Are L1 = {a^n b^n | n < 4 } and L2 = {a^n b^n | n < 10^10^10 }, regular languages?

Question

Is L1 = {a^n b^n | n < 4 }, a regular language ?

In my opinion, it is regular, as I could draw an FSA for it, however, in class, my professor had taken an example, L2 = {a^n b^n | n < 10^10^10 } and said, this is not regular...

so, my question is, if I can draw fsa for L1, I can even draw for L2...why did prof. say, it is not regular? because, both the languages, L1 and L2, are finite... I had just taken L1 language on my own to just think over the question... L1 was not discussed in class... Also, I have read, that all finite languages are regular... so both of these should be, in my opinion... :)

if anyone can clarify, I would be grateful. Thank you very much, in advance.

I think the idea is that while 10^10^10 is finite as it is "1" followed by 10 Billion zeros! As there are believed to be only about 10^82 atoms in the universe, constructing a FSA will be mildly impossible. — John Hascall, Jan 21 '16 at 14:05
@JohnHascall No, that's a specious argument. This is firmly in the realm of theory - whether or not one could actually physically construct the FSM is wholly irrelevant. — Esoteric Screen Name, Jan 21 '16 at 14:18
@JohnHascall thank you. so, in your opinion, is L1 is regular ? please clarify... as I am a college student, new in this subject. I think, L1 is regular, but I want some confirmation too :) — Aarjavee S. Kamdar, Jan 21 '16 at 14:24
a^n = a raised to n... there are 4 possibilities for a^n, viz. a^0, a^1, a^2, a^3, when n < 4. — Aarjavee S. Kamdar, Jan 21 '16 at 15:06
@EsotericScreenName Thank you, I agree with you. My question is exactly in theoretical perspective :) — Aarjavee S. Kamdar, Jan 21 '16 at 15:07
According to [wiki](https://en.wikipedia.org/wiki/Regular_grammar), both L1 and L2 are _extended_ regular grammars, but not _strictly_ regular grammars. — gudok, Jan 21 '16 at 15:17
If I understand you, the only valid strings in your language are: "a^0 b^0", "a^1 b^1", "a^2 b^2", "a^3 b^3", so yes, that seems regular. — John Hascall, Jan 21 '16 at 15:38
yep.. that's right :)... I meant exactly that... :) L1 = {epsilon, ab, aabb, aaabbb} — Aarjavee S. Kamdar, Jan 21 '16 at 15:40

score 4 · Accepted Answer · answered Jan 21 '16 at 15:33

Every language that has a finite number of strings is regular. So both L1 and L2 are regular. Because if a language has a finite number of strings we can construct the following NFA where ε denotes the empty transition:

 ------ first string
|      
ε
|
 ------ second string
|
ε
|
 ------ ...
|
.
.
.
|
 ------ last string

Are L1 = {a^n b^n | n < 4 } and L2 = {a^n b^n | n < 10^10^10 }, regular languages?

1 Answers1