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--- lfa:2022:lab08-the-pumping-lemma [2022/11/26 23:03]
alexandra.udrescu01
+++ lfa:2022:lab08-the-pumping-lemma [2022/12/09 08:55] (current)
pdmatei
@@ Line 4: / Line 4: @@
 ** Pumping Lemma **
-Let L be an infinite regular language, Then, for $\forall$ w $\in$ L, $\exists$ n $\in$ $\mathbf{N}$, |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.
+Let L be an infinite regular language. Then, for $\forall w \in L$, $\exists n \in \mathbf{N}$, $ |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$.
 </note>
 <note important>
+** Complement of Pumping Lemma **
+Let L be an infinite language. If $\forall n \in \mathbf{N}$, $\exists w_{n} \in L $ with $ |w| \ge n $ such that regardless of how $ w_{n} $ is split into $ 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.
 </note>
@@ Line 15: / Line 18: @@
 **8.1.1.** Show that the pumping lemma holds for finite languages.
+<hidden Solution>
+<note>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
+</note>
+</hidden>
 **8.1.2.*** Find a language which is not regular for which the pumping lemma holds.
@@ Line 24: / Line 39: @@
 **8.2.1.** $  L = \{ \: A^n B^m \: | \: 0 \leq n \leq m \: \} $
-**8.2.2.** $  L = \{ \: w \in \{A,B\}^* \: | \: \#A(w) = \#B(w) \: \} $
+<hidden Solution> <note>
+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
+</note>
+</hidden>
+**8.2.2.** $  L = \{ \: w \in \{A,B\}^* \: | \: \#_A(w) = \#_B(w) \: \} $
+<hidden Solution> <note>
+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
+</note>
+</hidden>
 **8.2.3.** $math[L = \{(01)^n(10)^n \mid n > 0 \} ]
+<hidden Solution> <note>
+$ 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
+</note>
+</hidden>
 **8.2.4.** $  L = \{ \: w \in \{A,B\}^* \: | \: \text{w is a palindrome} \: \} $
+<hidden Solution> <note>
+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
+</note>
+</hidden>
 **8.2.5.** $  L = \{ \: w \in \{0\}^* \: | \: \text{the length of w is a prime number} \: \} $
+<hidden Solution> <note>
+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
+</note>
+</hidden>
 **8.2.6.** $  L = \{ \: w \in \{0\}^* \: | \: \text{the length of w is a power of two} \: \} $
-**8.2.7.** $  L = \{ \: ww^R  \: | \: w\in \{0,1\}^* \}
-$
+<hidden Solution> <note>
+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 => |y| \lt n \lt 2^p => 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
+</note>
+</hidden>
+**8.2.7.** $  L = \{ \: ww^R  \: | \: w\in \{0,1\}^* \} $
+<hidden Solution> <note>
+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
+</note>
+</hidden>
@@ Line 41: / Line 271: @@
 **8.3.1.** Using the pumping lemma, prove that $  L = \{ \: A^nB^m \: | \: n \neq m \}$ is not a regular language.
+<hidden Solution>
+<note>
+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.
+</note>
+</hidden>
 /*