ASSIGNMENT ONE SOLUTIONS
MATH 4805 / COMP 4805 / MATH 5605
(1) (a) (0 + 1)∗ 010 (finite automata below).
(b) First observe that the following regular expression generates the binary strings
with an even number of 0s and an odd number of 1s
r = (00 + 11 + (01 + 10)(00 + 11)∗ (01 + 10))∗ .
Next observe that the following regular expression generates the binary strings
with an even number of 0s and an odd number of 1s with the additional property
that no prefix has even number of 0s and an odd number of 1s
s = (1 + (01 + 10)(00 + 11)∗ 0) .
Any string that has an even number of 0s and an odd number of 1s must have a
longest prefix that has an even number of 0s and an even number of 1s, followed
by a suffix that has an even number of 0s and an odd number of 1s. Therefore,
a regular expression for the language is rs (finite automata below).
(2) (a) Connect passwords “Must be between 6 and 8 characters long” so there are a
finite number of possibilities. Thus, connect passwords are a finite language, and
hence a recognizable language.
(b) Consider the following languages that are recognizable (or equivalently, regular).
• The finite automaton below accepts strings that have length at least six.
• Given any a, b, c ∈ A the following regular expression generates strings
that have the substring abc: A∗ abcA∗ . Therefore, by the closure of regular
languages under union, the strings containing a substring of length three
in any fixed login name is a regular language. Therefore, by the closure
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•
•
•
•
of regular language under complement, the strings that do not contain a
substring of length three in any fixed login name is a regular language.
The following regular expression generates strings that have a substring
that is a postal code: A∗ U DU DU DA∗ . Therefore, by the closure of regular
language under complement, the strings that do not contain a substring
that is a postal code is a regular language.
The following regular expression generates strings that have a substring
that is a license plate: A∗ U U U DDDA∗ . Therefore, by the closure of
regular language under complement, the strings that do not contain a
substring that is a license plate is a regular language.
In the previous three bullets, replace the initial regular expressions by (i)
A∗ cbaA∗ , (ii) A∗ DU DU DU A∗ , and (iii) A∗ DDDU U U A∗ to prove that
the strings whose reverse does not contain the given substrings is a regular
language, respectively.
Next we consider avoiding the desired substrings in ww instead of the reverse of w. In the previous three bullets, replace the initial regular expressions by (i) A∗ abcA∗ +bcA∗ a+cA∗ ab, (ii) A∗ U DU DU DA∗ +DA∗ U DU DU +
U DA∗ U DU D + DU DA∗ U DU + U DU DA∗ U D + DU DU DA∗ U , and (iii)
A∗ U U U DDDA∗ +DA∗ U U U DD+DDA∗ U U U D+DDDA∗ U U U +U DDDA∗ U U +
U U DDDA∗ U to prove that the strings whose concatenation with themselves (ie ww) does not contain the given substrings is a regular language,
respectively. Observe that if w is a string of length at least six (as per
the first bullet) and s is a string of length at most six (as in the previous
three bullets), then some string in w∗ contains s as a substring if and only
if ww contains s as a substring. Therefore, the strings of length at least
six whose Kleene closure does not contain the given substrings is a regular
language by using the same logic above.
The following non-deterministic finite automaton accepts strings that contain a symbol in Q.
Therefore, by the closure of regular language under complement, the strings
that do not contain a symbol in Q is a regular language.
• Let a1 , a2 , a3 ∈ A. The following regular expression generates strings that
contain only these three symbols: (a1 + a2 + a3 )∗ . By considering all
choices for a1 , a2 , and a3 and by the closure of regular languages under
union, the set of strings containing at most three distinct symbols is a
regular language. By the closure of regular languages under complement,
the set of strings that contain at least four distinct symbols is regular.
• The following regular expression generates strings whose first six symbols
contain at least one symbol from L∪U , one symbol from D, and one symbol
from P . In this regular expression, the terms are simply the permutations
of the multiset {A, A, A, (L+U ), D, P } (ensuring that the first six symbols
contain the desired three types of symbols) followed by A∗ . To be complete,
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the full regular expression appears below.
