An age-old puzzle that's challenged kids and pancake chefs for ages:

The goal of the puzzle is to move an entire stack of disks from one rod/plate to another with the following restrictions:

- Only one disk can be moved at a time.
- Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack
- No disk may be placed on top of a smaller disk.

So how can we go about solving this programatically? The key insight is that this problem has optimal substructure. This means that the solution for a larger problem, say for 4 disks is composed of the solutions for smaller subproblems: 1,2 and 3 disks.

```
def moveTower(height,fromPole, toPole, withPole):
if height >= 1:
moveTower(height-1,fromPole,withPole,toPole)
moveDisk(fromPole,toPole)
moveTower(height-1,withPole,toPole,fromPole)
def moveDisk(fp,tp):
print("moving disk from",fp,"to",tp)
moveTower(3,"A","B","C")
```

We first move the first `n-1`

disks from pole A to B by calling moveTower recursively but for the stack of n-1 disks. Then we can move disk `n`

from pole A to C. We can then do the same recurisive call and move the `n-1`

disks from B to C.

The recursive solution seems a bit expensive, with the two recursive calls leading to a $O(2^n)$ runtime. Can we do a bit better?

### Top-Down Memoization

Our recursive search right now is solving the same subproblem over and over again. If we saved or **memoized** the result, we could possibly trade off some of our runtime by using more memory and essentially caching the solutions to each subproblem.

```
memo = []
lookup = ["ABC","ACB","BAC","BCA","CAB","CBA"]
def topDown(height,fromPole, toPole, withPole):
global memo
memo = [[-1 for x in range(6)] for y in range(height+1)]
memo[0][0] = ""
for x in range(6):
memo[0][x] = ""
return topDownHelper(height,fromPole, toPole, withPole)
def topDownHelper(height,fromPole, toPole, withPole):
global memo
move = fromPole + toPole + withPole
index = 0
while move != lookup[index]:
index += 1
if memo[height][index] < 0:
memo[height][index] = topDownHelper(height-1,fromPole,withPole,toPole)\
+ fromPole +'->'+ toPole +', ' \
+ topDownHelper(height-1,withPole,toPole,fromPole)
return memo[height][index]
print topDown(3,"A","B","C")
```

### Bottom-up Dynamic Programming

Another approach would be to intelligently build up our solution, solving for 1 disk, then 2 disks, etc... By going in the opposite direction of our recursive solution, we save time by not solving the same subproblems. This is a technique dubbed **dynamic programming**. Since we are building up our solution successively, we only need to remember the solutions that are needed immediately. This saves us some valuable space.

```
moves = ["A->B","A->C","B->A","B->C","C->A","C->B"]
prevLookup = [1,0,3,2,5,4]
prev2Lookup = [5,3,4,1,2,0]
def bottomUp(height,fromPole, toPole, withPole):
prev = ["" for x in range(6)]
curr = ["" for x in range(6)]
for i in range(height):
for j in range(len(moves)):
prevStr = prev[prevLookup[j]] + ', ' if prev[prevLookup[j]] != "" else ""
nextStr = ', ' + prev[prev2Lookup[j]] if prev[prev2Lookup[j]] != "" else ""
curr[j] = prevStr + moves[j] + nextStr
curr,prev = prev,curr
return prev
print bottomUp(4,"A","B","C")[0]
```

### Performance

Doing a quick check on each implementation's running time, we see that the recursive solution explodes exponentially, while our memoized/dynamic programming implementations grow linearly. Clearly dynamic programming is the way to go!