如何在固定颜色显示器中找到最红的颜色？

Question

我正在为一台名为 Amiga 的旧计算机编写 UI（有些人可能还记得这些）。他们的显示模式具有固定数量的颜色 - 2、4、8、16 等，您可以使用 0 到 255 的 RGB 设置其中任何一个。

所以在 4 色显示器上你可以：

颜色 0 R=0，G=0，B=0（黑色）
颜色 1 R=255, G=255, B=255（白色）
颜色 2 R=128，G=128，B=128（中灰色）
颜色 3 R=255，G=0，B=0（红色）

您可以在 'normal' 显示器上拥有的最大颜色数是 255（有一些特殊的显示模式可以提高这个值，但我不需要担心这些）。

我正在尝试编写一些 C 代码来读取显示器的颜色列表并找到最红、最绿和最蓝的颜色，但我无法理解所涉及的比较。

如果我有 3 个变量 - 红色、绿色和蓝色 - 所有包含当前颜色 rgb 值的整数和另外 3 个 - stored_red、stored_stored_green 和 stored_blue.

我如何编写一个函数来记住相对于其他颜色而言最红的颜色编号？

我试过：

If ((red>stored_red) && ((blue <stored_blue) || (green<stored_green)))

但这行不通。我想我需要研究比率，但就是想不通数学。

Answer 1

有不同的方法来检查最大值 "redness"/"greenness" 等，这里第一个简单的方法是我最初使用的答案：

/* for all palette (color table) entries */
if(red > stored_red) {
    max_red_index = current_index;
    stored_red = red;
}
if(blue > stored_blue) {
    max_blue_index = current_index;
    stored_blue = blue;
}

当然绿色也一样。
这只会给出具有最大分量的颜色，但在@Chris Turner 的评论之后我认为这可能不是你想要的。

另一种方法是检查红色与其他颜色的比例（从现在开始我将只使用红色）：

redness = red*2/(blue + green + 1)

这给"redness"一个1到510之间的数字，不考虑亮度。例如 R-G-B 50-10-10 比 255-100-100 (2) 更红 (4)。

许多其他公式也是可能的，例如

redness = red*red/(blue + green + 1)

也会考虑亮度，认为比 50-10-10 红 255-100-100。

要使用这些公式，只需将上面的 red 变量设置为红色公式的结果即可。

Answer 2

您应该计算红色的百分比以找到最红的颜色。

这里是一个使用RGB结构的程序，具有比较两个RGB结构是否为"reddness"的函数。如果两种颜色的红色百分比相同，则 .r 字段中具有最大值的颜色被视为 "redder"。另一个函数采用指向 RGB 结构的指针数组和 returns 指向被视为 "reddest".

的 RGB 三元组的指针

注意在计算百分比时必须注意检查RGB元组是否全为零(0, 0, 0)；这可能会导致尝试除以零。感谢@alain 在我的原始代码中发现了这个错误。

#include <stdio.h>
#include <stdbool.h>

struct RGB {
    int r;
    int g;
    int b;
};

bool is_redder_pct(struct RGB *triple1, struct RGB *triple2);
struct RGB * max_red(struct RGB *triples, size_t num_triples);

int main(void)
{
    struct RGB colors[] = { { .r = 125, .g = 0, .b = 0 },
                            { .r = 150, .g = 255, .b = 0 },
                            { .r = 100, .g = 20, .b = 21 },
                            { .r = 255, .g = 21, .b = 22 },
                            { .r = 0, .g = 0, .b = 0 },
                            { .r = 255, .g = 255, .b = 255 },
                            { .r = 128, .g = 128, .b = 128 },
                            { .r = 255, .g = 0, .b = 0 } };
    size_t num_colors = sizeof colors / sizeof *colors;

    struct RGB *reddest = max_red(colors, num_colors);

    printf("The reddest color is: (%d, %d, %d)\n",
           reddest->r, reddest->g, reddest->b);

    return 0;
}

/* Returns true if triple1 is at least as red as triple2 */
bool is_redder_pct(struct RGB *triple1, struct RGB *triple2)
{
    bool ret_val;
    int triple1_sum = triple1->r + triple1->g + triple1->b;
    int triple2_sum = triple2->r + triple2->g + triple2->b;

