Cs419 lec5 lexical analysis using dfa

WELCOME TO A
JOURNEY TO
CS419
Dr. Hussien Sharaf
Computer Science Department
dr.sharaf@from-masr.com

FINITE AUTOMATA FA
 Its goal is to act as a recognizer for specific
language/pattern.
 Any problem can be presented in form of
decidable problem that can be answered
by Yes/No.
 Hence FA (machine with limited memory)
can solve any problem.
 FA used to be hard wired devices like
controller of a lift.
Dr. Hussien M. Sharaf 2

IMPLEMENTATION OF FINITE
AUTOMATA IN CODE
 Possible ways:
1. Use the position in the code (nested within
tests) to maintain the state implicitly.
2. Use a variable to maintain the current state
and write transitions as doubly nested case
statements inside a loop, where the first
case statement tests current state and the
nested second level tests the input
character, given the state or vice versa.

EXAMPLE FOR METHOD 1
The position in the code to maintain the state
Implicitly.

CODE 1

EXAMPLE FOR METHOD 2
 A variable to maintain the current state.

CODE 2

 There is a fixed number of states and the DFA can
only be in one state at a time.
DFA = “a 5-tuple “ (Q, Σ, , q0, F)
Q: {q0, q1, q2, …} is set of states.
Σ: {a, b, …} set of alphabet.
(delta): A transition function, which is a total
function from Q x Σ to Q, this function:
 Takes a state and input symbol as arguments.
 Returns a single state.
: Q x Σ→Q
q0 Q is the start state.
F Q is the set of final/accepting states.
DETERMINISTIC FINITE AUTOMATA DFA

 : Q x Σ→Q
 Maps from domain of (states, letters) to
range of states.
TRANSITION FUNCTION 
(q0, a)
(q2, b)
(q1, b)
q1
q2
q3

 Lets name it Conditional link instead of transition
 λ: Q x Q → Σ
 Maps from domain of (states, states) to range
of letters.
RENAMING FUNCTION  TO λ (LAMDA)
Σ
(q0, q1)
(q2, q3)
(q1, q3)
a
b
c
Allows more than one link
from domain to codomain
Not recommended

EXAMPLE1.1
 Build an FA that accepts only aab
 L = {aab}
S1
-
S3
a
S2
a b
+
S4
S5
b
b
a
a,b
a b
S1 S2 S5
S2 S3 S5
S3 S5 S4
S4 S5 S5
a, b

EXAMPLE1.2
 Build a DFA that accepts only aab
 DFA that is not well defined.
S1
-
S3
a
S2
a b
+
S4
a b
S1 S2 ?
S2 S3 ?
S3 ? ?
S4 ? ?

EX2
 (a+b)*
a, b
±

FA ACCEPTING NOTHING
1. FA with no final states
a
-
a,b
b
2. FA with disconnected graph. Start state does not
have a path to the final state.
a
-
a,b
b
+
b

EX3
 All words with even count of letters.
((a+b)(a+b))*
1± 2
a, b
a, b

EX4.1
 All words that start with “a”
a(a+b)*
1-
2
b
a 3 + a,b
1-
2b
a 3 +
a,b
a,b
Does not accept all inputs

EX4.2
 All words that start with “a”
a(a+b)*
4+
1-
2b
a
3 +
a,b
a,b
a,b
Special accept state for string “a”, might give
better performance in hardware implementation

ASSIGNMENT2
 Compilers_DFA_Sheet_2.pdf
 Deadline is 7 March-2013

EX5
 All words that start with triple letter
(aaa+bbb)(a+b)*
1-
2a 3
a,b
4
b 5
b
6+
b
a a

EX6
{aa, ba, baba, aaaa, bbba, abba, aaabaa, …}
All words with even count of letters and ends with
“a”. (a+b)a ((a+b)a (b(a+b)a)* )*
-
a,b
+
a,b
5
b
a
b

EX7
{aa, ba, baba, aaaa, ab, bb, bababa, aaba, …}
All words with even count of letters having “a” in
an even position from the start, where the first
letter is letter number one. (a+b)a((a+b)a)*
-
a,b

EX8 COMPILERS CONSTRUCTION PAGE 50
 Consider the following FA:
•Accepts identifiers where
symbols are not acceptded with
error handling.
•An identifier must start with a
letter.
letter [a-zA-Z]
Other =~letter
Other# = ~(letter|digit)
—
letter
digit
letter
error
other
Other#
any

 Integer constants
—
digit
error
- digit
digit
other
other

 Decimal/floating constants
— +
digit
error
- digit
digit
other
other
+
digit.
digit
other

THANK YOU

Cs419 lec5 lexical analysis using dfa

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Cs419 lec5 lexical analysis using dfa