Document 6528705

CS2MJ3 – Fall 2012. Sample solutions to the assignment 1. Most questions have more than one
solution. Total of this assignment is 153pts. 100% of this assignment is 125pts. There are 28
bonus pts. Each assignment is worth 7%.
If you think your solution has been marked wrongly, write a short memo stating where
marking in wrong and what you think is right, and resubmit to me during class, office
hours, or just slip under the door to my office.
1.[5]
Let M = (Q, Σ, δ, s0, F) be a deterministic finite automaton. Give a deterministic finite
state automaton M’ such that L(M’) = L(M) - {ε}.
Solution:
Let M=(Q,Σ,δ,s0,F) be a DFA such that L=L(M). If ε∉L(M), then M’=M. If ε∈L(M) then s0∈F.
Define M’= (Q∪{q0},Σ,δ1,q0,F), where q0∉Q, and δ1 is defined as follows:
δ1(q0,a) = δ(s0,a) for each a∈Σ, δ1(s,a) = δ(s,a) for each s∈Q.
Clearly L(M’)=L(M)-{ε}.
2.[5]
Let L = {a,b} and L1* = L*. Prove that L ⊆ L1.
Solution:
First we prove that a ∈ L1. Suppose that a ∉ L1. This means that a ∉ L1*, a contradiction since a ∈
L ⊆ L* = L1*. Identically we prove b ∈ L1. Thus L ⊆ L1.
An immediate extension of this result (identical proof!) is:
for every alphabet Σ and any language L, if L* = Σ*, then Σ ⊆ L.
3.[10] Consider the automaton below.
Construct an equivalent regular expression using the idea of generalised nondeterministic
automata. Provide all steps.
b
a
C
b
a
b
a
b
a
1
2
3
4
4.[15]
Consider the following regular expression: a*b*∪(a*bb∪ab*)a*b.
a.
Using a standard construction construct an equivalent non-deterministic finite
state automaton.
b.
Transform the automaton from 4a into an equivalent deterministic finite state
Solution:
a.[5]
Step 1 [5]. We construct a nondeterministic automaton with - moves:
Q={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22},
initial_state=1, F={22}, ={a,b}, and  is defined below

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
a

{2,8}
{4}
{2}
{22}
{4}
{9}
{6}
{6,11}
b
{3}
{5}
{7}
{10}
{13}
{12}
{14}
{15}
{16}
{17}
{17}
{14}
{18}
{20}
{18}
{19}
{21}
{22}
5
Step 2[5]. Now we get rid of - moves:
Q={1,2,3,4,5,6,7,8,9}, initial_state=1, F={1,2,4,6,9}, and  is defined below

1
2
3
4
5
6
7
8
9
a
{2,7}
{2}
{5}
b
{3}
{3,4}
{5}
{4}
{6}
{8}
{8}
{7,9}
{9}
Step 3[5]. We now transform non deterministic automaton into deterministic one.
Q={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19}, initial_state=1,
F={1,2,4,5,6,9,10,12,13,14,15,17,18}, ={a,b}, garbage=19 and  is defined below

1
2
3
4
5
6
7
8
9
10
11
12
11
13
14
15
16
17
18
19
a
2
5
19
7
5
11
7
8
19
7
11
19
11
7
19
8
19
19
19
19
b
3
4
8
6
12
10
9
16
19
13
14
15
14
13
19
17
19
18
18
19
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5.[19] a.
b.
c.
Is a language { anbkcn+k | n≥0, k≥0 } regular? Prove your answer.
Is the language {a 2m+3n+2 | m,n ≥ 0} regular? Prove your answer.
Is the language {a n! | n ≥ 0} regular? Prove your answer.
Solution:
6[5].
a.[5]
L is not regular.
Suppose it is regular, so Pumping Lemma hold. Let n the number from Pumping
Lemma, and let x = anbkcn+k. Clearly |x|n. Let x = uvw, |uv|n, |v|1. Since
|uv|n, v=ai, for some i, where 1in.
Hence, y=uviw=an+ibkcn+k  L, a contradiction, n+i+k > n+k.
b.[5]
L is regular as L=(aa)*(aaa)*aa.
c.[5]
L is not regular.
Suppose it is regular, so Pumping Lemma hold. Let n the number from Pumping
Lemma, and let x = an!. Clearly |x|=n!  n. Let x = uvw, |uv|n, |v|1. Let y=uv2w.
Since |uv|n then |v|n. Hence |y|=|uvw|+|v|  n!+n.
