THE KEPLER CONJECTURE

arXiv:math/9811078v1 [math.MG] 11 Nov 1998

THE KEPLER CONJECTURE

Thomas C. Hales

1. Introduction and Review Theorem (Kepler conjecture). No packing of √ congruent spheres in three dimensions has density greater than the density π/ 18 of the face-centered cubic packing. The proof provides an explicit list of good local packings. For any combinatorial structure other than those appearing in the face-centered cubic and hexagonal √ close packings, the local density is bounded by a constant strictly less than π/ 18 ≈ 0.740480. For the structures arising in this paper, the bound on score that we obtain is approximately 0.4429, which corresponds to a density of about 0.740476. In the other papers, we have been less explicit about the constant. The optimization techniques that we use are general methods in global optimization including linear programming, branch and bound algorithms, and interval arithmetic. It may be possible to solve other global optimization problems in discrete geometry by similar methods. 1.1. The steps. This theorem comes as the result of this paper and several that precede it [F], [I], [II], [III], [IV], [V]. This paper is a direct continuation of [IV]. It does not depend directly on [III] or [V]. It completes the proof of the fourth step of the Kepler conjecture, as outlined in [I]. At this point in the argument almost all of the discrete geometry has been completed. This paper describes the lengthy computer calculations that have been carried out to finish the proof. Sections 2 and 3 describe the combinatorial classification of all planar maps that are of relevance to this problem. The computer algorithm that is used to classify the planar maps has already been described in [III]. This paper merely describes the parameters that are used to initialize the algorithm and then states the results obtained by the computers. Section 3 eliminates the pathological cases that require individual attention. Section 4 specifies a refined linear programming problem. The detailed structures version – 7/31/98 Research supported in part by the NSF Typeset by AMS-TEX 1

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of exceptional clusters are encoded into this linear program. The refined linear program tends to give remarkably good bounds on the density. The appendix contains a long list of inequalities that are used. Analyzing the results of the linear programs, we find that there is a short list of planar maps that require further attention. These are treated individually in Sections 5, 6, and 7. In most cases this amounts to nothing more than adding a few additional linear inequalities to the linear program. Almost all the hard work needed to prove the Kepler conjecture has been carried out in earlier papers. The surprising result – if there is one to be found – is that the Kepler conjecture can be solved by systems of linear inequalities. In fact, the nonlinearities of the problem can be confined to low dimensions. It is this confinement of the nonlinearities that makes the problem accessible to computerbased interval methods. Many of the interval verifications appearing in the appendix were carried out by S. Ferguson. I express my sincere thanks for his fundamental contributions to this project.

2. Parameters 2.1. The paper [III] gives a general algorithm for generating all planar maps that are relevant to the Kepler conjecture. We refer to it as the classification algorithm. It has been implemented in [H1]. This algorithm depends on various parameters. The purpose of Section 2 is to describe and justify the parameters that enter into the classification of planar maps. The parameters are given. vertexCountMin=1, vertexCountMax=100, scoreTarget=8, squanderTarget=14.8, squanderVertex[p,q]=

p=0 1 2 3 4 5 6

q=0 x x x x 4.139 0.55 6.339

1 2 3 4 5 x x 7.135 10.649 x x 6.95 7.135 x x 8.5 4.756 12.981 x x 3.642 8.334 x x x 3.781 x x x x 11.22 x x x x x x x x x

x=14.8, scoreFace={1, 1, 1, 1, 0, −1.03, −2.06, −3.03, −3.03}, squanderFace={0, 0, 0, 2.378, 4.896, 7.414, 9.932, 10.916},

6 x x x x x x x

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faceCountMax=6, faceCountMaxAtExceptionalVertex=5, vertexAdjustment={x, x, x, 1.4, 1.5}, excludePent= true, vertexCountMin. A lower bound on the number of vertices of height at most 2.51 in the decomposition star is 1. If there are no vertices of height at most 2.51, there is a single standard region, and the score of the truncated cluster is 4πφ0 < 0. vertexCountMax. There are at most 100 vertices of height 2.51 in the decomposition star. The projections of these vertices to the unit sphere are separated by arclengths at least α = 2 arcsin(1/2.51). By a theorem of L. Fejes T´oth, each Delaunay triangle on the unit sphere with vertices at these points has area ≥ 3β − π, with cos β = 1/(1 + sec α). There are at most 4π/(3β − π) < 40 triangles or at most 2 + 40/2 vertices. See [H2] for details. scoreTarget. Any decomposition star scoring less than 8 pt may be discarded. squanderTarget. Decomposition stars squandering more than 14.8 pt > (4πζ − 8) pt may be discarded. squanderVertex[p,q]. This array gives lower bounds on what is squandered (as a multiple of pt) by the standard regions around a vertex that has p triangles, q quadrilaterals, and no exceptional regions. These constants were computed in Section III.5.1. The entries marked x do not occur. scoreFace,squanderFace. A standard region is a polygon or one of the exceptions in Section IV.4.4. Each of these exceptions has a polygonal hull, which is a polygon obtained by removing the internal edges and vertices from the exceptional region. It will be shown in Section 3.3 that the polygonal hull can be treated as a polygonal face in the classification algorithm. This allows us to assume, for the purposes of classification, that all standard regions are polygons. By the results of [IV], each polygon is at most an octagon. The constants in the arrays scoreFace and squanderFace are the bounds on what is scored and squandered by a face as a function of the number of sides. These constants are derived in Theorem IV.4.4. 2.2. Face Counts. faceCountMax is an upper bound on the number of faces at a vertex of the planar map. The bound of six on the number of faces at a vertex is established by the lemma that follows. faceCountMaxAtExceptionalVertex is an upper bound on the number of faces at a vertex at which there is at least one exceptional face. The lemma allows one case with six faces, but we set the parameter at five. This is justified in Section 3.2, where the case with six faces is folded into the classification algorithm in a different way. The proof of the lemma will use the inequalities III.10

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(Group 3, inequalities 4–8) and III.4.1 several times. Given p quasi-regular tetrahedra and q quad clusters at a vertex, subject to the constraint that the dihedral angles at the vertex are at most d, let τ LP(p, q, d) denote a linear programming lower bound on what these p + q standard clusters squander. Similarly, let σLP(p, q, d) be a linear programming upper bound on the score. Lemma. Suppose that there are six or more standard clusters around a vertex. Then either (1) all the clusters are quasi-regular tetrahedra and the vertex has type (6, 0), or (2) there are five quasi-regular tetrahedra, no quad clusters, and one exceptional cluster. Proof. Assume (1) does not occur. If there are no exceptional standard regions at the vertex, then the result follows from the inequalities of III.5.2 and IV.4.5: τ (D∗ ) > t5 + min τ (p, q) ≥ 4.896 pt + 11.22 pt ≥ (4πζ − 8) pt. p+2q>6

Without loss of generality, we assume there is an exceptional region at the vertex. If there are seven or more clusters, then the dihedral angles around the vertex cannot be 2π: 1.153 + 6(0.8638) > 2π. (See III.4.3, and III.10.G1.3.) The lower bound 1.153 on dihedral angles was stated for quad clusters, but the proof was written to apply to exceptional clusters as well. We now assume six standard clusters at the vertex. There are several cases according to the number k of triangular regions at the vertex. (k ≤ 2) If there are at least four nontriangular regions at the vertex, then the dihedral angle around the vertex is at least 4(1.153) + 2(0.8638) > 2π, which is impossible. (k = 3) If there are three nontriangular regions at the vertex, then we squander at least 2t4 + t5 + τ LP(3, 0, 2π − 3(1.153)) > (4πζ − 8) pt. (k = 4) If there are two exceptional regions at the vertex, we squander at least 2t5 + τ LP(4, 0, 2π − 2(1.153)) > (4πζ − 8) pt. If there are two nontriangular regions at the vertex, we have t5 + τ LP(4, 1, 2π − 1.153) > (4πζ − 8) pt. We are left with the case of five triangular regions and one exceptional region. 

THE KEPLER CONJECTURE

2.3.

5

vertexAdjustment.

Let v1 , . . . , vk be vertices of exceptional regions such that no two are adjacent and no two are opposite vertices of a quadrilateral. (We may actually extend the results to the case of opposite vertices of a quadrilateral by using various interval calculations (A.2.3), provided the edge between the opposite vertices does not break the quad cluster into two flat quarters.) Suppose that there are five faces at each vertex vi and that at vi there are ai triangles, with 0 ≤ ai ≤ 4. Let F1 , . . . , Fr be the faces with a vertex in {v1 , . . . , vk }. Assume that Fi is an fi -gon. The vertex adjustments are constants, for 0 ≤ i ≤ 4, such that the (clusters over the) faces Fj around the vi squander at least 4 X j=1

squanderFace[fj ] +

k X

vertexAdjustment[ai ].

i=1

For example, vertexAdjustment[i] = 14.8 pt, for i = 0, 1, 2, so that any decomposition star with one of these vertices squanders at least (4πζ − 8) pt. The constants for i = 0, 1 are derived in Section 2.4, that for i = 2 in Section 2.6, that for i = 3 in Section 2.7, and that for i = 4 in Section 2.8. 2.4. vertexAdjustment[0], vertexAdjustment[1]. Lemma. Consider a vertex that has five standard clusters around it. Assume that at most one of them is a quasi-regular tetrahedron. Then the score of the decomposition star is less than 8 pt. Proof. If there are a quasi-regular tetrahedra, b quad clusters, and c exceptional clusters, then the clusters squander at least c t5 + τ LP(a, b, 2π − c(1.153)). If there is a vertex that is not a corner of any of these clusters, the standard clusters around it squander an additional 0.55 pt (III.5.2). We run the linear programs for each (a, b, c) and find that the bound is always greater than (4πζ − 8) pt. (a, b, c) lower bound (0, 5, 0) 22.27 pt (III.5.2) (0, b, c ≥ 1) t5 + 4t4 ≈ 14.41 pt (1, 4, 0) 17.62 pt (III.5.2) (1, 3, 1) t5 + 12.58 pt (τ LP) (1, 2, 2) 2t5 + 7.53 pt (τ LP) (1, b, c ≥ 3) 3t5 + t4 

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2.5. Some flat quarters. Set ξV = 0.003521, ξΓ = 0.01561, ξΓ′ = 0.00935. They are the penalties that result from erasing an upright quarter of Voronoi type, an upright quarter of compression type, and an upright quarter of compression type with diagonal ≥ 2.57. (See IV.A10 , IV.A11 .) In the next lemma, we score domains        σ ˆ=      

a flat quarter by any of the functions on the given √ Γ, η234 , η456 ≤ 2, √ vor, η234 ≥ 2, vor0 , y4 ≥ 2.6, y1 ≥ 2.2, vor0 ,

vor0 ,

y4 ≥ 2.7, √ η456 ≥ 2.

Lemma. σ ˆ is an upper bound on the functions III.3.10(a)–(f ). That is, each function in III.3.10 is dominated by some choice of σ ˆ. Proof. The only case in doubt is III.3.10(d): ν(Q1 ) + ν(Q2 ) + vorx (S). This is established by the following lemma.  We consider the context (3, 1) that occurs when two upright quarters in the Qsystem lie over a flat quarter. Let (0, v) be the upright diagonal, and assume that (0, v1 , v2 , v3 ) is the flat quarter, with diagonal (v2 , v3 ). Let σ denote the score of the upright quarters and other anchored simplex lying over the flat quarter. Lemma. σ ≤ min(0, vor0 ). Proof. The bound of 0 is established in [II] and [F]. By F.4.7.5, if |v| ≥ 2.69, then the upright quarters satisfy ν < vor0 +0.01(π/2 − dih) so the upright quarters can be erased.* Thus we assume without loss of generality that |v| ≤ 2.69. We have

√ |v| ≥ E(S(2, 2, 2, 2.51, 2.51, 2 2), 2, 2, 2) > 2.6.

*The results of this paper rely on a large number of computer interval calculations. They are listed in an appendix, arranged according to section number.

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If |v1 − v2 | ≤ 2.1, or |v1 − v3 | ≤ 2.1, then

√ |v| ≥ E(S(2, 2, 2, 2.1, 2.51, 2 2), 2, 2, 2) > 2.72,

contrary to assumption. So take |v1 − v2 | ≥ 2.1 and |v1 − v3 | ≥ 2.1. Under these conditions we have the interval calculation ν(Q) < vor0 (Q) where Q is the upright quarter.  Remark. If we have an upright diagonal enclosed over a masked flat quarter in the context (4, 1), then there are 3 upright quarters. By the same argument as in the lemma, the two quarters over the masked flat quarter score ≤ vor0 . The third quarter can be erased with penalty ξV . Define the central vertex v of a flat quarter to be the vertex for which (0, v) is the edge opposite the diagonal. Lemma. µ < vor0 +0.0268 for all flat quarters. If the central vertex has height ≤ 2.17, then µ < vor0 +0.02. Proof. This is an interval calculation.  We measure what is squandered by a flat quarter by τˆ = sol ζpt − σ ˆ. Lemma. (1) Let v be a corner of an exceptional cluster at which the dihedral angle is at most 1.32. Then the vertex v is the central vertex of a flat quarter in the exceptional ˆ = vor0 √ region. Moreover, the flat quarter squanders at least 3.07 pt. If σ (η456 ≥ 2), we may use the stronger constant 3.07 pt + ξV + 2ξΓ′ . Proof. Let S = S(y1 , . . . , y6 ) be the simplex inside the exceptional cluster centered at√v, with y1 = |v|. The inequality dih ≤ 1.32 gives the interval calculation y4 ≤ 2 2, so S is a quarter. The result now follows by interval arithmetic.  Recall that Section IV.4.4 defines an integer n(R) that is equal to the number of sides if the region is a polygon. Lemma. Let R be an exceptional cluster with a dihedral angle ≤ 1.32 at a vertex v. Then R squanders > tn + 1.47 pt, where n = n(R). Proof. In most cases we establish the stronger bound tn + 1.5 pt. In the proof of Theorem IV.4.4, we erase all upright diagonals, except those completely surrounded by anchored simplices. The contribution to tn from the flat quarter Q at v in that proof is D(3, 1) (IV.4.5, IV.5.5.1). Note that ǫτ (Q) = 0 here because there are no deformations. If we replace D(3, 1) with 3.07 pt, then we obtain the bound. Now suppose the upright diagonal is completely surrounded by anchored simplices.

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Analyzing the constants of Section IV.5.11, we see that DLP (n, k)−D(n, k) > 1.5 pt. except when (n, k) = (4, 1). Here we have four anchored simplices around an upright diagonal. Three of them are quarters. We erase and take a penalty. Two possibilities arise. If the upright diagonal is enclosed over the flat quarter, its height is ≥ 2.6 by geometric √ considerations and the top face of the flat quarter has circumradius at least 2. The penalty is 2ξΓ′ + ξV , so the bound holds by the last statement of the previous lemma. If, on the other hand, the upright diagonal is not enclosed over the flat diagonal, the penalty is ξΓ + 2ξV . In this case, we obtain the weaker bound 1.47 pt + tn : 3.07 pt > D(3, 1) + 1.47 pt + ξΓ + 2ξV .  Remark. If there are r nonadjacent vertices with dihedral angles ≤ 1.32, we find that R squanders tn + r(1.47) pt. In fact, in the proof of the lemma, each D(3, 1) is replaced with 3.07 pt. The only questionable case occurs when two or more of the vertices are anchors of the same upright diagonal (a loop). Referring to Section IV.5.11, we have the following observations about various contexts.

(4, 1) can mask only one flat quarter and it is treated in the lemma. (4, 2) can mask only one flat quarter and DLP (4, 2) − D(4, 2) > 1.47 pt. (4, 3) cannot mask any flat quarters. (5, 0) can mask two flat quarters. Erase the five upright quarters, and take a penalty 4ξV + ξΓ . We get D(3, 2) + 2(3.07) pt > t5 + 4ξV + ξΓ + 2(1.47) pt. (5, 1) can mask two flat quarters, and DLP (5, 1) − D(5, 1) > 2(1.47) pt. 2.6. vertexAdjustment[2]. Lemma. Consider a vertex v with five faces around it. If only two are triangles, then the others are quadrilaterals. Proof. The constants 0.55 pt, 4.52 pt used throughout the proof come from III.5.3; tn comes from IV.4.4. (e = 3): First, assume that there are three exceptional regions around vertex v. They must be pentagons (2t5 + t6 > (4πζ − 8) pt). The aggregate of the five faces is an 11-gon (or less). If there is a vertex not on this aggregate, use 3t5 + 0.55 pt >

THE KEPLER CONJECTURE

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(4πζ − 8) pt. So there are at most nine quasi-regular tetrahedra. The score is at most 9 pt + 3s5 < 8 pt. The argument if there is a quad, pentagon, and hexagon is the same (t4 + t6 = 2t5 , s4 + s6 = 2s5 ). (e = 2): Assume next that there are two pentagons and a quadrilateral around the vertex. The aggregate of the two pentagons, quadrilateral, and two triangles is a 10-gon (at most). There must be a vertex not on the aggregate of five faces, for otherwise the score is at most 10 pt + 2s5 < 8 pt. The dihedral angle of one of the pentagons is at most 1.32. For otherwise, τ LP(2, 1, 2π − 2(1.32)) + 2t5 + 0.55 pt > (4πζ − 8) pt. Thus, one of the two pentagons has a flat quarter. Let us show that this case of the lemma follows once we show that any pentagon with a dihedral angle less than 1.32 squanders at least 5.66 pt. If both pentagons have a dihedral angle < 1.32 this is clear 2(5.66) pt + τ LP(2, 1, 2π − 2(1.153)) + 0.55 pt > (4πζ − 8) pt. If there is one pentagon with angle > 1.32, we then have 5.66 pt + τ LP(2, 1, 2π − 1.153 − 1.32) + t5 + 0.55 pt > (4πζ − 8) pt. To obtain the bound 5.66 pt, we argue as follows. If there are five anchored simplices surrounding a vertex, we have the bound by Table IV.5.11. If the configura− tion S+ 4 or S3 occurs, we squander at least 0.4 (IV.3.8, IV.3.7). So if there are any upright diagonals in the pentagon that carry a penalty, we may assume they have four anchors. If there are no penalties, Lemma 2.5 gives 3.07 pt + D(4, 1) > 5.66 pt. We do not need to deal with penalties from S+ 3 in the score of the flat quarter at v because all penalties from a flat quarter are applied to the adjacent subregion (see IV.5.5). The only remaining possibility is four anchored simplices surrounding an upright diagonal. Unless there are three upright quarters, the bound follows from IV.5.11. If there are three upright quarters, erasing gives penalty π0 = 3ξΓ , and 3.07 pt + D(4, 1) − π0 > 5.66 pt. This proves the lemma for two pentagons and a quadrilateral. (e = 1): Assume finally that there is one exceptional region at the vertex. If the exceptional region is a hexagon (or more), we are done t6 + τ LP(2, 2, 2π − 1.153) > (4πζ −8) pt. Assume it is a pentagon. The aggregate of the five regions at the vertex is a 9-gon (at most). If there are no more than 11 quasi-regular tetrahedra outside the aggregate, the score is at most (1 + 2(4.52)) pt+ s5 + σLP(2, 2, 2π − 1.153) < 8 pt (Prop III.5.2). So we may assume that there are at least three vertices not on the aggregate. If the dihedral angle of the exceptional cluster is greater than 1.32, we have τ LP(2, 2, 2π − 1.32) + 3(0.55) pt + t5 > (4πζ − 8) pt; and if it is less than 1.32, we have τ LP(2, 2, 2π − 1.153) + 3(0.55)pt + 1.47 pt + t5 > (4πζ − 8) pt. 

