雷 - 八叉树路口算法 [英] Ray - Octree intersection algorithms

问题描述

我正在寻找一个良好的光线，八叉树交集算法，它带给我的光线穿过一个迭代的方式叶子。我打算在CPU上实现它，因为我不想深入CUDA只是尚未：）

I'm looking for a good ray-octree intersection algorithm, which gives me the leafs the ray passes through in an iterative way. I'm planning on implementing it on the CPU, since I do not want to dive into CUDA just yet :)

目前，我的体素raycaster少了点3D DDA（Amanatides /佑版）上XxYxZ体素的非分层阵列。你可以想像，这是pretty的成本时，有一个很大的空白，这表现在下面的图片（亮红色=更多的工作:)）：

At the moment, my Voxel raycaster just does 3D DDA (Amanatides/Woo version) on a non-hierarchic array of XxYxZ voxels. You can imagine that this is pretty costly when there's a lot of empty space, as demonstrated in the following picture (brighter red = more work :) ):

我已经想通了，有两种类型的完成这个任务的算法：自下而上，它从叶子的作品回来向上，和自上而下 ，这是basicly深度优先搜索。

I've already figured out that there are two kinds of algorithms for this task: bottom-up, which works from the leafs back upwards, and top-down, which is basicly depth-first search.

我已经找到Revelles算法从2000年，所谓的 一个高效参数算法八叉树遍历 的，这看起来很有趣，但很老。这是一个顶向下的算法。

I've already found Revelles' algorithm from 2000, called An efficient parametric algorithm for octree traversal, which looks interesting, but is quite old. This is a top-down algorithm.

最流行的自下而上的方法似乎是的 K。宋，A DDA八叉树遍历算法进行光线追踪，Eurographics'91，北荷兰 - 爱思唯尔，书号0444 89096 3，P。 73-85 的问题是，大多数DDA八叉树的遍历算法，预计八叉树是相等的深度，这是我不希望 - 。空子树应该只是一个空指针或类似的东西。

The most popular bottom-up approach seems to be K. Sung, A DDA Octree Traversal Algorithm for Ray Tracing, Eurographics'91, North Holland-Elsevier, ISBN 0444 89096 3, p. 73-85. Problem is that most DDA Octree traversal algorithms expect the octree to be of equal depth, which I do not want - empty subtrees should just be a null pointer or something similar.

在最近的文献中有关疏体素八叉树我已经成功地阅读，（最显着的莱恩对SVO的的工作，他们似乎都基于某种DDA（Amanatides /胡的风格）。

In the more recent literature about Sparse Voxel Octrees I've managed to read through, (most notably Laine's work on SVO's, they all seem to be based on some kind of GPU-implemented version of DDA (Amanatides/Woo style).

现在，这里是我的问题：没有任何人有实施基本的经验，没有多余的装饰线，八叉树交集算法？你有什么建议？

Now, here's my question: does anybody have any experience implementing a basic, no-frills ray-octree intersection algorithm? What would you recommend?

推荐答案

有关的记录，这是我实现Revelles论文我最终使用：

For the record, this is my implementation of the Revelles paper I ended up using:

#include "octree_traversal.h"

using namespace std;

unsigned char a; // because an unsigned char is 8 bits

int first_node(double tx0, double ty0, double tz0, double txm, double tym, double tzm){
unsigned char answer = 0;   // initialize to 00000000
// select the entry plane and set bits
if(tx0 > ty0){
    if(tx0 > tz0){ // PLANE YZ
        if(tym < tx0) answer|=2;    // set bit at position 1
        if(tzm < tx0) answer|=1;    // set bit at position 0
        return (int) answer;
    }
}
else {
    if(ty0 > tz0){ // PLANE XZ
        if(txm < ty0) answer|=4;    // set bit at position 2
        if(tzm < ty0) answer|=1;    // set bit at position 0
        return (int) answer;
    }
}
// PLANE XY
if(txm < tz0) answer|=4;    // set bit at position 2
if(tym < tz0) answer|=2;    // set bit at position 1
return (int) answer;
}

