Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere for N = 2,..., 9

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1 Original Paper Forma, 1, 197 5, 006 Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere for N =,..., 9 Teruhisa SUGIMOTO 1 * and Masaharu TANEMURA 1, 1 The Institute of Statistical Mathematics, Minami-Azabu, Minato-ku, Tokyo , Japan Department of Statistical Science, The Graduate University for Advanced Studies, Minami-Azabu, Minato-ku, Tokyo , Japan * address: sugimoto@ism.ac.jp (Received January 00, 005; Accepted February 00, 005) Keyword: Spherical Cap, Sphere, Packing, Covering, Minkowski Set Abstract. Let C i (i = 1,..., N) be the i-th open spherical cap of angular radius r and let M i be its center under the condition that none of the spherical caps contains the center of another one in its interior. We consider the upper bound, r N (not the lower bound!) of r of the case in which the whole spherical surface of a unit sphere is completely covered with N congruent open spherical caps under the condition, sequentially for i =,..., N 1, that i 1 M i is set on the perimeter of C i 1, and that each area of the set ( ν= 1 C ν ) C i becomes maximum. In this study, for N =,..., 9, we found out that the solutions of the above covering and the solutions of Tammes problem were strictly correspondent. 1. Introduction How must N congruent non-overlapping spherical caps be packed on the surface of a unit sphere so that the angular diameter of spherical caps will be as great as possible? This packing problem is also called the Tammes problem and mathematically proved solutions were known for N = 1,..., 1, and 4 (SCHÜTTE and VAN DER WAERDEN, 1951; DANZER, 1963; FEJES TÓTH, 1969, 197; TESHIMA and OGAWA, 000). On the other hand, the problem How must the covering of a unit sphere be formed by N congruent spherical caps so that the angular radius of the spherical caps will be as small as possible? is also important. It can be considered that this problem is dual to the problem of packing of Tammes (FEJES TÓTH, 1969). Among the problems of packing and covering on the spherical surface, the Tammes problem is the most famous. However, the systematic method of attaining these solutions has not been given. In this paper, we would like to think of the covering in connection with the packing problem. Therefore, we consider the covering of the spherical caps such that none of them contains the center of another one in its interior. Such a set of centers is said to be a Minkowski set (FEJES TÓTH, 1999). Hereafter, we call the condition of Minkowski set of centers Minkowski condition. In addition, the covering which satisfies the Minkowski condition is called Minkowski covering. If angular radii of spherical caps which cover 197

2 198 T. SUGIMOTO and M. TANEMURA the unit sphere under the Minkowski condition are concentrically reduced to half, the resulting spherical caps do not overlap. Then, our purpose in this paper is to obtain the upper bound (not the lower bound!) of angular radius of spherical caps which cover the unit sphere under the Minkowski condition. Suppose we have N congruent open spherical caps with angular radius r on the surface S of the unit sphere and suppose that these spherical caps cover the whole spherical surface without any gap and that the Minkowski covering is realized. Further we suppose the spherical caps are put on S sequentially in the manner which is described just below. Let C i be the i-th open spherical cap and let M i be its center (i = 1,..., N). Our problem is to calculate the upper bound of r for the sequential covering, such that N congruent open spherical caps cover the whole spherical surface S under the condition that M i is set on the i 1 perimeter of C i 1, and that each area of set ( ν = 1 C ν ) C i become maximum in sequence for i =,..., N 1 (SUGIMOTO and TANEMURA, 00, 003, 004). In this paper, we calculate the upper bound of r for N =,..., 9 theoretically; the case N = 1 is self-evident. It is shown in this paper that the solutions of our problem are strictly correspondent to those of the Tammes problem for N =,..., 9. Further, it should be said that our method is a systematic and a different approach to the Tammes problem from the works by SCHÜTTE and VAN DER WAERDEN (1951), etc. In Sec., to solve our problem, we consider the properties of spherical caps under the Minkowski condition. Then, we explain the procedure of our sequential covering and define the upper bouds r N and r N 1. In Sec. 3 we give the solutions of our problem successively for N =,..., 9. Our conclusion is summarized in Sec. 4.. Sequential Covering Throughout this paper we assume that the center of the unit sphere is the origin O = (0, 0, 0). Hereafter, we represent the surface of this unit sphere by the symbol S. In the following, open spherical caps are simply written as spherical caps unless otherwise stated. We define the geodesic arc between an arbitrary pair of points T i and T j on S as the inferior arc of the great circle determined by T i and T j. Then the spherical distance between T i and T j is defined by the length of geodesic arc of this pair of points..1. Relationship between the kissing number and the angular radius of spherical caps First of all, we consider how many centers of congruent spherical caps can be placed on the perimeter of a spherical cap under the Minkowski condition. This problem is related to the kissing number of spherical caps. In the plane, one circle can contact simultaneously with six other congruent circles. Then the kissing number is always six in the plane. On the sphere, on the contrary, the kissing number of a circle (spherical cap) on the spherical surface changes with its angular radius. Its maximum value is five as we will see below. Here, we define a half-cap as the spherical cap whose angular radius is r/ and which is concentric with that of the original cap. When the kissing number is k, there can k half-caps contact with the central half-cap and there is no space for another half-cap to enter. At this instance, let us increase r until the peripheries of k half-caps contact with one another. Then, if the centers of two half-caps in contact are joined by a geodesic arc, there arise k spherical equilateral triangles of side-length r and of inner angle σ = π/k around the central

