在GLSL中优化光线追踪着色器

我已经编码了一个基于体素化的raytracer，它可以按预期工作，但是速度很慢。

当前，raytracer代码如下：

#version 430 
//normalized positon from (-1, -1) to (1, 1)
in vec2 f_coord;

out vec4 fragment_color;

struct Voxel
{
    vec4 position;
    vec4 normal;
    vec4 color;
};

struct Node
{
    //children of the current node
    int children[8];
};

layout(std430, binding = 0) buffer voxel_buffer
{
    //last layer of the tree, the leafs
    Voxel voxels[];
};
layout(std430, binding = 1) buffer buffer_index
{
    uint index;
};
layout(std430, binding = 2) buffer tree_buffer
{
    //tree structure       
    Node tree[];
};
layout(std430, binding = 3) buffer tree_index
{
    uint t_index;
};

uniform vec3 camera_pos; //position of the camera
uniform float aspect_ratio; // aspect ratio of the window
uniform float cube_dim; //Dimenions of the voxelization cube
uniform int voxel_resolution; //Side length of the cube in voxels

#define EPSILON 0.01
// Detect whether a position is inside of the voxel with size size located at corner
bool inBoxBounds(vec3 corner, float size, vec3 position)
{
    bool inside = true;
    position-=corner;//coordinate of the position relative to the box coordinate system
    //Test that all coordinates are inside the box, if any is outisde, the point is out the box
    for(int i=0; i<3; i++)
    {
        inside = inside && (position[i] > -EPSILON);
        inside = inside && (position[i] < size+EPSILON);
    }

    return inside;
}

//Get the distance to a box or infinity if the box cannot be hit
float boxIntersection(vec3 origin, vec3 dir, vec3 corner0, float size)
{
    dir = normalize(dir);
    vec3 corner1 = corner0 + vec3(size,size,size);//Oposite corner of the box

    float coeffs[6];
    //Calculate the intersaction coefficients with te 6 bonding planes 
    coeffs[0] = (corner0.x - origin.x)/(dir.x);
    coeffs[1] = (corner0.y - origin.y)/(dir.y);
    coeffs[2] = (corner0.z - origin.z)/(dir.z);

    coeffs[3] = (corner1.x - origin.x)/(dir.x);
    coeffs[4] = (corner1.y - origin.y)/(dir.y);
    coeffs[5] = (corner1.z - origin.z)/(dir.z);
    //by default the distance to the box is infinity
    float t = 1.f/0.f;

    for(uint i=0; i<6; i++){
        //if the distance to a boxis negative, we set it to infinity as we cannot travel in the negative direction
        coeffs[i] = coeffs[i] < 0 ? 1.f/0.f : coeffs[i];
        //The distance is the minumum of the previous calculated distance and the current distance
        t = inBoxBounds(corner0,size,origin+dir*coeffs[i]) ? min(coeffs[i],t) : t;
    }

    return t;
}

#define MAX_TREE_HEIGHT 11
int nodes[MAX_TREE_HEIGHT];
int levels[MAX_TREE_HEIGHT];
vec3 positions[MAX_TREE_HEIGHT];
int sp=0;

void push(int node, int level, vec3 corner)
{
    nodes[sp] = node;
    levels[sp] = level;
    positions[sp] = corner;
    sp++;
}

void main()
{   
    int count = 0; //count the iterations of the algorithm
    vec3 r = vec3(f_coord.x, f_coord.y, 1.f/tan(radians(40))); //direction of the ray
    r.y/=aspect_ratio; //modify the direction based on the windows aspect ratio
    vec3 dir = r;
    r += vec3(0,0,-1.f/tan(radians(40))) + camera_pos; //put the ray at the camera position

    fragment_color = vec4(0);
    int max_level = int(log2(voxel_resolution));//height of the tree
    push(0,0,vec3(-cube_dim));//set the stack
    float tc = 1.f; //initial color value, to be decreased whenever a voxel is hit
    //tree variables
    int level=0;
    int node=0;
    vec3 corner;

    do
    {
        //pop from stack
        sp--;
        node = nodes[sp];
        level = levels[sp];
        corner = positions[sp];

        //set the size of the current voxel 
        float size = cube_dim / pow(2,level);
        //set the corners of the children
        vec3 corners[] =
            {corner,                        corner+vec3(0,0,size),
            corner+vec3(0, size,0),         corner+vec3(0,size,size),
            corner+vec3(size,0,0),          corner+vec3(size,0,size),
            corner+vec3(size,size,0),       corner+vec3(size,size,size)};

        float coeffs[8];
        for(int child=0; child<8; child++)
        {
            //Test non zero childs, zero childs are empty and thus should be discarded
            coeffs[child] = tree[node].children[child]>0?
                //Get the distance to your child if it's not empty or infinity if it's empty
                boxIntersection(r, dir, corners[child], size) : 1.f/0.f;
        }
        int indices[8] = {0,1,2,3,4,5,6,7};
        //sort the children from closest to farthest
        for(uint i=0; i<8; i++)
        {
            for(uint j=i; j<8; j++)
            {
                if((coeffs[j] < coeffs[i]))
                {
                    float swap = coeffs[i];
                    coeffs[i] = coeffs[j];
                    coeffs[j] = swap;

                    int iSwap = indices[i];
                    indices[i] = indices[j];
                    indices[j] = iSwap;

                    vec3 vSwap = corners[i];
                    corners[i] = corners[j];
                    corners[j] = vSwap;
                }
            }
        }
        //push to stack
        for(uint i=7; i>=0; i--)
        {
            if(!isinf(coeffs[i]))
            {
                push(tree[node].children[indices[i]],
                    level+1, corners[i]);
            }
        }
        count++;
    }while(level < (max_level-1) && sp>0);
    //set color
    fragment_color = vec4(count)/100;
}

由于可能尚不清楚，请允许我解释一下。

我们检查以大立方体开头的射线盒交点。如果我们击中它，我们将测试组成它的8个立方体的相交。

如果我们找到任何一个，我们将检查组成该立方体的8个立方体的交集。

在2D模式下，外观如下：

在这种情况下，我们有4层，首先检查大框，然后检查红色框，然后检查绿色框，最后检查蓝色框。

打印出以颜色执行的光线跟踪步骤的次数（这是我提供的代码段的作用）

结果如下图：

如您所见，着色器在大多数情况下不会执行超过100次迭代。

但是，此着色器平均需要20万微秒才能在gtx 1070中执行。

由于问题不是执行次数，因此我的问题很可能是线程执行。

有谁知道我如何优化此代码？最大的瓶颈似乎是使用堆栈。

如果我在不压入堆栈的情况下运行相同的代码（生成错误的输出），则运行时性能将提高10倍