description of the language accepted by S->SS|bS|a

Question

I want to know what language is generated by this CFG

S → SS | bS | a

I have obtained some strings but cannot find a pattern

abbaaaa
aaaaaaa
ba
aaaaa
aaaaaaabaaaabbaa
babaaabaaaba
bbbababaababaa
baabaa
baa
aaaaaabbaaabbba

Nikolay Handzhiyski · Accepted Answer · 2021-12-02T07:30:02.693

1

This grammar does not generate epsilon;
The minimum length string generated by S is "a". Also last character generated by S is also a;
bS means that if there is b in the string then after it there must be more characters (because the grammar does not generate epsilon). These characters could be more b, but eventually a must follow (as noted up). That is n times of b, where n >= 0;
Because of SS the language allows repetitions (an infinite number of times (k)) of this zero or more b characters followed by one a.

The language L should be this (where Sigma is the alphabet):

This could be written with regular expression like this:

edited Dec 02 '21 at 07:30

answered Dec 01 '21 at 07:32

The language is produced by the regular expression `(a|b)*a`; I'd write the set simply as `{ ω a | ω ∈ Σ* }`. Not that yours is wrong, but all those subscripted exponents seem unnecessary. – rici Dec 01 '21 at 17:44
It could be. I found it to be more detailed and more related to the explanations in the text. The shortest regular expression I think is `(b*a)+`. – Nikolay Handzhiyski Dec 01 '21 at 18:08

1 Answers1