Properties of tangent lines in 3D (Degeneracies may be hard to characterize) Problem 2. Do 4 balls admit an infinite number of tangents? Test if an algebraic system of degree 12 [MPT01] has infinitely many real solutions { x 4 + 17 xy 3 + . . . = 0 5 y 3 + 3 x 2 y + . . . = 0 Characterize the configurations of 4 balls that admit infinitely many tangents
Lines tangent to balls Number of tangents to 4 balls: � 12 or ∞ [MPT01, DMPT03] [ST02] Configurations with ∞ many tangents: 4 unit balls: aligned centers [MPT01] 4 arbitrary balls?
Lines tangent to balls 4 balls admit infinitely many tangents iff aligned centers and at least 1 common tangent [Borcea, Goaoc, Lazard, Petitjean, Discrete & Comput. Geom. , 2006]
Lines tangent to balls O Sketch of proof (1/2) p − → p · v = 0 Line represented by ( p, v ) with v
Lines tangent to balls O Sketch of proof (1/2) p − → p · v = 0 Line represented by ( p, v ) with v Tangency condition with spheres centered at c i of radii r i p 2 = r 2 [MPT01] (with c 1 = O ) 1 | c 2 | 2 + r 2 1 − r 2 ( c 2 · v ) 2 0 c T 1 0 1 0 1 2 2 | c 3 | 2 + r 2 A − 2 | v | 2 A p = | v | 2 1 − r 2 ( c 3 · v ) 2 c T @ @ @ A 3 3 | c 4 | 2 + r 2 1 − r 2 ( c 4 · v ) 2 c T 4 4 { { { M Φ 2 ( v ) Φ 0
Lines tangent to balls O Sketch of proof (1/2) p − → p · v = 0 Line represented by ( p, v ) with v Tangency condition with spheres centered at c i of radii r i p 2 = r 2 [MPT01] (with c 1 = O ) 1 | c 2 | 2 + r 2 1 − r 2 ( c 2 · v ) 2 0 c T 1 0 1 0 1 2 2 | c 3 | 2 + r 2 A − 2 | v | 2 A p = | v | 2 1 − r 2 ( c 3 · v ) 2 c T @ @ @ A 3 3 | c 4 | 2 + r 2 1 − r 2 ( c 4 · v ) 2 c T 4 4 { { { M Φ 2 ( v ) Φ 0 Idea: For each variable, eliminate all other variables Infinite number of solutions ⇔ all the coefficients of the uivariate polynomials are 0 Deduce the geometric characterization
Lines tangent to balls O Sketch of proof (1/2) p − → p · v = 0 Line represented by ( p, v ) with v Tangency condition with spheres centered at c i of radii r i p 2 = r 2 [MPT01] (with c 1 = O ) 1 | c 2 | 2 + r 2 1 − r 2 ( c 2 · v ) 2 0 c T 1 0 1 0 1 2 2 | c 3 | 2 + r 2 A − 2 | v | 2 A p = | v | 2 1 − r 2 ( c 3 · v ) 2 c T @ @ @ A 3 3 | c 4 | 2 + r 2 1 − r 2 ( c 4 · v ) 2 c T 4 4 { { { M Φ 2 ( v ) Φ 0 Idea: For each variable, eliminate all other variables Infinite number of solutions ⇔ all the coefficients of the uivariate polynomials are 0 Deduce the geometric characterization In practice, we did not succeed . . .
