Automata and Formal Languages - CM0081
Finite Automata and Regular Expressions
Andrés Sicard-Ramírez
Universidad EAFIT
Semester 2017-2
Introduction
DFA
NFA
𝜀-NFA
RE
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
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From Finite Automata to Regular Expressions
Theorem (Hopcroft, Motwani and Ullman [2007], Theorem 3.4)
If 𝐿 = 𝐿(𝐴) for some DFA 𝐴, then there is a regular expression 𝑅
such that 𝐿 = 𝐿(𝑅).
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
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From Finite Automata to Regular Expressions
Inductive proof on the number of states of the automaton
Let the states of 𝐴 be {1, 2, … , 𝑛} with 1 the start state.
∗
Hopcroft, Motwani and Ullman [2007, Fig. 3.2].
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
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From Finite Automata to Regular Expressions
Inductive proof on the number of states of the automaton
Let the states of 𝐴 be {1, 2, … , 𝑛} with 1 the start state.
𝑘
𝑅𝑖𝑗
: Regular expression describing the set of labels of all paths
in 𝐴 from state 𝑖 to state 𝑗 such that the path has no
intermediate node whose number is greater than 𝑘.∗
∗
Hopcroft, Motwani and Ullman [2007, Fig. 3.2].
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From Finite Automata to Regular Expressions
Basis step: Proof for 𝑘 = 0 (i.e. no intermediate states) and 𝑖 ≠ 𝑗
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From Finite Automata to Regular Expressions
Basis step: Proof for 𝑘 = 0 (i.e. no intermediate states) and 𝑖 ≠ 𝑗
No transition from state 𝑖 to state 𝑗
𝑖
𝑗
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
0
𝑅𝑖𝑗
=∅
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From Finite Automata to Regular Expressions
Basis step: Proof for 𝑘 = 0 (i.e. no intermediate states) and 𝑖 ≠ 𝑗
No transition from state 𝑖 to state 𝑗
𝑗
𝑖
0
𝑅𝑖𝑗
=∅
One transition from state 𝑖 to state 𝑗
𝑖
a
𝑗
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
0
𝑅𝑖𝑗
=a
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From Finite Automata to Regular Expressions
Basis step: Proof for 𝑘 = 0 (i.e. no intermediate states) and 𝑖 ≠ 𝑗
No transition from state 𝑖 to state 𝑗
𝑗
𝑖
0
𝑅𝑖𝑗
=∅
One transition from state 𝑖 to state 𝑗
𝑖
a
𝑗
0
𝑅𝑖𝑗
=a
Various transitions from state 𝑖 to state 𝑗
a1
0
…
𝑅𝑖𝑗
= a1 + ⋯ + an
𝑗
𝑖
a𝑛
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From Finite Automata to Regular Expressions
Basis step: Proof for 𝑘 = 0 (i.e. no intermediate states) and 𝑖 = 𝑗
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From Finite Automata to Regular Expressions
Basis step: Proof for 𝑘 = 0 (i.e. no intermediate states) and 𝑖 = 𝑗
No loops
𝑖
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
0
𝑅𝑖𝑖
=𝜀
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From Finite Automata to Regular Expressions
Basis step: Proof for 𝑘 = 0 (i.e. no intermediate states) and 𝑖 = 𝑗
No loops
0
𝑅𝑖𝑖
=𝜀
𝑖
One loop
a
𝑖
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
0
𝑅𝑖𝑖
=𝜀+a
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From Finite Automata to Regular Expressions
Basis step: Proof for 𝑘 = 0 (i.e. no intermediate states) and 𝑖 = 𝑗
No loops
0
𝑅𝑖𝑖
=𝜀
𝑖
One loop
a
𝑖
0
𝑅𝑖𝑖
=𝜀+a
0
0
Question: Why not define 𝑅𝑖𝑖
by 𝑅𝑖𝑖
= 𝜀 + a∗ ?
