Hi Jonathan,
[log in to unmask]" type="cite">The valid rotations are 1,2,3,4,5,6,7,8 so there has to be something there even though it doesn't bring anything meaningful to the alfabet.   Maybe instead of trying to make each base symbol and complete rectangle of fills and rotations (or matrix) of symbol but rather just have table with the ones that are valid.  We have to look up the valid fills and valid rotations per base symbol anyways.  So I'm not seing the advantage here. 
Each base symbol has an 8-bit number for the valid fills and a 16-bit number for the valid rotations.  So each base symbol needs 3-bytes to determine validity.

Here is a table from draft-slevinski-signwriting-text that deals with validity in the ISWA 2010.
http://signpuddle.net/download/draft-slevinski-signwriting-text-03.html#rfc.appendix.B.5

Sum Binary Set
1 100000 {1}
2 010000 {2}
3 110000 {1, 2}
7 111000 {1, 2, 3}
15 111100 {1, 2, 3, 4}
31 111110 {1, 2, 3, 4, 5}
63 111111 {1, 2, 3, 4, 5, 6}
187 11011101 {1, 2, 4, 5, 6, 8}
255 11111111 {1, 2, 3, 4, 5, 6, 7, 8}
511 1111111110000000 {1, 2, 3, 4, 5, 6, 7, 8, 9}
48059 1101110111011101 {1, 2, 4, 5, 6, 8, 9, 10, 12, 13, 14, 16}
65535 1111111111111111 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}


For the punctuation in question, it has valid fills of 1 thru 4 ( sum value of 15) and it has valid rotations of 1 thru 8 (sum value of 255).  So to send the information for the base symbol we would need S387,15,255.  For any symbol key, we can use the associated fill and rotation numbers to determine if the key is valid.
 

[log in to unmask]" type="cite">Maybe instead of trying to make each base symbol and complete rectangle of fills and rotations (or matrix) of symbol but rather just have table with the ones that are valid.  We have to look up the valid fills and valid rotations per base symbol anyways.  So I'm not seing the advantage here. 
The advantage is that we do not need to have a bit for each and every cell.  With 6 fills and 16 columns, we would need a 96-bit number which is 12-bytes.  So to send the information for the base symbol we would need S387,71777214294589695 or we could use 6 16-bit numbers which would be S387,255,255,255,255,0,0

Setting each individual cell would require 4 times as much data per base symbol.  I chose the smaller representation because I was trying to reduce the size as much as possible.

Regards,
-Steve
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