1. The Chinese University of Hong Kong Fall 2011
CSCI 3130: Automata theory and formal languages
Non-regular languages
Andrej Bogdanov
http://www.cse.cuhk.edu.hk/~andrejb/csc3130

2. A non-regular language
• An example
L = {0n1n: n ≥ 0} is not regular.
• 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

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 language
0 M
x 0 0 0 0
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 language
0 M
x 0 0 0 0
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 language
0 M
0 0 0 0
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:
an
z a1 ak ak+1 … an-1
…
an+1…am
• 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

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:
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.
• So to prove L is not regular, it is enough to show:
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 you
1 choose n choose z ∈ L
2 write z = uvw (|uv| ≤ n,|v| ≥ 1) choose i
you win if uviw ∉ L

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

13. Example
Donald you
1 choose n choose z ∈ L
2 write z = uvw (|uv| ≤ n,|v| ≥ 1) choose i
you win if uviw ∉ L
L = {0n1n: n ≥ 0}
Donald you
1 choose n z = 0n1n
2 write z = uvw i=2
000000000000000111111111111111 uv2w = 0n+k1n ∉ L
u v w
0000000000000000000111111111111111
u v v w

14. Example
LDUP = {0n10n1: n ≥ 0}
Donald you
1 choose n z = 0n10n1
2 write z = uvw i=2
uv2w = 0n+k10n1 ∉ L
000000000000001000000000000001
u v w
0000000000000000001000000000000001
u v v w

15. Which of these are regular?
L1 = {x: x has same number of 0s and 1s} Σ = {0, 1}
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}

16. Example
L1 = {x: x has same number of 0s and 1s}
Donald you
1 choose n z = 0n1n
2 write z = uvw i=2
uv2w = 0n+k1n ∉ L3
000000000000000111111111111111
u v w
0000000000000000000111111111111111
u v v w

17. Example
L2 = {x: x = 0m1n, m > n ≥ 0}
Donald you
1 choose n z = 0n+11n
2 write z = uvw i=0
uv0w = 0j+l1n+1 ∉ L2
000000000000000111111111111111
u v w
00000000000111111111111111
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 z = (01)n1n
n=2 z = 010111
has n 01s and n 11s
n=3 z = 010101111

19. Example
L3 = {x: x has same number of 01s and 11s}
Donald you win!
1 choose n z = (01)n1n
2 write z = uvw i=0
010101010101010111111111 Taking out v will kill
u v w
at least one 01,
010101010101010111111111 but it does not kill any 11s
or u v w
or 010101010101010111111111 so uv0w ∉ L3
u v w

20. Example
L4 = {x: x has same number of 01s and 01s}
Donald you
1 choose n z = (01)n(10)n
n=1 z = 0110
n=2 z = 01011010
n=3 z = 010101101010

21. Example
L4 = {x: x has same number of 01s and 10s}
is regular!
Donald you
1 choose n z = (01)n(10)n
2 write z = uvw i=20
5 01 patterns
5 10 patterns
010101101010
u v w
6 01 patterns 4 01 patterns
6 10 patterns 4 10 patterns
01010101101010 0101101010
u v v w u w

22. Example
1 0 one more 10
0
r0 r1
1 1
q0
0 0 1 one more 01
1
s0 s1
0
L4 = {x: x has same number of 01s and 10s}

23. 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
1 choose n z = 1p: p > n is prime
2 write z = uvw = 1a1b1c i=?+c
a
uviw = 1a1ib1c
= 1(a+c)+ib
111111111111111111111111111111
u = 1a v = 1b w = 1c = 1(a+c)+(a+c)b
= 1(a+c)(b+1)
= 1composite ∉ L5