However the shortest string in L, which is longer than x is a(n+1)!, and
|a(n+1)!| = (n+1)! = (n+1)n! > n!+n, so y L, a contradiction.
Eliminate all useless productions from the below grammar. Give all the steps.
S → a | aA ∣ B | C
A → aB | ε
B → Aa
C → cCD
D → ddd
What language does this grammar generate?
Solution:
Step 1. There is only one type 1 useless symbol, namely: C
After erasing it we get:
S 6 a | aA * B
A 6 aB | ε
B 6 Aa
D 6 ddd
Step 2. There is only one type 2 useless symbol, namely: D
After erasing it we get:
S 6 a | aA * B
A 6 aB | ε
B 6 Aa
7
The grammar generates the language with odd number of a=s, i.e. { an | n is odd }
7[5].
Eliminate all ε-productions from
S → ABaC
A → BC
B→b|ε
C→D|ε
D→d
Give all the steps.
Solution:
In the first step we conclude that B, C can eventually generate ε, so they are nullable.
Then, since A 6 BC we conclude that A is also nullable. It is easy to observe that only A,B,C are
nullable. Hence:
S 6 ABaC | BaC | AaC | ABa | aC | Ba | Aa | a
A 6 BC | B | C
B6b
C6D
D6d
8.[5]
Eliminate all unit-productions from
S → Aa | N
A → a | bc |B
B → A | bb
Give all the steps.
Solution:
S 6 N is useless. We have A 6 B and B 6 A, so actually A is the same as B. Therefore we
can change:
A 6 a | bc | B
B 6 A | bb
into just
A 6 a | bc | bb
Hence the final grammar is:
S 6 Aa
A 6 a | bc | bb
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9. [10] Convert the grammar
S → abAB
A → bAB | ε
B → BAa | A | ε
into Chomsky normal form. Give all the steps.
Solution:
First we have to get rid of all ε-productions. The obtained grammar is:
S 6 abAB | abA | abB |ab
A 6 bAB | bA | bB |b
B 6 BAa | Aa | Ba | a | A
Next we have to get rid of all unit-productions. The result is:
S 6 abAB | abA | abB |ab
A 6 bAB | bA | bB | b
B 6 BAa | Aa | Ba | a | bAB | bA | bB |b
Now, we may produce a Chomsky normal form:
S6 XaXbAB | XaXbA | XaXbB |XaXb
Xa 6 a
Xb 6 b
XbAB 6XaXAB
XbA 6XbA
XbB 6XbB
XAB 6AB
A 6 XBXAB | XbA | XbB | b
B 6 BAXa | AXa | BXa | a | XBXAB | XaA | XbB | b
10.[10]
Convert the grammar
S → ABb | a
A → aaA | b
B → bAb | bBB | ABB | a
into Greibach normal form.
Solution:
First substitute A by aaA | b in B 6 ABB. The result is
i.e.
B 6 bAb | bBB | a | aaABB | bBB,
B 6 bAb | bBB | a | aaABB
Now we substitute A by aaA | b in S → ABb. The result is
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S → aaABb | bAAb | a
Hence now we have:
S → aaABb | bAAb | a
A → aaA | b
B 6 bAb | bBB | a | aaABB
which is almost Graibach. We just need to add Ca 6 a, Cb 6 b, and replace a and b that are not
first by Ca and Cb, respectively.
Final result:
S → aCaABCb | bAACb | a
A → aCaA | b
B 6 bACb | bBB | a | aCaABB
Ca 6 a
Cb 6 b
11.[10] Construct a pushdown automaton accepting: { anbm ∣ 2n = 3m, n≥0, m≥0 }.
Solution:
For a string aibj to be in { anbm ∣ 2n = 3m, n≥0, m≥0 }, it must be true that i=3k and j=2k for
some k$0. Thus, for each symbol ‘a’ of the input, we push two copies of symbol ‘a’ into the
stack (this can be done in two steps, using the states s1, s2). Then, for each symbol ‘b’ we read
from the input, we match it with three a=s in the stack (this has to be done in three steps, using the
states q1, q2, q3). The automaton looks as follows:
10
For an alternative definition of pushdown automaton (more popular, red stuff in class notes), we
have:
A = (Q, Σ, Γ, s, $, δ, F), where
Q = {s, q1, q2, q3, q4, r}, Σ={a,b}, Γ= {a,b,$}, F = {r} and δ is defined as follows:
δ(s,ε,$) = { (r,ε)} (just to accept empty string)
δ(s,a,$) = {(s,aa$)} (we assume here that the top of the stack is the first symbol on the
left, so ‘a’ in this case)
δ(s,a,a) = {(s,aa)}
δ(s,b,a) = {(q1, ε )}
δ(q1,ε,a) = { (q2,ε)}
δ(q2,ε,a) = { (q3,ε)}
δ(q3,b,a) = { (q4,ε)}
δ(q3,ε,$) = { (r,ε)}
δ(q4,b,a) = { (q1,ε)}
12.[10]
Construct a pushdown automaton accepting: { aibjck ∣ i,j,k≥0, i+k=j }.