3 Packing and Minkowski Covering of Congruent Spherical Caps 199 Table 1. Relationship between k and r. k Range of r 1 (π /3, π] (π cos 1 (1/3), π /3] 3 (π /, π cos 1 (1/3)] 4 (tan 1, π /] 5 (0, tan 1 ] half-cap. Thus by applying the spherical cosine theorem to one of the spherical equilateral triangles, we have r = cos 1 cosσ. 1 cosσ () 1 By substituting σ = π/k into (1), we can calculate the upper bound value of r for a given kissing number k (>1). Note that, for k = 1, it is clear that the upper bound value is π. For k 6, Eq. (1) cannot have a solution. Namely, for k = 6, we get at once r = cos 1 1 = 0, and for k > 6, we get the inequality cosσ/(1 cosσ) > 1 indicating that no real value solution exists for r. Therefore the maximum value of the kissing number of a spherical cap (circle on the spherical surface) is five. As a result, we obtain the relationship between k and r as is shown in Table 1 (SUGIMOTO and TANEMURA, 001, 00, 003; SUGIMOTO, 00). For example, from Table 1, the kissing number k is four when the angular diameter of a spherical cap is in the range (tan 1, π/]. Note that inclusion of the upper bound in the range of r in Table 1 is correspondent to the Minkowski condition. Next, let us consider the packing with half-caps and the covering with corresponding spherical caps (of angular radius r) where each half-cap is concentric with correspondent spherical caps. Then it is observed that this covering satisfies the Minkowski condition; namely, this is the Minkowski covering. For the covering that the centers of spherical caps are chosen on the perimeters of other spherical caps under the Minkowski condition, we see from Table 1 that, for example, four spherical caps can be placed on the perimeter of a spherical cap when angular radius is in the range (tan 1, π/]. We note that the discussions of this subsection are valid for both of open and closed spherical caps... Overlapping area of congruent spherical caps In order to solve our problem, we consider the overlapping area of congruent spherical caps under the Minkowski condition. Assume C i and C j be two congruent spherical caps, of angular radius r, which are mutually overlapping under the Minkowski condition, and let A ij = A(C i C j ) be the overlapping area where A(X) is the area of X. Further, when T i and T j are the points on S, let d s (T i, T j ) denote the spherical distance between points T i and T j. Especially, we denote s ij = d s (M i, M j ) as the spherical distance between centers of C i and C j (r s ij r). Then let

4 00 T. SUGIMOTO and M. TANEMURA h ij be the spherical distance between cross points of perimeters of C i and C j. By using the spherical cosine formula about a spherical right triangle, we have h ij = cos 1 Then the overlapping area of C i and C j is given by cos r. cos( sij ) A ij ij r 1 cos h cos = cos cos r 4cos sin r 1 1 cos hij tanr sin hij + π. ( ) For the detailed derivation of (), see Appendix A.. It is obvious that A ij is a continuous function of r and s ij (SUGIMOTO and TANEMURA, 001, 00; SUGIMOTO, 00). Then, we get the following Lemma (SUGIMOTO and TANEMURA, 004). Lemma. If the range of angular radius r is 0 < r π/3, then the overlapping area A ij of the set C i C j is a monotone decreasing function of s ij = d s (M i, M j ) when r is fixed. Then, A ij is maximum when s ij = r. Proof. At first, we show that the range of r should be 0 < r π/3 in order A ij to be a function of s ij. Under the Minkowski condition, if π/3 < r π, the area of the set C i C j is always constant 4π cosr and, if r > π, two spherical caps cannot be put on the spherical surface S. Therefore, the range of angular radius is limited to 0 < r π/3. Let G(a) and G(b) be the set C i C j for s ij = a and b (a < b), respectively. We assume that C i and C j contact with each other first. Let e be the geodesic arc between fixed centers of these spherical caps. Next, let C i be fixed and let us move C j by moving M j along e toward M i. Then, it is obvious that G(a) G(b) holds during this movement. Therefore, for r fixed, A ij is a monotone decreasing function of s ij. Thus, A ij is maximum when s ij = r. On the contrary, A ij is a monotone increasing function of h ij when r is fixed. Further, when s ij = r, we find that the area of set C i C j is minimum. Let N be the number of spherical caps when the whole spherical surface is completely covered. And, let C i be the perimeter of C i (i = 1,..., N). Then we define: Wi = Wi 1 Ci max A( Wi 1 Ci), i =, K, N 1; W1 = C1. () 3 Mi Ci 1 In other words, W i is the union of W i 1 and C i satisfying the condition that the area A(W i 1 C i ) is maximum with the restriction M i C i 1 (SUGIMOTO and TANEMURA, i 00, 003). Hereafter, we call that ν =1 C ν is in an extreme state when the set of spherical caps C 1,..., C i possesses the property (3). We are always necessary to examine

5 Packing and Minkowski Covering of Congruent Spherical Caps 01 Fig. 1. The sketch of W and W C 3. i whether ν =1 C ν is in an extreme state in the sequential covering procedure of our problem mentioned in Sec. 1. For this purpose, we calculate the area A(W i 1 C i ) by using () and the area formula of spherical triangle. Here, we consider W defined in (3). From Lemma, the area A(C 1 C ) will become maximum when M, the center of C, is put on C 1. Therefore, for i =, the set W = ν =1 C ν is in an extreme state when s 1 = r for 0 < r π/3 as shown in Fig. 1(a). 3 Next, we consider W 3. In order to make the situation that ν =1 C ν is in an extreme state (A(W C 3 ) is maximum with the restriction M 3 C ), from the definition of (3), we can assume N 4. Therefore, the whole spherical surface must be covered by four or more spherical caps under the Minkowski condition. Then, we present the following theorem (SUGIMOTO and TANEMURA, 003). Theorem 1. If the range of angular radius r is 0 < r π cos 1 3 (1/3), then ν =1 C ν is in an extreme state when s 1 = s 13 = s 3 = r. Proof. We first examine the range of r. From the consideration of kissing number in the foregoing Subsection, four or more spherical caps cannot be placed on S under the Minkowski condition when r > π cos 1 (1/3). Therefore, the range of r should be 0 < r π cos 1 (1/3). From Lemma, ν =1 C ν is in an extreme state when s 1 = r (see Fig. 1(a)). Namely, at this time, ν =1 C ν is identical to W. Then, we define T 1 and T as the two cross points of perimeters C 1 and C. Now, from (3), the center M 3 is set on the perimeter of C outside