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2.7. vertexAdjustment[3]. When there are three triangles at a vertex with an exceptional, we establish that the vertex adjustment is 1.4 pt. Erase all upright quarters except those in loops. The result follows from Lemma 2.5 and the subsequent remark if the dihedral angle at the vertex of an exceptional region is ≤ 1.32. Assume the angles on the exceptional regions are ≥ 1.32. If there are three triangles, a quad, and an exceptional at the vertex, then we have τ LP(3, 1, 2π − 1.32) > 1.4 pt + t4 . The final case is three triangles and two exceptionals at the vertex. There can be no heptagons and at most one hexagon 2t6 = t5 + t7 > (4πζ − 8) pt. In these calculations, the flat quarters are scored by σ ˆ . By the Inequality 5.5.1, each flat quarter contributes D(3, 1) to tn . The vertex adjustment measures what the flat quarter and surrounding quasi-regular tetrahedra squander in excess of D(3, 1). Most of the calculations now follow directly from the interval calculations A.2.7. A few additional comments are needed if there is a loop masking a flat quarter. If the constant DLP (n, k) − D(n, k) > r(1.4) pt, where r is the greatest possible number of masked flat quarters within the loop, then the result follows. This gives every case except (n, k) = (4, 1), (5, 0). If (n, k) = (4, 1), we erase and take the penalty π0 = 2ξV + ξΓ whenever dih ≥ 1.32 (cf. Remark 4.3). This penalty appears in A.2.7. If (n, k) = (5, 0), there are various possibilities. If there is one masked flat quarter, use DLP (5, 0) − D(5, 0) > 1.4 pt. If there are two masked flat quarters, erase with penalty π0 = 2ξV + ξΓ /2 at each flat quarter. If dih ≤ 1.32, use 3.07 pt − D(3, 1) > 1.4 pt + π0 . If dih ≥ 1.32, use the interval calculations A.2.7 with penalties. 2.8. vertexAdjustment[4]. If there are four triangles and one exceptional at a vertex, we establish the vertex adjustment of 1.5 pt. These calculations are similar to those of Section 2.7, and are based on the interval calculations A.2.8. Details are left to the reader. It is simpler because there is only one exceptional region, and it is not necessary to distinguish the case dih ≥ 1.32 from dih ≤ 1.32. This completes the justification of the vertex adjustments used in the classification algorithm.

3. The Classification 3.1 excludePent=true. This parameter tells the classification algorithm that a particular unusual pentagonal subregion has been treated separately and need not been considered in the classification. The purpose of this section is to treat that subregion. The unusual configuration is a vertex that has only two subregions, a triangle and a pentagon. The aggregate of the two is a quadrilateral.

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Let v1 , v2 , . . . , v5 be the five corners of the pentagonal cluster R, where v1 is a vertex at only two standard clusters, R and a quasi-regular tetrahedron, S = (0, v1 , v2 , v5 ). Since the four edges (v2 , v3 ), (v3 , v4 ), (v4 , v5 ), and (v5 , v2 ) have length less than 2.51, the aggregate of R and S will resemble a quad cluster in many respects. √ Lemma. One of the edges (v1 , v3 ), (v1 , v4 ) has length less than 2 2. Both of the edges have lengths less than 3.02. Also, |v1 | ≥ 2.3. Proof. This is a standard exercise in geometric considerations (F.1, III.2). We deform the figure using pivots to a configuration v2 , . . . , v5 at height 2, and |vi − vj | = 2.51, (i, j) = (2, 3), (3, 4), (4, 5), (5, 2). We scale v1 until |v1 | = 2.51. We √ can also take the distance from v1 to v5 and v2 to be √ 2. If we have |v1 − v3 | ≥ 2 2, then we stretch the edge |v1 − v4 | until |v1 − v3 | = 2√2. The resulting configuration is rigid. Pick coordinates to find that |v1 − v4 | < 2 2. If we have |v1 − v3 | ≥ 2.51, follow a similar procedure to reduce to the rigid configuration |v1 − v3 | = 2.51, to find that |v1 − v4 | < 3.02. The estimate |v1 | ≥ 2.3 is similar.  There are restrictive bounds on the dihedral angles of the simplices (0, v1 , vi , vj ) along the edge (0, v1 ). The quasi-regular tetrahedron has a dihedral angle of at most 1.875 (III.10.1.2). The dihedral angles of the simplices (0, v1 , v2 , v3 ), (0, v1 , v5 , v4 ) adjacent to it are at most 1.63. The dihedral angle of the remaining simplex (0, v1 , v3 , v4 ) is at most 1.51. (These constants come from 1.624 and 1.507 in Section 4.8.) This leads to lower bounds as well. The quasi-regular tetrahedron has a dihedral angle that is at least 2π − 2(1.63) − 1.51 > 1.51. The dihedral angles adjacent to the quasi-regular tetrahedron is at least 2π − 1.63 − 1.51 − 1.875 > 1.26. The remaining dihedral angle is at least 2π − 1.875 − 2(1.63) > 1.14. Lemma. A decomposition star with this configuration squanders > (4πζ − 8) pt. Proof. We will show that the pentagonal exceptional and quasi-regular tetrahedron squander at least 11.16 pt. Let P denote the aggregate cluster formed by these two standard clusters. First we show how the lemma follows from this bound. There are no other exceptionals (11.16 pt+t5 > (4πζ −8) pt), and every vertex not on P has type (5, 0), by Proposition III.5.2. In particular, there are no quad clusters. There are at most 4 quasi-regular tetrahedra at every corner of P , because the τ LP(5, 0, 2π − 1.32) > 6.02 pt. (The 1.32 comes from that fact that P has no flat quarters because v is too short to be enclosed over one [F.1.3], Lemma 2.5.) The only triangulation with these properties is obtained by removing one edge from the icosahedron (Exercise). This implies that there are two opposite corners of P each having four√quasi-regular tetrahedra. Since the diagonals of P have lengths greater than 2 2, the results of Section 2.8 show that these eight quasiregular tetrahedra squander at least 2(1.5) pt. There are two adjacent vertices of

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type (5, 0) whose tetrahedra are distinct from these eight quasi-regular tetrahedra. They give an additional 2(0.55) pt. Now (11.16 + 2(1.5) + 2(0.55)) pt > (4πζ − 8) pt. We prove the bound 11.16 pt. Let P Sij be the simplex (0, v1 , vi , vj ), for (i, j) = (2, 3), (3, 4), (4, 5), (2, 5). We have (4) dih(Sij ) = 2π. Suppose one of the edges √ (v1 , v3 ) or (v1 , v4 ) has length ≥ 2 2. Say (v1 , v3 ). We have τ (S25 ) − 0.2529 dih(S25 ) > −0.3442,

τ0 (S23 ) − 0.2529 dih(S23 ) > −0.1787,

τˆ(S45 ) − 0.2529 dih(S45 ) > −0.2137,

τ0 (S34 ) − 0.2529 dih(S34 ) > −0.1371. The penalty is at most 5ξΓ (Section IV.A10 ). Summing, we find X (4)

τx (Sij ) > 2π(0.2529) − 0.3442 − 0.1787 − 0.2137 − 0.1371 > 11.16 pt + 5ξΓ ,

where τx = τ, τˆ, τ0 as appropriate. √ Now suppose (v1 , v3 ) and (v1 , v4 ) have length ≤ 2 2. If there is an upright diagonal that is not enclosed over either flat quarter, the penalty is at most 3ξΓ + 2ξV . Otherwise, the penalty is smaller: 4ξΓ′ + ξV . We have X (4)

τ (Sij ) > 2π(0.2529) − 0.3442 − 2(0.2137) − 0.1371 > 11.16 pt + 3ξΓ + 2ξV .

 3.2. Six regions. We return to the case of 5 quasi-regular tetrahedra and one exceptional cluster around a vertex that was left unresolved in Section 2.2. When there is an exceptional region at a vertex of degree six, we claim that the exceptional region must be a pentagon. If the exceptional region is a heptagon or more, we squander t7 + τ LP(5, 0, 2π − 1.153) > (4πζ − 8) pt. If the exceptional region is a hexagon, we squander t6 +τ LP(5, 0, 2π−1.153) > t9 . Also, s6 +σLP(5, 0, 2π −1.153) < s9 . The aggregate of the six clusters is a nonagon. The argument of Section IV.4.6 extends to this context to give the bound of 8 pt. Thus, we can assume that the exceptional region is a pentagon. The six standard clusters score at most s5 + σLP(5, 0, 2π − 1.153) < −3.03 pt, and squander at least t5 + τ LP(5, 0, 2π − 1.153) > t8 . The constants −3.03 pt and t8 are the bounds for an octagon in scoreFace and squanderFace. We note that there can be at most one vertex of degree six with an exceptional region. Indeed, if there are two, then they

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13

must both be vertices of the same pentagon: t8 + t5 > (4πζ − 8) pt. Such a vertex on the octagonal aggregate leads to t8 + τ LP(4, 0, 2π − 1.32 − 0.8638) > (4πζ − 8) pt,

or

t8 + 1.47 pt + τ LP(4, 0, 2π − 1.153 − 0.8638) > (4πζ − 8) pt, t8 + τ LP(5, 0, 2π − 1.153) > (4πζ − 8) pt.

We conclude that this case can be fed into the classification algorithm as an octagonal face. We return to this case briefly in Section 3.10, after all relevant planar graphs are classified. 3.3. Nonpolygonal standard regions. In the next section we will turn to the classification of planar maps. Certain combinatorial complications arise if the standard regions are not polygons. Because of these complications it is best to classify the planar maps arising from the polygonal hulls of standard regions. That is, we suppress the internal structure of the standard regions that are not simple polygons. We know from Section IV.4.4 what the possibilities for the standard regions are. If we have, for example, the eight sided figure whose hull is a pentagon, it will be treated as a pentagon in the combinatorial classification. Moreover, the constants t5 and s5 from Section IV.4.4 may be used in the determination of the possibilities, because we have t8 > t5 , s8 + 1 pt < s5 . This allows us to run the combinatorial classification assuming that every face is a polygon. After the classification is complete, we will go back and interpret various faces as one of the cases of Diagram IV.4.4. (See Section 3.6, 3.7, and 3.8.) 3.4. Classification. We have now explained and justified all the parameters in Section 2.1. There are a few additional properties of the planar map that we make use of in the classification. The graph has no loops or multiple joins. Two degree four vertices cannot be adjacent. A cycle of four edges bounds a quadrilateral face, or there is an edge between two opposite corners, or it bounds a region with one vertex (for example, a vertex of type (4, 0) or (2, 1)). A cycle of 3 edges bounds a triangle. (See I.4, I.5.) The classification algorithm constructs all the planar maps of decomposition stars that have the potential of scoring > 8 pt, as implied by the parameters of Section 2.1. As we construct the planar maps face by face, the lower bound on whatever is squandered increases monotonically. It is this property that insures the termination of the algorithm. As soon as a case reaches (4πζ − 8) pt squandered, it may be discarded. Details of the algorithm appear in III. We have made no comprehensive effort to make the list as short as possible, because extraneous cases or redundancies are eliminated quickly in Section 3.5 by

14

THOMAS C. HALES

linear programming. The archive [H1] contains a list of all the possibilities obtained in the classification. In summary, the archive contains the following planar maps (quad) 1749 planar maps (pent) 2459 planar maps (hex) 429 planar maps (hept) 413 planar maps (oct) 44 planar maps 5094 total These planar maps are listed according to the standard region whose hull has the greatest number of sides. So for example, the planar maps in the hexagon list have various triangles, quadrilaterals, pentagons, and at least one hexagon. The planar maps will be referred to as PM(4, n), n = 1, . . . , 1749, PM(5, n), n = 1, . . . , 2459, etc. If we eliminate all these possibilities, the Kepler Conjecture will be established. The quadrilaterals have been treated in full by III, IV. The decomposition stars of the hcp and fcc packings score 8 pt. They are local maxima in the strong sense that any deformation of a quasi-regular tetrahedron or quad cluster from the regular ones appearing in the fcc and hcp brings a drop in score below 8 pt [II]. There is one case that is more difficult than the others. It is the pentahedral prism, treated by S. Ferguson in [V]. The next few sections will eliminate the easiest cases. Section 3.5 eliminates all but 189 cases by linear programming. Section 3.6 treats the nonpolygonal region n(R) = 8 with pentagonal hull. Section 3.7 treats nonpolygonal regions with n(R) = 8 and hexagonal hull. Section 3.8 treats n(R) = 7 with pentagonal hull. After Section 3.8 we may assume that all standard regions are polygons. Section + 3.9 eliminates the configurations S− 3 , S4 of upright quarters. Section 3.10 treats the case of six vertices, from Section 3.2. 3.5. Linear programs. Following the methods outlined in III, we obtain linear programming bounds on the scores of the 5094 planar maps generated by the classification algorithm. In all but 189 cases, the bound is less than 8 pt. There is one other case that was eliminated by branch and bound methods. According to the numbering of planar maps in the archive [H1], we are left with the following cases. (pentagon): {2, 45, 263, 273, 274, 275, 288, 290, 302, . . . , 2321, 2351} (150 cases) (hexagon): {59, 70, 104, 111, 129, 130, 131, 146, 226, 248, 250, 251, 256, 281, 296, 302, 303, 305, 310, 363, 368, 385} (22 cases) (heptagon): {27, 36, 46, 54, 70, 75} (6 cases) (octagon): {1, 2, 4, 5, 6, 7, 8, 12, 14, 15, 16} (11 cases)

THE KEPLER CONJECTURE

15

The linear program uses all the linear inequalities from Sphere Packings III for quasi-regular tetrahedra and quad clusters (Section 10 group 1, group 3, group 5, Prop. 4.1, 4.2, 4.3). It uses the linear relations among solid angles and dihedral angles. It uses the bounds from the arrays scoreFace, squanderFace, and vertexAdjustment. It does not use any of the linear inequalities such as III.10 group 5, or III.A.6 that depend on special conditions being met. It does not use any of the inequalities that assume the quad cluster has a special structure. It does not use, for example, the inequalities for the octahedral case of the quad cluster in III.A.3. As with all these calculations, code and details can be found in [H1]. If P the linear program associated with a particular planar map is expressed as max σi such that A x ≤ b, we can determine bounds on the variables xj as P follows. Define a new linear program max c x such that A x ≤ b, and σi ≥ 8 pt. If c x = xj or −xj , or in fact, any linear combination of variables, then solving the linear program gives a bound on xj or any linear combination c x. The bounds on xj obtained by this method will be called the LP-bounds on a variable. We write LP-min(xj ) ≤ xj ≤ LP-max(xj ) for the linear programming bounds that follow in this manner. The LP-bounds depend on the system of linear equations A x ≤ b that we use, although this is not indicated in the notation. 3.6. A pentagonal hull with n = 8. The next few sections treat the nonpolygonal standard regions illustrated in Diagram IV.4.4. In this standard cluster, the aggregate of the octagonal cluster and the quasi-regular tetrahedron has a pentagonal hull. Let P denote this aggregated cluster. We have bounds pt + s8 on what is scored and t8 on what is squandered. There is no other exceptional cluster t8 + t5 > (4πζ − 8) pt. We add the inequalities τ > t8 and σ < pt + s8 to the exceptional face. With these changes we rerun the linear programs for the 149 remaining pentagonal planar maps. This time through, all but one case has a bound less than 8 pt. The one case, PM(5, 575), can be described at the planar map with 15 triangles obtained by deleted all five triangles at a vertex from the icosahedral planar map. The property of this triangulation that we exploit is that it has three triangles at every vertex of the pentagonal region. Let v be the vertex of the pentagonal region that is a corner of the quasi-regular tetrahedron in the aggregate P . The four quasi-regular tetrahedra at v give t8 + τ LP(4, 0, 2π − 2(1.153)) > (4πζ − 8) pt. 3.7. n = 8, hexagonal hull. We treat the two cases from Diagram IV.4.4 that have a hexagonal hull. One can be described as a hexagonal cluster with an enclosed vertex that has height at most 2.51 and distance at least 2.51 from each corner of the hexagon. The other is described as a hexagonal cluster with an enclosed vertex of height at most 2.51,

16

THOMAS C. HALES

but this time with distance less than 2.51 from one of the corners of the hexagonal cluster. The argument for the case n = 8 with hexagonal hull is similar to the argument of Section 3.7. Add the inequalities τ (R) > t8 and σ(R) < s8 for each hexagonal region. Rerunning the 20 remaining hexagonal cases with these changes brings all but two planar map under 8 pt. The planar maps that remain are PM(6, 59) and PM(6, 131). The planar maps are shown. Diagram 3.7

59

131

We branch and bound on the 8 shaded regions of each case shown in the Diagram. See Section III.A for a description of the how the branch-and-bound method is implemented and for a list of the inequalities. All 28 cases for PM(6, 59) and PM(6, 131) are under 8 pt. 3.8. n = 7, pentagonal hull. We treat the two cases illustrated in Diagram IV.4.4 that have a pentagonal hull. One can be described as a pentagon with an enclosed vertex that has height at most 2.51 and distance at least 2.51 from each corner of the pentagon. The other is described as a pentagon with an enclosed vertex of height at most 2.51, but this time with distance less than 2.51 from one of the corners of the pentagon. In discussing various maps, we let vi be the vertices of the clusters, and we set yi = |vi | and yij = |vi − vj |. If F is a standard region, subregion, cluster, or subcluster, we let dihF,i denote the dihedral angle of F along (0, vi ). The subscript F is dropped, when there is no great danger of ambiguity. Add the equations τ (R) > t7 , σ(R) < s7 for each pentagonal face. There is no other exceptional region, because t5 + t7 > (4πζ − 8) pt. With these changes, of the 150 pentagonal planar maps of Section 3.5, all but one of the linear programs give a bound under 8 pt. The case PM(5, 575) that remains is the one that occurred in

THE KEPLER CONJECTURE

17

Section 3.6. It is the planar map obtained by removing the five triangles around a vertex from an icosahedron. We treat the case PM(5, 575). Let v12 be the vertex enclosed over the pentagon. We let v1 , . . . , v5 be the five corners of the pentagon. Break the pentagon into five simplices along (0, v12 ): Si = (0, v12 , vi , vi+1 ). We have LP-max(yi ) ≤ 2.168, and LP-max(dihF,i ) ≤ 2.89, for i = 1, 2, 3, 4, 5. In particular, the pentagonal region is convex. An LP-lower bound on the score of the pentagonal region is −0.2345. Since −0.4339 is less than this the lower bound, the configuration S− 3 does not occur. Similarly, since −0.25 is less than the lower bound, S+ 4 does not occur (IV.3.7, IV.3.8). Suppose that there is a loop in context (4, 2). Again the score is less than the LP-lower bound, showing that this does not occur: s7 + ZLP (4, 2) − Z(4, 2) < −0.2345. The constants come from Table IV.5.11 and Theorem IV.4.4. The LP-upper bound on what is squandered by the subregion is 0.644. If we branch and bound on the triangular faces, this upper bound can be improved to 0.6079. If there is a loop other than (4, 2) and (4, 1), we find t7 + DLP (n, k) − D(n, k) > 0.644. We conclude that all loops have context (4, 1). Case 1. The vertex v12 has distance at least 2.51 from the five corners of the pentagon. The penalty to switch the pentagon to a pure vor0 score is at most 5ξΓ (see IV.A10 ). There cannot be two flat quarters because then √ √ |v12 | > E(S(2, 2, 2, 2.51, 2 2, 2 2), 2.51, 2.51, 2.51) > 2.51. √ (Case 1-a) Suppose there is one flat quarter, |v1 − v4 | ≤ 2 2. There is a lower bound of 1.2 on the dihedral angles of the simplices (0, v12 , vi , vi+1 ). This is obtained as follows. The proof relies on the convexity of the quadrilateral region. We leave it to the reader to verify that the following pivots can be made to preserve convexity. Disregard all vertices except v1 , v2 , v3 , v4 , v12 . We give the argument that dih(0, v12 , v1 , v4 ) > 1.2. The others are similar. Disregard the length |v1 − v4 |. We show that sd := dih(0, v12 , v1 , v2 ) + dih(0, v12 , v2 , v3 )+ = dih(0, v12 , v3 , v4 ) < 2π − 1.2. Lift v12 so |v12 | = 2.51. Maximize sd by taking |v1 − v2 | = |v2 − v3 | = |v3 − v4 | = 2.51. Fixing v3 and v4 , pivot v1 around (0, v) toward v4 , dragging v2 toward v

18

THOMAS C. HALES

until |v2 − v12 | = 2.51. Similarly, we obtain |v3 − v12 | = 2.51. We now have sd ≤ 3(1.63) < 2π − 1.2, by an interval calculation. Return to the original figure and move v12 without increasing |v12 | until each simplex (0, v12 , vi , vi+1 ) has an edge (v12 , vj ) of length 2.51. Interval calculations show that the four simplices around v12 squander 2π(0.2529) − 3(0.1376) − 0.12 > (4πζ − 8) pt + 5ξΓ . (Case 1-b) Assume there are no flat quarters. A linear programming bound on P the perimeter |vi − vi+1 | is 1.407. We have arc(2, 2, x)′′ = 2x/(16 − x2 )3/2 > 0. The arclength of the perimeter is therefore at most 2 arc(2, 2, 2.51) + 2 arc(2, 2, 2) + arc(2, 2, 2.387) < 2π. There is a well-defined interior of the spherical pentagon, a component of area < 2π. If we deform by decreasing the perimeter, the component of area < 2π does not change. Disregard all vertices but v1 , . . . , v5 , v12 . If a vertex vi satisfies |vi − v12 | > 2.51, deform vi as in Section IV.4.9 until |vi − vi−1 | = |vi − vi+1 | = 2, or |vi − v12 | = 2.51. If at any time, four of the edges realize the bound |vi − vi+1 | = 2, we have reached an impossible situation, because it leads to the contradiction (see A.3.8) 2π =

(5) X

dih < 1.51 + 4(1.16).

