1. Non-regular languages
The Chinese University of Hong Kong
Fall 2011
CSCI 3130: Automata theory and formal languages
Non-regular languages
Andrej Bogdanov

2. A non-regular language
An example
We reason by contradiction:
Suppose we have managed to construct a DFA M for L
We argue something must be wrong with this DFA
In particular, M must accept some strings outside L
L = {0n1n: n ≥ 0} is not regular.

3. A non-regular language
imaginary DFA for L with n states M x
What happens when we run M on input x = 0n+11n+1?
M better accept, because x  L

4. A non-regular languageM x
0r1n+1
What happens when we run M on input x = 0n+11n+1?
M better accept, because x  L
But since M has n states, it must revisit at least one of its states while reading 0n+1

5. Pigeonhole principle
Suppose you are tossing n + 1 balls into n bins. Then two balls end up in the same bin.
Here, balls are 0s, bins are states:
If you have a DFA with n states and it reads n + 1 consecutive 0s, then it must end up in the same state twice.

6. A non-regular languageM x
0r1n+1
What happens when we run M on input x = 0n+11n+1?
M better accept, because x  L2
But since M has n states, it must revisit at least one of its states while reading 0n+1
But then the DFA must contain a loop with 0s

7. A non-regular languageM
0r1n+1
The DFA will then also accept strings that go around the loop multiple times
But such strings have more 0s than 1s, so they are not in L2!

8. General method for showing non-regularity
Every regular language L has a property:
For every sufficiently long input z in L, there is a “middle part” in z that, even if repeated any number of times, keeps the input inside L an z a1 ak
ak+1
an-1 … …
an+1…am

9. Pumping lemma for regular languages
Pumping lemma: For every regular language L
There exists a number n such that for every string z in L, we can write z = u v w where  |uv| ≤ n  |v| ≥ 1  For every i ≥ 0, the string u vi w is in L. v z … … u w

10. Arguing non-regularity
If L is regular, then:
So to prove L is not regular, it is enough to show:
There exists n such that for every z in L, we can write z = u v w where |uv| ≤ n, |v| ≥ 1 and  For every i ≥ 0, the string u vi w is in L.
For every n there exists z in L, such that for every way of writing z = u v w where |uv| ≤ n and |v| ≥ 1, the string u vi w is not in L for some i ≥ 0.

11. Proving non regularity
For every n there exists z in L, such that for every way of writing z = u v w where |uv| ≤ n and |v| ≥ 1, the string u vi w is not in L for some i ≥ 0.
This is a game between you and an imagined adversary (say Donald)
Donald
choose n write z = uvw (|uv| ≤ n,|v| ≥ 1)
you
choose z  L choose i you win if uviw  L 1 2

12. Arguing non-regularity
You need to give a strategy that, regardless of what the adversary does, always wins you the game
Donald
choose n write z = uvw (|uv| ≤ n,|v| ≥ 1)
you
choose z  L choose i you win if uviw  L 1 2

13. Example L = {0n1n: n ≥ 0} Donald
choose n write z = uvw (|uv| ≤ n,|v| ≥ 1)
you
choose z  L choose i you win if uviw  L 1 2
L = {0n1n: n ≥ 0}
Donald
you 1
choose n
z = 0n1n 2
write z = uvw
i = 2
uv2w = 0n+k1n  L u v w u v w

14. Example LDUP = {0n10n1: n ≥ 0} Donald you choose n z = 0n10n12
write z = uvw
i = 2
uv2w = 0n+k10n1  L u v w u v w

15. Which of these are regular?
L1 = {x: x has same number of 0s and 1s}
L2 = {x: x = 0n1m, n > m ≥ 0}
L3 = {x: x has same number of patterns 01 and 10}
L4 = {x: x has same number of patterns 01 and 10}
L5 = {x: x has different number of 0s and 1s}
S = {0, 1}

16. Example L1 = {x: x has same number of 0s and 1s} Donald you choose n
z = 0n1n 2
write z = uvw
i = 2
uv2w = 0n+k1n  L3 u v w u v w

17. Example L2 = {x: x = 0m1n, m > n ≥ 0} Donald you choose n
z = 0n+11n 2
write z = uvw
i = 0
uv0w = 0j+l1n+1  L2 u v w u w

18. Example L3 = {x: x has same number of 01s and 11s}
Donald
you 1
choose n
z = (01)n(11)n
n = 1
z = 0111
 L4
has too many 11s
What we have in mind:
n = 1
z = 011
n = 2
z =
n = 3
z =
z = (01)n1n
has n 01s and n 11s

19. Example L3 = {x: x has same number of 01s and 11s} Donald you win!
choose n
z = (01)n1n 2
write z = uvw
i = 0
Taking out v will kill
at least one 01,
but it does not kill any 11s u v w u v w or
so uv0w  L3

20. Example L4 = {x: x has same number of 01s and 01s} Donald you choose n
z = (01)n(10)n
n = 1
z = 0110
n = 2
z =
n = 3
z =

21. Example is regular! L4 = {x: x has same number of 01s and 10s} Donald
you 1
choose n
z = (01)n(10)n 2
write z = uvw
i = 2
i = 0 u v w
5 01 patterns
5 10 patterns u v w
6 01 patterns
6 10 patterns u w
4 01 patterns
4 10 patterns

22. Example L4 = {x: x has same number of 01s and 10s} one more 10r0 r1 1 1 q0
one more 01 1 1 s0 s1
L4 = {x: x has same number of 01s and 10s}

23. there is an easier way! Example
L4 = {x: x has different number of 0s than 1s}
Donald
you 1
choose n
z = ?
there is an easier way!
L1 = {x: x has same number of 0s and 1s} = L4
If L4 is regular, then L1 = L4 is also regular
But L1 is not regular, so L4 cannot be regular

24. Extra example L5 = {1p: p is prime} Donald you choose n
z = 1p: p > n is prime 2
write z = uvw = 1a1b1c i
= ?
= a + c
uviw = 1a1ib1c
= 1(a+c)+ib
= 1(a+c)+(a+c)b
u = 1a
v = 1b
w = 1c
= 1(a+c)(b+1)
= 1composite  L5