Alternating tree Automata and Parity games Based on chapter 9 Logic and infinite games, edited by Gradel, Thomas and Wilke, LNCS 2500. Lior Schneider, 17.5.17

topics Defining Alternating tree automata and its behavior Parity games equality to Ata The word problem complexity The complementation problem complexity The Emptiness problem complexity

perliminaries Through all of this lecture we will fix a set of Propositional variables 𝑃. Propositional variables are the atomic formulas of propositional logic Transition system: 𝑆,𝑅,𝜆 : 𝑆 : the states 𝑅⊆𝑆𝑥𝑆 : relations 𝜆:𝑃→𝒫(𝑆) : assigns a set of states to every propositional variable Transition systems are known as Kripke structure: 𝑀=(𝑆,𝐼,𝑅,𝐿) 𝑃= 𝑝,𝑞 ; 𝑆= 𝑠 1 , 𝑠 2 , 𝑠 3 ;𝐼= 𝑠 1 ; 𝑅= 𝑠 1 , 𝑠 2 , 𝑠 2 , 𝑠 1 , 𝑠 2 , 𝑠 3 , 𝑠 3 , 𝑠 3 𝐿= 𝑠 1 , 𝑝,𝑞 , 𝑠 2 , 𝑞 , 𝑠 3 , 𝑝 M produces the path: 𝑠 1 𝑠 2 𝑠 1 𝑠 2 𝑠 3 𝑠 3 .. The word: {𝑝,𝑞},{𝑞},{𝑝,𝑞},{𝑞},{𝑝},{𝑝},{𝑝},… 𝑠 1 {𝑝,𝑞} 𝑠 2 {𝑞} 𝑠 3 {𝑝}

perliminaries 𝑠 1 {𝑝,𝑞} 𝑠 2 {𝑞} Transition system: For every variable 𝑝∈𝑃 and every state 𝑠∈𝜆(𝑝), we say that 𝑝 is true in 𝑠, and for 𝑠∈𝑆\ 𝜆(𝑝), we say that 𝑝 is false in 𝑠. 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑠∈𝑆, 𝑠𝑅 = {𝑠′ ∈ 𝑆 | (𝑠, 𝑠′) ∈ 𝑅} 𝑎𝑛𝑑 𝑅𝑠 = {𝑠′ ∈ 𝑆 | (𝑠′, 𝑠) ∈ 𝑅}. Pointed transition system: (𝒮, 𝑠 𝐼 ) We call a transition system 𝒮= (𝑆,𝑅, 𝜆) (𝑟𝑒𝑠𝑝 (𝒮, 𝑠 𝐼 )) finite iff 𝑆 is finite and 𝜆(𝑝) =∅ for just finitely many 𝑝∈𝑃. 𝑠 3 {𝑝}

perliminaries transition conditions 𝑇 𝐶 𝑄 𝑜𝑣𝑒𝑟 𝑄 𝑠𝑒𝑡 𝑜𝑓 𝑠𝑡𝑎𝑡𝑒𝑠 : The symbols 0 and 1 are transition condition For every 𝑝∈𝑃, 𝑝 and ¬𝑝 are transition conditions. For every 𝑞∈𝑄, 𝑞, ∎𝑞, ∆𝑞 are transition conditions. For every 𝑞 1 , 𝑞 2 ∈𝑄, 𝑞 1 ∧ 𝑞 2 𝑎𝑛𝑑 𝑞 1 ∨ 𝑞 2 are transition conditions. Note that this definition does not allow transition conditions like 𝑞 1 ∧∎ 𝑞 2 𝑜𝑟 p∧ q for p ∈ P and q ∈ Q.

perliminaries Alternating Tree Automata: 𝒜= 𝑄, 𝑞 𝐼 , 𝛿,Ω 𝒜= 𝑄, 𝑞 𝐼 , 𝛿,Ω 𝑄 is a finite set of states of the automaton. 𝑞 𝐼 ∈𝑄 is a state called the initial state. 𝛿 :𝑄→ 𝑇 𝐶 𝑄 is called transition function. Ω :𝑄→𝜔 is called priority function.

Alternating Tree Automata behavior An instance of Ata is a tuple 𝑞,𝑠 𝜖𝑄×𝑆. The behavior of an Ata defined by the current instance. The automaton operates 𝛿 𝑞 𝑎𝑛𝑑 𝑎𝑐𝑡𝑠 𝑎𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔𝑙𝑦: (𝑞 𝐼 ,𝑠 𝐼 ) 𝛿 𝑞 𝐼 = 0,1 , 𝑝, ¬𝑝 𝛿 𝑞 𝐼 =∎/∆𝑞′∈𝑄 stops (𝑞′,𝑠 1 ) 𝛿 𝑞 𝐼 = 𝑞 ′ , 𝑞′∈𝑄 𝑅 𝑚𝑎𝑛𝑦 𝑡𝑖𝑚𝑒𝑠 𝛿 𝑞 𝐼 = 𝑞 1 ∧/∨ 𝑞 2 ; 𝑞 1 , 𝑞 2 ∈𝑄 (𝑞′,𝑠 3 ) (𝑞′,𝑠 𝐼 ) (𝑞′,𝑠 2 ) ( 𝑞 2 ,𝑠 𝐼 ) ( 𝑞 1 ,𝑠 𝐼 )

