swildner@channelz.gun.de (Sascha Wildner) (03/21/91)
Hello world!
Is there any way to get those 4096 colors of the Amiga in a row so that the
difference between color[n] and color[n+1] is only one modification of either
R, G or B? Don't understand what I mean?
R G B
Color[0] = 0 0 0
Color[1] = 0 0 1
Color[2] = 0 1 1
Color[3] = 0 1 0 would be correct, while
R G B
Color[0] = 0 0 0
Color[1] = 0 1 1 wouldn't be, because Color[1] differs from Color[0] by two
modifications (G & B).
Is there a way to accomplish this for all 4096 colors?
--
+---------------------------------------------------------------+
| Sascha Wildner, Am Druvendriesch 27, W-5030 Huerth 6, Germany |
| |
| Phone : +49 2233 15571 Zerberus: S.WILDNER@TBC.ZER |
| MagicNet: DINO-BOX:BILBO UUCP : swildner@channelz.gun.de |
| Fido : 2:241/5008 |
+---------------------------------------------------------------+tomb@hplsla.HP.COM (Tom Bruhns) (03/22/91)
swildner@channelz.gun.de (Sascha Wildner) writes: >Is there any way to get those 4096 colors of the Amiga in a row so that the >difference between color[n] and color[n+1] is only one modification of either >R, G or B? The explanation did not make it absolutely clear: is a transition in a given color allowed to be more than one level at a time (in what you want)? If so, one answer to your question is to gray-code the 12 bits; in gray (or maybe Gray ;-) code, only one bit changes each step, all the way around a circle till you get back where you started. I believe this also would be allowed in HAM (though I have to admit I'm only marginally familiar with HAM's limitations). You should be able to find documentation on Gray code in any library with a techinical section. e.g.: Bin Gray Bin Gray 0000 0000 1000 1100 0001 0001 1001 1101 0010 0011 1010 1111 0011 0010 1011 1110 0100 0110 1100 1010 0101 0111 1101 1011 0110 0101 1110 1001 0111 0100 1111 1000 A symmetry should be apparent in the Gray code. You can generate Gray code from straight binary through appropriate XOR op's, I believe. Gray is used for things like optical encoders to insure that transitions from one word to the next don't generate outputs that appear to come from a "long ways away".
metahawk@itsgw.rpi.edu (Wayne G Rigby) (03/22/91)
In article <swildner.3264@channelz.gun.de> swildner@channelz.gun.de (Sascha Wildner) writes: >Hello world! >Is there any way to get those 4096 colors of the Amiga in a row so that the >difference between color[n] and color[n+1] is only one modification of either >R, G or B? Don't understand what I mean? > > R G B >Color[0] = 0 0 0 >Color[1] = 0 0 1 >Color[2] = 0 1 1 >Color[3] = 0 1 0 would be correct, while If you mean that you want to line up all of the Amiga's 4096 colors in Gray Code order. Gray code is a binary coding scheme in which only one bit changes from one number to the next. The general algorithm for converting a binary number to a Gray number is as follows: Convert 10011 binary to Gray code. This is the same number shifted by one bit. | binary | 1 V = 1 0 + 1 = 1 MOD 2 = 1 0 + 0 = 0 MOD 2 = 0 1 + 0 = 1 MOD 2 = 1 1 + 1 = 2 MOD 2 = 0 So 10011 in binary is 11010 in Gray code. And 10100 in binary is 11110 in Gray code. The rest of the real numbers are left up to the reader to convert. A description Gray code is in Introduction to Computer Engineering by Franco P. Preparata ------------------------------------ and many other computer texts. > > R G B >Color[0] = 0 0 0 >Color[1] = 0 1 1 wouldn't be, because Color[1] differs from Color[0] by two > modifications (G & B). > >Is there a way to accomplish this for all 4096 colors? >-- > > > +---------------------------------------------------------------+ > | Sascha Wildner, Am Druvendriesch 27, W-5030 Huerth 6, Germany | > | | > | Phone : +49 2233 15571 Zerberus: S.WILDNER@TBC.ZER | > | MagicNet: DINO-BOX:BILBO UUCP : swildner@channelz.gun.de | > | Fido : 2:241/5008 | > +---------------------------------------------------------------+ Wayne Rigby metahawk@rpi.edu
pashdown@javelin.es.com (Pete Ashdown) (03/22/91)
swildner@channelz.gun.de (Sascha Wildner) writes: >Hello world! >Is there any way to get those 4096 colors of the Amiga in a row so that the >difference between color[n] and color[n+1] is only one modification of either >R, G or B? Don't understand what I mean? Umm, I believe this is what is commonly known as "HAM" mode. -- "Why can't I be you?" - Robert Smith "Why can't he be you?" - Patsy Cline "Why can't you be you?" - `Seven Faces of Eve' Pete Ashdown pashdown@javelin.sim.es.com ...uunet!javelin.sim.es.com!pashdown