Question

Which of the following languages are empty, finite but not empty, decidable but not

finite, partially decidable but not decidable, or not partially decidable? Remember that
encode(T) for a Turing machine T is an encoding of T as a string and encode(x) for a string
x is its encoding.
a) {encode(M)encode(x) : Turing machine M halts on string x}.
b) {encode(M)encode(x) : Turing machine M does not halt on string x}.
c) {encode(M) : Turing machine M halts on blank tape}.
d) {encode(M) : Turing machine M does not halt on blank tape}.
e) {encode(M) : Turing machine M has at least 10 states}.
f) {encode(M) : Turing machine M uses at least 100 tape squares during
its computation on blank tape}.
g) {encode(x) : string x is a palindrome, that is, x = x
R}.

Answer 1

The languages presented vary from partially decidable to not partially decidable, with some being finite and others infinite. Language a) and c) are partially decidable but not decidable, b) and d) are not partially decidable, e) is finite but not empty and decidable, f) is partially decidable but not decidable, and g) is decidable but not finite.

Step-by-step explanation:

In computational theory, languages can be categorized based on their properties, such as whether they are empty, finite, decidable, partially decidable, or not even partially decidable. We will analyze each language mentioned in the question to determine its category.

• a) The language {encode(M)encode(x) : Turing machine M halts on string x} is partially decidable but not decidable. This is because the halting problem is undecidable; however, one can construct a Turing machine that halts if another Turing machine halts, thus making it partially decidable.
• b) The language {encode(M)encode(x) : Turing machine M does not halt on string x} is not partially decidable. This language is related to the complement of the halting problem, which is known to be undecidable and not recognizable.
• c) The language {encode(M) : Turing machine M halts on a blank tape} is partially decidable but not decidable, as it is a variant of the halting problem.
• d) The language {encode(M) : Turing machine M does not halt on a blank tape} is not partially decidable, similar to language b).
• e) The language {encode(M) : Turing machine M has at least 10 states} is finite but not empty. It is straightforward to check if a given Turing machine has at least 10 states, making it decidable as well.
• f) The language {encode(M) : Turing machine M uses at least 100 tape squares during its computation on a blank tape} is partially decidable but not decidable. It involves computation analysis which can be unbounded, leading to indecidability, but one can recognize machines that satisfy the condition.
• g) The language {encode(x) : string x is a palindrome} is decidable but not finite. A Turing machine can verify palindromes in a straightforward and finite procedure, thus being a decidable language.