    /* if triple1 is black, triple1 is not redder than triple2 */
    if (triple1_sum == 0) {
        ret_val = false;

    /* otherwise, if triple2 is black, triple1 is redder than triple2 */
    } else if (triple2_sum == 0) {
        ret_val = true;

    /* otherwise the percentages are calculated in a comparison */    
    } else {
        ret_val = triple1->r / (triple1_sum * 1.0)
            >= triple2->r / (triple2_sum * 1.0);
    }

    return ret_val;
}

/* Returns a pointer to the RGB struct in the array TRIPLES
 * that compares "reddest" */
struct RGB * max_red(struct RGB *triples, size_t num_triples)
{
    struct RGB *max = &triples[0];

    for (size_t i = 1; i < num_triples; i++) {
        struct RGB *curr = &triples[i];
        if (is_redder_pct(curr, max) && curr->r > max->r) {
            max = curr;
        }
    }

    return max;
}

程序输出：

The reddest color is: (255, 0, 0)

Answer 3

下面是基于CompuPhase - Colour metric的一个"a low-cost approximation"色差，但是避开了平方根函数（从而测出色差的平方，方便比较），并对结果进行缩放以避免整数除法导致的精度损失。它按照从 "most red" 到 "least red".

的顺序对颜色元组数组进行排序

比较中灰色比黑色更红，黑色比白色更红，这看起来相当随意！

#include <stdio.h>
#include <stdlib.h>

/*
 * Sorts an array of RGB color values (each component from 0 to 255)
 * in descending order of redness.
 *
 * Uses a low-cost approximation of color distance from
 * <https://www.compuphase.com/cmetric.htm>.  It uses a weighted Euclidean
 * distance function to measure the distance between colors, where the weight
 * factors depend on the mean level of "red":
 *
 * rmean = (c1.r + c2.r) / 2
 * dr = c1.r - c2.r
 * dg = c1.g - c2.g
 * db = c1.b - c2.b
 *
 * dc = sqrt(((2 + (rmean / 256)) * dr * dr) + (4 * dg * dg) +
 *           ((2 + ((255 - rmean) / 256)) * db * db))
 *
 * Uses a modified version of the above which returns the square of the
 * distance between two colors scaled by a silly amount to avoid loss of
 * precision caused by integer division.
 */

struct rgb {
    unsigned char r;    /* red range 0..255 */
    unsigned char g;    /* green range 0..255 */
    unsigned char b;    /* blue range 0..255 */
};

/* distance squared between two colors, scaled by some silly amount. */
long color_dist_squared(const struct rgb *c1, const struct rgb *c2)
{
    long rsum = c1->r + c2->r;
    long dr = (long)c1->r - (long)c2->r;
    long dg = (long)c1->g - (long)c2->g;
    long db = (long)c1->b - (long)c2->b;

    return (((1024 + rsum) * dr * dr) + (2048 * dg * dg) +
            ((1534 - rsum) * db * db));
}

/* distance squared from pure red, scaled by some silly amount. */
long antiredness_squared(const struct rgb *c)
{
    const struct rgb pure_red = { .r = 255, .g = 0, .b = 0 };

    return color_dist_squared(&pure_red, c);
}

/*
 * qsort() comparison function.
 * a and b point to struct rgb values.
 * Returns 1 if *a is more anti-red (less red) than *b.
 * Returns 0 if *a and *b are equally (anti-)red.
 * Returns -1 if *a is less anti-red (more red) than *b.
 */
int compar_antiredness(const void *a, const void *b)
{
    const struct rgb *ca = (const struct rgb *)a;
    const struct rgb *cb = (const struct rgb *)b;
    long ara = antiredness_squared(ca);
    long arb = antiredness_squared(cb);
    long diff = ara - arb;

