Fachbereich Mathematik
Kord Eickmeyer
Summer Term 2015
Exercise Sheet no. 1 – Formal Foundations of Computer Science
(E1.1) [warm-up: bracketing and regular expressions]
Give all α ∈ REG({0, 1}) which, after removing all parentheses, lead to 0+10∗. Which
is the one we mean by 0+10∗? What are the languages L(α) for these α?
Solution. Possible bracketings are:
(i) (0+1)(0∗)
(ii) ((0+1)0)∗
(iii) (0+((10)∗))
(iv) (0+(1(0∗)))
(v) (0+(10))∗
Of these, (0+(1(0∗))) is what we usually mean.
(E1.2) [a sample DFA]
Consider the following DFA over the alphabet Σ = {0, 1}:
1
0
q0
1
q1
0
(i) Write this automaton formally as a tuple (Q, Σ, δ, q0 , F ), i.e. give the sets Q and
F , the initial state q0 and the transition function δ.
(ii) Describe in words the language recognised by this automaton.
(iii) Give a regular expression describing the same language.
Solution.
(i) A = (Q, Σ, δ, q0 , F ) with
Q := {q0 , q1 },
Σ := {0, 1},
0 1
δ := q0 q0 q1
q1 q0 q1
F := {q1 }.
(ii) The automaton accepts exactly those words whose last letter is a 1.
(iii) L(A) = L((0+1)∗1).
(E1.3) [regular languages]
For each of these languages, give a regular expression Ei and a deterministic finite automaton Ai such that Li = L(Ei ) = L(Ai ):
L1 := {w ∈ {a, b, c}∗ |w|a ≥ 1 and |w|b ≥ 2},
L2 := {w ∈ {0, 1}∗ there is at most one pair of adjacent 1s in w},
L3 := {w ∈ {0, 1}∗ every pair of adjacent 0s appears before any pair of adjacent 1s}.
Solution. Let α := (a+b)∗. Then
L1 = L(αaαbαbα + αbαaαbα + αbαbαaα).
An automaton for L1 is
a
a
a, b
≥1, 0
b
≥1, 1
b
a
a
0, 0
≥1, ≥2
b
b
0, 1
a
b
0, ≥2
Here, a state label 0, ≥2 means that words for which the automaton is in this state have
0 a’s and at least two b’s.
(E1.4) [converting NFAs to DFAs]
Sketch a transition diagram for the NFA A := ({p, q, r, s, t}, {0, 1}, δ, p, {s, t}), with δ
as below. Convert it into a DFA, sketch the resulting transition diagram. Informally
describe the language it accepts.
δ
:=
p
q
r
s t
0 {p, q} {r, s} {p, r} ∅ ∅
1 {p}
{t}
{t} ∅ ∅
(E1.5) [syntactic sugar for regular expressions]
In applications, regular expressions are often allowed to contain additional forms of subexpressions. Let Σ be a finite alphabet and define the set EREG(Σ) of extended regular
expressions as
• REG(Σ) ⊆ EREG(Σ),
• E+ ∈ EREG(Σ) for every E ∈ EREG(Σ), and
• E{m, n} ∈ EREG(Σ) for every E ∈ EREG(Σ) and m ≤ n ∈
N.
The inteded meaning of the additional constructs E+ and E{m, n} is that
L(E+) := {w1 , . . . , wk k ≥ 1 and w1 , . . . , wk , ∈ L(E)},
L(E{m, n}) := {w1 , . . . , wk m ≤ k ≤ n and w1 , . . . , wk ∈ L(E)}.
Show that these extensions do not increase the class of of definable languages: For every
α ∈ EREG(Σ) there is an α0 ∈ REG(Σ) such that L(α) = L(α0 ).
(E1.6) [outlook: languages which are not regular]
Consider the alphabet Σ = {(, ), a, ◦} and the language L ⊆ Σ∗ defined as the smallest
language for which
• a∈L
• for every α1 , . . . , αk ∈ L also (α1 ◦ · · · ◦ αk ) ∈ L.
In other words, L contains all correctly bracketed expressions:
L = {a, (a ◦ a), (a ◦ a ◦ a), (a ◦ (a ◦ a)), ((a ◦ a) ◦ a), . . .}.
We will see later that L is not a regular language. Though we lack the tools to prove
this now, it is instructive to try constructing a regular expression or an automaton for L
to see where you get stuck.
If we restrict the nesting depth in the bracketed expressions, the resulting languages do
become regular: Define
L0 := {a},
Ln+1 := Ln ∪ {(α1 ◦ . . . ◦ αk ) | α1 , . . . , αk ∈ Ln }.
Then each Ln is regular. For each n ≥ 0, give an automaton A with L(A) = Ln .