(This inequality relies on the observation, which we leave to the reader, that in any such assembly, pivots can by applied to bring |v12 − vi | = 2.51 for at least one edge of each of the five simplices.) The vertex v12 may be moved without increasing |v12 | so that eventually by these deformations (and reindexing if necessary) we have |v12 − vi | = 2.51, i = 1, 3, 4. (If we have i = 1, 2, 3, the two dihedral angles along (0, v2 ) satisfy < 2(1.51) < π, so the deformations can continue.) There are two cases. In both cases |vi − v12 | = 2.51, for i = 1, 3, 4. (a) (b)

|v12 − v2 | = |v12 − v5 | = 2.51,

|v12 − v2 | = 2.51,

|v4 − v5 | = |v5 − v1 | = 2,

Case (a) follows from interval calculations X

τ0 ≥ 2π(0.2529) − 5(0.1453) > 0.644 + 7ξΓ .

In case (b), we have again 2π(0.2529) − 5(0.1453).

THE KEPLER CONJECTURE

19

In this interval calculation we have assumed that |v12 − v5 | < 3.488. Otherwise, setting S = (v12 , v4 , v5 , v1 ), we have ∆(S) < ∆(3.4882 , 4, 4, 8, 2.512, 2.512 ) < 0, √ and the simplex does not exist. (|v4 − v1 | ≥ 2 2 because there are no flat quarters.) This completes case 1. Case 2. The vertex v12 has distance at most 2.51 from the vertex v1 and distance at least 2.51 from the others. Let (0, v13 ) be the upright diagonal of a loop (4, 1). The vertices of the loop are not v2 , v3 , v4 , v5 with v12 enclosed over (0, v2 , v5 , v13 ) by Lemma IV.3.6. The vertices of the loop are not v2 , v3 , v4 , v5 with v12 enclosed over (0, v1 , v2 , v5 ) because this would lead to a contradiction y12 ≥ E(S(2, 2, 2, 2.51, 2.51, 3.2), 2.51, 2.51, 2) > 2.51, or y12 ≥ E(S(2, 2, 2, 2.51, 2.51, 3.2), 2, 2.51, 2) > 2.51. We get a contradiction for the same reasons unless (v1 , v12 ) is an edge of some upright quarter of every loop of type (4, 1). We consider two cases. (2-a) There is a flat quarter along an edge other than (v1 , v12 ). (That is, the central vertex is v2 , v3 , v4 , or v5 .) (2-b) Every flat quarter has central vertex v1 . Case 2-a. We erase all upright quarters including those in loops, taking penalties as required. There cannot be two flat quarters by geometric considerations √ √ E(S(2, 2, 2, 2 2, 2 2, 2.51), 2.51, 2.51, 2) > 2.51 √ √ E(S(2, 2, 2, 2 2, 2 2, 2.51), 2, 2.51, 2.51) > 2.51 The penalty is at most 7ξΓ . We show that the region (with upright quarters erased) squanders > 7ξΓ + 0.644. We assume that the central vertex is v2 (case 2-a-i) or v3 (case 2-a-ii). In case 2-a-i, we have three types of simplices around v12 , characterized by the bounds on their edge lengths. Let (0, v12 , v1 , v5 ) have type A, (0, v12 , v5 , v4 ) and (0, v12 , v4 , v3 ) have type B, and let (0, v12 , v3 , v1 ) have type C. In case 2-a-ii there are also three types. Let (0, v12 , v1 , v2 ) and (0, v12 , v1 , v5 ) have type A, (0, v12 , v5 , v4 ) type B, and (0, v12 , v2 , v4 ) type D. Upper bounds on the dihedral angles along the edge (0, v12 ) are given in A.3.8. These upper bounds come as a result of a pivot argument similar to that establishing the bound 1.2 in Case 1-a. These upper bounds imply the following lower bounds. In case 2-a-i, dih > 1.33 (A), dih > 1.21 (B), dih > 1.63 (C),

20

THOMAS C. HALES

and in case 2-a-ii, dih > 1.37 (A), dih > 1.25 (B), dih > 1.51 (D), In every case the dihedral angle is at least 1.21. In case 2-a-i, the interval calculations of A.3.8 lead to a lower bound on what is squandered by the four simplices around (0, v12 ). Again, we move v12 without decreasing the score until each simplex (0, v12 , vi , vi+1 ) has an edge satisfying |v12 − vj | ≤ 2.51. Interval calculations give X (4)

τ0 > 2π(0.2529) − 0.2391 − 2(0.1376) − 0.266 > 0.808.

In case 2-a-ii, we have X (4)

τ0 > 2π(0.2529) − 2(0.2391) − 0.1376 − 0.12 > 0.853.

So we squander more than 7ξΓ + 0.644, as claimed. Case 2-b. We now assume that there are no flat quarters with central vertex v2 , . . . , v5 . We claim that v12 is not enclosed over (0, v1 , v2 , v3 ) or (0, v1 , v5 , v4 ). In fact, if v12 is enclosed over (0, v1 , v2 , v3 ), then we reach the contradiction π < dih(0, v12 , v1 , v2 ) + dih(0, v12 , v2 , v3 ) < 1.63 + 1.51.

We claim that v12 is not enclosed over (0, v5 , v1 , v2 ). Let S1 = (0, v12 , v1 , v2 ), and S2 = (0, v12 , v1 , v5 ). We have the linear programming bound y4 (S1 ) + y4 (S2 ) = |v1 − v2 | + |v1 − v5 | < 4.804. An interval calculation gives (A.3.8.2-b) X (2)

dih(Si ) ≤

X (2)

(dih(Si ) + 0.5(0.4804/2 − y4 (Si ))) < π.

So v12 is not enclosed over (0, v1 , v2 , v5 ). Erase all upright quarters, taking penalties as required. Replace all flat quarters with vor0 -scoring taking penalties as required. (Any flat quarters has v1 as its central vertex.) We move v12 keeping |v12 | fixed and not decreasing |v12 − v1 |. The only effect this has on the score comes through the quoins along (0, v1 , v12 ). Stretching |v12 − v1 | shrinks the quoins and increases the score. (The sign of the derivative of the quoin with respect to the top edge is computed in the proof of Lemma IV.4.9.)

THE KEPLER CONJECTURE

21

If we stretch |v12 − v1 | to length 2.51, we are done by case 1 and case 2-a. (If deformations produce a flat quarter, use case 2-a, otherwise use case 1.) By the claims, we can eventually arrange (reindexing if necessary) so that (i)

|v12 − v3 | = |v12 − v4 | = 2.51,

or

(ii) |v12 − v3 | = |v12 − v5 | = 2.51. We combine this with the deformations of Section IV.4 so that in case (i) we may also assume that if |v5 − v12 | > 2.51, then |v4 − v5 | = |v5 − v1 | = 2 and that if |v2 − v12 | > 2.51, then |v1 − v2 | = |v2 − v3 | = 2. In case (ii) we may also assume that if |v4 − v12 | > 2.51, then |v3 − v4 | = |v4 − v5 | = 2 and that if |v2 − v12 | > 2.51, then |v1 − v2 | = |v2 − v3 | = 2. Break the pentagon into subregions by cutting along the edges (v12 , vi ) that satisfy |v12 − vi | ≤ 2.51. So for example in case (i), we cut along (v12 , v3 ), (v12 , v4 ), (v12 , v1 ), and possibly along (v12 , v2 ) and (v12 , v5 ). This breaks the pentagon into triangular and quadrilateral regions. In case (ii), if |v4 − v12 | > 2.51, then the argument used in Case 1 to show that |v4 − v12 | < 3.488 applies here as well. In case (i) or (ii), if |v12 − v2 | > 2.51, then for similar reasons, we may assume ∆(|v12 − v2 |2 , 4, 4, 8, 2.512, |v12 − v1 |2 ) ≥ 0. This justifies the hypotheses for the interval calculations in A.3.8. We conclude that X τ0 ≥ 2π(0.2529) − 3(0.1453) − 2(0.2391) > 0.6749.

Recall that the LP-upper bound on what is squandered is 0.6079. If the penalty is less than 0.067 = 0.6749 − 0.6079, we are done.

We have ruled out the existence of all loops except (4, 1). Note that a flat quarter with central vertex v1 gives penalty at most 0.02 by a lemma of Section 2.5. If there is at most one such a flat quarter and at most one loop, we are done: 3ξΓ + 0.02 < 0.067. Assume there are two loops of context (4, 1). They both lie along the edge (v1 , v12 ), which precludes any unmasked flat quarters. If one of the upright diagonals has height ≥ 2.696, then the penalty is at most 3ξΓ +3ξV < 0.067. Assume both heights are at most 2.696. The total dihedral angle of the exceptional cluster along (0, v1 ) is at least four times that of one of the flat quarters along (0, v1 ), or 4(0.74) by an interval calculation. This is contrary to the linear programming upper bound (0.289) on the dihedral angle along (0, v1 ). This completes Case 2. This shows that heptagons with pentagonal hulls do not occur.

22

THOMAS C. HALES

+ 3.9. S− 3 , S4 .

We use the bounds on the score and on what is squandered from Sections IV.3.7, − IV.3.8: S+ 4 (σ < −0.25, τ > −0.4), S3 (σ < −0.4339, τ > 0.5606). If we have the − configuration S3 , there is only one exceptional region (0.5606 + t5 > (4πζ − 8) pt). The configuration with six standard regions around a vertex from Section 3.2 does not occur because the five quasi-regular tetrahedra in the configuration squander > 6 pt, giving 6 pt + 0.5606 > (4πζ − 8) pt. We add the inequalities τ > 0.5606 and σ < −0.4339 to the exceptional region. All 189 linear programming bounds are under 8 pt when these changes are made. The configuration S+ 4 requires more work. Add the inequalities σ < −0.25, τ > 0.4 at the exceptional regions. The configuration S+ 4 does not appear in a pentagon. Rerun the other 22 + 6 + 11 linear programs of Section 3.5 with these changes. All but 7 cases are under 8 pt. They are PM(6, 59), PM(6, 70), PM(6, 131), PM(7, 36), PM(7, 75), PM(8, 8), PM(8, 14). The following argument eliminates these 7 cases. Let v1 and v2 be the corners of S+ 4 that are on the large gap. For the remaining 7 cases we solve the linear programming problem maximizing |v1 | + |v2 | subject to all the usual constraints, together with the additional constraint that the score of the decomposition star is at least (4πζ − 8) pt. Since we do not know how S+ 4 is situated with respect to the planar map there are several cases involved here. We find that |v1 | + |v2 | < 4.6 in every case. The LP-lower bound on the score of the exceptional cluster is −0.31547. We solve the linear programming problem involving just the four quarters of S+ 4 maximizing the dihedral angle α of the gap subject to the constraint that the score of the four quarters is at least −0.31547. The solution gives α < 1.743. This is inconsistent with an interval calculation, which asserts α > 1.78. Thus, the decomposition star scores less than 8 pt. 3.10. Type (n, k) = (5, 1). We return briefly to the case of six clusters around a vertex discussed in Section 3.2. In the classification they were aggregated into an octahedron. We take each of the remaining six cases with an octahedron, and replace the octahedron with a pentagon and six triangles around a new vertex. There are eight ways of doing this. All eight ways in each of the six cases gives an LP bound under 8 pt. This completes the case from Section 3.2. Now turn to the decomposition stars with an upright diagonal with five anchors. Five quarters around a common upright diagonal in a pentagonal region can certainly occur. We claim that any other upright diagonal with five anchors leads to a decomposition star that squanders ≥ (4πζ − 8) pt. In fact, the only other possible context is (5, 1), and there is squanders DLP (5, 1) > 0.5941, by IV.5.11. It scores < −0.376.

THE KEPLER CONJECTURE

23

If this appears in an octagon, we squander > DLP (5, 1) + D(5, 1) > (4πζ − 8) pt. If this appears in a heptagon, we squander > DLP (5, 1) + D(4, 1) + 0.55 pt > (4πζ − 8) pt, because there must be a vertex that is not a corner of the heptagonal cluster. It cannot appear on a pentagon. If it appears on a hexagonal region, it squanders DLP (5, 1)+D(3, 1) > 0.65995 and scores ZLP (5, 1) < −0.376. Add these constraints to the linear program of the remaining 22 hexagonal cases (Section 3.5). The LPbound on the score with these changes is < 8 pt. 3.11. Summary. We summarize some of the results obtained up to this point. The worst pathologies have now been eliminated. Consider a decomposition star scoring ≥ 8 pt that has an exceptional region. Then 1. The exceptional region is a polygon with n = 5, 6, 7, or 8 sides. − 2. There are no S+ 4 or S3 configurations (even in the original decomposition star before anything is erased).

3. The planar map is one of the 189 cases obtained in Section 3.5. 4. If an upright quarter with at least 5 anchors appears, it must have context (5, 0) and lie over a pentagonal standard region.

4. Refined Linear Programming Bounds 4.1. Internal structures. From the results of Section 3, we know that every standard region is a polygon, + and there are no S− 3 or S4 configurations of exceptional regions. We describe linear programming arguments that reduce the list of possibilities from 189 to 88. We do this by separating each planar map into a number of subcases according to the detailed structure of the exceptional cluster. We list the structures and afterward we give an extended explanation of what the diagrams mean.

24

THOMAS C. HALES

Diagram 4.1

0.008

2(0.008)

0.008

THE KEPLER CONJECTURE

25

The possibilities are listed in the diagram only up to symmetry by the dihedral group action on the polygon. We do not prove the completeness of the list, but its completeness can be seen by inspection, in view of the comments that follow here and in Section 4.2. The conventions for generating the possibilities are different for the pentagons and hexagons than for the heptagons and octagons. We describe the pentagons and − + hexagons first. There are no S+ 4 or S3 configurations. We erase S3 . If there is one loop we leave the upright diagonal in the figure. If there are two loops (so that both necessarily have context (4, 1)), we erase one and keep the other. We draw all edges √ from an enclosed upright diagonal to its anchors, and all edges of length [2.51, 2 2] that are not masked by upright quarters. Since the only remaining upright quarters belong to loops, the four simplices around a loop are anchored simplices and the edge opposite the diagonal has length at most 3.2. When the figure is a heptagon or octagon, we proceed differently. We erase all S+ 3 configurations and all loops (either context (4, 1) or (4, 2)) and draw only the flat quarters. An undrawn diagonal of the polygon has length at least 2.51. Thus, in these cases much less internal structure is represented. In the cases where S+ 3 or loops have been erased, a number indicating a penalty accompanies the diagram. These penalties are derived in the next section. 4.2. Penalties. Erasing an upright quarter of compression type gives ξΓ and one of Voronoi type gives ξV . We take the worst possible penalty. It is at most nξΓ in an n-gon. If there is a masked flat quarter, the penalty is only 2ξV from the two upright quarters along the flat quarter. We note in this connection that both edges of a polygon along a flat quarter lie on upright quarters, or neither does. If an upright diagonal appears enclosed over a flat quarter, the flat quarter is part of a loop with context (4, 1), for a penalty 2ξΓ′ + ξV . This is smaller than the penalty obtained from a loop with context (4, 1), when the upright diagonal is not enclosed over the flat quarter ξΓ + 2ξV . So we calculate the worst-case penalties under the assumption that the upright diagonals are not enclosed over flat quarters. A loop of context (4, 1) gives ξΓ + 2ξV or 3ξΓ . A loop of context (4, 2) gives 2ξΓ or 2ξV . If we erase S+ 3 , there is a penalty of 0.008 (or 0 if it masks a flat quarter.) This is dominated by the penalty 3ξΓ of context (4, 1). Suppose we have an octagonal standard region. We claim that a loop does not occur in context (4, 2). If there are at most three vertices that are not corners of

26

THOMAS C. HALES

the octagonal cluster, then there are at most 12 quasi-regular tetrahedra, and the score is at most s8 + 12 pt < 8 pt. Assume there are more than three vertices that are not corners of the octagonal cluster. We squander t8 + (DLP (4, 2) − D(4, 2)) + 4τLP (5, 0) > (4πζ − 8) pt. As a consequence, context (4, 2) does not occur. So there are at most 2 upright diagonals and at most 6 quarters, and the penalty is at most 6ξΓ . Let f be the number of flat quarters This leads to  6ξ , f = 0, 1,   Γ   4ξ + 2ξ , f = 2, Γ V π0 =  2ξΓ + 4ξV , f = 3,    0, f = 4.

The 0 is justified by a parity argument. Each upright quarter occurs in a pair at each masked flat quarter. But there is an odd number of quarters along the upright diagonal, so no penalty at all can occur. Suppose we have a heptagonal standard region. Three loops are a geometric impossibility. Assume there are at most two upright diagonals. If there is no context (4, 2), then we have the following bounds on the penalty  6ξΓ ,     4ξ + 2ξ , Γ V π0 =  3ξΓ ,    ξΓ + 2ξV ,

f = 0, f = 1, f = 2, f = 3.

