[Haskell-beginners] tower of hanoi problem

[Dudley Brooks]

An aside about the Towers of Hanoi which does not answer the OP's 
question but is interesting:  A recursive solution is a great method for 
having a computer play the game.  But human beings might want to play it 
also, and humans don't think recursively -- we can't keep track of the 
intermediate results and where we are in the process.  We think 
iteratively. It turns out that if you actually physically play the game 
you will discover that there is a very simple iterative pattern to the 
moves.  I won't tell it, so that people can have the fun of discovering it.

Disclosure:  I "discovered" this while teaching the game to a child 
friend who had no particular mathematical or computer-related 
inclination ... and it was actually *she* who discovered it.  She 
couldn't articulate it in words ... but she could *do* it.

Here's a "Chinese proverb" definition (or example) of recursion: A 
journey of 10,000 miles starts with a single step ... followed by being 
able to take one more step no matter how many steps you have already taken.

On 2/14/15 7:24 AM, Mike Meyer wrote:
> On Sat, Feb 14, 2015 at 6:43 AM, Roelof Wobben <r.wobben at home.nl 
> <mailto:r.wobben at home.nl>> wrote:
>
>     Hello,
>
>     I try to answer your questions. Maybe then it is clear where My
>     thinking gets the wrong turn.
>
>
>         · Divide your input into two cases: the terminal one that ends
>         the recursion and the one you recurse on.
>
>
>     I think the terminal one is when there is no moves anymore so when
>     a and b are empty and c holds all the disk.
>
> I'd say the terminal one is one move, not no moves. I even gave it 
> away with the next comment:
>
>         · Decide how to handle the terminal case. In this case, you'll
>         return a list consisting of one move. 
>
>     There is a problem. I thought I must return all the moves then.
>
>
> Ah, I think we miscommunciated about what "terminal" means. I meant 
> the case that terminates the recursion, not the problem. That result 
> will be combined with the values returned by other calls to create the 
> ultimate return value having all the moves.
>
> So when hanoi is called with only 1 disk to move, you just return the 
> list consisting of that move. If it's called with more than 1 disk, 
> you have to make multiple recursive calls with fewer disks than it was 
> called with. I don't think you can do this without special handling 
> for the single disk case, so you might as well make that one your 
> terminating case. Handling the no-disk case would be part of the 
> initial error checking.
>
>         · Decide how to divide the non-terminal case into at least two
>         sub-problems that are all closer to the terminal case by some
>         measurement.
>
>     two subproblems ?  I can only think about one and that is looking
>     what the next move can be.
>
>
> There's gotta be at least two subproblems, because you have to create 
> smaller problems for the recursive calls. If you can make the problem 
> smaller without creating two subpropblems - well, then your problem 
> can just goes away!
>
> For hanoi, you don't need to worry about what the next move will be. 
> Assume that you have a working hanoi routine that can handle fewer 
> disks that you currently have. How can you use that to solve your 
> current problem?  It should be obvious that you have to move some 
> number less than all of them to one peg, then move the rest to the 
> other peg, and finally move the first stack you moved on top of the 
> second stack. So you need to figure out how many to move and where to 
> move them to so you don't make any illegal moves in the process.
>
>         · Write the code that makes the non-terminal calls and
>         combines their values to create your return value.
>
>     Oke, when I have the right answers to the former questions, I can
>     think about how to solve this one.
>
>
> Yup.
>
>         In cases where your function returns a list, the combination
>         is almost always appending one case to the other.
>
>     I also think so , so it will be old case ++ new case because
>     otherwise it will display in the wrong order.
>
>
> Well, the solution I outlined has more than one case to deal with, as 
> you have to make three recursive calls.
>
>
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