用于“高对比度"的快速公式.曲线 [英] Fast formula for a "high contrast" curve

问题描述

我的内循环包含一个计算，分析表明存在问题.

My inner loop contains a calculation that profiling shows to be problematic.

该想法是取灰度像素x(0 ＜＝ x ＜1)，并增加其对比度".我的要求很宽松，只有以下几点:

The idea is to take a greyscale pixel x (0 <= x <= 1), and "increase its contrast". My requirements are fairly loose, just the following:

• 对于x< .5，0< = f(x)< x
• 对于x> .5，x
• f(0)= 0
• f(x)= 1-f(1-x)，即它应该是对称的"
• 最好，函数应该平滑.

因此图必须看起来像这样:

So the graph must look something like this:

.

我有两种实现方式(它们的结果不同，但都一致):

I have two implementations (their results differ but both are conformant):

float cosContrastize(float i) {
    return .5 - cos(x * pi) / 2;
}

float mulContrastize(float i) {
    if (i < .5) return i * i * 2;
    i = 1 - i;
    return 1 - i * i * 2;
}

因此，我要求对这些实现之一进行微优化，或者要求使用自己的原始更快的公式.

So I request either a microoptimization for one of these implementations, or an original, faster formula of your own.

也许你们当中的一位甚至可以旋转这些位；)

Maybe one of you can even twiddle the bits ;)

推荐答案

通常，您可以简单地设置阈值，但是我认为这太愚蠢了:

Trivially you could simply threshold, but I imagine this is too dumb:

return i < 0.5 ? 0.0 : 1.0;

由于您提到对比度增加"，因此我假设输入值为亮度值.如果是这样，并且它们是离散的(也许是8位值)，则可以使用查找表来快速完成此操作.

Since you mention 'increasing contrast' I assume the input values are luminance values. If so, and they are discrete (perhaps it's an 8-bit value), you could use a lookup table to do this quite quickly.

您的"mulContrastize"看起来相当快.一种优化是使用整数数学.再次说，您的输入值实际上可以作为[0..255]中的8位无符号值传递. (再次，可能是一个很好的假设?)您可以做类似...的事情.

Your 'mulContrastize' looks reasonably quick. One optimization would be to use integer math. Let's say, again, your input values could actually be passed as an 8-bit unsigned value in [0..255]. (Again, possibly a fine assumption?) You could do something roughly like...

int mulContrastize(int i) {
  if (i < 128) return (i * i) >> 7; 
  // The shift is really: * 2 / 256
  i = 255 - i;
  return 255 - ((i * i) >> 7);

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