Lecture Closure Properties for Regular Languages

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Lecture Closure Properties for Regular
Languages
CSCE 355 Foundations of Computation

• Topics
• Strings distinguishing states
• Equivalence relation

October 13, 2009
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CSCE 355 Fall 2OO9 Quiz 7

• Fill in the underlined values for s and i in the
  following proof that the language L 0m1n m lt
  n is not pumpable
• "Given p gt 0, let s (some string depending on
  p)
• Clearly, s e L and lsl gt p.
• Given strings x, y, z such that xyz s, ' lxyl
  lt p,
• and lyl gt 0, let i (some integer gt 0).
• Then clearly, xyiz e L."
• If your answers work, then no explanation is
  needed.

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Proposition 7.1. If L and M are regular
languages, then so is L ? M.
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Proposition 7.2. If L is regular, then L is
regular.

• Proof. Let A (Q,S, d, q0, F) be a DFA for L.
• Define B (Q, S, d, q0, ???).

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Proof Approaches

• Transform regular expressions
• Transform a DFA
• Transform an NFA or e-NFA.

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• Recall wR is the reversal of the string w.
• LR wR w is in L
• If L abb, abc, e What is LR ?
• Note (wR)R w
• But is (LR)R L ?

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Proposition 7.3. If L is regular, then so is LR.

• Proof based on transforming any regular
  expression for L to one for LR.
• Define a recursive construction
• Basis steps
• Ø R Ø,
• R ,
• For any symbol a ? S, aR a, Recursive
  steps for any two languages L and M,
• (L U M) R LR U MR ,
• (L M) R MR U LR ,
• (L) R (LR )

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Proof of (L U M) R LR U MR
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Find rR where r b(a bc)

• For example,
• lets use the rules to find rR where r b(a
  bc).
• (b(a bc)) R
• ((a bc)) RbR
• ((a bc) R)b
• (aR (bc) R)b
• (a (c) RbR)b
• (a (cR)b)b
• (a cb)b.

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Proposition 7.4. If L and M are regular, then so
is L nM.
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Corollary 7.5. If L and M are regular (and over
the same alphabet), then L -M is regular.
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string homomorphisms

• A string homomorphism is a function on strings
  that works by substituting a particular string
  for each symbol
• Definition 7.6. Let S and T be alphabets. A
  string homomorphism (or just a homomorphism) from
  S to T is a function h that takes any string w
  ? S and produces a string in T (that is, if w ?
  S, then h(w) ? T)
• such that h preserves concatenation, i.e., if w
  and x are any strings in S, then h(wx)
  h(w)h(x).

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Example

• Note h(w) h(w1w2 wn)
• h(w1)h(w2 wn)
• h(w1)h(w2) h(wn)
• For example, let S a, b, c and let T 0,
  1. Define the homomorphism h by
• h(a) 01,
• h(b) 110, and
• h(c)
• Then h(abaccab)

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• Definition 7.7. Let S and T be alphabets, and let
  h be a homomorphism from S to T. For any
  language L ? S, we define the language h(L) ? T
  as h(L) h(w) w is in L.

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Image of L under h, h(L)

• Suppose L is the language denoted by (ab)c
• Then the image of L under h (the same h)

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Inverse image of M under h, h-1(M)

• h-1(M) w ? S h(w) is in M.

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• Proposition 7.8. Let H be a homomorphism h S ?
  T and let L, and M be languages over S and T
  respectively.
• If L is regular, then so is h(L).
• If M is regular, then so is h-1(M).

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Proof of If L is regular, then so is h(L) by
transforming reg expr.
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• h(b(a bc)) h(b(a bc))
• (h(b)h(a bc))
• (h(b)(h(a) h(bc))
• (h(b)(h(a) h(b)h(c)))
• (h(b)(h(a) h(b)h(c)))
• (110(01 110()))
• (110(01 110))
• (11001 110110).

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