successful instance 𝑠 1 {𝑝,𝑞} 𝑠 2 {𝑞} 𝑠 3 {𝑝} We define a successful instance 𝑞,𝑠 accordingly: Recall: For every variable 𝑝∈𝑃 and every state 𝑠∈𝜆(𝑝), we say that 𝑝 is true in 𝑠, and for 𝑠∈𝑆\ 𝜆(𝑝), we say that 𝑝 is false in 𝑠. 𝛿 𝑞 =𝑝 𝑎𝑛𝑑 𝑝 𝑖𝑠 𝑡𝑟𝑢𝑒 𝑖𝑛 𝑠 (𝑝∈𝜆(𝑠)) 𝛿 𝑞 =¬𝑝 𝑎𝑛𝑑 𝑝 𝑖𝑠 𝑓𝑎𝑙𝑠𝑒 𝑖𝑛 𝑠 (𝑝∉𝜆(𝑠)) 𝛿 𝑞 =1 If 𝛿(𝑞) =𝑞′ The instance (𝑞, 𝑠) is successful iff (𝑞′, 𝑠) is successful. If 𝛿(𝑞) = 𝑞 1 ∧ 𝑞 2 , then the instance (𝑞, 𝑠) succeeds iff both the instances ( 𝑞 1 , 𝑠) and ( 𝑞 2 , 𝑠) succeed. If 𝛿(𝑞)= 𝑞 1 ∨ 𝑞 2 , then the instance succeeds iff at least one of the instances ( 𝑞 1 , 𝑠) and ( 𝑞 2 , 𝑠) succeeds. If 𝛿(𝑞) =∎𝑞′, then the instance (𝑞, 𝑠) succeeds iff for every 𝑠′∈𝑠𝑅 the instance (𝑞′, 𝑠′) succeed. If 𝛿(𝑞) =∆𝑞′, then the instance (𝑞, 𝑠) succeeds iff for at least one 𝑠′∈𝑠𝑅 the instance (𝑞′, 𝑠′) succeed. Finally: The automaton accepts the transition system (𝓢, 𝒔 𝑰 ) iff the initial instance ( 𝒒 𝑰 , 𝒔 𝑰 ) succeeds.

Formalization problem If we have an infinite tree, how can we determine if an instance is successful or not? 𝛿 𝑞 =𝑞 We will convert the automaton into a parity game! (𝑞,𝑠) 𝛿 𝑞 =𝑞

Parity game conversion Let 𝒜= 𝑄, 𝑞 𝐼 ,𝛿,Ω , 𝒮= 𝑆, 𝑠 𝑖 , we will examine a word 𝑣∈ 𝑄×𝑆 ∗ and a letter 𝑞,𝑠 ∈𝑄×𝑆. (notation of 𝑣(𝑞,𝑠) word concatenation with letter) We will define 𝒜 on 𝒮 behavior as the least language 𝑉⊆ 𝑄×𝑆 ∗ such that 𝑞 𝐼 , 𝑠 𝐼 ∈𝑉 and ∀𝑣 𝑞,𝑠 ∈𝑉: 𝛿 𝑞 = 𝑞 ′ →𝑣 𝑞,𝑠 𝑞 ′ ,𝑠 ∈𝑉 𝛿 𝑞 = 𝑞 1 ∧ 𝑞 2 𝑜𝑟 𝑞 1 ∨ 𝑞 2 →𝑣 𝑞,𝑠 𝑞 1 ,𝑠 ∈𝑉 and 𝑣 𝑞,𝑠 𝑞 2 ,𝑠 ∈𝑉 𝛿 𝑞 =∎𝑞′ 𝑜𝑟 ∆𝑞′→𝑣 𝑞,𝑠 𝑞′,𝑠′ ∈𝑉

Parity game conversion The Arena 𝑉 0 , 𝑉 1 ,𝐸 : Moves 𝐸≜ 𝑣, 𝑣 𝑞,𝑠 𝑣 𝑞,𝑠 ∈𝑉 } a word and concatenated word such that the concatenated word is in V. The parity game priority mapping function will be Ω 𝑣 𝑞,𝑠 ≜Ω 𝑞 ∀𝑣 𝑞,𝑠 ∈𝑉 𝒜 𝑎𝑐𝑐𝑒𝑝𝑡𝑠 𝒮 𝑖𝑓𝑓 ∃ winning strategy for player 0 in the equivalent parity game 𝒢=( 𝑉 0 , 𝑉 1 ,𝐸 , Ω, 𝑞 𝐼 , 𝑠 𝐼 ) ℒ 𝒜 =𝑎𝑙𝑙 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑒𝑑 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛 𝑠𝑦𝑠𝑡𝑒𝑚 𝑡ℎ𝑎𝑡 𝒜 𝑎𝑐𝑐𝑒𝑝𝑡𝑠 𝑽 𝟏 𝑽 𝟎 𝛿 𝑞 =1 𝛿 𝑞 =0 𝛿 𝑞 =𝑝∧𝑠∈𝜆(𝑝) 𝛿 𝑞 =𝑝∧𝑠∉𝜆(𝑝) 𝛿 𝑞 =¬𝑝∧𝑠∉𝜆(𝑝) 𝛿 𝑞 =¬𝑝∧𝑠∈𝜆(𝑝) 𝛿 𝑞 =𝑞′ 𝛿 𝑞 = 𝑞 1 ∧ 𝑞 2 𝛿 𝑞 = 𝑞 1 ∨ 𝑞 2 𝛿 𝑞 =∎𝑞′ 𝛿 𝑞 =Δ𝑞′

