最近在看recast&detour源码的时候有遇到许多数学上的算法问题,特此记录,以便以后查看。
介绍
https://www.codeproject.com/Articles/15573/D-Polygon-Collision-Detection
应用分离轴定理 SAT ,看是否能找到分离轴,如果能找到那么就是不相交。否则相交。
利用点积的几何意义: 投影
源码
判断 polya 和 polyb 两个多边形是否相交
/// All vertices are projected onto the xz-plane, so the y-values are ignored. bool dtOverlapPolyPoly2D(const float* polya, const int npolya, const float* polyb, const int npolyb) { const float eps = 1e-4f; for (int i = 0, j = npolya-1; i < npolya; j=i++) { const float* va = &polya[j*3]; const float* vb = &polya[i*3]; // 与边垂直的向量,作为分离轴 const float n[3] = { vb[2]-va[2], 0, -(vb[0]-va[0]) }; float amin,amax,bmin,bmax; projectPoly(n, polya, npolya, amin,amax); projectPoly(n, polyb, npolyb, bmin,bmax); if (!overlapRange(amin,amax, bmin,bmax, eps)) { // Found separating axis return false; } } for (int i = 0, j = npolyb-1; i < npolyb; j=i++) { const float* va = &polyb[j*3]; const float* vb = &polyb[i*3]; const float n[3] = { vb[2]-va[2], 0, -(vb[0]-va[0]) }; float amin,amax,bmin,bmax; projectPoly(n, polya, npolya, amin,amax); projectPoly(n, polyb, npolyb, bmin,bmax); if (!overlapRange(amin,amax, bmin,bmax, eps)) { // Found separating axis return false; } } return true; }
计算多边形在某条轴上的相对投影范围
static void projectPoly(const float* axis, const float* poly, const int npoly, float& rmin, float& rmax) { // 求最大和最小的点积值 相当于 多边形在 轴上的投影范围(真正的这个范围需要除以|axis|,因为都乘了无所谓) rmin = rmax = dtVdot2D(axis, &poly[0]); for (int i = 1; i < npoly; ++i) { const float d = dtVdot2D(axis, &poly[i*3]); rmin = dtMin(rmin, d); rmax = dtMax(rmax, d); } }
判断是否区域有重叠
inline bool overlapRange(const float amin, const float amax, const float bmin, const float bmax, const float eps) { return ((amin+eps) > bmax || (amax-eps) < bmin) ? false : true; }