91Ó°ÊÓ

The cyclic shift also called rotation or conjugation and a language is defined as $\left\{\mathrm{yx}\right|\mathrm{xy}∈\mathrm{A}\}$. The context-free languages are closed under this operation. The rotation closure, conjugation, and cyclic shift contains a language which rotates and seems same which is belongs to $\mathrm{A}$.

By using pushdown deterministic automata, a pushdown automaton for the original language, construct one for the cyclic shift. The new automaton operates in two stages, corresponding to the $\mathrm{yy}$and the $\mathrm{xx}$ part of the word $\mathrm{yxyx}$ where $yxyx$is in the original language.

In the first stage, whenever the automaton would like to pop a non-terminal $\mathrm{AA}$, it can instead push a non-terminal $\mathrm{A}\text{'}\mathrm{A}\text{'}$, the idea is that at the end of the first stage, the stack would contain, in reverse order, the symbols that are found in the stack after reading $\mathrm{xx}$by the original automaton. In the second stage (the switch is non-deterministic), instead of pushing a non-terminal $\mathrm{AA}$, are allowed to pop a non-terminal $\mathrm{A}\text{'}\mathrm{A}\text{'}$.

If the original automaton can indeed generate the stack upon reading $\mathrm{xx}$, then the new one would be able to exactly pop the entire stack.

Here suppose we are given a pushdown deterministic automaton with alphabet $∑∑$, set of states, set of accepting states, non-terminals, initial state, and a set of allowable transitions. Each allowable transition is of the form $(\mathrm{q},\mathrm{a},\mathrm{A},\mathrm{q}\text{'},\mathrm{Î\pm })(\mathrm{q},\mathrm{a},\mathrm{A},\mathrm{q}\text{'},\mathrm{Î\pm }),$meaning that when in state, upon and the new PDA has a new non-terminal $\mathrm{A}\text{'}\mathrm{A}\text{'}$for each.

For every two states

$\mathrm{q},\mathrm{q}\text{'}∈\mathrm{Qq},\mathrm{q}\text{'}∈\mathrm{Q}$, and

there are two states:

$\begin{array}{l}(\mathrm{q},\mathrm{q}\text{'},1),(\mathrm{q},\mathrm{q}\text{'},2,\mathrm{A})\\ (\mathrm{q},\mathrm{q}\text{'},1),(\mathrm{q},\mathrm{q}\text{'},2,\mathrm{A}).\end{array}$

The starting states from the actual starting state is chosen non-deterministically among them via direct transitions are $(\mathrm{q},\mathrm{q},1)(\mathrm{q},\mathrm{q},1).$

For each transition $(\mathrm{q},\mathrm{a},\mathrm{A},\mathrm{q}\text{'},\mathrm{Î\pm })(\mathrm{q},\mathrm{a},\mathrm{A},\mathrm{q}\text{'},\mathrm{Î\pm }),$there are corresponding transitions $((\mathrm{q},\mathrm{q}\text{'}\text{'},1),\mathrm{a},\mathrm{A},(\mathrm{q}\text{'},\mathrm{q}\text{'}\text{'},1),\mathrm{Î\pm })((\mathrm{q},\mathrm{q}\text{'},1),\mathrm{a},\mathrm{A},(\mathrm{q}\text{'}\text{'},\mathrm{q}²,1),\mathrm{Î\pm })$ and $((\mathrm{q},\mathrm{q}\text{'}\text{'},2,\mathrm{B}),\mathrm{a},\mathrm{A},(\mathrm{q}\text{'},\mathrm{q}\text{'}\text{'},2,\mathrm{B}),\mathrm{Î\pm })((\mathrm{q},\mathrm{q}\text{'}\text{'},2,\mathrm{B}),\mathrm{a},\mathrm{A},(\mathrm{q}\text{'},\mathrm{q}\text{'}\text{'},2,\mathrm{B}),\mathrm{Î\pm }).$There are other transitions as well.

For each transition , $(\mathrm{q},\mathrm{a},\mathrm{A},\mathrm{q}\text{'},\mathrm{Î\pm })(\mathrm{q},\mathrm{a},\mathrm{A},\mathrm{q}\text{'},\mathrm{Î\pm }),$there are transitions $((\mathrm{q},\mathrm{q}\text{'}\text{'},1),\mathrm{a},\mathrm{A},(\mathrm{q}\text{'},\mathrm{q}\text{'}\text{'},1),\mathrm{Î\pm })((\mathrm{q},\mathrm{q}\text{'},1),\mathrm{a},\mathrm{A},(\mathrm{q}\text{'}\text{'},\mathrm{q}²,1),\mathrm{Î\pm })$ there are transitions $((\mathrm{q},\mathrm{q}\text{'}\text{'},1),∈,\mathrm{A},(\mathrm{q}\text{'},\mathrm{q}\text{'}\text{'},2,∈),\mathrm{A})((\mathrm{q},\mathrm{q}\text{'}\text{'},1),∈,\mathrm{A},(\mathrm{q},\mathrm{q}\text{'}\text{'},2,∈),\mathrm{A}),$

Accept a word $yxyx$ where $\mathrm{yxyx}$ is accepted by the original pushdown deterministic automaton. A state $(\mathrm{q},\mathrm{q},1)(\mathrm{q},\mathrm{q},1).$means that we're at stage at state and the original pushdown deterministic automaton is at state after reading xx. A state $(q,q\text{'},2,A)(q,q\text{'},2,A)$is similar, where $\mathrm{AA}$corresponds to the last $\mathrm{A}\text{'}\mathrm{A}\text{'}$that was popped. At stage 1, we are allowed to push instead of popping $\mathrm{AA}$. We do that for each non-terminal that is produced while processing xx, but only popped while processing $\mathrm{A}\text{'}\mathrm{A}\text{'}$. At stage two, we are allowed to pop A′A′ instead of pushing $\mathrm{AA}$. If we do this, then we have to remember that the top-of-stock is really $\mathrm{AA}$; this only applies when there are no "temporary" things on the stack, which in the simulated pushdown deterministic automaton is the same as the top-of-stack being or of the form $\mathrm{B}\text{'}\mathrm{B}\text{'}$.

Here is a simple example. Consider an automaton that pushes $\mathrm{AA}$for each xx, and pops AA for each $\mathrm{yy}$. The new automaton accepts words of two forms. For words of the first form, stage one consists of pushing kk times stage two consists of popping kk times $\mathrm{A}\text{'}\mathrm{A}\text{'}$, pushing times $\mathrm{AA}$, and popping two times AA. For words of the second form, we first push kk times $\mathrm{AA}$, then pop kk times AA, push one $\left\{\mathrm{yx}\right|\mathrm{xy}∈\mathrm{A}\}$time A′A′, transition to stage 2, and pop n times $\mathrm{A}\text{'}\mathrm{A}\text{'}$.

and so square brackets are allowed when pushing$\mathrm{AA}$ , since the top-of-stack is $\mathrm{BB}$ then we know that$\mathrm{BB}$is also the "real" top-of-stack, and so round parentheses are allowed even though the shadow top-of-stack is$\mathrm{AA}$ and Finally, we consume and pop $\mathrm{C}\text{'}\mathrm{C}\text{'}$

Hence, the rotational closure of language A to be$\mathrm{RC}\left(\mathrm{A}\right)=\left\{\mathrm{yx}\right|\mathrm{xy}∈\mathrm{A}\}$ is closed under rotational closure.