Imagine a set of stacked files, such as papers on your desk. Occasionally, you will need to retrieve

one of these files, which you will find by "sequential search": looking at the first paper in the

stack, then the second, and so on until you find the document you require. A sensible sequential search algorithm is to place the most recently retrieved file at the top of the stack, the idea being.

that files accessed more often will "rise to the top" and thus require less searching in the long run.

For simplicity's sake, imagine such a scenario with just three files, labeled $A, B, C$ .

$$

\begin{array}{l}{\text { (a) Let } X_{n} \text { represent the sequence of the entire stack after the } n \text { th search. For example, if the files }} \\ {\text { are initially stacked } A \text { on top of } B \text { on top of } C, \text { then } X_{0}=A B C \text { . Determine the state space for }} \\ {\text { this chain. }}\end{array}

$$

$$

(b)X_{0}=A B C, \text { list all possible states for } X_{1} .[\text {Hint} : \text { One of the three files will be selected and }

$$rise to the front of the stack. Is every arrangement listed in (a) possible, starting from $A B C ? ]$

$$

\begin{array}{l}{\text { (c) Suppose that, at any given time, there is probability } p_{A} \text { that file } A \text { must be retrieved, } p_{B} \text { that }} \\ {\text { file } B \text { must be retrieved, and similarly for } p_{C}\left(=1-p_{A}-p_{B}\right) . \text { Determine all of the }} \\ {\text { non-zero one-step transition probabilities. }}\end{array}

$$