AAA(L + U )DP A∗ + AAA(L + U )P DA∗ + AAAD(L + U )P A∗ + AAADP (L + U )A∗ + AAAP (L + U )DA∗ +
AAAP D(L + U )A∗ + AA(L + U )ADP A∗ + AA(L + U )AP DA∗ + AA(L + U )DAP A∗ + AA(L + U )DP AA∗ +
AA(L + U )P ADA∗ + AA(L + U )P DAA∗ + AADA(L + U )P A∗ + AADAP (L + U )A∗ + AAD(L + U )AP A∗ +
AAD(L + U )P AA∗ + AADP A(L + U )A∗ + AADP (L + U )AA∗ + AAP A(L + U )DA∗ + AAP AD(L + U )A∗ +
AAP (L + U )ADA∗ + AAP (L + U )DAA∗ + AAP DA(L + U )A∗ + AAP D(L + U )AA∗ + A(L + U )AADP A∗ +
A(L + U )AAP DA∗ + A(L + U )ADAP A∗ + A(L + U )ADP AA∗ + A(L + U )AP ADA∗ + A(L + U )AP DAA∗ +
A(L + U )DAAP A∗ + A(L + U )DAP AA∗ + A(L + U )DP AAA∗ + A(L + U )P AADA∗ + A(L + U )P ADAA∗ +
A(L + U )P DAAA∗ + ADAA(L + U )P A∗ + ADAAP (L + U )A∗ + ADA(L + U )AP A∗ + ADA(L + U )P AA∗ +
ADAP A(L + U )A∗ + ADAP (L + U )AA∗ + AD(L + U )AAP A∗ + AD(L + U )AP AA∗ + AD(L + U )P AAA∗ +
ADP AA(L + U )A∗ + ADP A(L + U )AA∗ + ADP (L + U )AAA∗ + AP AA(L + U )DA∗ + AP AAD(L + U )A∗ +
AP A(L + U )ADA∗ + AP A(L + U )DAA∗ + AP ADA(L + U )A∗ + AP AD(L + U )AA∗ + AP (L + U )AADA∗ +
AP (L + U )ADAA∗ + AP (L + U )DAAA∗ + AP DAA(L + U )A∗ + AP DA(L + U )AA∗ + AP D(L + U )AAA∗ +
(L + U )AAADP A∗ + (L + U )AAAP DA∗ + (L + U )AADAP A∗ + (L + U )AADP AA∗ + (L + U )AAP ADA∗ +
(L + U )AAP DAA∗ + (L + U )ADAAP A∗ + (L + U )ADAP AA∗ + (L + U )ADP AAA∗ + (L + U )AP AADA∗ +
(L + U )AP ADAA∗ + (L + U )AP DAAA∗ + (L + U )DAAAP A∗ + (L + U )DAAP AA∗ + (L + U )DAP AAA∗ +
(L + U )DP AAAA∗ + (L + U )P AAADA∗ + (L + U )P AADAA∗ + (L + U )P ADAAA∗ + (L + U )P DAAAA∗ +
DAAA(L + U )P A∗ + DAAAP (L + U )A∗ + DAA(L + U )AP A∗ + DAA(L + U )P AA∗ + DAAP A(L + U )A∗ +
DAAP (L + U )AA∗ + DA(L + U )AAP A∗ + DA(L + U )AP AA∗ + DA(L + U )P AAA∗ + DAP AA(L + U )A∗ +
DAP A(L + U )AA∗ + DAP (L + U )AAA∗ + D(L + U )AAAP A∗ + D(L + U )AAP AA∗ + D(L + U )AP AAA∗ +
D(L + U )P AAAA∗ + DP AAA(L + U )A∗ + DP AA(L + U )AA∗ + DP A(L + U )AAA∗ + DP (L + U )AAAA∗ +
P AAA(L + U )DA∗ + P AAAD(L + U )A∗ + P AA(L + U )ADA∗ + P AA(L + U )DAA∗ + P AADA(L + U )A∗ +
P AAD(L + U )AA∗ + P A(L + U )AADA∗ + P A(L + U )ADAA∗ + P A(L + U )DAAA∗ + P ADAA(L + U )A∗ +
P ADA(L + U )AA∗ + P AD(L + U )AAA∗ + P (L + U )AAADA∗ + P (L + U )AADAA∗ + P (L + U )ADAAA∗ +
P (L + U )DAAAA∗ + P DAAA(L + U )A∗ + P DAA(L + U )AA∗ + P DA(L + U )AAA∗ + P D(L + U )AAAA∗
For the overall proof, observe that each bullet above proves that certain languages are regular. Moreover, the intersection of these languages are precisely
the passwords defined in the question. Therefore, passwords are a regular language by the closure of regular languages under intersection.
(3) (a) No, Lp is not recognizable. Consider the string w = 1n 01n , where n is the
pumping constant for Lp . Observe that w ∈ Lp and |w| ≥ n. Consider any
x, y, z ∈ {0, 1}∗ such that (i) w = xyz, (ii) |xy| ≤ n, and (iii) |y| > 0. By these
constraints, x = 1a , y = 1b , and z = 1c 01n where a ≥ 0, b > 0, and a + b + c = n.
