# Tower of Hanoi

The Tower of Hanoi puzzle was invented by the French mathematician Edouard Lucas in 1883. He was inspired by a legend that tells of a Hindu temple where the puzzle was presented to young priests. At the beginning of time, the priests were given three poles and a stack of 64 gold disks, each disk a little smaller than the one beneath it. Their assignment was to transfer all 64 disks from one of the three poles to another, with two important constraints. They could only move one disk at a time, and they could never place a larger disk on top of a smaller one. The priests worked very efficiently, day and night, moving one disk every second. When they finished their work, the legend said, the temple would crumble into dust and the world would vanish.

Although the legend is interesting, you need not worry about the world ending any time soon. The number of moves required to correctly move a tower of 64 disks is $2^{64}-1 = 18,446,744,073,709,551,615$. At a rate of one move per second, that is $584,942,417,355$ years! Clearly there is more to this puzzle than meets the eye.

The animation below demonstrates a solution to the puzzle with four discs. Notice that, as the rules specify, the disks on each peg are stacked so that smaller disks are always on top of the larger disks. If you have not tried to solve this puzzle before, you should try it now. You do not need fancy disks and poles–a pile of books or pieces of paper will work.

How do we go about solving this problem recursively? How would you go about solving this problem at all? What is our base case? Let’s think about this problem from the bottom up. Suppose you have a tower of five disks, originally on peg one. If you already knew how to move a tower of four disks to peg two, you could then easily move the bottom disk to peg three, and then move the tower of four from peg two to peg three. But what if you do not know how to move a tower of height four? Suppose that you knew how to move a tower of height three to peg three; then it would be easy to move the fourth disk to peg two and move the three from peg three on top of it. But what if you do not know how to move a tower of three? How about moving a tower of two disks to peg two and then moving the third disk to peg three, and then moving the tower of height two on top of it? But what if you still do not know how to do this? Surely you would agree that moving a single disk to peg three is easy enough, trivial you might even say. This sounds like a base case in the making.

Here is a high-level outline of how to move a tower from the starting pole, to the goal pole, using an intermediate pole:

- Move a tower of height-1 to an intermediate pole, using the final pole.
- Move the remaining disk to the final pole.
- Move the tower of height-1 from the intermediate pole to the final pole using the original pole.

As long as we always obey the rule that the larger disks remain on the bottom of the stack, we can use the three steps above recursively, treating any larger disks as though they were not even there. The only thing missing from the outline above is the identification of a base case. The simplest Tower of Hanoi problem is a tower of one disk. In this case, we need move only a single disk to its final destination. A tower of one disk will be our base case. In addition, the steps outlined above move us toward the base case by reducing the height of the tower in steps 1 and 3. Below we present a possible Python solution to the Tower of Hanoi puzzle.

```
def move_tower(height, from_pole, to_pole, with_pole):
if height >= 1:
move_tower(height - 1, from_pole, with_pole, to_pole)
move_disk(from_pole, to_pole)
move_tower(height - 1, with_pole, to_pole, from_pole)
```

Notice that the code above is almost identical
to the English description. The key to the simplicity of the algorithm
is that we make two different recursive calls, the first to move all but the bottom disk on the
initial tower to an intermediate pole. Before we make a second recursive call, we simply move the
bottom disk to its final resting place. Finally we move the tower
from the intermediate pole to the top of the largest disk. The base case
is detected when the tower height is 0; in this case there is nothing to
do, so the `move_tower`

function simply returns. The important thing to
remember about handling the base case this way is that simply returning
from `move_tower`

is what finally allows the `move_disk`

function to be
called.

If we implement this simple `move_disk`

function, we can then illustrate the required moves to solve the problem:

```
def move_disk(from_pole, to_pole):
print('moving disk from {} to {}'.format(from_pole, to_pole))
```

Now, calling `move_tower`

with the arguments `3, 'A', 'B', 'C'`

will give us the output:

```
moving disk from A to B
moving disk from A to C
moving disk from B to C
moving disk from A to B
moving disk from C to A
moving disk from C to B
moving disk from A to B
```

Now that you have seen the code for both `move_tower`

and `move_disk`

, you
may be wondering why we do not have a data structure that explicitly
keeps track of what disks are on what poles. Here is a hint: if you were
going to explicitly keep track of the disks, you would probably use
three `Stack`

objects, one for each pole. The answer is that Python
provides the stacks that we need implicitly through the call stack.