2.2k views

Consider the following languages:

• $L1=\left\{ww \mid w \in \{a,b\}^*\right\}$
• $L2=\left\{ww^R \mid w \in \{a,b\}^*, w^R \text{ is the reverse of w} \right\}$
• $L3=\left\{0^{2i} \mid \text{ i is an integer} \right\}$
• $L4= \left\{ 0^{i^2} \mid \text{ i is an integer} \right\}$

Which of the languages are regular?

1. Only $L1$ and $L2$
2. Only $L2, L3$ and $L4$
3. Only $L3$ and $L4$
4. Only $L3$

edited | 2.2k views

$L1=\{ww \mid w \in \{a,b\}^*\}\qquad$CSL

$L2=\{ww^R \mid w \in \{a,b\}^*,w^R \text{ is the reverse of w}\}\qquad$Palindrome, so CFL

$L3=\{0^{2i} \mid i \text{ is an integer}\}\qquad$Linear power and regular expression can be stated as $(00)^*$

$L4=\{0^{i^2} \mid i \text{ is an integer}\}\qquad$Non-linear power, so CSL

by Boss (16.5k points)
edited
+1
As I tis given that set of strings in L3 will be integer....how does it behave for negative numbers?
0
What about negative no  in integers
0
why we can't consider L4 as 0* ?
0
coz $L_4$ will not contain $000$...it only contains exponential number of 0s.
0

@Satbir

Is L4 really a CSL or it should be Recursive Enumerable?

0
it is CSL
0

@satbir

Is this argument correct

Pallindrome, so CSL

I know since its palindrome. Hence, you can use stack but saying that way doesn't seem logically correct.

Don' you think so?

0

Its correct.

all Palindrome strings can be detected with help of 1 stack.

hence it is a CFL

all CFL are CSL

so it should be correct.

Why do you think it is incorrect ?

L1 -> This is CSL

L2 -> This is CFL.

L3 -> This is regular. Regular expression (00)*

L4 -> This is CSL.

by Boss (41.9k points)
+1 vote
ans is D. only L3 is regular
by Loyal (8.1k points)
0
Plz explain how L3 is regular..?

L1, L2 are not Regular because we can't compare strings in Finite Automata. L1 is CSL, L2 is CFL

L3 is Regular because strings generated by L3 are even no of 0's.

L4 is not Regular because we can't design FA.

by Junior (983 points)
0
L = w.(w reverse)* | w € {a, b}*?

Please solve this... Is it regular or non regular??