AN ALGORITHM FOR FINDING THE DISTANCE BETWEEN TWO ELLIPSES. Ik-Sung Kim. 1. Introduction and preliminaries

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1 Commun. Korean Math. Soc. 21 (2006), No. 3, pp AN ALGORITHM FOR FINDING THE DISTANCE BETWEEN TWO ELLIPSES Ik-Sung Kim Abstract. We are interested in the distance problem between two objects in three dimensional Euclidean space. There are many distance problems for various types of objects including line segments, boxes, polygons, circles, disks, etc. In this paper we present an iterative algorithm for finding the distance between two given ellipses. Numerical examples are given. 1. Introduction and preliminaries The distance problem between two given objects in three dimensional space can be found often in computer-aided geometric design systems. Further, it is important to propose an efficient algorithm for finding the distance between two objects. There are many distance problems for various types of objects including line segments [5], boxes [6], polygons [8], circles [7], disks [1], etc. In the literature, many problems already have been studied and various numerical techniques to compute the optimal distance have been given. In this paper we consider the problem of finding the distance between two given ellipses. The representation of an ellipse in three dimensional space can be given by using a geometric transformation of a standard ellipse in the xy plane. This may simplify the distance function between the two ellipses. Thus, our problem is reduced to the distance problem between one standard ellipse and the other ellipse. We can present an iterative algorithm which is mainly based on computing the distance between a given point and a standard ellipse. Received April 19, Mathematics Subject Classification: 65D10. Key words and phrases: distance between two ellipses.

2 560 Ik-Sung Kim The standard form of the equation of an ellipse centered with origin in the xy plane can be represented by x 2 (1.1) a 2 + y2 b 2 = 1, where a and b are the semimajor axis and the semiminor axis respectively. Further, the parametric form of this standard ellipse can be given by (1.2) x = a cos t, y = b sin t, where π t π. Also, one may characterize an ellipse in three dimensional Euclidean space by giving the three points c, p and q, where c is its center and p denotes one of the ends of the major axis and q denotes one of the ends of the minor axis. Let E be an arbitrary ellipse defined by its center c = (c 1, c 2, c 3 ) T and the two points p = (p 1, p 2, p 3 ) T and q = (q 1, q 2, q 3 ) T. Then E can be represented by a geometric transformation(rotation and translation) of a standard ellipse. Let us consider a relative standard ellipse E s with respect to E: (1.3) E s : in its parametric form x 2 a 2 + y2 b 2 = 1 (1.4) x = a cos t, y = b sin t, where (1.5) a = p c = (p 1 c 1 ) 2 + (p 2 c 2 ) 2 + (p 3 c 3 ) 2, (1.6) b = q c = (q 1 c 1 ) 2 + (q 2 c 2 ) 2 + (q 3 c 3 ) 2. Then, by using a 3 3 rotation matrix R = (r ij ) given by (1.7) (r 11, r 21, r 31 ) T = Re 1 = R(1, 0, 0) T = p c p c, (r 12, r 22, r 32 ) T = Re 2 = R(0, 1, 0) T = q c q c, (r 13, r 23, r 33 ) T = Re 3 = R(0, 0, 1) T (p c) (q c) = (p c) (q c). the coordinate w = ( x, ỹ, z) T of a point on E can be represented by (1.8) w = Rw + c, where the coordinate w = (x, y, 0) T is the corresponding coordinate of a point of a standard ellipse E s with respect to w.

3 An algorithm for finding the distance Distance between a point and a standard ellipse Let there be a point s = (s 1, s 2, s 3 ) T and a standard ellipse defined by (2.1) E s : in its parametric form x 2 a 2 + y2 b 2 = 1 (2.2) x = a cos t, y = b sin t. We consider the problem of computing the distance between s and E s and finding the optimal point lying on E s. Suppose that Q(s, E s ; t) is a function of a parameter t with respect to s and E s defined by (2.3) Q(s, E s ; t) = (s 1 a cos t) 2 + (s 2 b sin t) 2. Then, by finding the optimal point ŵ = (ˆx, ŷ, 0) T = (a cos ˆt, b sin ˆt, 0) T on E s such that (2.4) m = Q(s, E s ; ˆt) = min Q(s, E s; t) π t π = min π t π (s 1 a cos t) 2 + (s 2 b sin t) 2 we can compute the distance d(s, E) given by d(s, E) = m + s 2 3. The necessary condition Q = 0 for a minimum induces the following t equation: (2.5) A sin t B cos t + C sin t cos t = 0 with A = as 1, B = bs 2 and C = (b 2 a 2 ). Thus, if C = 0 in (2.5), then we can easily find t = ˆt by ( ) B (2.6) ˆt = arctan. A In case of C = c 2 2 c2 1 0 the equation (2.5) induces the following two equations: (2.7) A sin t B cos t + C sin t cos t = 0 (0 t π) and (2.8) A sin t + B cos t + C sin t cos t = 0 ( π t 0).

