On the Helly Number for Hyperplane Transversals to Unit Balls Boris Aronov, Jacob E. Goodmany, Richard Pollackz, Rephael Wengerx
Abstract
We prove some results about the Hadwiger problem of nding the Helly number for line transversals of disjoint unit disks in the plane, and about its higher-dimensional generalization to hyperplane transversals of unit balls in d-dimensional Euclidean space. These include (a) a proof of the fact that the Helly number remains 5 even for arbitrarily large sets of disjoint unit disks|thus correcting a 40-year-old error; (b) a lower bound of d +3 on the Helly number for hyperplane transversals to suitably separated families of unit balls in Rd; and (c) a new proof of Danzer's theorem that the Helly number for unit disks in the plane is 5.
1 Introduction In 1955, Hadwiger [11] posed the problem of determining the smallest number k with the property that if every collection of k members of a family of n k pairwise disjoint unit disks in the plane are met by a line, then all the Polytechnic University, Brooklyn, NY 11201, U.S.A. (
[email protected]). Supported in part by a Sloan Research Fellowship. y City College, City University of New York, New York, NY 10031, U.S.A. (
[email protected]). Supported in part by NSF Grant DMS-9322475, NSA Grant MDA904-98-I-0032, and PSC-CUNY Grant 668416. z Courant Institute of Mathematical Sciences, New York University, New York, NY 10012, U.S.A. (
[email protected]). Supported in part by NSF Grants CCR9711240 and CCR-9732101 and NSA Grant MDA904-98-I-0505. x Ohio State University, Columbus, OH 43210, U.S.A. (
[email protected]). Supported in part by NSA Grant MDA904-97-I-0019.
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disks are met by a line; i.e., the problem of determining the so-called Helly number (if one exists) for line transversals to disjoint unit disks in the plane. He pointed out, by means of an example consisting of disks centered at the vertices of a regular pentagon and almost touching, that k = 4 would not suce. It is this problem, along with its generalization to higher dimensions, that we consider in this paper. Hadwiger's problem was solved two years later in the planar case by Danzer [3], who gave an intricate case analysis to show that k = 5 works. In 1958, in a paper in which he extended Danzer's result to parallelograms, Grunbaum [9] asserted the following strengthened version of Danzer's theorem: \For families of disjoint, congruent circles containing at least six members, T (4) implies T ." (The notation T (k), which has since come to be widely used, means that every k members of the family have a (line) transversal; similarly, \T " means that all do.) He provided only an outline of the proof. It turns out, however, that this assertion is incorrect: even for arbitrarily large families of unit disks, T (5) is necessary to get T . We give an example in Section 2 below. (This corrects an error that has become embedded in the literature during the past forty years; see [4], [5], and [7], for example.) As Grunbaum points out [10], this means that the situation is now much cleaner than before, in the following sense: In [9], he had conjectured that T (5) ) T for any family of disjoint translates of a compact convex set in the plane. This was nally proven in 1989 by Tverberg [14], thus generalizing Danzer's theorem on unit disks. It appeared, however, that the circular disk was exceptional among all planar convex sets in that T (4) was sucient for T for its translates (in a family of sucient size). One sees now that this is not the case after all. In [3], Danzer also considered brie y two variants of the problem in higher dimensions: one for line transversals of pairwise disjoint unit balls (which we will not touch on here), the other for hyperplane transversals. In the latter case he observed that in R3, no Helly number exists for pairwise disjoint unit balls. In his example, however, the balls are permitted to have line transversals; it is this that allows the result to fail, just as the existence of a Helly number for line transversals in the plane would fail if the disks were permitted to have points in common | see [3]. As demonstrated in a number of recent papers [1, 2, 6, 8, 12, 13, 15], the appropriate generalization of \pairwise disjoint," when considering k-dimensional transversals, is \(k?1)separated": A family of convex sets in Rd is (k?1)-separated if no k + 1 of them have a (k?1)-transversal; thus \0-separated" means pairwise disjoint, 2
