Exercises 38-45 involve the Reve's puzzle, the variation of...

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 . Show that if $R(n)$ is the number of moves used by the Frame-Stewart algorithm to solve the Reve's puzzle with $n$ disks, where $k$ is chosen to be the smallest integer with $n \leq k(k+1) / 2,$ then $R(n)$ satisfies the recurrence relation $R(n)=2 R(n-k)+2^{k}-1,$ with $R(0)=0$ and $R(1)=1$.

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 .
Show that if $R(n)$ is the number of moves used by the Frame-Stewart algorithm to solve the Reve's puzzle with $n$ disks, where $k$ is chosen to be the smallest integer with $n \leq k(k+1) / 2,$ then $R(n)$ satisfies the recurrence relation $R(n)=2 R(n-k)+2^{k}-1,$ with $R(0)=0$ and $R(1)=1$.

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