If an upright diagonal has context (4, 2), then  f = 0, 1,   5ξΓ , π0 = 3ξΓ + 2ξV , f = 2,   ξΓ + 4ξV , f = 3. This gives the bounds used in the diagrams of cases. 4.3. Variable relations. We are now ready to describe an augmentation of the linear programs that were presented in Section 3.5 to reduce to 189 cases. Take the decompositions of Section 4.1 and generate all the decompositions that can be obtained from them by the action of the dihedral group on the vertices. This gives 22 pentagonal cases, 117 hexagonal cases, 29 heptagonal cases, and

THE KEPLER CONJECTURE

27

47 octagonal cases. We let PM(5, c, p) denote the planar map PM(5, c) with the additional structure on the pentagon from case p = 1, . . . , 22. Similarly, we have PM(n, c, p) extending PM(n, c), for n = 5, 6, 7, 8, where (n, c) ranges over the 189 pairs of Section 3.5. We associate a new linear program with each PM(n, c, p). This linear program comes by augmenting the linear program of Section 3.5 with many new inequalities. They are listed in Sections 4.3–4.10. Let R be a standard region. Let σR be a linear programming variable denoting the score of the cluster over R, scored in the sense of [F] without any erasures. Let σF be linear programming variables for each subregion F of R. Here a subregion has a different meaning from IV. Here we mean the polygons in one of the decompositions of a standard region in Diagram 4.1. If the subregion is a flat quarter, σF represents the function σ ˆ , defined in Section 2.5. If the subregion is an upright quarter, σF represents the score ν from [F]. If the subregion is an anchored simplex that is not an upright quarter, σF is a variable representing the analytic Voronoi function vor if the simplex has type SC , and it represents vor0 otherwise. (The types SA , SB , SC are defined in Section IV.2.5.) Whether or not the simplex has type SC , we have σF ≤ 0. In fact, if vor0 scoring is used, we note that there are no quoins, and φ(1, t0 ) < 0. If the subregion is triangular, if no vertex represents an upright diagonal, and if the subregion is not a quarter, then σF represents the score, which will be vor or vor0 depending on whether the simplex has type SA . In either case, σF ≤ vor0 In most other cases, σF represents vor0 . However, if R is a heptagon or octagon, and F has ≥ 4 sides, then σF is vor0 except on simplices of type SA , where it is the analytic Voronoi function. If R is a pentagon or hexagon, and F is a quadrilateral that is not adjacent to a flat quarter, then we have σF represent the actual score of the subcluster over the subregion. In this case the score has a well-defined meaning for the quadrilateral, because it is not possible for an upright quarter in the Qsystem to straddle the quadrilateral region and an adjacent region. Consequently, any erasing that was done can be associated with the subregion without ambiguity. By the results of [II], we have σF ≤ 0. We also have σF ≤ vor0 . The linear programming variables σF , for F in a standard region R, satisfy the inequality X σR < π0 + σF , F

where π0 is the penalty exhibited in Diagram 4.1. When the standard region has a single subregion this relation becomes σR < π0 + σF . We emphasize that in this case the variables σR and σF have different meanings, although they are associated with the same region. The penalty is part of σR , but is not part of σF .

28

THOMAS C. HALES

We also have linear programming variables solF , dihF,i , where i indexes the vertices of F . Let n(F ) be the number of vertices of the polygon F . These variables are related by the linear inequalities solF = −(n(F ) − 2)π + X dihi = dihF,i ,

X

dihF,i

F

2π =

X

dihF,i ,

F

τF + σF = solF ζpt. The sum of the dihedral angles at a vertex i is dihi , the dihedral angle of the standard region at i, if i indexes a vertex of the standard region. The sum is 2π, if i indexes an enclosed upright diagonal. 4.4. Flat Quarters. Section A.4.4 of the appendix lists various inequalities for flat quarters that enter the linear program. 4.5. Upright Quarters. Section A.4.5 of the appendix lists various inequalities for upright quarters that enter the linear program. (4.5.1–4.5.4)

(See the appendix.)

We have a few other inequalities that hold for anchored simplices that are not upright quarters. The variables σF have the interpretation described in Section 4.3.

(4.5.5)

 √ y1 ∈ [2.51, 2 2]   0, σF < −0.05, y1 ∈ [2.51, 2.696],   −0.119, y1 ∈ [2.51, 2.696],

η126 ≥

√ 2,

√ if y4 ∈ [2.51, 2 2]. The bound of 0 is established in Section 4.3. (Even when simplex has type SC , the bound −0.05 is based on the upper bound vor0 .)

(4.5.6)

σF
0.486.

If there are two flat quarters and no upright quarters, there is a quadrilateral subregion F . It satisfies

(4.6.3)

σF < −0.168, τF > 0.352.

These are twice the constants appearing in (4.6.11).

30

THOMAS C. HALES

√ If there is an edge of length between 2.51 and 2 2 running between two opposite corners of the hexagonal cluster, and if there are no flat or upright quarters on one side, leaving a quadrilateral region F , then F satisfies

(4.6.4)

σF < −0.075, τF > 0.176.

If the hexagonal cluster has an upright diagonal in context (4, 2), and if there are no flat quarters then the hexagonal cluster R satisfies

(4.6.5)

σR < −0.297, τR > 0.504.

If the hexagonal cluster has an upright diagonal in context (4, 2), and if there is one flat quarter, let {F } be the set of four subregions around the upright diagonal. (That is, take all subregions except for the flat quarter.) X

σF < −0.253,

X

τF > 0.4686.

(4)

(4)

(4.6.6)

If the hexagonal cluster has an upright diagonal in context (4, 2), and if there are two flat quarters, let {F } be the set of four subregions around the upright diagonal. (That is, take all subregions except for the flat quarters.) X

σF < −0.2,

X

τF > 0.3992.

(4)

(4)

(4.6.7)

If the hexagonal cluster has an upright diagonal in context (4, 1), and if there are no flat quarters, let {F } be the set of four subregions around the upright diagonal. Assume that the√edge opposite the upright diagonal on the anchored simplex has length at least 2 2. (That is, there are only four subregions.) X

σF < −0.2187

X

τF > 0.518.

(4)

(4.6.8)

(4)

THE KEPLER CONJECTURE

31

In this same context, let F be the pentagonal subregion. It satisfies

(4.6.9)

σF < −0.137, τF > 0.31.

If the hexagonal cluster has an upright diagonal in context (4, 1), and if there is one flat quarter, let {F } be the set of four subregions around the upright diagonal. Assume that the√edge opposite the upright diagonal on the anchored simplex has length at least 2 2. (That is, there are only five subregions.) X σF < −0.1657, (4)

(4.6.10)

X

τF > 0.384.

(4)

In this same context, let F be the quadrilateral subregion. It satisfies

(4.6.11)

σF < −0.084, τF > 0.176.

Proposition. Inequalities 4.6.1–4.6.11 are valid. Proof. We prove the inequalities in reverse order 4.6.11– 4.6.1. The constants 0.009 and 0.05925 from IV.A13 bounding what a flat quarter scores and squanders will be used repeatedly. Proof of (4.6.10) and (4.6.11). Break the quadrilateral cluster into two simplices S and S ′ along the long edge of the anchored simplex S. The anchored simplex S satisfies τ (S) ≥ 0, σ(S) ≤ 0. The other simplex satisfies τ (S ′ ) > 0.176 and σ(S ′ ) < −0.084 by an interval calculation. This gives (4.6.11). For (4.6.10), we combine these bounds with the linear programming bound on the four anchored simplices around the upright diagonal. From the inequalities IV.A2 –IV.A7 , IV.A22 , IV.A24 , we find that they score < −0.0817 and squander > 0.208. Adding these to the bounds from (4.6.11), we obtain (4.6.10). Proof of (4.6.8) and (4.6.9). The pentagon is a union of an anchored simplex and a quadrilateral region. LP-bounds similar to those in the previous paragraph and based on the inequalities of IV show that the loop scores at most −0.0817 and squanders at least 0.208. If we show that the quadrilateral satisfies σF < −0.137, τF > 0.31, then Inequalities (4.6.8)√and (4.6.9) follow. If by deformations a diagonal of the quadrilateral drops to 2 2, then the result follows from A.4.6.8 and IV.A13 . By this we may now assume that the quadrilateral has the form (a1 , 2, a2 , 2, a3 , 2, a4 , b4 ),

a2 , a3 ∈ {2, 2.51}.

32

THOMAS C. HALES

If the diagonals drop under 3.2 and max(a2 , a3 ) = 2.51, again the result follows from A.4.6.8 and IV.A13 . If the diagonals drop under 3.2 and a2 = a3 =√2, then the result follows from IV.A19 . So finally we attain by deformations b4 = 2 2 with both diagonals greater than 3.2. But this does not exist, because ∆(4, 4, 4, 3.22, 4, 8, 3.22) < 0.

Proof of (4.6.5), (4.6.6), and (4.6.7). Inequalities 4.6.7 are derived in Section IV.5.11. Inequalities 4.6.5, 4.6.6 are LP-bounds based on IV.A2 –IV.A7 , IV.A22 . Proof of√(4.6.4). Deform as in IV. If at any point a diagonal of the quadrilateral drops to 2 2, then the result follows from IV.A13 and VI.A.4.6.11: vor0 < 0.009 − 0.084 = −0.075, τ0 > 0 + 0.176 = 0.176, Continue deformations until the quadrilateral has the form (a1 , 2, a2 , 2, a3 , 2, a4 , b4 ),

a2 , a3 ∈ {2, 2.51}.

There is necessarily a diagonal of length ≤ 3.2, because ∆(4, 4, 3.22 , 8, 4, 3.22) < 0. Suppose the diagonal between vertices v2 and v4 has length at most 3.2. If a2 = 2.51 or a3 = 2.51, the result follows from IV.A13 and VI.A.4.6.11. Take a2 = a3 = 2. Inequality 4.6.4 now follows from IV.A19 . Proof of (4.6.3). We prove that the quadrilateral satisfies σF < −0.168 τF > 0.352.

There are two types of quadrilaterals. In (a), there are two flat quarters whose central vertices are opposite corners of the hexagon. In (b), the flat quarters share a vertex. We consider case (a) first. Case (a). We deform the quadrilateral as in [IV]. If at any point there is a diagonal of length at most 3.2, the result follows from A.4.6.10 and A.4.6.11. Otherwise, the deformations give us a quadrilateral (a1 , 2, a2 , 2.51, a3 , 2, a4 , 2),

ai ∈ {2, 2.51}.

The tcc approximation now gives the result (see IV.5.3). Case (b). Label the vertices of the quadrilateral v1 , . . . , v4 , where (v1 , v2 ) and (v1 , v4 ) are the diagonals of the flat quarter. Again, we deform the quadrilateral. If at any point of the deformation, we find that |v1 − v3 | ≤ 3.2, the result follows from

THE KEPLER CONJECTURE

33

√ A.4.6.10, A.4.6.11. If during the deformation |v2 − v4 | ≤ 2 2, the result follows from IV.A13 and the interval calculations A.4.6.3. If the diagonal (v2 , v4 ) has length at least 3.2 throughout the deformation, we eventually obtain a quadrilateral of the form (a1 , 2.51, a2, 2, a3 , 2, a4 , 2.51), ai ∈ {2, 2.51}. But this does not exist: ∆(4, 4, 3.22 , 2.512 , 2.512 , 3.22 ) < 0. √ We may assume that |v2 − v4 | ∈ [2 2, 3.2]. The result now follows from interval calculations A.4.6.3. Proof of (4.6.2). This case requires more effort. We show that vor0 < −0.221 τ0 > 0.486

Label the corners (v1 , . . . , v5 ) cyclically with (v1 , v5 ) the diagonal of the flat quarter in the hexagonal cluster. We use the deformation theory of IV. The proof appears in steps (1), . . . , (6). (1) If during the deformations, |v1 − v4 | ≤ 3.2 or |v2 − v5 | ≤ 3.2, the result follows from Inequalities (4.6.9) and (4.6.11). We may assume this does not occur. √ (2) If an edge (v1 , v3 ), (v2 , v4 ), or (v3 , v5 ) drops to 2 2, continue with deforma√ tions that do not further decrease this diagonal. If |v1 − v3 | = |v3 − v5 | = 2 2, then the result follows from IV.A13 and interval calculations A.4.6.2.a. √ If we have |v1 − v3 | = 2 2, deform the figure to the form (a1 , 2, a2 , 2, a3 , 2, a4 , 2, a5 , 2.51),

a2 , a4 , a5 ∈ {2, 2.51}.

Once it is in this form, break the flat quarter (0, v1 , v2 , v3 ) from the cluster and deform v3 until a3 ∈ {2, 2.51}. The result follows from an interval calculation A.4.6.2.b. We handle a boundary case of the preceding calculation separately. After breaking the flat quarter off, we have the cluster √ (a1 , 2 2, a3 , 2, a4 , 2, a5 , 2.51),

a3 , a4 , a5 ∈ {2, 2.51}.

If |v1 − v4 | = 3.2, we break the quadrilateral cluster into two pieces along this diagonal and use interval calculations A.4.6.2.c to conclude the result. This completes √ the analysis of the case |v1 − v3 | = 2 2. (3) If |v2 − v4 | ≤ 3.2, then deform until the cluster has the form (a1 , 2, a2 , 2, a3 , 2, a4 , 2, a5 , 2.51),

a1 , a3 , a5 ∈ {2, 2.51}.

34

THOMAS C. HALES

Then cut along the special simplex to produce a quadrilateral. Disregarding cases already treated by A.4.6.2.c, we can deform it to √ (a1 , 2, a2 , 2 2, a4 , 2, a5 , 2.51),

ai ∈ {2, 2.51},

with diagonals at least 3.2. The result now follows from the interval calculations A.4.6.2.d. In summary of (1), (2), (3), we find that by disregarding cases already considered, we may deform the cluster into the form (a1 , 2, a2 , 2, a3 , 2, a4 , 2, a5 , 2.51), √ √ |v1 − v3 | > 2 2, |v3 − v5 | > 2 2, |v2 − v4 | > 3.2.

ai ∈ {2, 2.51},

(4) Assume |v1 −v3 |, |v3 −v5 | ≤ 3.2. If max(a1 , a3 , a5 ) = 2.51, we invoke A.4.6.2.b and IV.A13 to prove the inequalities. So we may assume a1 = a3 = a5 = 2. The result now follows from interval calculations A.4.6.2.e. This completes the case |v1 − v3 |, |v3 − v5 | ≤ 3.2. (5) Assume |v1 − v3 |, |v3 − v5 | ≥ 3.2. We deform to (a1 , 2, a2 , 2, a3 , 2, a4 , 2, a5 , 2.51),

ai ∈ {2, 2.51}.

If a2 = 2.51 and a1 = a3 = 2, then the simplex does not exist by Section IV.5.6. Similarly, a4 = 2.51, a5 = a3 = 2 does not exist. The tcc bound gives the result except when a2 = a4 = 2. The condition |v2 − v4 | ≥ 3.2 forces a3 = 2. These remaining cases are treated with the interval calculations A.4.6.2.f. (6) Assume |v1 − v3 | ≤ 3.2 and |v3 − v5 | ≥ 3.2. This case follows from deformations, A.4.6.2.b, and A.4.6.2.g. This completes the proof of Inequalities (4.6.2). Proof of (4.6.1). Label the corners of the hexagon v1 , . . . , v6 . The proof to this inequality is similar to the other cases. We deform the cluster by the method of IV until it breaks into pieces that are small enough to be estimated by interval calculations. If a diagonal between opposite corners has length at most 3.2, then the hexagon breaks into two quadrilaterals and the result follows from Inequality 4.6.9. If a flat quarter is formed during the course of deformation, then the result follows from Inequality 4.6.2 and IV.A13 . Deform until the hexagon has the form (a1 , 2, a2 , 2, . . . , a6 , 2),

ai ∈ {2, 2.51}.

We may also assume that the hexagon is convex (Section IV.4.11). If there are no special simplices, we consider the tcc-bound. The tcc-bound implies Inequality 4.6.1, except when ai = 2, for all i. But if this occurs, the perimeter of the convex spherical polygon is 6 arc(2, 2, 2) = 2π. Thus, there is a

THE KEPLER CONJECTURE

35

pair of antipodal points on the hexagon. The hexagon degenerates to a lune with vertices at the antipodal points. This means that some of the angles of the hexagon are π. One of the tccs has the form C(2, 1.6, π), in the notation of Lemma IV.4.11. With this extra bit of information, the tcc bound implies (4.6.1). √ If there is one special simplex, say |v5 − v1 | ∈ [2 2, 3.2], we remove it. The score of the special simplex is (Inequalities IV.A13 ) vor0 < 0,

τ0 > 0.05925,

vor0 < 0.0461,

τ0 > 0,

if max(|v1 |, |v5 |) = 2.51,

if |v1 | = |v5 | = 2,

The resulting pentagon can be deformed. If by deformations, we obtain |v2 − v5 | = 3.2 or |v1 − v4 | = 3.2, the result follows from Inequalities (4.6.9) and interval √ calculations A.4.6.1.a. If |v5 − v1 | = 2 2, we use (4.6.2) and IV.A13 unless |v1 | = |v5 | = 2. If |v1 | = |v5 | = 2, we use the interval calculations A.4.6.1.b. If a second special simplex forms during the deformations, the result follows from the interval calculations A.4.6.1.c. The final case of (4.6.1) to consider is that of two special simplices. We divide this into two cases. (a) The central vertices of the specials are v2 and v6 . (b) The central vertices are opposite v1 and v4 . In case (a), the result follows by deformations and interval calculations A.4.6.1.d. In case (b), the result follows by deformations and interval calculations A.4.6.1.e. This completes the proof of inequalities (4.6.1) and the proof of the Proposition.  4.7. Pentagonal Regions. There are a few inequalities that arise for pentagonal regions. These are listed in Appendix A.4.7.

(4.7.1)

(See Appendix)

Proposition. If the pentagonal region has no flat quarters and no upright quarters, the subregion F is a pentagon. It satisfies

(4.7.2)

σF < −0.128,

τF > 0.36925.

We recall that σR < σF + π0 , where σR is a variable representing the score of the standard cluster, and π0 = 0.008 is the penalty from Diagram 4.1. Proof. The proof is by deformations and interval calculations. If a deformation produces a new flat quarter, then the result follows from IV.A √13 and Inequality (4.6.9). So we may assume that all diagonals remain at least 2 2. If all diagonals

36

THOMAS C. HALES

remain at least 3.2, the result follows from the tcc-bound on the pentagon. Thus, we assume that some diagonal is at most 3.2. We deform the cluster into the form (a1 , 2, a2 , 2, a3 , 2, a4 , 2, a5 , 2),

|vi | = ai ∈ {2, 2.51}.

Assume that |v1 − v3 | ≤ 3.2. If max(a1 , a3 ) = 2.51, the result follows from IV.A13 and Inequality (4.6.9). Assume a1 = a3 = 2. There is a diagonal of the quadrilateral of length at most 3.23 because ∆(3.232 , 4, 4, 3.232, 4, 3.22) < 0. The result follows from A.4.7.2.  4.8. Dihedral Bounds. In Section 4.1, we gave the possible decompositions of a standard cluster into smaller pieces. Appendix A.4.8 gives lower and upper bounds on the dihedral angles along each edge (0, v) of each of the smaller pieces. 4.9. Additional Inequalities. Depending on the lengths of the edges y5 , y6 , y4 , there are additional inequalities that hold. Generally these inequalities are for a fragment of a subregion. These inequalities are listed in A.4.9. 4.10. Miscellaneous inequalities. If the region F is scored by vor0 , we also add the inequalities of Appendix III.A for vor0 . Recall that vor0 is defined by a formula of the form (F.3.7) vor0 = sol φ0 +

X

Adih −

X

4δoct quo .

φ0 is a constant. We have σF ≤ vor0 , if F is not a flat quarter. We introduce a new variable Adih for each vertex of F and two new variables quo for each edge between adjacent vertices (one for each Rogers simplex along the edge). These variables are subject to the collection of linear inequalities of Appendix III.A.4. To write down the linear inequalities from Appendix III.A.4 for Adih, it is necessary to have bounds on the heights of vertices and on the dihedral angles of F . The LP-bounds on these variables are used for these bounds. This argument gives improved bounds on Adih by determining new LP-bounds on the yi and dihF,i variables and using these new bounds in the inequalities constraining Adih. We will make use of this argument several times. We refer to this procedure as updating the Adih variables. Improvements in the bounds on AdihF,i are possible in this way because Adih is a nonlinear function of yi and dihF,i .

