Expression for this language - expression

Question

A language to generate all strings that have more a's than b's (not necessarily only one more, as with the nonterminal A for the language EQUAL, but any number more a's than b's).

Answer 1

First, let us show that the language is not regular by using the Pumping Lemma for Regular Languages. Assume this language were regular. Then, by the pumping lemma, we'd know that any string in the language of length at least p could be written as uxv such that |ux| <= p, |x| > 0 and for all nonnegative integers n, u(x^n)v is also in the language. Choose as our representative string b^p a^(p+1). This string is clearly in our language because it has more a's than b's. Rewriting this as uxv with |ux| <= p means that u and x consist only of b's. Because |x| > 0, choosing n > 1 causes us to increase the number of b's alone, which means there are not more a's than b's in the resulting string. This is a contradiction, so our only assumption - that the language is a regular language - must be false.
Next, let's see if the language is context-free. Notice that the language with more a's than b's is similar to the one with the same number of a's and b's, which we recognize as a canonical context-free language:
S -> SS | aSb | bSa | e
We can modify this grammar as follows:
whenever we allow one a, allow any number of a instead
instead of allowing the empty string, require at least one a be in the string
This suggests the grammar
S -> ST | TS | ASb | bSA | A
T -> TT | ATb | bTA | e
A -> aA | a
Note: we introduced nonterminal A to capture the idea of allowing any number of a's in place of one, and we added the nonterminal T to assist with the "at least one" a concept. We needed a new nonterminal here because otherwise every time we used the production S -> SS we'd be requiring extra a's in the grammar that strings in the language might not have.
To show this grammar is correct we can show:
it only generates strings in the language, that is, all strings it generates have more a than b
it generates all strings in the language, that is, for any string with more a's than b's, this grammar derives that string
To show the first part, simply note that any production that adds b also adds at least one a, and that to eliminate all instances of S requires using the production S -> a exactly once, which ensures at least one more a is in the final string than there are b's.
To show the second part, consider this procedure to generate any string:
if the string is of the form a*, then use S -> A -> a*.
if the string starts with a and ends with b, use S -> ASb -> aSb to peel off the outermost a and b. Then, solve the subproblem of size (N - 2) using this same procedure. Notice that we have removed the same number of a and b, so if the original string had more a than b, so does the string in the subproblem.
if the string starts with b and ends with a, use S -> bSA -> bSa to peel off the outermost a and b. Then, solve the subproblem of size (N - 2) using this same procedure. Notice that we have removed the same number of a and b, so if the original string had more a than b, so does the string in the subproblem.
if the string starts and ends with the same symbol, find any prefix of the string such that it and the complementary suffix both have at least as many a's as b's. This must be possible since the string has more a's than b's and, since the word starts and ends with the same symbol, neither the prefix nor the suffix needs to be empty. Then, use either S -> TS or S -> ST, depending on whether the prefix or suffix may contain the same numbers of a's and b's, and solve the subproblems.
Rigorously proving the claim in step 3 of this procedure is left as an exercise:
If a string consisting of a's and b's begins and ends with the same symbol and contains more a's than b's, the string can be split into a prefix and a suffix each of which has at least as many a's as b's.