Question

Is it possible to come up with a linear grammar with unequal number of 0s and 1s?

Such as 0100, 01100, 111,1,0, 100101001…

I know there is a context-free grammar for this, but is there a linear grammar?

Thanks.

Answer 1

A grammar is regular if and only if it is either left regular or right regular. The left regular grammars are equivalent to the left linear grammars. The right regular grammars are equivalent to the right linear grammars. Therefore, if a regular grammar exists that generates the indicated language, then it is either right or left regular, and hence equivalent to either a left or right linear grammar.

edit₁:

Note that there’s no regular grammar generating the indicated language L_UNEQ. To see this, consider the fact that L_EQ = { w : n_a(w) = n_b(w)} is the complement of L_UNEQ. Because the regular languages are closed under complementation and L_EQ is not a regular language, L_UNEQ is not a regular language.

edit₂:

I believe the pumping lemma for linear languages can be used to show that the indicated language L_UNEQ is not linear. Here is what I’ve come up with. I’m fairly confident it’s correct. My primary concern is that you were asked – presumably – for a linear language generating the indicated language; however, I came to the conclusion that there is no such grammar.

Assume L_UNEQ is linear. By the pumping lemma for linear languages, there exists an n > 0 depending on L_UNEQ such that for all z ∈ L_UNEQ, z can be written uvwxy where:

• |vx| > 0,
• |uvxy| ≤ n, and
• uvⁱwxⁱy ∈ L_UNEQ for all i ≥ 0.

Let n be the constant guaranteed be the pumping lemma. Consider the string

• z = aⁿb^{(n! + 2n)}aⁿ

Since z ∈ L_UNEQ, it can be decomposed into substrings uvwxy satisfying the constraints of the pumping lemma such that, for all i ≥ 0, the string

• uvⁱwxⁱy = a^|u|a^i|v|a^{(n – |u| – |v|)}b^{(n! + 2n)}a^{(n – |x| – |y|)}a^i|x|a^|y|

is a member of L_UNEQ. Since 1 ≤ |vx| ≤ n, |vx| divides n!. Hence, (n!|vx|^-1 + 1) is a natural number. Setting i to (n!|vx|^-1 + 1) gives the string

• z‘ = uv^(n!|vx|-1 + 1)wx^(n!|vx|-1 + 1)y = a^|u|a^(n!|vx|-1 + 1)|v|a^{(n – |u| – |v|)}b^{(n! + 2n)}a^{(n – |x| – |y|)}a^(n!|vx|-1 + 1)|x|a^|y|

Simplifying the pumped string gives us an equal number of a‘s and b‘s:

• n_a(z‘) = 2n – |vx| + (n!|vx|^-1 + 1)|vx| = 2n + n!

Since (2n + n!) is equivalent to the number of b‘s in the pumped string, z‘ ∉ L_UNEQ. But this contradicts the assumption that L_UNEQ is a linear language. Hence, L_UNEQ is not a linear language.