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Unit-5: ω-regular properties
- B. Srivathsan
Chennai Mathematical Institute
NPTEL-course July - November 2015
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Module 4: Simple properties of NBA
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Unit-5: -regular properties B. Srivathsan Chennai Mathematical - - PowerPoint PPT Presentation
Unit-5: -regular properties B. Srivathsan Chennai Mathematical - - PowerPoint PPT Presentation
Unit-5: -regular properties B. Srivathsan Chennai Mathematical Institute NPTEL-course July - November 2015 1 / 19 Module 4: Simple properties of NBA 2 / 19 Determinization Product construction Emptiness Complementation Union 3 / 19
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Determinization Product construction Emptiness Complementation Union
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Deterministic Büchi Automata
Words where b occurs infinitely often q0 q1 b a a b
◮ Single initial state ◮ From every state - on an alphabet, there is a unique transition
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Question: Can every NBA be converted to an equivalent DBA?
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(a + b)∗bω: a occurs only finitely often
q0 q1 b a,b b
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(a + b)∗bω: a occurs only finitely often
q0 q1 b a,b b
◮ Automaton has to guess the point from where only b occurs ◮ A deterministic Büchi automaton cannot make this guess
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(a + b)∗bω: a occurs only finitely often
q0 q1 b a,b b
◮ Automaton has to guess the point from where only b occurs ◮ A deterministic Büchi automaton cannot make this guess
The above language cannot be accepted by a DBA
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(a + b)∗bω: a occurs only finitely often
q0 q1 b a,b b
◮ Automaton has to guess the point from where only b occurs ◮ A deterministic Büchi automaton cannot make this guess
The above language cannot be accepted by a DBA
Theorem 4.50 (Page 190) of Principles of Model Checking, Baier and Katoen. MIT Press (2008)
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Determinization
DBA less powerful than NBA
Product construction Emptiness Complementation Union
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p0 p1 p2 a b a b a b q0 q1 b a a b Word (ab)ω is accepted by both automata
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p0 p1 p2 a b a b a b q0 q1 b a a b Word (ab)ω is accepted by both automata
Coming next: The synchronous product construction
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p0 p1 p2 a b a b a b q0 q1 b a a b
〈p0,q0〉 〈p1,q0〉 〈p2,q1〉 〈p0,q1〉 a b a b a b a b
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p0 p1 p2 a b a b a b q0 q1 b a a b
〈p0,q0〉 〈p1,q0〉 〈p2,q1〉 〈p0,q1〉 a b a b a b a b
〈p1,q1〉 is not present
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p0 p1 p2 a b a b a b q0 q1 b a a b
〈p0,q0〉 〈p1,q0〉 〈p2,q1〉 〈p0,q1〉 a b a b a b a b
〈p1,q1〉 is not present No accepting state!
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p0 p1 p2 a b a b a b q0 q1 b a a b
〈p0,q0〉 〈p1,q0〉 〈p2,q1〉 〈p0,q1〉 a b a b a b a b
〈p1,q1〉 is not present No accepting state!
But intersection of the two automata is not empty
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◮ Need to modify the product construction ◮ Track accepting states of both automata ◮ Ensure that both automata visit accepting states infinitely
- ften
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p0 p1 p2 a b a b a b q0 q1 b a a b
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p0 p1 p2 a b a b a b q0 q1 b a a b
p0 q0 1 p1 q0 1 p2 q1 1 p0 q1 1 p0 q0 2 p1 q0 2 p2 q1 2 p0 q1 2 p0 q0 3 p1 q0 3 p2 q1 3 p0 q1 3
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p0 p1 p2 a b a b a b q0 q1 b a a b
p0 q0 1 p1 q0 1 p2 q1 1 p0 q1 1 p0 q0 2 p1 q0 2 p2 q1 2 p0 q1 2 p0 q0 3 p1 q0 3 p2 q1 3 p0 q1 3 a b
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p0 p1 p2 a b a b a b q0 q1 b a a b
p0 q0 1 p1 q0 1 p2 q1 1 p0 q1 1 p0 q0 2 p1 q0 2 p2 q1 2 p0 q1 2 p0 q0 3 p1 q0 3 p2 q1 3 p0 q1 3 a b a b
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p0 p1 p2 a b a b a b q0 q1 b a a b
p0 q0 1 p1 q0 1 p2 q1 1 p0 q1 1 p0 q0 2 p1 q0 2 p2 q1 2 p0 q1 2 p0 q0 3 p1 q0 3 p2 q1 3 p0 q1 3 a b a b a b
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p0 p1 p2 a b a b a b q0 q1 b a a b
p0 q0 1 p1 q0 1 p2 q1 1 p0 q1 1 p0 q0 2 p1 q0 2 p2 q1 2 p0 q1 2 p0 q0 3 p1 q0 3 p2 q1 3 p0 q1 3 a b a b a b a b
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p0 p1 p2 a b a b a b q0 q1 b a a b
p0 q0 1 p1 q0 1 p2 q1 1 p0 q1 1 p0 q0 2 p1 q0 2 p2 q1 2 p0 q1 2 p0 q0 3 p1 q0 3 p2 q1 3 p0 q1 3 a b a b a b a b a b
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p0 p1 p2 a b a b a b q0 q1 b a a b
p0 q0 1 p1 q0 1 p2 q1 1 p0 q1 1 p0 q0 2 p1 q0 2 p2 q1 2 p0 q1 2 p0 q0 3 p1 q0 3 p2 q1 3 p0 q1 3 a b a b a b a b a b a b
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p0 p1 p2 a b a b a b q0 q1 b a a b
p0 q0 1 p1 q0 1 p2 q1 1 p0 q1 1 p0 q0 2 p1 q0 2 p2 q1 2 p0 q1 2 p0 q0 3 p1 q0 3 p2 q1 3 p0 q1 3 a b a b a b a b a b a b b a
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p0 p1 p2 a b a b a b q0 q1 b a a b
p0 q0 1 p1 q0 1 p2 q1 1 p0 q1 1 p0 q0 2 p1 q0 2 p2 q1 2 p0 q1 2 p0 q0 3 p1 q0 3 p2 q1 3 p0 q1 3 a b a b a b a b a b a b b a b a
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p0 p1 p2 a b a b a b q0 q1 b a a b
p0 q0 1 p1 q0 1 p2 q1 1 p0 q1 1 p1 q0 2 p2 q1 2 p0 q0 3 p2 q1 3 a b a b a b a b a b a b b a b a
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p0 p1 p2 a b a b a b q0 q1 b a a b
a
p0 q0 1 p1 q0 1 p2 q1 1 p0 q1 1 p1 q0 2 p2 q1 2 p0 q0 3 p2 q1 3 a b a b a b a b a b a b b a b a 12/19
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p0 p1 p2 a b a b a b q0 q1 b a a b
a
p0 q0 1 p1 q0 1 p2 q1 1 p0 q1 1 p1 q0 2 p2 q1 2 p0 q0 3 p2 q1 3 a b a b a b a b a b a b b a b a
Word is accepted by product ↔ it is accepted by both component automata
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p0 p1 b a b a q0 b
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p0 p1 b a b a q0 b
p0 q0 1 p1 q0 1 p1 q0 2 p1 q0 3 b b b b
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Determinization
DBA less powerful than NBA
Product construction
Language intersection
Emptiness Complementation Union
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Determinization
DBA less powerful than NBA
Product construction
Language intersection
Emptiness
Next unit ...
Complementation Union
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p0 p1 b a b a Language: b occurs infinitely often
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p0 p1 b a b a Language: b occurs infinitely often p0 p1 b a b a Language: a occurs infinitely often
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p0 p1 b a b a Language: b occurs infinitely often p0 p1 b a b a Language: a occurs infinitely often Not the complement!
(ab)ω present in both
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Challenges
◮ Mere interchange of accepting states does not work ◮ Moreoever, NBA are more expressive than DBA
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Complementation
Theorem
Given an NBA , there is an algorithm to compute the NBA accepting the complement language ( )c
Proof out of scope of this course
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p0 p1 p2 a b a b a b q0 q1 b a a b
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p0 p1 p2 a b a b a b q0 q1 b a a b For union, take the disjoint union of the two NBA
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Determinization
DBA less powerful than NBA
Product construction
Language intersection
Emptiness
Next unit ...
Complementation Union
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