and in the case where one is interested in the existence of a hyperplane transversal, \(d?2)-separated" means that no d of the sets have a transversal of codimension greater than 1. If one assumes that the balls in question are (d?2)-separated, Danzer's remark does not apply, and there is no a priori reason why a Helly number should not exist. It is this problem, of nding the Helly number for separated unit balls (and, more generally, for separated translates of a single compact convex set) that we set out initially to investigate. While we were unable to nd an upper bound for the Helly number, or even to prove that one exists in dimension greater than 2, we did discover a lower bound. This is presented in Section 3. There, we rst prove a theorem about the strict monotonicity (as a function of d) of the Helly number, if one exists, for hyperplane transversals of (d?2)-separated unit balls in Rd. Starting with the example in Section 2, this then allows us to construct, for each d 2, an example of d + 3 unit balls in Rd, no d having a (d?2)-transversal, for which every d + 2 have a hyperplane transversal, yet all d + 3 do not. This shows, in particular, that the Helly number for hyperplane transversals of (d?2)-separated unit balls in Rd, if one exists, must be at least d + 3. Finally, in Section 4, we present a new (and more transparent) proof of Danzer's planar theorem, based on an analysis of the individual connected components of the space of line transversals to a family of disjoint unit disks. This proof does away with the intricate case analysis of Danzer's original argument, and we remain hopeful that its methods may prove generalizable to higher dimensions, and may eventually lead to an upper bound on the Helly number for hyperplane transversals. In any case, we conjecture that such a Helly number does indeed exist: Conjecture 1 For each d > 2, there is a k(d) such that for hyperplane transversals to families of (d?2)-separated unit balls in Rd, T (k(d)) ) T . For background and additional information on Helly's theorem and geometric transversals, we refer the reader to the excellent survey [4] and to several surveys containing more recent results: [7, 5, 16, 17].
2
T (4)
6) T
The following example shows that T (4) does not suce for T , even in the case of six or more disjoint unit disks. Theorem 1 For each n 6, there exist n pairwise disjoint unit disks in R2 such that any four have a common transversal but some ve do not. 3
1
2
3
5
4
6
Figure 1: Six unit disks.
Proof: We rst describe the construction for n = 6, and then indicate how
to extend it to n > 6. Begin with six unit disks centered at (say) (0; 0), (3; 0), (10; 1), (12; 1). Then translate the fth and sixth up and down (resp.) a distance , for suitably small (:001 will do), and translate the third and fourth up and down (resp.) a distance 2 . Figure 1 shows the new disks (with the size of exaggerated). Notice that the disks are now pairwise disjoint. One checks easily (by calculating the distances from various centers to the lines tangent to various pairs of disks) that every four disks are met by a line; disks 3; 4; 5; 6, for example, are met by a vertical line. On the other hand, since a vertical line is the only common transversal to 3 ; 4; 5; 6, disks 1; 3; 4; 5; 6 have no common transversal. By making successively smaller, we may add an arbitrarily large ( nite) number of additional disks with centers at (say) (?3; 0), (?6; 0), : : : , while maintaining T (4) but avoiding T . 2
3 A lower bound for the hyperplane transversal Helly number in Rd Danzer remarks in [3] that if kd is the smallest k for which T (k) ) T for line transversals of disjoint unit balls in Rd, then trivially kd+1 kd for d 2. For hyperplane transversals to suitably separated unit balls, it turns out that the Helly number is, in fact, strictly monotone:
Theorem 2 If there is is a collection of n (d?2)-separated unit balls in Rd
for which T (k) holds, but not T (k + 1), then there is a collection of n + 1 (d?1)-separated unit balls in Rd+1 for which T (k+1) holds, but not T (k +2).
