aa r X i v : . [ m a t h . M G ] J a n A Computer Program for Borsuk’s Conjecture
Chuanming Zong
Abstract.
In 1933, Borsuk proposed the following problem: Can every bounded setin E n be divided into n +1 subsets of smaller diameters? This problem has been stud-ied by many authors, and a lot of partial results have been discovered. In particular,Kahn and Kalai’s counterexamples surprised the mathematical community in 1993.Nevertheless, the problem is still far away from being completely resolved. This pa-per presents a broad review on related subjects and, based on a novel reformulation,introduces a computer proof program to deal with this well-known problem.
1. Borsuk’s Conjecture
Let X be a subset of the n -dimensional Euclidean space E n with diameter d ( X ) = sup x , y ∈ X k x , y k , (1 . k x , y k denotes the Euclidean distance between x and y . As usual, let ∂ ( X ) and int( X )denote the boundary and the interior of X , respectively.In 1933, K. Borsuk [4] studied the continuous maps between metric spaces. As a corollary of hismain result it was shown that, whenever an n -dimensional Euclidean ball is divided into n subsets,at least one of these subsets has the same diameter of the ball. Then, at the end of the paper, heproposed the following problem:
Borsuk’s Problem.
Can every bounded set in E n be divided into n +1 subsets of smaller diameter? Usually, the positive statement of this problem is referred as
Borsuk’s Conjecture , though Borsukhimself only proposed it as an open problem. For convenience, let b ( X ) denote the smallest numbersuch that X can be partitioned into b ( X ) subsets of smaller diameter. Then, Borsuk’s conjecturecan be reformulated as following. Borsuk’s Conjecture.
For every bounded subset X of the n -dimensional Euclidean space E n wehave b ( X ) ≤ n + 1 .
2. Reductions and Positive Results
Definition 1.
Assume that X is a subset of E n . Then we define b X = { λ x + (1 − λ ) x : x i ∈ K, ≤ λ ≤ } . Usually, we call b X the convex hull of X . In particular, we call a compact subset K of E n an n -dimensional convex body if it has nonempty interior and satisfying K = b K .It is obvious that X ⊆ b X and d ( X ) = d ( b X ). Therefore, to solve Borsuk’s problem, it is sufficientto deal with all the convex bodies K . Definition 2.
Assume that C is an n -dimensional convex body and u is a unit vector in E n . Then C has two tangent hyperplanes H and H with norm u . Let d ( C, u ) denote the distance between H and H . If there is a constant c such that d ( C, u ) = c holds for all unit vectors u , we will call C a convex body of constant width. Clearly, balls are convex bodies of constant width. In addition, Reuleaux triangles and Meissnerbodies are particular examples. In fact, convex bodies of constant width in E n is a fascinatingfield of research. There are hundreds of papers on this subject. Many well-known mathematicianssuch as W. Blaschke, M. Fujiwara, H. Lebesgue, K. Reidemeister, L. A. Santal´o and W. S¨uss havemade contribution to this field. Nevertheless, up to now some basic problems about convex bodiesof constant width are still open (see [6, 29]). The next result is useful for Borsuk’s problem. Lemma 1 (P ´ al [31] , Lebesgue [26] ). For every bounded set X in E n there is a convex body C of constant width satisfying both X ⊆ C and d ( X ) = d ( C ) . o v v v v v v u u u TH Figure 1.
A regular triangle can be embedded into a Reuleaux triangle,the Reuleaux triangle can be embedded into a regular hexagon, and theregular hexagon can be divided into three subsets of smaller diameter.
Based on this lemma, to prove Borsuk’s conjecture it is sufficient to deal with all the convexbodies of unit constant width. On the other hand, by continuity argument, one can deduce thatevery convex body of unit constant width can be inscribed into a regular hexagon that the distancebetween the opposite sides is 1. In other words, the hexagon has edge length 2 / √
3. Then, it iseasy to see that the hexagon can be divided into three parts of diameter √ /
2, as shown by Figure1. Thus one obtains the following theorem.
Theorem 1 (Bonnesen and Fenchel [3] ). Every two-dimensional set of diameter d can bedivided into three subsets of diameter at most √ d . Remark 1.
