Question

我遇到了这款具有抗锯齿功能的FxAA着色器，似乎运行良好。但是，不知何故无法理解逻辑。谁能解释一下？

[[FX]]

// Samplers
sampler2D buf0 = sampler_state {
    Address = Clamp;
    Filter = None;
};

context FXAA {
    VertexShader = compile GLSL VS_FSQUAD;
    PixelShader = compile GLSL FS_FXAA;
}



[[VS_FSQUAD]]

uniform mat4 projMat;
attribute vec3 vertPos;
varying vec2 texCoords;

void main(void) {
    texCoords = vertPos.xy; 
    gl_Position = projMat * vec4( vertPos, 1 );
}


[[FS_FXAA]]

uniform sampler2D buf0;
uniform vec2 frameBufSize;
varying vec2 texCoords;

void main( void ) {
    //gl_FragColor.xyz = texture2D(buf0,texCoords).xyz;
    //return;

    float FXAA_SPAN_MAX = 8.0;
    float FXAA_REDUCE_MUL = 1.0/8.0;
    float FXAA_REDUCE_MIN = 1.0/128.0;

    vec3 rgbNW=texture2D(buf0,texCoords+(vec2(-1.0,-1.0)/frameBufSize)).xyz;
    vec3 rgbNE=texture2D(buf0,texCoords+(vec2(1.0,-1.0)/frameBufSize)).xyz;
    vec3 rgbSW=texture2D(buf0,texCoords+(vec2(-1.0,1.0)/frameBufSize)).xyz;
    vec3 rgbSE=texture2D(buf0,texCoords+(vec2(1.0,1.0)/frameBufSize)).xyz;
    vec3 rgbM=texture2D(buf0,texCoords).xyz;

    vec3 luma=vec3(0.299, 0.587, 0.114);
    float lumaNW = dot(rgbNW, luma);
    float lumaNE = dot(rgbNE, luma);
    float lumaSW = dot(rgbSW, luma);
    float lumaSE = dot(rgbSE, luma);
    float lumaM  = dot(rgbM,  luma);

    float lumaMin = min(lumaM, min(min(lumaNW, lumaNE), min(lumaSW, lumaSE)));
    float lumaMax = max(lumaM, max(max(lumaNW, lumaNE), max(lumaSW, lumaSE)));

    vec2 dir;
    dir.x = -((lumaNW + lumaNE) - (lumaSW + lumaSE));
    dir.y =  ((lumaNW + lumaSW) - (lumaNE + lumaSE));

    float dirReduce = max(
        (lumaNW + lumaNE + lumaSW + lumaSE) * (0.25 * FXAA_REDUCE_MUL),
        FXAA_REDUCE_MIN);

    float rcpDirMin = 1.0/(min(abs(dir.x), abs(dir.y)) + dirReduce);

    dir = min(vec2( FXAA_SPAN_MAX,  FXAA_SPAN_MAX),
          max(vec2(-FXAA_SPAN_MAX, -FXAA_SPAN_MAX),
          dir * rcpDirMin)) / frameBufSize;

    vec3 rgbA = (1.0/2.0) * (
        texture2D(buf0, texCoords.xy + dir * (1.0/3.0 - 0.5)).xyz +
        texture2D(buf0, texCoords.xy + dir * (2.0/3.0 - 0.5)).xyz);
    vec3 rgbB = rgbA * (1.0/2.0) + (1.0/4.0) * (
        texture2D(buf0, texCoords.xy + dir * (0.0/3.0 - 0.5)).xyz +
        texture2D(buf0, texCoords.xy + dir * (3.0/3.0 - 0.5)).xyz);
    float lumaB = dot(rgbB, luma);

    if((lumaB < lumaMin) || (lumaB > lumaMax)){
        gl_FragColor.xyz=rgbA;
    }else{
        gl_FragColor.xyz=rgbB;
    }
}

Answer 1

FxAA是一种对图像执行抗锯齿的过滤算法。与其他AA技术相反，它应用于图像的像素，而不是在绘制图像的基元时。在像游戏这样的3D应用程序中，它作为后期处理步骤应用于渲染场景之上。

基本思想是：寻找垂直和水平边缘。如果在边缘的末端，则在正交方向上模糊。

这是主题的good description和original paper。

这个简单的FxAA如何运作？

1 个答案: