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 = 𝑠1 , 𝑠2 , 𝑠3 ; 𝐼 = 𝑠1 ;

𝑅=
𝑠1 , 𝑠2 , 𝑠2 , 𝑠1 , 𝑠2 , 𝑠3 , 𝑠3 , 𝑠3

𝐿=
𝑠1 , 𝑝, 𝑞 , 𝑠2 , 𝑞 , 𝑠3 , 𝑝

M produces the path: 𝑠1 𝑠2 𝑠1 𝑠2 𝑠3 𝑠3 . .

The word: {𝑝, 𝑞}, {𝑞}, {𝑝, 𝑞}, {𝑞}, {𝑝}, {𝑝}, {𝑝}, …
𝑠1
{𝑝, 𝑞}
𝑠2
{𝑞}
𝑠3
{𝑝}
perliminaries

𝑠1
{𝑝, 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, 𝑠𝐼 )

We call a transition system 𝒮 = (𝑆, 𝑅, 𝜆) (𝑟𝑒𝑠𝑝 (𝒮, 𝑠𝐼 )) finite iff 𝑆 is finite and
𝜆(𝑝) = ∅ for just finitely many 𝑝 ∈ 𝑃.
𝑠2
{𝑞}
𝑠3
{𝑝}
perliminaries


transition conditions 𝑇𝐶 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 𝑞1 , 𝑞2 ∈ 𝑄, 𝑞1 ∧ 𝑞2 𝑎𝑛𝑑 𝑞1 ∨ 𝑞2 are transition conditions.
Note that this definition does not allow transition conditions like 𝑞1
∧ ∎𝑞2 𝑜𝑟 p∧ q for p ∈ P and q ∈ Q.
perliminaries

Alternating Tree Automata:

𝒜 = 𝑄, 𝑞𝐼 , 𝛿, Ω

𝑄 is a finite set of states of the automaton.

𝑞𝐼 ∈ 𝑄 is a state called the initial state.

𝛿 ∶ 𝑄 → 𝑇𝐶 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:
(𝑞𝐼 , 𝑠𝐼 )
𝛿 𝑞𝐼 = 0,1 , 𝑝, ¬𝑝
𝛿 𝑞𝐼 = ∎/∆𝑞′ ∈ 𝑄
(𝑞′, 𝑠1 )
stops
𝛿 𝑞𝐼 = 𝑞 ′ , 𝑞′ ∈ 𝑄
𝑅 𝑚𝑎𝑛𝑦 𝑡𝑖𝑚𝑒𝑠
𝛿 𝑞𝐼 = 𝑞1 ∧/∨ 𝑞2 ; 𝑞1 , 𝑞2 ∈ 𝑄
(𝑞′, 𝑠𝐼 )
(𝑞′, 𝑠3 )
(𝑞′, 𝑠2 )
(𝑞2 , 𝑠𝐼 )
(𝑞1 , 𝑠𝐼 )
successful instance

𝑠1
{𝑝, 𝑞}
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) = 𝑞1 ∧ 𝑞2 , then the instance (𝑞, 𝑠) succeeds iff both the instances (𝑞1 , 𝑠)
and (𝑞2 , 𝑠) succeed.

If 𝛿(𝑞) = 𝑞1 ∨ 𝑞2 , then the instance succeeds iff at least one of the instances (𝑞1 , 𝑠)
and (𝑞2 , 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, 𝒔𝑰 ) iff the initial
instance (𝒒𝑰 , 𝒔𝑰 ) succeeds.
𝑠2
{𝑞}
𝑠3
{𝑝}
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, 𝑞𝐼 , 𝛿, Ω , 𝒮 = 𝑆, 𝑠𝑖 , 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
𝑞𝐼 , 𝑠𝐼 ∈ 𝑉 and ∀𝑣 𝑞, 𝑠 ∈ 𝑉:

𝛿 𝑞 = 𝑞 ′ → 𝑣 𝑞, 𝑠 𝑞 ′ , 𝑠 ∈ 𝑉

𝛿 𝑞 = 𝑞1 ∧ 𝑞2 𝑜𝑟 𝑞1 ∨ 𝑞2 → 𝑣 𝑞, 𝑠 𝑞1 , 𝑠 ∈ 𝑉 and 𝑣 𝑞, 𝑠 𝑞2 , 𝑠 ∈ 𝑉

𝛿 𝑞 = ∎𝑞′ 𝑜𝑟 ∆𝑞′ → 𝑣 𝑞, 𝑠 𝑞′, 𝑠′ ∈ 𝑉
∗
such that
Parity game conversion

The Arena 𝑉0 , 𝑉1 , 𝐸 :
𝑽𝟎
𝑽𝟏
𝛿 𝑞 =0
𝛿 𝑞 =1
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 = 𝑞1 ∨ 𝑞2
𝛿 𝑞 = 𝑞1 ∧ 𝑞2
𝛿 𝑞 = Δ𝑞′
𝛿 𝑞 = ∎𝑞′
 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
∈𝑉

𝒜 𝑎𝑐𝑐𝑒𝑝𝑡𝑠 𝒮 𝑖𝑓𝑓 ∃ winning strategy for player 0 in the equivalent parity game
𝒢 = ( 𝑉0 , 𝑉1 , 𝐸 , Ω, 𝑞𝐼 , 𝑠𝐼 )

ℒ 𝒜 = 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑒𝑑 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛 𝑠𝑦𝑠𝑡𝑒𝑚 𝑡ℎ𝑎𝑡 𝒜 𝑎𝑐𝑐𝑒𝑝𝑡𝑠
≜ Ω 𝑞 ∀𝑣 𝑞, 𝑠
Examples

𝑄 = 𝑞𝐼 ; 𝛿 𝑞𝐼 = ∎𝑞𝐼 ; Ω 𝑞𝐼 = 0 𝑤ℎ𝑒𝑟𝑒 𝒮 𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡𝑒𝑑 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛 𝑠𝑦𝑠𝑡𝑒𝑚.

Priority, loacation

0,1
(𝑞𝐼 , 𝑠𝐼 )
Notice that player 0 doesn’t have any
Loacation but the only priority is 0
Hence player 1 looses for every game
𝛿 𝑞 = ∎𝑞′ ∈ 𝑄
0,1
(𝑞𝐼 , 𝑠1 )
𝑅 𝑚𝑎𝑛𝑦 𝑡𝑖𝑚𝑒𝑠
0,1
(finite and infinite), therefor the
Automaton accepts any pointed
transition system.
(𝑞𝐼 , 𝑠3 )
0,1
(𝑞𝐼 , 𝑠2 )
𝑽𝟎
𝑽𝟏
𝛿 𝑞 =0
𝛿 𝑞 =1
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 = 𝑞1 ∨ 𝑞2
𝛿 𝑞 = 𝑞1 ∧ 𝑞2
𝛿 𝑞 = Δ𝑞′
𝛿 𝑞 = ∎𝑞′
Examples

𝑄 = 𝑞𝐼 ; 𝛿 𝑞𝐼 = ∎𝑞𝐼 ; Ω 𝑞𝐼 = 1 where 𝒮 is any pointed transition system with
some infinite path starting at 𝑠𝐼 .
1,1

Priority, loacation
(𝑞𝐼 , 𝑠𝐼 )
𝛿 𝑞 = ∎𝑞′ ∈ 𝑄
Notice that player 0 still doesn’t have any
1,1
Loacation but the only priority is 1
𝑅 𝑚𝑎𝑛𝑦 𝑡𝑖𝑚𝑒𝑠
(𝑞𝐼 , 𝑠1 )
Hence player 1 wins for every infinite
1,1

Game by choosing the infinite path
(𝑞𝐼 , 𝑠3 )
1,1
If 𝒮 is any pointed transition system with
(𝑞𝐼 , 𝑠2 )
no infinite path, player 1 will loose.
therefor the Automaton accepts any pointed transition system with no
infinite paths.
𝑽𝟎
𝑽𝟏
𝛿 𝑞 =0
𝛿 𝑞 =1
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 = 𝑞1 ∨ 𝑞2
𝛿 𝑞 = 𝑞1 ∧ 𝑞2
𝛿 𝑞 = Δ𝑞′
𝛿 𝑞 = ∎𝑞′
Examples

𝑄 = 𝑞𝐼 ; 𝛿 𝑞𝐼 = Δ𝑞𝐼 ; Ω 𝑞𝐼 = 1 𝑤ℎ𝑒𝑟𝑒 𝒮 𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡𝑒𝑑 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛 𝑠𝑦𝑠𝑡𝑒𝑚.

Priority, loacation
1,0
(𝑞𝐼 , 𝑠𝐼 )

All the location are player 0
𝛿 𝑞 = Δ𝑞′ ∈ 𝑄
1,0
hence Player 0 can choose a finite (𝑞 , 𝑠 ) 𝑅 𝑚𝑎𝑛𝑦 𝑡𝑖𝑚𝑒𝑠
𝐼 1
1,0
path and therefore
(𝑞𝐼 , 𝑠3 )
1,0
the automaton Doesn’t accept
(𝑞𝐼 , 𝑠2 )
any 𝒮
𝑽𝟎
𝑽𝟏
𝛿 𝑞 =0
𝛿 𝑞 =1
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 = 𝑞1 ∨ 𝑞2
𝛿 𝑞 = 𝑞1 ∧ 𝑞2
𝛿 𝑞 = Δ𝑞′
𝛿 𝑞 = ∎𝑞′
Current disadvantages



While 𝒮 may be finite our parity game 𝒢 may be infinite (for example 𝛿 𝑞𝐼
= 𝑞𝐼 )
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, 𝑞𝐼 , 𝛿, Ω , 𝒮 = 𝑆, 𝑠𝑖 , Let 𝑉 ⊆ 𝑄 × 𝑆 and E ⊆ 𝑉 × 𝑉 be the smallest graph
with 𝑞𝐼 , 𝑠𝐼 ∈ 𝑉 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 ∀ 𝑞, 𝑠 ∈ 𝑉 𝑤𝑒 ℎ𝑎𝑣𝑒:

𝛿 𝑞 = 𝑞′ → 𝑞′ , 𝑠 ∈ [𝑉] 𝑎𝑛𝑑 𝑞, 𝑠 𝑞′ , 𝑠 ∈ [𝐸]

𝛿 𝑞 = 𝑞1 ∧ 𝑞2 𝑜𝑟 𝑞1 ∨ 𝑞2 → 𝑞1 , 𝑠 , 𝑞2 , 𝑠 ∈ [𝑉] and 𝑞, 𝑠 , 𝑞1 , 𝑠 ,

𝛿 𝑞 = ∎𝑞′ 𝑜𝑟 ∆𝑞′ → 𝑞 ′ , 𝑠 ′ ∈ 𝑉 𝑎𝑛𝑑

We split 𝑉 𝑖𝑛𝑡𝑜 𝑉0 𝑎𝑛𝑑 𝑉1 𝑎𝑠 𝑒𝑎𝑟𝑙𝑖𝑒𝑟 𝑎𝑛𝑑 𝑡ℎ𝑒 Ω 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒.

Previous definition:

We will define 𝒜 on 𝒮 behavior as the least language 𝑉 ⊆ 𝑄 × 𝑆
∀𝑣 𝑞, 𝑠 ∈ 𝑉:
𝑞, 𝑠 , 𝑞′ , 𝑠 ′
𝑞, 𝑠 𝑞2 , 𝑠
∈ [𝐸]
∈ 𝐸 ∀𝑠 ′ ∈ sR

𝛿 𝑞 = 𝑞 ′ → 𝑣 𝑞, 𝑠 𝑞 ′ , 𝑠 ∈ 𝑉

𝛿 𝑞 = 𝑞1 ∧ 𝑞2 𝑜𝑟 𝑞1 ∨ 𝑞2 → 𝑣 𝑞, 𝑠 𝑞1 , 𝑠 ∈ 𝑉 and 𝑣 𝑞, 𝑠 𝑞2 , 𝑠 ∈ 𝑉

𝛿 𝑞 = ∎𝑞′ 𝑜𝑟 ∆𝑞′ → 𝑣 𝑞, 𝑠 𝑞′, 𝑠′ ∈ 𝑉
∗
such that 𝑞𝐼 , 𝑠𝐼 ∈ 𝑉 and
Alternative Formal Definition for the
Arena

Our new Arena: 𝒢 =

Definition of the function []: 𝑄 × S + → (𝑄 × 𝑆) for each word 𝑄 × S
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
𝑉0 , 𝑉1 , 𝐸 , Ω, 𝑞𝐼 , 𝑠𝐼 .
+
it
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 ∶ [𝑉0 ] → [𝑉] be a winning strategy for Player 0 in 𝒢′

We define a mapping 𝑓0 ∶ 𝑉0 → 𝑉 𝑏𝑦 𝑓0 𝑣 ≜ 𝑣𝑓0′ 𝑣 .

Let 𝜋 be a play in 𝒢 which is consistent with 𝑓0 .

Then, [𝜋] is consistent with 𝑓′0 , and thus, [𝜋] and 𝜋 are won by Player 0.

Consequently, 𝑓0 is a winning strategy for Player 0 in 𝒢.

Clearly, we can apply a symmetric argument if 𝑓1′ ∶ [𝑉1 ] → [𝑉] is a winning
strategy for Player 1 in 𝒢′.
Complex Transition Conditions

Recall:


transition conditions 𝑇𝐶 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 𝑞1 , 𝑞2 ∈ 𝑄, 𝑞1 ∧ 𝑞2 𝑎𝑛𝑑 𝑞1 ∨ 𝑞2 are transition conditions.
Note that this definition does not allow transition conditions like 𝑞1 ∧ ∎𝑞2 𝑜𝑟 p∧ q
for p ∈ P and q ∈ Q.

Example of such 𝜑: “Change the inner state to 𝑞1 if 𝑝 is true, otherwise
change the inner state to 𝑞2 ”

Formally: 𝜑 = (𝑞1 ∧ 𝑝) ∨ (𝑞2 ∧ ￢𝑝)

So we will use some new states: 𝑞𝜑 , 𝑞𝑞1 ∧𝑝 , 𝑞𝑞1 ∧¬𝑝 , 𝑞𝑝 , 𝑞¬𝑝
Complex Transition Conditions


(𝑞𝐼 , 𝑠𝐼 )
𝛿 𝑞𝜑 ≜ 𝑞𝑞1 ∧𝑝 ∨ 𝑞𝑞2 ∧¬𝑝
𝛿 𝑞𝑞1 ∧𝑝 ≜ 𝑞1 ∧ 𝑞𝑝

𝛿 𝑞𝑞2 ∧¬𝑝 ≜ 𝑞2 ∧ 𝑞¬𝑝

𝛿 𝑞𝑝 ≜ 𝑝

𝛿 𝑞¬𝑝 ≜ ¬𝑝

Example:

We set: 𝛿 𝑞1 = 𝛿 𝑞2 = ∎𝑞𝜑 ; Ω 𝑞1 = 2; Ω 𝑄\q1 = 1; 𝑞𝐼 = 𝑞𝜑

Priority, loacation
1,0
𝛿 𝑞𝜑 = 𝑞𝑞1 ∧𝑝 ∨ 𝑞𝑞2 ∧¬𝑝
1,1
1,1
(𝑞𝑞1 ∧𝑝 , 𝑠𝐼 )
(𝑞𝑞2 ∧¬𝑝 , 𝑠𝐼 )
1,?
2,1
(𝑞𝑝 , 𝑠𝐼 )
(𝑞1 , 𝑠𝐼 )
1,0
1,0
(𝑞𝜑 , 𝑠′1 ) (𝑞𝜑 , 𝑠′2 )
 The automaton accepts some pointed transition system
𝒮 iff every infinite path starting from 𝑠𝐼 contains infinitely many states in which
p is true. There will be infinite amount of successful
𝑞1 , 𝑠 ′ 𝑖𝑛𝑠𝑡𝑎𝑛𝑐𝑒𝑠 𝑤𝑖𝑡ℎ 𝑒𝑣𝑒𝑛 𝑝𝑟𝑖𝑜𝑟𝑖𝑡𝑦.
(𝑞2 , 𝑠𝐼 ) (𝑞¬𝑝 , 𝑠𝐼 )
1,0
p
1,1
1,1
1,0
(𝑞𝜑 , 𝑠′1 ) (𝑞𝜑 , 𝑠′2 )
¬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, 𝑞𝐼 , 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
and in space 𝒪 𝑑 𝑄 𝑆 log 𝑄 𝑆

The word problem is in 𝑈𝑃 ∩ 𝑐𝑜 − 𝑈𝑃
𝑅 +1
𝑄 𝑆 +1
𝑑
2
𝑑
2
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
𝑄 𝑆 +1
𝑅 +1
𝑑
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
2+ 𝑑 2

Roughly we get 𝒪( 𝑆

With 𝑑 = 2 we get 𝑆 3 , with 𝑑 = 20 we get |𝑆|^12 – not so nice
)
𝑑
2
)
2
𝑑
2
The Word problem

There is an algorithm which computes in time: 𝒪 𝑑 𝑄
𝑅 +1
𝑄 𝑆 +1
𝑑
𝑑
2
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 𝑞𝐼

Every state in 𝑆 except 𝑠𝐼 has at least one predecessor (𝑠𝐼 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
For a fixed 𝑞:
𝑞,𝑠 ∈[𝑉] |
𝑞, 𝑠 𝐸 |
The Word problem



𝐸 =
𝑞,𝑠 ∈[𝑉] |
𝑞∈𝑄
𝑞, 𝑠 𝐸 |
For a fixed 𝑞:

If 𝛿(𝑞) ∈ {0, 1} 𝑜𝑟 𝛿(𝑞) ∈ {𝑝, ￢𝑝}, then (𝑞, 𝑠) has no successor:

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) = 𝑞1 ∧ 𝑞2 𝑜𝑟 𝛿(𝑞) = 𝑞1 ∨ 𝑞2 for some 𝑞1 , 𝑞2 ∈ 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)
𝑞,𝑠 ∈[𝑉] |
𝑞, 𝑠 𝐸 |=0
The Word problem

There is an algorithm which computes in time: 𝒪 𝑑 𝑄
𝑅 +1
𝑄 𝑆 +1
𝑑
𝑑
2
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, 𝑞𝐼 , 𝛿, Ω) be an alternating tree automaton. There is an
alternating tree automaton 𝒜 = (𝑄, 𝑞𝐼 , 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, 𝑞𝐼 , 𝛿, Ω) be an alternating tree automaton. There is an
alternating tree automaton 𝒜 = (𝑄, 𝑞𝐼 , 𝛿, Ω) such that 𝒜 accepts the
complement of the language of 𝒜.

Let 𝒮 be a pointed transition system and 𝒢 = ( 𝑉0 , 𝑉1 , 𝐸 , Ω, 𝑞𝐼 , 𝑠𝐼 ) 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 = 𝑉0 ∪ 𝑉1 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 = ( V1 ∪ 𝑉 ′ , 𝑉0 \V ′ , 𝐸 , Ω, 𝑞𝐼 , 𝑠𝐼 )

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 𝑓0 ∶ 𝑉0 → 𝑉 be a winning strategy for Player 0 in 𝒢. There is a unique
extension 𝑓′0 ∶ 𝑉1 ∪ 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 𝑓1 : 𝑉1 → 𝑉 be a winning strategy for Player 1 in 𝒢. Clearly, the
restriction of 𝑓1 𝑡𝑜 𝑉1 \ 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, 𝑞𝐼 , 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, 𝑞𝐼 , 𝛿, Ω) there is an
automaton 𝒜′ = (𝑄, 𝑞𝐼 , 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 → 𝑇𝐶 𝑄 𝑓𝑜𝑟 𝑞

The Emptiness Problem

Lemma: For every alternating tree automaton 𝒜 = (𝑄, 𝑞𝐼 , 𝛿, Ω) there is an
automaton 𝒜′ = (𝑄, 𝑞𝐼 , 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 = (𝑉0 , 𝑉1 , 𝐸), Ω, (𝑞𝐼 , 𝑠𝐼 ) be the parity game to determine whether
𝒜 accepts 𝒮.

Let 𝑉 ′ = {𝑣(𝑞, 𝑠) ∈ 𝑉1 | 𝛿(𝑞) = 𝑞′ ∧ 𝑞′ 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑞′ ∈ 𝑄}. The locations in 𝑉′ have
exactly one successor. The parity game 𝒢 ′ = V0 ∪ 𝑉 ′ , 𝑉1 \V ′ , 𝐸 , Ω, 𝑞𝐼 , 𝑠𝐼
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, 𝑞𝐼 , 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 = (𝑉𝜗 , 𝐸𝜗 ) where 𝑉𝜗 ⊆ 𝑄, 𝐸 ⊆ 𝑉𝜗 × 𝑉𝜗 and:

A tile with port is a tuple (𝜗, 𝑞) s.t 𝜗 = 𝐸𝜗 , 𝑉𝜗 𝑎𝑛𝑑 𝑞 ∈ 𝑉𝜗 ∩ 𝑄Δ
The Emptiness Problem

A tile with port is a tuple (𝜗, 𝑞) s.t 𝜗 = 𝐸𝜗 , 𝑉𝜗 𝑎𝑛𝑑 𝑞 ∈ 𝑉𝜗 ∩ 𝑄Δ

𝑄Δ ≜ {𝑞 ∈ Q|∃𝑞 ′ ∈ 𝑄: 𝛿 𝑞 = Δ𝑞′ }

𝑄∎ ≜ {𝑞 ∈ Q|∃𝑞 ′ ∈ 𝑄: 𝛿 𝑞 = ∎𝑞′ }

We also define 𝑞, 𝑉:

For subsets 𝑉 ⊆ 𝑄 𝑤𝑒 𝑑𝑒𝑓𝑖𝑛𝑒 𝑉 ≜ {𝑞|𝑞 ∈ V}
𝑞 ≜ 𝑞 ′ 𝑖𝑓 𝛿 𝑞 = Δ𝑞 ′ 𝑜𝑟 𝛿 𝑞 = ∎𝑞′

All tiles will be: Θ, all tiles with port will be: Θ𝑝

We will call a tile with port 𝜗0 = 𝑉𝜗0 , 𝐸𝜗0 , 𝑞0 and a tile 𝜗1 = (𝑉𝜗1 , 𝐸𝜗1 ) (it can
be a tile with port as well) concatenable iff 𝑞0 ∈ 𝑉𝜗1 𝑎𝑛𝑑 𝑉𝜗0 ∩ 𝑄∎ ⊆ 𝑉𝜗1

i.e: the port must be able to make delta\square transition into 𝑉𝜗1 state and
that the state intersection of 𝑉𝜗0 𝑎𝑛𝑑 𝑄∎ that make delta\square transition
must be contained in 𝑉𝜗1
The Emptiness Problem


Let 𝑔 = (𝜗1 , 𝑞1 ), (𝜗2 , 𝑞2 ), ・・・ ∈ Θ𝜔 be an infinite sequence of tiles with port
where (𝜗𝑖 , 𝑞𝑖 ) 𝑎𝑛𝑑 (𝜗𝑖+1 , 𝑞𝑖+1 ) are concatenable for every 𝑖 ∈ 𝜔. We define the
graph of 𝒈 in a usual way:

𝑉≜

𝐸≜
𝑖∈𝜔
𝑖 × 𝑉𝑖
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
4

Theorem: There is a deterministic parity 𝜔 − 𝑎𝑢𝑡𝑜𝑚𝑎𝑡𝑜𝑛 𝐶 with 2𝒪( 𝑄
states and priorities bounded by 𝒪( 𝑄 4 ) which accepts a sequence of
concatenable tiles 𝑔 ∈ Θ𝜔 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 (𝑞1 , 𝑖1 ), 𝑞2 , 𝑖2 be
two states of B, and let (𝑉, 𝐸, 𝑞) be a tile with port. There is a transition
𝑞1 , 𝑖1 , 𝑉, 𝐸, 𝑞 , 𝑞2 , 𝑖2 in B iff:
log 𝑄 )
The Emptiness Problem
non-deterministic parity ω-automaton
Where 𝑚 = 𝑄
2
we get: 2𝒪( 𝑄
4
log 𝑄 )
states and priorities bounded by 𝒪( 𝑄 4 )
done