Lines tangent to balls O Sketch of proof (1/2) p − → p · v = 0 Line represented by ( p, v ) with v Tangency condition with spheres centered at c i of radii r i p 2 = r 2 [MPT01] (with c 1 = O ) 1 | c 2 | 2 + r 2 1 − r 2 ( c 2 · v ) 2 0 c T 1 0 1 0 1 2 2 | c 3 | 2 + r 2 A − 2 | v | 2 A p = | v | 2 1 − r 2 ( c 3 · v ) 2 c T @ @ @ A 3 3 | c 4 | 2 + r 2 1 − r 2 ( c 4 · v ) 2 c T 4 4 { { { M Φ 2 ( v ) Φ 0 Proof, case 1: Suppose non-coplanar centers (M inversible) Prove that the 4 balls admit finitely many tangents
Lines tangent to balls O Sketch of proof (1/2) p − → p · v = 0 Line represented by ( p, v ) with v Tangency condition with spheres centered at c i of radii r i p 2 = r 2 [MPT01] (with c 1 = O ) 1 | c 2 | 2 + r 2 1 − r 2 ( c 2 · v ) 2 0 c T 1 0 1 0 1 2 2 | c 3 | 2 + r 2 A − 2 | v | 2 A p = | v | 2 1 − r 2 ( c 3 · v ) 2 c T @ @ @ A 3 3 | c 4 | 2 + r 2 1 − r 2 ( c 4 · v ) 2 c T 4 4 { { { M Φ 2 ( v ) Φ 0 Proof, case 1: Suppose non-coplanar centers (M inversible) Prove that the 4 balls admit finitely many tangents p = M − 1 � � 1 1 Express p in terms of v : 2 Φ 0 + 2 | v | 2 Φ 2 ( v ) p · v = 0 and p 2 = r 2 Substitute p in 1
Lines tangent to balls O Sketch of proof (2/2) { p → − M − 1 � Φ 2 ( v ) + | v | 2 Φ 0 � �� · v = 0 (cubic) v � 2 = 4 r 2 � M − 1 � � Φ 2 ( v ) + | v | 2 Φ 0 �� 1 | v | 4 (quartic)
Lines tangent to balls O Sketch of proof (2/2) { p − → M − 1 � Φ 2 ( v ) + | v | 2 Φ 0 � �� · v = 0 (cubic) v � 2 = 4 r 2 � M − 1 � � Φ 2 ( v ) + | v | 2 Φ 0 �� 1 | v | 4 (quartic) Suppose for a contradiction that there are ∞ many solutions v ∈ P 2 ( R ) The solutions form a curve Γ in P 2 ( R )
Lines tangent to balls O Sketch of proof (2/2) { p − → M − 1 � Φ 2 ( v ) + | v | 2 Φ 0 � �� · v = 0 (cubic) v � 2 = 4 r 2 � M − 1 � � Φ 2 ( v ) + | v | 2 Φ 0 �� 1 | v | 4 (quartic) Suppose for a contradiction that there are ∞ many solutions v ∈ P 2 ( R ) The solutions form a curve Γ in P 2 ( R ) In P 2 ( C ) , Γ intersect any curve and, in particular, the conic | v | 2 = 0
Lines tangent to balls O Sketch of proof (2/2) { p → − M − 1 � Φ 2 ( v ) + | v | 2 Φ 0 � �� · v = 0 (cubic) v � 2 = 4 r 2 � M − 1 � � Φ 2 ( v ) + | v | 2 Φ 0 �� 1 | v | 4 (quartic) Suppose for a contradiction that there are ∞ many solutions v ∈ P 2 ( R ) The solutions form a curve Γ in P 2 ( R ) In P 2 ( C ) , Γ intersect any curve and, in particular, the conic | v | 2 = 0 { � 2 = 0 M − 1 Φ 2 ( v ) is on the conic {| x | 2 = 0 } � � M − 1 Φ 2 ( v ) � M − 1 Φ 2 ( v ) � � · v = 0 M − 1 Φ 2 ( v ) is on the tangent to this conic at v M − 1 Φ 2 ( v ) ≡ v ❀ v is real and | v | 2 = 0 ❀ v = 0 , contradiction
Properties of tangent lines in 3D Problem 3. Conjecture: 4 disjoint unit balls (nonaligned centers) admit at most 8 tangents
Properties of tangent lines in 3D Problem 3. Conjecture: 4 disjoint unit balls (nonaligned centers) admit at most 8 tangents Constant size problem 5 variables: the line 6 parameters: the ball centers 5 equations: tangency and line conditions 7 inequalities: the balls are disjoint & not aligned
Properties of tangent lines in 3D Problem 3. Conjecture: 4 disjoint unit balls (nonaligned centers) admit at most 8 tangents Constant size problem 5 variables: the line 6 parameters: the ball centers 5 equations: tangency and line conditions 7 inequalities: the balls are disjoint & not aligned Tried a few years ago and failed, but . . .