Proof: Since the hypothesis remains valid if the given balls are enlarged very slightly, we may assume without loss of generality that for any k of them there is a hyperplane cutting the interior of each. 4
n+1
Mn+1 n 2
1
Mn
M1 ?1
M2 ?2
H0
?n
Figure 2: The n + 1 lifted balls. Suppose the given balls, ?1 ; : : : ; ?n , have centers at (x11; : : : ; x1d); : : : ; (xn1 ; : : : ; xnd ), respectively. Embed Rd in Rd+1 as the hyperplane of the rst d coordinates, and call it H0. Let 1 ; : : : ; n+1 be the unit balls centered at (x11; : : : ; x1d; M1); : : : ; (xn1; : : : ; xnd ; Md ); (xn1; : : : ; xnd ; Md+1 ), respectively, where the Mi are chosen in the order M1 ; : : : ; Mn+1 , and each is chosen suciently larger than its predecessor , in a manner to be determined below; see Figure 2. (Notice that the center of ?n is used twice.) We must show that these n + 1 balls satisfy all three conditions. (i) They are (d?1)-separated : Suppose i1 ; : : : ; id+1 are any d +1 of the balls, with i1 < : : : < id+1 . Since ?i1 ; : : : ; ?id have no (d?2)-transversal, a (d?1)- at meeting i1 ; : : : ; id cannot be vertical. Hence if Mid+1 is chosen suciently large, the ball id+1 will lie above all such hyperplanes. (If, for Mid+1 arbitrarily large, there were a hyperplane through i1 ; : : : ; id+1 , by compactness we could nd a sequence of hyperplanes converging to a vertical one, each meeting i1 ; : : : ; id ; hence the limiting one would meet i1 ; : : : ; id as well, which is impossible.) (ii) Any k + 1 of the balls 1 ; : : : ; n+1 have a hyperplane transversal : This is clear if both n and n+1 are included among them: just take the vertical extension of a (d?1)-transversal to the k distinct projections into the hyperplane H0. If not, suppose the balls are i1 ; : : : ; ik+1 with 5
. Let F be a (d?1)- at in the space H0 cutting the interior of each of ?i1 ; : : : ; ?ik , and let H be the hyperplane obtained by extending F vertically. If H is tilted by a suciently small amount, the resulting hyperplane H 0 will still meet all of the lifted balls i1 ; : : : ; ik ; if Mik+1 is large enough, such an H 0 can be chosen to meet ik+1 as well. (iii) Some k +2 of the balls 1; : : : ; n+1 have no hyperplane transversal : Suppose ?i1 ; : : : ; ?ik+1 have no (d?1)-transversal in H0. We claim that i1 ; : : : ; ik+1 ; n+1 will have no d-transversal in Rd+1 if Mn+1 is chosen large enough. If, for each choice of Mn+1 , no matter how large, we can nd a hyperplane through n+1 meeting all of i1 ; : : : ; ik+1 , then (by the compactness of the space of hyperplanes meeting a ball containing all of ?1 ; : : : ; ?n ) there is a sequence H1; H2; : : : of hyperplanes converging to a vertical hyperplane H , such that each Hi meets all of i1 ; : : : ; ik+1 . It follows that H does as well, so that H \ H0 is a (d?1)- at in H0 meeting all of ?i1 ; : : : ; ?ik+1 , contrary to hypothesis. Hence for Mn+1 large enough, i1 ; : : : ; ik+1 ; n+1 will have no d-transversal. 2 i1 < : : : < ik+1
Corollary 1 The Helly number for hyperplane transversals to families of d ? 2 or more (d?2)-separated unit balls in Rd is at least d + 3. Proof: This is immediate from Theorems 1 and 2. 2
4 A new proof of Danzer's theorem on unit disks We begin with a de nition and two lemmas.
De nition 1 A family C1; C2; C3 of pairwise disjoint convex bodies (compact convex sets with interior) is said to pin a line L if L supports each Ci , but no line in the neighborhood of L meets every Ci . Lemma 1 Suppose C1; C2; C3 pin L, with C1 on one side of L and C2; C3 on the other. If xi 2 Ci \ L for each i, then x1 lies between x2 and x3 . Proof: It is clear that if x1 can be separated by a point x 2 L from x2 and
, then L can be rotated around x through an arbitrarily small angle, to a new position in which it is still meets every Ci. 2
x3
In the sequel, we will use the symbol L(C1; : : : ; Cn) for the space of line transversals of the convex sets C1; : : : ; Cn. 6
Lemma 2 Suppose convex bodies C1; : : : ; Cn, n 3, in R2 have a common transversal, and xi 2 Ci for each i. Let Ci (t), t < 1, be the contraction
of Ci about xi by the factor t. If, for every positive t < 1, the convex sets C1(t); : : : ; Cn(t) have no common transversal, then every connected component of L(C1; : : : ; Cn) consists of a single line pinned by (at least) three of the sets Ci .