4 562 Ik-Sung Kim Thus, if we let v = cos t in (2.7) and (2.8), then the corresponding values of sin t can be given by { 1 v 2 if 0 t π sin t = 1 v 2 if π t 0, and the equation (2.5) leads to the following quartic equation: ( ) ( A A (2.9) v v B 2 C 2 ) ( ) ( ) A A 2 + C C 2 v 2 2 v = 0. C C By solving the equation (2.9) for v we have some solutions v j (j = 1, 2,..., s) such that 1 v j 1 and 1 s 4, and get the value t j for each v j such that cos t j = v j (0 t j π). Thus, t has two values µ 1 j = t j and µ 2 j = t j for each t j, and the corresponding values of sin t can be given by sin µ 1 j = 1 (v j ) 2 (j = 1, 2,..., s), (2.10) sin µ 2 j = 1 (v j ) 2 (j = 1, 2,..., s). In this case we can choose t = ˆt = µ m l (2.11) for the solution of (2.5) such that (s 1 a cos µ m l ) 2 + (s 2 b sin µ m l ) 2 [ = min (s1 a cos µ i j) 2 + (s 2 b sin µ i j) 2]. i=1,2 j=1,2,...,s Thus, we can find the optimal point ŵ = (ˆx, ŷ, 0) T = (a cos ˆt, b sin ˆt, 0) T on E s. 3. Distance between two ellipses Let E α and E β be the two given ellipses in the three dimensional space. Suppose that E α is defined by its center c α = (c α 1, cα 2, cα 3 )T and the two points p α = (p α 1, pα 2, pα 3 )T and q α = (q1 α, qα 2, qα 3 )T and E β is defined by c β = (c β 1, cβ 2, cβ 3 )T, p β = (p β 1, pβ 2, pβ 3 )T and q β = (q β 1, qβ 2, qβ 3 )T. Then we can give the corresponding two standard ellipses Es α and Es β defined by (3.1) E α s : w α =(a α cos t α, b α sin t α, 0) T, π t α π E β s : w β =(a β cos t β, b β sin t β, 0) T, π t β π,

5 An algorithm for finding the distance 563 where (3.2) a α = p α c α = (3.3) b α = q α c α = (3.4) a β = p β c β = (3.5) b β = q β c β = (p α 1 cα 1 )2 + (p α 2 cα 2 )2 + (p α 3 cα 3 )2, (q α 1 cα 1 )2 + (q α 2 cα 2 )2 + (q α 3 cα 3 )2, (p β 1 cβ 1 )2 + (p β 2 cβ 2 )2 + (p β 3 cβ 3 )2, (q β 1 cβ 1 )2 + (q β 2 cβ 2 )2 + (q β 3 cβ 3 )2. Also, the coordinate w α = ( x α, ỹ α, z α ) T of a point of E α and the coordinate w β = ( xβ, ỹβ, z β ) T of a point of E β can be represented by E α : w α = R α w α + c α (3.6) = R α (a α cos t α, b α sin t α, 0) T + (c α 1, c α 2, c α 3 ) T, π t α π E β : wβ = R β w β + c β = R β (a β cos t β, b β sin t β, 0) T + (c β 1, cβ 2, cβ 3 )T, π t β π, where the two 3 3 rotation matrices R α = (r α ij ) and Rβ = (r β ij ) are given by the following: For k = α or β (3.7) (r k 11, r k 21, r k 31) T = R k e 1 = R k (1, 0, 0) T = pk c k p k c k, (r k 12, r k 22, r k 32) T = R k e 2 = R k (0, 1, 0) T = qk c k q k c k, (r 13 k, r k 23, r k 33) T = R k e 3 = R k (0, 0, 1) T = (pk c k ) (q k c k ) (p k c k ) (q k c k ). Furthermore, since Euclidean metric is invariant under R α = (r α ij ) and R β = (r β ij ) the distance between Eα and E β can be given by d(e α, E β ) = min u E α v E β u v = min (R α w α + c α ) (R β w β + c β ) π t α π π t β π