Properties of tangent lines in 3D Problem 3. Conjecture: 4 disjoint unit balls (nonaligned centers) admit at most 8 tangents Constant size problem 2 5 variables: the line 6 parameters: the ball centers 5 equations: tangency and line conditions 2 7 inequalities: the balls are disjoint & not aligned Tried a few years ago and failed, but . . . { M − 1 ` Φ 2 ( v ) + | v | 2 Φ 0 ` ´´ · v = 0 ˛ 2 = 4 | v | 4 ˛ ˛ M − 1 ` Φ 2 ( v ) + | v | 2 Φ 0 ´˛
Properties of tangent lines in 3D Problem 3. Conjecture: 4 disjoint unit balls (nonaligned centers) admit at most 8 tangents Constant size problem 2 5 variables: the line 6 parameters: the ball centers 5 equations: tangency and line conditions 2 7 inequalities: the balls are disjoint & not aligned Tried a few years ago and failed, but . . . { degree 3/7, 68 monomials M − 1 ` Φ 2 ( v ) + | v | 2 Φ 0 ` ´´ · v = 0 ˛ 2 = 4 | v | 4 degree 4/12, 763 monomials ˛ M − 1 ` ˛ Φ 2 ( v ) + | v | 2 Φ 0 ´˛
Properties of tangent lines in 3D Problem 3. Conjecture: 4 disjoint unit balls (nonaligned centers) admit at most 8 tangents Constant size problem 2 5 variables: the line 6 parameters: the ball centers 5 equations: tangency and line conditions 2 7 inequalities: the balls are disjoint & not aligned Tried a few years ago and failed, but . . . { degree 3/7, 68 monomials M − 1 ` Φ 2 ( v ) + | v | 2 Φ 0 ` ´´ · v = 0 ˛ 2 = 4 | v | 4 degree 4/12, 763 monomials ˛ M − 1 ` ˛ Φ 2 ( v ) + | v | 2 Φ 0 ´˛ Discriminant variety & a point per connected component of the complement
Properties of tangent lines in 3D Problem 3. Conjecture: 4 disjoint unit balls (nonaligned centers) admit at most 8 tangents Constant size problem 2 5 variables: the line 6 parameters: the ball centers 5 equations: tangency and line conditions 2 7 inequalities: the balls are disjoint & not aligned Tried a few years ago and failed, but . . . { degree 3/7, 68 monomials M − 1 ` Φ 2 ( v ) + | v | 2 Φ 0 ` ´´ · v = 0 ˛ 2 = 4 | v | 4 degree 4/12, 763 monomials ˛ M − 1 ` ˛ Φ 2 ( v ) + | v | 2 Φ 0 ´˛ Discriminant variety & a point per connected component of the complement Still too big . . . so far
Plan I. Computer algebra & theorems in geometry I.a. Properties of tangent lines in 3D I.b. Voronoi diagrams of three lines II. Avoiding algebraic numbers Intersections of two quadrics
Voronoi diagrams Voronoi diagrams, Dirichlet tessellations, Medial axes In 2D: Well understood for points and polygons In 3D: Well understood for points
Voronoi diagrams Voronoi diagrams, Dirichlet tessellations, Medial axes In 2D: Well understood for points and polygons In 3D: Well understood for points What about Voronoi diagrams of polyhedra? of lines?
Voronoi diagrams of lines Complexity: n lines: Ω( n 2 ) , O ( n 3+ ǫ ) (conjectured near quadratic) [Sharir94] n lines with c possible directions: O ( n 2+ ǫ ) [KS03] Algorithms: Exact: naive (compute, for each line, a cell of an arrangement of quadrics) Approximation: many Applications: Medial axis of polyhedra
Voronoi diagrams of 3 lines 3 lines in general position (not parallel to a common plane and pairwise skew) The topology of the Voronoi diagram is invariant Trisector: 4 infinite branches of a smooth quartic or a cubic and a line 2D cells: 2 connected components of a hyperb. paraboloid bounded by 3 branches and 1 branch of the trisector, resp. [Everett, Lazard, Lazard, Safey El Din, ACM Symp. on Comp. Geom. , 2007]
