cs419 lec6 lexical analysis using nfa
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Compilers
WELCOME TO A JOURNEY TO
CS419 Lecture 6
Scanning using Non-deterministic Finite automata (NFA) cont.
Cairo UniversityFCI
Dr. Hussien SharafComputer Science Departmentdr.sharaf@from-masr.com
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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.
Dr. Hussien M. Sharaf
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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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Dr. Hussien M. Sharaf
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
Dr. Hussien M. Sharaf
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
Dr. Hussien M. SharafLAMBDA TRANSITIONS
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a a
1q 3qa0q
2q a
input tape head does not move
Dr. Hussien M. SharafLAMBDA TRANSITIONS
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a a
1q 3qa0q
2q a
“accept”
String is acceptedaa
all input is consumed
Dr. Hussien M. SharafLAMBDA TRANSITIONS
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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
Dr. Hussien M. SharafLAMBDA TRANSITIONS
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a a
1q 3qa0q
2q a
(read head doesn’t move)
a
Dr. Hussien M. SharafLAMBDA TRANSITIONS
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a a
1q 3qa0q
2q a
“reject”
String is rejectedaaa
a
Input cannot be consumed
Automaton halts
Dr. Hussien M. SharafLAMBDA TRANSITIONS
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Language accepted: }{aaL
1q 3qa0q
2q a
Dr. Hussien M. SharafLAMBDA TRANSITIONS
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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
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