# Answered Essay: Given an arbitrary alphabet Σ ={a,a2, ,an), we can impose a total ordering on it in the sense that we can define

Given an arbitrary alphabet Σ ={a,a2, ,an), we can impose a total ordering on it in the sense that we can define

We know that $\sum*$ is the set of all words formed by by alphabets in $\sum$ .

a)

For a operation to be regular on a set, the set must be closed under that operation.

Let $w \in \sum*\ and\ SORT(w)=w'$ . Clearly, $w' \in \sum*\ \ \ \forall\ w \in \sum*$ as $w'$ is just a permutation of $w$. This means, $\sum*$ is closed under operation $SORT$. Thus,

$SORT(\sum *)$ is regular.

b)

Let, $\sum = \{0,1\}\ \ and\ \ \ L = \{00,10\}$ . Now, $SORT(10)=01 \notin L$ . Thus,

$SORT(L) \nsubseteq L$

c)

Let $w \in L$. From definition of sort we have,

$SORT(w) = w_{\sigma(1)}w_{\sigma(2)}...w_{\sigma(k)}\ \ s.t.\ \ w_{\sigma(i)}.

Clearly

$SORT(SORT(w)) = SORT(w_{\sigma(1)}w_{\sigma(2)}...w_{\sigma(k)})$

$= w_{\sigma(1)}w_{\sigma(2)}...w_{\sigma(k)})\ \ \ [as\ already\ \ w_{\sigma(i)}
$= SORT(w)$

g)

No, $SORT$doesn’t always preserve regularity. This is because for a operation to be regular on a set, the set must be closed under that operation. However $SORT$ doesn’t always follow this property. For example –

Let, $\sum = \{0,1\}\ \ and\ \ \ L = \{00,10\}$ . Now, $SORT(10)=01 \notin L$ . Thus,

$SORT$ doesn’t preserve regularity.

h)

Decidable languages are the decision problems which are algorithmically solvable. Since, $SORT$ is an operation for which many well-known algorithms are present (Bubble Sort, Insertion Sort etc) , we can say that

$SORT$ preseves decidability.

i)

Undecidable languages are the decision problems which are not algorithmically solvable. Since, $SORT$ is an operation for which many well-known algorithms are present (Bubble Sort, Insertion Sort etc) , we can say that

$SORT$ doesn’t preseve non-decidability.

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