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c++ - GJK 算法陷入不同 Voronoi 区域案例的循环中

转载 作者:塔克拉玛干 更新时间:2023-11-03 05:59:27 26 4
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我目前正在尝试实现 https://mollyrocket.com/849 中介绍的简化 GJK 算法进入我的 C++ 游戏。

但是,我在第二维和第三维遇到了奇怪的行为:算法有时(每秒调用多次时很常见)会卡在案例循环中。例如,调试消息一遍又一遍地将以下内容打印到 std::cout:

3ACxAB
4AB
3ABxAC
4ABD
4AD

如果您查看我的代码,您会发现这些行代表算法允许的情况。例如。 3ACxAB 表示单纯形当前是一个三角形,并且原点位于面的 voronoi 区域,在叉积 AC x AB 的方向(可以解释为如三角形“上方”或“下方”)。 4AB 表示单纯形是一个四面体,原点在边 AB 的 voronoi 区域。

A 始终是新添加的点。在代码中,A 始终是 simplex 的最大索引。 (^simplex[1]` 如果它是一条线,如果是三角形则为 2,如果是 a) 四面体则为 3。

即使经过几天的错误搜索(我找到了一些,但仍然存在一个或多个错误),该算法仍无法正常工作。

你看到代码有什么问题了吗?因为我和我的两个 friend 都没有。

PS:我没有从 Casey 的视频中复制任何计算(例如方向 vector 的叉积)。看完之后,我自己下定决心,所以这里可能存在潜在的问题,尤其是在三维方面,凯西故意没有说。


我的支持功能:

//hullA/B: convex hull of A resp. B; baseA/B: location of A/B
Vector3f gjkSupport(Vector3f direction,
std::vector<GLfloat> hullA, std::vector<GLfloat> baseA,
std::vector<GLfloat> hullB, std::vector<GLfloat> baseB) {
//Initialize
GLfloat maxDotP = -std::numeric_limits<GLfloat>::max();
Vector3f furthestPointA, furthestPointB;
//Get furthest point in given direction out of hullA by getting the maximum dot
//product of the direction vector and a hull vertex's position vector
for (GLuint i = 0; i < hullA.size(); i += 3) {
Vector3f current (hullA[i]+baseA[0], hullA[i+1]+baseA[1], hullA[i+2]+baseA[2]);
// * = dot product
GLfloat dotP = direction * current;
if (dotP > maxDotP) {
maxDotP = dotP;
furthestPointA = current;
}
}
maxDotP = -std::numeric_limits<GLfloat>::max();
//Get furthest point in negative of the given direction out of hullB
for (GLuint i = 0; i < hullB.size(); i += 3) {
Vector3f current (hullB[i]+baseB[0], hullB[i+1]+baseB[1], hullB[i+2]+baseB[2]);
GLfloat dotP = -direction * current;
if (dotP > maxDotP) {
maxDotP = dotP;
furthestPointB = current;
}
}
//Furthest Minkowski Difference point is difference of d*A[i]-(-d)*B[j]
return furthestPointA - furthestPointB;
}

我的单纯形函数:

bool gjkSimplex(std::vector<Vector3f> &simplex, Vector3f &direction) {
GLuint simplexSize = simplex.size();
std::cout << simplexSize;
switch (simplexSize) {
//If the simplex is a line segment
case 2:
//Point is closest feature
if ((simplex[0]-simplex[1])*-simplex[1] < 0) {
std::cout << "A";
simplex = {simplex[1]};
//direction = A0
direction = -simplex[1];
//Line is closest feature
} else {
std::cout << "AB";
//direction = AB x (A0 x AB)
// ^ = cross product
direction = (simplex[0]-simplex[1]) ^ ((-simplex[1]) ^ (simplex[0]-simplex[1]));
}
break;
//If the simplex is a triangle
case 3:
//Point is closest feature
if ((simplex[0]-simplex[2])*(-simplex[2]) < 0 && (simplex[1]-simplex[2])*(-simplex[2]) < 0) {
std::cout << "A";
//direction = A0
direction = -simplex[2];
simplex = {simplex[1]};
//Line to second-latest point is closest feature
} else if ((((simplex[0]-simplex[2])^(simplex[1]-simplex[2]))^(simplex[1]-simplex[2]))*-simplex[2] > 0) {
std::cout << "AB";
//direction = AB x (A0 x AB)
direction = (simplex[1]-simplex[2]) ^ ((-simplex[2]) ^ (simplex[1]-simplex[2]));
simplex = {simplex[1], simplex[2]};
//Line to oldest point is closest feature
} else if (((simplex[0]-simplex[2])^((simplex[0]-simplex[2])^(simplex[1]-simplex[2])))*-simplex[2] > 0) {
std::cout << "AC";
//direction = AC x (A0 x AC)
direction = (simplex[0]-simplex[2]) ^ ((-simplex[2]) ^ (simplex[0]-simplex[2]));
simplex = {simplex[0], simplex[2]};
//Face is closest feature
} else {
//Origin is in direction AC x AB
if (((simplex[1]-simplex[2]) ^ (simplex[0]-simplex[2])) * (-simplex[2]) < 0) {
std::cout << "ACxAB";
//direction = AC x AB
direction = (simplex[0]-simplex[2]) ^ (simplex[1]-simplex[2]);
//origin is in direction AB x AC (other side of the face)
} else {
std::cout << "ABxAC";
//direction = AB x AC
direction = (simplex[1]-simplex[2]) ^ (simplex[0]-simplex[2]);
simplex = {simplex[1], simplex[0], simplex[2]};
}
}
break;
//If the simplex is a tetrahedron
case 4:
//Newest point is closest feature
if ((simplex[0]-simplex[3])*(-simplex[3]) < 0 && (simplex[1]-simplex[3])*(-simplex[3]) < 0 &&
(simplex[2]-simplex[3])*(-simplex[3]) < 0) {
std::cout << "A";
//direction = A0
direction = -simplex[3];
simplex = {simplex[3]};
//Edge between newest and second-newest point is closest feature
} else if ((((simplex[2]-simplex[3]) ^ ((simplex[1]-simplex[3]) ^ (simplex[2]-simplex[3]))) * (-simplex[2]) < 0) &&
((((simplex[1]-simplex[3]) ^ (simplex[0]-simplex[3])) ^ (simplex[2]-simplex[3])) * (-simplex[2]) < 0)) {
std::cout << "AB";
//direction = AB x (A0 x AB)
direction = (simplex[2]-simplex[3]) ^ ((-simplex[3]) ^ (simplex[2]-simplex[3]));
simplex = {simplex[2], simplex[3]};
//Edge between newest and third-newest vertex is closest feature
} else if ((((simplex[1]-simplex[3]) ^ ((simplex[0]-simplex[3]) ^ (simplex[1]-simplex[3]))) * (-simplex[2]) < 0) &&
((((simplex[0]-simplex[3]) ^ (simplex[2]-simplex[3])) ^ (simplex[1]-simplex[3])) * (-simplex[2]) < 0)) {
std::cout << "AC";
//direction = AC x (A0 x AC)
direction = (simplex[1]-simplex[3]) ^ ((-simplex[3]) ^ (simplex[1]-simplex[3]));
simplex = {simplex[1], simplex[3]};
//Edge between newest and oldest point is closest feature
} else if ((((simplex[0]-simplex[3]) ^ ((simplex[2]-simplex[3]) ^ (simplex[0]-simplex[3]))) * (-simplex[2]) < 0) &&
((((simplex[2]-simplex[3]) ^ (simplex[1]-simplex[3])) ^ (simplex[0]-simplex[3])) * (-simplex[2]) < 0)) {
std::cout << "AD";
//direction = AD x (A0 x AD)
direction = (simplex[0]-simplex[3]) ^ ((-simplex[3]) ^ (simplex[0]-simplex[3]));
simplex = {simplex[0], simplex[3]};
//Face between the three newest points is closest feature
} else if (((simplex[1]-simplex[3]) ^ (simplex[2]-simplex[3])) * (-simplex[3]) > 0) {
std::cout << "ABC";
//direction = AC x AB (outer normal of face)
direction = (simplex[1]-simplex[3]) ^ (simplex[2]-simplex[3]);
simplex = {simplex[1], simplex[3], simplex[2]};
//Face between newest, second-newest and oldest point is closest feature
} else if (((simplex[2]-simplex[3]) ^ (simplex[0]-simplex[3])) * (-simplex[3]) > 0) {
std::cout << "ABD";
//direction = AB x AD
direction = (simplex[2]-simplex[3]) ^ (simplex[0]-simplex[3]);
simplex = {simplex[0], simplex[2], simplex[3]};
//Face between newest, second-oldest and oldest point is closest feature
} else if (((simplex[0]-simplex[3]) ^ (simplex[1]-simplex[3])) * (-simplex[3]) > 0) {
std::cout << "ACD";
//direction = AD x AC
direction = (simplex[0]-simplex[3]) ^ (simplex[1]-simplex[3]);
simplex = {simplex[0], simplex[3], simplex[1]};
//Origin is encased by simplex
} else {
//Collision detected
std::cout << "ABCD";
return true;
}
break;
default:
direction = {1,1,1};
simplex = {};
break;
}
std::cout << "\n";
return false;
};

GJK 主循环:

//Narrow Phase collision function using GJK
bool SolidObject::collidesWith(SolidObject *object) {
//Initialize by using an arbitrary direction
Vector3f direction (1,1,1);
std::vector<Vector3f> simplex;
Vector3f point = gjkSupport(direction,
this->meshes[0].getConvexHull(), this->base, object->meshes[0].getConvexHull(), object->base);
simplex = {point};
//Set direction to the negative of the resulting point
direction = -point;

bool originInSimplex = false;
while (!originInSimplex) {
//Get furthest point in new direction
point = gjkSupport(direction,
this->meshes[0].getConvexHull(), this->base, object->meshes[0].getConvexHull(), object->base);
//The furthest point in the negative direction is not in the opposing octant
// => no collision
if (point*direction < 0) {
return false;
}
//Add point to the simplex
simplex.push_back(point);
//Update simplex and direction, and return whether the simplex contains the origin
originInSimplex = gjkSimplex(simplex, direction);
}
std::cout << "\n";
return true;
}

最佳答案

在三角形的情况下:

//Point is closest feature
if ((simplex[0]-simplex[2])*(-simplex[2]) < 0 && (simplex[1]-simplex[2])*(-simplex[2]) < 0) {
std::cout << "A";
//direction = A0
direction = -simplex[2];
simplex = {simplex[1]};
}

应该是

simplex = {simplex[2]};

在四面体的情况下:

所有边检查都使用simplex[2] 执行点生成,但它们应该使用最新的点simplex[3]

我认为你的第一个边缘检查在第二个条件下使用了错误的脸,所以而不是

if ((((simplex[2]-simplex[3]) ^ ((simplex[1]-simplex[3]) ^ (simplex[2]-simplex[3]))) * (-simplex[2]) < 0) &&
((((simplex[1]-simplex[3]) ^ (simplex[0]-simplex[3])) ^ (simplex[2]-simplex[3])) * (-simplex[2]) < 0)) {
std::cout << "AB";
//direction = AB x (A0 x AB)
direction = (simplex[2]-simplex[3]) ^ ((-simplex[3]) ^ (simplex[2]-simplex[3]));
simplex = {simplex[2], simplex[3]};
}

应该是

if ((((simplex[2]-simplex[3]) ^ ((simplex[1]-simplex[3]) ^ (simplex[2]-simplex[3]))) * (-simplex[3]) < 0) &&
((((simplex[2]-simplex[3]) ^ (simplex[0]-simplex[3])) ^ (simplex[2]-simplex[3])) * (-simplex[3]) < 0)) {
std::cout << "AB";
//direction = AB x (A0 x AB)
direction = (simplex[2]-simplex[3]) ^ ((-simplex[3]) ^ (simplex[2]-simplex[3]));
simplex = {simplex[2], simplex[3]};
}

第二个边检查的第二个条件也一样,应该是:

((((simplex[1]-simplex[3]) ^ (simplex[2]-simplex[3])) ^ (simplex[1]-simplex[3])) * (-simplex[3]) < 0)

和第三次边缘检查的第二个条件:

((((simplex[0]-simplex[3]) ^ (simplex[1]-simplex[3])) ^ (simplex[0]-simplex[3])) * (-simplex[3]) < 0)

我还修复了三角形检查的输出单纯形。输入单纯形的最后一点应该始终是输出单纯形的最后一点。此外,点的顺序应与计算的方向一致。

这里是完整的固定函数:

