OctTree can check if a box collides with a box in the OctTree.

This commit is contained in:
William Moberg
2015-12-08 17:12:58 +01:00
parent 32d2aa6a93
commit a7fb77d998
2 changed files with 65 additions and 43 deletions
+58 -40
View File
@@ -48,7 +48,7 @@ OctTree::OctTree(const AABB& octTreeBounds, int subDivisions)
minPos.x = parentMin.x;
maxPos.x = parentCenter.x;
}
//If child is 2,3,6,7
if (bits.test(1)) {
minPos.y = parentCenter.y;
@@ -80,14 +80,25 @@ OctTree::~OctTree()
}
}
bool OctTree::RayCollides(const Ray& ray, Output& data) const
bool OctTree::BoxCollides(const AABB& boxToTest, AABB& outBoxIntersected) const
{
data.CollideDistance = -1;
return rayCollides(ray, data);
if (hasChildren()) {
for (int i : childIndicesContainingBox(boxToTest)) {
if (m_Children[i]->BoxCollides(boxToTest, outBoxIntersected))
return true;
}
} else {
for (const auto& objBox : m_ContainingBoxes) {
if (Collision::AABBVsAABB(boxToTest, objBox)) {
outBoxIntersected = objBox;
return true;
}
}
}
return false;
}
//Currently all Nodes must have exactly 0 or 8 children, and objectdata should only exist in the last bottom nodes.
bool OctTree::rayCollides(const Ray& ray, Output& data) const
bool OctTree::RayCollides(const Ray& ray, Output& data) const
{
//If the node AABB is missed, everything it contains is missed.
if (Collision::RayAABBIntr(ray, m_Box)) {
@@ -102,7 +113,7 @@ bool OctTree::rayCollides(const Ray& ray, Output& data) const
std::sort(childInfos.begin(), childInfos.end(), isFirstLower);
//Loop through the children, starting with the one closest to the ray origin. I.e the first to be hit.
for (const ChildInfo& info : childInfos) {
if (m_Children[info.Index]->rayCollides(ray, data)) {
if (m_Children[info.Index]->RayCollides(ray, data)) {
return true;
}
}
@@ -127,39 +138,8 @@ bool OctTree::rayCollides(const Ray& ray, Output& data) const
void OctTree::AddBox(const AABB& box)
{
if (hasChildren()) {
int minInd = childIndexContainingPoint(box.MinCorner());
int maxInd = childIndexContainingPoint(box.MaxCorner());
//Because of the predictable ordering of the child indices,
//the number of bits set when xor:ing the indices will determine the number of children containing the box.
std::bitset<3> bits(minInd ^ maxInd);
switch (bits.count()) {
case 0: //Box contained completely in one child.
m_Children[minInd]->AddBox(box);
break;
case 1: //Two children.
m_Children[minInd]->AddBox(box);
m_Children[maxInd]->AddBox(box);
break;
case 2: //Four children.
//Bit-hax to calculate the right 4 cildren containing the box.
//This works because of the childrens index determine what part of
//the dimensions they are responsible for (which octant).
bits.flip();
//At this point the bits necessarily have exactly one bit set.
for (int c = 0; c < 8; ++c) {
//If the child index have the same bit set as the bits, add box to it.
if (bits.to_ulong() & c) {
m_Children[c]->AddBox(box);
}
}
break;
case 3: //Eight children.
for (OctTree*& c : m_Children) {
c->AddBox(box);
}
break;
default:
break;
for (auto i : childIndicesContainingBox(box)) {
m_Children[i]->AddBox(box);
}
} else {
m_ContainingBoxes.push_back(box);
@@ -196,6 +176,44 @@ int OctTree::childIndexContainingPoint(const glm::vec3& point) const
return (1 << 2) * (point.x >= c.x) | (1 << 1) * (point.y >= c.y) | (point.z >= c.z);
}
std::vector<int> OctTree::childIndicesContainingBox(const AABB& box) const
{
int minInd = childIndexContainingPoint(box.MinCorner());
int maxInd = childIndexContainingPoint(box.MaxCorner());
//Because of the predictable ordering of the child indices,
//the number of bits set when xor:ing the indices will determine the number of children containing the box.
std::bitset<3> bits(minInd ^ maxInd);
switch (bits.count()) {
//Box contained completely in one child.
case 0:
return{ minInd };
//Two children.
case 1:
return{ minInd, maxInd };
//Four children.
case 2:
{
std::vector<int> ret;
//Bit-hax to calculate the right 4 cildren containing the box.
//This works because of the childrens index determine what part of
//the dimensions they are responsible for (which octant).
bits.flip();
//At this point the bits necessarily have exactly one bit set.
for (int c = 0; c < 8; ++c) {
//If the child index have the same bit set as the bits, add box to it.
if (bits.to_ulong() & c) {
ret.push_back(c);
}
}
return ret;
}
case 3: //Eight children.
return{ 0,1,2,3,4,5,6,7 };
default:
return std::vector<int>();
}
}
inline bool OctTree::hasChildren() const
{
return m_Children[0] != nullptr;