6 0 T. SUGIMOTO and M. TANEMURA of C 1 and we need to consider the area A(W C 3 ). At this time, s 3 = r. Hence, from Lemma, the area A 3 = A(C C 3 ) is always fixed and maximum for any M 3 C. Here, as shown in Fig. 1(b), let T 3 be the fixed point on C such that d s (T 1, T 3 ) = r and that it is outside C 1. First, we put M 3 at T 3. Next, let us move M 3 along C toward T 1. Then, as shown in Fig. 1(c), we see the relation A(W C 3 ) = A((C 1 C 3 ) (C ) c ) + A 3 holds. Therefore, under the condition M 3 C, A(W C 3 ) is maximum when A((C 1 C 3 ) (C ) c ) attains its maximum. Then, we seek for the position of M 3 when A((C 1 C 3 ) (C ) c ) is maximum. Let G(a) and G(b) be the set (C 1 C 3 ) (C ) c for s 13 = a and b (a < b), respectively. Then, it is obvious that G(a) G(b) holds. Therefore, for r fixed, A((C 1 C 3 ) (C ) c ) is a monotone decreasing function of s 13. Hence, A((C 1 C 3 ) (C ) c ) is maximum when M 3 is put at T 1. Therefore, A(W C 3 ) attains its maximum when M 3 is selected on T 1. Then, 3 s 13 = r holds. Thus, as shown in Fig. 1(d), ν =1 C ν is in an extreme state when s 1 = s 13 = s 3 = r for 0 < r π cos 1 (1/3)..3. Procedure of sequential covering As mentioned in Sec. 1, our problem is to calculate the upper bound of r for the sequential covering, such that N congruent open spherical caps cover the whole spherical surface under the condition that M i is set on the perimeter C i 1, and that each area A(W i 1 C i ) becomes maximum in sequence for i =,..., N 1. Note that, although N spherical caps are needed in our problem, N 1 spherical caps are used in the sequential covering since we want to make the situation that N 1 ν = 1 C ν is in an extreme state. First, before beginning covering, the angular radius r of spherical caps is chosen sufficiently small so that N 1 ν = 1 C ν cannot cover the whole spherical surface in the Minkowski covering. Note that, in the result which will be obtained in the procedure below, the set W N 1 does not cover the whole spherical surface. Algorithmically, the procedure of sequential covering is the followings (SUGIMOTO and TANEMURA, 003): STEP 1: The center M 1 of the first spherical cap C 1 is put at (x, y, z) = (0, 0, 1). Then, from (3), W 1 = C 1 holds. STEP : The center M of C is put at a certain point on the perimeter C 1. As a result, ν =1 C ν is in an extreme state (A(W 1 C ) is maximum with the restriction M C 1 ) since s 1 = r. If (N 1) 3, go to the next step; otherwise the procedure ends. 3 STEP 3: The center M 3 of C 3 is put on one of the cross points of C 1 and C. Then ν =1 C ν is in an extreme state since s 1 = s 13 = s 3 = r. If (N 1) 4, put i = 4 and go to the next step; otherwise the procedure ends. STEP 4: First, the center M i of C i is placed at a certain point on the acceptable part of C i 1 which is outside the other spherical caps. Next, move M i among the acceptable points on C i 1 and compute A(W i 1 C i ) for respective points of M i. Then, M i is fixed at the position where A(W i 1 C i ) attains its maximum (i.e. the set W i is formed on S). Go to STEP 5. STEP 5: If i = N 1, the procedure ends; otherwise put i i + 1 and go to STEP 4. Therefore, our sequential covering satisfies the condition that, sequentially for i i =,..., N 1, each A(W i 1 C i ) is maximum with the restriction M i C i 1 ( ν =1 C ν is in an extreme state).

7 Packing and Minkowski Covering of Congruent Spherical Caps Upper bounds r N and r N 1 We define r N as the upper bound of r which is mentioned at the top of Subsec..3. Next, we define another upper bound of r, r N 1, such that the set N 1 ν = 1 C ν which contains W N cannot cover S under the Minkowski condition. Then r N 1 should be equal to the spherical distance of the largest interval in the uncovered region (W N ) c of S. It is because, when the angular radius r is equal to r N 1, the set N 1 ν = 1 C ν which contains W N can cover S except for a finite number of points or a line segment under our sequential covering. Therefore, N 1 ν = 1 C ν is in an extreme state if and only if at least one endpoint of the interval, which has the above mentioned spherical distance r N 1, comes on the perimeter C N. Further, when there are two or more uncovered points, the spherical distance of any pair of these uncovered points is less or equal to r N 1 since the largest interval is assumed to be r N 1. Then, we can put the center M N of C N at one of the uncoverd points. At this moment, we see that N ν =1 C ν which contains W N 1 covers S without any gap. Then, we notice the fact that r N is equal to r N 1. Therefore, for our problem, it is necessary to know the spherical distance of the largest interval in the uncovered region (W N ) c. In our cases in Sec. 3, it becomes important to consider triangles or quadrangles as the shape of (W N ) c in the final steps of sequential covering. For this purpose, we investigate the farthest pair of points in spherical triangle and quadrangle so that it is useful for later considerations. First, we consider the spherical triangle. Let T 1, T, and T 3 be the points on S and let the spherical triangle T 1 T T 3 be such that the side T 1 T of the triangle T 1 T T 3 is an inferior arc of the great circle determined by T 1 and T : namely the side T 1 T is the geodesic arc and satisfies 0 < d s (T 1, T ) π. Next, we define the point H as the middle point of the geodesic arc T 1 T. Then, we get the position of H as follows: T1 + T H = 1 cos cos T, T ( (( 1 ) ) ) where a bold symbols H, T 1, and T are unit vectors from the origin O to the points H, T 1, and T on the unit sphere, respectively, and where (T 1, T ) is the inner product of vectors T 1 and T. Then, we get the following theorem. Theorem. If the spherical triangle T 1 T T 3 satisfies the condition π/ d s (T 1, H) = d s (H, T ) = d s (T 1, T )/ d s (H, T 3 ) > 0, then the farthest pair of points inside the spherical triangle T 1 T T 3 is the pair of T 1 and T. Proof. Let C be the closed spherical cap with its center at H and with its angular radius as d s (T 1, H). From the condition that d s (T 1, H) d s (H, T 3 ), T 3 is inside C. In order to prove Theorem, we need to consider three cases: (I) d s (T 1, H) = d s (H, T 3 ) = π/ and T 3 C ; (II) d s (T 1, H) = d s (H, T 3 ) < π/ and T 3 C ; (III) d s (T 1, H) π/ and T 3 C. First, for the case (I), let T 1, T, and H be (0, 1, 0), (0, 1, 0) and (0, 0, 1), respectively. Namely, the geodesic arc T 1 T is the half of the great circle. Therefore, the perimeter of C is the equator of unit sphere. Then, all the sides of spherical triangle T 1 T T 3 are the geodesic arcs.

