Today

1
Today

• Chapter 1
• RE Regular Languages,
• nonregular languages
• RL pumping lemma
• Chapter 2
• Context-Free Languages (CFLs)

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Regular Expressions (Def. 1.26)

• Given an alphabet ?, R is a regular expression
  if
• R a, with a??
• R ?
• R ?
• R (R1?R2), with R1 and R2 regular expressions
• R (R1?R2), with R1 and R2 regular expressions
• R (R1), with R1 a regular expression

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Thm 1.28 RL RE

• We need to prove both ways
• If a language is described by a regular
  expression,then it is regular (Lemma 1.29)(Last
  week we saw how we can convert a
  regularexpression R into an NFA M such that
  L(R)L(M))
• Today we do the second partIf a language is
  regular, then it can be described bya regular
  expression (Lemma 1.32)

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Generalized NFA

• Generalized nondeterministic finite automaton
  M(Q, ?, ?, qstart, qaccept) with
• Q finite set of states
• ? the input alphabet
• qstart the start state
• qaccept the accept state
• ?(Q\qaccept)?(Q\qstart) ? R the transition
  function
• (R is the set of regular expressions over ?)

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Example GNFA
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Characteristics of GNFAs ?

• ?(Q\qaccept)?(Q\qstart) ? R

The interior Q\qaccept,qstart is fully
connected by ? From qstart only outgoing
transitions To qaccept only ingoing
transitions Impossible qi?qj transitions are
?(qi,qj) ?
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Proof Idea of Lemma 1.32
Proof idea (given a DFA M) Construct an
equivalent GNFA M with k?2 states Reduce
one-by-one the internal states until k2 This
GNFA will be of the form This regular expression
R will be such that L(R) L(M)
R
qS
qA
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DFA M ? Equivalent GNFA M
Let M have k states Qq1,,qk
- Add two states qaccept and qstart
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Remove Internal state of GNFA
If the GNFA M has more than 2 states,
ripinternal qrip to get equivalent GNFA M
by - Removing state qrip QQ\qrip - Changing
the transition function ? by ?(qi,qj)
?(qi,qj) ? (?(qi,qrip)(?(qi,qj))?(qrip,qj))
for every qi?Q\qaccept and qj?Q\qstart
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Proof Lemma 1.32
Let M be DFA with k states Create equivalent
GNFA M with k2 states Reduce in k steps M to
M with 2 states The resulting GNFA describes a
single regular expressions R The regular
language L(M) equals the languageL(R) of the
regular expression R
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Recap Regular Languages Regular Expressions
Let R be a regular expression, then there exists
an NFA M such that L(R) L(M) The language
L(M) of a DFA M is equivalent toa language L(M)
of a GNFA M, which canbe converted to a
two-state M The transition qstart ?R?
qacceptof M obeys L(R) L(M) Hence RE ?
NFA DFA ? GNFA ? RE
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Nonregular Languages 1.4

• Which languages cannot be recognized by finite
  automata?
• Example L 0n1n n?N
• Playing around with FA convinces you that the
  finiteness of FA is problematic for all n?N
• The problem occurs between the 0n and the 1n
• Informal the memory of a FA is limited by the
  the number of states Q

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Repeating DFA Paths
Consider an accepting DFA M with size Q On a
string of length p, p1 states get visited For
p?Q, there must be a j such that the
computational path looks like q1,,qj,,qj,,qk
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Repeating DFA Paths
The action of the DFA in qj is always the
same. If we repeat (or ignore) the qj,,qj part,
the newpath will again be an accepting path
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Line of Reasoning

• Proof by contradiction
• Assume that L is regular
• Hence, there is a DFA M that recognizes L
• For strings of length ? Q the DFA M has to
  repeat itself
• Show that M will accept strings outside L
• Conclude that the assumption was wrong

Note that we use the simple DFA, not the more
elaborate (but equivalent) NFA or GNFA
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Pumping Lemma (Thm 1.37)
For every regular language L, there is a pumping
length p, such that for any string s?L and
s?p, we can write sxyz with1) x yi z ? L for
every i?0,1,2,2) y ? 13) xy ? p Note
that 1) implies that xz ? L 2) says that y
cannot be the empty string ?Condition 3) is not
always used
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Formal Proof of Pumping Lemma
Let M (Q,?,?,q1,F) with Q q1,,qp Let s
s1sn?L(M) with s n ? p Computational path of
M on s is thesequence r1,,rn1 ? Qn1 withr1
q1, rn1?F and rt1 ?(rt,st) for 1?t?n Because
n1 ? p1, there are two statessuch that rj rk
(with jltk and k ? p1) Let x s1sj1, y
sjsk1, and z sksn1 x takes M from q1r1 to
rj, y takes M from rj to rj, and z takes M from
rj to rn1?F As a result xyiz takes M from q1 to
rn1?F (i ? 0)
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Formal Proof of Pumping Lemma
Let M (Q,?,?,q1,F) with Q q1,,qp Let s
s1sn?L(M) with s n ? p Computational path of
M on s is thesequence r1,,rn1 ? Qn1 withr1
q1, rn1?F and rt1 ?(rt,st) for 1?t?n Because
n1 ? p1, there are two termssuch that rj rk
(with jltk and k ? p1) Let x s1sj1, y
sjsk1, and z sksn1 x takes M from q1r1 to
rj, y takes M from rj to rj, and z takes M from
rj to rn1?F As a result xyiz takes M from q1 to
rn1?F (i ? 0)
y ? 1 and xy ? p
x yi z ? L(M) for every i?0,1,2,
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Pumping 0n1n (Ex. 1.38)
Assume that B 0n1n n?0 is regular Let p be
the pumping length, and s 0p1p ? B P.L. s
xyz 0p1p, with xyiz ? B for all i?0 Three
options for y 1) y0k, hence xyyz 0pk1p ?
B 2) y1k, hence xyyz 0p1kp ? B 3) y0k1l,
hence xyyz 0p1l0k1p ? B Conclusion The
pumping lemma does not hold,the language B is
not regular.
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F ww w?0,1 (Ex. 1.40)
Let p be the pumping length, and take s
0p10p1 Let s xyz 0p10p1 with condition 3)
xy?p Only one option y0k, with xyyz
0pk10p1 ? F Without 3) this would have been a
pain.
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Intersecting Regular Languages
Let C w of 0s in w equals of 1s in
w Problem If xyz?C with y?C, then xyiz?C Idea
If C is regular and F is regular, then the
intersection C?F has to be regular as well
Solution Assume that C is regular Take the
regular F 0n1m n,m?N, then for the
intersection C?F 0n1n n?N But we know
that C?F is not regular Conclusion C is not
regular
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Pumping Down E 0i1j i?j
Problem pumping up s0p1p with y0k givesxyyz
0pk1p, xy3z 0p2k1p, which are all in
E(hence do not give contradictions) Solution
pump down to xz 0pk1p. Overall for s xyz
0p1p (with xy?p) y0k, hence xz 0pk1p ?
E Contradiction E is not regular
End Chapter One