Voronoi diagrams of 3 lines Complete characterization of the Voronoi of three lines ❀ fundamental properties & point-location algorithms Th. All 4 branches are monotonic in Y Th. Localizing and ordering points on a branch can be done by evaluating the sign of linear forms with rational coefficients [Everett, Lazard, Lazard, Safey El Din, ACM Symp. on Comp. Geom. , 2007]
Voronoi diagrams of 3 lines Idea of the proof • Trisector of 3 lines in general position: non-singular with 4 simple real points at ∞ ❀ homeomorphic to 4 lines with no intersection point • Topology of the Voronoi diagram is invariant by continuous deformation over the set of all triplets of 3 lines in gen. position which is connected • Study of an example
Voronoi diagrams of 3 lines Idea of the proof • Trisector of 3 lines in general position: non-singular with 4 simple real points at ∞ ❀ homeomorphic to 4 lines with no intersection point • Topology of the Voronoi diagram is invariant by continuous deformation over the set of all triplets of 3 lines in gen. position which is connected • Study of an example
Voronoi diagrams of 3 lines Bisector of two skew lines: Hyperbolic paraboloid (HP)
Voronoi diagrams of 3 lines Trisector of three lines Intersection of 2 HP bisectors: H 1 , 2 ∩ H 1 , 3 Intersection of 2 quadrics: smooth quartic, nodal quartic, a cubic and a line, 2 conics, 4 lines, etc. Idea of the proof Determine the type of H 1 , 2 ∩ H 1 , 3 Deduce that the trisector is non-singular
Voronoi diagrams of 3 lines Type of the intersection of two quadrics H 1 , 2 , H 1 , 3
Voronoi diagrams of 3 lines Type of the intersection of two quadrics H 1 , 2 , H 1 , 3 Q 1 , 2 , Q 1 , 3 : 4 × 4 matrices associated with H 1 , 2 , H 1 , 3 Pencil of quadrics/matrices: { P ( λ ) = Q 1 , 2 + λQ 1 , 3 | λ ∈ R }
Voronoi diagrams of 3 lines Type of the intersection of two quadrics H 1 , 2 , H 1 , 3 Q 1 , 2 , Q 1 , 3 : 4 × 4 matrices associated with H 1 , 2 , H 1 , 3 Pencil of quadrics/matrices: { P ( λ ) = Q 1 , 2 + λQ 1 , 3 | λ ∈ R } Characteristic polynomial D ( λ )= det P ( λ ) , degree 4 in λ
Voronoi diagrams of 3 lines Type of the intersection of two quadrics H 1 , 2 , H 1 , 3 Q 1 , 2 , Q 1 , 3 : 4 × 4 matrices associated with H 1 , 2 , H 1 , 3 Pencil of quadrics/matrices: { P ( λ ) = Q 1 , 2 + λQ 1 , 3 | λ ∈ R } Characteristic polynomial D ( λ )= det P ( λ ) , degree 4 in λ When D ( λ ) has no multiple root: H 1 , 2 ∩ H 1 , 3 is a smooth quartic or ∅ [Segre 1883]
Voronoi diagrams of 3 lines Type of the intersection of two quadrics H 1 , 2 , H 1 , 3 Q 1 , 2 , Q 1 , 3 : 4 × 4 matrices associated with H 1 , 2 , H 1 , 3 Pencil of quadrics/matrices: { P ( λ ) = Q 1 , 2 + λQ 1 , 3 | λ ∈ R } Characteristic polynomial D ( λ )= det P ( λ ) , degree 4 in λ When D ( λ ) has no multiple root: H 1 , 2 ∩ H 1 , 3 is a smooth quartic or ∅ [Segre 1883] When D ( λ ) has multiple root(s): [DLLP08] The real type of H 1 , 2 ∩ H 1 , 3 follows from: number of multiples roots, λ i , of D ( λ ) , number of positive/negative eigenvalues of P ( λ i )