8 04 T. SUGIMOTO and M. TANEMURA Furthermore, the spherical triangle T 1 T T 3 is formed with two sectors T 1 HT 3 and T T 3 H, and is contained in C. Therefore, it is clear that the spherical distance d s (T 1, T ) = π is the largest distance. Next, for the case (II), d s (T 1, T 3 ) is shorter than the length of arc T 1 T 3 of perimeter C and the geodesic arc T 1 T 3 is inside of C. Therefore, the spherical triangle T 1 HT 3 is contained in the sector T 1 HT 3. Hence, it is entirely contained in C. Similarly, we find that the spherical triangle T T 3 H is contained in C. Thus, the closed spherical cap C entirely contains the spherical triangle T 1 T T 3. The proof for the case (III) is the same procedure as in the case (II). Therefore, in the cases (II) and (III), the spherical distance d s (T 1, T ) is the largest distance. Thus, the farthest pair of points in spherical triangle T 1 T T 3 is the pair of T 1 and T. Note that if the condition of Theorem is not satisfied, the claim of Theorem does not hold. For example, when the spherical triangle is an equilateral triangle of side lengths between π/ and π, its height is larger than the sides: namely the farthest pair of points in spherical equilateral triangle is not the pairs of spherical triangular vertices. Next, we consider the case of quadrangle. Let us given the spherical triangle T 1 T T 3 which satisfies the condition of Theorem. Then, we add the point T 4 on S to the opposite side of T 3 against the great circle which passes T 1 and T. By joining T 4 with T 1 and T by the geodesic arcs, respectively, we make the spherical quadrangle T 1 T 4 T T 3 that is formed with two spherical triangles T 1 T T 3 and T 1 T 4 T. Then, we have the following Corollary from Theorem. Corollary. If the spherical quadrangle T 1 T 4 T T 3 satisfies the following conditions ds T1, T ds T3, T4, π ds( T1, T) ds( T1, H)= ds( H, T)= ds H, T3 0, > d T, H d H, T 0, s > 1 s 4 then the farthest pair of points in the spherical quadrangle T 1 T 4 T T 3 is the pair of T 1 and T. Proof. Let C be the closed spherical cap whose center is H and whose angular radius is d s (T 1, H). From Theorem, the spherical triangle T 1 T T 3 is entirely contained in C and its farthest pair is the pair of T 1 and T. Similarly, the spherical triangle T 1 T 4 T is contained in C and its farthest pair is the pair of T 1 and T. Therefore, the spherical quadrangle T 1 T 4 T T 3 is entirely contained in C whose angular diameter is determined by d s (T 1, T ). Thus, the pair of T 1 and T is the farthest pair of points in spherical quadrangle T 1 T 4 T T 3. Now, we consider the meanings of the upper bounds r N and r N 1 in an illustrative example. Figure shows the distinction between two upper bounds r N and r N 1. In Fig.

9 Packing and Minkowski Covering of Congruent Spherical Caps 05 Fig.. Upper bounds r N and r N 1. (a), the shaded region is the covered region W N and the white region is the uncovered region (W N ) c. At this time, the shape of (W N ) c is the quadrangle T 1 T 4 T T 3 on S. We note that the sides of uncovered region (W N ) c are not geodesic arcs but are perimeters of spherical caps. Here, we suppose that the spherical quadrilateral T 1 T 4 T T 3 (the quadrilateral enclosed by dotted-and-dashed segments in Fig. (a)) satisfies the condition of Corollary of Theorem. Then, it is obvious that the quadrangle T 1 T 4 T T 3 (the quadrangle enclosed by solid segments in Fig. (a)) on S is inside the spherical quadrilateral T 1 T 4 T T 3. Therefore, from Corollary of Theorem, we find that the pair of T 1 and T is the farthest pair of points in the spherical quadrilateral T 1 T 4 T T 3 and in the quadrangle T 1 T 4 T T 3 on S.

10 06 T. SUGIMOTO and M. TANEMURA Namely, the spherical distance of the largest interval in (W N ) c is d s (T 1, T ). Then, when the radii r of N 1 spherical caps are all equal to d s (T 1, T ) and the center M N 1 is put on T 1 or T, the set N 1 ν = 1 C ν which contains W N can cover S except for a point (T or T 1 ). Therefore, the upper bound, r N 1, of r such that the set N 1 ν = 1 C ν which contains W N cannot cover S under the Minkowski condition is d s (T 1, T ). At this time, as shown in Fig. (b), if T C N is selected as M N 1, the set N 1 ν = 1 C ν is in an extreme state where T 1 is the unique uncovered point. Therefore, when M N is put at T 1 C N 1, the set N ν =1 C ν which contains W N 1 covers S without any gap. Thus, as shown in Fig. (c), our upper bound r N is equal to d s (T 1, T ). Namely, in Fig., we see the fact that r N = r N 1 = d s (T 1, T ). Here, we note the advantage of using the upper bound r N 1. The value of r N 1 is easier to calculate than r N, since it is better to examine the extreme situation where the spherical surface S cannot be covered by N 1 spherical caps than the situation where S is covered by N caps. Then, at the last stage of the process of covering, we only need to observe the situation where a few uncovered regions remain since our covering is sequential. Moreover, the covering of our problem is finished in fact when N 1 spherical caps cover S except for a finite number of points or a line segment since open spherical caps are considered. In such a case, as shown in the cases below, the position of the center of the N-th spherical cap is almost unique. Although r N 1 and r N are not necessarily coincident, the value of r N 1 will give a strong candidate for r N. 3. Results 3.1. N =, 3, 4, 5 and 6 For cases of N =,..., 6, we find that r N is equal to the upper bound of the range of r for k in Table 1. For N =, r (i.e. r N for N = ) is equal to π. It is because that the radius of the first spherical cap C 1 should be equal to r 1 = π in order C 1 to cover the whole S except for a point. From the STEP 1 as described in Subsec..3, the center M 1 of C 1 is put at the south pole (0, 0, 1). In this case, the north pole (0, 0, 1) is open for r 1 = π. Then it is obvious that we can put M, the center of the second spherical cap C, at the north pole. At this time, S is covered without any gap by C 1 and C which satisfy the Minkowski condition. Thus we see r = r 1 = π. We note that this value π is the upper bound of the range of r for k = 1 as shown in Table 1. For N = 3, r 3 should be equal to the upper bound of the range of r for k =. It is because that, when r = π/3, the set ν =1 C ν under the condition M C 1 can cover S except for a point. First, let M 1 be the south pole as described above. Then, if M is put at the point (sin(π/3), 0, cos(π/3)) = ( 3 /, 0, 1/) on C 1, the unique uncovered point P of S will be ( 3 /, 0, 1/). Then, ν =1 C ν is in an extreme state. Therefore, we get r = π/3, and 3 we can put the center M 3 of C 3 at P. At this time, ν =1 C ν which contains W covers the whole of S under the Minkowski condition. It is obvious that the position of centers is the trisection point of a great circle. Thus the correspondent angular radius r 3 is equal to π/3. For N = 4, r 4 should be equal to the upper bound of the range of r for k = 3. The reason is the following. First, we can assume that M 3 is put on one of the cross points of perimeters C 1 and C under the condition M C 1. It is because that, from Theorem 1 in Subsec.

