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cs3102: Theory of Computation Class 3: Nondeterminism

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PS1 is due now!

cs3102: Theory of ComputationClass 3: Nondeterminism

Spring 2010

University of Virginia

David Evans

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Deterministic machine:

at every step, there is only one choice

Nondeterministic machine:

at some steps, there may be more than one choice.

Omnipotent: Machine splits into a new machine for each choice; if any machine accepts, accept.

Omniscient: Whenever it has to make a choice, machine always guesses right.

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A deterministic finite automaton is a 5-tuple:

Qfinite set (“states”)

- finite set (“alphabet”)
- : Q Qtransition function
q0 Qstart state

F Qset of accepting states

How do we need to change this to support nondeterminism?

A nondeterministic finite automaton is a 5-tuple:

Qfinite set (“states”)

- finite set (“alphabet”)
- transition function
q0 Qstart state

F Qset of accepting states

Output of transition function is a set of states, not just one state.

Is an NFA more powerful than a DFA?

Languages that can be recognized by an A

Languages that can be recognized by a B

A and B are non-comparable.

Languages that can be recognized by an A

B is less powerful than A:

SomeAcan recognize every language a B can recognize.

There is some language that can be recognized by an A but not by any B.

Languages that can be recognized by a B

Languages that can be recognized by an A

Languages that can be recognized by a B

A and B are equal:

every language that can be recognized by an A can be recognized by some B

every language that can be recognized by a B can be recognized by some A

Easy part: is there any language a DFA can recognize that cannot be recognized by an NFA?

Hard part: is there any language an NFA can recognized that cannot be recognized by a DFA?

A nondeterministic finite automaton is a 5-tuple:

Qfinite set (“states”)

- finite set (“alphabet”)
- q0 Qstart state
F Qset of accepting states

Proof by construction:

Construct the NFA N corresponding to any DFA A = (Q, , , q0,F):

N = (Q, , ’, q0,F) where

q Q, ’(q, ) = reject

a ,’(q, a) = { (q, a) }

Easy part: is there any language a DFA can recognize that cannot be recognized by an NFA?

Hard part: is there any language an NFA can recognized that cannot be recognized by a DFA?

No: NFAs are at least as powerful as DFAs.

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How many states may be needed for the DFA corresponding to and NFA with N states?

How much bigger is the transition function for the DFA corresponding to and NFA with N states?

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This matches most email addresses. Matching the full spec of all possible email addresses is left as exercise.

Base: Language

a { a }

{ }

{ }

Induction:R1, R2 are regular expressions

R1 R2 L(R1) L(R2)

R1 R2 { xy | xL(R1) and yL(R2) }

R1*{ xn | xL(R1) and n ≥ 0 }

Is there a Regular Expression that describes the same language as any NFA?

Is there some NFA that describes the same language as any Regular Expression?

Base: Language

a { a }

{ }

{ }

Induction:R1, R2 are regular expressions

R1 R2 L(R1) L(R2)

R1 R2 { xy | xL(R1) and yL(R2) }

R1*{ xn | xL(R1) and n ≥ 0 }

Proof by construction.

Trivial to draw NFAs for

Each of these languages.

Base: Language

a { a }

{ }

{ }

Proof by construction.

Trivial to draw NFAs for

Each of these languages.

Induction:R1, R2 are regular expressions

R1 R2 L(R1) L(R2)

R1 R2 { xy | xL(R1) and yL(R2) }

R1*{ xn | xL(R1) and n ≥ 0 }

Proof by induction and construction.

Assume there are NFAs A1 and A2 that recognize the languages L(R1) and L(R2). Show how to construct an NFA that recognizes the produced language for each part.

- PS2 will be posted by tomorrow
- Finish reading Chapter 1
- Thursday: how to prove a language is non-regular