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
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From Finite Automata to Regular Expressions
Basis step: Proof for 𝑘 = 0 (i.e. no intermediate states) and 𝑖 = 𝑗
No loops
0
𝑅𝑖𝑖
=𝜀
𝑖
One loop
a
0
𝑅𝑖𝑖
=𝜀+a
𝑖
0
0
Question: Why not define 𝑅𝑖𝑖
by 𝑅𝑖𝑖
= 𝜀 + a∗ ?
Various loops
a1
𝑖
…
a𝑛
0
𝑅𝑖𝑖
= 𝜀 + a1 + ⋯ + an
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From Finite Automata to Regular Expressions
Inductive step: Proof for 𝑘
𝑘−1
Inductive hypothesis 𝑅𝑖𝑗
: path from state 𝑖 to state 𝑗 that goes
through no state higher than 𝑘 − 1.
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From Finite Automata to Regular Expressions
Inductive step: Proof for 𝑘
𝑘−1
Inductive hypothesis 𝑅𝑖𝑗
: path from state 𝑖 to state 𝑗 that goes
through no state higher than 𝑘 − 1.
The path does not through state 𝑘 at all
𝑖
𝑘
𝑗
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
𝑘
𝑘−1
𝑅𝑖𝑗
= 𝑅𝑖𝑗
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From Finite Automata to Regular Expressions
Inductive step: Proof for 𝑘
𝑘−1
Inductive hypothesis 𝑅𝑖𝑗
: path from state 𝑖 to state 𝑗 that goes
through no state higher than 𝑘 − 1.
The path does not through state 𝑘 at all
𝑖
𝑘
𝑘−1
𝑅𝑖𝑗
= 𝑅𝑖𝑗
𝑗
𝑘
The path goes through state 𝑘 at least once
𝑖
𝑘
𝑘
𝑘
𝑗
𝑘
𝑘−1
𝑘−1 ∗ 𝑘−1
𝑅𝑖𝑗
= 𝑅𝑖𝑘
(𝑅𝑘𝑘
) 𝑅𝑘𝑗
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From Finite Automata to Regular Expressions
Inductive step: Proof for 𝑘
𝑘−1
Inductive hypothesis 𝑅𝑖𝑗
: path from state 𝑖 to state 𝑗 that goes
through no state higher than 𝑘 − 1.
The path does not through state 𝑘 at all
𝑖
𝑘
𝑘−1
𝑅𝑖𝑗
= 𝑅𝑖𝑗
𝑗
𝑘
The path goes through state 𝑘 at least once
𝑖
𝑘
𝑘
𝑘
𝑗
𝑘
𝑘−1
𝑘−1 ∗ 𝑘−1
𝑅𝑖𝑗
= 𝑅𝑖𝑘
(𝑅𝑘𝑘
) 𝑅𝑘𝑗
From the previous cases:
𝑘
𝑘−1
𝑘−1
𝑘−1 ∗ 𝑘−1
𝑅𝑖𝑗
= 𝑅𝑖𝑗
+ 𝑅𝑖𝑘
(𝑅𝑘𝑘
) 𝑅𝑘𝑗
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From Finite Automata to Regular Expressions
Given that the states of 𝐴 are {1, 2, … , 𝑛} with 1 the start state,
then
𝑛
𝑛
𝐿(𝐴) = 𝐿(𝑅1𝑓
+ ⋯ + 𝑅1𝑓
) with 𝑓𝑖 ∈ 𝐹 .
1
𝑚
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From Finite Automata to Regular Expressions
Example
To convert the DFA 𝐴 to a regular expression.
1
start
1
0
2
0, 1
𝐿(𝐴) = {𝑤 ∈ {0, 1}∗ ∣ 𝑤 has at least one 0}.
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From Finite Automata to Regular Expressions
Example (cont.)
1
start
1
0
2
0, 1
0
𝑅𝑖𝑗
0
𝑅𝑖𝑗
Regexp
0
𝑅11
𝜀+1
0
𝑅12
0
0
𝑅21
∅
0
𝑅22
𝜀+0+1
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From Finite Automata to Regular Expressions
Example (cont.)