11 Packing and Minkowski Covering of Congruent Spherical Caps 07 3., ν =1 C ν is in an extreme state when s 1 = s 13 = s 3 = r. If r is equal to the spherical distance between the cross points of C 1 and C, the spherical surface S except for a point is covered by the set W 3. As before, let M 1 be the south pole and let M be at (sinr, 0, cosr). At this time, let us assume that the angular radius r is equal to the upper bound of the range r for k = 3 in Table 1. Thus, when r = π cos 1 (1/3), M 3 is selected as one of the trisection point of C 1 and let it be (( 1/)sinr, ( 3 /)sinr, cosr) = ( /3, / 3, 1/3). Then, it is easy to see that the point P = (( 1/)sinr, ( 3 /)sinr, cosr) = ( /3, / 3, 1/3) is the unique uncovered point on S. We note, at the same time, that the coordinates of M turns out to be ( /3, 0, 1/3). Therefore, we get r 3 = π cos 1 (1/3), and we can put M 4 4 on that point P. As a result, the set ν =1 C ν which contains W 3 can cover the whole of S when r = r 3 and we get finally r 4 = π cos 1 (1/3). Then, we find that the position of centers of these four spherical caps is in accord with the vertices of regular tetrahedron. For N = 5, before deriving r 5, r 4 = π/ is shown first. When r = π/, from Theorem 1, the centers M 1, M, and M 3 are put, for example, at (0, 0, 1), (1, 0, 0), and (0, 1, 0), 3 respectively, in order for ν =1 C ν to be in an extreme state. Then, the uncovered region (W 3 ) c is the spherical equilateral triangle of side-length π/ and vertices (0, 0, 1), ( 1, 0, 0), and (0, 1, 0). Therefore, from the discussion in Subsec..4, we find that the largest spherical distance in (W 3 ) c is equal to π/; namely r 4 = π/. Next, when M 4 is put on the point 4 (0, 0, 1) C 3 or ( 1, 0, 0) C 3 (in this paper, we choose ( 1, 0, 0)), the set ν =1 C ν is in an extreme state. As a result, the set W 4 can cover S except for a line segment which is an inferior great circle connecting (0, 1, 0) and (0, 0, 1) and whose length is π/. Therefore, when M 5 is put at any point on this line segment except for points (0, 1, 0) and (0, 0, 1), we 5 find that ν =1 C ν covers S without gap under the Minkowski condition. Thus, r 5 = r 4 = π/. Namely, r 5 is equal to the upper bound of the range of r for k = 4. For N = 6, we can assume that r 6 is equal to the result of the case N = 5. First, similar to the case N = 5, we put the centers of C 1, C, C 3, and C 4 of radius r = π/ at (0, 0, 1), (1, 0, 0), (0, 1, 0), and ( 1, 0, 0), respectively. If we put M 5 on (0, 0, 1) or (0, 1, 0) (in the 5 following, we assume (0, 1, 0) is chosen as M 5 ), the set ν =1 C ν is in an extreme state and the spherical surface S except for a point, namely the point (0, 0, 1), can be covered by the 6 set W 5. As a result, the set ν =1 C ν containing W 5 covers the whole of S when we put M 6 on (0, 0 1). Thus, our assumption r 5 = r 6 = π/ is confirmed. Then, we find that the position of centers of these spherical caps for N = 6 is in accord with the vertices of regular octahedron. Therefore, if all spherical caps of our covering for N = 6 are replaced by halfcaps (the spherical cap whose angular radius is r/ and which is concentric with that of the original cap), all of those half-caps contact other four half-caps and there is no space for those half-caps to move. 3.. N = 7 From the considerations for the cases N 6 and Subsec..1, we can assume that, for N = 7, the angular radius r should satisfy the inequalities tan 1 < r < π/. The reason is the following. First, the center M 1 of the first spherical cap C 1 is set at the south pole (0, 0, 1) as before. Now, we remind the cases of N 6 in Subsec. 3.1, in order to investigate the position of centers of spherical caps for N = 7. For N =, M is placed on the perimeter C 1. For N = 3, M and M 3 are both on C 1. For N = 4, the centers M, M 3, and M 4, are placed

12 08 T. SUGIMOTO and M. TANEMURA on C 1. For N = 5 and 6, there are four centers of other spherical caps on C 1. Then, from these facts and from Subsec..1, we can consider two cases for N = 7 as follows: four centers M, M 3, M 4, and M 5 are on C 1 ; and five centers M, M 3, M 4, M 5, and M 6 are on C 1. For N = 7, we find that the second case is excluded because of the following reason. When five centers M, M 3, M 4, M 5, and M 6 are placed on C 1 according to our sequential covering, the range of angular radius r must be 0 < r tan 1 from the consideration in Subsec..1. Here, we consider the case that r is equal to tan Then, it is obvious that the spherical distance between M 1 : (0, 0, 1) and any point of the set which is covered by our six spherical caps is smaller than tan 1, while the length of longitude line joining south and north poles is larger than tan 1. Thus, the union of our six spherical caps would leave a big open area (whose size is at least comparable to the area of spherical cap of angular radius π tan 1 ) near the north pole of the unit sphere. Furthermore, the open area near the north pole would become still bigger for 0 < r < tan 1 6. Hence, the set ν =1 C ν cannot cover S at all. Thus, we should exclude the second case. Therefore, we have to consider the first case that four centers M, M 3, M 4, and M 5 of spherical caps are placed on C 1. Then, below, we investigate the range of r under this condition. From the consideration in Subsec..1, when r π/, it is possible to put four spherical caps on the perimeter of a spherical cap. Especially, when the range of r is (tan 1, π/], four spherical caps can be placed on the perimeter of a spherical cap and, at the same time, five spherical caps cannot be placed on the perimeter of a spherical cap. Further, from the setup of our problem, we want to make r the biggest possible. Therefore, we can assume that r is larger than tan 1. On the other hand, if r π/, we cannot cover S by seven spherical caps without breaking the Minkowski condition due to the results of cases N 6. Thus, r should be in the range tan 1 < r < π/. We note that the equality sign does not enter in these inequalities. Let (x, y, z) be the coordinates of cross points where the perimeters of C a (the coordinates of the center: (a 1, a, a 3 )) and C b (the coordinates of the center: (b 1, b, b 3 )) intersect. By solving the simultaneous equations ax+ ay+ az= cos r, 1 3 bx+ by+ bz= cos r, 1 3 x + y + z = 1, ( 4) we will have the coordinates of the cross points. Thus, in the case where the centers of two spherical caps are put respectively at (0, 0, 1) and (sinr, 0, cosr), we get z = cos r, sin r x cos r z = cos r, x + y + z = 1. ( 5) Let the points K 1 = (x 1, y 1, z 1 ) and K = (x, y, z ) be the solutions of simultaneous equation (5). Solving (5), we obtain