Clearly, the constant √ ≈ . . . . is optimal.In 1945, H. Hadwiger [15] claimed a proof for Borsuk’s conjecture based on Lemma 1. Soonafterwards, he realized that his proof was relied on the assumption that convex bodies of constantwidth have regular boundaries, which is apparently wrong. In fact, he was able to prove thefollowing result. Theorem 2 (Hadwiger [16] ). Every n -dimensional convex body with smooth boundary can bedivided into n + 1 subsets of smaller diameter. Let B denote the n -dimensional unit ball centered at the origin o and assume that K is an n -dimensional convex body with a smooth boundary and o ∈ int( K ). Let x be a boundary pointof K , let u ( x ) denote the unit norm of K at x , and define f ( x ) = u ( x ) − x (2 . ∂ ( K ) to ∂ ( B ). Assume that Y , Y , . . . , Y n +1 are n + 1 subsets of ∂ ( B ) satisfyingboth ∂ ( B ) = n +1 [ i =1 Y i (2 . d ( Y i ) < , i = 1 , , . . . , n + 1 . (2 . Let X , X , . . . , X n +1 be subsets of ∂ ( K ) satisfying f ( X i ) = Y i (2 . Z i to be the convex hull of o ∪ X i . Then, it can be easily shown that d ( Z i ) < d ( K ) (2 . i = 1, 2, . . . , n + 1. Hadwiger’s theorem is proved.In 1947, J. Perkal [33] claimed that Borsuk’s conjecture was correct in E . However, he did notgive the proof idea. In 1955, by modifying Hadwiger’s approach, H. G. Eggleston [8] presenteda detailed proof for the three-dimensional case of the conjecture. Later, different proofs werediscovered by B. Gr¨unbaum [13], A. Happes and P. R´ev´esz [19], and A. Happes [18]. Theorem 3 (Perkal [33] , Eggleston [8] ). Every three-dimensional bounded set can be dividedinto four subsets of smaller diameters.
In fact, the reduction idea can be extend to three dimensions (see Gr¨unbaum [13]). First, everyset of diameter one is a subset of a convex body of diameter one. Second, every convex body ofdiameter one is a subset of a set of constant width one. Third, every set of unit constant widthcan be embedded into a regular octahedron whose opposite facets are distance one apart. Fourth,by cutting off three suitable small pyramids from the octahedron one obtains a suitable truncatedoctahedron, as shown by Figure 2. Finally, the truncated octahedron can be divided into fourpolytopes of diameter less than 0 . . Thus, Theorem 3 is proved.
Figure 2.
The truncated octahedron, which contains the set of unit con-stant width, can be divided into four subsets of diameter smaller than one.
Remark 2.
It is easy to see that the constant 0 . b ( K ) ≤ n + 1 for every n-dimensional smooth convex body K (Theorem 2). In 1955, H. Lenz [27] showed that, on the one hand, b ( K ) ≤ n if K has a smoothboundary but nonconstant width, while on the other hand, b ( K ) ≥ n + 1 for all sets of constantwidth. In 1971, A. S. Riesling [34] showed that b ( K ) ≤ n + 1 for every n-dimensional centrallysymmetric convex body K . In 1971, C. A. Rogers [35] proved that b ( K ) ≤ n + 1 when K isinvariant under the symmetry group of a regular n-dimensional simplex.
3. Counterexamples
In 1993, J. Kahn and G. Kalai [22] made the following counterintuitive discovery.
Theorem 4 (Kahn and Kalai [22] ). For every integer n ≥ , there exists a subset X n of the n -dimensional Euclidean space such that b ( X n ) ≥ . √ n . Remark 3.
