Now consider the following variant of the Tower of Hanoi problem. To the original problem,
let us add a new condition: Each disk move now must be performed only in the clockwise
direction — ie., A ? B, B ? C, or C ? A. So, for example, if I have to move a disk from A
to C then that will involve a minimum of two moves (first, from A to B, and then from B to
C); but from C to A needs only one move. All the old conditions from the original version of
the problem still apply (i.e., you cannot move a larger disk on top of a smaller disk).
Under this new problem formulation:
• Let Q(n) denote the minimum number of moves required to transfer n disks from A ? B
using C.
• Let R(n) denote the minimum number of moves to transfer n disks from A ? C
clockwise using B.
Prove that:
(1)
$Q(n) = \begin{cases} 1 & , n = 1\\ 2R(n-1) + 1 & , n > 1 \end{cases}$
(2)
$R(n) = \begin{cases} 2 & , n = 1\\ Q(n) + Q(n-1) + 1 & , n > 1 \end{cases}$
Hint 1: To provide the proof you may need to first come up with a corresponding algorithm
under the new clockwise condition. That should give you the mathematical recurrence shown
above. Please illustrate your algorithm in the form of a figure following the example of the
figure shown above in the question, for the original version of the problem.
Hint: If R(n) is the number of moves required to move n disks from peg A to peg C (using
B), then how many moves will be needed to move n disks from peg B to peg A (using C)?