Voronoi diagrams of 3 lines Proof. We compute the characteristic polynomial D ( λ ) its discriminant ∆ We determine the type of H 1 , 2 ∩ H 1 , 3 when ∆ = 0 (when ∆ � = 0 , H 1 , 2 ∩ H 1 , 3 is a non-singular quartic)
Voronoi diagrams of 3 lines The setting Any two skew lines can be moved into the lines Z = 1 , Y = a X and { Y = aX Z ℓ 1 Z = 1 Z = − 1 , Y = − a X ℓ 3 ( α, β, 1) O ( x, y, 0) Y Y = − ax { ℓ 2 Z = − 1 X The 3 lines are not parallel to a common plane ⇒ Third line: defined by point ( x, y, 0) and vector ( α, β, 1)
Voronoi diagrams of 3 lines α 2 + β 2 + 1 a 2 λ 4 ` ´ D ( λ ) = 2 aβ 2 + ayβ + aα x − β α + 2 a + 2 aα 2 − β α a 2 ´ λ 3 ` − 2 a +( β 2 +6 a 2 β 2 − 2 β xa 3 − 6 β α a 3 +6 yβ a 2 − 6 aβ α − 2 aβ x +6 α xa 2 + y 2 a 2 − 2 aα y + x 2 a 2 − 2 yα a 3 + 6 a 2 α 2 + a 4 α 2 + 4 a 2 ) λ 2 xa − ya 2 − 2 β a 2 − β + 2 aα + α a 3 ´ ` − 2 ( xa − y − β + aα ) λ 1 + a 2 ´ ( xa − y − β + aα ) 2 ` + ∆ =16 a 4 ( a x − y − β + a α ) 2 ( y + a x − a α − β ) 2 ∗ gros facteur By our general position assumption, 16 a 4 ( a x − y − β + a α ) 2 ( y + a x − a α − β ) 2 > 0
Voronoi diagrams of 3 lines determine the type of the trisector when Key problem: gros facteur = 8 a 8 α 4 y 2 + 7 a 4 β 2 x 4 − 4 aβ 3 x + 16 a 8 β 4 x 4 + 32 a 4 α 2 y 2 + 2 a 6 α 2 β 4 x 2 + 38 a 8 α 2 x 2 + 2 y 4 β 2 a 4 α 2 + 44 a 8 α 2 β 2 x 2 + 2 a 6 α 4 y 2 β 2 + a 2 α 2 β 4 y 2 − 8 y 4 a 4 + 3 a 8 α 4 β 2 − 8 a 8 α 4 + 20 y 2 β 2 a 2 + β 6 x 2 a 2 + 20 a 4 β 2 α 2 + 38 y 2 β 2 a 4 + 2 α 2 y 4 a 6 x 2 + 8 a 4 β 2 + 32 a 4 y 2 − 8 a 4 x 2 − 8 a 4 β 4 + 28 α 2 y 4 a 5 xβ + 10 a 7 α 3 β 2 y +28 a 7 β 2 x 4 yα − 4 aβ 5 x +48 β xa 5 +16 β xa 3 +32 yα a 5 − 8 β 5 xa 3 + 48 β 3 xa 5 +32 a 7 β 3 x +8 β 3 xa 3 +32 β a 7 x + a 12 α 6 − 28 β α 3 a 8 yx +20 β 2 α 2 a 8 + 8 β 2 α a 7 y − 4 β 2 α a 5 y + 14 β 4 α a 3 y + 10 a 5 α 2 β 3 x + 2 a 7 α 2 β 3 x + 32 a 5 α 3 y + 2 a 5 α 3 β 2 y − 4 a 7 α 2 β x +8 a 5 α 2 β x − 4 β 3 α 3 a 6 xy +48 a 7 α 3 y + x 4 a 4+10 aβ 4 α y + 8 aβ 2 α y + 28 a 5 β 4 x 2 α y − 8 a 8 x 4 + 48 ya 7 α − 4 a 6 β x 3 α y 3 − 2 a 6 β x 5 α y + 2 a 4 β 4 x 2 y 2 + 22 a 6 α 2 y 2 β 2 + 32 a 2 α 2 y 2 β 2 − 8 a 10 α 2 + 20 a 6 α 2 x 2 + 16 a 4 α 2 + 20 a 6 y 2 β 2 + 38 a 6 β 2 α 2 + 16 a 6 α 2 y 2 + 16 a 6 β 2 x 2 + 14 y 4 a 5 xβ + 16 a 10 x 2 − 8 a 6 y 4 +8 a 6 α 4 y 2 +22 a 6 α 2 β 2 x 2 +44 a 4 α 2 y 2 β 2 +3 a 4 α 2 β 4 +16 a 4 +32 x 2 a 8 + 8 a 6 β 4 x 2 + 8 a 8 α 2 + a 8 y 4 − 4 a 5 x 5 β + β 6 − 28 a 4 β 3 yα x + 8 a 5 α 3 y 3 x 2 + 3 β 4 y 2 a 2 + 8 β 4 a 4 x 2 − 12 a 6 β 3 xα y + 24 β a 6 yα x + 32 a 6 β 2 + 8 a 6 y 2 + 8 a 6 x 2 + β 4 − 8 β 4 a 2 +6 β 2 x 2 a 2 +16 α 2 y 2 a 2 − 8 y 5 a 5 α +3 y 4 a 6 x 2 +8 a 6 β 2 x 4 +32 a 6 α 2 − 40 a 7 β 2 x 2 yα + 32 β 2 x 2 a 8 − 8 a 2 β 2 + 32 a 6 − 8 a 5 β 5 x 3 + 16 y 2 a 2