Clearly, 1 . √ n is much larger than n + 1 when n is sufficiently large. In particular,whenever n > . √ n > n + 1 , which gives counterexamples to Borsuk’s conjecture in high dimensions.Kahn and Kalai’s counterexamples were indeed surprising. However, the proof idea is verynatural, once understood. In 1981, D. Larman [23] raised the following combinatorial problem: Larman’s Problem:
Let A be a family of subsets of { , , . . . , n } such that every two membersof A overlap in at least k elements. Can A be divided into n subfamilies A , A , ..., A n such thatevery two members of A i overlap in at least k + 1 elements ?Given A as in the statement of Larman’s problem, let ℓ ( A , n, k ) denote the smallest number m for which there exist subfamilies A , A , ..., A m such that A = m [ i =1 A i and every two members of A i overlap in at least k + 1 elements. Then an affirmative answer toLarman’s problem for the integer n implies that ℓ ( A , n, k ) ≤ n .At first glance, it is not easy to notice a connection between Borsuk’s problem and Larman’s.However, they are closely related. Assume in what follows that every member of the family A hascardinality h .Denote by T n the mapping from A to E n defined by T n ( A ) = ( x , x , ..., x n ) (3 . x i = (cid:26) , i A, , i ∈ A. Note that two members A and A ′ of A overlap in exactly j elements if and only if k T n ( A ) − T n ( A ′ ) k = p h − j ) . (3 . T n ( A ′ ) = { T n ( A ) : A ∈ A ′ } for any subfamily A ′ of A , we see that d ( T n ( A ′ )) ≤ p h − k ) (3 . A ′ overlap in exactly k elements. Butthen b ( T n ( A )) = ℓ ( A , n, k ) , (3 . n implies that ℓ ( A , n, k ) ≤ n + 1.Clearly now, if for a given n we can find A and k as above such that ℓ ( A , n, k ) > n + 1 , (3 . n and Borsuk’s problem for dimension n will be “no”. This is just the starting point of Kahn and Kalai’s work.In combinatorics, the structures of finite sets were comparatively well-studied. In 1981, P. Frankland R. M. Wilson [10] proved the following two lemmas. Lemma 2.
Let p be a prime and F be a family of (2 p − -element subsets of { , , ..., n } such that card { F ∩ F ′ } 6 = p − for every two distinct members F, F ′ ∈ F . Then card {F} ≤ (cid:18) np − (cid:19) . Lemma 3.
For p a prime, let m ( p ) be the maximum number of p -element subsets of { , , ..., p } such that no two of them overlap in p elements. Then m ( p ) ≤ (cid:18) pp (cid:19) . Based on Larman’s reformulation of Borsuk’s problem and Frankl and Wilson’s lemmas, J. Kahnand G. Kalai were able to deduce Theorem 4.
Remark 4.
Afterwards, Kahn and Kalai’s breakthrough was simplified by N. Alon [30] andimproved by several authors, in particular by Hinrichs and Richter [20] to n ≥ b ( X ) depending only on the dimension n of the set X areknown. In fact, all of them were discovered before Kahn and Kalai’s counterexamples. In 1961, L.Danzer [7] showed that b ( X ) < s ( n + 2) (2 + √ n − . (3 . b ( X ) ≤ n − + 1 . (3 . b ( X ) ≤ n (4 + log n ) (cid:18) (cid:19) n . (3 .
4. Borsuk’s Problem in Metric Spaces
Let R n be an n -dimensional linear space over real numbers and let σ be a metric defined on R n .Then M n = { R n , σ } , the space together with the metric, is an n -dimensional metric space. It isnatural to consider Borsuk’s problem in general metric spaces.It is well-known in metric geometry that, if σ is a metric defined on R n , then the set C = { x : x ∈ R n , σ ( o , x ) ≤ } (4 . C is a centrally symmetric convex body centered at o and x and y are two points of R n , defining σ ( x , y ) to be the smallest positive number ρ suchthat ρ ( x − y ) ∈ C , one can easily verify that σ ( x , y ) is a metric defined on R n . Therefore, in R n ,there is an one-to-one correspondence between metrics and the centrally symmetric convex bodiescentered at the origin. For example, the Euclidean metric corresponding to the unit ball, the ℓ metric corresponding to a cross polytope, and the ℓ ∞ metric corresponding to a unit cube.In 1957, according to Gr¨unbaum [13], it was proved by E. Shamir that, if M is a metric planesuch that its unit domain is not a parallelogram, then every bounded set can be separated into threesubsets of smaller diameter; if M is a metric plane such that its unit domain is a parallelogram,then every bounded set can be separated into four subsets of smaller diameter .In 1957, H. Hadwiger[17] made the following related conjecture: Hadwiger’s covering conjecture.