Proof: If L belongs to a connected component of L(C1; : : : ; Cn), and L
supports two (or fewer) of the sets Ci, then we can rotate L continuously, about an appropriate point, so that at each stage it cuts all the sets Ci in their interiors. If L supports three of the sets Ci but is not pinned by them, the same holds true. The conclusion follows. 2
Theorem 3 If n 5 unit disks in R2 are pairwise disjoint, and every ve have a common transversal, then all n do.
Proof: We begin by a reduction similar to that used by Tverberg in [14]. For each ordered triple (P; Q; R) of distinct points, we de ne d(P jQR) to be the distance from P to the line containing Q and R. If the points P1 ; : : : ; Pn are chosen generically, we can guarantee that no three of the points are collinear, and that d(P jQR) 6= d(S jT U ) unless P = S and fQ; Rg = fT ; U g. We now separate out the following proposition. Proposition 1 If, for some k 3, T (k) ) T (k + 1) for every set of n 5 pairwise disjoint unit disks, then T (k) ) T . Proof: Suppose there is a counterexample to the conclusion, consisting of unit disks with centers at P1 ; : : : ; Pn . So jPiPj j > 2 for all j = 6 i, any k disks
have a common transversal, yet all n lack a common transversal. Enlarge the radius of each disk to 1+ , for suitable > 0. The disks are still pairwise disjoint, any k still have a common transversal, which we can now assume is not tangent to any of the disks, and all n still lack one. (For the last assertion, we may argue as follows: If, for a sequence i ! 0, all n disks have a common transversal, then by compactness there is a subsequence of transversals approaching a limit; this limit would then be a transversal to all of the original disks.) Now perturb the centers of the disks to some generic position. The resulting disks still satisfy the same three conditions. In addition, by the remark above, no three of the centers are collinear and the point-to-line distances de ned above are all distinct. We now shrink the disks uniformly about their centers. The contracted system of disks still has the disjointness property, as well as the property 7
6
L
4
3
2 1
5
Figure 3: A possible placement. that all n disks have no common transversal. At some point, however, the property that any k have a common transversal will cease to hold if we shrink any further, say for disks 1; : : : ; k with centers P1; : : : ; Pk (resp.). By Lemma 2, every component of the transversal space T (1; : : : ; k ) must consist of a single line, pinned by three of the disks. But by the genericity of the centers, this cannot happen simultaneously for more than one triple. Hence T (1; : : : ; k ) consists of just a single line L, pinned by (say) disks 1 ; 2; 3. It follows that if we can show that 1 ; : : : ; k ; i have a common transversal for every i = k + 1; : : : ; n, this transversal must be L, from which it would follow that 1; : : : ; n all have a common transversal, namely L, yielding a contradiction. This shows that it is enough to prove the result for the case n = k + 1, i.e., to show that T (k) ) T (k + 1). 2 Now back to the proof of Theorem 3, which Proposition 1 shows is enough to carry out for the case n = 6. Suppose, then, that the line L, constructed as in Proposition 1, which is the unique common transversal of 1; 2; 3; 4; 5, does not meet 6. Without loss of generality, let us think of L as horizontal, with 1 tangent to L from below and 2 and 3 tangent to L from above; since the point of tangency of 1 lies strictly between the other two by Lemma 1, let us think of them as occurring in the left-to-right order 3 , 1, 2 , as in Figure 3. By Theorem 1 of [15], whose planar case says that the lines meeting a planar family of disjoint convex sets in a particular order form a connected set, the space L(1 ; 2; 3) consists of at most three connected components. We consider three cases, depending on this number. Case 1: L(1; 2; 3) consists of a single component. Since 1, 2, and 3 pin L, this must be fLg alone. Since L(1 ; 2; 3; 4; 6) 6= ;, L must cut 6 , which gives a contradiction. Case 2: L(1 ; 2; 3) has precisely two components, fLg and (say) W 0 . (W 0 is not exactly a pencil, but rather what we will call a wedge of lines, for 8
W L
0
0
2
3
L
1
