CSE303
Q3 PRACTICE
SOLUTIONS
YES/NO questions
Circle the correct answer. Write SHORT justification.
1. For any finite language
L
there is a deterministic automata
M
, such
that
L
=
L
(
M
).
Justify
: Any finite language is regular
y
2. Any regular language is finite.
Justify
:
L
=
a
*
is infinite
n
3. Any finite language is regular.
Justify
:
L
=
S
{
L
w
:
w
∈
L
}
, each
L
w
is regular and regular lan
guages are closed under finite union.
y
4. Given
L
1
, L
2
regular languages over Σ, then (
L
1
∪
(Σ
*

L
1
))
L
2
is
regular.
Justify
: closure of regular languages over union and complement
y
5. For any
M
,
L
(
M
) =
S
{
R
(1
, j, n
) :
q
j
∈
F
}
, where
R
(1
, j, n
) is the
set of all strings in Σ
*
that may drive
M
from state initial state to
state
q
j
without passing through any intermediate state numbered
n
+ 1 or greater, where
n
is the number of states of
M
.
Justify
: only when
M
is a finite automaton
n
6. The Generalized Finite Automaton accepts regular expressions.
Justify
: accepts regular expressions
y
7. There is an algorithm that for any finite automata
M
computes a
regular expression
r
, such that
L
(
M
) =
r
.
Justify
: defined in the proof of Main Theorem
y
8. Pumping Lemma says that we can always prove that a language is
regular.
Justify
: it gives certain characterization of infinite regular languages
n
9. Pumping Lemma proves that a language is not regular.
Justify
: PL is usually used to prove that an infinite language is not
regular
n
10.
L
=
{
a
n
:
n
≥
0
}
is not regular.
Justify
:
L
=
a
*
n
11.
L
=
{
b
n
a
n
:
n
≥
0
}
is not regular.
Justify
:proved using Pumping Lemma
y
12.
L
=
{
a
2
n
:
n
≥
0
}
is regular.
Justify
:
L
= (
aa
)
*
y
1
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13. Let
L
be a regular language, and
L
1
⊆
L
, then
L
1
is regular.
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 Spring '08
 Ko,K
 Formal language, Regular expression, Regular language, Nondeterministic finite state machine

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