Question

Let A be the language containing only the single string s, where

$\begin{array}{l}\mathbf{s}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{}\mathbf{if}\mathbf{}\mathbf{life}\mathbf{}\mathbf{never}\mathbf{}\mathbf{will}\mathbf{}\mathbf{be}\mathbf{}\mathbf{found}\mathbf{}\mathbf{on}\mathbf{}\mathbf{Mars}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{1}\mathbf{}\mathbf{if}\mathbf{}\mathbf{life}\mathbf{}\mathbf{will}\mathbf{}\mathbf{be}\mathbf{}\mathbf{found}\mathbf{}\mathbf{on}\mathbf{}\mathbf{Mars}\mathbf{}\mathbf{someday}\mathbf{.}\mathbf{)}\end{array}$

Is decidable? Why or why not? For the purposes of this problem, assume that the question of whether life will be found on Mars has an unambiguous YES or NO answer.

Short answer

Yes, the problem$A$ is decidable.

Step-by-step solution

Step 1: Decidable languages.

Decidable languages are closed under complementation. To design a machine for the complement of a language, and simulate the machine for language on an input. If it accepts then accept and vice versa. Decidable languages are closed under inverse homeomorphisms.

Step 2: Decidable languages is not closed under homomorphism.

a).

The language$\mathrm{A}$is one of the two languages $\left\{0\right\}\mathrm{or}\left\{1\right\}.$In either case, the language is finite and hence decidable. If they are not able to determine which of these two languages is $\mathrm{A}$, you won’t be able to describe the decider for$\mathrm{A}$. However, take two Turing machines, one of which is$\mathrm{A}$’s decider.

$\begin{array}{l}\mathrm{s}=(0\mathrm{if}\mathrm{life}\mathrm{never}\mathrm{will}\mathrm{b}\mathrm{efound}\mathrm{on}\mathrm{Mars}.\\ 1\mathrm{if}\mathrm{life}\mathrm{will}\mathrm{be}\mathrm{found}\mathrm{on}\mathrm{Mars}\mathrm{some}\mathrm{day}.)\end{array}$

Here, The given language contain number of string which has two languages they are$\left\{0\right\}\mathrm{or}\left\{1\right\}.$where$\left\{0\right\}$is comes when. And$\left\{1\right\}$is when and as given in the question the language given is unambiguous. It means various parse tree can be form by that strings of the language which is evaluate by the grammar. If in the answer may be$\left\{0\right\}\mathrm{or}\left\{1\right\}.$anything may come either of the case the problem is decidable because its solution is present. Or any problem is said to be decidable if and only if when its solution is present or it can be solved by Turing machine concept. Where in the two Turing machine one is $A$’s decider. Because in the Turing machine is having turn around capability it means the

Turing machine can move right and left side directions and in Turing machine also has the capability of read and write in the infinite length of tape. the language is finite and hence decidable. If they are not able to determine which of these two languages is $A$, you won’t be able to describe the decider for $A$.

Hence, this language is finite so, the problem$\mathrm{A}$ is decidable.