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Exercises $38-45$ involve the Reve's puzzle, the variation of the Tower of Hanoi puzzle with four pegs and $n$ disks. Before presenting these exercises, we describe the Frame-Stewart algorithm for moving the disks from peg 1 to peg 4 so that no disk is ever on top of a smaller one. This algorithm, given the number of disks $n$ as input, depends on a choice of an integer $k$ with $1 \leq k \leq n .$ When there is only one disk, move it from peg 1 to peg 4 and stop. For $n>1$ , the algorithm proceeds recursively, using these three steps. Recursively move the stack of the $n-k$ smallest disks from peg 1 to peg $2,$ using all four pegs. Next move the stack of the $k$ largest disks from peg 1 to peg $4,$ using the three-peg algorithm from the Tower of Hanoi puzzle without using the peg holding the $n-k$ smallest disks. Finally, recursively move the smallest $n-k$ disks to peg $4,$ using all four pegs. Frame and Stewart showed that to produce the fewest moves using their algorithm, $k$ should be chosen to be the smallest integer such that $n$ does not exceed $t_{k}=k(k+1) / 2,$ the $k$ th triangular number, that is, $t_{k-1}<n \leq t_{k}$ . The long-standing conjecture, known as Frame's conjecture, that this algorithm uses the fewest number of moves required to solve the puzzle, was proved by Thierry Bousch in 2014 .
Describe the moves made by the Frame-Stewart algorithm, with $k$ chosen so that the fewest moves are required, for
$\begin{array}{llll}{\text { a) } 5 \text { disks. }} & {\text { b) } 6 \text { disks. }} & {\text { c) } 7 \text { disks. }} & {\text { d) } 8 \text { disks. }}\end{array}$