COSC 341 Lecture 9

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COSC 341Lecture 9

• Let ? be an alphabet.
• The following are equal - the languages
• accepted by NFA-?s ( non-deterministic finite
  automata)
• accepted by DFAs
• represented by regular expressions
• generated by regular grammars.
• Let
• RE the collection of all languages over ?
  represented by regular expressions
• NFA the languages accepted by NFA-?s
• RG the set of languages generated by regular
  grammars.
• Stephen Cole Kleene proved that
• RE NFA RG.

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RE ? NFA
Claim if L is a language represented by a
regular expression, then there is some NFA-? that
accepts L. Justification The building blocks of
RE are ?, ?, and a for every a ? ?. Here are
NFA-?s to accept them
q0
q1
?
q0
q1
a
q0
q1
More complex languages in RE are formed by union,
concatenation, and Kleene star.
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Example
An NFA-? for (a?b)
a
a
b
b
a
a?b
?
?
?
?
b
(a?b)
?
?
?
a
?
?
?
?
b
?
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NFA ? RE
Claim If L is accepted by some NFA-?, then L is
represented by some regular expression. Justificat
ion An expression graph is like a state diagram
but it can have regular expressions as labels on
arcs. An NFA-? is an expression graph. An
expression graph can be reduced to one with just
two nodes (proof by induction). If we reduce an
NFA-? in this way, the arc label then corresponds
to the regular expression representing
it. Example
p
s
q
r
is shrunk to
prs
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Other example
Find a regular expression for L accepted by
c
c
1
2
3
b
b
First get 2 graphs each with 1 accept
state G1 G2 Reducing G1 in two steps
gives with reg. exp. b, while from G2 we get
reg. exp. bccb.
c
c
1
2
3
b
b
c
c
1
2
3
b
b
1
b
cc
1
3
b
So full r.e. is b?(bccb).
b
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Regular grammars

• Recall that a regular grammar has rules of form
• X ? aY, or
• X ? a, or
• X ? ?.
• Easy to rewrite regular grammar so that has no
  rules of form X ? a. Just use new nonterminal A
  to replace X ? a by the pair of rules
• X ? aA
• A ? ?.
• This is a bad idea if youre worried about
  ?-rules and CNF, but a good trick for proving
  that
• RG NFA.
• (See page 206 of the textbook.)

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RG ? NFA

• Let G be a regular grammar without X ? a rules.
  Build the NFA (Q, ?, ?, S, F)
• Q the set of nonterminals of G
• initial state is start symbol S
• F set of non-terminals A with A ? ? in G
• B ? ?(A,a) iff the rule A ? aB is in G.
• Example
• G S ? aS ? aA
• A ? bA ? bB
• B ? ?

a
b
S
A
B
a
b
Both accept the language ab.
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NFA ? RG
Let M be any NFA. From M build regular grammar G
by taking Q as non-terminals, q0 as start symbol,
? as terminals. Whenever B ? ?(A,a), add the rule
A ? aB, and if A ? F, add A ? ?. Example
M
b
b
q0
q1
q2
a, b
G q0 ? aq0 ? bq0 ? bq1 q1 ? bq2 q2 ? ? The
language consists of all strings ending in bb,
with regular expression (a?b)bb.