Examples 𝑄= 𝑞 𝐼 ; 𝛿 𝑞 𝐼 = ∎ 𝑞 𝐼 ;Ω 𝑞 𝐼 =0 𝑤ℎ𝑒𝑟𝑒 𝒮 𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡𝑒𝑑 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛 𝑠𝑦𝑠𝑡𝑒𝑚. Priority, loacation Notice that player 0 doesn’t have any Loacation but the only priority is 0 Hence player 1 looses for every game (finite and infinite), therefor the Automaton accepts any pointed transition system. 0,1 (𝑞 𝐼 ,𝑠 𝐼 ) 𝛿 𝑞 =∎𝑞′∈𝑄 0,1 ( 𝑞 𝐼 ,𝑠 1 ) 𝑅 𝑚𝑎𝑛𝑦 𝑡𝑖𝑚𝑒𝑠 0,1 ( 𝑞 𝐼 ,𝑠 3 ) 0,1 ( 𝑞 𝐼 ,𝑠 2 ) 𝑽 𝟏 𝑽 𝟎 𝛿 𝑞 =1 𝛿 𝑞 =0 𝛿 𝑞 =𝑝∧𝑠∈𝜆(𝑝) 𝛿 𝑞 =𝑝∧𝑠∉𝜆(𝑝) 𝛿 𝑞 =¬𝑝∧𝑠∉𝜆(𝑝) 𝛿 𝑞 =¬𝑝∧𝑠∈𝜆(𝑝) 𝛿 𝑞 =𝑞′ 𝛿 𝑞 = 𝑞 1 ∧ 𝑞 2 𝛿 𝑞 = 𝑞 1 ∨ 𝑞 2 𝛿 𝑞 =∎𝑞′ 𝛿 𝑞 =Δ𝑞′

Examples 𝑄= 𝑞 𝐼 ; 𝛿 𝑞 𝐼 = ∎ 𝑞 𝐼 ;Ω 𝑞 𝐼 =1 where 𝒮 is any pointed transition system with some infinite path starting at 𝑠 𝐼 . Priority, loacation Notice that player 0 still doesn’t have any Loacation but the only priority is 1 Hence player 1 wins for every infinite Game by choosing the infinite path If 𝒮 is any pointed transition system with no infinite path, player 1 will loose. therefor the Automaton accepts any pointed transition system with no infinite paths. 1,1 (𝑞 𝐼 ,𝑠 𝐼 ) 𝛿 𝑞 =∎𝑞′∈𝑄 1,1 ( 𝑞 𝐼 ,𝑠 1 ) 𝑅 𝑚𝑎𝑛𝑦 𝑡𝑖𝑚𝑒𝑠 1,1 ( 𝑞 𝐼 ,𝑠 3 ) 1,1 ( 𝑞 𝐼 ,𝑠 2 ) 𝑽 𝟏 𝑽 𝟎 𝛿 𝑞 =1 𝛿 𝑞 =0 𝛿 𝑞 =𝑝∧𝑠∈𝜆(𝑝) 𝛿 𝑞 =𝑝∧𝑠∉𝜆(𝑝) 𝛿 𝑞 =¬𝑝∧𝑠∉𝜆(𝑝) 𝛿 𝑞 =¬𝑝∧𝑠∈𝜆(𝑝) 𝛿 𝑞 =𝑞′ 𝛿 𝑞 = 𝑞 1 ∧ 𝑞 2 𝛿 𝑞 = 𝑞 1 ∨ 𝑞 2 𝛿 𝑞 =∎𝑞′ 𝛿 𝑞 =Δ𝑞′

Examples 𝑄= 𝑞 𝐼 ; 𝛿 𝑞 𝐼 =Δ 𝑞 𝐼 ;Ω 𝑞 𝐼 =1 𝑤ℎ𝑒𝑟𝑒 𝒮 𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡𝑒𝑑 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛 𝑠𝑦𝑠𝑡𝑒𝑚. Priority, loacation All the location are player 0 hence Player 0 can choose a finite path and therefore the automaton Doesn’t accept any 𝒮 1,0 (𝑞 𝐼 ,𝑠 𝐼 ) 𝛿 𝑞 =Δ𝑞′∈𝑄 1,0 ( 𝑞 𝐼 ,𝑠 1 ) 𝑅 𝑚𝑎𝑛𝑦 𝑡𝑖𝑚𝑒𝑠 1,0 ( 𝑞 𝐼 ,𝑠 3 ) 𝑽 𝟏 𝑽 𝟎 𝛿 𝑞 =1 𝛿 𝑞 =0 𝛿 𝑞 =𝑝∧𝑠∈𝜆(𝑝) 𝛿 𝑞 =𝑝∧𝑠∉𝜆(𝑝) 𝛿 𝑞 =¬𝑝∧𝑠∉𝜆(𝑝) 𝛿 𝑞 =¬𝑝∧𝑠∈𝜆(𝑝) 𝛿 𝑞 =𝑞′ 𝛿 𝑞 = 𝑞 1 ∧ 𝑞 2 𝛿 𝑞 = 𝑞 1 ∨ 𝑞 2 𝛿 𝑞 =∎𝑞′ 𝛿 𝑞 =Δ𝑞′ 1,0 ( 𝑞 𝐼 ,𝑠 2 )