Solution:
The basic idea for this problem is to use different states to enforce the order of the occurrences of
the letters a, b and c, and use the stack to meet the requirement of i+k=j.
11
For an alternative definition of pushdown automaton (more popular, red stuff in class notes), we
have:
A = (Q, Σ, Γ, s, $, δ, F), where
Q = {s, q, p, r}, Σ={a,b,c}, Γ= {a,b,c,$}, F ={r} and δ is defined as follows:
δ(s,a,$) = {(s,a$)}
δ(s,b,$) = {(q,b$)}
δ(s,a,a) = {(s,aa)}
δ(s,b,a) = {(q,ε)}
δ(q,b,a) = {(q,ε)}
δ(q,b,$) = {(q,b$)}
δ(q,b,b) = {(p,bb)}
δ(p,c,b) = {(p,ε)}
δ(p,ε,$) = {(r,ε)}
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13.[10]
Construct a pushdown automaton that accept the language generated by the
following grammar:
S → aS | aSbS ∣ε
You may use either the construction from the textbook or the one presented in class.
Solution:
Let L be the language generated by a given grammar and let L= = L - {ε}.
We may use the simpler construction presented in class.
After removing ε-productions and transforming the obtained result into the Graibach normal
form, we may get the following grammar (construction is not unique):
S 6 aS | aSCbS | a | aCbS | aSCb | aCb
Cb 6 b
Now, the construction from the class results in (colours in automata correspond to colours
in grammar, ε treated separately – q0 is final):
13
For an alternative definition of pushdown automaton (more popular, red stuff in class notes), we
have:
A = ({p,r},{a,b},{S, Cb}, p,S ,δ, F), where F ={r} and δ is the following:
δ(p,a,S ) = {(p,SCbS), (p,S), (p,CbS), (p,SCb), (p,Cb),(p,ε)}
δ(p,b,Cb) = { (p,ε)}
δ(p,ε,S ) = {(r,ε)}.
14.[10] Show that the following language is not context-free
{ aibj ∣ i ≤ j2 }.
Solution:
Assume L = { anbj ∣ n≤j2 } is context free, then pumping lemma hold.
Let p be the number in pumping lemma. Let z = aqbp and q=p2. Clearly, |z| ≥p and z∈L.
Therefore, z=uvwxy and |vx| ≥1, |vwx| ≤p, and (∀i ≥0) uviwxiy ∈L. Let us consider two cases,
Case 1) vx = ak, i.e. containing no b’s. Then, z’ = uv2wx2y = aq+kbp, and k ≥1.
Therefore, z’ ∉L, contradiction.
Case 2) vx = ajbk, and k ≥1, i.e. containing at least one b. Then, consider z’ = uv0wx0y.
#a(z’) = q-j =p2-j, and 1≤j ≤p-1 (#a(z’) is the number of ‘a’ in z’)
#b(z’) = p-k, and 1≤k≤p.
So, (#b(z’))2 = (p-k)2 = p2 -2kp +k2
Clearly kp-j >k2 - kp, since kp-j ≥p-(p-1)=1, but k2 - kp = k(k-p) ≤0. Now, it is easy to
verify that, p2-j >p2 -2kp +k2. Therefore, z’ ∉L, contradiction.
Bonus questions.
15.[8]
Transform the pushdown automaton from Example 2.14 (page 115, Figure 2.15 of
the textbook) into an equivalent context free grammar using the construction from
Lemma 27 of the textbook.