In the n -dimensional Euclidean space E n , every convex body K can be covered by n translates of λK , where λ is a positive number satisfying λ < . It is easy to show that λK can be replaced by int( K ). The conjecture is simple sounding and itstwo-dimensional case had been proved by F. W. Levi [28] before the conjecture was made. In fact,he showed that there is a positive number λ < such that every parallelogram P can be covered byfour translates of λP and every other convex domain K can be covered by three translates of λK .Hadwiger’s conjecture has been studied by many authors including K. Bezdek, V. G. Boltjanski,I. T. Gohberg, M. Lassak, H. Martini, C. A. Rogers, V. Soltan and C. Zong. Many partial results are known. For example, any n -dimensional convex body K with smooth boundary can be coveredby n + 1 translates of λK , where λ is a suitable positive number satisfying λ <
1. However, up tonow, no complete solution is known for any other dimension.Assume that X is a bounded set in the metric space M n with metric σ and let b X denote itsclosed convex hull. For convenience, let b σ ( X ) denote the smallest number k such that X canbe divided into k subsets of smaller diameter with respect to σ and let h ( b X ) denote the smallestnumber of translates of λ b X which can cover b X , where λ is any positive number satisfying λ < b σ ( X ) ≤ h ( b X ) (4 . X in R n . In 1965, V. G. Boltyanski and I. T. Gohberg[1] proposed the following two problems related to Borsuk’s conjecture. Problem 1.
Is it true that b σ ( X ) ≤ n holds for all bounded sets X in R n and all metrics σ on R n ? Problem 2.
Assume that C is the centrally symmetric convex body determined by the metric σ in R n . Is it true that b σ ( X ) ≤ h ( C )holds for all bounded sets X in R n ?Clearly, the counterexamples to Borsuk’s conjecture listed in Section 3 did provide negativeanswer to Problem 2 in high dimensions. In 2008, C. Zong [39] discovered a particular set X anda centrally symmetric convex body C in R satisfying both b σ ( X ) = 5 (4 . h ( C ) = 4 , (4 . Theorem 5.
In three-dimensional ℓ p space b ℓ p ( X ) ≤ holds for all bounded sets X . Remark 5.
Clearly, Hadwiger’s conjecture implies Theorem 5. However, the conjecture is stillopen in three dimensions. A computer proof programm was proposed by C. Zong [40] in 2010.The centrally symmetric case was proved by M. Lassak [25] in 1984.Let C p denote the unit domain of the three-dimensional ℓ p space. Let τ ( p ) denote the smallestnumber such that there exists a parallelepiped P satisfying P ⊆ C p ⊆ τ ( p ) P. (4 . τ ( p ) ≤
2, where the equality holds if and only if p = 1. Then theorem 5 canbe deduced by considering two cases with respect to p = 1 and p > Theorem 6. In n -dimensional ℓ p spaces, let C p denote the unit domain, for every boundedcentrally symmetric set X we have b ℓ p ( X ) ≤ h ( C p ) ≤ n if p = 1 , n + 1 if < p < ∞ , n if p = ∞ . Let C be the unit domain of an n -dimensional metric space M n = { R n , σ } . For every boundedcentrally symmetric set X, one can deduce that b σ ( X ) ≤ b σ ( C ) ≤ h ( C ) . (4 . Then theorem 6 can be shown by considering three cases with respect to p = 1, 1 < p < ∞ and p = ∞ .
5. A Reformulation for Borsuk’s Problem
Let B denote the n -dimensional unit ball centered at the origin of E n and let K n denote the spaceof all n -dimensional convex bodies associated with the Hausdorff metric δ H ( · ), where δ H ( K , K ) = min { r : K ⊂ K + rB, K ⊂ K + rB } . (5 . Definition 3.
Let m be a fixed positive integer. For an n -dimensional convex body K we define f m ( K ) to be the smallest positive number θ such that K can be divided into m subsets X , X , . . . , X m satisfying d ( X i ) ≤ θd ( K ) , i = 1 , , . . . , m. Assume that K and K are n -dimensional convex bodies satisfying d ( K ) ≥ , (5 . d ( K ) ≥ , (5 . δ H ( K , K ) ≤ ǫ, (5 . ǫ is a small positive number. Clearly by (5.4) we have d ( K ) − ǫ ≤ d ( K ) ≤ d ( K ) + 2 ǫ (5 . K ⊆ K + ǫB. (5 . K can be divided into m subsets X , X , . . . , X m such that d ( X i ) ≤ θ d ( K ) (5 . i = 1 , , . . . , m , where θ = f m ( K ) <
1. Then, by (5.6) we have K = m [ i =1 ( K ∩ ( X i + ǫB )) (5 . d ( K ∩ ( X i + ǫB )) ≤ d ( X i + ǫB ) ≤ θ d ( K ) + 2 ǫ ≤ θ ( d ( K ) + 2 ǫ ) + 2 ǫ ≤ ( θ + 2 ǫ ) d ( K ) . (5 . | f m ( K ) − f m ( K ) | ≤ ǫ (5 . Lemma 4.
