Towers of Hanoi #2

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Problem Description

In a monastery in Benares India there are three diamond towers holding 64 disks made of gold. The disks are each of a different size and have holes in the middle so that they slide over the towers and sit in a stack. When they started, 1500 years ago, all 64 disks were all on the first tower arranged with the largest on the bottom and the smallest on the top. The monks' job is to move all of the disks to the third tower following these three rules:

1. Each disk sits over a tower except when it is being moved. 
2. No disk may ever rest on a smaller disk.
3. Only one disk at a time may be moved.

The monks are very good at this after all this time and can move about 1 disk per second.  When they complete their task, THE WORLD WILL END! The question is, how much time  do we have left? 

Techniques & Background

Here is version 2 of the Towers problem.  Version 2 extends the code developed in version 1 by knowing which moves must be made in what order to solve the problem   Version 3, which  adds graphics code to show solutions or allows the user to solve the puzzle by dragging disks from tower to tower, will be presented next time.    If you haven't studied Towers1, and are interested in understanding the solution, please do so first. 

In version 2, we  added a TTowers object to hold the towers and the disks.  Right now, they are just numbers, you can think of the disk number a its diameter.  When we get to the graphics version we will expand this idea (TTowers will be contain array of TPeg objects and  TPeg will contain an array of TDisk objects).  Here's the definition for now:

TTowers = class(Tobject)
    Peg:array[1..3, 1.. maxdisks] of integer;
    NbrDisks:array[1..3] of integer; {current nbr of disks on each peg}
    procedure SetDisks(NewNbrDisks:integer);   {initialize ttowers with a number of disks}
    function MoveOne(Frompeg, topeg:integer):integer;  {return disk nbr of disk moved}
end;

The only slightly tricky part is in the MoveOne function.  It is called with FromPeg and ToPeg parameters so we must decide which peg is on top of FromPeg and move that number to the next disk position on ToPeg.  The xth disk of FromPeg would be referenced as Peg[FromPeg, x]. so the  top disk of  FromPeg can be referenced  by Peg[FromPeg, NbrDisks[FromPeg]].   If you can follow that, then you can understand the rest of MoveOne.  

 Running/Exploring the Program 

Suggestions for Further Explorations

  
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