Voronoi diagrams of 3 lines − 16 β xα ya 2 − 8 a 8 y 2 + 7 β 4 x 2 a 2 + 16 a 8 β 2 + 16 a 3 β 4 x 2 α y + a 12 α 4 + 28 a 5 β 2 x 2 α y − 56 a 6 β 3 x 3 α y +54 a 4 α 2 y 2 x 2 β 2 +16 a 8 − 8 a 9 α 3 y 3 − 8 α 3 y 5 a 5 − 40 a 5 α 2 y 2 β x − 48 a 4 α 3 y 3 xβ + 44 a 6 y 2 α 2 x 2 − 12 y 3 xa 4 β α − 4 yα a 7 x 2 + 8 a 3 yx 2 α + 8 a 3 y 2 xβ − 4 y 2 xa 5 β + 32 a 3 α y 3 + 48 y 3 α a 5 − 4 x 3 a 3 β + 20 x 2 a 4 y 2 + 8 x 3 a 5 β + 10 a 9 α 3 yx 2 + 14 a 9 α 4 β x + 10 y 3 β 2 a 5 α − 4 a 3 β 5 x 3 − 8 a 10 α 4 + 2 β 3 y 2 a 5 x + 28 a 7 α 2 y 2 xβ + 8 a 7 y 3 α + 8 yα 3 a 9 + 48 a 7 β 3 x 3 + 8 a 6 α 2 y 4 +44 a 6 β 2 x 2 y 2 +22 a 8 α 2 y 2 x 2 − 8 a 9 α 5 y +8 yx 2 a 9 α − 32 x 2 a 9 yβ 2 α + 32 a 10 β 2 α 2 x 2 + 20 a 10 α 2 x 2 + 3 a 10 α 4 x 2 + 20 y 2 x 2 a 8 + 32 x 2 a 8 y 2 β 2 + 3 y 4 β 2 a 4 +8 a 8 β 2 x 4 − 32 a 5 β 2 x 2 y 3 α +2 a 5 x 2 y 3 α +12 a 5 β 3 x 3 y 2 +32 a 9 β 3 x 3 + 32 a 9 x 3 β +2 a 9 α 2 x 3 β +54 a 8 β 2 x 2 α 2 y 2 − 48 a 8 β 3 x 3 α y +6 a 8 α 4 y 4+7 a 8 α 2 y 4 − 28 a 8 α 3 y 3 β x + 28 a 9 α 2 y 2 β x + β 2 a 10 α 4 x 2 + 2 β 2 a 8 α 2 x 4 − 2 a 10 α 5 yxβ − 24 a 10 α 3 yxβ +10 a 11 α 4 xβ − 4 a 9 y 3 α − 24 a 8 β xα y 3+8 a 9 y 2 β x − 12 a 8 x 3 β α y + 16 a 9 α 4 xβ y 2 + 8 a 9 x 3 α 2 β 3 − 32 a 7 x 2 α 3 β 2 y − 20 a 9 β 2 x 2 α 3 y − 4 a 9 α 5 y 3 + 16 a 9 yα + 12 a 7 x 3 α 2 β 3 + 28 a 7 α 4 xβ y 2 − 24 a 4 β x 3 α y + 32 a 4 y 2 α 2 x 2 + 2 a 8 x 2 α 4 y 2 + 8 α 2 xa 11 β + 80 a 6 x 2 α 2 y 2 β 2 − 8 a 3 β 3 x 3 − 4 α 3 x 3 a 8 β y + 32 a 3 α 3 y 3 + 16 β 2 x 2 a 10 + 10 a 5 x 4 yα − 20 a 3 α 2 y 2 xβ 3 − 12 a 6 α 3 yxβ +
Voronoi diagrams of 3 lines 8 a 3 α 3 y 3 β 2 10 y 2 β 3 a 3 x 2 y 3 β 2 a 3 α 16 β xa 10 yα 10 β xa 7 y 4 + + + + − 12 y 3 a 5 α 3 β 2 − 8 y 3 a 7 α 5 − 4 y 5 a 7 α 3 + 6 y 2 α 2 a 10 − 4 a 11 α 5 y + a 10 α 6 y 2 + 7 y 2 α 4 a 10 − 32 β 3 xa 5 α 2 y 2 − 28 y 3 xa 6 β α − 56 y 3 xa 6 β α 3 − 20 a 7 β 2 x 2 α y 3 + 8 a 7 β 3 x 3 y 2 + 16 a 7 α 2 y 4 xβ − 8 a 6 x 4 − 4 a 7 y 5 α + 2 a 7 x 3 y 2 β + 10 a 7 x 2 α y 3 + 12 a 7 x 2 α 3 y 3 − 32 a 7 x 3 α 2 y 2 β − 4 yα 3 a 11 + 16 y 4 a 6 α 4 − 20 a 5 α 2 y 2 x 3 β + 3 a 6 y 2 x 4+3 a 8 x 4 α 2 − 8 a 7 x 5 β − 8 a 7 β 3 x 5+6 a 4 β 4 x 4+48 a 5 α 3 y 3+16 a 4 α 4 y 4+ 8 a 4 y 4 α 2 + 38 a 6 y 2 x 2 + 16 yα a 5 β 2 x 4 + 48 β x 3 a 7 + 16 a 6 β 4 x 4 + a 6 β 2 x 6 − 2 β 5 α a 2 xy + y 4 β 2 a 6 x 2 +10 a 5 x 3 y 2 β +14 a 7 x 4 yα +2 a 6 β 2 x 4 y 2 + a 6 α 2 y 2 x 4 − 4 a 4 β 3 xα y 3 − 28 a 4 β 3 x 3 α y − 28 a 6 β x 3 α y − 4 β 3 x 5 a 5 +22 a 4 y 2 β 2 x 2 + y 6 a 6 + α 2 y 6 a 6 − 2 β xα y 5 a 6 + x 6 a 6 + 10 β x 3 a 7 α 2 + 2 yα 3 a 7 x 2 − 32 a 3 α 2 y 2 β x + 28 a 3 β 2 x 2 α y − 24 a 2 β 3 yα x is zero Degree 18 in 5 variables with 253 monomials
Voronoi diagrams of 3 lines Technique: heavy use of computer algebra tools • Prove that the gros facteur is never negative (RAGLib) • Manipulation of Gr¨ obner bases (GB) ❀ gros facteur = 0 only if ( y + aα )( ax + β ) = 0 • Classification of quadric intersections ❀ gros facteur = 0 iff the trisector is a cubic and line without intersection
Voronoi diagrams of 3 lines Lemma. The gros facteur (and ∆ ) are never negative
Voronoi diagrams of 3 lines Lemma. The gros facteur (and ∆ ) are never negative Proof. Compute at least a point in every connected component of the semi-algebraic set ∆( a, x, y, α, β ) < 0
Voronoi diagrams of 3 lines Lemma. The gros facteur (and ∆ ) are never negative Proof. Compute at least a point in every connected component of the semi-algebraic set ∆( a, x, y, α, β ) < 0 Major problem in computer algebra Recent advances provide practical solutions
