Skip to main content
Commonmark migration
Source Link

Is $L=\{ xyx \mid x,y \in \{a,b\}^* \text {and } |x| \ge 1 \}$ context-free?

 

If yes, please explain how we can write grammar or create a PDA for it. If not a CFL, then prove it through pumping lemma.

I have tried to apply the pumping lemma with $w = a^nb^naba^nb^n$ as the word in $L$, but without success.

Is $L=\{ xyx \mid x,y \in \{a,b\}^* \text {and } |x| \ge 1 \}$ context-free?

 

If yes, please explain how we can write grammar or create a PDA for it. If not a CFL, then prove it through pumping lemma.

I have tried to apply the pumping lemma with $w = a^nb^naba^nb^n$ as the word in $L$, but without success.

Is $L=\{ xyx \mid x,y \in \{a,b\}^* \text {and } |x| \ge 1 \}$ context-free?

If yes, please explain how we can write grammar or create a PDA for it. If not a CFL, then prove it through pumping lemma.

I have tried to apply the pumping lemma with $w = a^nb^naba^nb^n$ as the word in $L$, but without success.

Question Protected by CommunityBot
edited tags
Link
Raphael
  • 72.9k
  • 30
  • 181
  • 393
Rollback to Revision 4
Source Link

Is $L=\{ aba \mid a,b \in \{0,1\}^* \text {and } |x| \ge 1 \}$$L=\{ xyx \mid x,y \in \{a,b\}^* \text {and } |x| \ge 1 \}$ context-free?

If yes, please explain how we can write grammar or create a PDA for it. If not a CFL, then prove it through pumping lemma.

I have tried to apply the pumping lemma with $w = a^nb^naba^nb^n$ as the word in $L$, but without success.

Is $L=\{ aba \mid a,b \in \{0,1\}^* \text {and } |x| \ge 1 \}$ context-free?

If yes, please explain how we can write grammar or create a PDA for it. If not a CFL, then prove it through pumping lemma.

I have tried to apply the pumping lemma with $w = a^nb^naba^nb^n$ as the word in $L$, but without success.

Is $L=\{ xyx \mid x,y \in \{a,b\}^* \text {and } |x| \ge 1 \}$ context-free?

If yes, please explain how we can write grammar or create a PDA for it. If not a CFL, then prove it through pumping lemma.

I have tried to apply the pumping lemma with $w = a^nb^naba^nb^n$ as the word in $L$, but without success.

modifying for more clarity
Source Link
Loading
Tweeted twitter.com/#!/StackCompSci/status/328462850295214081
added 196 characters in body; edited title
Source Link
Loading
formatting using Latex
Link
Loading
formatting using Latex
Link
Loading
Source Link
Loading