2. Languages

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2. Languages
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2.1 Strings and Languages

• A string over a set X is a finite sequence of
  element from X.
• Strings are the fundamental objects used in the
  definition of languages
• The strings are built by a set of elements called
  alphabet
• An alphabet consists of a finite set of
  indivisible objects.
• The alphabet of a language is donated S

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• Definition 2.1.1
• Let S be an alphabet.
• S, the set of string over S, is defined
  recursively as follows
• i) Basis ? ? S.
• ii) Recursive step if w ? S and a ? S,
  then wa ? S
• iii) Closure w? S only if it can be obtained
  from ? by a finite number of applications
  of the recursive step.
• For any non empty alphabet S, S contains
  infinitely many elements.
• If S a,
• ? S contains strings
• ? , a, aa, aaa, .
• length of a string w, no. of elements (length
  (w))
• ? If S contain n elements,
• ? nk string of length k is S

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• Example 2.1.1
• Let S a, b, c
• S
• length 0 ?
• length 1 a b c
• length 2 aa ab ac ba bb bc ca cbcc
• length 3 aaa aab aac aba
• baa
• caa
• Definition 2.1.2
• A language over an alphabet S is a subset of S
• idea concat. 2 strings ? new string

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• Definition 2.1.3
• Let u, v ? S . The concat. of u and v , ? uv ,
  is a binary operation on
• S defined as follow
• i) Basis If length (v) 0, then u ? and
  uv U
• ii) Recursive step Let v be a string with
  length (v) n gt 0. Then v wa ,
• for some string w with length n-1, and a ? S
  , and uv (uw)a
• Let u ab, v ca, and w bb, then uv abca,
  vw cabb
• (uv)w abcabb u(vw) abcabb
• The result of concat. Is independent of the
  order.
• Property ? associability

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• Theorem 2.1.4 proves this.
• Let u, v, w ? S. Then (uv)w u(vw)
• Proof by induction on the length of the string
  w.
• Basis length (w) 0 ? w ? , (uv)w uv by the
  definition of concat.
• on the other hand, u(vw) u(v) uv
• Induction Hypothesis Assume that (uv)w u(vw)
  for all string w of
• length n or less
• Inductive step We need to prove , ?
• (uv)w u(vw) , ?strings w , length n1
• Let w is a string, then w xa for some string x
  of length a ? S
• (uv)w (uv)(xa) , subs. w xa
• ((uv)x)a) , def. of concat
• (u (vx))a , inductive hypothesis
• u( (vx)a) , def. of
  concat
• u(v (xa)) , def. of concat
• u(vw) , subst. xa w

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• Definition 2.1.5
• Let u be a string in S. The reversed of u , uR ,
  defined as follows
• i) Basis length (u) 0 , then u ? , and ?R
  ?
• ii) Recursive step If length (u) n gt 0 ,
  then u wa for some string
• w with length n-1 and some a ? S , and
  uR awR
• Theorem 2.1.6
• Let u, v ? S, then (uv) R vRuR
• Proof
• Basis length (v) 0 , then v ? and (uv)R
  uR. Similarly , vRuR ?RuR , ?string v of length
  n
• Inductive step
• Prove ? (uv)R vRuR , for any string v of length
  n1
• Let v be a string of length n1. then v wa ,
  where w is a string of
• length n , a ? S .

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• (uv)R (u(wa))R
• ((uw)a)R , associativity of concat.
• a(uw)R , def. of reversed
• a(wRuR) , inductive hyp.
• (awR)uR ,assoc. of concat
• (wa)RuR , def. of reversed
• vRuR

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2.2 Finite Spec. of Languages.

• The spec. of a languages requires an unambiguous
  description of the strings
• of the languages.
• Example
• The language, L of strings over a, b in which
  each string begin with a
• and has even length is determined by

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• i) Basis aa, ab ? L
• ii) Recursive step If a ? L, then uaa, uba,
  ubb ? L.
• iii) Closure A string u ? L only if can be
  obtained from the basis element let a finite
  member of application of the recursive step.
• The strings in L are built by joining 2 elements
  to the Right-hand side of
• a previously constructed string. The basis
  ensures that each string in L
• begin with an a. Adding substrings of length 2
  maintain even parity.
• Definition 2.2.1
• The concat. Of languages X and Y, XY is the
  language
• XY uv u ? X and v ? Y
• The concat. of X with itself n times, Xn .
• X0 ?

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• Example
• Let X a,b,c and Y abb,ba. Then
• XY aabb, babb, cabb, aba, bba, cba
• X0 ?
• X1 X a, b, c
• X2 X X aa, ab, ac, .. cc
• X3 XXX aaa, aab, ccc
• Definition 2.2.2
• Let X be a set. Then
• X 8Ui0Xi and X 8Ui1Xi
• all strings over nonnull strings
• alphabet X over X
• ? X XX

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• Example 2.2.4
• L a,b bba,b consists of the strings
  over a,b that contains the substring bb
• Example 2.2.5
• L aa a,b U a,bbb
• L1 bb and L2 ?, bb, bbbb be languages
  over b
• L aa, bb, ab, ba , even length over a,b
• L a,b

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• 2.3 REGULAR SETS AND EXPRESSION
• A set is regular if it can be generated from the
  empty set, the set
• containing null string, and the elements of the
  alphabet using union,
• concat., and the Kleens Star operations.
• Definition 2.3.1
• Let S be an alphabet. The regular sets over S are
  defined recursively as follows
• i) Basis ø, ? and a , ? a? S are regular
  sets over S
• ii) Recursive step let X and Y be regular sets
  over S . The sets
• X ? Y
• XY
• X
• are regular sets over S.
• iii) Closure X is a regular set over S only if
  it can be obtained from the basis elements by a
  finite number of applications of the recursive
  step.

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• Example L a, bbb a,b
• bb is a regular set over a, b
• a , b are regular set.
• Using Kleene Star operation ? a, b , the set
  of all strings over a, b
• By concat. b b bb is regular
• Applying twice concat. ? a, b bb ab
• Exercise
• Write the set of strings that begin and end with
  an a and contain at least one
• b is regular over a, b.
• Hint
• The strings in this set could be described
  intuitively as an a, followed by
• any string, followed by a b , followed by any
  string, followed by an a.
• a a, bb a, ba

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• Definition 2.3.2
• Let S be an alphabet. The regular expression over
  S are defined recursively
• as follows
• i) Basis ø, ?, and a , for every a? S , are
  regular expressions over S .
• ii) Recursive step Let u and v be regular
  expressions over S . The expressions
• (u ? v)
• (uv)
• (u)
• are regular expressions aver S .
• iii) Closure u is a regular expression over S
  only if it can be obtained from the basis element
  by a finite.

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• Employing Kleene star , concat. and union
• a regular set of
• a, bbb a, b
• Can be written as
• (a ? b) bb (a ? b)
• and set of
• a a, bb a, ba
• ? a (a ? b)b (a ? b)a
• The set bawab w? (a, b) is regular over a,
  b

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• The following table demonstrates the recursive
  generation of a regular set
• and corresponding regular expression.

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• Exercise
• Write a regular expression that represent the
  regular sets with
• strings containing over a, b
• 1) aa
• 2) bb
• 3) aa and bb
• 4) exactly two bs
• 5) two or more bs
• Ans
• 1) (a ?b) aa(a?b)
• 2) (a ?b) bb(a?b)
• 3) (a ?b) aa bb(a?b)
• 4) aba? ba
• 5) i) abab(a ?b)
• ii) (a ?b)baba
• iii) (a ?b)b(a ?b)b(a ?b)