Current disadvantages While 𝒮 may be finite our parity game 𝒢 may be infinite (for example 𝛿 𝑞 𝐼 = 𝑞 𝐼 ) therefor we have the need to define the behavior in more convenient way. This would be helpful for our next slides

Alternative Formal Definition for the Arena The same 𝒜= 𝑄, 𝑞 𝐼 ,𝛿,Ω , 𝒮= 𝑆, 𝑠 𝑖 , Let 𝑉 ⊆𝑄×𝑆 and E ⊆ 𝑉 × 𝑉 be the smallest graph with 𝑞 𝐼 , 𝑠 𝐼 ∈ 𝑉 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 ∀ 𝑞,𝑠 ∈ 𝑉 𝑤𝑒 ℎ𝑎𝑣𝑒: 𝛿 𝑞 = 𝑞 ′ → 𝑞 ′ ,𝑠 ∈[𝑉] 𝑎𝑛𝑑 𝑞,𝑠 𝑞 ′ ,𝑠 ∈[𝐸] 𝛿 𝑞 = 𝑞 1 ∧ 𝑞 2 𝑜𝑟 𝑞 1 ∨ 𝑞 2 → 𝑞 1 ,𝑠 , 𝑞 2 ,𝑠 ∈[𝑉] and 𝑞,𝑠 , 𝑞 1 ,𝑠 , 𝑞,𝑠 𝑞 2 ,𝑠 ∈[𝐸] 𝛿 𝑞 =∎ 𝑞 ′ 𝑜𝑟 ∆ 𝑞 ′ → 𝑞 ′ , 𝑠 ′ ∈ 𝑉 𝑎𝑛𝑑 𝑞,𝑠 , 𝑞 ′ , 𝑠 ′ ∈ 𝐸 ∀ 𝑠 ′ ∈sR We split 𝑉 𝑖𝑛𝑡𝑜 𝑉 0 𝑎𝑛𝑑 𝑉 1 𝑎𝑠 𝑒𝑎𝑟𝑙𝑖𝑒𝑟 𝑎𝑛𝑑 𝑡ℎ𝑒 Ω 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒. Previous definition: We will define 𝒜 on 𝒮 behavior as the least language 𝑉⊆ 𝑄×𝑆 ∗ such that 𝑞 𝐼 , 𝑠 𝐼 ∈𝑉 and ∀𝑣 𝑞,𝑠 ∈𝑉: 𝛿 𝑞 = 𝑞 ′ →𝑣 𝑞,𝑠 𝑞 ′ ,𝑠 ∈𝑉 𝛿 𝑞 = 𝑞 1 ∧ 𝑞 2 𝑜𝑟 𝑞 1 ∨ 𝑞 2 →𝑣 𝑞,𝑠 𝑞 1 ,𝑠 ∈𝑉 and 𝑣 𝑞,𝑠 𝑞 2 ,𝑠 ∈𝑉 𝛿 𝑞 =∎𝑞′ 𝑜𝑟 ∆𝑞′→𝑣 𝑞,𝑠 𝑞′,𝑠′ ∈𝑉

Alternative Formal Definition for the Arena Our new Arena: 𝒢= 𝑉 0 , 𝑉 1 , 𝐸 , Ω, 𝑞 𝐼 , 𝑠 𝐼 . Definition of the function []: 𝑄×S + →(𝑄×𝑆) for each word 𝑄×S + it assigns the value of its last letter 𝑞,𝑠 Note that [] maps location from V to [V], preserve edges and priorities, therefor 𝒢′ is created by applying [] on 𝒢

Alternative Formal Definition for the Arena Theorem: Player 0 has a winning strategy in 𝒢 iff Player 0 has a winning strategy in 𝒢‘ Proof: Let 𝑓′ 0 :[ 𝑉 0 ]→[𝑉] be a winning strategy for Player 0 in 𝒢′ We define a mapping 𝑓 0 : 𝑉 0 →𝑉 𝑏𝑦 𝑓 0 𝑣 ≜ 𝑣 𝑓 0 ′ 𝑣 . Let 𝜋 be a play in 𝒢 which is consistent with 𝑓 0 . Then, [𝜋] is consistent with 𝑓′ 0 , and thus, [𝜋] and 𝜋 are won by Player 0. Consequently, 𝑓 0 is a winning strategy for Player 0 in 𝒢. Clearly, we can apply a symmetric argument if 𝑓 1 ′ :[ 𝑉 1 ]→[𝑉] is a winning strategy for Player 1 in 𝒢′.

