MA/CSSE 474Theory of Computation

Regular Expressions Intro

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Still more language ambiguity!

Regular Languages

Regular Language

Regular Expression

Finite State Machine

Describes

Accepts

Regular Expressions

The regular expressions over an alphabet are the strings that can be obtained as follows:

1. is a regular expression.2. is a regular expression.3. Every element of is a regular expression.4. If , are regular expressions, then so is .5. If , are regular expressions, then so is .6. If is a regular expression, then so is *.7. is a regular expression, then so is +.8. If is a regular expression, then so is ().

#7 is here for convenience only (syntactic sugar); many authors do not include + in the list of r.e. builders.

Regular Expression Examples

If = {a, b}, the following are regular expressions:

a(a b)*(abba )+(a bab)

Regular Expressions Define Languages

Define L, a semantic interpretation function for regular expressions (Let and be arbitrary regular expressions over alphabet ).

1. L() = .2. L() = {}.3. If c , L(c) = {c}.4. L() = L() L(). 5. L( ) = L() L(). 6. L(*) = (L())*. 7. L(+) = L(*) = L() (L())*. If L() is equal to , then

L(+) is also equal to . Otherwise L(+) is the language that is formed by concatenating together one or more strings drawn from L().

8. L(()) = L().

The Role of the Rules

• Rules 1, 3, 4, 5, and 6 give the language its power to define sets.

• Rule 8 has as its only role grouping other operators. • Rules 2 and 7 appear to add functionality to the

regular expression language, but they don’t.

2. is a regular expression.

7. is a regular expression, then so is +.

Operator Precedence in Regular Expressions

Regular ArithmeticExpressions Expressions

Highest Kleene star and + exponentiation

concatenation multiplication

Lowest union addition

a b* c d* x y2 + i j2

Analyzing a Regular Expression

L((a b)*b) = L((a b)*) L(b)

= (L((a b)))* L(b)

= (L(a) L(b))* L(b)

= ({a} {b})* {b}

= {a, b}* {b}.

From English to reg exps

L = {w {a, b}*: |w| is even}

L = {w {0, 1}*: w is a binary representation of a multiple of 4}

L = {w {a, b}*: w contains an odd number of a’s}

The Details Matter

a* b* (a b)*

(ab)* a*b*

More Regular Expression Examples

L ( (aa*) ) =

L ( (a )* ) =

L = {w {a, b}*: there is no more than one b in w}

L = {w {a, b}* : no two consecutive letters in w are the same}

The Details Matter

L1 = {w {a, b}* : every a is immediately followed a b}

A regular expression for L1:

A FSM for L1:

L2 = {w {a, b}* : every a has a matching b somewhere}

A regular expression for L2:

A FSM for L2:

Kleene’s Theorem Finite state machines and regular expressions define the same class of languages.

To prove this, we must show:

Theorem: Any language that can be defined by a regular expression can be accepted by some FSM and so is regular.

Theorem: Every regular language (i.e., every language that can be accepted by some DFSM) can be defined with a regular expression.

For Every Regular Expression There is a Corresponding FSM

We’ll show this by construction. An FSM for:

:

A single element c of :

:

Union

If is the regular expression and if both L() and L() are regular:

Concatenation

If is the regular expression and if both L() and L() are regular:

Kleene Star

If is the regular expression * and if L() is regular:

An Example

(b ab)*

An FSM for b An FSM for a An FSM for b

An FSM for ab:

An Example

(b ab)*

An FSM for (b ab):

An Example

(b ab)*

An FSM for (b ab)*:

For Every FSM There is a Corresponding Regular Expression

• We’ll show this by construction. The construction is different than the textbook's.

• Let M = ({q1, …, qn}, , , q1, A) be a DFSM.Define Rijk to be the set of all strings x * such that

• (qi,x) |-M (qj, ), and

• if (qi,y) |-M (q𝓁, ), for any prefix y of x (except y= and y=x), then 𝓁 k

• That is, Rijk is the set of all strings that take us from qi to qj

without passing through any intermediate states numbered higher than k.• In this case, "passing through" means both entering

and leaving.• Note that either i or j (or both) may be greater than k.

*

*

Example: Rijk

• Rijk is the set of all strings that take us from qi to qj without

passing through any intermediate states numbered higher than k.• In this case, "passing through" means both entering

and leaving.• Note that either i or j (or both) may be greater than k.

R000

R010

R011

R021

R022

R232

R233

DFAReg. Exp. construction• Rijk is the set of all strings that take M from qi to qj

without passing through any intermediate states numbered higher than k.

• Examples: Rijn is

• Also note that L(M) is the union of R1jn over all qj in A.

• We will show that for all i,j{1, …, n} and all k {0, …, n}, Rijk is defined by a regular expression.– We already know that the union of languages defined by

reg. exps. is defined by a reg. exp.

DFAReg. Exp. continued• Rijk is the set of all strings that take M from qi to qj without passing

through any intermediate states numbered higher than k.

It can be computed recursively:• Base cases (k = 0):

– If i j, Rij0 = {a : (qi, a) = qj}

– If i = j, Rii0 = {a : (qi, a) = qi} {}

• Recursive case (k > 0): Rijk is Rijk-1 Rikk-1(Rkkk-1)*Rkjk-1

• We show by induction that each Rijk is defined by some regular expression rijk.

DFAReg. Exp. Proof pt. 1

• Base case definition (k = 0):– If i j, Rij0 = {a : (qi, a) = qj}– If i = j, Rii0 = {a : (qi, a) = qi} {}

• Base case proof:Rij0 is a finite set of symbols, each of which is either or a single symbol from . So Rij0 can be defined by the reg. exp. rij0 = a1a2…ap (or a1a2…ap if i=j),where {a1, a2, …,ap} is the set of all symbols a such that (qi, a) = qj.

• Note that if M has no direct transitions from qi to qj, then rij0 is (it is if i=j and no "loop" on that state).

DFAReg. Exp. Proof pt. 2• Recursive definition (k > 0):

Rijk is Rijk-1 Rikk-1(Rkkk-1)*Rkjk-1

• Induction hypothesis: For each 𝓁 and 𝓂, there is a regular expression r𝓁𝓂k-1 such that L(r𝓁𝓂k-1 )= R𝓁𝓂k-1.

• Induction step. By the recursive parts of the definition of regular expressions and the languages they define, and by the above recursive defintion of Rijk : Rijk = L(rijk-1 rikk-1(rkkk-1)*rkjk-1)

DFAReg. Exp. Proof pt. 3• We showed by induction that each Rijk is

defined by some regular expression rijk.

• In particular, for all qjA, there is a regular expression r1jn that defines R1jn.

• Then L(M) = L(r1j1n … r1jpn ),

where A = {qj1, …, qjp

}

An ExampleStart q1 q2 q3

0

01

1

0,1

k=0 k=1 k=2r11k (00)*

r12k 0 0 0(00)*

r13k 1 1 0*1

r21k 0 0 0(00)*

r22k 00 (00)*

r23k 1 1 01 0*1

r31k (0 1)(00)*0

r32k 0 1 0 1 (0 1)(00)*

r33k (0 1)0*1