THE KEPLER CONJECTURE

37

One other bound that we have not explicitly mentioned is the bound σ(R) < sn . For heptagons and octagons, this is a better bound than the one used in Section 2. In heptagons and octagons, if we have a subregion with four or more sides, then σF < Z(n, k) and τF > D(n, k). (See IV.5.5.1, IV.5.5.2, and the interpretation of variables in Section 4.3.) Information about the internal structure of a exceptional cluster gives improvements to the vertex adjustments 1.4 pt and 1.5 pt. The vertex adjustments contribute to the bound on the score through the bound (4.10.1)

4 X

squanderFace[fj ] +

j=1

k X

vertexAdjustment[ai ]

i=1

from Section 2.3. Assume that at the vertex v there are four tetrahedra and an exceptional cluster, and that the exceptional cluster has a flat quarter with central vertex v. The calculations of Section 2.8 show that the four quarters and exceptional squander at least 1.5 pt. If there is no flat quarter (masked or unmasked) whose first edge lies along (0, v), then the four quasi-regular tetrahedra at v squander at least 1.5 pt. These are stronger inequalities than those used in Section 2.3, because the adjustment is spread over a smaller region. We can make similar improvements in vertexAdjustment[3]. 4.11. Linear programming results. The LP problem in Section 3.5 has been augmented by all these additional inequalities. New linear programming bounds are obtained by running these in all possible cases. If the planar map has more than one exceptional region, then one exceptional region is selected for the refined inequalities and the others use the unrefined bounds of IV.4.4. All but 88 planar maps give LP bounds under 8 pt. (For each planar map PM(n, c) that remains, there can be various PM(n, c, p) that give an LP-bound over 8 pt.) 4.12. Upright Diagonals. Many of the planar graphs that remain exhibit particularly poor behavior on upright diagonals. This section gives a few bounds to make that explicit. We divide the upright simplices into two domains depending on the height of the upright diagonal, using y1 = 2.696 as the break point. We refer the tests in this section as the 2.696-tests. So if we write in later sections that the diagonal is at most 2.696, or that a certain configuration of upright diagonals leads to a score under 8 pt by the 2.696-tests, it means that the result is a consequence of the inequalities listed here. If an upright diagonal has height ≤ 2.696, many additional inequalities hold. See Inequalities 4.5.4.

38

THOMAS C. HALES

Suppose an upright diagonal has height ≥ 2.696. We have νΓ < vor0 (or equivalently, Γ < octavor0 ) (IV.A10 ), so that upright quarters of compression type can be erased. This is not the case for upright quarters of Voronoi type. We have the weaker inequality (IV.A11 ) ν < vor0 +ξV ,

ξV = 0.003521.

Nevertheless, ν < vor0 holds unless y3 , y5 ≥ 2.45 or y2 , y6 ≥ 2.45 (IV.A11 ). If we can show that these conditions do not occur in a given decomposition star, then the upright quarters can be erased. (If there are masked flat quarters, they become scored by σ ˆ .) When the quarters are erased, the proof that the decomposition star scores less than 8 pt follows from the proof for the structure obtained by erasing. To make it easier to show that the conditions (say y2 , y6 ≥ 2.45) cannot occur, we list some inequalities that hold under this hypothesis. See A.4.12.1–A.4.12.9. To show that the height of an upright quarter is less than 2.696, it is sufficient to apply the relevant inequalities from the list (A.4.12.1– A.4.12.9) and verify that the score of the decomposition star drops below 8 pt. The inequalities (A.4.12.1– A.4.12.4) have been designed with pentagonal exceptional clusters in mind. The additional inequalities (A.4.12.5–A.4.12.9) have been designed for subclusters in hexagonal exceptional clusters. In A.4.12.3 we use the analytic Voronoi function. Under normal conditions, this would be scored by the truncated Voronoi function, except on anchored simplices of type SC . Nevertheless, in this particular context, the use of the analytic Voronoi function is justified by the following lemma. (In [IV], we switched to the truncated Voronoi from the analytic function because of the possibility that the simplex is bounded by a flat quarter that is not in the Q-system.) Lemma. In the context of A.4.12.3, the flat quarter in the pentagonal cluster is in the Q-system. Proof.

√ √ E(S(2, 2, 2.45, 2 2, 2.51, 2.51), 2, 2, 2) > 2 2,

so nothing is enclosed over the flat quarter. √ √ E(S(2, 2, 2, 2 2, 2.51, 2.51), 2.51, 2.45, 2) > 2 2, so no edge between vertices of the packing can cross inside the anchored simplex. This implies that the flat quarter does not have a conflicting diagonal and is not part of an isolated pair.  Similar arguments show that there is not a simplex with negative orientation along the top face of the anchored simplex. Similar arguments justify the use of the analytic Voronoi function in hexagonal clusters on anchored simplices provided √ the length of the edge opposite the upright diagonal is between 2.51 and 2 2.

THE KEPLER CONJECTURE

39

4.13. Octagons. There are only two planar maps with an octagon that arise. The first case is PM(8, 14, p), where p is an octagon with two flat quarters. The central vertices of the flat quarters are opposite corners of the octagon (the two with four triangles). To eliminate this case, we first check that there is no loop of context (4, 2). In fact, if there were, we would have, by Table IV.5.11, the upper bound s8 + ZLP (4, 2) − Z(4, 2) < −0.31398 on the score of the octahedral cluster. With this bound the LP-bound on the score of the decomposition star is less that 8 pt. Since the two flat quarters are at opposite corners, the only penalties come from loops in context (4, 1) that mask flat quarters. We conclude that the penalty is at most 2(ξΓ + 2ξV ) = 0.045304. If F is the face representing the hexagonal subregion, we have τF > D(6, 2). (Recall from the interpretation of the variables τF in Section 4.3 that the variable τF does not include penalties.) With these additional inequalities we branch and bound on the shaded triangular faces of the diagram and find that the score is less than 8 pt. Diagram 4.7

The case PM(8, 16, p) is easily treated. This is the planar map illustrated in the second frame of the diagram. The branch and bound on the shaded triangles in the diagram gives an upper bound of 8 pt.

40

THOMAS C. HALES

5. Heptagons Diagram 5.1 c=27

c=36 11

6

c=46

1

5

1

7 10

12

9

4

12

13

11

8

c=75

5

7

2

3

6

11

10

11

2

10

7

4

12 8

There are four planar maps that have an LP-bound over 8 pt. If we add branchand-bound inequalities from the indicated shaded triangles, then PM(7, 36) drops under 8 pt. In each remaining case, there is one or two flat quarters. In these cases, we make free use of the branch-and-bound inequalities on the triangles shaded in Diagram 5.1. There is a total of 10 cases if we include all the internal structures for the heptagon PM(7, c, p). The arguments to eliminate the remaining cases follow a similar pattern. We take the case PM(7, 27, p6) as an illustration of the argument. Here p6 is the structure on the heptagon that has a flat quarter centered at the vertex v6 , and no others. 1. Case PM(7, 27, p6). The complement of the flat quarter Q centered at vertex 6 is a hexagonal subregion F . If R is the heptagonal region, a linear programming optimization shows that the lower bound on σR − σQ is -0.157793. If we combine the linear programming optimization with the branch and bound inequalities on the shaded triangular faces in the diagram for PM(7, 27), we obtain the better lower bound of σR − σQ > −0.14014. If there is no loop masking the flat quarter, then the paper [IV] writes F as a union of subregions summing to at most Z(6, 1) < −0.157793, which contradicts our bound. Therefore, there is a loop masking the flat quarter. Erase it, and take the penalty ξΓ + 2ξV . We have LP-max(y6 ) < 2.0579. Since LP-max(y6 ) < 2.2, we have y7,11 > 2.7, by Section 2.5. In IV, the complementary hexagon is expressed as a union of subregions summing to at most Z(6, 1) + ξΓ + 2ξV < −0.14014, which contradicts the LP bound. When there are two flat quarters Q1 , Q2 , the argument is readily adapted. One shows that LP-min(σR − σQ1 − σQ2 ) > Z(5, 2) + π0 .

THE KEPLER CONJECTURE

41

This can be divided into subcases following the model of PM(7, 27, p6 ), depending on which diagonals of the flat quarters have length ≥ 2.7. The penalty may be adjusted accordingly. (In every case y1 ≤ 2.2.) This approach eliminates all heptagonal cases. There are small variations in case PM(7, 46). First of all, it was helpful to update the Adih variables, in the sense of Section 4.10. In the cases involving two flat quarters, it was helpful to establish a penalty of 3ξΓ = 0.04683 instead of the penalty 4ξΓ + 2ξV proposed in Section 4.1. This smaller penalty results if neither flat quarter S = S(y1 , . . . , y6 ) is masked by a loop. For this, it suffices to show that y1 ≤ 2.2 and y4 ≤ 2.7. These linear programming bounds are readily verified. Finally, the branch-and-bound inequalities for all triangular faces were used. Logs of the linear programming sessions that established these results appear in [H1]. This completes the discussion of heptagons.

6. Pentagons and Hexagons This section gives a few comments about the linear programming arguments that have been used to eliminate the final pentagonal cases. There are essentially no new arguments and only a couple of new inequalities, so our discussion here will be brief. Logs recording the linear programming sessions that eliminated these final cases can be found at [H1]. The classification algorithm yields 2459 pentagonal planar maps. The linear programming bound (Section 3.5) eliminates all but 150 cases. The refined linear programming bound (Section 4.11) eliminates all but 69 cases. These 69 cases are treated case by case. The classification algorithm yields 429 hexagonal planar maps. The linear programming bound (Section 3.5) eliminates all but 22 cases. The refined linear programming bound (Section 4.11) eliminates all but 13 cases. These 13 cases are treated case by case. The arguments are much the same in each case. The typical case first uses the weak or strong 2.696-test to show that all upright diagonals have 2.696. Next, the branch and bound inequalities are used. If r1 -triangular faces and r2 quadrilateral faces are selected for the branch and bound argument, then there are 2r1 +2r2 linear programs. (The four cases for each quad cluster correspond to the four scoring types described in III.A.1.) To keep the calculations brief, we usually limit the number of faces by r1 + 2r2 ≤ 10. We add more faces as needed on the cases that fail. Many of the branch and bound inequalities assume the hypothesis that the heights of the vertices of a quasi-regular tetrahedron are at most 2.13. It is easy to check whether this condition is satisfied by linear programming. When this condition is satisfied, the results from the branch and bound inequalities tend to be much better.

42

THOMAS C. HALES

When there are two exceptional faces in the planar graph, some improvements are possible. We have the crude linear program that uses the bounds of IV.4.4, and the refined linear programming bounds that make use of the internal structure of one of the exceptional regions. In this step of the optimization, we make use of the internal structure on both exceptional faces. When there is a subregion of an exceptional cluster that has five or more sides, we update the Adih variables by the procedure described in Section 4.10. Consider an upright diagonal with with three upright quarters, that is, context (4, 1). If the 2.696-test shows that the height of the upright diagonal is at most 2.696, and if an upright quarter shares both faces along the upright diagonal with other upright quarters, then we may assume that the upright quarter has compres√ sion type. For otherwise, there is a face of circumradius at least 2, and hence two upright quarters of compression type. The inequality (6.1)

octavor < octavor0 −0.008, √ if y1 ∈ [2.51, 2.696], and η126 ≥ 2 shows that the upright quarters can be erased without penalty because ξΓ < 2(0.008). If erased, the case is treated as part of a different case. This allows the inequality for νΓ in Section 4.4.1 to be used. Furthermore, it can often be concluded that all three upright quarters have compression type. For this, we use the Inequalities 4.7.1 and 4.5.5, which can often √ be used to show that if the anchored simplex has a face of circumradius at least 2, then the linear programming bound on the score is less than 8 pt. Inequality 4.7.1 holds for a quadrilateral subregion if certain conditions are satis√ fied. One of the conditions is y4 ∈ [2 2, 3.0], where y4 is a diagonal of the subregion. Since this diagonal is not one of the linear programming variables, these bounds cannot be verified directly from the linear program. Instead we use the the interval calculation (6.2)

dih > 1.678, if y4 ≥ 3.0, and y2 + y3 + y5 + y6 < 8.77,

which relates the desired bound y4 ≤ 3 to the linear programming variables dih, y2 , y3 , y5 , and y6 . There are a few other interval-based inequalities that are used in particular cases. See A.6.3. They use compression scoring. Before the inequality is used on a flat quarter, it√is necessary to verify that the quarter satisfies y1 ≤ 2.2, y4 ≤ 2.7, η234 , η456 ≤ 2, to insure that the flat quarter has compression type. The circumradius is not a linear-programming variable, so its upper bound must be deduced from edge-length information. All of the pentagonal cases are eliminated by the repetitive application of these procedures through linear programming. See the logs in [H1] for details about how these inequalities are applied in the individual cases. Th two most difficult cases are discussed further in an Appendix. This completes the proof of the Kepler conjecture. 

THE KEPLER CONJECTURE

43

References [F] S. Ferguson, T. Hales, A Formulation of the Kepler Conjecture, math.MG/9811072. [I] Thomas C. Hales, Sphere Packings I, Discrete and Computational Geometry, 17 (1997), 1-51, math.MG/9811073. [II] Thomas C. Hales, Sphere Packings II, Discrete and Computational Geometry, 18 (1997), 135-149, math.MG/9811074. [III] Thomas C. Hales, Sphere Packings III, math.MG/9811075. [IV] Thomas C. Hales, Sphere Packings IV, math.MG/9811076. [V] S. Ferguson, Sphere Packings V, thesis, University of Michigan, 1997, math.MG/9811077. [H1] Thomas C. Hales, Packings, http://www.math.lsa.umich.edu/∼hales/packings.html [H2] Thomas C. Hales, Remarks on the Density of Sphere Packings, Combinatorica, 13 (2) (1993) 181-197.

44

THOMAS C. HALES

Appendix. Calculations Interval calculations are arranged according to the section in which they appear. Each inequality is accompanied by one or more reference numbers. These identification numbers are needed to find further details about the calculation in [H1]. Most of the verifications were completed by Samuel Ferguson. The verifications that he did are marked with a dagger (†). The computer-based proofs are carried out by interval arithmetic as in our earlier papers. A few inequalities such as those of Sections A.2.7 and A.2.8 are not standard in the sense that they have substantially higher dimension that the others. Linear programming methods were used to break these problems into six-dimensional pieces according to the suggestion of [III,Appendix 2]. Edge lengths whose bounds are not specified are assumed to be between 2 and 2.51. The first edge of an upright quarter is its diagonal. The fourth edge of a flat quarter is its diagonal. σ ˆ is the function of Section 2.5 and τˆ = sol ζpt − σ ˆ. Section A.2.3. The interval calculations here show that the vertex Adjustments can be applied to opposite vertices of a quadrilateral unless the edge between those vertices forms a flat quarter. By the arguments in the text, we may assume that the dihedral angles of the exceptional regions at those vertices are at least 1.32. Also, the three quasi-regular tetrahedra at the vertex squander at least 1.5 pt by a linear programming bound, if the angle of the quad cluster is at least 1.55. Thus, we assume that the dihedral angles at opposite vertices of the quad cluster are at most 1.55. A linear program also gives τ + 0.316 dih > 0.3864 for a quasi-regular tetrahedron. If we give bounds of the form τx + 0.316 dih > b, for the part of the quad cluster around a vertex, where τx is the appropriate squander function, then we obtain X

τx > −0.316(2π − 1.32) + b + 3(0.3864)

for a lower bound on what is squandered. If the two opposite vertices give at least 2(1.4) pt + 0.1317, then the use of the vertex adjustment at the two opposite vertices is justified. The following inequalities give the desired result. √ (912536613) τµ + 0.316 dih > 0.5765, if dih ≤ 1.55, y4 ∈ [2.51, 2 2]. √ (640248153) τ0 + 0.316 dih > 0.5765 if dih ≤ 1.55, y4 ≥ 2 2. √ τν + 0.316 dih2 > 0.2778, if y1 ∈ [2.51, 2 2], (594902677) Section A.2.5. √ ν < vor0 +0.01(π/2 − dih), if y1 ∈ [2.696, 2 2]. ν < vor0 , if y1 ∈ [2.6, 2.696], y4 ∈ [2.1, 2.51]. √ µ < vor0 +0.0268, if y4 ∈ [2.51, 2 2]. √ µ < vor0 +0.02, if y1 ∈ [2, 2.17], y4 ∈ [2.51, 2 2]. √ dih > 1.32, if y4 = 2 2. τˆ > 3.07 pt, for all flat quarters satisfying dih ≤ 1.32. √ √ τ0 > 3.07 pt + ξV + 2ξΓ′ , if y4 ∈ [2.51, 2 2], η456 ≥ 2, dih ≤ 1.32. – The Kepler Conjecture (VI) – printed March 6, 2008

(269048407) (553285469) (293389410) (695069283) (814398901) (352079526) (179025673)

THE KEPLER CONJECTURE

45

Section A.2.7†. If the circumradius of a quasi-regular tetrahedron is ≥ 1.41, then by [I.9.17], τ > 1.8 pt, and many of the inequalities hold. In Sections A.2.7 and A.2.8, let S1 , . . . , S5 be 5 simplices arranged around a common edge (0, v), with |v| ∈ [2, 2.51]. Let yi (Sj ) be the edges, with y1 (Sj ) = |v| for all j, y3 (Sj ) = y2 (Sj+1 ), and y5 (Sj ) = y6 (Sj+1 P). where the subscripts j are extended modulo 5. In Sections A.2.7 and A.2.8, dih(Sj ) ≤ 2π. Set π0 = 2ξV + ξΓ if σ ˆ = vor0 in the cases (y4 ≥ 2.6, y1 ≥ 2.2) and (y4 ≥ 2.7). Set π0 = 0, otherwise. √ (551665569) τ (S1 ) + τ (S2 ) + τ (S4 ) > 1.4 pt, if y4 (S3 ), y4 (S5 ) ≥ 2 2. √ τ (S1 ) + τ (S2 ) + τ (S3 ) > 1.4 pt, if y4 (S4 ), y4 (S5 ) ≥ 2 2. (824762926) √ τ (S1 ) + τ (S2 ) + (ˆ τ (S3 ) − π0 ) + τ (S4 ) > 1.4 pt + D(3, 1), if y4 (S3 ) ∈ [2.51, 2 2], y4 (S5 ) ≥ 2.51, dih(S5 ) > 1.32, (675785884) √ τ (S1 ) + τ (S2 ) + τ (S3 ) + (ˆ τ (S4 ) − π0 ) > 1.4 pt + D(3, 1), if y4 (S4 ) ∈ [2.51, 2 2], y4 (S5 ) ≥ 2.51, dih(S5 ) > 1.32. (193592217) Section A.2.8†. As in A.2.7, the quasi-regular tetrahedra are P generally compression scored. Define π0 as in Section A.2.7. The constraint (5) dih(Sj ) = 2π is assumed. √ (325738864) τ (S1 ) + τ (S2 ) + τ (S3 ) + τ (S4 ) > 1.5 pt, if y4 (S5 ) ≥ 2 2. √ τ (S5 )−π0 ) > 1.5 pt+D(3, 1), if y4 (S5 ) ∈ [2.51, 2 2]. τ (S1 )+τ (S2 )+τ (S3 )+τ (S4 )+(ˆ (314974315) Section A.3.1. τ − 0.2529 dih > −0.3442, if y1 ∈ [2.3, 2.51], and dih ≥ 1.51. (572068135) √ τ0 − 0.2529 dih > −0.1787, if y1 ∈ [2.3, 2.51], y6 ∈ [2 2, 3.02], 1.26 ≤ dih ≤ 1.63. (723700608) √ (560470084) τˆ − 0.2529 dih2 > −0.2137, if y2 ∈ [2.3, 2.51], y4 ∈ [2.51, 2 2], τ0 −0.2529 dih > −0.1371, if y1 ∈ [2.3, 2.51], y5 , y6 ∈ [2.51, 3.02], 1.14 ≤ dih ≤ 1.51. (535502975) Section A.3.8. A. dih < 1.63, if y6 ≥ 2.51, y2 , y3 ∈ [2, 2.168]. (821707685) B. dih < 1.51, if y5 = 2.51, y6 ≥ 2.51, y2 , y3 ∈ [2, 2.168]. (115383627) √ C. dih < 1.93, if y6 ≥ 2.51, y4 = 2 2, y2 , y3 ∈ [2, 2.168]. (576221766) √ D. dih < 1.77, if y5 = 2.51, y6 ≥ 2.51, y4 = 2 2, y2 , y3 ∈ [2, 2.168]. (122081309) τ0 − 0.2529 dih > −0.2391, if y6 ≥ 2.51, dih ≥ 1.2, y2 , y3 ∈ [2, 2.168]. (644534985) τ0 −0.2529 dih > −0.1376, if y5 = 2.51, y6 ≥ 2.51, dih ≥ 1.2, and y2 , y3 ∈ [2, 2.168]. (467530297) √ τ0 − 0.2529 dih > −0.266, if y6 ≥ 2.51, y4 ∈ [2.51, 2 2], dih ≥ 1.2, y2 , y3 ∈ [2, 2.168]. (603910880) √ τ0 − 0.2529 dih > −0.12, if y5 = 2.51, y6 ≥ 2.51, y4 ∈ [2.51, 2 2], dih ≥ 1.2, y2 , y3 ∈ [2, 2.168]. (135427691) dih < 1.16, if y5 = 2.51, y6 ≥ 2.51, y4 = 2, y2 , y3 ∈ [2, 2.168]. (60314528) τ0 − 0.2529 dih > −0.1453, if y2 , y3 ∈ [2, 2.168], y5 ∈ [2.51, 3.488], y6 = 2.51. (312132053)

46

THOMAS C. HALES

Section A.3.8, Case 2-b. dih2 > 0.74, if y1 ∈ [2.51, 2.696], y2 , y3 ∈ [2, 2.168]. (751442360) 2 2 2 τ0 − 0.2529 dih > −0.2391, if ∆(y5 , 4, 4, 8, 2.51 , y6 ) ≥ 0, y2 , y3 ∈ [2, 2.168], y5 ∈ [2.51, 3.488]. (893059266) dih +0.5(2.402 − y4 ) < π/2, if y5 ≥ 2.51, y2 , y3 ∈ [2, 2.168]. (690646028) Section A.3.9. √ dih > 1.78, if y4 = 3.2, y1 ∈ [2.51, 2 2], y2 + y3 ≤ 4.6.

