1. Properties of Regular Languages
Reading: Chapter 4

2. Pumping Lemma for Regular Languages
Lemma: (the pumping lemma)
Let M be a DFA with |Q| = n states, and let x be a string in L(M), where |x| >= n. Then x = uvw, where u,v, and w are all in Σ* and:
1<= |uv| <= n
|v| >= 1
uviw is in L(M), for all i >= 0
Note this statement of the pumping lemma is slightly different than the books’.

3. Pumping Lemma for Regular Languages
Proof:
Let M be a DFA where |Q| = n, and let x be a string in L(M) where |x|>=n. Furthermore, let x = a1a2 … am where δ(q0, a1a2 … ap) = qjp.
a1 a2 a3 … am
qj0 qj1 qj2 qj3… qjm m>=n and qj0 is q0
Consider the first n symbols, and first n+1 states on the above path:
a1 a2 a3 … an
qj0 qj1 qj2 qj3… qjn
Since |Q| = n, it follows from the pigeon-hole principle that js = jt for some 0<=s<t<=n, i.e., some state appears on this path twice (perhaps many states appear more than once, but at least one does).

4. … … … same cycle q0 qj1 qjs qjt qjn q0 qjs = qjt qjn a1 a2 as as+1 at
… … …
same
cycle q0
qj1
qjs
qjt a1 a2 as
as+1 at
at+1
qjn an q0
qjs = qjt
qjn
a1…as
at+1…an
as+1…at

5. Since 0<=s<t<=n and uv = a1…at it follows that:
Let:
u = a1…as
v = as+1…at
Since 0<=s<t<=n and uv = a1…at it follows that:
1<= |v| and therefore 1<=|uv|
|uv|<=n and therefore 1 <= |uv| <= n
In addition, let:
w = at+1…am
It follows that uviw = a1…as(as+1…at)iat+1…am is in L(M), for all i >= 0. 
In other words, when processing the accepted string x, the cycle was traversed once, but could have been traversed as many times as desired, and the resulting string would still be accepted.

6. q0 0,1 q4 q2 q1 q3 1 Example: n = 5 x = 0001000 is in L(M)
q0 q1 q2 q3 q1 q2 q3 q4
first n+1 states
u = 0
v = 001
w = uviw is in L(M), i.e., 0(001)i000 is in L(M), for all i>= 0 q0
0,1 q4 q2 q1 q3 1

7. Note this does not mean that every string accepted by the DFA has this form:
001 is in L(M) but is not of the form 0(001)i000
Similarly, this doesn’t even mean that every long string accepted by the DFA has this form:
is in L(M), is very long, but is not of the form 0(001)i000
Note, however, in this latter case could be similarly decomposed. q0
0,1 q4 q2 q1 q3 1

8. Note: It may be the case that no x in L(M) has |x| >= n.q0
0,1 q1 q2

9. Example:
n = 4
x = bbbab is in L(M) b b b a b
|x| = q0 q0 q0 q0 q1 q3
u = ε
v = b
w = bbab
(b)ibbab is in L(M), for all i>=0 a b a q0 q1 b b a q2 q3
a,b

10. Example:
n = 4
x = bbbab is in L(M) b b b a b
|x| = q0 q0 q0 q0 q1 q3
u = b
v = b
w = bab
b(b)ibab is in L(M), for all i>=0 a b a q0 q1 b b a q2 q3
a,b

11. Example:
n = 4
x = bbbab is in L(M) b b b a b
|x| = q0 q0 q0 q0 q1 q3
u = bb
v = b
w = ab
bb(b)iab is in L(M), for all i>=0 a b a q0 q1 b b a q2 q3
a,b

12. Example:
n = 4
x = bbbab is in L(M) b b b a b
|x| = q0 q0 q0 q0 q1 q3
u = b
v = bb
w = ab
b(bb)iab is in L(M), for all i>=0 a b a q0 q1 b b a q2 q3
a,b

13. NonRegularity Example
Theorem: The language:
L = {0k1k | k>=0} (1)
is not regular.
Proof: (by contradiction) Suppose that L is regular. Then there exists a DFA M such that:
L = L(M) (2)
We will show that M accepts some strings not in L, contradicting (2).
Suppose that M has n states, and consider a string x=0m1m, where m>>n.
By (1), x is in L.
By (2), x is also in L(M).
Note that L basically corresponds to balanced parenthesis.

