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--- lfa:2022:lab08-the-pumping-lemma [2022/11/27 00:24]
alexandra.udrescu01
+++ lfa:2022:lab08-the-pumping-lemma [2022/12/09 08:55] (current)
pdmatei
@@ Line 18: / Line 18: @@
 **8.1.1.** Show that the pumping lemma holds for finite languages.
 <hidden Solution>
@@ Line 28: / Line 29: @@
 </hidden>
 **8.1.2.*** Find a language which is not regular for which the pumping lemma holds.
@@ Line 36: / Line 38: @@
 **8.2.1.** $  L = \{ \: A^n B^m \: | \: 0 \leq n \leq m \: \} $
 <hidden Solution> <note>
@@ Line 60: / Line 64: @@
 </hidden>
-**8.2.2.** $  L = \{ \: w \in \{A,B\}^* \: | \: \#A(w) = \#B(w) \: \} $
+**8.2.2.** $  L = \{ \: w \in \{A,B\}^* \: | \: \#_A(w) = \#_B(w) \: \} $
 <hidden Solution> <note>
@@ Line 84: / Line 90: @@
 </hidden>
 **8.2.3.** $math[L = \{(01)^n(10)^n \mid n > 0 \} ]
 <hidden Solution> <note>
@@ Line 110: / Line 119: @@
 $ w_k = (01)^a((01)^b0)^k1(01)^{n-b-a-1}(10)^n $
-Pick k = 2 => $ w_k = (01)^a(01)^b0(01)^b01(01)^{n-b-a-1}(10)^n \notin L $ because it has 2 conseccutive "0"
+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:**
@@ Line 122: / Line 131: @@
 $ w_k = (01)^a0(1(01)^b)^k(01)^{n-b-a-1}(10)^n $
-Pick k = 3 => $ w_k = (01)^a01(01)^b1(01)^b1(01)^b(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_k = (01)^a0111(01)^{n-b-a-1}(10)^n \notin L $
+  * b = 0 => $ w_3 = (01)^a0111(01)^{n-b-a-1}(10)^n \notin L $
-  * b > 0 => $ w_k = (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"
+  * 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:**
@@ Line 136: / Line 145: @@
 $ w_k = (01)^a0(1(01)^b0)^k1(01)^{n-b-a-2}(10)^n $
-Pick k = 0=> $ w_k = (01)^a01(01)^{n-b-a-2}(10)^n = (01)^{n-b-1}(10)^n \notin L$
+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
@@ Line 142: / Line 151: @@
 </hidden>
 **8.2.4.** $  L = \{ \: w \in \{A,B\}^* \: | \: \text{w is a palindrome} \: \} $
 <hidden Solution> <note>
@@ Line 168: / Line 180: @@
 **8.2.5.** $  L = \{ \: w \in \{0\}^* \: | \: \text{the length of w is a prime number} \: \} $
@@ Line 186: / Line 200: @@
 $ w_k = 0^{p+(k-1)b} $
-Pick $ k = p+1 $ => $ w_{p+1} = 0^{p+cp} = 0^{(p+1)c} \notin L $
+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
@@ Line 192: / Line 206: @@
 </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>