The functional f m ( K ) is continuous on K n . In particular, when d ( K ) ≥ , d ( K ) ≥ and δ H ( K , K ) ≤ ǫ, we have | f m ( K ) − f m ( K ) | ≤ ǫ. In 1958, H. G. Eggleston [9] proved the following result.
Lemma 5.
Assume that K is an n -dimensional convex body of unit constant width. First, itsinsphere S and circumsphere S are concentric. Let r and R be the radii of S and S , respectively,then we have − p n/ (2 n + 2) ≤ r ≤ R ≤ p n/ (2 n + 2) . By Lemma 2 and Lemma 5, to solve Borsuk’s problem in E n , it is sufficient to deal with all theconvex bodies K satisfying B ⊆ K ⊆ r n B, (5 . r n = p n/ (2 n + 2)1 − p n/ (2 n + 2) . (5 . For convenience, we denote the set of all n -dimensional convex bodies satisfying this condition by D n . Clearly it is a compact connected subset of K n .With this preparation, Borsuk’s problem can be reformulated as following. Borsuk’s Problem.
Is there a positive α n < f n +1 ( K ) ≤ α n holds for all K ∈ D n ? Remark 6.
In any metric space, Borsuk’s corresponding problem can be reformulated in a similarway.
6. A Computer Program for Borsuk’s Conjecture
Definition 4.
Let β be a given positive number, and let K , K , · · · , K ̟ ( β ) be ̟ ( β ) convexbodies in D n . If for each K ∈ D n we always can find a corresponding K i satisfying δ H ( K, K i ) ≤ β, we call N = { K , K , · · · , K ̟ ( β ) } a β -net in D n . Remark 7.
Writing B ( K i , β ) = (cid:8) K ∈ K n : δ H ( K, K i ) ≤ β (cid:9) , it is easy to show that N = { K , K , · · · , K ̟ ( β ) } is a β -net in D n if and only if D n ⊆ ̟ ( β ) [ i =1 B ( K i , β ) . Let Z n be the integer lattice in E n , let κ be a small positive number, and let P n denote the setof all lattice polytopes of κ Z n which are elements of D n . Assume that K is a convex body in D n with boundary ∂ ( K ). For each x ∈ ∂ ( K ) we choose g ( x ) to be one of its nearest lattice points of κ Z n and define P = conv [ x ∈ ∂ ( K ) g ( x ) . (6 . δ H ( K, P ) ≤ √ nκ. (6 . κ Z n in c D n form a √ nκ -net in D n , where c D n denotes the setof all lattice polytopes P satisfying(1 − √ nκ ) B ⊆ P ⊆ ( r n + √ nκ ) B. (6 . A Possible Proof ProgramStep 1.
Assume that Borsuk’s conjecture is true in E n . Based on some particular examples, onecan guess a possible constant α n such that f n +1 ( K ) ≤ α n (6 . K ∈ D n . Step 2.
Taking κ = 1 − α n √ n (6 . κ Z n , then for every convex body K in D n there is a lattice polytope P in c D n satisfying δ H ( P, K ) ≤
14 (1 − α n ) . (6 . Step 3.
Enumerate all the lattice polytopes in c D n . The number of the lattice polytopes is huge.The enumeration can be done only by a computer. For example, by deleting the lattice verticessuccessively. Step 4.
For each lattice polytope P , by trying suitable patterns with the help of computer toverify that f n +1 ( P ) ≤ α n . (6 . Conclusion.
By Lemma 4, (6.6) and (6.7), one has f n +1 ( K ) ≤ f n +1 ( P ) + 12 (1 − α n ) ≤ α n + 12 (1 − α n ) = 12 (1 + α n ) < . Then, the theorem will follow.
Example 1. In E , we may try α = 0 . r = ( √
10 + 2) / κ = 0 . Acknowledgements.
This work is supported by the National Natural Science Foundation ofChina (NSFC11921001) and the National Key Research and Development Program of China(2018YFA0704701).
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