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algorithm - Ray - 八叉树相交算法

转载 作者:塔克拉玛干 更新时间:2023-11-03 02:17:34 28 4
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我正在寻找一个好的光线八叉树相交算法,它以迭代的方式为我提供光线穿过的叶子。我计划在 CPU 上实现它,因为我还不想深入研究 CUDA :)

目前,我的 Voxel raycaster 仅在 XxYxZ 体素的非分层阵列上执行 3D DDA(Amanatides/Woo 版本)。您可以想象,当有很多空白空间时,这是相当昂贵的,如下图所示(更亮的红色 = 更多的工作 :)):

Workload for dumb 3D DDA - red = more work

我已经知道有两种算法可以完成这项任务:自下而上,它从叶子向后向上工作,以及自上而下 ,这基本上是深度优先搜索。

我已经从 2000 年找到了 Revelles 的算法,称为 An efficient parametric algorithm for octree traversal ,看起来很有趣,但是很老了。这是一个自上而下的算法。

最流行的自下而上方法似乎是K。 Sung,用于光线追踪的 DDA 八叉树遍历算法,Eurographics'91,North Holland-Elsevier,ISBN 0444 89096 3,p. 73-85. 问题是大多数 DDA 八叉树遍历算法都期望八叉树具有相同的深度,我不希望这样 - 空子树应该只是一个空指针或类似的东西。

在我设法通读的关于稀疏体素八叉树的最新文献中,(最著名的是 Laine's work on SVO's ,它们似乎都基于某种 GPU 实现的 DDA 版本(Amanatides/Woo 风格)。

现在,这是我的问题:有人有任何实现基本的、简洁的射线八叉树相交算法的经验吗?你会推荐什么?

最佳答案

郑重声明,这是我最终使用的 Revelles 论文的实现:

#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->center[0] * 2 - ray.origin[0];//camera origin fix
ray.direction[0] = - ray.direction[0];
a |= 4 ; //bitwise OR (latest bits are XYZ)
}
if(ray.direction[1] < 0){
ray.origin[1] = octree->center[1] * 2 - ray.origin[1];
ray.direction[1] = - ray.direction[1];
a |= 2 ;
}
if(ray.direction[2] < 0){
ray.origin[2] = octree->center[2] * 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);
}
}

关于algorithm - Ray - 八叉树相交算法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10228690/

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