bool gjkSimplex(std::vector<Vector3f> &simplex, Vector3f &direction) {
GLuint simplexSize = simplex.size();
std::cout << simplexSize;
switch (simplexSize) {
//If the simplex is a line segment
case 2:
//Point is closest feature
if ((simplex[0]-simplex[1])*-simplex[1] < 0) {
std::cout << "A";
//direction = A0
direction = -simplex[1];
simplex = {simplex[1]};
//Line is closest feature
} else {
std::cout << "AB";
//direction = AB x (A0 x AB)
// ^ = cross product
direction = (simplex[0]-simplex[1]) ^ ((-simplex[1]) ^ (simplex[0]-simplex[1]));
}
break;
//If the simplex is a triangle
case 3:
//Point is closest feature
if ((simplex[0]-simplex[2])*(-simplex[2]) < 0 && (simplex[1]-simplex[2])*(-simplex[2]) < 0) {
std::cout << "A";
//direction = A0
direction = -simplex[2];
simplex = {simplex[2]};
//Line to second-latest point is closest feature
} else if ((((simplex[0]-simplex[2])^(simplex[1]-simplex[2]))^(simplex[1]-simplex[2]))*-simplex[2] > 0) {
std::cout << "AB";
//direction = AB x (A0 x AB)
direction = (simplex[1]-simplex[2]) ^ ((-simplex[2]) ^ (simplex[1]-simplex[2]));
simplex = {simplex[1], simplex[2]};
//Line to oldest point is closest feature
} else if (((simplex[0]-simplex[2])^((simplex[0]-simplex[2])^(simplex[1]-simplex[2])))*-simplex[2] > 0) {
std::cout << "AC";
//direction = AC x (A0 x AC)
direction = (simplex[0]-simplex[2]) ^ ((-simplex[2]) ^ (simplex[0]-simplex[2]));
simplex = {simplex[0], simplex[2]};
//Face is closest feature
} else {
//Origin is in direction AC x AB
if (((simplex[1]-simplex[2]) ^ (simplex[0]-simplex[2])) * (-simplex[2]) < 0) {
std::cout << "ACxAB";
//direction = AC x AB
direction = (simplex[0]-simplex[2]) ^ (simplex[1]-simplex[2]);
//origin is in direction AB x AC (other side of the face)
} else {
std::cout << "ABxAC";
//direction = AB x AC
direction = (simplex[1]-simplex[2]) ^ (simplex[0]-simplex[2]);
simplex = {simplex[1], simplex[0], simplex[2]};
}
}
break;
//If the simplex is a tetrahedron
case 4:
//Newest point is closest feature
if ((simplex[0]-simplex[3])*(-simplex[3]) < 0 && (simplex[1]-simplex[3])*(-simplex[3]) < 0 &&
(simplex[2]-simplex[3])*(-simplex[3]) < 0) {
std::cout << "A";
//direction = A0
direction = -simplex[3];
simplex = {simplex[3]};
//Edge between newest and second-newest point is closest feature
} else if ((((simplex[2]-simplex[3]) ^ ((simplex[1]-simplex[3]) ^ (simplex[2]-simplex[3]))) * (-simplex[3]) < 0) &&
((((simplex[2]-simplex[3]) ^ (simplex[0]-simplex[3])) ^ (simplex[2]-simplex[3])) * (-simplex[3]) < 0)) {
std::cout << "AB";
//direction = AB x (A0 x AB)
direction = (simplex[2]-simplex[3]) ^ ((-simplex[3]) ^ (simplex[2]-simplex[3]));
simplex = {simplex[2], simplex[3]};
//Edge between newest and third-newest vertex is closest feature
} else if ((((simplex[1]-simplex[3]) ^ ((simplex[0]-simplex[3]) ^ (simplex[1]-simplex[3]))) * (-simplex[3]) < 0) &&
((((simplex[1]-simplex[3]) ^ (simplex[2]-simplex[3])) ^ (simplex[1]-simplex[3])) * (-simplex[3]) < 0)) {
std::cout << "AC";
//direction = AC x (A0 x AC)
direction = (simplex[1]-simplex[3]) ^ ((-simplex[3]) ^ (simplex[1]-simplex[3]));
simplex = {simplex[1], simplex[3]};
//Edge between newest and oldest point is closest feature
} else if ((((simplex[0]-simplex[3]) ^ ((simplex[2]-simplex[3]) ^ (simplex[0]-simplex[3]))) * (-simplex[3]) < 0) &&
((((simplex[0]-simplex[3]) ^ (simplex[1]-simplex[3])) ^ (simplex[0]-simplex[3])) * (-simplex[3]) < 0)) {
std::cout << "AD";
//direction = AD x (A0 x AD)
direction = (simplex[0]-simplex[3]) ^ ((-simplex[3]) ^ (simplex[0]-simplex[3]));
simplex = {simplex[0], simplex[3]};
//Face between the three newest points is closest feature
} else if (((simplex[1]-simplex[3]) ^ (simplex[2]-simplex[3])) * (-simplex[3]) > 0) {
std::cout << "ABC";
//direction = AC x AB (outer normal of face)
direction = (simplex[1]-simplex[3]) ^ (simplex[2]-simplex[3]);
simplex = {simplex[1], simplex[2], simplex[3]};
//Face between newest, second-newest and oldest point is closest feature
} else if (((simplex[2]-simplex[3]) ^ (simplex[0]-simplex[3])) * (-simplex[3]) > 0) {
std::cout << "ABD";
//direction = AB x AD
direction = (simplex[2]-simplex[3]) ^ (simplex[0]-simplex[3]);
simplex = {simplex[2], simplex[0], simplex[3]};
//Face between newest, second-oldest and oldest point is closest feature
} else if (((simplex[0]-simplex[3]) ^ (simplex[1]-simplex[3])) * (-simplex[3]) > 0) {
std::cout << "ACD";
//direction = AD x AC
direction = (simplex[0]-simplex[3]) ^ (simplex[1]-simplex[3]);
simplex = {simplex[0], simplex[1], simplex[3]};
//Origin is encased by simplex
} else {
//Collision detected
std::cout << "ABCD";
return true;
}
break;
default:
direction = {1,1,1};
simplex = {};
break;
}
std::cout << "\n";
return false;
};

关于c++ - GJK 算法陷入不同 Voronoi 区域案例的循环中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25750552/

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