What is the union of two recursive languages?

Introduction

Recursive languages are a class of formal languages in the field of computer science, specifically in the theory of computation. A language is said to be recursive, or decidable, if there exists a Turing machine which will accept every input string that is in the language and reject every input string that is not in the language. In other words, for a recursive language, there is always a definite 'yes' or 'no' answer for whether a given string is in the language.

The concept of union in the context of languages refers to the operation that combines the strings of two languages into a single language. If L1 and L2 are two languages, the union of L1 and L2, denoted as L1 ∪ L2, is the set of all strings that are in L1 or in L2 (or in both).

The Union of Recursive Languages

The union of two recursive languages is recursive. This statement can be understood by considering two recursive languages L1 and L2. Since L1 and L2 are recursive, there exist two Turing machines M1 and M2 that accept L1 and L2 respectively.

To show that L1 ∪ L2 is recursive, we need to construct a Turing machine M that accepts L1 ∪ L2. The Turing machine M works as follows: M takes an input w. It simulates M1 on w and M2 on w in parallel. If either M1 or M2 accepts, M accepts. If both M1 and M2 reject, M rejects. Since M always halts (because M1 and M2 always halt), L1 ∪ L2 is recursive.

Formal Proof

Let's provide a formal proof for the statement.

Let L1 and L2 be recursive languages. Then there exist total Turing machines M1 and M2 such that L(M1) = L1 and L(M2) = L2.

We construct a Turing machine M as follows:

M = "On input w,

1. Run M1 and M2 on w in parallel.
2. If either M1 or M2 accepts, accept.
3. If both M1 and M2 reject, reject."

Since M1 and M2 are total, they always halt. Therefore, M always halts and L(M) = L1 ∪ L2. Hence, L1 ∪ L2 is recursive.

Examples

Let's consider two recursive languages L1 and L2.

L1 = {0^n 1^n | n ≥ 0} is the set of all strings with n zeros followed by n ones. A Turing machine M1 that accepts L1 can be designed to count the number of zeros and ones and accept if they are equal.

L2 = {0^n | n is a prime number} is the set of all strings with a prime number of zeros. A Turing machine M2 that accepts L2 can be designed to count the number of zeros and check if the count is a prime number.

The union of L1 and L2, L1 ∪ L2 = {0^n 1^n | n ≥ 0} ∪ {0^n | n is a prime number}, is the set of all strings with n zeros followed by n ones or a prime number of zeros. A Turing machine M that accepts L1 ∪ L2 can be designed to simulate M1 and M2 on the input in parallel and accept if either M1 or M2 accepts.

Conclusion

In conclusion, the union of two recursive languages is recursive. This is an important concept in the theory of computation as it helps in understanding the properties of recursive languages and the operations that can be performed on them. The proof of this statement involves the construction of a Turing machine that simulates two other Turing machines in parallel, demonstrating the power and flexibility of Turing machines in the theory of computation.

Diagram

No diagram is necessary for this answer.

Summary

The union of two recursive languages is recursive. This means that if we have two recursive languages, we can combine them into a single language using the union operation. The union of two languages is the set of all strings that are in either of the two languages.

Analogy

Think of two boxes, one containing blue balls and the other containing red balls. The union of the two boxes would be a new box that contains all the blue balls and all the red balls. Similarly, the union of two recursive languages combines all the strings from both languages into a single language.