14. Also, since x = 0m1m it follows that uv is a substring of 0m.
Since x is very long, i.e., |x| = 2*m >> n, it follows from the pumping lemma that:
x = uvw
1 <= |uv| <= n
1 <= |v|, and
uviw is in L(M), for all i >= 0
Since 1<=|uv|<=n and n<<m, it follows that 1<=|uv|<=m.
Also, since x = 0m1m it follows that uv is a substring of 0m.
In other words v=0j, for some j>=1.
Since uviw is in L(M), for all i>=0, it follows that 0m+cj1m is in L(M), for all c>=1.
But by (2), 0m+cj1m is in L, for any c>=1, a contradiction. 
By the way, 0n1n corresponds to balanced parenthasis.
What does this say about DFAs?
They can’t count arbitrarily high.
But can we do this with a program?
Of course. Count the number of 0’s, count the number of 1’s, and then compare them with an…IF STATEMENT!
Told you if-statements would be helpful!
So what does this all mean…are DFAs a good model of general computing?
Of course not, they can’t even count!
A very primitive device.

15. NonRegularity Example
Theorem: The language:
L = {0k1k2k | k>=0} (1)
is not regular.
Proof: (by contradiction) Suppose that L is regular. Then there exists a DFA M such that:
L = L(M) (2)
We will show that M accepts some strings not in L, contradicting (2).
Suppose that M has n states, and consider a string x=0m1m2m, where m>>n.
By (1), x is in L.
By (2), x is also in L(M).

16. Also, since x = 0m1m2m it follows that uv is a substring of 0m.
Since is very long, i.e., |x| = 3*m >> n, it follows from the pumping lemma that:
x = uvw
1 <= |uv| <= n
1 <= |v|, and
uviw is in L(M), for all i >= 0
Since 1<=|uv|<=n and n<<m, it follows that 1<=|uv|<=m.
Also, since x = 0m1m2m it follows that uv is a substring of 0m.
In other words v=0j, for some j>=1.
Since uviw is in L(M), for all i>=0, it follows that 0m+cj1m2m is in L(M), for all c>=1.
But by (2), 0m+cj1m2m is in L, for any c>=1, a contradiction. 
Note that the above proof is almost identical to the previous proof.

17. NonRegularity Example
Theorem: The language:
L = {0m1n2m+n | m,n >=0} (1)
is not regular.
Proof: (by contradiction) Suppose that L is regular. Then there exists a DFA M such that:
L = L(M) (2)
We will show that M accepts some strings not in L, contradicting (2).
Suppose that M has n states, and consider a string x=0m1n2m+n, where m>>n.
By (1), x is in L.
By (2), x is also in L(M).

18. Since |x| = m >> n, it follows from the pumping lemma that:
x = uvw
1 <= |uv| <= n
1 <= |v|, and
uviw is in L(M), for all i >= 0
Since 1<=|uv|<=n and n<<m, it follows that 1<=|uv|<=m.
Also, since x = 0m1n2m+n it follows that uv is a substring of 0m.
In other words v=0j, for some j>=1.
Since uviw is in L(M), for all i>=0, it follows that 0m+cj1m2m+n is in L(M), for all c>=1. In other words v can be “pumped” as many times as we like, and we still get a string in L(M).
But by (2), 0m+cj1n2m+n is in L, for any c>=1, a contradiction. 
Note that the above proof is almost identical to the previous proof.