1
0
0
0 ∗ 0
𝑅𝑖𝑗
= 𝑅𝑖𝑗
+ 𝑅𝑖1
(𝑅11
) 𝑅1𝑗
0
𝑅𝑖𝑗
Regexp
1
𝑅𝑖𝑗
Regexp
0
𝑅11
𝜀+1
1
𝑅11
(𝜀 + 1) + (𝜀 + 1)(𝜀 + 1)∗ (𝜀 + 1)
0
𝑅12
0
1
𝑅12
0 + (𝜀 + 1)(𝜀 + 1)∗ 0
0
𝑅21
∅
1
𝑅21
∅ + ∅(𝜀 + 1)∗ (𝜀 + 1)
0
𝑅22
𝜀+0+1
1
𝑅22
𝜀 + 0 + 1 + ∅(𝜀 + 1)∗ 0
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From Finite Automata to Regular Expressions
Example (cont.)
Some simplifications for regular expressions
Let 𝐿 and 𝑀 be regular expression variables.
(𝜀 + 𝐿)∗ = 𝐿∗
(𝜀 + 𝐿)𝐿∗ = 𝐿∗
𝐿 + 𝑀 ∗𝐿 = 𝑀 ∗𝐿
𝐿∅ = ∅𝐿 = ∅
(∅ is the annihilator for concatenation)
𝐿 + ∅ = ∅ + 𝐿 = 𝐿 (∅ is the identity for union)
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From Finite Automata to Regular Expressions
Example (cont.)
1
0
0
0 ∗ 0
𝑅𝑖𝑗
= 𝑅𝑖𝑗
+ 𝑅𝑖1
(𝑅11
) 𝑅1𝑗
1
𝑅𝑖𝑗
Regexp
Simplified
1
𝑅11
(𝜀 + 1) + (𝜀 + 1)(𝜀 + 1)∗ (𝜀 + 1)
1∗
1
𝑅12
0 + (𝜀 + 1)(𝜀 + 1)∗ 0
1∗ 0
1
𝑅21
∅ + ∅(𝜀 + 1)∗ (𝜀 + 1)
∅
1
𝑅22
𝜀 + 0 + 1 + ∅(𝜀 + 1)∗ 0
𝜀+0+1
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From Finite Automata to Regular Expressions
Example (cont.)
2
1
1
1 ∗ 1
𝑅𝑖𝑗
= 𝑅𝑖𝑗
+ 𝑅𝑖2
(𝑅22
) 𝑅2𝑗
1
𝑅𝑖𝑗
Regexp
2
𝑅𝑖𝑗
1
𝑅11
1
𝑅12
1
𝑅21
1
𝑅22
1∗
2
𝑅11
2
𝑅12
2
𝑅21
2
𝑅22
1∗ 0
∅
𝜀+0+1
Regexp
1∗
+
1∗ 0(𝜀
+ 0 + 1)∗ ∅
1∗ 0 + 1∗ 0(𝜀 + 0 + 1)∗ (𝜀 + 0 + 1)
∅ + (𝜀 + 0 + 1)(𝜀 + 0 + 1)∗ ∅
𝜀 + 0 + 1 + (𝜀 + 0 + 1)(𝜀 + 0 + 1)∗ (𝜀 + 0 + 1)
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From Finite Automata to Regular Expressions
Example (cont.)
2
1
1
1 ∗ 1
𝑅𝑖𝑗
= 𝑅𝑖𝑗
+ 𝑅𝑖2
(𝑅22
) 𝑅2𝑗
2
𝑅𝑖𝑗
2
𝑅11
2
𝑅12
2
𝑅21
2
𝑅22
Regexp
1∗
+
1∗ 0(𝜀
+0+
Simplified
1)∗ ∅
1∗
1∗ 0 + 1∗ 0(𝜀 + 0 + 1)∗ (𝜀 + 0 + 1)
1∗ 0(0 + 1)∗
∅ + (𝜀 + 0 + 1)(𝜀 + 0 + 1)∗ ∅
∅
𝜀 + 0 + 1 + (𝜀 + 0 + 1)(𝜀 + 0 + 1)∗ (𝜀 + 0 + 1)
(0 + 1)∗
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From Finite Automata to Regular Expressions
Example (cont.)
The regular expression equivalent to the automaton 𝐴 is
2
𝑅12
= 1∗ 0(0 + 1)∗ .
