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whoareyou

Fri Mar 22, 2013 4:12 pm


Towers of Hanoi Question

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Hello,

I've been looking through the solution to the Towers of Hanoi problem for a long time, and even tracing the function using the call stack output on http://www.cs.cmu.edu/~cburch/survey/recurse/hanoiex.html.

I understand the first call, MoveTower(5, A, B, C), but in the second call and the calls thereafter, the destination and temporary peg switch. Why is this?


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jbking

Fri Mar 22, 2013 5:16 pm


Re: Towers of Hanoi Question

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Because to make that move requires one to make other moves first.  Remember the rules of which disks can be moved and when.


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whoareyou

Fri Mar 22, 2013 5:27 pm


RE:Towers of Hanoi Question

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I still don't get it :(. Can you explain in more detail?


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Panphobia

Sat Mar 23, 2013 1:13 pm


Re: Towers of Hanoi Question

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move n−1 discs from A to B. This leaves disc n alone on peg A
move disc n from A to C
move n−1 discs from B to C so they sit on disc n


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whoareyou

Sat Mar 23, 2013 6:12 pm


RE:Towers of Hanoi Question

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I still can't see why the destination and temporary pegs keep switching.


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Panphobia

Sat Mar 23, 2013 6:54 pm


Re: Towers of Hanoi Question

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public static void Towers(int n, int from, int to, int temp){
        if(n==0)return;
        Towers(n-1,from,temp,to);
        System.out.println("Move top disc from "+from+" to "+to);
        Towers(n-1,temp,to,from);
    }
if you passed Towers(3,1,2,3);
it would look like this through the output
[code]Move top disc from 1 to 2
Move top disc from 1 to 3
Move top disc from 2 to 3
Move top disc from 1 to 2
Move top disc from 3 to 1
Move top disc from 3 to 2
Move top disc from 1 to 2[/code]
and then like this graphically, with the actual pegs
[code](0)  _|_         |          |
    __|__        |          |
   ___|___       |          |
  ____|____  ____|____  ____|____

(1.1) |          |          |
    __|__        |          |
   ___|___      _|_         |
  ____|____  ____|____  ____|____ (A -> B)

(1.2) |          |          |
      |          |          |
   ___|___      _|_       __|__
  ____|____  ____|____  ____|____ (A -> C)

(1.3) |          |          |
      |          |         _|_
   ___|___       |        __|__
  ____|____  ____|____  ____|____ (A -> C)



(2.1) |          |          |
      |          |         _|_
      |       ___|___     __|__
  ____|____  ____|____  ____|____ (A -> B)



(3.1) |          |          |
      |          |          |
     _|_      ___|___     __|__
  ____|____  ____|____  ____|____ (C -> A)

(3.2) |          |          |
      |        __|__        |
     _|_      ___|___       |
  ____|____  ____|____  ____|____ (C -> B)

(3.3) |         _|_         |
      |        __|__        |
      |       ___|___       |
  ____|____  ____|____  ____|____ (A -> B)[/code]


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whoareyou

Sat Mar 23, 2013 6:58 pm


RE:Towers of Hanoi Question

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I realize that the algorithm is actually doing that and I can follow what the algorithm is doing. But my question is WHY is it necessary to switch the destination and temporary pegs. I know that it does it in the algorithm, but why does it make sense to do that?


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Panphobia

Sat Mar 23, 2013 7:08 pm


RE:Towers of Hanoi Question

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Basically, you do not want the exact same peg to keep on having a disc being added to it, that would defeat the purpose of the towers of hanoi, just search on how the Lucas' Tower/Towers of Hanoi is played, you know there are always 2^n-1 amount of moves for the least amount of moves, and to play optimally, what needs to be done? where do the discs need to be put? see if I instead used the function [code]public static void Towers(int n, int from, int to, int temp){
        if(n==0)return;
        Towers(n-1,from,to,temp);
        System.out.println("Move top disc from "+from+" to "+to);
        w++;
        Towers(n-1,from,to,temp);
    }[/code]
with the same values being passed I would get this [code]Move top disc from 1 to 2
Move top disc from 1 to 2
Move top disc from 1 to 2
Move top disc from 1 to 2
Move top disc from 1 to 2
Move top disc from 1 to 2
Move top disc from 1 to 2[/code]