13 Packing and Minkowski Covering of Congruent Spherical Caps 09 Fig. 3. The curve of A(W 3 C 4 ) when M 4 is moved on the arc K 4 K of C 3. Here, the arc K 4 K is divided into 100 equal intervals and the area A(W 3 C 4 ) is calculated on 101 end points of the intervals. Note that the curve of r corresponds to the case of r = tan 1. The values of r of other curves are described in the text. Fig. 4. The curve of A((W 4 C 5 ) c ) when M 5 is moved on the arc K 6 K 5 of C 4. The similar computation method as in Fig. 3 is taken. See the legend in Fig. 3. ( ) + cos r cos r 1 cos r 1 cos r 1 ( x1, y1, z1)=,, cos r, 6 sin r sin r + cos r cos r 1 cos r 1 cos r 1 ( x, y, z)=,, cos r. 7 sin r sin r Here, the center M 1 of C 1 is set at the south pole (0, 0, 1) as already mentioned. We put the centers M and M 3 of C and C 3 each, at K 1 and (sinr, 0, cosr), respectively. Note that K 1 and K are the points where the perimeters of C 3 and C 1 intersect. We also note here that the locations of centers of the second and the third spherical caps defined as above are different from the cases N = 4, 5, 6 because of the convenience of our computation. We will use this convention for all cases of N 7 hereafter. Then, taking into account Theorem 1

14 10 T. SUGIMOTO and M. TANEMURA Fig. 5. The kite K 8 K 4 K 6 K 7 on unit sphere. 3 in Subsec.., ν =1 C ν is in an extreme state. At this time, similarly, we calculate the points where the perimeters of C and C 1 intersect. Among these cross points, let K 3 = (x 3, y 3, z 3 ) be the point which is outside C 3. Let K 4 = (x 4, y 4, z 4 ) be one of the cross points between C and C 3 and let it not be the south pole (0, 0, 1). The explicit expressions of coordinates of cross points K 3 and K 4 are shown in Appendix A.1. Let us place the center M 4 at K 4 and move it to K along the arc K 4 K of C 3. We note that, during this movement, the distance s 43 between M 4 and M 3 is equal to the angular radius r and the overlapping area A 43 of C 4 and C 3 is invariant while the area A(W 3 C 4 ) is variable. Then, we want to know the position of M 4 where the area A(W 3 C 4 ) is maximum. Therefore, we calculate the area A(W 3 C 4 ) against the moving point M 4 numerically for several fixed values of r among tan 1 r < π/. In order to use the result later, the computation for r = tan 1 is also performed. The results are shown in Fig. 3. In this figure, the horizontal axis is the position of M 4 on the arc K 4 K of C 3 and the vertical axis is the area A(W 3 C 4 ). In this computation, the arc K 4 K is divided into 100 equal intervals and the area A(W 3 C 4 ) is calculated on 101 end points of the intervals. Hereafter, the similar computations are performed for determination of centers of spherical caps (see Figs. 4, 7, and 10). As a result, for the four values of r in Fig. 3, we find that this curve of A(W 3 C 4 ) is symmetrical at the center of the arc K 4 K (it is evident from the spherical symmetry) and A(W 3 C 4 ) is maximum at both ends. The same fact as above would hold for every values of r in the range tan 1 4 r < π/. Therefore, we expect that ν =1 C ν is in an extreme state if and only if M 4 is put at K or K 4 for tan 1 r < π/. To make sure, we 4 shall check that these points K and K 4 satisfy the condition that ν =1 C ν is in an extreme state after obtaining the exact values of the angular radius r at the last paragraph in this subsection. At the moment, we choose M 4 on the point K. Then, let K 5 = (x 5, y 5, z 5 ) be one of the cross points of perimeters C 4 and C 1, and let it be outside C 3. Further, let K 6 = (x 6, y 6, z 6 ) be one of the cross points of C 4 and C 3, and let it not be the south pole (0, 0, 1). The explicit expressions of cross points K 5 and K 6 are given in Appendix A.1. Next, the center M 5 is put at a certain point on the arc K 6 K 5 of C 4. Then, we need to calculate the area A(W 4 C 5 ) when M 5 is moved on the arc K 6 K 5 of C 4. However, in order to simplify calculation, we pay attention to the area A((W 4 C 5 ) c ). It is because, for i, we find the relation that the area A(W i 1 C i ) is maximum with the restriction