Complex Transition Conditions Recall: transition conditions 𝑇 𝐶 𝑄 𝑜𝑣𝑒𝑟 𝑄 𝑠𝑒𝑡 𝑜𝑓 𝑠𝑡𝑎𝑡𝑒𝑠 : The symbols 0 and 1 are transition condition For every 𝑝∈𝑃, 𝑝 and ¬𝑝 are transition conditions. For every 𝑞∈𝑄, 𝑞, ∎𝑞, ∆𝑞 are transition conditions. For every 𝑞 1 , 𝑞 2 ∈𝑄, 𝑞 1 ∧ 𝑞 2 𝑎𝑛𝑑 𝑞 1 ∨ 𝑞 2 are transition conditions. Note that this definition does not allow transition conditions like 𝑞 1 ∧∎ 𝑞 2 𝑜𝑟 p∧ q for p ∈ P and q ∈ Q. Example of such 𝜑: “Change the inner state to 𝑞 1 if 𝑝 is true, otherwise change the inner state to 𝑞 2 ” Formally: 𝜑=( 𝑞 1 ∧𝑝)∨( 𝑞 2 ∧￢𝑝) So we will use some new states: 𝑞 𝜑 , 𝑞 𝑞 1 ∧𝑝 , 𝑞 𝑞 1 ∧¬𝑝 , 𝑞 𝑝 , 𝑞 ¬𝑝

Complex Transition Conditions (𝑞 𝐼 ,𝑠 𝐼 ) 1,0 𝛿 𝑞 𝜑 ≜ 𝑞 𝑞 1 ∧𝑝 ∨ 𝑞 𝑞 2 ∧¬𝑝 𝛿 𝑞 𝑞 1 ∧𝑝 ≜ 𝑞 1 ∧ 𝑞 𝑝 𝛿 𝑞 𝑞 2 ∧¬𝑝 ≜ 𝑞 2 ∧ 𝑞 ¬𝑝 𝛿 𝑞 𝑝 ≜𝑝 𝛿 𝑞 ¬𝑝 ≜¬𝑝 Example: We set: 𝛿 𝑞 1 =𝛿 𝑞 2 =∎ 𝑞 𝜑 ;Ω 𝑞 1 =2;Ω 𝑄\ q 1 =1; 𝑞 𝐼 = 𝑞 𝜑 Priority, loacation The automaton accepts some pointed transition system 𝒮 iff every infinite path starting from 𝑠 𝐼 contains infinitely many states in which p is true. There will be infinite amount of successful 𝑞 1 , 𝑠 ′ 𝑖𝑛𝑠𝑡𝑎𝑛𝑐𝑒𝑠 𝑤𝑖𝑡ℎ 𝑒𝑣𝑒𝑛 𝑝𝑟𝑖𝑜𝑟𝑖𝑡𝑦. 𝛿 𝑞 𝜑 = 𝑞 𝑞 1 ∧𝑝 ∨ 𝑞 𝑞 2 ∧¬𝑝 1,1 1,1 ( 𝑞 𝑞 1 ∧𝑝 ,𝑠 𝐼 ) ( 𝑞 𝑞 2 ∧¬𝑝 ,𝑠 𝐼 ) 2,1 1,? 1,1 1,1 (𝑞 1 ,𝑠 𝐼 ) ( 𝑞 𝑝 ,𝑠 𝐼 ) (𝑞 2 ,𝑠 𝐼 ) ( 𝑞 ¬𝑝 ,𝑠 𝐼 ) 1,0 1,0 1,0 1,0 (𝑞 𝜑 ,𝑠′ 1 ) (𝑞 𝜑 ,𝑠′ 2 ) (𝑞 𝜑 ,𝑠′ 2 ) p (𝑞 𝜑 ,𝑠′ 1 ) ¬p

The Word problem Means to decide whether a given alternating tree automaton 𝒜 accepts a given finite pointed transition system 𝒮. Motivation: In Section 10, this result will be used to determine the complexity of the model checking problem for the modal μ-calculus. (which we will not elaborate on) A naïve approach to solve the word problem would be to compute the whole behavior of the Ata. This is actually impossible to solve it that way because it can be infinite behavior. Luckily we learnt some tools to deal with it. We will reduce the equivalent parity game into a finite one which will be a great help for us

The Word problem Theorem: Let 𝒜=(𝑄, 𝑞 𝐼 , 𝛿,Ω) be an alternating tree automaton with d different non-zero priorities and let 𝒮 be a finite pointed transition system: There is an algorithm which computes in time: 𝒪 𝑑 𝑄 𝑅 +1 𝑄 𝑆 +1 𝑑 2 𝑑 2 and in space 𝒪 𝑑 𝑄 𝑆 log 𝑄 𝑆 The word problem is in 𝑈𝑃∩𝑐𝑜−𝑈𝑃

The Word problem The word problem is in 𝑈𝑃∩𝑐𝑜−𝑈𝑃:  UP ("Unambiguous Non-deterministic Polynomial-time") decision problems solvable in polynomial time on a unambiguous Turing machine with at most one accepting path for each input. 𝑃⊆𝑈𝑃⊆𝑁𝑃 A language is in UP if a given answer can be verified in polynomial time, and the verifier machine only accepts at most one answer for each problem instance. UP contains parity game problem and specifically the question weather player 0 has a winning strategy is in 𝑈𝑃∩𝑐𝑜−𝑈𝑃 as explained in chapter 6.

