1. Lecture5 Non-regular languages
 2004 SDU

2. Are all languages regular?
Answer: No!
One argument:
# of DFA’s on ?
# of languages on ?
the latter is larger than the former.
 2004 SDU

3. Pumping Lemma
First technique to show that specific given languages are not regular.
The idea:
Each regular language satisfies the pumping lemma, so, if a language does not, then, it is non-regular!
 2004 SDU

4. Statement of Pumping Lemma
If A is a regular language, then there is a number p (pumping length) where, if s is any string in A of length at least p, then s must be able to be divided into three pieces, s = xyz, satisfying the following conditions:
1. for each i  0, xyiz  A,
2. |y| > 0, and
3. |xy|  p.
 2004 SDU

5. Describing the pumping lemma (contd.)
Take string
with
There is a walk with label: w1 w2 wn s1
sn-1 sn s0
 2004 SDU

6. Describing the pumping lemma (contd.)
, number of states.
If string has length
Then, a state is repeated in the walk
......
......
 2004 SDU

7. Describing the pumping lemma (contd.)
Write w=xyz ,
here si is the first repeat state,
and sl is the second apperance of si
y=wi+1…wl
sl-1
si+1 wl
wi+1 w1 wi w2
......
...... wn s0 s1
wl+1
si(sl) sn
z=wl+1…wn
x=w1…wi
 2004 SDU

8. Describing the pumping lemma (contd.)
Observations : |xy| = |s0,…,si,si+1…,sl-1,si| - 1  (p+1)-1=p
|y| = |si,…,sl-1,si|-1  2-1=1
y=wi+1…wl
sl-1
si+1 wl
wi+1 w1 wi w2
...... s0 s1
si(sl)
x=w1…wi
 2004 SDU

9. Describing the pumping lemma (contd.)
Observation: the string xz is accepted. w1
...... wi w2
...... wn s0 s1
wl+1 sn
si(sl)
z=wl+1…wn
x=w1…wi
 2004 SDU

10. Describing the pumping lemma (contd.)
Observation: the string xyyz is accepted.
y=wi+1…wl
sl-1
si+1
wi+1 wl wn
wl+1 w1 w2
...... wi
...... s0 s1
si(sl) sn
z=wl+1…wn
x=w1…wi
 2004 SDU

11. Proof of pumping lemma
Idea: If a string w of length m is accepted by a DFA with p states, and m  p, then there is a repeat state (let i be the first one) in the accepting state sequence.
s0w1 s1…wisiwi+1si+1…wlsl(=si)wl+1sl+1…wmsm z y x
|xy| = |w1,…,wl| = |s0,…,si, si+1, …, si| - 1  (p+1)-1 = p
|y| = |wi+1, …,wl | = |si, si+1, …, si| - 1  2-1=1
xz (= w1…wiwl+1…wm): s0, …, si, sl+1, …, sm
xyyz( = w1…wiwi+1…wlwi+1…wlwl+1…wm): s0,…, si, si+1, .., sl, si+1, …, sl, sl+1, …,sm
 2004 SDU

12. An exmaple 1 1,+ + + s1 s4 s2 1 1 s0 + Let p = 4. Input: 1+1+1 x = “1”
y = “+1”
z = “+1”
Actually, we can also choose p as 3 even 2, why?
Consider input 111, 11, or 1+1.
Can they be pumped?
What is the minimum pumping length?
 2004 SDU

13. Application of Pumping Lemma
Example 1: A = { 0n1n | n  0} is not regular.
Proof. By contradiction. Assume, on the contrary, that A is regular.
Then the pumping lemma holds for A. Let p be a pumping length.
Let w = 0p1p. Then w is in A and is longer than p. Then by pumping
Lemma, w can be divided into w=xyz such that
For each i  0, xyiz  A
|y| > 0, and
|xy|  p.
Then x and y are all 0’s. So, xyyz has more 0’s than 1’s and is in A.
A contradiction. So, A is not regular. y
 2004 SDU

14. Application of Pumping Lemma
Example 2: B={ w | w{0,1}* and has an equal number of 0’s and 1’s}
is not regular.
Proof. By contradiction. Assume, on the contrary, that B is regular.
Then the pumping lemma holds for B. Let p be a pumping length.
Let w = 0p1p. Then w is in B and is longer than p. Then by pumping
Lemma, w can be divided into w=xyz such that
For each i  0, xyiz  B
|y| > 0, and
|xy|  p.
Then x and y are all 0’s. So, xyyz has more 0’s than 1’s and is in B.
A contradiction. So, B is not regular.
Another proof. Let B’=B0*1*. If B is regular, then B’=A is regular,
a contradiction.
 2004 SDU

15. Application of Pumping Lemma
Example 3: C = {ww | w {0,1}*} is not regular.
Proof. By contradiction. Assume, on the contrary, that C is regular.
Then the pumping lemma holds for C. Let p be a pumping length.
Let w = 0p10p1. Then w is in C and is longer than p. Then by pumping
Lemma, w can be divided into w=xyz such that
For each i  0, xyiz  C
|y| > 0, and
|xy|  p.
Then x and y are all 0’s. Obviously xyyz cannot be represented as ww.
A contradiction to that xyyz  C. So, C is not regular.
 2004 SDU

16. Application of Pumping Lemma
Example 4: D={1a | a = n2 for some n  0} is not regular.
Proof. By contradiction. Assume, on the contrary, that D is regular.
Then the pumping lemma holds for D. Let p be a pumping length.
Let w = 1b, b=p2. Then w is in D and is longer than p.
Then by pumping Lemma, w can be divided into w=xyz such that
For each i  0, xyiz  D
|y| > 0, and
|xy|  p.
Then xyyz=1c, where p2 < c = p2 + |y|  p2 + p, then c cannot be
expressed as n2 for any n. A contradiction to that xyyz  D.
So, D is not regular.
 2004 SDU

17. Application of Pumping Lemam
Example 5: E={0i1j | i > j} is not regular.
Proof. By contradiction. Assume, on the contrary, that E is regular.
Then the pumping lemma holds for E. Let p be a pumping length.
Let w = 0p1p-1. Then w is in E and is longer than p.
Then by pumping Lemma, w can be divided into w=xyz such that
For each i  0, xyiz  E
|y| > 0, and
|xy|  p.
Then both x and y are strings of all 0’s.
Then xz = 0x1p-1, where x  p-1. A contradiction to that xz  E.
So, E is not regular.
 2004 SDU