Automata and Formal Languages
Closure Properties of DFAs
Sipser pages 44 - 47
Lecture 9 Tim Sheard 1
Closure properties of DFAs
Languages captured by DFA’s are closed under
• Union
• Concatenation
• Kleene Star
• Complement
• Intersection
That is to say if L1 and L2 are recognized by a DFA,
then there exists another DFA, L3, such that
1. L3 = complement L1 { x | x ∉ L1 }
2. L3 = L1 ∪ L2 { x | x ∈ L1 or x ∈ L2 }
3. L3 = L1 ∩ L2 { x | x ∈ L1 and x ∈ L2 }
4. L3 = L1*
5. L3 = L1 • L2 (The first 3 are easy, we’ll wait on 4 and 5)
Complement
Complementation
Take a DFA for L and change the status - final
or non-final - of all its states. The resulting
DFA will accept exactly those strings that the
first one rejects. It is, therefore, a DFA for the
Complent(L).
Thus, the complement of DFA recognizable
language is DFA recognizable.
Complement Example
Contains a “0”
Contains only “1”
2nd Complement Example
Just “abc”
Anything but “abc”
As a Haskell Program
compDFA :: (Ord q) => DFA q s -> DFA q s
compDFA m = DFA (states m)
(symbols m)
(trans m)
(start m)
new
where new = [ s
| s <- states m
, not(elem s (accept m))]
Intersection
The intersection L ∩ M of two DFA
recognizable languages must be
recognizable by a DFA too. A constructive
way to show this is to construct a new
DFA from 2 old ones.
Constructive Proof
The proof is based on a construction that given two
DFAs A and B, produces a third DFA C such that
L(C) = L(A) ∩ L(B). The states of C are pairs
(p,q) , where p is a state of A and q is a state
of B. A transition labeled a leads from (p,q) to
(p',q') iff there are transitions
p
→ p '
a
q
→
a
q'
in A and B. The start state is the pair of original
start states; the final states are pairs of original
final states. The transition function
δA∩Β(q,a) = ( δA(q,a), δB(q,a) )
This is called the product construction.
Example 1 aa+aaa+aaaa
a+aa+aaa
What is the
intersection?
Make a new DFA
where states of the
new DFA are pairs of
states form the old
ones
Automata and Formal Languages
Lecture 9 Tim Sheard 10
Reachable states only
Intersection
{a,aa,aaa} ∩ {aa,aaa,aaaa}
Example 2
1
p 0 q A – string contains a 0
0,1
0
1
r s
0,1 B – string contains a 1
C – string contains a
0 and a 1
Automata and Formal Languages
Contains a “0”
Contains both a
“1” and a “0”
Contains a “1”
Lecture 9 Tim Sheard 13
As a Haskell Program
intersectDFA (DFA bigQ1 s1 d1 q10 f1)
(DFA bigQ2 s2 d2 q20 f2)
= DFA [(q1,q2) | q1 <- bigQ1, q2 <- bigQ2]
s1
(\ (q1,q2) a -> (d1 q1 a, d2 q2 a))
(q10,q20)
[(q1,q2) | q1 <- f1, q2 <- f2])
Difference
The identity:
L - M = L ∩ C(M)
reduces the closure under set-theoretical
difference operator to closure under
complementation and intersection.
M={aa,aaa,aaaa}
Example Difference
L={a,aa,aaa}
L - M = L ∩ C(M)
- =
Union
• The union of the languages of two DFAs (over the
same alphabet) is recognizable by another DFA.
• We reuse the product construction of the intersection
proof, but widen what is in the final states of the
constructed result.
Let A=(Qa,Σ,Ta,sa,Fa) and
B = (Qb,Σ,Tb,sb,Fb)
Then: A ∪ B =((Qa×Qb),Σ,δ,(sa,sb),Final)
Final = { (p,q) | p ∈ Fa, q ∈ Qb} ∪
{ (p,q) | p ∈ Qa, q ∈ Fb}
δ((a,b),x) = ( Τa(a,x), Τb(b,y) )
Automata and Formal Languages
B={bc}
A={ab}
A ∪ B ={ab,bc}
Lecture 9 Tim Sheard 18
Only reachable from start
As a Haskell Program
unionDFA (DFA bigQ1 s1 d1 q10 f1)
(DFA bigQ2 s2 d2 q20 f2)
= DFA [(q1,q2) | q1 <- bigQ1, q2 <- bigQ2]
s1
(\ (q1,q2) a -> (d1 q1 a, d2 q2 a))
(q10,q20)
([(q1,q2) | q1 <- f1, q2 <- bigQ2] ++
[(q1,q2) | q1 <- bigQ1, q2 <- f2])
Example Closure Construction
Given a language L, let L' be the set of all
prefixes of even length strings which
belong to L. We prove that if L is
regular then L' is also regular.
It is easy to show that prefix(L) is regular
when L is (How?). We also know that the
language Even of even length strings is
regular (How?). All we need now is to
note that
L' = Even ∩ prefix(L)
and use closure under intersection.
What’s next
We have given constructions for showing
that DFAs are closed under
1. Complement
2. Intersection
3. Difference
4. Union
In order to establish the closure properties
of
1. Reversal
2. Kleene star
3. Concatenation
We will need to introduce a new
computational system.
