[next] [prev] [up] Date: Wed, 15 Dec 93 13:15:45 -0500
~~~ [prev] [up] From: Dale I. Newfield <dn1l+@andrew.cmu.edu >
[next] [prev] [up] Subject: Re: Description of Tangle, Part 2
Just to make sure everyone knows what we are talking about, here is a
message from the archives:
Excerpts from mail: 25-Apr-92 Description of Tangle, Part 2 by Chris
Worrell@eql.caltec 
> Annotating Don.Woods diagram  (which is in the correct orientation)
>               2       3
>         ---------------------
>         |     @       #     |
>         |     @       #     |
>       1 |$$    @      # %%%%| 4
>         |  $    @    %#%    |
>         |   $    @ %% #     |
>         |    $    %@  #     |
>         |    $  %%  @@#     |
>         |    %%%      #@@   |
>       4 |%%%% $       #  @@@| 2
>         |     $       #     |
>         |     $       #     |
>         ---------------------
>               1        3
>  
> The duplicate piece in each tangle is:
>                 1       2       3       4
> Tangle 1        Blue    Red     Yellow  Green
> Tangle 2        Yellow  Blue    Green   Red
> Tangle 3        Green   Yellow  Blue    Red
> Tangle 4        Red     Green   Yellow  Blue
>  
> All 4 Tangles are the same puzzle, just colored differently.
> Each has all 24 color permutations, plus a duplicate.

I had kind of hoped that the connectivity on the different puzzles was
different, instead of just the colors.

(Actually, the sequence I sent before was slightly wrong--here is the
one I actually used. Using Don's format)
>Don used the sequence:                 Dale used:
> 
>         1   3   5   7   9               1   2   6  10  15
>         2   4   6   8  10               3   4   7  11  16
>        11  12  13  14  15               5   8  12  17  20
>        16  17  18  19  20               9  13  18  21  23
>        21  22  23  24  25              14  19  22  24  25

But yes, Don's fillpattern still gets more constraints in earlier--here
is the number of constraints at each step
Don's: 0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2
Mine:  0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2
As you can see, I had my 1's clustered more toward the beginning, which
is non-optimal.

Assuming that there is only a change in color(and not in connectivity),
as was posted by Chris in april of 92, I would think modifying code to
attempt the 10x10 would be fairly simple...(seeing as my code went poof
sometime last year, when a disk crashed(not that it was
complicated))...wanna try?

(Thanks for the pointers to the Apr 92 discussion)
I agree with the concensus expressed in the archives that this puzzle is
inherently "not that great" because no non-brute-force method has been
found/seems to exist.

-Dale


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