15 Packing and Minkowski Covering of Congruent Spherical Caps 11 i M i C i 1 ( ν =1 C ν is in an extreme state) is the same as that the area A((W i 1 C i ) c ) is maximum with the restriction M i C i 1. Therefore, for i = 5, we calculate A((W 4 C 5 ) c ) against the moving point M 5 numerically for several fixed values of r among tan 1 r < π/. Here, the computation is performed as in the determination of M 4. Figure 4 shows the results. In this figure, the horizontal axis is the position of M 5 on the arc K 6 K 5 of C 4 and the vertical axis is the area A((W 4 C 5 ) c ). As a result, for the four values of r in Fig. 4, the curve of A((W 4 C 5 ) c ) is symmetrical at the center of the arc K 6 K 5 (it is evident from the spherical symmetry) and A((W 4 C 5 ) c ) is maximum when M 5 is placed on K 6 or K 5. The same fact as above would hold for every values of r in the range tan 1 r < π/. Therefore, we expect 5 that ν =1 C ν is in an extreme state if and only if M 5 is put at K 5 or K 6 for tan 1 r < π/. To make sure, we shall check whether the points K 5 and K 6 are such points after obtaining the exact values of the angular radius r at the last paragraph in this subsection like the case of M 4. Here, we choose M 5 on the point K 5. Then, let K 7 = (x 7, y 7, z 7 ) be one of the cross points of the perimeters C 5 and C 4, and let it be outside of C 1. Similarly, let K 8 = (x 8, y 8, z 8 ) be one of the cross points of C 5 and C, and let it be outside of C 1. The exact coordinates of K 7 and K 8 are also given in Appendix A.1. At the time when C 5 is put on the sphere, the shape of the uncovered region (W 5 ) c becomes a kite on the unit sphere (see Fig. 5). We note that the sides of the kite K 8 K 4 K 6 K 7 are not geodesic arcs but are perimeters of spherical caps. Then, we see that there are four centers M, M 3, M 4, and M 5 on the perimeter C 1. From the configuration of the vertices K 8, K 4, K 6 and K 7 of the kite K 8 K 4 K 6 K 7, for the range tan 1 r < π/, we find that there hold always the following relations of the spherical distance between each vertices. d K, K d K, K, s d K, K d K, K, s ( 8 4)= s( 8 7) ( 6 4)= s( 6 7) < < < < d K, K d K, K d K, K, s 6 4 s 8 4 s 6 8 d K, K d K, K d K, K. s 6 4 s 4 7 s 6 8 () 8 Note that, as defined in Subsec.., d s (K i, K j ) is the spherical distance between points K i = (x i, y i, z i ) and K j = (x j, y j, z j ); namely ( 9) 1 s i j i j i j i j d ( K, K ) = cos x x + y y + z z. Refer to Appendix A.1 for the explicit coordinates of K 4, K 6, K 7, and K 8. We produced the relations (8) by using mathematical software. Especially four inequalities in (8) are obtained numerically. So far, we have arranged five spherical caps. Next, let us consider the sixth and seventh spherical caps. As mentioned in Subsec..4, if we take the angular radius of spherical caps to be equal to the largest spherical distance in the kite K 8 K 4 K 6 K 7, the sixth spherical cap C 6

16 1 T. SUGIMOTO and M. TANEMURA Fig. 6. (a) Our sequential covering for N = 7. (b) Our solution of Tammes problem for N = 7. Both viewpoints are (0, 0, 10). In this example, the coordinates of the centers are respectively (0, 0, 1), ( , 0.968, ), ( , 0, ), ( , 0.968, ), ( , , ), ( , , ), and ( , , ). can cover the region except for a point in the kite under the Minkowski condition. Therefore, we find that r 6 is the largest spherical distance in the kite. It is obvious that the kite K 8 K 4 K 6 K 7 on the sphere is inside the spherical quadrilateral K 8 K 4 K 6 K 7 for fixed vertices K 8, K 4, K 6, and K 7 (see Fig. 5). From (8) and the coordinates of K 8, K 4, K 6, and K 7 in Appendix A.1, we find numerically that four vertices of the spherical quadrilateral K 8 K 4 K 6 K 7 satisfies the relations d s (K 6, K 8 ) > d s (K 4, K 7 ), and π/ d s (K 6, H) = d s (H, K 8 ) = d s (K 6, K 8 )/ > d s (H, K 4 ) = d s (H, K 7 ) > 0 for tan 1 r < π/. Note that H is the middle point of the geodesic arc K 6 K 8. Therefore, from Corollary of Theorem, the farthest pair of points in the spherical quadrilateral K 8 K 4 K 6 K 7 is the pair of K 6 and K 8 ; namely d s (K 6, K 8 ) is the largest spherical distance in the kite K 8 K 4 K 6 K 7 and is equal to r 6. Then, from (9) and the coordinates of K 6 and K 8 in Appendix A.1, we have r = d ( K, K )= cos s In addition, we note K 8 C 5. Thus, ν = cos r 8cos r 18cos r cos r+ 8cos r cos r+ 1 ( 10) 6 C ν is in an extreme state (the set ν =1 C ν covers S except for the point K 6 ) if and only if M 6 is put on the point K 8. At this time, K 6 is the unique uncovered point. Equation (10) is solved against r by using mathematical software. As a result, the value of the upper bound for N = 7 is obtained r r = = 1 π cos cos rad. ( 11) 7 Therefore, when M 6 and M 7 are put at K 8 and K 6, respectively, then ν =1 C ν which contains W 6 covers the whole of S (see Fig. 6(a)). Namely, our sequential covering for N = 7 is completed. Here, we check whether the position of points K and K 5 for M 4 and M 5 satisfy the 4 5 condition that ν =1 C ν and ν =1 C ν are in an extreme state, respectively. For that purpose,

17 Packing and Minkowski Covering of Congruent Spherical Caps 13 Fig. 7. The change of A((W 5 C 6 ) c ) when M 6 is moved on the arc K 7 K 8 of C 5. The similar computation method as in Fig. 3 is taken. See the legend in Fig. 3. when r is equal to the value of (11), we examine the position where the area A(W i 1 C i ) is maximum with the restriction M i C i 1 (i = 4 and 5). First, about M 4, we calculate numerically the area A(W 3 C 4 ) against the moving point M 4 for r = cos 1 (1 (4/ 3 )cos(7π/18)). As mentioned above, we checked numerically that 4 A(W 3 C 4 ) is maximum with the restriction M 4 C 3 ( ν =1 C ν is in an extreme) if and only if M 4 is put at K or K 4 for r = cos 1 (1 (4/ 3 )cos(7π/18)). The fact is indicated by the curve of r corresponding to r = cos 1 (1 (4/ 3 )cos (7π/18)) in Fig. 3. Next, about M 5, we use the fact that A(W 4 C 5 ) is maximum with the restriction M 5 C 4 is the same as that A((W 4 C 5 ) c ) is maximum with the restriction M 5 C 4. As mentioned above, we checked numerically that A((W 4 C 5 ) c 5 ) is maximum with the restriction M 5 C 4 ( ν =1 C ν is in an extreme) if and only if M 5 is put at K 5 or K 6 for r = cos 1 (1 (4/ 3 )cos (7π/18)). The result is graphically presented by the curve of r corresponding to r = cos 1 (1 (4/ 3 )cos (7π/18)) in Fig. 4. Thus, our choice for M 4 and M 5 is justified N = 8 It is expected that the solution r 8 for N = 8 should not be larger than r 7. Then, we assume r r 7. Further, at the moment, we assume tan 1 r. If our answer for N = 8 is not obtained by this assumption tan 1 r, we will consider the range of r < tan 1 next. First, as in the case of N = 7, the centers M 1, M, and M 3 are placed at (0, 0, 1), K 1, and (sinr, 0, cosr), 3 respectively. Since, from Theorem 1 in Subsec.., the set ν =1 C ν is in an extreme state when the centers M 1, M, and M 3 satisfy the relations s 1 = s 13 = s 3 = r. Next, from the considerations for determinations of M 4 and M 5 in Subsec. 3., we can assume that the 4 allocation of points K and K 5 for M 4 and M 5 respectively satisfy the condition that ν =1 C ν and C ν each are in an extreme state. To make sure, we shall check that K and K 5 are such points after obtaining the exact values of r 8. Therefore, we use the same positions of the first five spherical caps for the case N = 7. When the fifth spherical cap C 5 is put on the sphere, in the same way as the foregoing Subsection, a quadrilateral kite K 8 K 4 K 6 K 7 on the sphere might be formed as the uncovered region. Hence, the relations (8) hold.