Figure 4: The case of two components. lack of a better term.) It follows from the fact that the disks are pairwise disjoint that all the transversals in a single connected component meet the disks in the same order (this is the easy direction of the planar case of Theorem 1 of [15]); without loss of generality, we can therefore suppose that every line in W 0 meets 1 ; 2; 3 in precisely that order (as we move along it in the upward direction). Since 2 is strictly to the right of 3, every line in W 0 , directed upward, points to the left. The unique line in W 0 making the smallest angle with L is the internal common tangent to 1 and 2 other than L; call this line L0 . (It is unique, since translating it up or down we would miss 1 or 2, respectively.) All of this is summarized in Figure 4. Since L does not cut 6 , while L(1 ; 2; 3; 4; 6) and L(1; 2; 3, 5 ; 6) are non-empty, it follows that 4 and 5 must meet lines in W 0 . Since 4 and 5 also meet L, and there is no room for them in the two (large) quadrants determined by L and L0 occupied by 1 and 2 (here is one place where the fact that all the disks have the same size is used!), it follows that 4 and 5 each span one or the other of the small quadrants determined by L and L0 , and in particular that they each meet L0, so that L0 is a common transversal to 1 ; 2; 3; 4; 5. This contradicts the assumption that the ve disks have L as their only transversal. Case 3: L(1 ; 2; 3) has precisely three components. This forces the disks 1 , 2 , and 3 to be close to each other, since now each lies between the other two along some line. Figure 5 shows two ways in which this can happen: (a) is a case in which 2 is practically on top of 1, and (b) is a case in which 2 and 3 are symmetrically placed over 1 . (Note that 9
W0 L0 L
W0 L0
W 00 L00
3
2
L
1 (a)
W 00 L00
2
3 1 (b)
Figure 5: The case of three components. (a) can also occur with 2 and 3 reversed; we can ignore this case, by symmetry.) Notice that at most one of the two wedges W 0 and W 00 can contain a line close to L; this happens in (a) (and in the symmetric case where 1 is very close to 3 rather than to 2). Notice also that if 2 is moved to the right, and 3 to the left, the result will only be to make W 0 and W 00 smaller than shown in Figure 5. As in Case 2, all the lines meeting the disks in the order 1 ; 2; 3 point from right to left when directed upward; call the component of L(1; 2; 3) consisting of those lines W 0 . Similarly, those lines meeting the disks in the order 1 ; 3; 2 point from left to right; call that component W 00 . If 6 were cut only by W 0, say, the situation would be exactly the same as in Case 2: the fact that L(1; 2; 3; 4; 6) and L(1 ; 2; 3; 5; 6) are non-empty would immediately imply that 4 and 5 must meet lines in W 0 , hence that they each meet L0 , which would give a contradiction. Therefore we may suppose that 6 is cut by both W 0 and W 00, and that 4 meets only one of the wedges, say W 0, and 5 only the other wedge, W 00. Since 1, 2, and 3 pin L, if we x 1 under L it is only the positions of 2 and 3 that matter. We will show that every possible placement of 2 and 3 leads to a contradiction. For this, it is enough to imagine 2 and 3 very close together, since separating them will only shrink the wedges W 0 and W 00, hence limit the possible placements of the remaining three disks. If 2 is practically on top of 1 , as in Case 3a, the two wedges W 0 and W 00 may be close enough so that 6 can meet both in two distinct ways: either above 2 and 3 , or below 1 , as shown by the disks labeled 6 and 60 (respectively) in Figure 6a. But 4 and 5 must meet L, in addition to meeting W 0 (hence L0) and W 00 (hence L00) respectively. The 10
W0 L0 L
40
W0 0 L
W 00 L00
6 3 5
2 1
4
L
60
5
W 00
L00
6 2
3 1
4
60
(a) Case 3a
(b) Case 3b
Figure 6: Placements of disks 4, 5, and 6. only possibilities are indicated by 4 or 40 (for 4 ) and 5 (for 5 ). In each case, however, it is clear that the hypothesis is violated for some ve disks, in fact in each case one can even nd three disks with no common transversal: 4; 5; 6, 40 ; 1; 6, 4; 3; 60, and 40 ; 2; 60. In Case 3b, where L0 is too steep to permit a placement of 4 to the left of 3 , 6 may still occupy either of the positions shown in Figure 6b. If 6 is over 2 and 3 , 4; 5; 6 have no common transversal; if below 1 (represented by 60 in Figure 6b), either 3; 4; 60 or 2; 5; 60 will have no common transversal. This completes the proof. 2
Remark 1 If one is willing to assume T (6) in place of T (5), then the proof
of T (7), and hence of T , becomes even easier: One sees immediately that the arguments for Cases 1 and 2 remain identical, while in Case 3 the same reduction shows that it is sucient to consider the situation where W 0 (hence L0 ) meets (say) 4 and 5 but not 6, W 00 (hence L00) meets 6 but not 4 or 5, and both W 0 and W 00 meet 7 . But then 1; 2; 3; 4; 6 and 7 have no transversal, and we are done!
