Download Closure Properties and Pumping Lemmas of Regular Languages and Context-Free Languages and more Study notes Computer Science in PDF only on Docsity!
CMPSCI 601: Recall From Last Time Lecture 4
Kleene’s Theorem: Let
be any language. Then the following are equivalent:
, for some DFA
.
, for some NFA without transitions
, for some NFA.
, for some regular expression.
is regular.
Myhill-Nerode Theorem: The language
is regular iff has a finite number of equivalence classes. Fur- thermore, this number of equivalence classes is equal to the number of states in the minimum-state DFA that ac- cepts
.
CMPSCI 601: Regular Language Closure Lecture 4
Closure Theorem for Regular Sets: Let
be
regular languages and let
and
be homomorphisms. Then the following languages are regular:
A homomorphism of strings is a function such that for
any strings and ,
. The set
is
defined as
.
Let
be a homomorphism.
If is a regular expression over
, we can compute a
regular expression
by induction on the definition of
regular expressions. Then we can prove by induction that
If is a DFA with alphabet
and
, we can
make a DFA for
as follows. States, start, and
final states are the same as . For every letter in
and
every state , define
to be
Then for any
, and
^
^
CMPSCI 601: Review of CFL’s Lecture 4
Definition: A context-free grammar (CFG) is a 4-
tuple
,
^ variables = nonterminals,
^ terminals,
^ rules = productions, ^ ,
^ ,
^ ^ are all finite.
^
Parse Tree:
V
T F F C D
( E ) E + T E + T T F F V L D 3 * ( x 1 + y 1 + z 1 )
F V L D
T
E
L D
Pumping Lemma for Regular Sets: Let
be a DFA. Let
. Let
s.t.
. Then
s.t. the following all hold:
^ ^ ^ ^
^ ^ ^ ^ , and
^ ^ ^ ^
Proof: Let
s.t.
.
^ ^ ^
By the Pigeonhole Principle,
. Thus,
for all
N. '
CFL Pumping Lemma: Let
be a CFL. Then there is a constant
, depending only on
, such that if
^ and
, then there exist strings
such that:
^ ^ , and
^ ^ , and
^ ^ , and
for all N ,
Proof: Let
be a CFG with
.
Let
be so large that for
s.t.
for some
, the parse tree for
has height
.
Let
,
.
The parse tree for
has height greater than
.
Thus, some path repeats a nonterminal,.
y
S
N
N
u v w x
N
Prop:
N
is not a CFL.
Proof:
The argument is almost identical. We let
where
is larger than the constant given by the CFL Pumping Lemma. So
with
^ ,
, and in
for all . Again,
neither nor can contain letters of two different types,
or is not in
(^)
. But then cannot
contain equal numbers of ’s,
’s, and (^) ’s, as only one or two types of letter have been added.
Any CFL satisfies the conclusion of the CFL Pumping Lemma, but it is not true that any non-CFL must fail to satisfy it. There are other tools that can show a language to be a non-CFL. These include stronger forms of the Pumping Lemma and more closure properties.
Let
be the set of strings in
(^)
that
have an equal number of ’s,
’s, and (^) ’s. You can use the CFL Pumping Lemma on this with the right choice of
, but far easier is using the fact that the intersection of
with
(^)
is the language
.
If
is a CFL and a regular language, then
must
be regular. Proving this, however, requires a different characterization of the CFL’s.
ε (^) r Z (^) 0 /
a/A aA/
b/B bB/
q s
Theorem 4.1 Let
be any language. Then the following are equivalent:
1.
, for some CFG . 2.
, for some PDA .
3.
is a context-free language.
Proof: We give only a sketch here – there are detailed proofs in [HMU], [LP], and [S].
To prove (1) implies (2), we can build a “bottom-up parser” or “top-down” parser, similar to those used in real-world compilers except that the latter are deterministic.
The proof that the language of any PDA is a CFL (that (2) implies (1)) is of less practical interest.
Given states and , let
be the set of strings that could
take the PDA from state with empty stack to state with
empty stack.
If we can define rules making each
a CFL we win, be-
cause the language of the PDA is the union of
% for all
final states
, where
is the start state. (So our grammar has a rule
% for each
.)
We have all rules of the form
, and a rule
whenever moves of the PDA warrant it.
Here I am skipping some assumptions on the PDA, and the (nontrivial) proof that any accepting run of the PDA
corresponds to a valid derivation in our grammar. '