    return (diff > 0) - (diff < 0);
}

int main(void)
{
    struct rgb colors[] = {
        { .r = 125, .g = 0, .b = 0 },
        { .r = 150, .g = 255, .b = 0 },
        { .r = 100, .g = 20, .b = 21 },
        { .r = 255, .g = 21, .b = 22 },
        { .r = 0, .g = 0, .b = 0 },
        { .r = 255, .g = 255, .b = 255 },
        { .r = 128, .g = 128, .b = 128 },
        { .r = 255, .g = 0, .b = 0 },
    };
    size_t num_colors = sizeof(colors) / sizeof(colors[0]);
    size_t i;

    printf("Unsorted colors:\n");
    for (i = 0; i < num_colors; i++) {
        printf("[%zu] R=%u, G=%u, B=%u\n", i, (unsigned)colors[i].r,
               (unsigned)colors[i].g, (unsigned)colors[i].b);
    }
    printf("\n");
    qsort(colors, num_colors, sizeof(colors[0]), compar_antiredness);
    printf("Colors sorted from most red to least red:\n");
    for (i = 0; i < num_colors; i++) {
        printf("[%zu] R=%u, G=%u, B=%u\n", i, (unsigned)colors[i].r,
               (unsigned)colors[i].g, (unsigned)colors[i].b);
    }
    return 0;
}

上面的输出：

Unsorted colors:
[0] R=125, G=0, B=0
[1] R=150, G=255, B=0
[2] R=100, G=20, B=21
[3] R=255, G=21, B=22
[4] R=0, G=0, B=0
[5] R=255, G=255, B=255
[6] R=128, G=128, B=128
[7] R=255, G=0, B=0

Colors sorted from most red to least red:
[0] R=255, G=0, B=0
[1] R=255, G=21, B=22
[2] R=125, G=0, B=0
[3] R=100, G=20, B=21
[4] R=128, G=128, B=128
[5] R=0, G=0, B=0
[6] R=150, G=255, B=0
[7] R=255, G=255, B=255

编辑：当然，使用以下函数从最绿到最不绿，或从最蓝到最不蓝排序同样容易：

/* distance squared from pure green, scaled by some silly amount. */
long antigreenness_squared(const struct rgb *c)
{
    const struct rgb pure_green = { .r = 0, .g = 255, .b = 0 };

    return color_dist_squared(&pure_green, c);
}

/*
 * qsort() comparison function.
 * a and b point to struct rgb values.
 * Returns 1 if *a is more anti-green (less green) than *b.
 * Returns 0 if *a and *b are equally (anti-)green.
 * Returns -1 if *a is less anti-green (more green) than *b.
 */
int compar_antigreenness(const void *a, const void *b)
{
    const struct rgb *ca = (const struct rgb *)a;
    const struct rgb *cb = (const struct rgb *)b;
    long aga = antigreenness_squared(ca);
    long agb = antigreenness_squared(cb);
    long diff = aga - agb;

    return (diff > 0) - (diff < 0);
}

/* distance squared from pure blue, scaled by some silly amount. */
long antiblueness_squared(const struct rgb *c)
{
    const struct rgb pure_blue = { .r = 0, .g = 0, .b = 255 };

    return color_dist_squared(&pure_blue, c);
}

/*
 * qsort() comparison function.
 * a and b point to struct rgb values.
 * Returns 1 if *a is more anti-blue (less blue) than *b.
 * Returns 0 if *a and *b are equally (anti-)blue.
 * Returns -1 if *a is less anti-blue (more blue) than *b.
 */
int compar_antiblueness(const void *a, const void *b)
{
    const struct rgb *ca = (const struct rgb *)a;
    const struct rgb *cb = (const struct rgb *)b;
    long aba = antiblueness_squared(ca);
    long abb = antiblueness_squared(cb);
    long diff = aba - abb;

    return (diff > 0) - (diff < 0);
}

如何在固定颜色显示器中找到最红的颜色？

How can you find reddest colour in a fixed colour display?

c

colors

amiga