References [1] T. Bisztriczky, On separated families of convex bodies. Arch. Math. 54 (1990), 193{199.
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[2] S. Cappell, J. E. Goodman, J. Pach, R. Pollack, M. Sharir, and R. Wenger, Common tangents and common transversals. Adv. in Math. 106 (1994), 198{215. [3] L. Danzer, U ber ein Problem aus der kombinatorischen Geometrie. Arch. Math. (Basel) 8 (1957), 347{351. [4] L. Danzer, B. Gru nbaum, and V. Klee, Helly's theorem and its relatives. In V. Klee, ed., Convexity, Proc. Sympos. Pure Math. 7, pp. 101{180. Amer. Math. Soc., Providence, 1963. [5] J. Eckhoff, Helly, Radon, and Caratheodory type theorems. In P. M. Gruber and J. M. Wills, eds., Handbook of Convex Geometry, pp. 389{448. North-Holland, Amsterdam, 1993. [6] J. E. Goodman and R. Pollack, Hadwiger's transversal theorem in higher dimensions. J. Amer. Math. Soc. 1 (1988), 301{309. [7] J. E. Goodman, R. Pollack, and R. Wenger, Geometric transversal theory. In J. Pach, ed., New Trends in Discrete and Computational Geometry, Algorithms and Combinatorics 10, pp. 163{198. SpringerVerlag, Berlin, 1993. [8] J. E. Goodman, R. Pollack, and R. Wenger, Bounding the number of geometric permutations induced by k-transversals. J. Combin. Theory Ser. A 75 (1996), 187{197. [9] B. Gru nbaum, On common transversals. Arch. Math. (Basel) 9 (1958), 465{469. [10] B. Gru nbaum, Personal communication, 1998. [11] H. Hadwiger, Ungeloste Probleme, No. 7. Elem. Math. 10 (1955), 111. [12] V. Klee, T. Lewis, and B. Von Hohenbalken, Appollonius revisited: Supporting spheres for sundered systems. Discrete Comput. Geom. 18 (1997), 385{395. [13] T. Lewis, B. Von Hohenbalken, and V. Klee, Common supports as xed points. Geom. Dedicata 60 (1996), 277{281. [14] H. Tverberg, Proof of Grunbaum's conjecture on common transversals for translates. Discrete Comput. Geom. 4 (1989), 191{203. 12
[15] R. Wenger, Geometric permutations and connected components. Tech. Rep. TR-90-50, DIMACS, 1990. [16] R. Wenger, Helly-type theorems and geometric transversals. In J. E. Goodman and J. O'Rourke, eds., Handbook of Discrete and Computational Geometry, pp. 63{82. CRC Press, Boca Raton, 1997. [17] R. Wenger, Progress in geometric transversal theory. In B. Chazelle, J. E. Goodman, and R. Pollack, eds., Advances in Discrete and Computational Geometry, pp. 375{393. Amer. Math. Soc., Providence, 1998.
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