ECS 120: Theory of Computation
Homework 3
Due: Oct. 19, beginning of class or 1pm in 2131 Kemper
Problem 1.
1.19 in Sipser.
0,1
0
1
0
0
1
1
q0
q1
q2
q3
1
0
1
0
1
0
0
1
q0
q1
q2
q3
q4
H
0,1
q0
q1
Problem 2.
1.48 in Sipser.
Soln:
A DFA cannot “count” the number of occurences of 01’a and 10’s, so we need a
trick. Notice that
any
string
s
∈ {
0
,
1
}
∗
is one and only one of the following, where
n
≥
1 and
i
j
≥
1:
(a)
ε
(b) 0
i
1
1
i
2
0
i
3
. . .
1
i
n
-
1
0
i
n
(begins and ends with a 0)
(c) 1
i
1
0
i
2
1
i
3
. . .
0
i
n
-
1
1
i
n
(begins and ends with a 1)
(d) 0
i
1
1
i
2
0
i
3
. . .
0
i
n
-
1
1
i
n
(begins with a 0 and ends with a 1)
(e) 1
i
1
0
i
2
1
i
3
. . .
1
i
n
-
1
0
i
n
(begins with a 1 and ends with a 0)
Now, strings of the form (a), (b) and (c) are in
D
, and strings of the form (d) and (e)
are not in
D
(just count the number of occurences of 01 and 10 in each). Therefore,
a string
x
∈ {
0
,
1
}
∗
is in
D
if and only if
x
starts and ends with the same character
or
x
=
ε
. So
D
is described by the regular expression
ε
∪
0
∪
1
∪
1(0
∪
1)
∗
1
∪
0(0
∪
1)
∗
0,
hence it is regular.
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- Fall '07
- Filkov
- Formal language, Regular expression, Nondeterministic finite state machine, NOEXT END
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