Operations on Languages

1
Operations on Languages

• Let L, L1, L2 be subsets of S
• Concatenation L1L2 xy x is in L1 and y is
  in L2
• Concatenating a language with itself L0 e
• Li LLi-1, for all i gt 1
• Kleene Closure L Li L0 U L1 U L2 U
• Positive Closure L Li L1 U L2 U
• Question Does L contain e?

2
Kleene closure
Say, L1 a, abc, ba, on S a,b,c Then, L2
aa, aabc, aba, abca, abcabc, abcba, baa,
baabc, baba L3 a, abc, ba. L2 L e, L1,
L2, L3, . . .
3
Regular Expressions

• Highlights
• A regular expression is used to specify a
  language, and it does so precisely.
• Regular expressions are very intuitive.
• Regular expressions are very useful in a variety
  of contexts.
• Given a regular expression, an NFA-e can be
  constructed from it automatically.
• Thus, so can an NFA, a DFA, and a corresponding
  program, all automatically!

4
Definition of a Regular Expression

• Let S be an alphabet. The regular expressions
  over S are
• Ø Represents the empty set
• e Represents the set e
• a Represents the set a, for any symbol a in S
• Let r and s be regular expressions that
  represent the sets R and S, respectively.
• rs Represents the set R U S (precedence 3)
• rs Represents the set RS (precedence 2)
• r Represents the set R (highest precedence)
• (r) Represents the set R (not an op, provides
  precedence)
• If r is a regular expression, then L(r) is used
  to denote the corresponding language.

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• Examples Let S 0, 1
• (0 1) All strings of 0s and 1s
• 0(0 1) All strings of 0s and 1s, beginning
  with a 0
• (0 1)1 All strings of 0s and 1s, ending
  with a 1
• (0 1)0(0 1) All strings of 0s and 1s
  containing at least one 0
• (0 1)0(0 1)0(0 1) All strings of 0s
  and 1s containing at least two 0s
• (0 1)0101 All strings of 0s and 1s
  containing at least two 0s
• (1 010) All strings of 0s and 1s
  containing an even number of 0s
• 1(0101) All strings of 0s and 1s
  containing an even number of 0s
• (1010)1 All strings of 0s and 1s
  containing an even number of 0s

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• Identities
• Øu uØ Ø Multiply by 0
• eu ue u Multiply by 1
• Ø e L Li L0 U L1 U L2 U
• e e e
• uv vu
• u Ø u
• u u u
• u (u)
• u(vw) uvuw
• (uv)w uwvw
• (uv)u u(vu)
• (uv) (uv)
• u(uv)
• (uvu)
• (uv)
• u(vu)
• (uv)u

7
Equivalence of Regular Expressionsand NFA-es

• Note
• Throughout the following, keep in mind that a
  string is accepted by an NFA-e if there exists a
  path from the start state to a final state.
• Lemma 1 Let r be a regular expression. Then
  there exists an NFA-e M such that L(M) L(r).
  Furthermore, M has exactly one final state with
  no transitions out of it.
• Proof (by induction on the number of operators,
  denoted by OP(r), in r).

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• Basis OP(r) 0
• Then r is either Ø, e, or a, for some symbol a
  in S
• For Ø
• For e
• For a

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• Inductive Hypothesis Suppose there exists a k ?
  0 such that for any regular expression r where 0
  ? OP(r) ? k, there exists an NFA-e such that L(M)
  L(r). Furthermore, suppose that M has exactly
  one final state.
• Inductive Step Let r be a regular expression
  with k 1 operators (OP(r) k 1), where k 1
  gt 1.
• Case 1) r r1 r2
• Since OP(r) k 1, it follows that 0lt OP(r1),
  OP(r2) lt k. By the inductive hypothesis there
  exist NFA-e machines M1 and M2 such that L(M1)
  L(r1) and L(M2) L(r2). Furthermore, both M1 and
  M2 have exactly one final state.
• Construct M as