The Word problem There is an algorithm which computes in time: 𝒪 𝑑 𝑄 𝑅 +1 𝑄 𝑆 +1 𝑑 2 𝑑 2 and in space 𝒪 𝑑 𝑄 𝑆 log 𝑄 𝑆 Upper bounds: Lets fix 𝒜 and examine 𝒮 complexity: 𝑅 ≤ 𝑆 2 ;𝑑 𝑄 𝑎𝑟𝑒 𝒜 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 there for we get:𝒪 𝑆 2 ( 𝑄 𝑆 +1 𝑑 2 ) Roughly we get 𝒪 ( 𝑆 2+ 𝑑 2 ) With 𝑑=2 we get 𝑆 3 , with 𝑑=20 we get |𝑆|^12 – not so nice

The Word problem There is an algorithm which computes in time: 𝒪 𝑑 𝑄 𝑅 +1 𝑄 𝑆 +1 𝑑 2 𝑑 2 and in space 𝒪 𝑑 𝑄 𝑆 log 𝑄 𝑆 Lets estimate 𝐸 𝑎𝑛𝑑 | 𝑉 |: As we know, 𝑉 ⊆𝑄×𝑆→ 𝑉 ≤|𝑄||𝑆| Let 𝑆 ′ ⊆𝑆 be the set of states in 𝒮 which are reachable from 𝑞 𝐼 Every state in 𝑆 except 𝑠 𝐼 has at least one predecessor ( 𝑠 𝐼 𝑚𝑎𝑦 ℎ𝑎𝑣𝑒 𝑛𝑜 𝑝𝑟𝑒), Hence, 𝑅 ≥ 𝑆 −1→ 𝑅 +1≥ 𝑆 To determine |[𝐸]|, we count the number of successors of every location in |[𝑉]|. The successors of a location (𝑞, 𝑠)∈[𝑉] are (𝑞, 𝑠)[𝐸]. 𝐸 = 𝑞∈𝑄 𝑞,𝑠 ∈[𝑉] | 𝑞,𝑠 𝐸 | For a fixed 𝑞:

The Word problem 𝐸 = 𝑞∈𝑄 𝑞,𝑠 ∈[𝑉] | 𝑞,𝑠 𝐸 | For a fixed 𝑞: To sum up: 𝐸 = 𝑞∈𝑄 𝑞,𝑠 ∈[𝑉] | 𝑞,𝑠 𝐸 | For a fixed 𝑞: If 𝛿(𝑞)∈{0, 1} 𝑜𝑟 𝛿(𝑞)∈{𝑝, ￢𝑝}, then (𝑞, 𝑠) has no successor: 𝑞,𝑠 ∈[𝑉] | 𝑞,𝑠 𝐸 | =0 If 𝛿(𝑞)∈𝑄, then every location (𝑞, 𝑠)∈ [𝑉] has exactly one successor: 𝑞,𝑠 ∈[𝑉] | 𝑞,𝑠 𝐸 | ≤ 𝑆 ′ ≤ 𝑅 +1 If 𝛿(𝑞)= 𝑞 1 ∧ 𝑞 2 𝑜𝑟 𝛿(𝑞)= 𝑞 1 ∨ 𝑞 2 for some 𝑞 1 , 𝑞 2 ∈𝑄, then we have: 𝑞,𝑠 ∈[𝑉] | 𝑞,𝑠 𝐸 | ≤2 𝑆 ′ ≤2 𝑅 +1 If 𝛿 𝑞 =∎ 𝑞 ′ 𝑜𝑟 ∆𝑞′, then ∀ 𝑞,𝑠 ∈ 𝑉 , 𝑞,𝑠 𝐸 = 𝑞 ′ ×𝑠𝑅= 𝑠𝑅 : 𝑞,𝑠 ∈[𝑉] | 𝑞,𝑠 𝐸 | ≤ 𝑠∈𝑆 𝑠𝑅 = 𝑅 To sum up: 𝑞,𝑠 ∈[𝑉] 𝑞,𝑠 𝐸 ≤ 2 𝑅 +1 , 𝐸 ≤2|𝑄|( 𝑅 +1)

The Word problem There is an algorithm which computes in time: 𝒪 𝑑 𝑄 𝑅 +1 𝑄 𝑆 +1 𝑑 2 𝑑 2 and in space 𝒪 𝑑 𝑄 𝑆 log 𝑄 𝑆 𝐸 ≤2|𝑄|( 𝑅 +1) Jurdzi´nski’s algorithm: For a parity game with 𝑛 vertices, 𝑚 edges and 𝑑≥2 𝑝𝑟𝑖𝑜𝑟𝑖𝑡𝑖𝑒𝑠 can be solved in time 𝒪 𝑑𝑚 𝑛 𝑑 2 𝑑 2 and space: 𝒪 𝑑𝑛 log 𝑛 And now its easy to apply: 𝑛= 𝑄 𝑆 ;𝑚=2 𝑄|( 𝑅 +1)