(161665083)

Section A.4.4. The following inequalities hold for flat quarters. In these inequalities the fourth edge is the diagonal.

− dih2 + 0.35y2 − 0.15y1 − 0.15y3 + 0.7022y5 − 0.17y4 > −0.0123,

dih2 − 0.13y2 + 0.631y1 + 0.31y3 − 0.58y5 + 0.413y4 + 0.025y6 > 2.63363, − dih1 + 0.714y1 − 0.221y2 − 0.221y3 + 0.92y4 − 0.221y5 − 0.221y6 > 0.3482, dih1 − 0.315y1 + 0.3972y2 + 0.3972y3−

0.715y4 + 0.3972y5 + 0.3972y6 > 2.37095, − sol − 0.187y1 − 0.187y2 − 0.187y3 + 0.1185y4 + 0.479y5 + 0.479y6 > 0.437235†, sol + 0.488y1 + 0.488y2 + 0.488y3 − 0.334y5 − 0.334y6 > 2.244†, −ˆ σ − 0.145y1 − 0.081y2 − 0.081y3 − 0.133y5 − 0.133y6 > −1.17401,

−ˆ σ − 0.12y1 − 0.081y2 − 0.081y3 − 0.113y5 − 0.113y6 + 0.029y4 > −0.94903, σ ˆ + 0.153y4 + 0.153y5 + 0.153y6 < 1.05382, σ ˆ + 0.419351 sol +0.19y1 + 0.19y2 + 0.19y3 < 1.449, σ ˆ + 0.419351 sol < −0.01465 + 0.0436y5 + 0.0436y6 + 0.079431 dih, σ ˆ < 0.0114, τˆ > 1.019 pt,

(4.4.1)

Inequalities (4.4.1) (867513567) If there are four quasi-regular tetrahedra {S1 , . . . , S4 } at the central vertex of the flat quarter Q, then (4.4.2†)

σ ˆ (Q) +

X

σ(Si ) < 0.114.

(4)

Inequalities 4.4.2

(867359387)

THE KEPLER CONJECTURE

47

Section A.4.5. In a quadrilateral cluster, with a given edge y4 as the diagonal, the other diagonal will be denoted y4′ . The following relations for upright quarters (scored by ν) hold. (We use the inequalities of III.A for upright quarters in quad clusters, which are scored by a different function.) In these inequalities the upright diagonal is the first edge. We include in this group, the inequalities IV.A2 , IV.A3 for upright quarters.

y1 > 2.51, √ y1 < 2 2, dih1 − 0.636y1 + 0.462y2 + 0.462y3 − 0.82y4 + 0.462y5 + 0.462y6 > 1.82419,

− dih1 + 0.55y1 − 0.214y2 − 0.214y3 + 1.24y4 − 0.214y5 − 0.214y6 > 0.75281, dih2 + 0.4y1 − 0.15y2 + 0.09y3 + 0.631y4 − 0.57y5 + 0.23y6 > 2.5481, − dih2 − 0.454y1 + 0.34y2 + 0.154y3 − 0.346y4 + 0.805y5 > −0.3429, dih3 + 0.4y1 − 0.15y3 + 0.09y2 + 0.631y4 − 0.57y6 + 0.23y5 > 2.5481, − dih3 − 0.454y1 + 0.34y3 + 0.154y2 − 0.346y4 + 0.805y6 > −0.3429, sol + 0.065y2 + 0.065y3 + 0.061y4 − 0.115y5 − 0.115y6 > 0.2618,

− sol − 0.293y1 − 0.03y2 − 0.03y3 + 0.12y4 + 0.325y5 + 0.325y6 > 0.2514, −ν − 0.0538y2 − 0.0538y3 − 0.083y4 − 0.0538y5 − 0.0538y6 > −0.5995, ν ≤ 0,

(Calculations F.3.13.3, F.3.13.4)

τν − 0.5945 pt > 0.

(A.4.5.1)

(498839271)

ν − 4.10113 dih1 < −4.3223,

ν − 0.80449 dih1 < −0.9871, ν − 0.70186 dih1 < −0.8756,

ν − 0.24573 dih1 < −0.3404,

ν − 0.00154 dih1 < −0.0024, ν + 0.07611 dih1 < 0.1196.

(A.4.5.2, IV.A2 )

48

THOMAS C. HALES

τν + 4.16523 dih1 > 4.42873, τν + 0.78701 dih1 > 1.01104, τν + 0.77627 dih1 > 0.99937, τν + 0.21916 dih1 > 0.34877, τν + 0.05107 dih1 > 0.11434, (A.4.5.3, IV.A3 )

τν − 0.07106 dih1 > −0.07749.

The following additional inequalities are known to hold if the upright diagonal has height at most 2.696. νΓ denotes the restriction of ν to a simplex of compression type. y1 < 2.696, dih1 − 0.49y1 + 0.44y2 + 0.44y3 − 0.82y4 + 0.44y5 + 0.44y6 > 2.0421, − dih1 + 0.495y1 − 0.214y2 − 0.214y3 + 1.05y4 − 0.214y5 − 0.214y6 > 0.2282, dih2 + 0.38y1 − 0.15y2 + 0.09y3 + 0.54y4 − 0.57y5 + 0.24y6 > 2.3398,

− dih2 − 0.375y1 + 0.33y2 + 0.11y3 − 0.36y4 + 0.72y5 + 0.034y6 > −0.36135, sol + 0.42y1 + 0.165y2 + 0.165y3 − 0.06y4 − 0.135y5 − 0.135y6 > 1.479,

− sol − 0.265y1 − 0.06y2 − 0.06y3 + 0.124y4 + 0.296y5 + 0.296y6 > 0.0997, −ν + 0.112y1 − 0.142y2 − 0.142y3 − 0.16y4 − 0.074y5 − 0.074y6 > −0.9029, ν + 0.07611 dih1 < 0.11,

νΓ − 0.015y1 − 0.16(y2 + y3 + y4 ) − 0.0738(y5 + y6 ) > −1.29285, τν − 0.07106 dih1 > −0.06429, τν > 0.0414.

(A.4.5.4) (319046543) In connection with Inequalities 4.5.4, we occasionally use the stronger constant 0.2345 instead of 0.2282. To justify this constant, we have checked using interval arithmetic that the the bound 0.2345 holds if y1 ≤ 2.68 or y4 ≤ 2.475. Further interval calculations show that the anchored simplices can be erased if they share an upright diagonal with such a quarter. Inequalities (4.5.5)

(365179082)

vor0 < −0.043/2, if y6 = 2.51, y1 ∈ [2.51, 2.696]. (368244553†) √ vor0 (S) + vor0 (S(2, y2 , y3 , y4 , 2, 2)) < −0.043, if y1 ∈ [2.51, 2.696], y4 ∈ [2 2, 3.2], y4′ ≥ 2.51. (820900672†) √ vor0 (S)+vor0 (S(2.51, y2 , y3 , y4 , 2, 2)) < −0.043, if y1 ∈ [2.51, 2.696], y4 ∈ [2 2, 3.2], y4′ ≥ 2.51. (961078136†) Inequalities (4.5.7), (The last of these was verified by S. Ferguson.) (424186517)

THE KEPLER CONJECTURE

Section A.4.6.1.a†. √ vor0 < −0.212 − 0.0461 + 0.137, if y4 = 3.2, y5 ∈ [2 2, 3.2]. √ τ0 > 0.54525 − 0 − 0.31. if y4 = 3.2, y5 ∈ [2 2, 3.2].

49

(725257062) (977272202)

Section A.4.6.1.b†. Let R be a pentagonal region with parameters √ (2, 2, a2 , 2, a3 , 2, a4 , 2, 2, 2 2),

a2 , a3 , a4 ∈ {2, 2.51}.

Since this is a pentagonal region, we may discard any edge-length combinations that produce ∆ < 0. Assume |v1 − v3 | ≥ 3.2, |v3 − v5 | ≥ 3.2. Under these conditions the following inequalities hold. vor0 +0.0461 < −0.212 (583626763) τ0 > 0.54525 (390951718) Section A.4.6.1.c†. Let R be a pentagonal region with parameters (a1 , 2, a2 , 2, a3 , 2, a4 , 2, a5 , b5 ),

ai ∈ {2, 2.51}.

Since this is a pentagonal region, we may discard any edge-length √ combinations that produce ∆ < 0. Assume |v1 − v3 |, |v3 − v5 |, b5 = |v5 − v1 | ∈ [2 2, 3.2]. Under these conditions the following inequalities hold. vor0 +0.0461 < −0.212 (621852152) τ0 > 0.54525 (207203174) Section A.4.6.1.d†. Let R be a hexagonal region with parameters (a1 , 2, a2 , 2, a3 , 2, a4 , 2, a5 , 2, a6 , 2),

ai ∈ {2, 2.51}.

Since this is a hexagonal region, we may discard√any edge-length combinations that produce ∆ < 0. Assume |v1 − v5 |, |v1 − v3 | ∈ [2 2, 3.2], |v1 − v4 | ≥ 3.2, |v3 − v5 | ≥ √ 2 2. Under these conditions the following inequalities hold. vor0 < −0.212. (368258024) τ0 > 0.54525. (564618342) Section A.4.6.1.e†. Let R be a hexagonal region with parameters (a1 , 2, a2 , 2, a3 , 2, a4 , 2, a5 , 2, a6 , 2),

ai ∈ {2, 2.51}.

Since this is a hexagonal region, we may discard any √ edge-length combinations that produce ∆ < 0. Assume |v2 − v6 |, |v3 − v5 | ∈ [2 2, 3.2], |v2 − v5 | ∈ [3.2, 3.78], |v3 − v6 | ≥ 3.2. Under these conditions the following inequalities hold. vor0 < −0.212. (498774382) τ0 > 0.54525. (544865225)

50

THOMAS C. HALES

Section A.4.6.2.a†. √ √ √ vor0 < −0.221 − 2(0.009), if y4 = 2 2, y5 ∈ [2 2, 3.2], and y6 ∈ [2.51, 2 2]. (234734606) √ √ √ τ0 > 0.486−2(0.05925), if y4 = 2 2, y5 ∈ [2 2, 3.2], y6 ∈ [2.51, 2 2]. (791682321) Section A.4.6.2.b†. Let Q be a quadrilateral region with parameters (a1 , b1 , a2 , 2, a3 , 2, a4 , 2.51),

a2 , a3 , a4 ∈ {2, 2.51}.

Since this is a quadrilateral region, we may discard any edge-length combinations √ that produce ∆ < 0. Assume |v − v | ≥ 3.2, |v − v | ≥ 2 2. b = |v1 − v2 | ∈ 1 3 2 4 1 √ [2 2, 3.2]. Under these conditions the following inequalities hold. vor0 (Q) < −0.221 − 0.009. (995351614) τ0 (Q) > 0.486 − 0.05925. (321843503) Section A.4.6.2.c. √ √ vor0 < −0.19 − (y5 − 2)0.14, if y4 = 2, y5 ∈ [2 2, 3.2], and y6 ∈ [3.2, 3.47]. (354217730) √ (595674181) τ0 > 0.281, if y4 = 2, y5 ∈ [2 2, 3.2], y6 ∈ [3.2, 3.23]. vor0 < −0.11, if y4 = 2, y5 = 2.51, y6 = 3.2. (547486831†) τ0 > 0.205, if y4 = 2, y5 = 2.51, y6 = 3.2. (683897354†) √ √ vor0 < 0.009 + (y5 − 2 2)0.14, if y5 ∈ [2 2, 3.2], y4 = y6 = 2. refno938003786 Section A.4.6.2.d†. Let Q be a quadrilateral region with parameters √ (a1 , 2, a2 , 2 2, a3 , 2, a4 , 2.51),

ai ∈ {2, 2.51}.

Since this is a quadrilateral region, we may discard any edge-length combinations that produce ∆ < 0. Assume |v1 − v3 | ≥ 3.2, |v2 − v4 | ≥ 3.2. Under these conditions the following inequalities hold. vor0 (Q) < −0.221 − 0.0461. (109046923) τ0 (Q) > 0.486. (642590101) Section A.4.6.2.e†. Let R be a pentagonal region with parameters (2, 2, a2 , 2, 2, 2, a4, 2, 2, 2.51),

a2 , a4 ∈ {2, 2.51}.

Since this is a pentagonal region, we may √ combinations that √ discard any edge-length produce ∆ < 0. Assume |v1 − v3 | ∈ [2 2, 3.2], |v3 − v5 | ∈ [2 2, 3.2]. Under these conditions the following inequalities hold. vor0 (R) < −0.221. (160800042) τ0 (R) > 0.486. (690272881)

THE KEPLER CONJECTURE

51

Section A.4.6.2.f†. Let R be a pentagonal region with parameters (a1 , 2, 2, 2, 2, 2, 2, 2, a5, 2.51),

a1 , a5 ∈ {2, 2.51}.

Since this is a pentagonal region, we may discard any edge-length combinations that produce ∆ < 0. Assume |v1 − v3 | ≥ 3.2, |v3 − v5 | ≥ 3.2. Assume dih(0, v3 , v4 , v5 ) + dih(0, v3 , v5 , v1 ) + dih(0, v3 , v1 , v2 ) ≥ dih(S(2, 2, 2, 3.2, 2, 2)) = arccos(−53/75).

Under these conditions the following inequalities hold. vor0 (R) < −0.221. τ0 (R) > 0.486.

(713930036) (724922588)

Section A.4.6.2.g†. Let R be a pentagonal region with parameters (2, 2, a2 , 2, 2, 2, a4, 2, a5 , 2.51),

a2 , a4 , a5 ∈ {2, 2.51}.

Since this is a pentagonal region, we may √ discard any edge-length combinations that produce ∆ < 0. Assume |v1 −v3 | ∈ [2 2, 3.2], |v1 −v4 | ≥ 3.2, |v3 −v5 | ≥ 3.2. Under these conditions the following inequalities hold. vor0 (R) < −0.221. (821730621) τ0 (R) > 0.486. (890642961) Section A.4.6.3†. √ √ vor0 < −0.168 − 0.009, y4 = 2 2, y5 , y6 ∈ [2.51, 2 2]. √ √ τ0 > 0.352 − 0.05925, y4 = 2 2, y5 , y6 ∈ [2.51, 2 2]. Let Q be a quadrilateral region with parameters (a1 , 2.51, a2, 2, a3 , 2, a4 , 2.51),

(341667126) (535906363)

ai ∈ {2, 2.51}.

√ Assume that |v2 − v4 | ∈ [2 2, 3.2], |v1 − v3 | ∈ [3.2, 3.46]. Note that ∆(4, 4, 8, 2.512, 2.512, 3.462 ) < 0. vor0 (Q) < −0.168. τ0 (Q) > 0.352.

(302085207) (411491283)

Section A.4.6.8†. √ (516537931) vor0 < −0.146, if y5 , y6 ∈ [2 2, 3.2]. √ ′ τ0 (S(y1 , . . . , y6 )) + τ0 (S(y1 , y2 , y3 , 2, y5 , 2)) > 0.31, if y5 , y6 ∈ [2 2, 3.2], and y2′ ∈ {2, 2.51}. (130008809)

52

THOMAS C. HALES

Section A.4.6.10†, A.4.6.11†. √ √ (531861442) vor0 < −0.084, if y5 ∈ [2.51, 2 2], y6 ∈ [2 2, 3.2]. √ √ vor0 < −0.084 − (y5 − 2)0.1, if y1 = y3 = y4 = 2, y6 = 2.51, y5 ∈ [2 2, 3.2]. (292827481) √ √ vor0 < 0.009 + (y5 − 2 2)0.1, if y5 ∈ [2 2, 3.2], y1 = y3 = y4 = y6 = 2. (710875528) √ √ (286122364) τ0 > 0.176, if y5 ∈ [2.51, 2 2], y6 ∈ [2 2, 3.2]. Section A.4.7.1. Consider a pentagonal region. If the pentagonal region has one flat quarter and no upright quarters, there is a quadrilateral region F . It satisfies σF < −0.075,

(4.6.4)

τF > 0.176.

Break the cluster into two simplices S = S(y1 , . . . , y6 ), S ′ = S(y1′ , y2 , y3√ , y4 , y5′ , y6′ ), ′ by drawing a diagonal of length y4 . Assume that the edge y5 ∈ [2.51, 2 2]. Let y4′ be the length of the diagonal that crosses y4 . σF < 2.1327 − 0.1y1 − 0.15y2 − 0.08y3 − 0.15y5 − 0.15y6 − 0.1y1′ − 0.17y5′ − 0.16y6′ ,

√ if dih(S) < 1.9, dih(S ′ ) < 2.0, y1 ∈ [2, 2.2], y4 ≥ 2 2,

σF < 2.02644 − 0.1y1 − 0.14(y2 + y3 ) − 0.15(y5 + y6 ) − 0.1y1′ − 0.12(y5′ + y6′ ), if y1 ∈ [2, 2.08],

y4 ≤ 3.

σF + 0.419351 sol < 0.4542 + 0.0238(y5 + y6 + y6′ ), √ if y4 , y4′ ≥ 2 2.

