Can a language be turing decidable if there exist enumerator that enumerates it?

In the field of computational complexity theory, particularly when discussing Turing machines and enumerators, it is essential to understand the concepts of decidability and enumerability. To address the question of whether a language can be Turing decidable if there exists an enumerator that enumerates it, we must consider the definitions and relationships between these concepts.

A language is said to be Turing decidable (or simply decidable) if there exists a Turing machine that halts and accepts for every string in the language and halts and rejects for every string not in the language. In other words, a Turing decidable language has a Turing machine that can determine, in a finite amount of time, whether any given string belongs to the language.

On the other hand, a language is Turing recognizable (or recursively enumerable) if there exists a Turing machine that halts and accepts for every string in the language, but for strings not in the language, the machine may either halt and reject or run indefinitely. Essentially, a Turing recognizable language has a Turing machine that can recognize members of the language, but it may not necessarily halt for non-members.

An enumerator is a type of Turing machine that generates, one by one, all the strings of a language. If a language is recursively enumerable, there exists an enumerator that can list all and only the strings in the language, though not necessarily in any particular order. The enumerator may run indefinitely, but every string in the language will eventually be listed.

To address the core question, let us analyze the relationship between enumerability and decidability. If a language is decidable, then it is also Turing recognizable. This is because if a Turing machine can decide membership in a language, it can also recognize membership by halting and accepting for strings in the language. Therefore, every decidable language is also recursively enumerable.

However, the converse is not necessarily true. Not every recursively enumerable language is decidable. A classic example of this is the Halting Problem, which is known to be recursively enumerable but not decidable. The Halting Problem consists of determining whether a given Turing machine will halt on a given input. While there exists an enumerator that can list all instances where a Turing machine halts on an input, there is no Turing machine that can decide this problem for all possible inputs.

Given this, let us consider the specific scenario where an enumerator exists for a language. If there exists an enumerator that enumerates a language, then the language is recursively enumerable by definition. The presence of an enumerator implies that there is a Turing machine capable of listing all the strings in the language. However, this does not necessarily imply that the language is decidable.

To determine whether the language is decidable, we need to ascertain if there exists a Turing machine that can decide membership in the language for any given string. In some cases, it is possible for a language to be both recursively enumerable and decidable. For instance, the language of all valid arithmetic expressions is both recursively enumerable and decidable, as there exists a Turing machine that can enumerate all valid expressions and another that can decide the validity of any given expression.

In contrast, consider the language of all Turing machines that halt on a given input. This language is recursively enumerable, as we can construct an enumerator that lists all such Turing machines by simulating each machine on the input and outputting those that halt. However, this language is not decidable, as the Halting Problem is undecidable.

To further illustrate, let us examine the language L of all prime numbers represented in binary. There exists an enumerator for L that generates all prime numbers in binary form. Additionally, there exists a Turing machine that can decide whether a given binary string represents a prime number by performing primality testing. Therefore, the language L is both recursively enumerable and decidable.

The existence of an enumerator for a language implies that the language is recursively enumerable. However, it does not necessarily mean that the language is decidable. To determine decidability, we must establish the existence of a Turing machine that can decide membership for any given string in the language. While all decidable languages are recursively enumerable, not all recursively enumerable languages are decidable. The distinction lies in the ability of the Turing machine to halt and provide a definitive answer for both membership and non-membership in the language.

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• Field: Cybersecurity
• Programme: EITC/IS/CCTF Computational Complexity Theory Fundamentals (go to the certification programme)
• Lesson: Turing Machines (go to related lesson)
• Topic: Enumerators (go to related topic)