1(v) = v’
If v ∈ V0:
For every v’∈ vE and v’ ∉ Xk
Then f1 is a a memoryless winning strategy for v ∈W’ (show by induction)
Xξ = Attr0(A, Zξ) – a zero attractor
v∈ Yk and v∈ Y  v∈ Zk+1  v∈ v∈ Xk+1

58. Summary What is a game, kinds of different games
we saw that its “enough” to solve parity games.
What is a strategy, a winning strategy
Are there simple Strategies?
saw: if “every Parity Game both Players win memoryless” then “every Regular Game both Players win forgetful”
Showed directly that Buchi games enjoy memoryless determincay.