Voronoi diagrams of 3 lines Lemma. The gros facteur (and ∆ ) are never negative Proof. Compute at least a point in every connected component of the semi-algebraic set ∆( a, x, y, α, β ) < 0 Major problem in computer algebra Recent advances provide practical solutions [Safey El Din] based on FGb/RS [Faug` ere, Rouillier] RAGLib On { ( a, x, y, α, β ) | ∆ < 0 } : returns ∅ (in 10 hours)
Voronoi diagrams of 3 lines Lemma. The gros facteur (and ∆ ) are never negative Proof. Compute at least a point in every connected component of the semi-algebraic set ∆( a, x, y, α, β ) < 0 Major problem in computer algebra Recent advances provide practical solutions [Safey El Din] based on FGb/RS [Faug` ere, Rouillier] RAGLib On { ( a, x, y, α, β ) | ∆ < 0 } : returns ∅ (in 10 hours) ∆ = 0 ⇔ the gros facteur and all its partial derivatives are 0
RAGLib Finding (at least) a point in every connected component of a semi-algebraic set ∆( a, x, y, α, β ) < 0 z = ∆( a, x, y, α, β )
RAGLib Finding (at least) a point in every connected component of a semi-algebraic set ∆( a, x, y, α, β ) < 0 • Compute all the generalized critical values of ∆ • Compute the largest negative critical value, z c z = ∆( a, x, y, α, β ) z c
RAGLib Finding (at least) a point in every connected component of a semi-algebraic set ∆( a, x, y, α, β ) < 0 • Compute all the generalized critical values of ∆ • Compute the largest negative critical value, z c z = ∆( a, x, y, α, β ) • Pick a value ˜ z in ( z c , 0) ˜ z z c
RAGLib Finding (at least) a point in every connected component of a semi-algebraic set ∆( a, x, y, α, β ) < 0 • Compute all the generalized critical values of ∆ • Compute the largest negative critical value, z c z = ∆( a, x, y, α, β ) • Pick a value ˜ z in ( z c , 0) • Compute (at least) a point in every connected component of ˜ z the algebraic set ∆( a, x, y, α, β ) = ˜ z z c
RAGLib Finding (at least) a point in every connected component of the algebraic set ∆( a, x, y, α, β ) = ˜ z { ∆( X ) = ˜ z Solve X × grad (∆)( X ) = 0 with X = ( a, x, y, α, β ) O ❀ At least a point in every cc of the semi-algebraic set ∆ < 0
Voronoi diagrams of 3 lines Recall ∆ = 0 ⇔ the gros facteur and all its partial derivatives are 0 ∆ = 0 only if ( y + a α ) ( a x + β ) = 0 Main Lemma
Voronoi diagrams of 3 lines Recall ∆ = 0 ⇔ the gros facteur and all its partial derivatives are 0 ∆ = 0 only if ( y + a α ) ( a x + β ) = 0 Main Lemma Proof: Prove { ∆ = 0 , ( y + a α ) ( a x + β ) � = 0 } has no real solution
Voronoi diagrams of 3 lines Recall ∆ = 0 ⇔ the gros facteur and all its partial derivatives are 0 ∆ = 0 only if ( y + a α ) ( a x + β ) = 0 Main Lemma Proof: Prove { ∆ = 0 , ( y + a α ) ( a x + β ) � = 0 } has no real solution Consider the system of: the gros facteur, all its partial derivatives, and
Voronoi diagrams of 3 lines Recall ∆ = 0 ⇔ the gros facteur and all its partial derivatives are 0 ∆ = 0 only if ( y + a α ) ( a x + β ) = 0 Main Lemma Proof: Prove { ∆ = 0 , ( y + a α ) ( a x + β ) � = 0 } has no real solution Consider the system of: the gros facteur, all its partial derivatives, and 1 − v ( a x + β ) = 0 1 − u ( y + a α ) = 0