(A.4.7.1)

The inequalities in (4.7.1) are verified in smaller pieces: √ vor0 < 1.01 − 0.1y1 − 0.05y2 − 0.05y3 − 0.15y5 − 0.15y6, if dih ≤ 1.9, y4 ≥ 2 2, and y1 ≤ 2.2. (131574415) √ vor0 < 1.1227 √ − 0.1y1 − 0.1y2 − 0.03y3 − 0.17y5 − 0.16y6, if dih ≤ 2, y4 ≥ 2 2, y5 ∈ [2.51, 2 2], and y2 + y3 ≤ 4.67. (929773933) √ vor0 < 1.0159 − 0.1y1 − 0.08(y2 + y3 ) + 0.04y4 − 0.15(y5 + y6 ), if y4 ∈ [2 2, 3], y1 ∈ [2, 2.08]. (223261160) √ vor0 < 1.01054 √ − 0.1y1 − 0.06(y2 + y3 ) − 0.04y4 − 0.12(y5 + y6 ), if y4 ∈ [2 2, 3], y5 ∈ [2.51, 2 2]. (135018647) Let Q be the quadrilateral subcluster expressed as a union of two simplices S and S ′ , as in Section + 0.0238(y5 + y6 + y6′ ), √ 4.7.1. vor0 (Q) + 0.419351 sol(Q) < 0.4542 √ ′ if y5 ∈ [2.51, 2 2] and both diagonals have length ≥ 2 2. (559676877)

THE KEPLER CONJECTURE

53

Section A.4.7.2. (These inequalities are closely related to IV.A21 .) ′ ′ vor0 (S(2,√ 2, 2, y4, 2, 2))+vor 0 (S(2, 2, 2, y4 , 2, 2))+vor0 (S(2, 2, 2, y4 , y4 , 2)) < −0.128, √ (587781327†) if y4 ∈ [2 2, 3.2], y4′ ∈ [2 2, 3.23]. ′ ′ τ0 (S(2,√2, 2, y4, 2, 2)) +√ τ0 (S(2, 2, 2, y4 , 2, 2)) + τ0 (S(2, 2, 2, y4 , y4 , 2)) > 0.36925, if (807067544†) y4 ∈ [2 2, 3.2], y4′ ∈ [2 2, 3.23]. √ (986970370†) τ0 (2, 2, y3 , y4 , 2, 2) < τ0 (2.51, 2, y3, y4 , 2, 2), if y4 ∈ [2 2, 3.06]. √ vor0 (2, 2, y3 , y4 , 2, 2) > vor0 (2.51, 2, y3, y4 , 2, 2), if y4 ∈ [2 2, 3.06]. (677910379†) vor0 < −0.128, if y1 = y2 = y4 = 2, y5 , y6 ∈ [3.06, 3.23]. (276168273) τ0 > 0.36925, if y1 = y2 = y3 = y4 = 2, y5 , y6 ∈ [3.06, 3.23]. (411203982) τ0 > 0.31, if y1 = y2 = y4 = 2, y3 = 2.51, y5 , y6 ∈ [3.06, 3.23]. (860823724) √ √ vor0 < −0.137 − (y5 − 2 2)0.14, if y1 = y4 = 2, y5 ∈ [2 2, 3.23], y6 ∈ [3.06, 3.23]. (353116955) √ √ τ0 > 0.31 + (y5 − 2 2)0.14, if y1 = y4 = 2, y5 ∈ [2 2, 3.23], y6 ∈ [3.105, 3.23]. (943315982) √ √ τ0 > 0.31 + (y5 − 2 2)0.14 + (y6 − 3.105)0.19, if y1 = y4 = 2, y5 ∈ [2 2, 3.23], y6 ∈ [3.06, 3.105]. (941799628) √ √ vor0 < 0.009 + (y5 − 2 2)0.14, if y1 = y4 = 2, y5 ∈ [2 2, 3.23], y2 = y6 = 2. (674284283) √ √ τ0 > 0.05925 + (y5 − 2 2)0.14, if y1 = y4 = 2, y5 ∈ [2 2, 3.23], y2 = y6 = 2. (775220784) √ τ0 > 0.05925, if y3 = 2.51, y5 = y6 = 2, y4 ∈ [2 2, 3.23]. (286076305) τ0 > −(y4 − 3.105)0.19, if y1 = 2.51, y2 = y3 = y5 = y6 = 2, y4 ∈ [3.06, 3.105]. (589319960) Section A.4.8†. We give lower and upper bounds on the dihedral angles along each edge (0, v) of each of the smaller pieces obtained in Section 4.1. The domains that we list are not disjoint. In general we consider an edge as belonging to the most restrictive domain that the information of the diagrams of 4.1 permit us to conclude that it lies in. The following chart summarizes the bounds. The dihedral angle is computed along the first edge. The chart is divided into three sections. In the first, there is no upright diagonal. In the second, the upright diagonal is the first edge. In the third, the upright diagonal is the third edge. The bounds in the second section have been established in IV.A8 . In the first group y1 , y2 , y3 ∈ [2, 2.51]. In the third row, the dihedral bound 1.624 holds for y6 on the larger interval [2.51, 3.02]. In the seventh row, the dihedral bound

54

THOMAS C. HALES

1.507 holds for y5 , y6 on the larger interval [2.51, 3.02]. y5 [2, 2.51] [2, 2.51] [2, 2.51] [2, 2.51] [2, 2.51] [2, 2.51] √ [2.51, 2√2] [2.51, 2√2] [2.51, 2√2] [2.51, 2 2]

y6 [2, 2.51] [2, 2.51] √ [2.51, 2√2] [2.51, 2√2] [2.51, 2√2] [2.51, 2√2] [2.51, 2√2] [2.51, 2√2] [2.51, 2√2] [2.51, 2 2]

y4 √ [2.51, √ 2 2] ≥2 2 [2, 2.51] √ [2.51, 2 2] ≥ 2.51 √ ≥2 2 [2, 2.51] √ [2.51, 2 2] ≥ 2.51 √ ≥2 2

dihmin 1.153 1.32 0.633 1.033 1.033 1.259 0.817 1.07 1.07 1.23

dihmax 2.28 2π 1.624 1.929 2π 2π 1.507 1.761 2π 2π

√ In the second group y1 ∈ [2.51, 2 2], y2 , y3 ∈ [2, 2.51]. y5 [2, 2.51] [2, 2.51] [2, 2.51] [2, 2.51]

y6 y4 [2, 2.51] [2, 2.51] √ [2, 2.51] [2.51, 2 2] [2, 2.51] ≥ 2.51 √ [2, 2.51] ≥2 2

dihmin 0.956 1.23 1.23 1.416

dihmax 2.184 π π π

√ In the third group y1 , y2 ∈ [2, 2.51], y3 ∈ [2.51, 2 2]. y5 y6 [2, 2.51] [2, 2.51] [2, 2.51] [2, 2.51] √ [2, 2.51] [2.51, 2√2] [2, 2.51] [2.51, 2 2] The table of dihedral angle bounds.

y4 dihmin [2, 2.51] 0.633 ≥ 2.51 1.033 [2, 2.51] 0 ≥ 2.51 0.777

dihmax 1.624 2π 1.381 2π (853728973)

Section A.4.9†. (The verifications in this section that do not involve the score were verified by S. Ferguson.) Depending on the lengths of the edges y5 , y6 , y4 , there are additional inequalities that hold. Generally these inequalities are for a fragment of a subregion. If v2 , v1 , v3 are consecutive corners, we form a simplex S(y1 , y2 , y3 , y4 , y5 , y6 ) = (0, v1 , v2 , v3 ). There is no score intrinsically associated with this simplex, unless the subregion is triangular. But we can state various useful bounds on the dihedral angles along the first edge (0, v1 ). A triple (a, b, c) preceding the inequality gives bounds on the edges (y5 , y6 , y4 ), respectively.

THE KEPLER CONJECTURE

55

In the first group, y1 , y2 , y3 ∈ [2, 2.51]. (y5 , y6 , y4 )

√ ([2, 2.51], [2, 2.51], ≥ 2 2) dih −0.372y1 + 0.465y2 + 0.465y3 + 0.465y5 + 0.465y6 > 4.885, √ √ ([2, 2.51], [2.51, 2 2], [2.51, 2 2]) 0.291y1 − 0.393y2 − 0.586y3 + 0.79y4 − 0.321y5 − 0.397y6 − dih < −2.47277, √ ([2, 2.51], [2.51, 2 2], ≥ 2.51) 0.291y1 − 0.393y2 − 0.586y3 − 0.321y5 − 0.397y6 − dih < −4.45567, √ √ ([2, 2.51], [2.51, 2 2], ≥ 2 2)

0.291y1 − 0.393y2 − 0.586y3 − 0.321y5 − 0.397y6 − dih < −4.71107, dih −0.214y1 + 0.4y2 + 0.58y3 + 0.155y5 + 0.395y6 > 4.52345, √ √ ([2.51, 2 2], [2.51, 2 2], [2, 2.51]) τF > D(3, 2),

(IV.A16 )

σF < Z(3, 2),

(IV.A16 )

− sol −0.492y1 − 0.492y2 − 0.492y3 + 0.43y4 +

0.038y5 + 0.038y6 < −2.71884, − σF − 0.058y1 − 0.105y2 − 0.105y3 − 0.115y4−

0.062y5 − 0.062y6 > −1.02014, σF + 0.419351 sol < 0.3085,

0.115y1 − 0.452y2 − 0.452y3 + 0.613y4 − 0.15y5 − 0.15y6 − dih < −2.177, √ √ √ ([2.51, 2 2], [2.51, 2 2], [2.51, 2 2]) 0.115y1 − 0.452y2 − 0.452y3 + 0.618y4 − 0.15y5 − 0.15y6 − dih < −2.17382, σF < −0.121,

τF > D(3, 3) = 0.21301, (IV.A17 ) √ √ ([2.51, 2 2], [2.51, 2 2], ≥ 2.51)

0.115y1 − 0.452y2 − 0.452y3 − 0.15y5 − 0.15y6 − dih < −3.725, √ √ √ ([2.51, 2 2], [2.51, 2 2], ≥ 2 2) 0.115y1 − 0.452y2 − 0.452y3 − 0.15y5 − 0.15y6 − dih < −3.927.

(A.4.9.1)

56

THOMAS C. HALES

√ In the next group y1 ∈ [2.51, 2 2], y2 , y3 ∈ [2, 2.51]. (y5 , y6 , y4 )

√ ([2, 2.51], [2, 2.51], [2.51, 2 2]) σF < 0,

cf. (4.5.5)

0.47y1 − 0.522y2 − 0.522y3 + 0.812y4 − 0.522y5 − 0.522y6 − dih < −2.82998,

([2, 2.51], [2, 2.51], ≥ 2.51) 0.47y1 − 0.522y2 − 0.522y3 − 0.522y5 − 0.522y6 − dih < −4.8681, √ ([2, 2.51], [2, 2.51], ≥ 2 2) 0.47y1 − 0.522y2 − 0.522y3 − 0.522y5 − 0.522y6 − dih < −5.1623.

(A.4.9.2)

√ In the next group y1 , y2 ∈ [2, 2.51], y3 ∈ [2.51, 2 2]. (y5 , y6 , y4 ) ([2, 2.51], [2, 2.51], ≥ 2.51)

− 0.4y3 + 0.15y1 − 0.09y2 − 0.631y6 − 0.23y5 − dih < −3.9788, √ ([2, 2.51], [2.51, 2 2], [2, 2.51]) 0.289y1 − 0.148y2 − 1.36y3 + 0.688y4 − 0.148y5 − 1.36y6 − dih < −6.3282, √ ([2, 2.51], [2.51, 2 2], ≥ 2.51) 0.289y1 − 0.148y2 − 0.723y3 − 0.148y5 − 0.723y6 − dih < −4.85746.

(A.4.9.3)

Inequalities (4.9.1) Inequalities (4.9.2) Inequalities (4.9.3)

(529738375) (456320257) (664959245)

Section A.4.10†. If v is a vertex of an exceptional cluster and there are exactly 4 quasi-regular tetrahedra along (0, v), then there are 5 vertices v1 , . . . , v5 adjacent to v. We have X (|v − vi | + |vi |) > 20.42. (4.10.2) (5)

If v1 and v5 are the two √ vertices on an exceptional cluster, and if the additional hypothesis |v1 − v5 | ≥ 2 2 holds, then (4.10.3)

X (5)

Inequalities (4.10.2) Inequalities (4.10.3)

(|v − vi | + |vi |) > 20.76. (615073260) (844430737)

THE KEPLER CONJECTURE

57

Section A.4.12.

(A.4.12.1)

ν < −0.055 and τν > 0.092,

√ if y1 ∈ [2.696, 2 2], y2 , y6 ∈ [2.45, 2.51]. (A.4.12.2)

(704795925)

σ ˆ < −0.039 and τˆ > 0.094,

√ if y2 ∈ [2.45, 2.51], y4 ∈ [2.51, 2 2]. (A.4.12.3)

(332919646)

vor < −0.197

and τV > 0.239,

vor0 < −0.136

and τ0 > 0.224,

√ √ if y1 ∈ [2.696, 2 2], y2 ∈ [2.45, 2.51], y6 ∈ [2.45, 2.51], and y4 ∈ [2.51, 2 2]. (335795137)

(A.4.12.4)

for a combination of anchored simplex S and special simplex S ′ , with y1 (S) ∈ √ √ [2.696, 2 2], y2 (S), y6 (S) ∈ [2.45, 2.51], y4 (S) ∈ [2 2, 3.2], and with cross-diagonal at least 2.51. Inequality A.4.12.4 can be verified by proving the following inequalities √ in√lower dimension. In the first four y1 ∈ [2.696, 2 2], y2 , y6 ∈ [2.45, 2.51], y4 ∈ [2 2, 3.2], and y4′ ≥ 2.51 (the cross-diagonal). vor0 (S(y1 , . . . , y6 )) + vor0 (S(2, y2 , y3 , y4 , 2, 2)) < −0.136 (967376139†) vor0 (S(y1 , . . . , y6 )) + vor0 (S(2.51, y2 , y3 , y4 , 2, 2)) < −0.136 (666869244†) τ0 (S(y1 , . . . , y6 )) + τ0 (S(2, y2 , y3 , y4 , 2, 2)) > 0.224 (268066802†) τ0 (S(y1 , . . . , y6 )) + τ0 (S(2.51, y2, y3 , y4 , 2, 2)) > 0.224 (508108214†) √ vor0 (S(y1 , . . . , y6 )) < −0.125, if y1 ∈ [2.696, 2 2], y2 , y6 ∈ [2.45, 2.51], y5 = 2.51. (322505397†) √ vor0 (S(y1 , . . . , y6 )) < 0.011, if y1 ∈ [2.696, 2 2], y5 = 2.51. (736616321†) √ τ0 (S(y1 , . . . , y6 )) > 0.17, if y1 ∈ [2.696, 2 2], y2 , y6 ∈ [2.45, 2.51], y5 = 2.51. (689417023†) √ (748466752†) τ0 (S(y1 , . . . , y6 )) > 0.054, if y1 ∈ [2.696, 2 2], y5 = 2.51. (A.4.12.5)

vor0 < −0.24 and τ0 > 0.346,

for an anchored simplex S and simplex S ′ with edge parameters (3, 2) in a hexagonal √ ′ ′ ′ cluster, with √ y2 (S) = y2 (S ), y3 (S) = y3 (S ), y4 (S) = y4 (S ), y1 (S) ∈ [2.696, 2 2], y4 (S) ∈ [2 2, 3.2], y2 (S), y6 (S) ∈ [2.45, 2.51], and √ max(y5 (S ′ ), y6 (S ′ )) ∈ [2.51, 2 2],

min(y5 (S ′ ), y6 (S ′ )) ∈ [2, 2.51].

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THOMAS C. HALES

This breaks into separate interval calculations for S and S ′ . Inequality (A.4.12.5) results from the following four inequalities vor0 (S) < −0.126 and τ0 (S) > 0.16 (369386367†) vor0 (S ′ ) < −0.114 and τ0 (S ′ ) > 0.186 (There are two cases for each, depending on which of y5 , y6 is longer.) (724943459†) (A.4.12.6)

vor0 < −0.089 √ if y1 ∈ [2.45, 2.51], y5 , y6 ∈ [2.51, 2 2].

and τ0 > 0.154,

(A.4.12.7)

and τ0 > 0.154,

vor0 < −0.089 √ if y1 ∈ [2.45, 2.51], y4 , y5 ∈ [2.51, 2 2].

(A.4.12.8)

(605071818†)

(642806938†)

vor0 < −0.149

and τ0 > 0.281, √ for a quadrilateral subcluster with both diagonals ≥ 2 2, and parameters

(a1 , b1 , a2 , b2 , a3 , b3 , a4 , b4 ), √ with a4 ∈ [2.45, 2.51], and b4 ∈ [2.51, 2 2]. (A.4.12.9)

(836331201†)

vor0 < −0.254 and τ0 > 0.42625,

for a combination of anchored simplex S and quadrilateral cluster Q. It is assumed √ 2.51]. The adjacent quadrilateral that y1 (S) ∈ [2.696, 2 2], y2 (S), y6 (S) ∈ [2.45, √ subcluster is assumed to have both diagonals ≥ 2 2, and parameters (a1 , b1 , a2 , b2 , a3 , b3 , a4 , b4 ), √ with b4 ∈ [2 2, 3.2]. The verification of this inequality reduces to separate inequalities for the anchored simplex and quadrilateral subcluster. For the anchored simplex we use the bounds vor0 (S ′ ) < −0.126, τ0 (S ′ ) > 0.16 that have already been established above. We then show that the quad cluster satisfies vor0 < −0.128 and τ0 > 0.26625. (327474205†) √ For this, use deformations to reduce either to the case where the diagonal is 2 2, or to √ the case where b1 = b2 = b3 = 2, a2 , a3 ∈ {2, 2.51}. When the diagonal (There are two is 2 2, the flat quarter can be scored by IV.A13 (0.009,0.05925). √ cases depending on which direction the diagonal of length 2 runs.) Section A.6. √ octavor < octavor0 −0.017, if y1 ∈ [2.51, 2.696], and η126 ≥ 2. Inequality (6.2)

(104506452) (601083647)

Section A.6.3. Γ < 0.3138 − 0.157y5, if y5 ∈ [2, 2.138], y4 ∈ [2.51, 2.6], (543730647) √ Γ < −0.06, if y2 ∈ [2.121, 2.145], y4 ∈ [2.51, 2 2], y5 ∈ [2.22, 2.238]. (163030624) √ Γ < 10−6 + 1.4 − 0.1y1 − 0.15(y2 + y3 + y5 + y6 ), if y4 ∈ [2, 2 2], y1 , y2 , y3 ∈ [2, 2.2], y5 , y6 ∈ [2, 2.35], (181462710)

THE KEPLER CONJECTURE

59

Appendix 2. Two Final Cases Two cases are unexpectedly difficult to eliminate, PM(5, 663), PM(6, 131). They are treated in this appendix. Several additional inequalities are used. When the exceptional region has upright quarters, the arguments from Section 6 work without difficulty. The case considered here is an exceptional region triangulated by flat quarters and simplices with edge parameters (3, 2). Flat Quarters. An LP-bound shows that the vertices have height at most 2.14. If the diagonal has length less than 2.77, then the circumradius √ of the face containing the origin and diagonal is at most η(2.14, 2.14, 2.77) < 2. This allows us to combine the cases defining σ ˆ into three cases. (1) The √ simplex has compression type. (2) The diagonal has length ≤ 2.7 and η456 ≥ 2. (3) The diagonal has length ≥ 2.7. In the last two cases, the scoring is by vor0 . √ vor < vor0 if y4 ∈ [2.7, 2 2], y1 , y2 , y3 ∈ [2, 2.14]. (463544803) √ (399326202) vor0 < −0.064, if y4 ∈ [2.51, 2.72], η456 ≥ 2. √ vor0 < 1.0612 − 0.08(y1 + y2 + y3 )− 0.142(y5 + y6 ), if y4 ∈ [2.7, 2 2]. (569240360) vor0 < −0.0713, if y4 ∈ [2.59, 2.64], y5 ∈ [2.47, 2.51], y6 ∈ [2.1, 3.51]. (252231882) √ (472436131) vor0 < −0.06, if y1 , y2 , y3 ∈ [2, 2.13], y4 ∈ [2.7, 2.74], η456 ≥ 2. √ vor0 < −0.058, if y4 ∈ [2.51, 2.747], η456 ≥ 2. (913534858) √ vor0 < −0.0498, if y4 ∈ [2.51, 2.77], η456 ≥ 2. (850226792) Flat Quarters of compression type. We assume that y1 , y2 , y3 ≤ 2.14 in the following inequalities. √ −Γ − 0.145y1 − 0.08(y2 + y3 ) − 0.133(y5 + y6 ) > −1.146, if y4 ∈ [2.51, 2 2]. (594246986) √ −Γ − 0.145y1 − 0.081(y2 + y3 ) − 0.16(y5 + y6 ) > −1.255, if y4 ∈ [2.51, 2 2], and y5 , y6 ∈ [2, 2.3]. (381970727) √ −Γ − 0.03y1 − 0.03(y2 + y3 ) − 0.094(y5 + y6 ) > −0.5361. if y4 ∈ [2.51, 2 2], y5 + y6 ≥ 4.3. (951798877) √ −Γ − 0.03y1 − 0.03(y2 + y3 ) − 0.16(y5 + y6 ) > −0.82 − 10−6 . if y4 ∈ [2.51, 2 2], y5 + y6 ≤ 4.3. (923397705) √ (312481617) Γ < −0.053 if y4 ∈ [2.51, 2 2], y5 ≥ 2.35. √ (292760488) Γ < −0.041, if y4 ∈ [2.51, 2 2], y5 ∈ [2.25, 2.51]. √ Γ + 0.419351 sol < 0.079431 dih +0.0436(y + y ) − 0.0294, if η 2, y4 ∈ 5 6 456 ≤ √ (155008179) [2.51, 2 2]. Γ < 1.1457 − 0.1(y1 + y2 + y3 )− 0.17y5 − 0.11y6 , if y1 , y2 , y3 ∈ [2, 2.13], y5 ∈ [2, 2.1], y6 ∈ [2.27, 2.43], y4 ∈ [2.51, 2.67]. (819450002) √ (495568072) 1.69y4 + y5 + y6 > 9.0659, if y4 ∈ [2.51, 2.7] and η456 ≥ 2. √ 1.69y4 + y5 + y6 > 9.044, if y4 ∈ [2.51, 2.77] and η456 ≥ 2. (838887715) √ (794413343) y5 + y6 > 4.4, if η456 ≥ 2, and y4 ≤ 2.72.

