Solutions by Bob
Problem 1.
(Exercise 2.5.2)
Consider the following
²
NFA.
²
a
b
c
→
p
{
q,r
}
∅
{
q
}
{
r
}
q
∅
{
p
}
{
r
}
{
p,q
}
*
r
∅
∅
∅
∅
(a) Compute the
²
closure of each state.
(b) Give all the strings of length three or less accepted by the automaton.
(c) Convert the automaton to a DFA.
(a)
eclose
(
p
) =
{
p,q,r
}
eclose
(
q
) =
{
q
}
eclose
(
r
) =
{
r
}
(b) First we check
²,a,b,c
and they’re all accepted. In addition, we have
p
∈
ˆ
δ
(
p,a
) and
p
∈
ˆ
δ
(
p,c
) which means if the ﬁrst bit is
a
or
c
, we can choose to go back to start state and
start over. Hence,
a
(
a
+
b
+
c
) and
c
(
a
+
b
+
c
) are all accepted. Now we check
ba,bb,bc
and they’re all accepted.
Now we know (
a
+
b
+
c
)(
a
+
b
+
c
) (all strings of length 2) are all accepted. For the
same reason as above, we have
a
(
a
+
b
+
c
)(
a
+
b
+
c
) and
c
(
a
+
b
+
c
)(
a
+
b
+
c
) be accepted.
However, there are no ways to accept
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 Spring '07
 Jiang

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