Provide a regular expression for the following expressions

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3240
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Provide a regular expression for the following
expressions

• All strings on a, b, c which contain no run of
  as of length greater than two
• Tip
• Come up with all sequences of a, b, c that have
  at most 2 as
• Then produce all permutations

It should end thought with a or aa
Produce all permutations
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Provide a regular expression for the following
expressions

• W is any string in 0,1 except 11 and 111
• Tip Enumerate any string up to length 3 and
  produce all strings of length greater than 3

What is the DFA?
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Provide a regular expression for the following
expressions
All strings of lowercase letters that begin and
end with a
a alone start end ends with a ?
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Regular expressions tips

• Common mistakes
• Do not write a regular expression for perl, sed
  They do not follow the formalities of typical
  regular expressions
• Symbols allowed , , ?, range
• Tips
• Expressions like 11,111 are wrong, which is
  the alphabet?

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Regural expressions to NFAs

• Produce an NFA for the following regular
  expression 0(001)1

NFA that accepts 0
NFA that accepts (001)
Final NFA
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NFA to DFA
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NFA to DFA
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DFA-minimization

• Group states into accepting and nonaccepting
• If a transition from a group G goes to more than
  one groups, then split G
• Repeat until no further splitting occurs

Accepting group 2,3
No-Accepting group 1
(d(1,a), d(2,a))(2,e) (d(1,b), d(2,b))(3,3)
(d(1,a), d(3,a))(2,e) (d(1,b), d(3,b))(3,3)
(d(2,a), d(3,a))(e,e) (d(2,b), d(3,b))(3,3)
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Advices for common mistakes

• Dont miss empty string (e)
• Dont miss two epsilon transitions to denote in
  DFA
• Dont use reg.exp. In transition edges.
• Dont miss final states (double-circle)
• Please write clearlyneatly!!

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Funny game

• We cannot claim that we have reached a
  contradiction just because the pumping lemma is
  violated for some specific values of p or xyz
• The pumping lemma is violated even for one w or i.

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Funny game

• The opponent picks p
• Given p, we pick a string w in L.
• Length gt p
• The opponent chooses the decomposition xyz,
  subject to xyltp, ygt1.
• We have to assume that the opponent makes the
  choice that will make it hardest for us to win
  the game
• We try to pick i in such a way that the pumped
  string is not in L.
• We win if we can do.

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HW3-1.29b

• Lwww. Assume that L is regular, so that the
  pumping lemma must hold.

• Opponent choose p
• We Whatever it is, we can choose s to be
  apbapbapb where wap.
• Opponent choose xyz decomposition
• We Whatever it is, y should consist of only as
  since xyp. The lemma says any pumped strings
  should belong to the language. However, with i0,
  the string ap-ybapbapb doesnt belong to L
  clearly. This is contradiction. Thus L is
  non-regular.

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Funny game

• We dont choose xyz decomposition, but the
  opponent does.
• Condition 3 is our friend. We dont want to
  violate it. We want to use it.
• We have to show the pumped string does not belong
  to the original language, not the chosen s.

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HW3-1.46a

• L0n1m0n m,n 0
• Assume that L is regular. Let the string s
  0p10p. Then y should consist of only 0s. With
  i0, the string 0p-y10p doesnt belong to L
  clearly. This is contradiction.
• Alternatively, let A010. Then AnB 0n10n
  C. If we assume A and B to be regular, then C
  must be regular. Proving that C is not regular
  will in turn prove that A is not regular because
  B is regular.

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Alternative ways

• Closure under simple operations
• Yes. Union, complement, so on.
• Subset relation?
• No
• Too few states for DNA?
• Not for proof

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HW3-1.46c

• Lnot palindrome
• Assume that L is regular, then its complement, L
  w w?0,1 is a palindrome, should be
  regular. Let the string s 0p10p xyz with
  xyp and y1. Then y should consist of only
  0s. With i0, the string 0p-y10p is not a
  palindrome. This is contradiction. Therefore, L
  is nonregular.
• Since L is nonregular, L is nonregular

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HW3-1.46d

• Lwtw
• Assume that L is regular. Let the string s
  0p100p1, where w0p1, t0. Since xyp, y must
  consist of only 0s. Let y be 0k where k1. Then
  the string 0p-kki100p1 for any i should belong
  to L.
• Does this belong to L?
• In this case, w should end with 1 since the whole
  string ends with 1. Therefore w must be
  0p-kki1. This is not true for i2.

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HW3-1.49b

• L0ky, y has at most k 0s
• Let the string s 1p0p1p. This makes y consist
  of only 1s. For i0, the number of 1s before 0
  becomes fewer than that of 1s after 0. This
  doesnt belong to the original language, thus we
  have contradiction.