英語(yǔ)翻譯
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 = 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...)