18 14 T. SUGIMOTO and M. TANEMURA Next, we search the position of M 6 which satisfies the condition that the area 6 A(W 5 C 6 ) is maximum with the restriction M 6 C 5 ( ν =1 C ν is in an extreme state). From the restriction M 6 C 5, M 6 is put at a certain point on the arc K 7 K 8 of C 5 and, during M 6 is moved on the arc K 7 K 8 of C 5, the area A(W 5 C 6 ) is calculated. At this time, in order to simplify calculation, we use the relation that the area A(W 5 C 6 ) is maximum with the restriction M 6 C 5 is the same as that the area A((W 5 C 6 ) c ) is maximum with the restriction M 6 C 5. Hence, we calculate the area A((W 5 C 6 ) c ) against the moving point M 6 numerically for several fixed values of r among tan 1 r < r 7. As a result, for the four values of r in Fig. 7, we find that A((W 5 C 6 ) c ) is maximum when M 6 is put on K 8. Figure 7 shows the graph of computational results. In this figure, the horizontal axis is the position of M 6 on the arc K 7 K 8 of C 5 and the vertical axis is the area A((W 5 C 6 ) c ). Then, we expect 6 that ν =1 C ν is in an extreme state if and only if M 6 is put at K 8 in the range tan 1 r < r 7. We shall check that K 8 is such a point after obtaining the exact values of the angular radius r as in the cases of M 4 and M 5. We choose here M 6 on the point K 8. Then, when M 6 is put at K 8, we notice that three cases are possible to be considered for the relation between r and d s (K 8, K 4 ): r = d s (K 8, K 4 ); r < d s (K 8, K 4 ); and r > d s (K 8, K 4 ). First, we consider the case r = d s (K 8, K 4 ). From (9) and the coordinates of K 4 and K 8 in Appendix A.1, we have r = d ( K, K )= cos s cos r+ cos r 9 cos r cos r cos r cos r cos r+ 1 ( 1) Equation (1) is the quartic equation of cosr and can be solved by using the algebraic formula of Ferrari. As a result, we get the following value of the angular radius r. 1 1 r = cos rad Then, we find the following relation among d s (K 8, K 4 ), d s (K 4, K 7 ), and r by using mathematical software. d K, K d K, K r. s = = 4 7 s Therefore, from (8) and (14), we see that the spherical triangle K 8 K 4 K 7 is equilateral and the uncovered region (W 6 ) c is the concave isosceles triangle K 4 K 6 K 7 on S inside the spherical isosceles triangle K 4 K 6 K 7. From the value of (13) and the coordinates of K 4, K 6, and K 7 in Appendix A.1, we find that the spherical isosceles triangle K 4 K 6 K 7 satisfies the relations π/ d s (K 4, H) = d s (H, K 7 ) = d s (K 4, K 7 )/ d s (H, K 6 ) > 0. Note that H is the middle point of the geodesic arc K 4 K 7. Therefore, from Theorem in Subsec..4, the largest spherical distance in (W 6 ) c 7 is d s (K 4, K 7 ). Hence, ν =1 C ν is in an extreme state (the set 7 ν =1 C ν covers S except for a point) if and only if r = d s (K 4, K 7 ) and M 7 is chosen on one of the points K 7 or K 4. Here we put M 7 on K 7 ; namely the spherical surface S is covered by

19 Packing and Minkowski Covering of Congruent Spherical Caps 15 Fig. 8. (a) Our sequential covering for N = 8. (b) Our solution of Tammes problem for N = 8. Both viewpoints are (0, 0, 10). In this example, the coordinates of the centers are respectively (0, 0, 1), (0.1999, , 0.610), (0.9658, 0, 0.610), (0.1999, , 0.610), ( , , 0.610), ( , , ), ( 0.873, , ), and (0.4864, , ). 7 the set ν =1 C ν except for the point K 4. Then, from the relation (14), we find r 7 = cos 1 ( 1/7 + /7). Thus, from the fact K 7 C 6, it is obvious that r 7 is equal to the upper bound r 8 for N = 8. Finally, M 8 is uniquely determined to be the uncovered point K 4, and 8 then the whole of S is covered by ν =1 C ν which contains W 7 (see Fig. 8(a)). Next, we consider the other two cases r < d s (K 8, K 4 ) and r > d s (K 8, K 4 ) by way of precaution. Here, we examine the relations among d s (K 8, K 4 ), d s (K 4, K 7 ), and r, and find numerically that the following relations hold by using mathematical software For tan r < cos +, ds K6, K4 r ds K4, K7 ds K8, K < < ( 15) 1 1 π For cos + r, ds K8, K4 ds K4, K7 r. 7 7 < < < < ( 16) In the case (15), since the angular radius r is smaller than d s (K 8, K 4 ), the uncovered region (W 6 ) c is reduced to the pentagon which is bounded by perimeters of spherical caps and d s (K 4, K 7 ) inside this pentagon is larger than r due to (8) and (15). As was described in Subsec..4, we expect the situation that the set W 7 covers S except for finite points. Therefore, in the uncovered pentagon (W 6 ) c, we take r 7 to be the spherical distance of the largest interval, such that at least one endpoint of the interval comes on the perimeter C 6. However, from the relation (15), we cannot take r to be the largest spherical distance in (W 6 ) c. Further, we find that an uncovered region is left on S when M 7 is put on the uncovered pentagon (W 6 ) c according to our sequential covering. Refer to the considerations of determination for M 7 and the fact that the uncovered region is left when W 7 is formed on S for tan 1 r < cos 1 ( 1/7 + /7) in the following Subsections. In the case (16), on the other hand, since the angular radius r is larger than d s (K 8, K 4 ), the uncovered region (W 6 ) c is reduced to the triangle which is bounded by perimeters of spherical caps and the spherical distance of the largest interval inside this triangle is smaller than r due to (8) and