Complementation problem Theorem: Let 𝒜=(𝑄, 𝑞 𝐼 , 𝛿,Ω) be an alternating tree automaton. There is an alternating tree automaton 𝒜 =(𝑄, 𝑞 𝐼 , 𝛿 , Ω ) such that 𝒜 accepts the complement of the language of 𝒜. Proof: Lets start with priority function Ω : ∀𝑞∈Q Ω 𝑞 =Ω+1 The transition function 𝛿: ∆ ∎ ∎ ∆

Complementation problem Theorem: Let 𝒜=(𝑄, 𝑞 𝐼 , 𝛿,Ω) be an alternating tree automaton. There is an alternating tree automaton 𝒜 =(𝑄, 𝑞 𝐼 , 𝛿 , Ω ) such that 𝒜 accepts the complement of the language of 𝒜. Let 𝒮 be a pointed transition system and 𝒢=( 𝑉 0 , 𝑉 1 ,𝐸 , Ω, 𝑞 𝐼 , 𝑠 𝐼 ) be the parity game which determines whether A accepts 𝒮. We show that 𝒜 accepts 𝒮 iff 𝒜 does not accept 𝒮. We examine the parity game 𝒢 which determines whether 𝒜 accepts 𝒮. Intuitively, we simply change the ownership of every location, and we increase every priority by 1. Let 𝑉= 𝑉 0 ∪ 𝑉 1 be the locations of 𝒢 and 𝒢 . Let 𝑉 ′ ⊆𝑉 be the locations 𝑣(𝑞, 𝑠)∈𝑉 with 𝛿(𝑞)=𝑞′ for some 𝑞′∈𝑄. We do not change the ownership of locations in V’. The automaton 𝒜 accepts 𝒮 iff there is winning strategy for Player 0 in the parity game: 𝒢 =( V 1 ∪ 𝑉 ′ , 𝑉 0 \ V ′ ,𝐸 , Ω , 𝑞 𝐼 , 𝑠 𝐼 ) Because parity games are determined, we have to show that there is a winning strategy for Player 0 in 𝒢 iff there is a winning strategy for Player 1 in 𝒢 .

Complementation problem Assuming a winning strategy for Player 0 in 𝒢 we will construct a winning strategy for Player 1 in 𝒢 : Let 𝑓 0 : 𝑉 0 →𝑉 be a winning strategy for Player 0 in 𝒢. There is a unique extension 𝑓′ 0 : 𝑉 1 ∪𝑉′→ 𝑉. Now, assume some play 𝜋 in 𝒢 which is consistent with 𝑓′ 0 . Then, 𝜋 in 𝒢 is consistent with 𝑓0, and thus 𝜋 is won by Player 1. Conversely, let 𝑓 1 : 𝑉 1 →𝑉 be a winning strategy for Player 1 in 𝒢 . Clearly, the restriction of 𝑓 1 𝑡𝑜 𝑉 1 \ 𝑉′ is a winning strategy for Player 0 in 𝒢. Consequently, Player 0 has a winning strategy in 𝒢 iff player 1 has a winning strategy in 𝒢 . We can also (but we will not) show that alternating tree automatons are closed under intersection and union.

The Emptiness Problem On this section: The Emptiness Problem complexity analysis

The Emptiness Problem Lets start by fixing 𝒜= 𝑄, 𝑞 𝐼 ,𝛿,Ω Definition of an inflating transition condition: We call a transition condition “inflating” if its in the form of 𝛿 𝑞 =𝑞∧𝑞 𝑜𝑟 𝑞∨𝑞 Lemma: For every alternating tree automaton 𝒜=(𝑄, 𝑞 𝐼 , 𝛿,Ω) there is an automaton 𝒜′=(𝑄, 𝑞 𝐼 , 𝛿′,Ω) with 𝐿(𝒜)=𝐿(𝒜′) such that for every 𝑞∈𝑄 𝛿(𝑞) is not inflated. Proof: very similar to the complement problem: First we will define 𝛿 ′ :𝑄→𝑇 𝐶 𝑄 𝑓𝑜𝑟 𝑞

The Emptiness Problem Lemma: For every alternating tree automaton 𝒜=(𝑄, 𝑞 𝐼 , 𝛿,Ω) there is an automaton 𝒜′=(𝑄, 𝑞 𝐼 , 𝛿′,Ω) with 𝐿(𝒜)=𝐿(𝒜′) such that for every 𝑞∈𝑄 𝛿(𝑞) is not inflated. Let 𝒮 be some pointed transition system. We want to show that 𝒜 accepts 𝒮 iff 𝒜’ accepts 𝒮. At first, we observe that 𝒜 has the same behavior 𝑉 on 𝒮 as 𝒜′. Let 𝒢=( 𝑉 0 , 𝑉 1 ,𝐸), Ω, ( 𝑞 𝐼 , 𝑠 𝐼 ) be the parity game to determine whether 𝒜 accepts 𝒮. Let 𝑉 ′ ={𝑣(𝑞, 𝑠)∈ 𝑉 1 | 𝛿(𝑞)=𝑞′∧𝑞′ 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑞′∈𝑄}. The locations in 𝑉′ have exactly one successor. The parity game 𝒢 ′= V 0 ∪ 𝑉 ′ , 𝑉 1 \ V ′ ,𝐸 ,Ω, 𝑞 𝐼 , 𝑠 𝐼 determines whether 𝒜′ accepts 𝒮. From now and on its exactly the same proof as the Complementation problem