M1
q1
f1
e
e
q0
e
e
M2
q2
f2
10

• Case 2) r r1r2
• Since OP(r) k1, it follows that 0lt OP(r1),
  OP(r2) lt k. By the inductive hypothesis there
  exist NFA-e machines M1 and M2 such that L(M1)
  L(r1) and L(M2) L(r2). Furthermore, both M1 and
  M2 have exactly one final state.
• Construct M as
• Case 3) r r1
• Since OP(r) k1, it follows that 0lt OP(r1) lt
  k. By the inductive hypothesis there exists an
  NFA-e machine M1 such that L(M1) L(r1).
  Furthermore, M1 has exactly one final state.
• Construct M as

e
M1
e
e
f1
q1
q0
e
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• Example
• r 0(01)
• r r1r2
• r1 0
• r2 (01)
• r2 r3
• r3 01
• r3 r4 r5
• r4 0
• r5 1

1
q0
12

• Example
• r 0(01)
• r r1r2
• r1 0
• r2 (01)
• r2 r3
• r3 01
• r3 r4 r5
• r4 0
• r5 1

1
q0
0
q2
13

• Example
• r 0(01)
• r r1r2
• r1 0
• r2 (01)
• r2 r3
• r3 01
• r3 r4 r5
• r4 0
• r5 1

e
e
q4
e
e
14

• Example
• r 0(01)
• r r1r2
• r1 0
• r2 (01)
• r2 r3
• r3 01
• r3 r4 r5
• r4 0
• r5 1

15

• Example
• r 0(01)
• r r1r2
• r1 0
• r2 (01)
• r2 r3
• r3 01
• r3 r4 r5
• r4 0
• r5 1

0
q8
q9
16

• Example
• r 0(01)
• r r1r2
• r1 0
• r2 (01)
• r2 r3
• r3 01
• r3 r4 r5
• r4 0
• r5 1

17
Definitions Required to Convert a DFAto a
Regular Expression

• Let M (Q, S, d, q1, F) be a DFA with state set
  Q q1, q2, , qn, and define
• Ri,j x x is in S and d(qi,x) qj
• Ri,j is the set of all strings that define a
  path in M from qi to qj.
• Note that states have been numbered starting at
  1!

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• Example
• R2,3 0, 001, 00101, 011,
• R1,4 01, 00101,
• R3,3 11, 100,

0
19

• Another definition
• Rki,j x x is in S and d(qi,x) qj, and
  for no u where 1 ? u lt x and
• x uv is it the case that d(qi,u) qp where
  pgtk
• In words Rki,j is the set of all strings that
  define a path in M from qi to qj but that pass
  through no state numbered greater than k. Note
  that it may be that igtk or jgtk.

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• Example
• R42,3 0, 1000, 011, R12,3 0
• 111 is not in R42,3 111 is not in R12,3
• 101 is not in R12,3
• R52,3 R2,3

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• Obeservations
• 1) Rni,j Ri,j
• 2) Rk-1i,j is a subset of Rki,j
• 3) L(M) Rn1,q R1,q
• 4) R0i,j Easily computed from the DFA!
• 5) Rki,j Rk-1i,k (Rk-1k,k) Rk-1k,j U Rk-1i,j

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• Notes on 5
• 5) Rki,j Rk-1i,k (Rk-1k,k) Rk-1k,j U Rk-1i,j
• Consider paths represented by the strings in
  Rki,j
• 
  
• IF x is a string in Rki,j then no state numbered
  gt k is passed through when processing x and
  either
• qk is not passed through, i.e., x is in Rk-1i,j
• qk is passed through one or more times, i.e., x
  is in Rk-1i,k (Rk-1k,k) Rk-1k,j

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• Lemma 2 Let M (Q, S, d, q1, F) be a DFA. Then
  there exists a regular expression r such that
  L(M) L(r).
• Proof
• First we will show (by induction on k) that for
  all i,j, and k, where 1 ? i,j ? n
• and 0 ? k ? n, that there exists a regular
  expression r such that L(r) Rki,j .
• Basis k0
• R0i,j contains single symbols, one for each
  transition from qi to qj, and possibly e if ij.
• case 1) No transitions from qi to qj and i ! j
• r0i,j Ø
• case 2) At least one (m ? 1) transition from qi
  to qj and i ! j
• r0i,j a1 a2 a3 am where d(qi, ap)
  qj,
• for all 1 ? p ? m