60

THOMAS C. HALES

Simplices with edge parameters (3,2). The hexagon is divided into two quarters and two simplices with edge parameters √ √ (3, 2), that is, there are two edges of length in [2.51, 2 2], say y5 , y6 ∈ [2.51, 2 2], y4 ∈ [2, 2.51]. These may be simplices of type SA . We have seen that the diagonal of the hexagon is at most 2.77. This allows the simplices to be divided into the following domains. (1) The simplex has type SA , so that y5 , y6 ∈ [2.51, 2.77], and the scoring is vor, the analytic Voronoi function. (2) The edge (say y6 ) of the face shared with the flat quarter has √ length y6 ≥ 2.77. (3) The edges have lengths y5 , y6 ∈ [2.51, 2.77], and η456 ≥ 2. In the last two cases, the scoring is by the truncated Voronoi function vor0 . − vor −0.058y1 − 0.08y2 − 0.08y3 − 0.16y4 − 0.21(y5 + y6 ) > −1.7531, in situation (1), if y1 , y2 , y3 ≤ 2.14. (378020227) − vor0 −0.058y1 −0.1y2 −0.1y3 −0.165y √4 −0.115y6 −0.12y5 > −1.38875, in situation (2), so that in particular y6 ∈ [2.77, 2 2], if y1 , y2 , y3 ≤ 2.14. (256893386) y4 + y5 + y6 > 7.206, in situation (3), (749955642) − vor0 −0.058y1 −0.05y2 −0.05y3 −0.16y4 −0.13y6 −0.13y5 > −1.24547, in situation (3), if y1 , y2 , y3 ≤ 2.14. (653849975) √ √ (480930831) vor0 < −0.077, if y4 ∈ [2.51, 2 2], y5 ∈ [2.77, 2 2]. vor +0.419351 sol < 0.289, if y4 , y5 ∈ [2.51, 2.77]. (271703736) vor0 < 1.798 − 0.1(y1 + y2 + y3 ) − 0.19y4 − 0.17(y5 + y6 ), if y5 , y6 ∈ [2.7, 2.77]. (900212351) √ If there is an enclosed vertex of height at most 2, then we can use √ vor < −0.078/2, y1 ∈ [2.51, 2 2], y4 ∈ [2.51, 2.6961]. (455329491) √ √ (857241493) vor(S, 2) < −0.078, if y5 , y6 ∈ [2.51, 2.6961], η456 ≥ 2.

THE KEPLER CONJECTURE

61

Appendix 3. Compatibility Notes It has been useful to make various changes in the program that was published in Sphere Packings I. This appendix makes a few comments about the global compatibility of the results and terminology from various papers. The definition of quasi-regular octahedron in Sphere Packings I is obsolete. The definition that is used appears in Section I of [F]. Also, there is an old definition of standard cluster for Delaunay stars that should be replaced with the standard cluster in a decomposition star in [F]. The Sections I.8.6.4, I.8.6.5, I.8.6.6, I.8.6.7 are no longer needed because of improvements in the numerical methods used to calculate the Voronoi function. Also, Lemma I.9.1.1 can now be verified quickly by computer, so the technical proof that is given is no longer needed. Lemmas I.9.17 and I.9.18 are proved by a long argument that is no longer necessary because of improvements in numerical methods. Many of the papers rely on the arctan formula for the dihedral angle, rather than the arccos formula that appears in I.8. dih(S) = π/2 + arctan(−∆4 /

p 4x1 ∆).

This leads to simple formulas for the derivatives of the dihedral angles that have been used extensively throughout the collection without explicit mention √ ∂2 dih = −y1 ∆3 /(u126 ∆), etc. For example, Lemma IV.4.9 uses these derivative formulas. In Sphere Packings II the notion of a small simplex is made obsolete by the constructions of [F]. Delaunay stars are replaced by decomposition stars. Restricted cells are replaced with V -cells in [F]. Simplices of compression type undergo a small change in meaning when the scoring functions are adjusted in [F]. In [II], in constructing the standard regions, we remove all arcs that do not bound a region, but in the classification of standard regions in [IV], these arcs are not removed. Lemma II.2.2 can be proved by simpler means. After the first paragraph of the proof, we observe that S = (v0 , v1 , v2 , w) has negative orientation along F = (v0 , v1 , v2 ). Hence S is a quasi-regular tetrahedron by I.3.4. Various lemmas are revised in [F] to account for the change in decomposition. (Lemma II.2.4, Section II.3.1, Lemma II.3.2, Theorem II.4.1.b). Several of the cases in II.4.5.2 are unnecessary in light of the revisions in [F]. The technical results in the appendix can now be obtained quickly by computer. The decomposition into subclusters proceeds along slightly different lines in [IV] and this paper. In [IV], loops are generally erased, but in this paper, additional precision is required, and the loops are generally retained. This means that the bounds on the score of a subcluster cannot be uncritically applied to the subclusters

62

THOMAS C. HALES

of this paper. A discussion of the loops must accompany the application of such bounds. The scoring function on the flat quarters in [IV] is slightly different for the scoring function σ ˆ on flat quarters in this paper. This is justified in Section 2.5, the section that introduces σ ˆ , where it is shown that σ ˆ dominates the possibilities of IV.3.10. A notational subtlety arises in Section 4.3. The various scoring schemes are combined into linear programming variables σR and σF for various subregions R and faces F . In the extreme case, when the subregion R is the full region F , the the variable σR , attached to a subregion, is not the same as the region σF , attached to the region. We will always use different subscripts to distinguish the two, but there is a danger of confusion. The definition of cluster in [V] is a collection of marked vertices (V.1.2). The corner cell in Section V.7.1 has no particular relation to the corner cells and truncated corner cells in [IV] and [VI]. Also, the paper [IV] and [VI] belong together, and the paper [V] is best read with [F] and III. When we say√that a simplex has compression type, it means the the scoring rule for η + (Q) ≤ 2 is used. Here η + is the function of Section F.3. To say that a simplex has compression type implies that the compression function is one term of the scoring function. But there will often be various correction terms, so that the scoring function need not be identical with the compression function. Similar comments apply to simplices of Voronoi type, which means precisely that η + (Q) > √ 2 2. It means loosely that the Voronoi function appears as part of the scoring function. In general, the terms Voronoi, Voronoi scoring, Voronoi function, and so forth are used loosely for objects related to the V -cells in the decomposition star. Perhaps the most highly context dependent term in this paper is the score. In its strict sense it means the scoring function defined in [F], but we often use it in a looser sense for various approximations to the score that are used throughout this series of papers. For example, after erasing upright quarters from an exceptional cluster, we write of the score of the exceptional cluster to mean the value of the truncated scoring function applied to a modified decomposition star. We hope that each particular passage makes the meaning clear. Similar warnings apply to the the term squander. The score is a function of a standard cluster, and it is occasionally extended to smaller objects such as subclusters. By an abuse of language, we often refer to the score of a standard region or subregion. This is short for the score of the cluster over the standard region, and so forth. According to the terminology of Section IV.3.2, a penalty is any upper bound to a particular function. When we discuss the penalty in a given situation, it will rarely be the least upper bound. Finally, we warn the reader that ordered pairs are used to denote many different things. There is the type (p, q) of a vertex in Section III.6. The notation (v, w) is

THE KEPLER CONJECTURE

63

used for an edge, (p, q) is used for the context of an upright diagonal in [F], and (n, k) denotes the edge parameters of a subregion in [IV].

64

THOMAS C. HALES

Index Each reference is to the closest numbered passage preceding the occurence of the index entry. The following papers are covered by this index. Overview of the Kepler Conjecture, math.MG/9811071 Thomas C. Hales A Formulation of the Kepler Conjecture, math.MG/9811072 Samuel P. Ferguson, Thomas C. Hales Sphere Packings III, math.MG/9811075 Thomas C. Hales Sphere Packings IV, math.MG/9811076 Thomas C. Hales Sphere Packings V, math.MG/9811077 Samuel P. Ferguson The Kepler Conjecture (VI), math.MG/9811078 Thomas C. Hales (a1 , b1 , . . . , an , bn ), IV.5.8 ⊥, (u, v, w)⊥ , IV.5.2 -0.4339 IV.3.7 -0.25 IV.3.8 -0.029 IV.3.8 0.003521 IV.3.8 0.0063 IV.3.9 0.008 (π0 ) IV.3.9 0.00935 IV.3.8 0.0114 IV.3.9 0.01561 IV.3.8 0.06688 IV.4.7 0.4 IV.3.8 0.4429 IV.1.1 0.55 pt III.5.3 0.5606 IV.3.7 1.255=2.51/2 14.8 pt III.3.2. 2.51, I.1 3.2 IV.4.2 8 pt IV.1.1 A(h) F.3.7, III.App.A4, IV.4.9 adjacent V.1.1.4 adjustment IV.5.5 adjustment vertex VI.2.3 analytic Voronoi function I.8.6.3, IV.1.1 anc F.4.5 anchored simplex IV.3.6 arc IV.2.8 B IV.4.9 base point F.1.5 branch and bound III.A.7 C(P ) IV.2.1 C(h, t) F.3.1 c0 , c1 , c2 ) IV.5.4 central vertex VI.2.5 circumradius F.1.4

THE KEPLER CONJECTURE

classification algorithm VI.2.1 clip II.2 close neighbor I.1, II.1 cluster V.1.2 compression F.3.1, I.1, I.8.5, II.1, IV.1.1, V.1.3.2 compression type F.3.8, II.3.1, IV.3.3 cone IV.2.1 conflicting diagonal F.1.7 context (p,q) F.3.8, IV.3.3 corner F.1.5, III.2.1, IV.1.2, V.1.2.2 corner cell IV.4.10 cross F.1 crown F.4.3 DLP VI.5.11 D∗ II.1 Dv IV.2.1 D(n, k) IV.4.5 Delaunay decomposition I.1 Delaunay star I.1 decomposition star F.3.0, V.1.2 deformation density II.1 density effective III.3.2 diagonal F.1, V.1.1.2 dih F.1, I.1, I.8.3, II.1 IV.1.1, IV.2.8 dih3 IV.2.8 dihmax I.1 dihmin I.1 dimension reduction I.8.7, V.6 distinguished edge IV.4.2 E F.1 edge (indexing) F.1 edge distinguished IV.4.2 edge parameters IV.4.5 effective density III.3.2 enclosed F.1.5, I.3.9, V.1.2 erase IV.3.2 exceptional IV.1.2 excludePent faceCountMax VI.2.2 faceCountMaxAtExceptionalVertex VI.2.2 fine decomposition IV.2.1 finished face III.8 flat quad cluster V.1.2.2 flat quarter F.1, V.1.1.2 floating point I.8.9 flute V.7.5.5 geometric considerations F.1 gma V.1.3.2

65

66

THOMAS C. HALES

graph of a cluster V.1.2 height F.1.5, V.1.2 hull IV.4.4 IEEE standard I.8.9, V.8 internal structures VI.4.1 interval arithmetic I.8.9, V.4 isolated pair F.1.10 K0 F.4.2 K F.4.2 k0 , k1 , k2 IV.5.5 LP-max LP-min large gap IV.3.6 local V-cell IV.2.1 local density V.1.3 loop IV.3.6 masked flat quarter IV.3.8 mixed quad cluster F.3.16, V.1.2.2 multiplicity of an edge IV.4.4 n(R) IV.4.4 octahedron F.1, V.1.1.3 octavor IV.App, V.1.3.2 octavor0 IV.App orientation F.2.0, F.2.1, I.8.2.3, V.1.1.5 origin F.1 overlap F.1, V.1.1 PM(n, k) III.App pattern (a1 , . . . , an ) I.5.0 penalty IV.3.2, IV.5.5 pentahedral (pentagonal) prism I.2.2, V.2 pivot F.1 pivot (axis) F.1 point F.3.12, V.1.3.9 polygonal hull IV.4.4 pt F.3.12, I.1, III.3.0, IV.1.1 pure Voronoi quad cluster V.1.2.2, V.7.5 Q0 Iv.2.2 ˆ F.3.8 Q Q-system F.1.7, V.1.1, V.1.1.4 qrtet V.1.1.2 quad cluster F.3.0, III.2.1, V.1.2.1 quarter F.1 quasi-regular face V.1.2 quasi-regular tetrahedron I.1, F.1, II.1 quo F.3.3 quoin F.3.3, V.7.5.5 R (Rogers simplex) I.8.6, IV.1.1 Rogers simplex I.8.6 Rogers’s bound III.3.0 Rogers’s lemma I.8.6.2

THE KEPLER CONJECTURE

rad I.1, I.8.2.3, II.1 restricted cell II.3.1 SA , SB , SC IV.2.5 S3+ IV.3.9 S3− IV.3.7 S4+ IV.3.8 Sˆi I.2 S IV.1.1 S(y1 , . . . , y6 ) F.1 sn IV.4.4 saturated I.1, II.1, V.1.1 sc V.1.4 score F.3.8, I.1, II.1, II.3, IV.1.1, V.1.3.1 score adjustment IV.5.5 scoreFace VI.2.1 scoreTarget VI.2.1 short edge F.1.5 sol F.3.0, I.1, I.8.4, II.1, IV.1.1, V.1.3.1 special IV.4.2 squander III.3.2, IV.1.1 squanderFace VI.2.1 squanderVertex VI.2.1 squanderTarget VI.2.1 squashed cluster V.7.5.2 standard cluster F.3.0, I.2, II.1, V.1.2.1 standard decomosition II.1, I.2 standard region I.2, II.1, F.1.5 structure, internal VI.4.1 subcluster IV.4.2 subdivision I.7, V.5 subregion IV.4.2 t0 = 1.255, F.3.7, IV.1.1 tn Iv.4.4 tcc IV.5.2 tip F.2.2, II.2, II.2.3, V.1.1.6 triangle (v1 , v2 )-triangle IV.2.4 truncated Voronoi function F.3.5, F.3.7, V.1.3.6 truncated corner cell IV.5.2 truncation V.7.5 type SA , SB , SC , IV.2.5 type (p, q) III.5 u I.8.1.3 u135 IV.4.9 unfinished face III.8 upright quarter F.1, V.1.1.2 V0 , V1 IV.4.9 VP (t0 ) IV.2.1 VR II.2

67

68

THOMAS C. HALES

VR (t) IV.3.1 V˜P (t0 ) IV.2.1 V-cell F.2.0, F.2.2, V.1.1, V.1.1.5, V.1.1.6 Voronoi type F.3.8, II.3.1, IV.3.3, vertex F.1, I.1, II.1 vertex adjustment VI.2.3 vertex central VI.2.5 vertexCountMax VI.2.1 vertexCountMin VI.2.1 visible IV.4.8 vor F.3.1, I.1, I.8.6.3, IV.1.1 vor0 F.3.7, IV.1.1 vor analytic V.1.3.2 vor(S, VR ) II.3.1 vor(S, t) F.3.1, IV.1.1 vorx IV.App.A4 xi I.8.1 yi I.8.1 ZLP VI.5.11 Z(n, k) IV.4.5 zn I.5.1, βψ IV.2.8 Γ F.3.1, I.8.5, IV.1.1 ∆i IV.4.9 ∆ I.8.1.1 δP′ IV.2.1 δef f F.3.13 δ(a, b, c) I.8.6.2 δP δP (v) IV.2.1, IV.2.3 δoct F.3.0, I.1, II.1, III.3.0, IV.1.1 ǫτ ǫσ IV.5.5 ζ III.3.0, IV.1.1 η + F.3.7 η0 F.4.1, IV.2.3 ηijk F.App η I.1, I.8.2, II.1, F.1.4, F.3.7 κ IV.3.3 λ IV.4.10 µ F.3.8 ν IV.3.3 ξΓ = 0.01561 IV.3.8 ξΓ,κ = ξκ + ξΓ IV.3.8 ξΓ′ = 0.00935 IV.3.8 ξV = 0.003521 IV.3.8 ξκ = −0.029 IV.3.8 πmax = 0.06688 IV.4.7 π0 IV.3.2 π0 (R, e) IV.5.5

THE KEPLER CONJECTURE

πσ IV.5.5 πτ IV.5.5 ρ I.8.2.1 σ F.3.7, F.3.8, IV.1.1 σ ˆ VI.2.5 σLP(p, q, d) VI.2.2 σ1 III.4.1 σR σF VI.4.2 σ3.2 III.4.1 σ ˜ IV.5.5 τ IV.1.1 τˆ VI.2.5 τ (S, t) IV.3.3 τ LP(p, q, d) VI.2.2 τ0 IV.3.3 τR , τF VI.4.2 τV IV.3.3 τΓ IV.3.3 τν IV.3.3 τx IV.A5 τ˜ IV.5.5 φ0 F.4.3, IV.1.1 φ(h, t) F.3.1, IV.1.1 χ0 IV.5.3 g χan 0 , χ0 IV.5.3 χ I.8.2, I.8.2.3

69