If i = 0, then xy i z = xz = 1a+c 01n = 1a+c 01a+b+c ∈
/ Lp since b > 0. Therefore,
Lp does not satisfy the pumping property, and hence Lp is not regular.
(b) Yes, Le is recognizable. Suppose w ∈ {0, 1}∗ and let n01 be its number of 01
substrings and let n10 be its number of 10 substrings. Observe that


n10 + 1 if w begins with 0 and w ends with 1
n01 = n10 − 1 if w begins with 1 and w ends with 0

n
otherwise
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where the final case includes (i) w = , (ii) w begins and ends with 0, and (iii)
w begins and ends with 1. Therefore, Le is generated by the regular expression
+ 0 + 1 + 0(0 + 1)∗ 0 + 1(0 + 1)∗ 1. Therefore, Le is regular, and hence Le is
recognizable.
(4) (a) Consider an arbitrary w = w1 w2 · · · wn ∈ La , where each wi ∈ {0, 1}. Observe
that w can be expressed as w = 0r 1s 2t where
r 6= 0 =⇒ (s = t).
Let x = , y = w1 , and z = w2 w3 · · · wn . Observe that these choices ensure (i)
w = xyz, (ii) |xy| ≤ n, and (iii) |y| > 0, where n is the pumping constant for
La . We wish to prove that (iv) xy i z ∈ La for all i ≥ 0. There are two cases to
consider. If r 6= 0, then xy i z = 0r−1+i 1s 2t for all i ≥ 0. In this case, s = t by
the implication given above. Therefore, xy i z = 0r−1+i 1s 2s ∈ L for all i ≥ 0. On
the other and, if r = 0, then xy i z = 1s−1+i 2t . Therefore, xy i z ∈ La for all i ≥ 0.
Therefore, La satisfies the pumping property.
(b) The difference between the pumping property and the generalized pumping property is that the former gives conditions on a prefix of length at most n in each
string, whereas the latter gives conditions on a substring of length at most n in
each string.
(c) Consider the string pws with p = 0, w = 1n 2n , and s = , where n is the
generalized pumping constant. Observe that pws = 01n 2n ∈ La and |w| ≥ n.
Consider any x, y, z ∈ {0, 1, 2}∗ such that (i) w = xyz, (ii) |xy| ≤ n, and (iii)
|y| > 0. By these constraints, x = 1a , y = 1b , and z = 1c 2n where a ≥ 0, b > 0,
and a + b + c = n. If i = 0, then pxy i zs = 0xz = 01a+c 2n = 01a+c 2a+b+c ∈
/ La
since b > 0. Therefore, La does not satisfy the generalized pumping property.
(d) If L is a recognizable language, then there exists some finite automata M =
(S, A, s0 , δ, F ) such that L(M ) = L. Let n = |S| be the number of states in
M . Consider a string pws ∈ L with |w| ≥ n. Let w = w1 w2 · · · wm where
each wi ∈ A and m ≥ n. Consider the states sj = δ ∗ (s0 , pw1 w2 · · · wj ) for
j = 0, 1, . . . , n. Since there are only |S| = n states, there must be a repetition
among s0 , s1 , . . . , sn . Let r and t be chosen such that sr = st and 0 ≤ r < t ≤ n.
Now consider the strings x = w0 w1 · · · wr , y = wr+1 wr+2 · · · w + t, and z =
wr+1 wr+2 · · · wt . Observe that the repeated state implies that δ ∗ (s0 , pxy i zs) =
δ ∗ (st , zs) for all i ≥ 0. Furthermore, the state δ ∗ (st , zs) ∈ F since pws =
pxyzs ∈ L. Therefore, pxy i zs ∈ L for all i ≥ 0.
(5) (a) The language Ls = {w ∈ A∗ | w has suffix s} is recognizable. Therefore, if L
is recognizable then so is L ∩ Ls by the closure of recognizable languages under
intersection. Observe that L ∩ Ls contains all strings in L that have suffix s.
Therefore, L ∩ Ls = L/s, and so L/s is recognizable.
(b) Let S = S1 × S2 , i = {(ii , i2 )}1, F = (F1 × (S2 \F2 )) ∪ (F2 × (S1 \F1 )), and define
δ : S × A → P (S) by δ((x, y), a) = {(δ1 (x, a), δ2 (y, a))} for all (x, y) ∈ S and
a ∈ A. Observe that M is essentially a (deterministic) finite-automata and that
it accepts the desired language.