19. By the way…
{0n1n | 0=<n} Is not regular
{0n1n | 0=<n<=k, for some fixed k} Is regular, for any fixed k.
For k=3:
L = {ε, 01, 0011, } q0 q1 q2 q3 1 1 1 1
0,1 1 1 q7 q6 q5 q4
0,1

20. The following theorems will all exploit the following observations…
Let Σ be an alphabet, and let |Σ| = k.
How many strings of length n are there? kn
For example, if Σ ={0,1} and n = 3, then there are 8 strings of length 3.

21. Theorem: Let M = (Q, Σ, δ, q0, F) be a DFA
Theorem: Let M = (Q, Σ, δ, q0, F) be a DFA. Then L(M) is non-empty iff there exists an x in L(M) such that |x| < |Q|.
Proof:
(if) Suppose there exists a string x in L(M) such that |x| < |Q|. Then clearly L(M) is non-empty.

22. Theorem: Let M = (Q, Σ, δ, q0, F) be a DFA
Theorem: Let M = (Q, Σ, δ, q0, F) be a DFA. Then L(M) is non-empty iff there exists an x in L(M) such that |x| < |Q|.
(only if) By contradiction.
Suppose that L(M) is non-empty, but that there exists no string x in L(M) such that |x| < |Q|.
It follows that |y| >= |Q|, for all y in L(M).
Let z be a string of shortest length in L(M). Then |z| >= |Q|.
By the pumping lemma z=uvw, |v|>=1 and uviw is in L(M) for all i>=0.
But then uv0w = uw is in L(M) and:
|uw| = |z| - |v|
<= |z| - 1 because |v| >= 1
< |z|
Since uw is in L(M), it follows that z is not a string of shortest length in L(M), a contradiction.

23. Corollary: Let M = (Q, Σ, δ, q0, F) be a DFA
Corollary: Let M = (Q, Σ, δ, q0, F) be a DFA. Then there is an algorithm to determine if L(M) is empty.
Proof:
From the theorem it follows that if L(M) is non-empty then there exists a string x where |x| < n and n = |Q| such that M accepts x. We can try running M on each string of length < n to see if any are accepted. If one is accepted then L(M) is non-empty, otherwise L(M) is empty. Since the number of states |Q| and the number of input symbols |Σ| are both fixed, it follows that there are at most a finite number of strings (of length less than |Q|) that need to be tested.
Note that a simple graph search algorithm works here too…in fact, the above algorithm is a graph search…

24. Theorem: Let M = (Q, Σ, δ, q0, F) be a DFA
Theorem: Let M = (Q, Σ, δ, q0, F) be a DFA. Then L(M) is finite iff |x| < |Q| for all x in L(M).
Proof:
(if) Suppose that |x| < |Q| for all x in L(M). Since the number of states |Q| and the number of input symbols |Σ| are both fixed, it follows that there are at most a finite number of strings of length less than |Q|. It follows that L(M) is finite (exercise: give an upper bound on the number of such strings).
(only if) By contradiction. Suppose that L(M) is finite, but that |x| >= |Q| for some x in L(M). From the pumping lemma it follows that x=uvw, |v|>=1 and uviw is in L(M) for all i>=0. But then L(M) would be infinite, a contradiction. 

25. Theorem: Let M = (Q, Σ, δ, q0, F) be a DFA
Theorem: Let M = (Q, Σ, δ, q0, F) be a DFA. Then L(M) is infinite iff there exists an x in L(M) such that |x| >= |Q|.
Proof:
(if) Suppose there exists an x in L(M) such that |x| >= |Q|. From the pumping lemma it follows that x=uvw, |v|>=1 and uviw is in L(M) for all i>=0. Therefore L(M) is infinite.
(only if) By contradiction. Suppose that L(M) is infinite, but that there is no x in L(M) such that |x| >= |Q|. It follows that each x in L(M) has length less than |Q|. Since the number of states |Q| and the number of input symbols |Σ| are both fixed, it follows that there are at most a finite number of strings of length less than |Q|. It follows that L(M) is finite. A contradiction. 
Note that the above also follows directly from the previous theorem.

