Introduction to Formal Language, Fall 2017
14-Mar-2017
Homework 1 Solutions
Instructor: Prof. Wen-Guey Tzeng
Scribe: Amir Rezapour
1. How many substrings aab are in wwR w, where w = aabbab?
Answer. aabbabbabbaaaabbab.
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2. Let L = {ab, aa, baa}. Which of the following strings are in L∗ : abaabaaabaa,
aaaabaaaa, baaaaabaaaab, baaaaabaa? Which strings are in L4 ?
Answer.
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aa aab aa aa , baa aa ab aa
3. For Σ = {a, b}, construct dfa’s that accept the sets consisting of all the strings with
at least two a’s.
Answer.
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4. Give a dfa for language L = {ab4 wb2 : w ∈ {a, b}∗ }.
Answer.
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1-1
5. Find a dfa for L = {w : (na (w) − nb (w)) mod 3 = 0}, where Σ = {a, b}. Answer.
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c = (Q, Σ, δ, q0 , Q − F ) are two dfa’s, then
6. Show that if M = (Q, Σ, δ, q0 , F ) and M
c).
L(M ) = L(M
Answer.
L(M ) = {w ∈ Σ∗ : δ ∗ (q0 , w) ∈ F }
c).
L(M ) = {w ∈ Σ∗ : δ ∗ (q0 , w) ∈
/ F } = {w ∈ Σ∗ : δ ∗ (q0 , w) ∈ Q − F } = L(M
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7. Construct a dfa for all strings on {0, 1} of length five or more in which the third
symbol from the right end is different from the leftmost symbol.
Answer.
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1-2
8. Let L be the language accepted by the automaton in Figure 1. Find a dfa that accepts
the language L2 − L.
Figure 1: Dfa
Answer.
Figure 2: Question 8
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9. In Figure 3, find δ ∗ (q0 , 1011) and δ ∗ (q1 , 01).
Figure 3: Nfa
Answer.
δ ∗ (q0 , 1011) → {q2 }
δ ∗ (q1 , 01) → {q1 }
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10. Design an nfa with no more than five states for the set {ababn : n ≥ 0}∪{aban : n ≥ 0}.
Answer.
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1-3
Figure 4: Nfa
11. Find an nfa with no more than three states that accepts the language L = {an : n ≥
1} ∪ {bm ak : m ≥ 0, k ≥ 0}.
Answer.
Figure 5: Nfa
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12. Convert the nfa defined by
δ(q0 , a) = {q0 , q1 }
δ(q1 , b) = {q1 , q2 }
δ(q2 , a) = {q2 }
(1)
δ(q0 , λ) = {q2 }
with initial state q0 and the final state q2 into an equivalent dfa.
Answer.
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13. Prove that for every nfa with an arbitrary number of final states there is an equivalent
nfa with only one final state. Can we make a similar claim for dfa’s?
Answer. We can create one final state qf inal and connect nfa’s final states to the
qf inal and set them to non-final state.
Since λ transition doesn’t exist in dfa, therefore we cannot always find the corresponding dfa with only one final state.
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1-4
14. Show that if L is regular, so is LR .
Answer.
Informally, the following procedure shows that we can always construct a finite accepter MLR that accepts LR for a given finite accepter ML that accepts L.
(a) Reverse all transitions in ML .
(b) Add a new initial stare qs and generate λ-transitions from qs to each of the final
states in ML .
(c) Turn all final states of ML into normal states of MLR and turn the initial states
of ML into a final state of MLR .
Formally, let ML = (QL , Σ, δL , q0 , FL ) be an NFA that accepts L. The NFA MLR below
accepts LR .
(a) MLR = (QL ∪ qs , Σ, δLR , qs, q0) and qs ∈
/ QL .
(b) δL (qi , a) = qj ←→ δLR (qj , a) = qi for all a ∈ Σ and qi , qj ∈ QL .
(c) δLR (qs , λ) = q for all q ∈ FL .
From (b), if w ∈ L(ML ), δL (q0 , w) = qj ∈ FL ←→ δL (qj , w) = q0 . From (c),
δLR (qs , λ) = qj for all qj ∈ FL . Thus, we have that for all w ∈ L(ML ), wR ∈ L(MLR ).
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15. From a language L, we create a new language, choplef t(L), by removing the leftmost
symbol of every string in L. Specifically,
choplef t(L) = {w : vw ∈ L, with|v| = 1}.
(2)
Show that if L is regular, then choplef t(L) is also regular.
Answer.
If L is regular, then there exist a dfa D that accepts L. Construct an nfa that simulates
D. Create a new start state qstart and link it to all states that D’s start state link to
with a λ transition. This nfa accepts choplef t(L), therefore choplef t(L) is regular.
Formally, let ML = (QL , Σ, δL , q0 , FL ) be an dfa that accepts L. The NFA Mchoplef t(L)
below accepts choplef t(L).
(a) Mchoplef t(L) = (QL ∪ qstart , Σ, δL , S, FL ) and qstart ∈
/ QL .
(b) δL (qi , a) = qj ←→ δL (qstart , λ) = qj for all a ∈ Σ and qi ∈ S.
From (b), Mchoplef t(L) accepts choplef t(L) = {w : vw ∈ L, with|v| = 1}.
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16. Minimize the number of the states in the following dfa.
Answer.
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1-5
1-6