6 564 Ik-Sung Kim or (3.9) = min π t α π π t β π = min π t α π π t β π d(e α, E β ) = R α ( w α (R α ) 1 R β w β (R α ) 1 (c β c α )) w α (R α ) 1 ( R β w β + (c β c α )) min u E α v E β u v ( = min w β (R β ) 1 R α w α + (c α c β )). π t β π π t α π We now describe an iteration algorithm for finding the optimal points ) T w α = ( x α, ỹ α, z ( xβ α and wβ =, ỹβ, z ) T β lying on E α and E β respectively. Our algorithm is mainly based on computing the distance between a point and a standard ellipse. Algorithm: Step 0. Compute a α, b α, a β, b β from (3.2), (3.3), (3.4) and (3.5). Define the rotation matrices R α and R β and their inverse matrices (R α ) 1 and (R β ) 1 from (3.7). Set j := 0. Step 1. Let t α 0 be given as an initial value for tα. Then, from (3.1) and (3.6) we can initialize w α 0 = (xα 0, yα 0, zα 0 )T = (a α cos t α 0, bα sin t α 0, 0)T and w α 0 = ( x α 0, ỹα 0, z α 0 ) T = R α w α 0 + cα. We consider the following minimization problem: To find w β 0 = (aβ cos t β 0, bβ sin t β 0, 0)T and w β 0 = ( xβ 0, ỹβ 0, z β 0 ) T = R β w β 0 +cβ such that (3.10) w β 0 w 0 α = min w β E β s w β (R β ) 1 ( R α w α 0 + (c α c β )). Thus, let s = (R β ) 1 ( R α w0 α + (cα c β ) ) and E s = Es β with a = a β and b = b β in (2.3). Then we can find the optimal point ŵ = ( x, ŷ, 0) = (a cos t, b sin t, 0) T on Es β such that (3.11) Q(s, E s ; t) = min π t π Q(s, E s; t). Set w β 0 = (a cos tβ 0, b sin tβ 0, 0)T = ŵ = (a cos t, b sin t, 0) T and w β 0 = R β w β 0 + cβ.

7 An algorithm for finding the distance 565 Step 2. Find w α j+1 = (xα j+1, yα j+1, zα j+1 )T = (a α cos t α j+1, bα sin t α j+1, ( T 0) T and w j+1 α = x α j+1, ỹα j+1, z j+1) α = R α wj+1 α + cα such that (3.12) w j+1 α w β j+1 = min ( w α (R α ) 1 R β w β w α Es α j + (cβ c )) α. ( ) That is, let s = (R α ) 1 R β w β j + (cβ c α ) and E s = Es α with a = a α and b = b α in (2.3). Then, the optimal point ŵ = ( x, ŷ, 0) = (a cos t, b sin t, 0) T on E α s such that (3.13) Q(s, E s ; t) = min π t π Q(s, E s; t). Set wj+1 α = (a cos tα j+1, b sin tα j+1, 0)T = ŵ = (a cos t, b sin t, 0) T and w j+1 α = Rα wj+1 α + cα. Further, find w β j+1 = (aβ cos t β j+1, bβ sin t β j+1 ), 0)T T = R β w β j+1 + cβ such that and w ( β j+1 = xβ j+1, ỹβ j+1, z β j+1 (3.14) w β j+1 w j+1 α = min ( w β (R β ) 1 R α w α w β Es β j + (c α c )) β. ( ) In other words, let s = (R β ) 1 R α wj α + (cα c β ) and E s = Es β with a = a β and b = b β in (2.3). Then we obtain the optimal point ŵ = ( x, ŷ, 0) = (a cos t, b sin t, 0) T on Es β such that (3.15) Q(s, E s ; t) = min π t π Q(s, E s; t). Set w β j+1 = ŵ = (a cos t, b sin t, 0) T and w β j+1 = Rβ w β j+1 + cβ. Define a distance function U between two optimal points w k α and w β k on Eα and E β respectively, given by U(k) = w k α w β k for k = 1, 2,.... And if U(j + 1) < U(j), then we set j := j + 1 and go back to step 1. Here, we see that for k = 0, 1, 2,... two optimal points w k α and w β k can be found after k iterations with an initial value w 0 α. Further, though the proof is not given the convergence of our algorithm may be guaranteed by a descent property : U(k + 1) U(k) for k = 0, 1, 2,.... Thus, ( ) ( ) two sequences w n α and wβ n of optimal points converge to the global optimal points w α and w β respectively, and the optimal distance between two ellipses E α and E β is given by d(e α, E β ) = w α w. β

