Table of Contents

8. Proving languages are not regular

Pumping Lemma

Let L be an infinite regular language. Then, $\exists n \in \mathbb{N}$: $ \forall w\ \ \text{s.t.}\ \ |w| \ge n $, $ w = xyz $, $ |xy| \le n $ and $ y \neq \varepsilon $, such that $ \forall k \ge 0, w_{k} = xy^{k}z \in L$.

Complement of Pumping Lemma

Let L be an infinite language. If $\forall n \in \mathbb{N}$, $\exists w_{n} \in L $ with $ |w| \ge n $ such that regardless of how $ w_{n} $ is split into ($ \forall x, y, z \in \Sigma^* $) $ w_{n} = xyz $ with $ |xy| \le n $ and $ y \neq \varepsilon $, $\exists k \ge 0 $ such that $ w_{n} = xy^{k}z \notin L $, then L in not a regular language.

8.1. The pumping lemma

8.1.1. Show that the pumping lemma holds for finite languages.

Solution

Solution

Let L be a finite language.

Pick $ n \gt max_{w \in L} |w| $ ⇒ $ \nexists w \in L $ with $ |w| \ge n $ ⇒ Pumping Lemma holds

8.1.2.* Find a language which is not regular for which the pumping lemma holds.

8.2. Languages which are not regular

Show that each of the languages from the list below is not regular.

8.2.1. $ L = \{ \: A^n B^m \: | \: 0 \leq n \leq m \: \} $

Solution

Solution

For a fixed n, pick a word $ w_n \in L $ and $ w_n = xyz $:

$ w_n = A^nB^{n+1} $

$ x = A^a $

$ y = A^b, b \ge 1 $

$ z = A^{n-b-a}B^{n+1} $

Find k such that $ xy^kz \notin L $:

$ w_k = A^{a+b*k+n-b-a}B^{n+1} = A^{n+(k-1)b}B^{n+1} $

Pick k = 3 ⇒ $ w_3 = A^{n+2b}B^{n+1} \notin L $ because $ b \ge 1 $

⇒ Complement of Pumping Lemma holds ⇒ L is not a regular language

8.2.2. $ L = \{ \: w \in \{A,B\}^* \: | \: \#_A(w) = \#_B(w) \: \} $

Solution

Solution

For a fixed n, pick a word $ w_n \in L $ and $ w_n = xyz $:

$ w_n = A^nB^n $

$ x = A^a $

$ y = A^b, b \ge 1 $

$ z = A^{n-b-a}B^n $

Find k such that $ xy^kz \notin L $:

$ w_k = A^{a+b*k+n-b-a}B^n = A^{n+(k-1)b}B^n $

Pick k = 2 ⇒ $ w_2 = A^{n+b}B^n \notin L $ because $ b \ge 1 $

⇒ Complement of Pumping Lemma holds ⇒ L is not a regular language

8.2.3. $ L = \{(01)^n(10)^n \mid n > 0 \} $

Solution

Solution

$ w_n = (01)^n(10)^n $

Case 1:

$ x = (01)^a $

$ y = (01)^b, b \ge 1 $

$ z = (01)^{n-b-a}(10)^n $

Pick k = 2 ⇒ $ w_2 = (01)^{n+b}(10)^n \notin L $

Case 2:

$ x = (01)^a $

$ y = (01)^b0, b \ge 0 $

$ z = 1(01)^{n-b-a-1}(10)^n $

$ w_k = (01)^a((01)^b0)^k1(01)^{n-b-a-1}(10)^n $

Pick k = 2 ⇒ $ w_2 = (01)^a(01)^b0(01)^b01(01)^{n-b-a-1}(10)^n \notin L $ because it has 2 conseccutive “0”

Case 3:

$ x = (01)^a0 $

$ y = 1(01)^b, b \ge 0 $

$ z = (01)^{n-b-a-1}(10)^n $

$ w_k = (01)^a0(1(01)^b)^k(01)^{n-b-a-1}(10)^n $

Pick k = 3 ⇒ $ w_3 = (01)^a01(01)^b1(01)^b1(01)^b(01)^{n-b-a-1}(10)^n $

  • b = 0 ⇒ $ w_3 = (01)^a0111(01)^{n-b-a-1}(10)^n \notin L $
  • b > 0 ⇒ $ w_3 = (01)^a01(01)^b1(01)^b1(01)^b(01)^{n-b-a-1}(10)^n \notin L $ because there are 2 sets of consecutive “1”

Case 4:

$ x = (01)^a0 $

$ y = 1(01)^b0, b \ge 0 $

$ z = 1(01)^{n-b-a-2}(10)^n $

$ w_k = (01)^a0(1(01)^b0)^k1(01)^{n-b-a-2}(10)^n $

Pick k = 0 ⇒ $ w_0 = (01)^a01(01)^{n-b-a-2}(10)^n = (01)^{n-b-1}(10)^n \notin L$

⇒ Complement of Pumping Lemma holds ⇒ L is not a regular language

8.2.4. $ L = \{ \: w \in \{A,B\}^* \: | \: \text{w is a palindrome} \: \} $

Solution

Solution

For a fixed n, pick a word $ w_n \in L $ and $ w_n = xyz $:

$ w_n = A^nBA^n $

$ x = A^a $

$ y = A^b, b \ge 1 $

$ z = A^{n-b-a}BA^n $

Find k such that $ xy^kz \notin L $:

Pick k = 2 ⇒ $ w_2 = A^{n+b}BA^n \notin L $ because $ b \ge 1 $

⇒ Complement of Pumping Lemma holds ⇒ L is not a regular language

8.2.5. $ L = \{ \: w \in \{0\}^* \: | \: \text{the length of w is a prime number} \: \} $

Solution

Solution

For a fixed n, pick a word $ w_n \in L $ and $ w_n = xyz $:

$ w_n = 0^p, p \gt n $, p prime

$ x = 0^a $

$ y = 0^b, b \ge 1 $

$ z = 0^{p-b-a} $

Find k such that $ xy^kz \notin L $:

$ w_k = 0^{p+(k-1)b} $

Pick $ k = p+1 $ ⇒ $ w_{p+1} = 0^{p+cp} = 0^{(p+1)c} \notin L $ because $ (p+1)c $ is not a prime number

⇒ Complement of Pumping Lemma holds ⇒ L is not a regular language

8.2.6. $ L = \{ \: w \in \{0\}^* \: | \: \text{the length of w is a power of two} \: \} $

Solution

Solution

For a fixed n, pick a word $ w_n \in L $ and $ w_n = xyz $:

$ w_n = 0^{2^p}, 2^p \gt n $

$ x = 0^a $

$ y = 0^b, b \ge 1 $

$ |xy| \le n \Rightarrow |y| \lt n \lt 2^p \Rightarrow 0 < b < 2^p $

$ z = 0^{2^p-b-a} $

Find k such that $ xy^kz \notin L $:

$ w_k = 0^{2^p+(k-1)b} $

Pick $ k = 2 $ ⇒ $ w_2 = 0^{2^p+b} \notin L $ because $ 2^p < 2^p + b < 2^p + 2^p = 2^{p+1} $

⇒ Complement of Pumping Lemma holds ⇒ L is not a regular language

8.2.7. $ L = \{ \: ww^R \: | \: w\in \{0,1\}^* \} $

Solution

Solution

For a fixed n, pick a word $ w_n \in L $ and $ w_n = xyz $:

$ w_n = 0^n110^n $

$ x = 0^a $

$ y = 0^b, b \ge 1 $

$ z = 0^{n-a-b}10^n $

Find k such that $ xy^kz \notin L $:

$ w_k = 0^{n+(k-1)b}110^n $

Pick $ k = 2 $ ⇒ $ w_2 = 0^{n+b}110^n \notin L $ because $ b \ge 1 $

⇒ Complement of Pumping Lemma holds ⇒ L is not a regular language

8.3. Combining the pumping lemma with closure properties

8.3.1. Using the pumping lemma, prove that $ L = \{ \: A^nB^m \: | \: n \neq m \}$ is not a regular language.

Solution

Solution

Assume that L is a regular language.

⇒ $ L(A^*B^*) \backslash L = \{ \: A^nB^n \: | \: n \in \mathbf{N} \} $ is a regular language because set difference is a closure property.

But we have previously proved that $ \{ \: A^nB^n \: | \: n \in \mathbf{N} \} $ is not regular.

⇒ Our assumtion was wrong ⇒ L is not a regular language.

Nice to try at home

Exercise I Show that $ \text{snd(L)}$ is a closure property for regular languages.

$ \text{snd(L)} = \{ \: w \: | \: xw \in L \: \text{, for some x such that |x| = |w|} \: \}$

Exercise II Prove that $ L = \{ \: A^nB^mC^{n-m} \: | \: n \geq m \geq 0 \: \}$ is not a regular language without using isomorphisms.