Image Dithering: How would I calculate quantization error and nearest colour to implement in a Floyd-Steinburg algorithm?

I intend to display (4, 8 or 16 bit per channel - no alpha) images on a 1 bit display in an embedded system. Images are stored in RGB tuples. My intention is to use Floyd-Steinburg, as it looks reasonably good, is more than quick enough and concise in code.

In reference to the WikiPedia article, I have two questions.

What would the best practice for expressing nearest colour be? Would the following work? (ignore that I'm returning a structure in c)

typedef rgb16_tag { unsigned short r, g, b } rgb16;

rgb16 nearest_1bit_colour(rgb16 p) {
    double c; rgb16 r;
    c  = ((double)(p.r + p.g + p.b + 3 * (1 << 15))) / ( 3.0 * (1 << 16));
    if (c>= 1.0) { 
       r.r = r.g = r.b = 1;
    } else {
       r.r = r.g = r.b = 0;
    }
    return r;
 }

and, Is the expression of quantization error done on a per channel basis? i.e. does this make sense?

rgb16 q, new, old, image[X开发者_StackOverflow中文版][Y];
int x, y;

... /* (somewhere in the nested loops) */
    old = image[x][y];
    new = nearest_1bit_colour(old);

    /* Repeat the following for each colour channel seperately. */
    q.{r,g,b} = old.{r,g,b} - new.{r,g,b};

    image[x+1][y].{r,g,b}   = image[x+1][y].{r,g,b} + 7/16 *   q.{r,g,b}
    image[x-1][y+1].{r,g,b} = image[x-1][y+1].{r,g,b} + 3/16 * q.{r,g,b}
    image[x][y+1].{r,g,b}   = image[x][y+1].{r,g,b} + 5/16 *   q.{r,g,b}
    image[x+1][y+1].{r,g,b} = image[x+1][y+1].{r,g,b} + 1/16 * q.{r,g,b}

I've seen two typical approaches to measuring the difference between two colors. The most common way is probably to just find the Euclidian distance between them through the color cube:

float r = i.r - j.r;
float g = i.g - j.g;
float b = i.b - j.b;
float diff = sqrtf( r * r + g + g + b * b );

The other is just to average the absolute differences, possibly weighting for luminance:

float diff = 0.30f * fabs( i.r - j.r ) +
             0.59f * fabs( i.g - j.g ) +
             0.11f * fabs( i.b - j.b );

As to your second question, yes. Accumulate the error separately in each channel.

Edit: Misread at first and missed that this was for a bi-level display. In that case, I'd suggest just using luminance:

float luminance = 0.30f * p.r + 0.59f * p.g + 0.11f * p.b;
if ( luminance > 0.5f * channelMax ) {
     // white
} else {
     // black
}

As you return an rgb16 value in nearest_1bit_colour and use it to compare it with other colors and you have to use white and black as returned colors, use 0 and 0xFFFF instead of 0 and 1 (which is black and a very dark gray). Additionally, I think you should compare c with 0.5 instead of 1.0:

if (c >= 0.5) { 
   r.r = r.g = r.b = 0xFFFF;
} else {
   r.r = r.g = r.b = 0;
}

Also, there might be pitfalls with (un)signedness:

q.{r,g,b} = old.{r,g,b} - new.{r,g,b};

This can get negative, so q shouldn't be of type rgb16 which apparently is unsigned short, but of type short instead.

Of course, the whole code is for 16-bit input data, for 4- or 8-bit input, you have to change it (or just convert 4-bit and 8-bit data to 16-bit data so you can use the same code).

As purely an integer based solution (my processor doesn't have a FPU), I think this might work.

#include <limits.h>
#include <assert.h>

typedef struct rgb16_tag { unsigned short r,g,b; } rgb16;
typedef struct rgb32_tag { unsigned long  r,g,b; } rgb32;

#define LUMINESCENSE_CONSTANT (ULONG_MAX >> (CHAR_BIT * sizeof (unsigned short)))

static const rgb32 luminescence_multiplier = {
    LUMINESCENSE_CONSTANT * 0.30f,
    LUMINESCENSE_CONSTANT * 0.59f,
    LUMINESCENSE_CONSTANT * 0.11f
};

int black_or_white( rgb16 p ) {
    unsigned long luminescence;

    assert((  luminescence_multiplier.r
            + luminescence_multiplier.g
            + luminescence_multiplier.b) < LUMINESCENSE_CONSTANT);

    luminescence =   p.r * luminescence_multiplier.r
                   + p.g * luminescence_multiplier.g
                   + p.b * luminescence_multiplier.b;

    return (luminescence > ULONG_MAX/2);  /* 1 == white; */
}

There is some excellent half-toning techniques HERE