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From Regular Expressions to Finite Automata
Theorem (Hopcroft, Motwani and Ullman [2007], Theorem 3.7)
Every language defined by a regular expression is also defined by a
finite automaton.
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From Regular Expressions to Finite Automata
Theorem (Hopcroft, Motwani and Ullman [2007], Theorem 3.7)
Every language defined by a regular expression is also defined by a
finite automaton.
Inductive proof
For each regular expression 𝑅 we construct an 𝜀-NFA 𝐸 such that
𝐿(𝑅) = 𝐿(𝐸) with:
1. exactly one accepting state,
2. no arcs into the initial state and
3. no arcs out of the accepting state.
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struct and -NFA A, s.t. L(A) = L(R).
From Regular
Expressions to Finite Automata
Proof: By structural induction:
Basis step: Automata for 𝜀 (a), ∅ (b) and a (c).∗
Basis: Automata for , ∅, and a.
ε
(a)
(b)
a
(c)
73
∗
Hopcroft, Motwani and Ullman [2007, Fig. 3.16].
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From Regular Expressions to Finite Automata
3.2.
FINITE
AUTOMATA
AND
Inductive
step:
Automaton
for REGULAR
𝑅 + 𝑆.∗ EXPRESSIONS
ε
ε
R
S
105
ε
ε
(a)
R
ε
S
(b)
∗
Hopcroft, Motwani and Ullman [2007, Fig.ε 3.17].
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ε
ε
From Regular Expressions to Finite Automata
S
Inductive step: Automaton for 𝑅𝑆.(∗a)
ε
R
S
(b)
ε
ε
R
ε
ε
(c)
Figure
3.17: The inductive step in the regular-expression-to--NFA construction
∗
Hopcroft, Motwani and Ullman [2007, Fig. 3.17].
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
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ε
R
S
From Regular Expressions to Finite Automata
(b)
Inductive step: Automaton for 𝑅∗ .∗
ε
ε
R
ε
ε
(c)
Figure 3.17: The inductive step in the regular-expression-to--NFA construction
automaton of Fig. 3.17(c). That automaton allows us to go either:
(a) Directly from the start state to the accepting state along a path
labeled . That path lets us accept , which is in L(R ) no matter
what
expression
is. [2007, Fig. 3.17].
∗
Hopcroft, Motwani
and R
Ullman
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From Regular Expressions to Finite Automata
Example
Convert the regular expression 𝑅 = (0 + 1)∗ 1(0 + 1) to an 𝜀-NFA 𝐸
such 𝐿(𝑅) = 𝐿(𝐸).
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From Regular Expressions to Finite Automata
Example
Convert the regular expression 𝑅 = (0 + 1)∗ 1(0 + 1) to an 𝜀-NFA 𝐸
such 𝐿(𝑅) = 𝐿(𝐸).
1. 𝜀-NFA for 0 + 1:
𝜀
0
𝜀
start
𝜀
1
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
𝜀
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From Regular Expressions to Finite Automata
Example (cont.)
2. 𝜀-NFA for (0 + 1)∗ :
start
𝜀
𝜀
𝜀
𝜀
0
1
𝜀
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
𝜀
𝜀
𝜀
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From Regular Expressions to Finite Automata
Example (cont.)
3. 𝜀-NFA for (0 + 1)∗ 1:
start
𝜀
𝜀
𝜀
𝜀
0
1
𝜀
𝜀
𝜀
𝜀
1
𝜀
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From Regular Expressions to Finite Automata
Example (cont.)
4. 𝜀-NFA for (0 + 1)∗ 1(0 + 1):
start
𝜀
𝜀
𝜀
𝜀
0
1
𝜀
𝜀
𝜀
𝜀
𝜀
Automata and Formal Languages - CM0081. Finite Automata and Regular Expressions
1
𝜀
𝜀
𝜀
0
1
𝜀
𝜀
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References
Hopcroft, J. E., Motwani, R. and Ullman, J. D. (2007). Introduction to
Automata theory, Languages, and Computation. 3rd ed. Pearson
Education.
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