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• case 3) No transitions from qi to qj and i j
• r0i,j e
• case 4) At least one (m ? 1) transition from qi
  to qj and i j
• r0i,j a1 a2 a3 am e where d(qi,
  ap) qj
• for all 1 ? p ? m
• Inductive Hypothesis
• Suppose that Rk-1i,j can be represented by the
  regular expression rk-1i,j for all
• 1 ? i,j ? n, and some k?1.
• Inductive Step
• Consider Rki,j Rk-1i,k (Rk-1k,k) Rk-1k,j U
  Rk-1i,j . By the inductive hypothesis there exist
  regular expressions rk-1i,k , rk-1k,k , rk-1k,j ,
  and rk-1i,j generating Rk-1i,k , Rk-1k,k ,
  Rk-1k,j , and Rk-1i,j , respectively. Thus, if
  we let
• rki,j rk-1i,k (rk-1k,k) rk-1k,j rk-1i,j
• then rki,j is a regular expression generating
  Rki,j ,i.e., L(rki,j) Rki,j .

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• Finally, if F qj1, qj2, , qjr, then
• rn1,j1 rn1,j2 rn1,jr
• is a regular expression generating L(M).
• Note not only does this prove that the regular
  expressions generate the regular languages, but
  it also provides an algorithm for computing it!

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• Example
• First table column is
• computed from the
• DFA.
• k 0 k 1 k 2
• rk1,1 e
• rk1,2 0
• rk1,3 1
• rk2,1 0
• rk2,2 e
• rk2,3 1
• rk3,1 Ø
• rk3,2 0 1

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• All remaining columns are computed from the
  previous column using the formula.
• r12,3 r02,1 (r01,1 ) r01,3 r02,3
• 0 (e) 1 1
• 01 1
• k 0 k 1 k 2
• rk1,1 e e
• rk1,2 0 0
• rk1,3 1 1
• rk2,1 0 0
• rk2,2 e e 00
• rk2,3 1 1 01
• rk3,1 Ø Ø
• rk3,2 0 1 0 1
• rk3,3 e e

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• r21,3 r11,2 (r12,2 ) r12,3 r11,3
• 0 (e 00) (1 01) 1
• 01
• k 0 k 1 k 2
• rk1,1 e e (00)
• rk1,2 0 0 0(00)
• rk1,3 1 1 01
• rk2,1 0 0 0(00)
• rk2,2 e e 00 (00)
• rk2,3 1 1 01 01
• rk3,1 Ø Ø (0 1)(00)0
• rk3,2 0 1 0 1 (0 1)(00)
• rk3,3 e e e (0 1)01

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• To complete the regular expression, we compute
• r31,2 r31,3
• k 0 k 1 k 2
• rk1,1 e e (00)
• rk1,2 0 0 0(00)
• rk1,3 1 1 01
• rk2,1 0 0 0(00)
• rk2,2 e e 00 (00)
• rk2,3 1 1 01 01
• rk3,1 Ø Ø (0 1)(00)0
• rk3,2 0 1 0 1 (0 1)(00)
• rk3,3 e e e (0 1)01

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• Theorem Let L be a language. Then there exists
  an a regular expression r such that L L(r) if
  and only if there exits a DFA M such that L
  L(M).
• Proof
• (if) Suppose there exists a DFA M such that L
  L(M). Then by Lemma 2 there exists a regular
  expression r such that L L(r).
• (only if) Suppose there exists a regular
  expression r such that L L(r). Then by Lemma 1
  there exists a DFA M such that L L(M).
• Corollary The regular expressions define the
  regular languages.
• Note The conversion from a regular expression to
  a DFA and a program accepting L(r) is now
  complete, and fully automated!