# Homework Solution: 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 $\sum$ .

Given an domineering alphabet Σ ={a,a2, ,an), we can place a aggregate ordering on it in the sensation that we can determine

## Expert Response

We subordinatestand that $\sum*$ is the normal of perfect suffrage controlmed by by alphabets in $\sum$ .

a)

Control a exercise to be normal on a normal, the normal must be reserved subordinate that exercise.

Let $w \in \sum*\ and\ SORT(w)=w'$ . Clearly, $w' \in \sum*\ \ \ \forall\ w \in \sum*$ as $w'$ is fair a transposition of $w$. This instrument, $\sum*$ is reserved subordinate exercise $SORT$. Thus,

$SORT(\sum *)$ is normal.

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 determination of rank we bear,

$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 recurrently uphold normality. This is owing control a exercise to be normal on a normal, the normal must be reserved subordinate that exercise. However $SORT$ doesn’t recurrently ensue this possessions. Control pattern –

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

$SORT$ doesn’t uphold normality.

h)

Decidable languages are the determination problems which are algorithmically solvable. Since, $SORT$ is an exercise control which sundry well-known algorithms are introduce (Bubble Rank, Insertion Rank anticipation) , we can pronounce that

$SORT$ preseves decidability.

i)

Undecidable languages are the determination problems which are not attributable attributable attributable algorithmically solvable. Since, $SORT$ is an exercise control which sundry well-known algorithms are introduce (Bubble Rank, Insertion Rank anticipation) , we can pronounce that

$SORT$ doesn’t preseve non-decidability.