Transcript Document

CSCI 2670
Introduction to Theory of
Computing
September 6, 2005
Announcements
• Homework due next Tuesday (9/13)
– 1.7(d,h), 1.16b, 1.21b (show the GNFA
steps),1.19a, 1.46c, 1.55(c,f,j)
• Old text numbers 1.5(d,g), 1.12b, 1.16b, 1.14a,
1.23d, 1.38(same instructions on (0010*1*, ,
and Σ*)
• If you ask me a question on a
homework problem by email, please
email me the problem in question
• Quiz tomorrow
– Section 1.2 & Formal definition of RE
Agenda
• Last week
– Section 1.2
– NFA’s
– Regular expression intro (Section 1.3)
• Today
– Continue regular expressions
• This week
– Sections 1.3 and 1.4
RE inductive definition
R is a regular expression if R is
Abuse of
1. a for some a  
notation.
2. ε
These should
be sets!
3. 
4. R1  R2 where R1 and R2 are both
regular expressions
5. R1  R2 where R1 and R2 are both
regular expressions
6. (R1*) where R1 is a regular expression
RE’s and regular languages
Theorem: A language is regular if and
only if some regular expression
describes it.
– i.e., every regular expression has a
corresponding DFA and vice versa
RE’s and regular languages
Lemma: If a language is described by a
regular expression, then it is regular.
– find an NFA corresponding to any
regular expression
– use inductive definition of RE’s
1. R=a for some a
q1
a
q2
N = {{q1,q2},,,q1,{q2}} where (q1,a)={q2}
and (r,x)= whenever r=q2 or x≠a
2. R=ε
q1
N = {{q1},,,q1,{q1}} where (q1,x)= for
all x
3. R=
q1
N = {{q1},,,q1,} where (q1,x)= for
all x
Remaining constructions
•
•
•
•
R = R1R2
R = R1R2
R = R1*
These were all shown to be regular
operators
– We know we can construct NFA’s for R
provided they exist for R1 and R2
Example
• R = 1
– R = (01)1
R1 = 0
0
R2 = 1
1
R3 = 01
0
ε
1
ε
R = Σ1
ε
ε
0
1
ε
ε
1
Example2
• R = 1(0ε)*
R1 = 1
1
R2 = 0ε
0
ε
ε
ε
ε
R = 1(0ε)*
1
0,1
R3 = *
ε
ε
ε
0
ε
ε
ε
0,1
ε
Equivalence of RE’s and DFA’s
• We have seen that every RE has a
corresponding NFA
– Therefore, every RE has a corresponding
DFA
– I.e, every RE describes a regular language
• We need to show that every regular
language can be described by a RE
• Begin by converting all DFA’s into
GNFA’s
– Generalized Non-deterministic Finite
Automata
GNFA’s
•
A GNFA is an NFA with the
following properties:
1. The start state has transition arrows
going to every other state, but no
arrows coming in from any other state
2. There is exactly one accept state and
there is an arrow from every other
state to this state, but no arrows to
any other state from the accept state
3. The start state is not the accept state
GNFA’s (continued)
4. Except for the start and accept
states, one arrow goes from every
state to every other state and also
from each state to itself
5. Instead of being labeled with symbols
from the alphabet, transitions are
labeled with regular expressions
Example GNFA
01
 0
1


0
10

Equivalence of DFA’s and RE’s
• First show every DFA can be
converted into a GNFA that accepts
the same language
• Then show that any GNFA has a
corresponding RE that accepts the
same language
Converting a DFA into a GNFA
• Add two new states
– New start state with an ε jump to the
original DFA’s start state
– New accept state with an ε jump from
each of the original DFA’s accept states
• This new state will be the only accept state
• All transition labels with multiple
labels are relabeled with the union of
the previous labels
• All pairs of states without transitions
get a transition labeled 
Converting a DFA to a GNFA
0
1
qs
ε
q1
q2
1
0
q3
ε
0,1
q4
0,1
Add two new states
qt
Converting a DFA to a GNFA
0
1
qs
ε
q1
q2
1
0
q3
ε
0,1
01
q4
qt
0,1
01
• All transition labels with multiple
labels are relabeled with the union of
the previous labels
Converting a DFA to a GNFA
0
1
qs
ε
q1
q2
1
0
q3
ε
01
q4
qt
01
• All pairs of states without transitions
get a transition labeled 
Converting a DFA to a GNFA
0
1
qs
ε
q1
q2
1
0
q3
ε
01
q4
01
• The resulting state diagram is a
GNFA
– All GNFA properties are satisfied
qt
Converting a DFA to a GNFA
0
1
qs
ε
q1
q2
1
0
q3
ε
01
q4
qt
01
• No step changed the strings accepted
by the machine
Converting a GNFA to a RE
• If the GNFA has two states, then the
label connecting the states is the RE
• Otherwise, remove one state at a
time without changing the language
accepted by the machine until the
GNFA has two states
Removing one state from a GNFA
a12
a12a13a33*a32
q2
q1
q 1’
a32
a13
q3
a33
q2 ’