1.2 Regular Expressions

1
1.2 Regular Expressions

• Regular Expression
• RE ? ?-NFA
• DFA ? RE

2
Regular Expressions

• A FA is a blueprint for constructing a machine
  recognizing a regular language.
• A regular expression is a user-friendly
  declarative way of describing a language.
• Example 0110
• Used in e.g.
• UNIX grep
• Perl programming language

3
Inductive Definition of RE

• R is a regular expression if R is
• Ø, the empty set
• ?, the empty string
• a, for some a ? S
• R1R2, where R1 and R2 are reg. exp.
• R1R2, where R1 and R2 are reg. exp.
• R1, where R1 is a regular expression

4
Language of a RE

• A regular expression R describes the language
  L(R)
• L(Ø) Ø
• L(?) ?
• L(a) a
• L(R1R2) L(R1)?L(R2)
• L(R1R2) L(R1)L(R2)
• L(R1) (L(R1))

union
concatenation
closure
5
Concatenation of Languages

• If L1 and L2 are languages, we can define the
  concatenation
• L1L2 w wxy, x?L1, y?L2
• Examples
• ab, bacd, dc ? abcd, abdc, bacd, badc
• Øab ? Ø

6
Exponentiation of a Language

• Li is the language L concatenated with itself i
  times.
• Recursive definition
• Base L0 ?
• Induction Li1 LLi
• Example
• ab, ba2 ? abab, abba, baab, baba
• Ø0 ? ?
• Ø2 ? ?

7
Kleene Closure

• L ??i0Li
• L0 ? L1 ? L2 ?
• Examples
• ab, ba ? ?, ab, ba, abab, abba,
• Ø ? ?
• ? ? ?

8
Example

• Order of precedence for operators
• ( ) gt Closure gt Concatenation gt Union
• L w 0 and 1 alternate in w, S0, 1, how
  to describe L in a regular expression?
• (01)(10)0(10)1(01)
• Or equivalently, (?1)(01)(0?)

9
Exercises

• S 0, 1
• What is the language for
• 01
• What is the regular expression for
• w w has at least one 1
• w w starts and ends with same symbol
• w w ? 5
• w every 3rd position of w is 1
• L L1 ? L2 ?
• L? (means an optional L)

10
Equivalence of RE and FA
?-NFA ? NFA ? DFA
RE
11
1.2 Regular Expressions

• Regular Expression
• RE ? ?-NFA
• DFA ? RE

12
From RE to ?-NFA

• For every regular expression R, we can construct
  an ?-NFA A, s.t. L(A) L(R).
• Proof by structural induction
• Ø
• ?
• a

13
From RE to ?-NFA

• RS
• RS
• R

14
Example (01)1(01)
15
Example 1.31 (ab)aba
16
1.2 Regular Expressions

• Regular Expression
• RE ? ?-NFA
• DFA ? RE

17
From DFA to RE

• For every DFA A(Q, S, ?, q0, F), there is a
  regular expression R, s.t. L(R) L(A).
• Proof
• Let Q 1, 2, n and q0 1
• Let Ri, jk be a RE describing the set of all
  label paths in A from state i to state j going
  through the states 1,k only.
• Then the union of all R1, jn (j?F) will be the
  final RE describing the language of A

18
From DFA to RE

• We computes the Ri, jk inductively
• Initially (k0)
• Ri, i ?ab, whenever ?(i, a)?(i, b)i,
  etc.
• Ri, j ab, if i?j and ?(i, a)?(i, b)j,
  etc.
• Ri, j Ø, if i?j and there are no direct
  transitions from state i to state j.
• Induction
• Ri, jk Ri, jk-1 Ri, kk-1 (Rk, kk-1)
  Rk, jk-1

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From DFA to RE

• Proof that Ri, jk is correctly computed
• A path that goes through the states 1,k only
  either
• never goes through state k, in which case the
  path's label is in the language of Ri, jk-1
• or goes through k one or more times. In this
  case
• Ri, kk-1 contains the portion of the path that
  goes from i to k for the first time.
• (Rk, kk-1) contains the portion of the path
  (possibly empty) from the first k visit to the
  last.
• Rk, jk-1 contains the portion of the path from
  the last k visit to state j.

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Example

• L(A) x0y x?1, y?0, 1

1
0
0,1
2
1
R1, 10 ?1
R1, 20 0
R2, 10 Ø
R2, 20 ?01
k 0
21
Example Contd

• We need the following simplification rules
• (?R) R
• RRS RS
• ØR RØ Ø (Annihilation)
• ØR RØ R (Identity)

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Example Contd
R1, 10 ?1
R1, 20 0
R2, 10 Ø
R2, 20 ?01
k 1
R1, 11 ?1(?1)(?1)(?1) 1
R1, 21 0(?1)(?1)0 10
R2, 11 ØØ(?1)(?1) Ø
R2, 21 ?01Ø(?1)0 ?01
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Example Contd
R1, 11 1
R1, 21 10
R2, 11 Ø
R2, 21 ?01
k 2
R1, 12 1 10(?01)Ø 1
R1, 22 1010(?01)(?01) 10(01)
R2, 12 Ø(?01)(?01)Ø Ø
R2, 22 ?01(?01)(?01)(?01) 01
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Example Contd

• The final regular expression for A is
• R1, 22 10(01)

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Observation

• for k 1 to n
• for i 1 to n
• for j 1 to n
• Bi, j Ri, j Ri, k (Rk, k) Rk, j
• end
• end
• R B (copy all the updates to R).
• End
• n3 expressions Ri, jk, and Ri, jk could have
  size 4n
• Alternative approach state elimination

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State Elimination Technique

• Label the edges of the FA with regular
  expressions instead of symbols.
• Add a new start state, qstart, with an ? edge to
  the old start state and a new accept state,
  qaccept, with ? arrows from the old accept
  states.
• eliminate all states except qstart and qaccept,
  the regular expression on the label is the answer.

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State Elimination Technique

• Eliminate qrip

R2
qrip
R3
qi
qj
R1
R4
R1(R2)R3R4
qi
qj
28
Example 1.35
29
Example 1.36
b
a
2
a
1
3
a
b
b
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Algebraic Laws for RE

• Regexs E and F are equivalent L(E) L(F).
• (E F) G E (F G)
• Union is associative
• E F F E
• Union is commutative
• Ø E E Ø E
• Ø is identity for union
• E E E
• Union is idempotent

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Algebraic Laws for RE

• (E F) G E (F G)
• Concatenation is associative
• ? E E ? E
• ? is right and left identity for concatenation
• Ø E E Ø Ø
• Ø is right and left annihilator for concatenation
• E (F G) E F E G
• Concatenation is left distributive over union
• (F G) E F E G E
• Concatenation is right distributive over union

32
Algebraic Laws for RE

• Ø ?
• ? ?
• (E) E
• Closure is idempotent
• E EE ?

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Assignment 2

• 1. Convert 0(100) to an ?-NFA
• 2. Eliminate the ? arcs in your answer
• 3. Convert the following FA into RE

0
s
q
0
1
1
0
t
p
1
1
0