Background

Charlie Darkbrown sits in another one of those boring Computer
Science lessons: At the moment the teacher just explains the standard
Tower of Hanoi problem, which bores Charlie to death!

The teacher points to the blackboard (Fig. 4) and says: "So here is the problem:

• There are three towers: A, B and C.
  
  
• There are n disks. The number n is constant while working the puzzle.
  
  
• All disks are different in size.
  
  
• The disks are initially stacked on tower A increasing in size from the top to the bottom.
  
  
• The goal of the puzzle is to transfer all of the disks from tower A to tower C.
  
  
• One disk at a time can be moved from the top of a tower either to an empty tower or to a tower with a larger disk on the top.

So your task is to write a
program that calculates the smallest number of disk moves necessary to
move all the disks from tower A to C."

Charlie: "This is incredibly boring—everybody knows that this can
be solved using a simple recursion.I deny to code something as simple
as this!"

The teacher sighs: "Well, Charlie, let's think about something for
you to do: For you there is a fourth tower D. Calculate the smallest
number of disk moves to move all the disks from tower A to tower D
using all four towers."

Charlie looks irritated: "Urgh. . . Well, I don't know an optimal algorithm for four towers. . . "

Problem

So the real problem is that problem solving does not belong to the
things Charlie is good at. Actually, the only thing Charlie is really
good at is "sitting next to someone who can do the job". And now guess
what — exactly! It is you who is sitting next to Charlie, and he is
already glaring at you.

Luckily, you know that the following algorithm works for n <=
12: At first k >= 1 disks on tower A are fixed and the remaining n-k
disks are moved from tower A to tower B using the algorithm for four
towers.Then the remaining k disks from tower A are moved to tower D
using the algorithm for three towers. At last the n - k disks from
tower B are moved to tower D again using the algorithm for four towers
(and thereby not moving any of the k disks already on tower D). Do this
for all k 2 ∈{1, .... , n} and find the k with the minimal number of
moves.

So for n = 3 and k = 2 you would first move 1 (3-2) disk from
tower A to tower B using the algorithm for four towers (one move). Then
you would move the remaining two disks from tower A to tower D using
the algorithm for three towers (three moves). And the last step would
be to move the disk from tower B to tower D using again the algorithm
for four towers (another move). Thus the solution for n = 3 and k = 2
is 5 moves. To be sure that this really is the best solution for n = 3
you need to check the other possible values 1 and 3 for k. (But, by the
way, 5 is optimal. . . )