Computing the Hausdorff Distance between Two B-Spline Curves. Zachi Shtain

Transcription

1 Computing the Hausdorff Distance between Two B-Spline Curves Zachi Shtain Based on the work of: Chen et al. 2010

2 Definition Given two curves C 1, C 2, their Hausdorff distance is defined as: H Where:, max max min,max min D C C p q p q 1 2 max min p q,max min p q are the one-sided pc qc qc Hausdorff distance pc qc qc pc p & q are known as the Hausdorff pair pc 2

3 3 Example

4 Point-Curve Projection Given a point p and a curve C(u), the point projection problem can be described: * min,, p C u p C u u a b If u* a, b condition:, we have the following necessary p C u*, C ' u * 0 4

5 Point-Curve Projection cont. Using the condition the problem turns into a root-finding problem of a polynomial equation Two steps for solving: Removing intervals which contain no solution Compute local extrema in each remaining intervals pc u*, C ' u* 0 5

6 Point-Curve Projection Example Given a Bezier curve defined by the control polygon {(-1, 0), (0, 2), (1, 0)} Find the location(s) on curve which have the minimum distance between the curve and the point (0, 0) 6

7 Point-Curve Projection Example cont. The inner product yields a Bezier curve defined by the control polygon: {-2, 4.667, , 2} pc u*, C ' u* 0 7

8 Point-Curve Projection Example cont. Using a subdivision method, we isolate the roots of the polynomial and get: t X Y

9 Point-Curve Projection Additional Examples 9

10 Locations of Hausdorff Distance The Hausdorff distance can occur at C 1 - discontinuity points (end points) The Hausdorff distance can occur at antipodal points of the two curves The Hausdorff distance can occur when one curve intersects the bisector of the second curve 10

11 Case 1 The Hausdorff distance can occur at two end points (one end point of each of the curves) 11

12 Case 2 The Hausdorff distance can occur at the end point of one the curves and at an inner point of the other curve, C2' v, C2' v 0 C1' v 0 C C v, 1' v 0,1 C 0 C v 0 C 1 C v C C v, 1 C 0 v

13 Case 3 The Hausdorff distance can occur at the two closest inner points C 1 u C2 v, C ' u 0 C u C v, C ' v uv, 0,1 13

14 Case 4 The Hausdorff distance can occur at two inner points 2 2 Ci r C j t Ci r C j s 0 r, s, t 0,1 C s C r C s, ' 0 i, j C i i j r C t, C ' t j i j 0 i 1, 2 j 14

15 Computing the Hausdorff Distance: The Algebraic Root-Finding Method Computing all of the roots of the non-linear equations of cases 2-4 Finding the resulting solution from these roots and the end points as well 15 C i i j r C t, C ' t j i j, C1 u C v 0, C2' v 0 1, C2' v 0 C1 u C2 v ' 0 0, C1' v 0 C 1 C v, 1' v 0 C1 u 2, C 2 ' v 0 0,1 2 2 Ci r C j t Ci r C j s 0 r, s, t 0,1 C r C s, C ' s 0 i, j 1, 2 C C v C C v C C v C v i j uv, 0,1

16 Avoiding Solving Case 4 Case 4 occurs only when the minimum distance occurs at one than more place Subdividing the given curves ensures only one place that the minimum distance will occur at Thus, the trivariate equation system of case 4 can be avoided 16

17 Improving the Computation It can be seen that the Hausdorff distance can be computed by solving the one-sided problem twice The problem turns into a min/max point to curve distance computation problem 17 D C, C max max min p q,max min p q H 1 2 pc qc qc pc

18 Outline of Algorithm First, utilize the elimination criteria in the geometric pruning method to prune the sub-intervals Second, subdivide the given curves in the remaining sub-intervals Termination conditions are also provided to speed up the process 18

19 General Geometric Pruning Method for Computing One-Sided Hausdorff Distnace Given two curves C 1 (u) and C 2 (v), we want to compute the squared one-sided Hausdorff distance from C 1 (u) to C 2 (v): where u 0,1 v 0,1 H C u, C v max min S u, v S u, v C u C v, C u C v

20 Properties used for Geometric Pruning Property #1: S u, v min S u, v v 0,1 S u, v max S u, v u 0,1,, H C u C v S u v The case where the Hausdorff distance occurs at two end points is detected using this property 20

21 Properties used for Geometric Pruning Property #2: cont. Property #3: u u 0,1 : S u, v min S u, v H C u, C v max S u, v v 0,1 0,1 v 0,1 v 0,1 : S u, v max S u, v H C u, C v min S u, v u 0,1 The above properties detect the case where the Hausdorff distance occurs at one end point 21

22 Bounding for the Geometric Pruning The geometric method need upper and lower bounds for elimination The bounds can be set as the minimum distances between the end points u min S u, v H C u, C v max S u, v v 0,1 0,1 22

