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warjournal
Maniac (V) Mad Scientist

From:
Insane since: Aug 2000

posted posted 04-03-2018 21:46

I've done it. Okay, mostly done it. Not quite perfect. After about a year
of playing with orthogonal latin squares, I created 8x8 orthos using pure
algorithm without back-tracking.

I repeat: No back-tracking.

code:
4 3 1 7 5 2 0 6    2 0 4 6 7 5 3 1
2 5 7 1 3 4 6 0    7 4 0 2 1 3 5 6
0 6 5 2 1 7 4 3    1 3 6 4 5 7 0 2
6 0 3 4 7 1 2 5    4 7 3 1 2 0 6 5
7 1 4 3 6 0 5 2    3 1 5 7 6 4 2 0
1 7 2 5 0 6 3 4    6 5 1 3 0 2 4 7
5 2 6 0 4 3 7 1    0 2 7 5 4 6 1 3
3 4 0 6 2 5 1 7    5 6 2 0 3 1 7 4



Still one little issue with the logic in my solver. But I understand it and should
be able to fix it. Once I do and get the hang of it, I should be able to take orthos
and multiply them bigger x2, x3, and x4.

This is a pretty big step forward.
Still a lot of work to do, but I'm that much closer to my end-game.
Woohoo!

warjournal
Maniac (V) Mad Scientist

From:
Insane since: Aug 2000

posted posted 04-14-2018 00:38

Ortho size=3 multiplied x3 = ortho of size 9:

code:
3 2 4 6 1 7 5 8 0  8 1 2 7 4 5 6 0 3
4 7 6 3 5 8 0 2 1  7 0 8 2 3 1 4 5 6
6 8 3 4 0 2 1 7 5  2 5 7 8 6 0 3 1 4
8 1 2 7 4 5 6 0 3  3 2 4 6 1 7 5 8 0
7 0 8 2 3 1 4 5 6  4 7 6 3 5 8 0 2 1
5 3 0 1 8 4 2 6 7  1 6 5 0 7 3 8 4 2
0 4 1 5 2 6 7 3 8  0 4 1 5 2 6 7 3 8
2 5 7 8 6 0 3 1 4  6 8 3 4 0 2 1 7 5
1 6 5 0 7 3 8 4 2  5 3 0 1 8 4 2 6 7


unique count=81



Meh.

Now that I understand it, too easy.
And boring.

I don't know. Maybe I can find something impressive
to do with this parlour trick.

warjournal
Maniac (V) Mad Scientist

From:
Insane since: Aug 2000

posted posted 04-15-2018 06:18

The above is boring. Why? Because it is perfectly fractal - and I mean
perfectly fractal. Seriously, to the point of being boring. I was angry,
frustrated, and disappointed.

Then I managed to break it. I am excited again.

code:
1 7 3 2 8 5 0 6 4  3 2 8 6 4 1 0 7 5 
3 1 7 4 2 8 5 0 6  2 7 3 4 1 8 6 5 0 
8 3 1 7 5 2 6 4 0  7 4 2 1 8 3 5 0 6 
5 0 6 3 1 7 4 2 8  4 1 6 7 5 0 2 8 3 
6 4 0 8 3 1 7 5 2  1 8 4 5 0 6 7 3 2 
0 6 4 1 7 3 2 8 5  8 3 1 0 6 5 4 2 7 
7 5 2 6 4 0 8 3 1  5 0 7 8 3 2 1 6 4 
2 8 5 0 6 4 1 7 3  0 6 5 3 2 7 8 4 1 
4 2 8 5 0 6 3 1 7  6 5 0 2 7 4 3 1 8 

unique count=81



The above is broken, but here is a little piece of it to show how boring it can be.
In the grid on the left, the top three lines:

code:
1 7 3  2 8 5  0 6 4
3 1 7  4 2 8  5 0 6
8 3 1  7 5 2  6 4 0



Boring! Imagine orthogonal latin squares that exhibit that pattern through-out.
Uninteresting parlour trick.

But I can break it, as shown in the full example, and I am excited again.

I'm off again.
Lots of work to do.
See ya'll in a few months or so.
And thanks for listening.

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