Date: Mon, 07 Mar 94 11:37:25 +0100
~~~ From: Jan de Ruiter <jandr@xirion.nl >
Subject: [no subject]

To anandrao@HK.Super.NET
Cc: cube-lovers@life.ai.mit.edu

On Fri, 18 Feb 1994, Jan de Ruiter wrote:
>
> Sorry about not reporting this earlier, but my search for solutions for
> Rubiks Tangle 10x10 confirms the finding of Don Woods: no solutions!
>
[snip]
> we could re-define the puzzle as follows:
> find which four pieces to duplicate in order to find solutions for
> the 10x10.
> If the number of solutions varies depending on the choice, you could
> find which four pieces to duplicate in order to find a set which has
> the minimum number of solutions for the 10x10.
^^^^^^^
The kind Mr. Rubik has already done that - the minimum is - ZERO!

Correct, but I think you know what I mean: minimum >= 1

The revised problem can be solved fairly easily using your program ( I
don't know, though, how long it takes to run to completion for the 10*10
case)

More than a week

- try to place only 99 tiles out of the 100 given tiles. You may
have several sub-solutions. It is then easy to determine for each of these
sub-solutions which tile you need to complete the 10*10 mosaic.

I am sorry, but I have to disagree on this. It is not that simple.

If you managed to place 99 pieces, you have already placed three or even
all four of the duplicated pieces (depending on which one is left over)
If you placed three, there are tree possibilities for the piece we need:
- it is nonexistent (illegal colour combinations): no solution
- it is one of the duplicated pieces: this means two of the four puzzles
will be identical which is OK, but not so nice, or
- it is any other piece: we found a good solution

If you placed all four duplicated pieces already, any solution you find
will not satisfy the conditions of the puzzle (i.e. precisely four duplicated
pieces).

And in both cases you have not solved:
which four pieces to duplicate in order to find solutions for the 10x10.
but:
which four pieces to duplicate in order to find solutions for the 10x10,
with the restriction that three of them must be identical to any three
taken from the set of four duplicates given by Rubik.

Solving the puzzle without this restriction requires a different approach.
I was thinking of starting the program with 5 of each (120 pieces), and after
placing the 5th duplicated piece remove the rest of the duplicates from
the remaining pieces, and see if this leads to a solution. As soon as back-
tracking removes the 5th duplicate, all other duplicates must be made
accessible again.

Jan