TITLE: Towers of Hanoi if big disks can go on top of small disks
QUESTION [6 upvotes]: The Tower of Hanoi puzzle is concerned with moving $n$ disks between three pegs so that a larger disk cannot be placed on top of a smaller disk. Based on a (now deleted) StackOverflow question, suppose that one can place larger disks above smaller ones.
One can represent the game state by a 3-tuple of ordered sets $(A, B, C)$.
For example, the solved state with $3$ disks is given by $([3,2,1], [], [])$:
1
2
3
* * *

Question: given an arbitrary game state, what is a minimal sequence of moves that reaches the solved state? (this thread suggests that reaching the solved state is always possible).

There is a unique solved state with all disks placed on the first peg in order (illustrated above).
Ideally, I am interested in an algorithm that reaches the solved state with fewest moves.
If describing such an algorithm is difficult, I would also be interested in the minimal number of moves required to reach an arbitrary game state from any other game state (the diameter of the game state graph).
Calculation by @PeterLang suggests that the diameter of the game state graph is given by [1, 4, 7, 10, 13, 16, 19, 22, 26, 29] for the number of disks ranging from 1 to 10. There appears to be only one OEIS sequence matching this pattern, and I have no clue if it should generalize.


Here is an example of solving the game with $3$ disks:
Given an initial state $([2], [1], [3])$, one can reach the solved state as follows:
                                      1        
           2         2      2         2     
2 1 3      1 3     3 1      3 1       3    
* * * => * * *  => * * * => * * * =>  * * *

with associated sequence of moves $[1 \to 2], [3\to 1], [2 \to 1], [2 \to 1]$.

I computed a graph $G$ with vertices corresponding to game states, so that an edge is drawn between two game states whenever one can be reached from another with a single legal move. Here is an example with two disks:

Surprisingly, the graph diameter seems to grow slowly. I wonder if it is always possible to reach the solved state in at most $1 + 3n$ moves, where $n$ is the number of disks (Peter's answer disproves this).
                 Graph Diameter  Number of Vertices
Number of Disks                                    
1                             1                   3
2                             4                  12
3                             7                  60
4                            10                 360
5                            13                2520

REPLY [3 votes]: As Peter Lang says, the number of states in the graph is $(n+2)!/2$. From each state, there are at most $6$ possible moves. So the diameter of the graph is at least
$$\log_6 \frac{(n+2)!}{2} = n \frac{\log n}{\log 6} + O(n).$$
On the other hand, Jaap Scherphuis's really clever Radix Sort argument (see comment thread above) shows that the diameter is at most
$$n+2 n \lceil \log_2 n \rceil = 2 n \frac{\log n}{\log 2} + O(n).$$
Thus, I would suggest focusing our efforts on finding the constant $c$ such that the diameter grows like $c n \log n$.

I can improve the lower bound a little bit. It never makes sense to move a disc from peg A to peg B and then move it back, nor is it ever a good idea to move a peg from A to B and then move it from B to C (because we could have just done that in one step, A to C). So, while we have $6$ choices for the first move, we only have $4$ reasonable choices for the moves after that. This improves the lower bond to
$$1+\log_4 \frac{(n+2)!}{12} = n \frac{\log n}{\log 4} + O(n).$$

I can also improve the upper bound, by using a variant of merge sort rather than radix sort. Let $F(n)$ be the largest number of steps to get from any position where all discs are on one peg to the position where the discs are sorted on another peg. Let $G(n)$ be the same, except that I ask for the discs to end on the same peg they started on. Since we can always use $n$ steps at the start to stack all our discs up, the diameter of the graph is $F(n) + O(n) = G(n) + O(n)$.
Now, start with all the discs on peg 1 (in some order). Using $F(\lfloor n/2 \rfloor)$ steps, take the top $n/2$ discs and put them, in upside down order, on peg 2, leaving the bottom $\lceil n/2 \rceil$ discs alone. Then take the discs that remain on peg 1 and move them to peg 3 in upside down order, leaving the discs which were moved earlier alone. Finally, merge both stacks back into peg 1, using $n$ steps. This shows
$$G(n) \leq F(\lfloor n/2 \rfloor) + F(\lceil n/2 \rfloor) + n.$$
Similarly, we have
$$F(n) \leq F(\lfloor n/2 \rfloor) + G(\lceil n/2 \rfloor) + n.$$
(This time, move half the pegs from 1 to 3 while sorting them, leave the others in place on 1 while sorting them, then merge unto 2.)
I believe this inductively gives us $F(n)$ and $G(n) \leq n \frac{\log n}{\log 2} + O(n)$.

There is still a factor of $2$ difference between $\tfrac{1}{\log 4}= \tfrac{1}{2 \log 2}$ and $\tfrac{1}{\log 2}$. Let's see if we can shrink it!