Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 121549

Questions about properties of the class of regular languages and individual languages.

0 votes
2 answers
626 views

Proof that $L=\{a^ncb^n| n \in \mathbb{N}\}$ is not regular

Prove that $L=\{a^ncb^n| n \in \mathbb{N}\}$ is not regular. Here is my try, I would really appreciate if someone could tell me if this is a correct proof. Proof: Lets assume L is regular. Then we k …
Frank's user avatar
  • 147
2 votes
Accepted

Proof that $L=\{a^ncb^n| n \in \mathbb{N}\}$ is not regular

Proof: Lets assume L is regular. Then we know that L must meet the requirements of the pumping lemma. So let p the pumping number. Let $w=a^pcb^p$. $w$ is obviously of the length p and is in L. Theref …
Frank's user avatar
  • 147
3 votes
1 answer
480 views

If $A$ is context-free then $A^*$ is regular

I am currently studying for my exam and I am having trouble to solve this question: Right or wrong: If $A$ is context-free then $A^*$ is regular. I think it's wrong because if $A$ is context-free it …
Frank's user avatar
  • 147
3 votes
Accepted

If $A$ is context-free then $A^*$ is regular

Let $A=\{a^nb^n \mid n \in \mathbb{N}\}$. Then we know that $A$ is non-regular and context-free. Also we can see that $A^*\cap a^*b^*=A$. Since $a^*b^*$ is a regular expression, we do know that it is …
Frank's user avatar
  • 147