Regular expressions-Theory of computation

Theory of Computation - RegularRegular
ExpressionsExpressions
Notation to specify a language
Declarative
Sort of like a programming language.
Fundamental in some languages like perl and
applications like grep or lex
Capable of describing the same thing as a NFA
The two are actually equivalent, so RE =
NFA = DFA
We can define an algebra for regular
expressions

Algebra for LanguagesAlgebra for Languages
 Previously we discussed these operators:
 Union (+)
 Concatenation (.)
 Kleene Star/Closure (*)

Definition of a Regular
Expression
 R is a regular expression if it is:
1. a for some a in the alphabet ∑, standing for the
language {a}
2. , standing for the language { }ε ε
3. Ø, standing for the empty language
4. R1+R2where R1 and R2 are regular expressions, and
+ signifies union (sometimes | is used)
5. R1R2 where R1 and R2 are regular expressions and
this signifies concatenation
6. R* where R is a regular expression and signifies
closure
7. (R) where R is a regular expression, then a
parenthesized R is also a regular expression

RE Examples:RE Examples:
 L(001) = {001}
 L(0+10*) = { 0, 1, 10, 100, 1000, 10000, … }
 L(0*10*) = {1, 01, 10, 010, 0010, …} i.e. {w | w has
exactly a single 1}
 L(∑∑)* = {w | w is a string of even length}
 L((0(0+1))*) = { , 00, 01, 0000, 0001, 0100, 0101, …}ε
 L((0+ )(1+ε ε)) = { , 0, 1, 01}ε
 L(1Ø) = Ø ; concatenating the empty set to any set
yields the empty set.
 R = Rε
 R+Ø = R

Equivalence of FA andEquivalence of FA and
RERE
 Finite Automata and Regular Expressions are
equivalent. To show this:
Show we can express a DFA as an equivalent
RE
Show we can express a RE as an NFA.

Turning a DFA into a RETurning a DFA into a RE
Theorem: If L=L(A) for some DFA , then there
is a regular expression R such that L=L(R).
State Elimination
We’ll see how to do this next, easier than
inductive construction, there is no
exponential number of expressions

DFA to RE: StateDFA to RE: State
EliminationElimination
Eliminates states of the automaton and
replaces the edges with regular expressions
that includes the behavior of the eliminated
states.
Eventually we get down to the situation
with just a start and final node, and this is
easy to express as a RE

DFA to RE via StateDFA to RE via State
Elimination (1)Elimination (1)
1. Starting with intermediate states and then
moving to accepting states, apply the state
elimination process to produce an
equivalent automaton with regular
expression labels on the edges.
• The result will be a one or two state
automaton with a start state and
accepting state.

DFA to RE StateDFA to RE State
Elimination (2):Elimination (2):
2. If the two states are different, we will have an
automaton that looks like the following:
Start
S
R
T
U
We can describe this automaton as:
(R+SU*T)*SU*

DFA to RE StateDFA to RE State
Elimination (3)Elimination (3)
3. If the start state is also an accepting state, then
we must also perform a state elimination from
the original automaton that gets rid of every
state but the start state. This leaves the
following:
Start
R
We can describe this automaton as simply R*.

DFA to RE State EliminationDFA to RE State Elimination
(4)(4)
4. If there are n accepting states, we must repeat
the above steps for each accepting states to get
n different regular expressions, R1, R2, … Rn.
For each repeat we turn any other accepting
state to non-accepting. The desired regular
expression for the automaton is then the union
of each of the n regular expressions: R1∪ R2…
∪ RN

DFARE Example
Convert the following to a RE
First convert the edges to RE’s:
3Start 1 2
1 1
0
0
0,1
3Start 1 2
1 1
0
0
0+1

DFADFA  RE Example (2)RE Example (2)
Eliminate State 1:
3Start 1 2
1 1
0
0
0+1
3Start 2
11
0+10 0+1
Answer: (0+10)*11(0+1)*

Converting a RE to anConverting a RE to an
AutomataAutomata
We have shown we can convert an automata to a
RE. To show equivalence we must also go the
other direction, convert a RE to an automaton.
We can do this easiest by converting a RE to an
NFA
Inductive construction
Start with a simple basis, use that to build
more complex parts of the NFA

RE toRE to NFA:NFA:
Basis:
R=a
R=ε
a
ε
R=Ø

R=S+T
S
T
ε
ε
ε
ε
R=ST
S T
ε
R=S*
S
ε
ε
ε
ε

RE to NFA Example
 Convert R= (ab+a)* to an NFA
We proceed in stages, starting from simple elements
and working our way up
a
a
b
b
ab
a bε

RE toRE to NFANFA Example (2):Example (2):
ab+a
a bε
a
ε
ε
ε
ε
(ab+a)*
a bε
a
ε
ε
ε
ε
εε
ε
ε

Algebraic Laws for RE’s
Just like we have an algebra for arithmetic, we
also have an algebra for regular expressions.
While there are some similarities to arithmetic
algebra, it is a bit different with regular
expressions.

Algebra for RE’s:Algebra for RE’s:
Commutative law for union:
L + M = M + L
Associative law for union:
(L + M) + N = L + (M + N)
Associative law for concatenation:
(LM)N = L(MN)
Note that there is no commutative law for
concatenation, i.e. LM ≠ ML

Algebra for RE’s (2):Algebra for RE’s (2):
The identity for union is:
L + Ø = Ø + L = L
The identity for concatenation is:
L = L = Lε ε
The annihilator for concatenation is:
ØL = LØ = Ø
Left distributive law:
L(M + N) = LM + LN
Right distributive law:
(M + N)L = LM + LN
Idempotent law:
L + L = L

Laws Involving Closure:Laws Involving Closure:
(L*)* = L*
i.e. closing an already closed expression does
not change the language
Ø* = ε
 *ε = ε
L+
= LL* = L*L
L* = L+
+ ε

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