Voronoi diagrams of 3 lines Recall ∆ = 0 ⇔ the gros facteur and all its partial derivatives are 0 ∆ = 0 only if ( y + a α ) ( a x + β ) = 0 Main Lemma Proof: Prove { ∆ = 0 , ( y + a α ) ( a x + β ) � = 0 } has no real solution Consider the system of: the gros facteur, all its partial derivatives, and 1 − v ( a x + β ) = 0 1 − u ( y + a α ) = 0 1 − w (1 + α 2 + β 2 ) = 0
Voronoi diagrams of 3 lines Recall ∆ = 0 ⇔ the gros facteur and all its partial derivatives are 0 ∆ = 0 only if ( y + a α ) ( a x + β ) = 0 Main Lemma Proof: Prove { ∆ = 0 , ( y + a α ) ( a x + β ) � = 0 } has no real solution Consider the system of: the gros facteur, all its partial derivatives, and 1 − v ( a x + β ) = 0 1 − u ( y + a α ) = 0 1 − w (1 + α 2 + β 2 ) = 0 1 − t Γ = 0 ´ 2 + 3 ( ax + β ) 2 + 3 a 2 ( y + aα ) 2 + 3 2 a ( yα − β x ) − a 2 + 1 1 + a 2 ´ 2 ` ` Γ =
Voronoi diagrams of 3 lines Recall ∆ = 0 ⇔ the gros facteur and all its partial derivatives are 0 ∆ = 0 only if ( y + a α ) ( a x + β ) = 0 Main Lemma Proof: Prove { ∆ = 0 , ( y + a α ) ( a x + β ) � = 0 } has no real solution Consider the system of: the gros facteur, all its partial derivatives, and 1 − v ( a x + β ) = 0 1 − u ( y + a α ) = 0 1 − w (1 + α 2 + β 2 ) = 0 1 − t Γ = 0 ´ 2 + 3 ( ax + β ) 2 + 3 a 2 ( y + aα ) 2 + 3 2 a ( yα − β x ) − a 2 + 1 1 + a 2 ´ 2 ` ` Γ = The Gr¨ obner basis is 1, hence the system has no solution (15mn with FGb [Faug` ere] in Maple)
Voronoi diagrams of 3 lines About the proof. Start with the system of the gros facteur and all its partial derivatives
Voronoi diagrams of 3 lines About the proof. Start with the system of the gros facteur and all its partial derivatives Compute reccursively Gr¨ obner bases If PQ is in basis B then study B ∪ { P } and B ∪ { 1 − uP }
Voronoi diagrams of 3 lines About the proof. Start with the system of the gros facteur and all its partial derivatives Compute reccursively Gr¨ obner bases If PQ is in basis B then study B ∪ { P } and B ∪ { 1 − uP } Set a = 2
Voronoi diagrams of 3 lines About the proof. Start with the system of the gros facteur and all its partial derivatives Compute reccursively Gr¨ obner bases If PQ is in basis B then study B ∪ { P } and B ∪ { 1 − uP } Set a = 2 By such manipulation, we add to the system 1 − w (1 + α 2 + β 2 ) = 0 1 − u ( y + a α ) = 0 1 − v ( a x + β ) = 0
Voronoi diagrams of 3 lines About the proof. Start with the system of the gros facteur and all its partial derivatives Compute reccursively Gr¨ obner bases If PQ is in basis B then study B ∪ { P } and B ∪ { 1 − uP } Set a = 2 By such manipulation, we add to the system 1 − w (1 + α 2 + β 2 ) = 0 1 − u ( y + a α ) = 0 1 − v ( a x + β ) = 0 and eventually obtain in the basis Γ 2 = (4 y α − 4 β x − 3) 2 + 3 (2 x + β ) 2 + 12 ( y + 2 α ) 2 + 75
Voronoi diagrams of 3 lines About the proof. Start with the system of the gros facteur and all its partial derivatives Compute reccursively Gr¨ obner bases If PQ is in basis B then study B ∪ { P } and B ∪ { 1 − uP } Set a = 2 By such manipulation, we add to the system 1 − w (1 + α 2 + β 2 ) = 0 1 − u ( y + a α ) = 0 1 − v ( a x + β ) = 0 and eventually obtain in the basis Γ 2 = (4 y α − 4 β x − 3) 2 + 3 (2 x + β ) 2 + 12 ( y + 2 α ) 2 + 75 Repeat for a = 3 , 5 and guess that ´ 2 + 3 ( ax + β ) 2 + 3 a 2 ( y + aα ) 2 + 3 2 a ( yα − β x ) − a 2 + 1 1 + a 2 ´ 2 ` ` Γ =
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