1This
i should have been stated as I in the question since it is a set of states.
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(c) If L is recognizable, then there exists a finite automata M = (S, A, i, δ, F ) such
that L(M ) = L. We will construct an -finite automata M 0 (S 0 , B, I, δ 0 , F ) as
follows. S 0 is a set of states that includes those in S as well as additional states
to be defined, and the set of initial states I = {i} includes only the single initial
state from M . The basic idea is to replace each arc labeled a in M by a path
whose arcs are labeled with the symbols in f (a)
Formally, define δ 0 : S 0 × B → P (S 0 ) as follows. If f (a) = for a ∈ A, then for
each s ∈ S let δ 0 (s, ) = {δ(s, a)}. If f (a) = b for a ∈ A and b ∈ B, then for
each s ∈ S let δ 0 (s, ) = {δ(s, a)}.
(6) (a) δ ∗ (s1 , 1) = s0 ∈
/ F and δ ∗ (s2 , 1) = s1 ∈ F so s1 M s2 .
(b) See the tables below.
Partition Table
s1 s2 s3 s4
x0 x0 x0
x1
x0
x1 x0
x0
for S/ ∼M
s5 s6
x0 x0 s 0
x1 s 1
x1
s2
x1 s 3
x0 x0 s 4
x1 s 5
Table 1
0
1
(0,4) (1,5) (4,0)
(1,2) (3,1) (2,0)
(1,3) (3,3) (2,6)
(1,5) (3,1) (2,6)
(1,6) (3,1) (2,4)
(2,3) (1,3) (0,6)
(2,5) (1,1) (0,6)
(2,6) (1,1) (0,4)
(3,5) (3,1) (6,6)
(3,6) (3,1) (6,4)
(5,6) (1,1) (6,4)
∗
(c) The language is {w ∈ {0, 1} | w has suffix 0 or 01}. See
below.
Table 2
0
1
(0,4) (1,5) (4,0)
(1,3) (3,3) (2,6)
(1,5) (3,1) (2,6)
(2,6) (1,1) (0,4)
(3,5) (3,1) (6,6)
the minimal automata
(7) The finite automaton is given by the transition table below. (The shortest strings
that distinguish between each pair of states are given in parentheses.)
5
0
1
→ {a}
{b, d}
{b}
()
∗{b}
{c}
{b, c}
(1)
∗{b, d}
{c}
{a, b, c}
(0)
{c}
{d}
{a}
(10)
∗{b, c}
{c, d} {a, b, c}
(11)
∗{a, b, c} {b, c, d} {a, b, c}
(01)
∗{d}
∅
{a}
(100)
∗{c, d}
{d}
{a}
(110)
∗{b, c, d} {c, d} {a, b, c}
(010)
∅
∅
∅
(1000)
In the above table → and ∗ denote the start state and final states, respectively.
(8) (a) See the table below.
(b) See the table below.
(c) See the table below.
state x
E(x)
E(x)· a E(x) · b
s1
{s1 , s2 }
{s4 }
∅
s2
{s2 }
{s4 }
∅
s3
{s1 , s2 , s3 } {s2 , s4 } {s1 }
s4
{s1 , s2 , s3 , s4 } {s2 , s4 } {s1 }
(d) See the NFA below.
E(E(x) · a) E(E(x) · b)
{s1 , s2 , s3 , s4 }
∅
{s1 , s2 , s3 , s4 }
∅
{s1 , s2 , s3 , s4 } {s1 , s2 }
{s1 , s2 , s3 , s4 } {s1 , s2 }
(9) Since L1 and L2 are recognizable languages there are finite automata that accept
these languages. Let M1 = (S1 , A1 , δ1 , F1 , i1 ) and M2 = (S2 , A2 , δ2 , F2 , i2 ) be finite
automata such that L(M1 ) = L1 and L(M2 ) = L2 . Consider the non-deterministic
finite automata M = (S, A, δ, F, i) such that S = S1 × S2 × {1, 2}, A = A1 ∪ A2 ,
F = F1 × F2 × {1}, i = {(i1 , i2 , 1)} and δ is defined as follows. If a1 ∈ A1 then,
δ((s1 , s2 , 1), a1 ) = {(δ1 (s1 , a1 ), s2 , 2)}
and if a2 ∈ A2 then,
δ((s1 , s2 , 2), a2 ) = {(s1 , δ2 (s2 , a2 ), 1)}
and for all other inputs, δ returns the empty set. Observe that M accepts the shuffle of
L1 and L2 . Therefore, the shuffle of two regular languages is also a regular language.
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