The Emptiness Problem Lets start by fixing 𝒜= 𝑄, 𝑞 𝐼 ,𝛿,Ω Definition of an inflating transition condition: We call a transition condition “inflating” if its in the form of 𝛿 𝑞 =𝑞∧𝑞 𝑜𝑟 𝑞∨𝑞 According to the lemma we can assume that our 𝒜 has no inflate transition conditions. Another notation: Tiles A tile over 𝑄 is a graph 𝜗=( 𝑉 𝜗 , 𝐸 𝜗 ) where 𝑉 𝜗 ⊆𝑄, 𝐸⊆ 𝑉 𝜗 × 𝑉 𝜗 and: A tile with port is a tuple (𝜗,𝑞) s.t 𝜗= 𝐸 𝜗 , 𝑉 𝜗 𝑎𝑛𝑑 𝑞∈ 𝑉 𝜗 ∩ 𝑄 Δ

The Emptiness Problem A tile with port is a tuple (𝜗,𝑞) s.t 𝜗= 𝐸 𝜗 , 𝑉 𝜗 𝑎𝑛𝑑 𝑞∈ 𝑉 𝜗 ∩ 𝑄 Δ 𝑄 Δ ≜{𝑞∈Q|∃ 𝑞 ′ ∈𝑄:𝛿 𝑞 =Δ𝑞′ } 𝑄 ∎ ≜{𝑞∈Q|∃ 𝑞 ′ ∈𝑄:𝛿 𝑞 =∎𝑞′ } We also define 𝑞 , 𝑉 : 𝑞 ≜ 𝑞 ′ 𝑖𝑓 𝛿 𝑞 =Δ 𝑞 ′ 𝑜𝑟 𝛿 𝑞 =∎𝑞′ For subsets 𝑉⊆𝑄 𝑤𝑒 𝑑𝑒𝑓𝑖𝑛𝑒 𝑉 ≜{ 𝑞 |𝑞∈V} All tiles will be: Θ, all tiles with port will be: Θ 𝑝 We will call a tile with port 𝜗 0 = 𝑉 𝜗0 , 𝐸 𝜗0 , 𝑞 0 and a tile 𝜗 1 =( 𝑉 𝜗1 , 𝐸 𝜗1 ) (it can be a tile with port as well) concatenable iff 𝑞 0 ∈ 𝑉 𝜗1 𝑎𝑛𝑑 𝑉 𝜗0 ∩ 𝑄 ∎ ⊆ 𝑉 𝜗1 i.e: the port must be able to make delta\square transition into 𝑉 𝜗1 state and that the state intersection of 𝑉 𝜗0 𝑎𝑛𝑑 𝑄 ∎ that make delta\square transition must be contained in 𝑉 𝜗1

The Emptiness Problem Let 𝑔=( 𝜗 1 , 𝑞 1 ), ( 𝜗 2 , 𝑞 2 ), ・・・∈ Θ 𝜔 be an infinite sequence of tiles with port where ( 𝜗 𝑖 , 𝑞 𝑖 ) 𝑎𝑛𝑑 ( 𝜗 𝑖+1 , 𝑞 𝑖+1 ) are concatenable for every 𝑖∈𝜔. We define the graph of 𝒈 in a usual way: 𝑉≜ 𝑖∈𝜔 𝑖 × 𝑉 𝑖 𝐸≜ We call an infinite path 𝜋 in (𝑉, 𝐸) even iff the maximal priority which occurs in 𝜋 infinitely often is even. We call the sequence 𝑔 even iff every infinite path 𝜋 𝑖𝑛 (𝑉, 𝐸) is even.

The Emptiness Problem Theorem: There is a deterministic parity 𝜔−𝑎𝑢𝑡𝑜𝑚𝑎𝑡𝑜𝑛 𝐶 with 2 𝒪( 𝑄 4 log 𝑄 ) states and priorities bounded by 𝒪( 𝑄 4 ) which accepts a sequence of concatenable tiles 𝑔∈ Θ 𝜔 iff 𝑔 is even. Proof concept: construct non-deterministic parity ω-automaton B → construct C by a determinization and a complementation of B B states are 𝑄× 0,…, 𝑄 . (m= 𝑄 |𝑄+1|) We specify the transition function 𝛿 by a set of triples. Let ( 𝑞 1 , 𝑖 1 ), 𝑞 2 , 𝑖 2 be two states of B, and let (𝑉, 𝐸, 𝑞) be a tile with port. There is a transition 𝑞 1 , 𝑖 1 , 𝑉, 𝐸, 𝑞 , 𝑞 2 , 𝑖 2 in B iff:

The Emptiness Problem non-deterministic parity ω-automaton Where 𝑚= 𝑄 2 we get: 2 𝒪( 𝑄 4 log 𝑄 ) states and priorities bounded by 𝒪( 𝑄 4 )

done