Solution:
To simplify notation we will write ‘i’ instead of ‘qi’, so Q=[1,2,3,4}. We start with productions
that are in fact automaton independent (steps 2 and 3 from the top of page 122):
For all i,j = 1,2,3:
Aij 6 Ai1 A1j | Ai2 A2j | Ai3 A3j
14
For all i=1,2,3:
Aii 6 ε
We have Γ = {0,$}. For u=$, we have:
δ(1, ε, ε ) = {(2, $ )} and δ(3, ε, $ ) = {(4, ε)}, so we add A14 6 A23
For u=0, we have :
δ(2, 0, ε ) = {(2, 0 )} and δ(2, 1, 0 ) = {(3, ε)}, so we add A13 6 0A221
and
δ(2, 0, ε ) = {(2, 0 )} and δ(3, 1, 0 ) = {(3, ε)}, so we add A23 6 0A231
16.[10]
Read pages 3-9 (27-29 on the right top) from Lecture Notes 3, or the similar pages
from any other appropriate textbook, and using this method (that does not use the
idea of ‘generalized automaton’) find a regular expression that generates the same
language as the automaton below.
a
C
1
b
b
a
a
2
b
3
15
k=0
k=1
k=2
k=3
r11k a ∪ε
a*
a*
r12k
b
a* b
a*ba*
a*ba* ∪a*ba*b ε ∪(a ∪ba*b) a*b * (a ∪ba*b) a*
r13k
∅
∅
a*ba*b
a*ba*b ∪a*ba*b ε ∪(a ∪ba*b) a*b
r21k
∅
∅
∅
a*b(ε ∪(a ∪ba*b) a*b * ba*
r22k a ∪ε
a*
a*
a* ∪a*b ε ∪(a ∪ba*b) a*b * (a ∪ba*b) a*
r23k
b
b
a* b
a*b ∪ a*b ε ∪(a ∪ba*b) a*b
r31k
b
ba*
ba*
ba* ∪ ε ∪(a ∪ba*b) a*b ε ∪(a ∪ba*b) a*b *ba*
r32k
a
a ∪ba*b
(a ∪ba*b) a*
(a ∪ba*b) a* ∪ ε ∪(a ∪ba*b) a*b ε ∪(a ∪ba*b)
a*b * (a ∪ba*b) a*
r33k
ε
ε
ε ∪(a ∪ba*b) a*b
a* ∪ a*ba*b ε ∪(a ∪ba*b) a*b * ba*
ε ∪(a ∪ba*b) a*b
*
*
*
L(M) = L(r113 ∪ r133 ) =
L( a* ∪ a*ba*b ε ∪(a ∪ba*b) a*b * ba* ∪a*ba*b ∪a*ba*b ε ∪(a ∪ba*b) a*b *
Or, by popular ‘abuse of notation’:
L(M) = a* ∪ a*ba*b ε ∪(a ∪ba*b) a*b * ba* ∪a*ba*b ∪a*ba*b ε ∪(a ∪ba*b) a*b *
17.[10]
Construct a pushdown automaton accepting: { anbm ∣ 2n≠3m, n≥0, m≥0 }.
Solution:
The idea here is to use the pushdown automaton for the language { anbm ∣ 2n = 3m, n≥0, m≥0}
from Question 11, to match every 2 b’s with 3 a=s. First observe that if the state q1 has been
rached them m>0. Moreover if the top of the stack is $ in any of the states q1, q2, or q4, then
2n>3m and if the top of the stack in q3 is ‘a’ and there is no symbol to read, then 2n<3m. The
case of m=0 (which implies n>0) needs to treated separately.
This lead us to the following automaton, where the black part is from the automaton accepting
the language { anbm ∣ 2n = 3m, n≥0,m≥0}, red part deals with m>0 and blue part with m=0.
16
.
For an alternative definition of pushdown automaton (more popular, red stuff in class notes), we
have:
A = (Q, Σ, Γ, s, $, δ, F), where
Q = {s, q1, q2, q3, q4, r, t}, Σ={a,b}, Γ= {a,b,$}, F = {r,t} and δ is defined as follows:
(the part from acceptance of { anbm ∣ 2n≠3m, n≥0, m≥0 }, and beginning of the case m=0).
δ(s,a,$) = {(s,aa$), (t,$)} (we assume here that the top of the stack is the first symbol on
the left, so ‘a’ in this case)
δ(s,a,a) = {(s,aa)}
δ(s,b,a) = {(q1, ε )}
δ(q1,ε,a) = { (q2,ε)}
δ(q2,ε,a) = { (q3,ε)}
δ(q3,b,a) = { (q4,ε)}
δ(q4,b,a) = { (q1,ε)}
(the part that deals with 2n<3m, m>0)
δ(q1,ε,$) = { (r,ε)}
δ(q2,ε,$) = { (r,ε)}
δ(q4,ε,$) = { (r,ε)}
17
(the part that deals with 2n>3m, m>0)
δ(q3, ε,a) = { (r,ε)}
(the remaining part that of m=0)
δ(t,a,$) = {(t, $ )}
18