CS321
Theory of Computation
Midterm, Fall 2014
Name:
1. (a) [5pt] Draw a DFA M with alphabet Σ = {a, b} such that
L(M ) = {ava : v ∈ Σ∗ }.
(b) [5pt] Draw an NFA N that accepts the following language.
L(N ) = {a}∗ ∪ {a}{b}∗ ∪ {ab}{c}∗
So for example aaaaa, abbbbb, abccc are in L(N ), but aba, aab,
and abcb are not in L(N ).
(c) [5pt] Draw a DFA or NFA M with alphabet Σ = {a, b} such that
L(M ) = {w : w ∈ Σ∗ , na (w) mod 2 = 0}.
2. (a) [5pt] Write a regular expression r such that
L(r) = {w : w = am bn , m + n is even}
.
r = (aa)*(bb)* + a(aa)*b(bb)*
(b) [5pt] Write a regular expression r such that
L(r) = {uv : u ∈ {a, b}∗ , v ∈ {a, b}∗ , na (u) is even , na (v) is even}.
r = (b*ab*ab*)*
This is just the language of all strings with an even number of
a's.
(c) [5pt] Prove that every finite language is regular.
Assume that L={w1,w2,...,w_n} is a finite language containing n
strings.
Consider the regular expression:
r = w1 + w2 + ... + w_n
From the definition of REs we have that
L(r) = {w1,w2,...,w_n} = L
This shows that L is regular since it has an RE.
A similar argument can be made using NFAs.
3. (a) [5pt] Prove that regular languages are closed under the concatenation operation.
Let L1 and L2 be two regular languages. We want to show that
L1L2 is also regular.
Let r1 and r2 be two REs such that L1 = L(r1) and L2=L(r2).
Now consider a new RE r = r1r2.
Based on the definitions associated with REs we know that:
L(r)=L(r1)L(r2)=L1L2.
This shows that r is an RE for L1L2 and hence that L1L2 is regular.
(b) [5pt] Let REPLACE be an operation on languages such that
REPLACE(L) is a new language that is identical to L, except
that each ‘a’ in a string is replaced with ‘cc’.
For example, if L = {a, ab, aabb, bbb} then REPLACE(L) = {cc, ccb, ccccbb, bbb}.
Show that the set of regular languages is closed under the REPLACE operation. (If you don’t understand the operation then
please ask the instructor.)
Assume that L is a regular language. We want to show that REPLACE(L) is
also regular.
Let r be an RE such that L=L(r).
Construct an RE r' by starting with r and then replacing each 'a' in r with 'cc'.
It is clear that L(r') = REPLACE(L). This is because r' will insert a 'cc' in a string
exactly when r would have inserted an 'a' in a string.
A similar argument could be made by starting with an NFA N such that L=L(N).
Now create a new NFA N' that starts with N but replaces any arc labeled by an
'a' with two consecutive arcs, each labeled by a 'c'. The picture below illustrates
the process for one transition, noting that q' is a new state that is introduced into
N'.
It should be clear that L(N') = REPLACE(L) since whenever N would have
required reading in an 'a', N' will require reading in 2 c's instead.
4. (a) [5pt] Let DFAL be the set of languages accepted by DFAs and
NFAL be the set of languages accepted by NFAs. Is it true that
DFAL ⊆ NFAL ? Give a brief argument in support of your answer.
We need to show that for any DFA M there is an NFA N
such that L(M)=L(N).
This is true because DFAs are special cases of NFAs. That
is, DFAs must satisfy conditions that general NFAs do not.
So every DFA is in fact an NFA.
(b) [5pt] Use the procedure taught in class or the book to convert the
following NFA to an equivalent DFA.
Start
q0
b
q4
λ
q1
b
b
q2
a
q5
b
q3
5. (a) [5pt] Consider the following NFA N .
b
a
Start
q0
q1
q2
b
a
b
a
b
q3
Draw an equivalent generalized transition graph that results by
removing q3 from N .
(b) [5pt] Consider the following generalized transition graph G.
ba∗
Start
(a + b)
q0
q1
b∗ a
bb
Using the pattern taught in class and the book, write a regular
expression that is equivalent to the generalized transition graph
G.
(ba*)*(a+b)(bb + b*a(ba*)*(a+b))*