Compilers

WELCOME TO A JOURNEY TO

CS419 Lecture 5

Scanning using deterministic Finite automata (DFA)

Cairo UniversityFCI

Dr. Hussien SharafComputer Science Departmentdr.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.

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EXAMPLE FOR METHOD 1

The position in the code to maintain the state

Implicitly.

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CODE 1

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EXAMPLE FOR METHOD 2

A variable to maintain the current state.

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CODE 2

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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→Qq0 Q is the start state.

F Q is the set of final/accepting states.

DETERMINISTIC FINITE AUTOMATA DFA

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: Q x Σ→Q Maps from domain of (states, letters)

to range of states.

TRANSITION FUNCTION

Q x Σ

Q

(q0, a)(q2, b)(q1, b)

q1

q2

q3

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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)

Q x QΣ

(q0, q1)(q2, q3)(q1, q3)

a

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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

aS2

a b+S4

S5

bb a a,b

a bS1 S2 S5

S2 S3 S5

S3 S5 S4

S4 S5 S5

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a, b

EXAMPLE1.2 Build a DFA that accepts only aab DFA that is not well defined.S1-

S3

aS2

a b+S4

a bS1 S2 ?

S2 S3 ?

S3 ? ?

S4 ? ?Dr. Hussien M. Sharaf 12

EX2

(a+b)*

a, b

±

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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

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EX3

All words with even count of letters.((a+b)(a+b))*

1±

2

a, b

a, b

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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

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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

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ASSIGNMENT2

Compilers_DFA_Sheet_2.pdf Deadline is 7 March-2013

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EX5

All words that start with triple letter(aaa+bbb)(a+b)*

1-

2a 3

a,b

4b 5b

6+

b

a a

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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

5b

a

b

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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

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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 =~letterOther# = ~(letter|digit)

— letter

digit

letter

error

other

Other#

any

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EX9 COMPILERS CONSTRUCTION PAGE 52

Integer constants

—

digit

error

- digit

digit

other

other

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EX10 COMPILERS CONSTRUCTION PAGE 52

Decimal/floating constants

— +

digit

error

- digit

digit

other

other

+digit.

digit

other

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THANK YOU

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