23 Bounding for the Geometric Pruning Remark We have u min S u v, v H C u, C v max S u, v u v 0,1 0,1 Where u(v) and v(u) are functions in v and u For the case of two Bezier Curves u v v v u u In B-Spline Cases, u(v) and v(u) can be set as piecewise linear functions 23 u min S u, v H C u, C v max S u, v v 0,1 0,1

24 Geometric Pruning Subdivision When all of the properties are not satisfied, the curve C 1 (u) should be divided to obtain: L, C R H C u v max H C u, C, H C u, C For each v, if the Hausdorff distance cannot occur in the interval L i, then it can be eliminated 24

25 Elimination Criteria for the Case of Two B-Splines Curves Given two B-Spline curves i j B u, B v m-th and n-th degree B-spline basis functions over the knot vectors: We obtain: 25 i j C1 u PB u C2 v Q B v u, v 0,1 n i i n j m j m U { 0,...,0, a, a,..., a, 1,...,1 } V { 0,...,0, b, b,..., b, 1,...,1} 2 3 n 2 3 n+1 n+1 m+1 m+1 n m 1 2 S u, v D B u B v r k r0 k0 r, k 2n 2m m

26 Elimination Criteria for the Case of Two B-Splines Curves cont. Suppose the different knots in the knot vector U are u, i 0... l i m The interval [0,1] can be partitioned into l subintervals The function S(u,v) can be represented in Bezier 26 Li u u i l i i1 m, m, form as: 2n m2 S u, v F Bˆ r0 k0 u u t u B v t u u i r k r, k 2n i 2m i i 1 m i m i um

27 Elimination Criteria for the Case of Two B-Splines Curves cont. When the function S(u,v) is in the Bezier form in the sub-interval L i, we subdivide at the middle to obtain two new intervals: u i m, u m u m, u i1 m Theorem: If F i,,, : 0,..., 1 0,..., i r k F r n k m 2nr k 2 the Hausdorff distance occurs in the interval u i m, u m u m, u can be eliminated i1 m and 27

28 Let Termination Conditions for the Subdivision Process Given a value of u, there exist a value of v such that d u min S u, v, u 0,1 v 0,1 d u When C 1 (u) & C 2 (v) are C 1 -continuous B-Spline curves d(u) is also C 1 -continuous except for a finite number of locations S u, v u 28

29 d u S u, v u Termination Conditions for the Subdivision Process cont. C u C v u, C ' v u 0 If the derivative of v(u) exists, then 2 C1u C2 vu, C1 ' u Otherwise: d ' u0 2 C1 u0 C2 v1, C1 ' u0 d ' u 2 C u C v, C ' u Finally, we set: C1 u C2 vu v u d ' u 2 C1 u C2 v u, ' ' ' _ d ' u max d ' u, d ' u v v

30 Termination Conditions for the Subdivision Process cont. From the following approximation formula: d u u d u d ' u u A trivial termination condition for a sub interval u, t ut 0,1 u could be simply set as: u 2 1 t t 2 u d ' 2 u 2 1 t t

31 Termination Conditions for the Subdivision Process cont. When the termination condition is satisfied, the Hausdorff distance can be approximated by the minimum distance between the point and the curve C 2 (v) u C 1 u 1 2 t t 2 31 u u 2 1 t t u d ' 2 u 2 1 t t 2

32 Algorithm Steps 1. Compute S(u,v) in B-Spline Form: 2. Put (S(u,v), L 1, L 2 ) into W 3. If the set W is empty go to step 8 4. Pop one element w = (S 1, w 1, w 2 ) from set W 5. If S 1 satisfies the elimination criteria, then go back to step 3; otherwise go to step 6 6. If the length of w 1 is small enough to satisfy the termination condition, put (S 1, w 1, w 2 ) into y and rerun to step 3. Otherwise go to step 7 32 n m 1 2 r k, S u v D B u B v r0 k0 r, k 2n 2m 7. Subdivide the interval w 1 into two subintervals and put the new pairs (S 11, w 11, w 21 ) & (S 12, w 12, w 22 ) into set W ; return to step 3 8. For each element w of set y, compute the corresponding solution of case 3 in the region w 1 xw 2 using the Newton-Raphson method 9. Put the solution with the maximum distance and the place where the Hausdorff distance occurs as the resulting output, C1 u C v C1 u C2 v ' 0 C u, C ' v uv, 0,1

33 Algorithm Results For two cubic Bezier curves defined by the control polygons: {(193, 226), (227, 230), (297, 134), (421, 135)} {(258, 481), (294, 438), (302, 268), (435, 213)} The Hausdorff distance is The Hausdorff pair (0.0685, 0) 33