Question

Show that the languages $\text{Missing left or extra right}$ and $\text{Missing left or extra right}$ are not context-free. (In fact, it can be shown that a language over the alphabet $a$ is context-free if and only if it is regular.)

Short answer

Both languages $\{a^n|\mathrm{n}$ is a prime number $\}$ and $\text{Missing or unrecognized delimiter for right}$ are not context-free. The proofs rely on the pumping lemma for context-free languages, which yields contradictions in both cases.

Step-by-step solution

Definitions and Preliminaries

First, we remind ourselves of the definition of a context-free language. A context-free language is a language generated by some context-free grammar. The crucial property here is that every context-free language satisfies the pumping lemma for context-free languages.

Apply the Pumping Lemma for Context-Free Languages

We state and use the pumping lemma for context-free languages, which states: For every context-free language L there exists some integer p (the pumping length) such that any string s in L with |s| ≥ p can be written as s = uvwxy with strings u, v, w, x, y such that: |vwx| ≤ p, |vx| ≥ 1 (meaning v and/or x must be non-empty), and for all integers i ≥ 0, u(v^i)w(x^i)y is in L. In other words, we can 'pump' v and x simultaneously an arbitrary number of times and still get strings in the language.

Prove the First Language isn't Context-Free

For the language $\{a^n\mid \mathrm{n}$ is a prime number $\}$, assume it is context-free. Let p be its pumping length and consider the string s = a^p (the length of s is p, which we can always choose to be a prime number by choosing a proper p). According to the pumping lemma, we should be able to divide s into uvwxy such that: |vwx| ≤ p, |vx| ≥ 1 and u(v^i)w(x^i)y is in the language for all i ≥ 0. However, when we pump v and x with i = 2, the length of the resulted string is more than p (since |vx| ≥ 1), hence it is not prime anymore. We thus reach a contradiction, because we can't find a proper division of a^p that satisfy all conditions, then the language cannot be context-free.

Prove the Second Language Isn't Context-Free

For the language $\{a^{n^2}\mid n\in \mathbb{N}\}$, assume it is context-free. Let p be its pumping length and consider the string s = a^{p^2}. Again, we can't divide a^{p^2} into uvwxy such that: |vwx| ≤ p, |vx| ≥ 1 and u(v^i)w(x^i)y is in the language for all i ≥ 0. When we pump v and x with i = 2, the length of the resulted string is more than p^2 and it is not a perfect square, which contradicts the conditions. Hence, the language cannot be context-free.

Key concepts

Pumping Lemma

The Pumping Lemma is a fascinating tool used in the theory of formal languages to explore the properties of context-free languages. It serves as a fundamental mechanism to prove that certain languages are not context-free. Here's how it works:- **Context-Free Languages**: These are languages that can be generated by context-free grammars which have a set of rules allowing certain sequences of symbols. But, they aren't as powerful in defining more complex patterns.- **Pumping Lemma Definition**: This lemma states that for any context-free language, there exists a certain "pumping length" called $p$. Any string $s$ from the language with a length of $|s|\ge p$ can be split into five parts: $uvwxy$.The key rules for these parts are:- The middle sections $vwx$ need to have a length less than or equal to $p$.- Either $v$ or $x$ (or both) must not be empty, meaning $|vx|\ge 1$.- You can multiply or "pump" these sections, maintaining $u(v^i)w(x^i)y$ in the language for all non-negative integers $i$.
This means you can repeat these segments any number of times and still have a valid string in your language. However, when the languages get complex, like those involving prime numbers or perfect squares, these conditions often lead to contradictions, showing the language isn't context-free.

Prime Numbers in Formal Languages

Prime numbers in formal languages possess unique characteristics, making them intriguing but tricky to work with in the context of formal languages.- **Language Definition**: The formal language $\{a^n\mid n\text{ is a prime number}\}$ is comprised of strings made with the letter $a$ repeated a prime number of times.- **Challenges**: Proving a language like this isn't context-free involves using the Pumping Lemma. Assume that it was context-free: - Choose a prime number $p$ large enough to serve as our pumping length. - Consider a string $s=a^p$.According to the Pumping Lemma, $s$ should be divided into $uvwxy$ following the rules provided. If we pump $v$ and $x$, resulting in multiplying their occurrences, the new string should also maintain its primality.- **Contradiction**: When $v$ and $x$ are pumped to have $i=2$, $u(v^2)w(x^2)y$, the length of the new string often becomes non-prime, which violates the original conditions of being a language of prime lengths. This contradiction implies the language is not context-free. In simpler terms, prime numbers have complex divisibility properties that create issues when attempting to apply the repetitive nature required by context-free structures.

Perfect Squares in Formal Languages

Perfect squares in formal languages are another fascinating topic often explored in computational theory. Like prime numbers, manipulating perfect squares within language structures can show inherent complexities.- **Language Description**: The language $\{a^{n^2}\mid n\in \mathbb{N}\}$ includes strings, each made by repeating the letter $a$ a perfect square number of times (e.g., 1, 4, 9).- **Conceptualization**: Assume it’s context-free: - Use the Pumping Lemma, with a pumping length $p$, and take a string $s=a^{p^2}$. - Following the lemma, $s$ should be splittable into $uvwxy$, adhering to its constraints.- **Contradiction Arises**: On pumping, say $i=2$, the new string $u(v^2)w(x^2)y$ changes the length dramatically: from a perfect square like $(p{)}^2$ to something random that's not necessarily square. Thus, the predictable square pattern breaks.This contradiction suggests that such a language doesn't follow the properties required by context-free languages. The specific growth pattern of perfect squares doesn't align well with the flexible structure demanded by context-free properties. This demonstrates another instance of limitations faced when using context-free grammars to model specific numeric patterns.