int new_node(double txm, int x, double tym, int y, double tzm, int z){
if(txm < tym){
    if(txm < tzm){return x;}  // YZ plane
}
else{
    if(tym < tzm){return y;} // XZ plane
}
return z; // XY plane;
}

void proc_subtree (double tx0, double ty0, double tz0, double tx1, double ty1, double tz1, Node* node){
float txm, tym, tzm;
int currNode;

if(tx1 < 0 || ty1 < 0 || tz1 < 0) return;
if(node->terminal){
    cout << "Reached leaf node " << node->debug_ID << endl;
    return;
}
else{ cout << "Reached node " << node->debug_ID << endl;}

txm = 0.5*(tx0 + tx1);
tym = 0.5*(ty0 + ty1);
tzm = 0.5*(tz0 + tz1);

currNode = first_node(tx0,ty0,tz0,txm,tym,tzm);
do{
    switch (currNode)
    {
    case 0: { 
        proc_subtree(tx0,ty0,tz0,txm,tym,tzm,node->children[a]);
        currNode = new_node(txm,4,tym,2,tzm,1);
        break;}
    case 1: { 
        proc_subtree(tx0,ty0,tzm,txm,tym,tz1,node->children[1^a]);
        currNode = new_node(txm,5,tym,3,tz1,8);
        break;}
    case 2: { 
        proc_subtree(tx0,tym,tz0,txm,ty1,tzm,node->children[2^a]);
        currNode = new_node(txm,6,ty1,8,tzm,3);
        break;}
    case 3: { 
        proc_subtree(tx0,tym,tzm,txm,ty1,tz1,node->children[3^a]);
        currNode = new_node(txm,7,ty1,8,tz1,8);
        break;}
    case 4: { 
        proc_subtree(txm,ty0,tz0,tx1,tym,tzm,node->children[4^a]);
        currNode = new_node(tx1,8,tym,6,tzm,5);
        break;}
    case 5: { 
        proc_subtree(txm,ty0,tzm,tx1,tym,tz1,node->children[5^a]);
        currNode = new_node(tx1,8,tym,7,tz1,8);
        break;}
    case 6: { 
        proc_subtree(txm,tym,tz0,tx1,ty1,tzm,node->children[6^a]);
        currNode = new_node(tx1,8,ty1,8,tzm,7);
        break;}
    case 7: { 
        proc_subtree(txm,tym,tzm,tx1,ty1,tz1,node->children[7^a]);
        currNode = 8;
        break;}
    }
} while (currNode<8);
}

void ray_octree_traversal(Octree* octree, Ray ray){
a = 0;

// fixes for rays with negative direction
if(ray.direction[0] < 0){
    ray.origin[0] = octree->size[0] - ray.origin[0];
    ray.direction[0] = - ray.direction[0];
    a |= 4 ; //bitwise OR (latest bits are XYZ)
}
if(ray.direction[1] < 0){
    ray.origin[1] = octree->size[1] - ray.origin[1];
    ray.direction[1] = - ray.direction[1];
    a |= 2 ; 
}
if(ray.direction[2] < 0){
    ray.origin[2] = octree->size[2] - ray.origin[2];
    ray.direction[2] = - ray.direction[2];
    a |= 1 ; 
}

double divx = 1 / ray.direction[0]; // IEEE stability fix
double divy = 1 / ray.direction[1];
double divz = 1 / ray.direction[2];

double tx0 = (octree->min[0] - ray.origin[0]) * divx;
double tx1 = (octree->max[0] - ray.origin[0]) * divx;
double ty0 = (octree->min[1] - ray.origin[1]) * divy;
double ty1 = (octree->max[1] - ray.origin[1]) * divy;
double tz0 = (octree->min[2] - ray.origin[2]) * divz;
double tz1 = (octree->max[2] - ray.origin[2]) * divz;

if( max(max(tx0,ty0),tz0) < min(min(tx1,ty1),tz1) ){
    proc_subtree(tx0,ty0,tz0,tx1,ty1,tz1,octree->root);
}
}

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