Cs419 lec6 lexical analysis using nfa

Compilers

WELCOME TO A JOURNEY TO

CS419 Lecture 6

Scanning using Non-deterministic Finite automata (NFA) cont.

Cairo UniversityFCI

Dr. Hussien SharafComputer Science [email protected]

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NON DETERMINISTIC FINITE AUTOMATA NFA There is a fixed number of states but we can be in multiple

states at one time.

NFA = “a 5-tuple “ (Q, Σ, , q0, F)Q A finite set of statesΣ A finite input alphabetq0 The initial/starting state, q0 is in QF A set of final/accepting states, which is a subset of Qδ A transition function, which is a total function from Q x Σ to 2Q , this function:

Takes a state and input symbol as arguments. Returns a set of states instead a single state as in

DFA.δ: (Q x Σ) –> 2Q -2Q is the power set of Q, the set of

all subsets of Q δ(q,s) is a function from Q x S to 2Q (but not to Q)

Dr. Hussien M. Sharaf

NFA

A finite automaton is deterministic if It has no edges/transitions labeled with

epsilon/lamda. For each state and for each symbol in

the alphabet, there is exactly one edge labeled with that symbol.

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1q 2q

3q

a

a

a

0q

}{aAlphabet =

NFA NFA travels all possible paths, and so it remains in many states

at once. As long as at least one of the paths results in an accepting state, the NFA accepts the input.


5

1q 2q

3q

a

a

a

0q

Two choices

}{aAlphabet =

NFA


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No transition

1q 2q

3q

a

a

a

0q

Two choicesNo transition

}{aAlphabet =

NFA


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An NFA accepts a string:if there is a computation of the NFAthat accepts the string

i.e., all the input string is processed and the automaton is in an accepting state

NFA


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a a

0q

1q 2q

3q

a

a

Acceptance Example 1

a

NFA


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a a

0q

1q 2q

3q

a

a

a

First ChoiceNFA


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a a

0q

1q 2q

3q

a

a

a “accept”

First Choice

All input is consumed

NFA


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a a

0q

1q 2q

3q

a

a

Second Choice

a

NFA


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a a

0q

1q 2qa

a

a

3q

Second Choice

“reject”

Input cannot be consumed

Automaton Halts

NFA


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aa is accepted by the NFA:

0q

1q 2q

3q

a

a

a

“accept”

0q

1q 2qa

a

a

3q “reject”

because this computationaccepts aa

this computationis ignored

NFA


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An NFA rejects a string:if there is no computation of the NFAthat accepts the string.

• All the input is consumed and the automaton is in a non final state

• The input cannot be consumed

OR

For each computation:

NFA

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a is rejected by the NFA:

0q

1q 2qa

a

a

3q “reject”

0q

1q 2qa

a

a

3q

“reject”

All possible computations lead to rejection

NFA


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aaais rejected by the NFA:

0q

1q 2q

3q

a

a

a

“reject”

0q

1q 2qa

a

a

3q “reject”

All possible computations lead to rejection

NFA


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1q 3qa0q

2q a


LAMBDA TRANSITIONS

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a a

1q 3qa0q

2q a

Dr. Hussien M. SharafLAMBDA TRANSITIONS

Acceptance Example 2

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a a

1q 3qa0q

2q a


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a a

1q 3qa0q

2q a

input tape head does not move


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a a

1q 3qa0q

2q a

“accept”

String is acceptedaa

all input is consumed


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a a

1q 3qa0q

2q a

Rejection Example 3

a

Dr. Hussien M. SharafDr. Hussien M. SharafLAMBDA TRANSITIONS

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a a

1q 3qa0q

2q a

a


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a a

1q 3qa0q

2q a

(read head doesn’t move)

a


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a a

1q 3qa0q

2q a

“reject”

String is rejectedaaa

a

Input cannot be consumed

Automaton halts


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Language accepted: }{aaL

1q 3qa0q

2q a


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Example 4

0q 1q 2qa b

3q


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a b

0q 1q 2qa b

3q


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0q 2qa b

3q

a b

1q


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a b

0q 1qa b

3q2q

“accept”


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0qa b

a b

Another String

a b

1q 2q 3q


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0qa b

a b a b

1q 2q 3q


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0qa b

a b a b

1q 2q 3q


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0qa b

a b a b

1q 2q 3q


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0qa b

a b a b

1q 2q 3q


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0qa b

a b a b

1q 2q 3q


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a b a b

0qa b

1q 2q 3q

“accept”


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ab

ababababababL ...,,,

Language accepted

0q 1q 2qa b

3q


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EXAMPLE 5

0q 1q 2q0

11,0


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{ }{ }*10=

...,101010,1010,10,λ=)(ML

0q 1q 2q0

11,0

Language accepted

(redundant state)


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DETERMINISTIC AND NONDETERMINISTIC AUTOMATA

Deterministic Finite Automata (DFA) One transition per input per state No -moves

Nondeterministic Finite Automata (NFA) Can have multiple transitions for one input

in a given state Can have -moves

THANK YOU

Dr. Hussien M. Sharaf 42

Cs419 lec6 lexical analysis using nfa

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