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 - i.e., 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:
Q(n) = {(1, n=1), (2R(n-1) + 1, n>1)}
R(n) = {(2, n=1), (Q(n) + Q(n-1) + 1, n>1)}
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)?