Towers of Hanoi - algorithm in Matlab

For more information, codes and details, visit: http://matrixlab-examples.com/tower-of-hanoi-algorithm.html

In this demonstration we’ll use 4 disks. The objective of the puzzle is to move all of the disks from tower A to tower C.

• • Two Easy Rules: Only one disk can be moved at a time and it can only be the top disk of any tower. Disks cannot be stacked on top of smaller disks.
• • We’re going to solve the puzzle by using recursion.

Recursion is a computer programming technique that involves the use of a procedure that calls itself one or several times until a specified condition is met.

• • If we want to move disk 4 from A to C, after several moves and at some point, we MUST be in this situation:

And now, we can accomplish our first goal: move the largest disk to its final destination.

But now, if we ignore tower C, we are just like we were at the beginning, but with just 3 disks left (instead of 4)! We’ve simplified the puzzle!

If we want to move disk 3 from B to C, after several moves and at some point, we MUST be in this situation:

And now, we can accomplish our second goal: move the second largest disk to its final destination.

Now, the position is trivial to finish… but we still can repeat the ideas above.

• • Let’s code that with recursion. We’ll call our function m (for move) and we’ll have four input parameters:

m(n, init, temp, fin)

where

n is the number of disks to move init is the initial tower temp is the temporary peg fin is the final tower

We have three situations to consider:

1.- m(n-1, init, fin, temp) we’ll move n-1 disks from A to B, with C as temporary peg.

2.- m(1, init, temp, fin) we’ll move 1 disk (our partial goal) from A to C, with B as temporary peg.

3.- m(n-1, temp, init, fin) we’ll move n-1 disks from B to C, with A as temporary peg.