Finding point-pairs. Find Closest Point from Dense Cloud

Transcription

1 Finding point-pais Given an a, find a coesponding b on the suface. Then one appoach would be to seach evey possible tiangle o suface point and then take the closest point. The key is to find a moe efficient way to do this Find Closest Point fom Dense Cloud Basic appoach is to divide space into egions. Suppose that we have one point b k * that is a possible match fo a point a k. The distance *= b k * - a k obviously acts as an uppe bound on the distance of the closest point to the suface. Given a egion R containing many possible points b j, if we can compute a lowe bound L on the distance fom a to any point in R, then we need only conside points inside R if L < *. 1

2 Find Closest Point fom Dense Cloud Thee ae many ways to implement this idea Simply patitioning space into many buckets Octees, KD tees, covaiance tees, etc. Appoaches to closest tiangle finding 1. (Simplest) Constuct linea list of tiangles and seach sequentially fo closest tiangle to each point. 2. (Only slightly hade) Constuct bounding sphees aound each tiangle and use these to educe the numbe of caeful checks equied. 3. (Faste if have lots of points) Constuct hieachical data stuctue to speed seach. 4. (Bette but hade) Rotate each level of the tee to align with data. 2

3 FindClosestPoint(a,[p,q,]) Many appoaches. One is to solve the system a p λ ( q p) + µ ( p) in a least squaes sense fo λ and µ. Then compute c = p+ λ ( q p) + µ ( p) If λ 0, µ 0, λ + µ 1, then c lies within the tiangle and is the closest point. Othewise, you need to find a point on the bode of the tiangle a Hint: Fo efficiency, wok out the least squaes poblem explicitly. You will have to solve a 2 x 2 linea system fo λ, µ p c q Finding closest point on tiangle E F D G A B p C Region λ<0 µ<0 λ+µ>1 Closest point A Yes Yes No p B No Yes No PojectOnSegment(c,p,q) C No Yes Yes q D No No Yes PojectOnSegment(c,q,) E Yes No Yes F Yes No No PojectOnSegment(c,,p) G No No No c q 3

4 p PojectOnSegment(c,p,q) c* q c ( c p) ( q p) λ = ( q p) ( q p) λ* = Max(0, Min( λ,1)) c* = p+ λ *( q p) a c q Bounding Sphee b Suppose you have a point p and ae tying to find the closest tiangle ( ak, bk, ck) to p. If you have aleady found a tiangle ( a, b, c ) with a point k j j j j on it, when do you need to check caefully fo some tiangle k? Answe: if qk is the cente of a sphee of adius ρ enclosing ( a, b, c ), then k k k you only need to check caefully if p q ρ < p. k k j 4

5 a c q Bounding Sphee b Assume edge ( ab, ) is the longest. Then the cente q of the sphee will obey ( b q) ( b q) = ( a q) ( a q) ( c q) ( c q) ( a q) ( a q) = 0 ( b a) ( c a) ( q a) Simple appoach: Ty q = a + b / 2. ( ) If inequality holds, then done. Else solve the system as thee equalities to get q. The adius ρ = q a. Hieachical cellula decompositions 5

6 Hieachical cellula decompositions Constucting tee of bounding sphees class BoundingSphee { public: Vec3 Cente; double Radius; Thing* Object; }; // Coodinates of cente // adius of sphee // some efeence to the thing // bounded 6

7 Constucting octee of bounding sphees class BoundingBoxTeeNode { Vec3 Cente; // splitting point Vec3 UB; // cones of box Vec3 LB; int HaveSubtees; int nsphees; double MaxRadius; // maximum adius of sphee in box BoundingBoxTeeNode* SubTees[2][2][2]; BoundingSphee** Sphees; : : BoundingBoxTeeNode(BoundingSphee** BS, int ns); ConstuctSubtees(); void FindClosestPoint(Vec3 v, double& bound, Vec3& closest); }; Constucting octee of bounding sphees BoundingBoxTeeNode(BoundingSphee** BS, int ns) { Sphees = BS; nsphees = ns; Cente = Centoid(Sphees, nsphees); MaxRadius = FindMaxRadius(Sphees,nSphees); UB = FindMaxCoodinates(Sphees,nSphees); LB = FindMinCoodinates(Sphees,nSphees); ConstuctSubtees(); }; 7

8 Constucting octee of bounding sphees ConstuctSubtees() { if (nsphees<= mincount length(ub-lb)<=mindiag) { HaveSubtees=0; etun; }; HaveSubtees = 1; int nnn,npn,npp,nnp,pnn,ppn,ppp,pnp; // numbe of sphees in each subtee SplitSot(Cente, Sphees, nnn,npn,npp,nnp,pnn,ppn,ppp,pnp); Subtees[0][0][0] = BoundingBoxTee(Sphees[0],nnn); Subtees[0][1][0] = BoundingBoxTee(Sphees[nnn],npn); Subtees[0][1][1] = BoundingBoxTee(Sphees[nnn+npn],npp); : : } Constucting octee of bounding sphees SplitSot(Vec3 SplittingPoint, BoundingSphee** Sphees, int& nnn, int& npn,,int& pnp) { // eode Sphees( ) into eight buckets accoding to // compaison of coodinates of Sphee(k)->Cente // with coodinates of splitting point. E.g., fist bucket has // Sphee(k)->Cente.x < SplittingPoint.x // Sphee(k)->Cente.y < SplittingPoint.y // Sphee(k)->Cente.z < SplittingPoint.z // This can be done in place by suitable exchanges. // Set nnn = numbe of sphees with all coodinates less than // splitting point, etc. } 8

9 Seaching an octee of bounding sphees p j Seaching an octee of bounding sphees void BoundingBoxTeeNode::FindClosestPoint (Vec3 v, double& bound, Vec3& closest) { double dist = bound + MaxRadius; if (v.x > UB.x+dist) etun; if (v.y > UB.y+dist) etun;. ; if (v.z < LB.z-dist) etun; if (HaveSubtees) { Subtees[0][0][0].FindClosestPoint(v,bound,closest); : Subtees[1][1][1].FindClosestPoint(v,bound,closest); } else fo (int i=0;i<nsphees;i++) UpdateClosest(Sphees[i],v,bound,closest); }; 9

10 Seaching an octee of bounding sphees void UpdateClosest(BoundingSphee* S, Vec3 v, double& bound, Vec3& closest) { double dist = v-s->cente;; if (dist - S->Radius > bound) etun; Vec3 cp = ClosestPointTo(*S->Object,v); dist = LengthOf(cp-v); if (dist<bound) { bound = dist; closest=cp;}; }; 10