1.5 byte processor

286

03:19:37

This is [not] the minimum amount of bits required to determine the color of a pixel (3 hexadecimals). You would need 6.144 kilobytes of space to store every combination of this setup. (color is always determined with 6 hexadecimals which is 3 bytes (24 bits))

published 8 years ago

hurz

8 years ago

Mimimum?

jason9

8 years ago

Minimum of you want a very diverse variety of colors. Normal pixels have each red/green/blue have anywhere from 0 to 255 brightness giving rise to the ability to create any color with very good accuracy. This circuit only gives each of the three prime colors anywhere from 0 to 15 brightness, or 0 to 240 with steps of 16, which makes it so that there is a much more limited number of colors to choose from, but there are still many. It could be done with only 9 bits rather than 12, but that would reduce the number of colors even further with each prime color having a value from 0 to 7. 6 bits gives only 4 possible values for each prime color or only 64 different colors to choose from.

BillyT

8 years ago

Dos only had 8 colours + flashing for each colour.

Benb0456

8 years ago

The main reason I didn't include 6 hexadecimals is because I couldn't fit 24 transistors and logic sources in the grid.

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