Assignment 1: solution set
Jalaj Upadhyay
1
Question 1
Let
M
:= (Σ
,Q,S,F,δ
)
be an automaton deﬁned over the alphabet set
Σ
using the transition table
δ
,
where
S
is the set of the starting states,
Q
is the set of all states, and
F
is the set of accepting states. For this
question, we assume
Σ =
{
0
,
1
}
.
(a)
Q
:=
{
q
0
,q
1
}
,
S
:=
{
q
0
}
and
F
=
Q
.
δ
(
q
0
,
0) =
q
0
,
δ
(
q
0
,
1) =
q
1
, and
δ
(
q
1
,
1) =
q
1
.
In order to prove the correctness of the above NFA, we prove both the direction. First, to see that all
words in
L
are accepted by
L
(
M
)
, note that we can break
L
as a union of
±
∪
0
+
∪
1
+
∪
0
+
1
+
. The ﬁrst part
of this union is trivial, the second part is accepted by
F
=
q
0
, the third part by a transition from
q
0
to
q
1
and
then staying thereafter in
q
1
. For the last part, we ﬁrst loop in
q
0
and then make a transition of the previous
form on seeing the ﬁrst
1
.
To show the converse, we prove the contrapositive. Any words in the complement of
L
has a substring
of the form
10
. Now, notice that once the NFA is fed in
1
, it moves to
q
1
, hence forth any encounter with
0
will result in crashing of NFA. Hence, if
w
∈
¯
L
, it is not accepted by
M
, proving the result.
(b)
Q
:=
{
q
0
,q
1
,q
2
,q
3
,q
4
,q
5
}
,
S
:=
{
q
0
}
and
F
=
{
q
2
,q
5
}
. The transition table is as follows:
δ
(
q
0
,
0) =
q
1
δ
(
q
0
,
1) =
q
3
δ
(
q
1
,
0) =
q
2
δ
(
q
1
,
1) =
q
4
δ
(
q
2
,
0) =
q
2
δ
(
q
2
,
0) =
q
5
δ
(
q
3
,
0) =
q
4
δ
(
q
4
,
0) =
q
5
δ
(
q
5
,
0) =
q
5
.
Note that there is encountering a