This presentation is based on the book Automata, Logic and infinite games, Gradel, Thomas and Wilke
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Player 0
Player 1
Player 0
Player 1
𝑣0
𝑣1
𝑣2
𝑣3
𝑣7
𝑣4
𝑣6
𝑣5
(Stewie)
(Spongebob)
𝑣0
𝑣1
𝑣2
𝜋 = 𝑣0 𝑣1 𝑣2 𝑣3 𝑣0 …
𝑣3
𝑣7
𝑣4
𝑣6
𝑣5
𝑣0
𝑣1
𝑣2
𝜋 = 𝑣0 𝑣1 𝑣2 𝑣3 𝑣0 …
𝑣3
𝑣7
𝑣4
𝑣6
𝑣5
𝑣0
𝑣1
𝑣2
𝜋 = 𝑣0 𝑣1 𝑣2 𝑣3 𝑣0 …
𝑣3
𝑣7
𝑣4
𝑣6
𝑣5
𝑣0
𝑣1
𝑣2
𝜋 = 𝑣0 𝑣1 𝑣2 𝑣3 𝑣0 …
𝑣3
𝑣7
𝑣4
𝑣6
𝑣5
𝑣0
𝑣1
𝑣2
𝜋 = 𝑣0 𝑣1 𝑣2 𝑣3 𝑣0 …
𝑣3
𝑣7
𝑣4
𝑣6
𝑣5
𝑣0
𝑣1
𝑣2
𝜋 = 𝑣0 𝑣1 𝑣2 𝑣3 𝑣0 …
𝑣3
𝑣7
𝑣4
𝑣6
𝑣5
(Stewie)
(Spongebob)
• In the previous example, if 𝑣0 𝑣1 𝑣2 𝑣3
(Stewie) wins
𝜔
∈ 𝑊𝑖𝑛 then player 0
𝑣7
𝑣0
𝑣1
𝑣2
1
1
0
1
0
1
𝑣6
2
𝑣5
2
𝑣3
𝑣4
𝜋 = 𝑣0 𝑣1 𝑣2 𝑣3 𝑣0 …
𝜒 𝜋 = 11011 …
𝑣7
𝑣0
𝑣1
𝑣2
1
1
0
1
0
1
𝑣6
2
𝑣5
2
𝑣3
𝑣4
𝜋 = 𝑣0 𝑣1 𝑣2 𝑣3 𝑣0 …
𝜒 𝜋 = 11011 …
𝑣7
𝑣0
𝑣1
𝑣2
1
1
0
1
0
1
𝑣6
2
𝑣5
2
𝑣3
𝑣4
𝜋 = 𝑣0 𝑣1 𝑣2 𝑣3 𝑣0 …
𝜒 𝜋 = 11011 …
𝑣7
𝑣0
𝑣1
𝑣2
1
1
0
1
0
1
𝑣6
2
𝑣5
2
𝑣3
𝑣4
𝜋 = 𝑣0 𝑣1 𝑣2 𝑣3 𝑣0 …
𝜒 𝜋 = 11011 …
𝑣7
𝑣0
𝑣1
𝑣2
1
1
0
1
0
1
𝑣6
2
𝑣5
2
𝑣3
𝑣4
𝜋 = 𝑣0 𝑣1 𝑣2 𝑣3 𝑣0 …
𝜒 𝜋 = 11011 …
𝑣7
Stewie Loses!
𝑣0
𝑣1
𝑣2
1
1
0
1
0
1
𝑣6
2
𝑣5
2
𝑣3
𝑣4
𝑣7
𝑣0
𝑣1
𝑣2
1
1
0
1
0
1
𝑣6
2
𝑣5
2
𝑣3
𝑣4
𝑣7
𝑣0
𝑣1
𝑣2
1
1
0
1
0
1
𝑣6
2
𝑣5
2
𝑣3
𝑣4
U
𝑣0
𝑣1
𝑣2
1
1
0
𝑈 = 𝑣0 , 𝑣1 , 𝑣2 , 𝑣3 , 𝑣7
𝑣7
1
0
1
𝑣6
2
𝑣5
2
𝑣3
𝑣4
𝑣0
𝑣1
𝑣2
1
1
0
𝑈 = 𝑣4 , 𝑣5 , 𝑣6 , 𝑣7
𝑣7
1
0
1
𝑣6
2
𝑣5
2
𝑣3
𝑣4
𝐴𝑡𝑡𝑟1 𝐺, 𝑣2
= 𝑣1 , 𝑣2
𝐴𝑡𝑡𝑟0 𝐺, 𝑣2
=𝑉
𝑣7
𝑣0
𝑣1
𝑣2
1
1
0
1
0
1
𝑣6
2
𝑣5
2
𝑣3
𝑣4
Every player’s dream
U
G
𝐴𝑡𝑡𝑟𝜎 𝐺, 𝑈
Take a break
• Formally:
the set of vertices can be partitioned into a
0-paradise and 1-paradise
𝒏
• 𝑛
• If the maximum parity of G is 𝟎 then G can be partitioned into
0-paradise and 1-paradise
• Player 1
𝑿 = 𝒗 ∈ 𝑽𝟎 | 𝒅𝒐𝒖𝒕 𝒗 = 𝟎 ⇒ 𝐀𝐭𝐭𝐫𝟏 𝑮, 𝑿 𝒊𝒔 𝟏-paradise
• Player 0
• Player 0 can win only if he avoids 1-paradises.
• 𝒀 = 𝑽\𝐀𝐭𝐭𝐫𝟏 𝑮, 𝑿 𝒊𝒔 𝟎-paradise
Not Quite!
𝜎 ≡ 𝑛 𝑚𝑜𝑑 2
• Claim: If 𝑍𝜎 is a 𝜎-paradise in 𝐺[𝑋𝜎 ] then 𝑍𝜎 ∪ 𝑋𝜎 is a 𝜎-paradise
in G
𝜎 ≡ 𝑛 𝑚𝑜𝑑 2
G
𝑋𝜎
𝑋
G𝜎
𝑋𝜎
𝑋𝜎
𝐺[𝑋𝜎 ]
𝐺[𝑋𝜎 ]
G[Z]
𝐺[𝑋𝜎 ]
Attr𝜎 𝐺 𝑋𝜎 , 𝑁
𝑁
𝑍𝜎
𝑍𝜎 G[Z]
Thanks