26. Theorem: Let M = (Q, Σ, δ, q0, F) be a DFA and let n = |Q|
Theorem: Let M = (Q, Σ, δ, q0, F) be a DFA and let n = |Q|. Then L(M) is infinite iff there exists an x in L(M) such that n <= |x| < 2n.
Proof: (left as an exercise; similar to the previous theorem).
Corollary: Let M = (Q, Σ, δ, q0, F) be a DFA. Then there is an algorithm to determine if L(M) is infinite.
Proof: (left as an exercise).

27. Closure Properties of Regular Languages
Consider various operations on languages:
= {x | x is in Σ* and x is not in L}
= Σ* - L
= {x | x is in L1 or L2}
= {x | x is in L1 and L2}
= {x | x is in L1 but not in L2}
L1L2 = {xy | x is in L1 and y is in L2}
L* = Li = L0 U L1 U L2 U…
L+ = Li = L1 U L2 U…
A closure property for the regular languages states that if a set of one or more languages is regular, then some operation on that set results in a regular language.

28. Theorem: Let M = (Q, Σ, δ, q0, F) be a DFA
Theorem: Let M = (Q, Σ, δ, q0, F) be a DFA. Then M’ = (Q, Σ, δ, q0, Q-F) is a DFA where L(M’) = Σ* - L(M).
Proof: (left as an exercise).
Corollary: The regular languages are closed under complementation, i.e., if L is a regular language then so is .
Does the above construction work for NFAs?

29. Theorem: The regular languages are closed with respect to union, concatenation and Kleene closure.
Proof: Let L1 and L2 be regular languages. Then there exist regular expressions r1 and r2 such that L(r1) = L1 and L(r2) = L2. But then r1 + r2 is a regular expression representing Similarly for concatenation and Kleene closure.

30. How about Intersection?
Lemma: Let L1 and L2 be subsets of Σ*. Then

31. Theorem: Let L1 and L2 be regular languages. Then is a regular language.
Proof: (algebraic proof) Let L1 and L2 be regular languages. Then:
is regular, because the regular languages are closed wrt complementation
is regular, because the regular languages are closed wrt union
But by the lemma: 

32. A direct construction of a DFA M such that :
Let:
M1 = (Q1, Σ, δ1,p0,F1), where Q1 = {p0, p1,…}
M2 = (Q2, Σ, δ2,q0,F2), where Q2 = {q0, q1,…}
where L(M1) = L1 and L(M2) = L2.
Construct M where:
Q = Q1 x Q2 Note that M has a state for each
= {[p0, q0], [p0, q1], [p0, q2],…} pair of states in M1 and M2
Σ = as with M1 and M2
F = F1 x F2
start state = [p0, q0]
δ([pi, qj], a) = [δ1(pi, a), δ2(qj, a)] for all [pi, qj] in Q and a in Σ

33. p0 p1 1 1 q0 q1 1 q2 0,1 Example: The construct gives: Q = Q1 x Q2
= {[p0, q0], [p0, q1], [p0, q2], [p1, q0], [p1, q1], [p1, q2]}
Σ = {0, 1}
F = {[p1, q2]}
start state = [p0, q0]
δ – Some examples: δ([p0, q2], 0) = [δ1(p0, 0), δ2(q2, 0)] = [p1, q2]
δ([p1, q2], 1) = [δ1(p1, 1), δ2(q2, 1)] = [p1, q2] p0 p1 1 1 q0 q1 1 q2
0,1

34. Question: What if Σ1 does not equal Σ2?
Final Result:
Question: What if Σ1 does not equal Σ2?
Try all three machines on: (what happens?)
0100
0101
0011
1010 1 1 1
p0, q0
p0, q1
p0, q2
p1, q0
p1, q1
p1, q2 1 1 1