8 566 Ik-Sung Kim The value of a distance function U(n) Initial values π 2π/3 π/ The number of iteration n Figure 1. Convergence of the distance function for different initial values. Finally, to test our algorithm we give an example for finding the distance between two ellipses E α and E β in the three dimensional space. We can see the convergence of the corresponding distance function U(k) for each initial value t α 0. In fact, it is certain that each function U(k) converges to the distance between two given ellipses E α and E β which is the global minimum. Example. Let there be two ellipses E α and E β. E α is characterized by its center c α = (c α 1, cα 2, cα 3 )T = (1, 1, 1) T and the two points p α = (p α 1, pα 2, pα 3 )T = (2, 3, 2) T and q α = (q1 α, qα 2, qα 3 )T = (5, 2, 5) T. Also, E β is defined by c β = (c β 1, cβ 2, cβ 3 )T = (3, 5, 5) T, p β = (p β 1, pβ 2, pβ 3 )T = (1, 2, 5) T and q β = (q β 1, qβ 2, qβ 3 )T = (6, 3, 2) T. We employ our algorithm for finding the distance between two ellipses E α and E β by using four different values t α 0 = π, tα 0 = 2π 3, tα 0 = π 3 and tα 0 = 0 as initial values for t α. Then we have the corresponding four initial optimal points w 0 α = (0, 1, 0)T, w 0 α = (3.9641, , )T, w 0 α = (4.9641, , ) T and w 0 α = (2, 3, 2)T. Moreover, using these initial optimal points we find the same optimal points w α = ( , , ) T and w β = (0.6134, , ) T in each case. We

9 An algorithm for finding the distance 567 can obtain them after the same 13 iterations for t α 0 = 2π 3 and tα 0 = π 3. Also the optimal points can be found after 10 iterations for each of t α 0 = π and t α 0 = 0. The distance is given by d(eα, E β ) = w α w β = Further, in Figure 1 we see the convergence of the distance function U(k) to the optimal distance between two ellipses in each case. References [1] H. A. Almohamad and S. Z. Selim, An algorithm for computing the distance between two circular disks, Appl. Math. Modelling. 27 (2003), [2] W. Gander and J. Hřebiček, Solving problems in scientific computing using maple and matlab, Springer-Verlag, New York, [3] W. Gander, G. H. Golub and R. Strebel, Least-squares fitting of circles and ellipses, BIT. 34 (1994), [4] M. Grossmann, Parametric curve fitting, Comp. J. 14 (1971), [5] V. Lumelsky, On the computation of distance between line segments, Inform. Process. Lett. 21 (1985), [6] W. Meyer, Distance between boxes:applications to collision detection and clipping, IEEE Int. Conf. Robot. Automat. (1986), [7] C. A. Neff, Finding the distance between two circles in three dimensional space, IBM J. Res. Develop. 34 (1990), [8] J. T. Schwartz, Finding the minimum distance between two convex polygons, Inform. Process. Lett. 13 (1981), [9] H. Späth, Orthogonal squared distance fitting with parabolas, Proceedings of IMACS-GAMM Inter. Symposium on Numerical Methods and Error Bounds, Oldenburg (1995), Department of Applied Mathematics Korea Maritime University Pusan , Korea ikim@hhu.ac.kr