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I worked *only* 9 hrs today, so taking a few minutes to do what I *truly* enjoy. Take 6 bit-planes of size=8 and all 6 must have the same 25% relationship between all of them. To me, this feels very correct and right on par. But, just to see, quickly cobbled some old code just to double-check. It took some code that interates over all possible permutations and outputs the ones that are orthogonal. Then I put the bit-planes over each other using bit-shifts. The important line looks like this: [code] nv = b0[row][col] | (b3[row][col]<<3) [/code] Then I ran that through my base(64) sequence of symbols (because 6 bits is 64 or 0-63). Then I counted up the unique symbols. Here is a small sample of what I got: [code] u f Q B a r 6 N n W J 2 j y F U g x 8 P s d K 5 Z o 1 G - k T C 4 L c t O 9 w h D S l + H 0 p Y M 7 q b A R e v V E z i 3 I X m 64 m X A R i z M 7 v e 3 I b q V E Y p O 9 + l 4 L h w H 0 t c D S C T s d G 1 g x 5 K - k P 8 Z o U F a r 2 J u f N 6 j y B Q n W 64 W n A R i z 6 N f u 3 I b q F U o Z O 9 + l K 5 x g H 0 t c T C S D s d G 1 w h L 4 - k P 8 p Y E V a r 2 J e v 7 M j y B Q X m 64 [/code] Each one has 64 unique symbols in base(64). This shows that (maybe not so clearly) that all of the bit-planes have the exact same 25% relationship between all of them. This shows that (maybe not so clearly) that you can order the bit-planes in any way that you want and still get valid output. Same relationship, any order. Ugh! Too many emotions to deal with right now. But I would like to thank the Doc for keeping the lights on. Still giving me a voice for the things that I truly do enjoy playing with. Omg, after approx 4 yrs I have generalized orthogonal latin squares (and MOLS) of size=n^2. Shaking the pillars of Heaven doesn't get much better than that.
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