In this section, we construct the poset diagrams by taking averages of the balls in B. We begin with the introduction of a vector space structure of the set of all balls,
including those with negative squared radii.
Vector space of balls. We follow Pedoe [Pedoe, 1988, Chapter IV], who introduced the vector space to study the geometry of circles in the plane or spheres in higher- dimensions. In particular, we represent the ball B(x, r) by the point b(x, r) = (x, kxk2− r2) in Rn+1. To have a bijection between the set of balls in Rn and the set of points
in Rn+1, we let r2 range over all real numbers or, equivalently, we let r range over all
non-negative real numbers and all positive multiples of the imaginary unit, a set we denote as √R. Borrowing the vector space structure of Rn+1, we have a vector space of balls in which addition and multiplication with scalars make sense. More formally, if B1, B2 are two balls with corresponding points b1, b2, and λ1, λ2 are real numbers, then
B0 = λ1B1+ λ2B2 is defined such that the corresponding point satisfies b0 = λ1b1+ λ2b2
in Rn+1. From the centers x
1, x2 and squared radii r21, r22 of B1, B2, we can compute the
center and squared radius of B0 as:
x0 = λ1x1+ λ2x2, (3.8)
r20 = kx0k2− λ1 kx1k2− r21 − λ2 kx2k2− r22 . (3.9)
Assuming λ1 + λ2 = 1, we can plug these expressions of the center and the squared
radius into the formula for the weighted distance of a point x ∈ Rn from B
0 and get
π0(x) = λ1π1(x) + λ2π2(x). (3.10)
Affine combinations. This is an interesting conclusion worth generalizing. For this purpose, we recall that a linear combination is a ball B0 = Pki=1λiBi. It is an affine
combination if P
iλi = 1, and it is a convex combination if, in addition, 0 ≤ λi for all
i. For affine combinations, the weighted distance satisfies a relation that generalizes (3.10):
Lemma 6 (Weighted Distance Lemma). Let B0 = Pki=1λiBi with Pki=1λi = 1. Then
π0(x) =
Pk
i=1λiπi(x), for every point x ∈ R n.
Proof. For k = 2, the claimed relation is the same as (3.10). For k > 2, we decompose the affine combination into two affine combinations of fewer than k balls each:
B00 = λ 1
2+...+λk (λ2B2+ . . . + λkBk) , (3.11)
Inductively, we get the claimed relation for the weighted distance from B00, and combin- ing it with (3.10), we get the claimed relation for the weighted distance from B0.
To understand the effect of a scalar multiplication on the weighted distance, we consider Borg = (0, 0), and note that µBorg = Borg for all µ ∈ R. Setting B0 = λB1, we
note that B0 = λB1+ (1 − λ)Borg. Using (3.10), we therefore get
π0(x) = λπ1(x) + (1 − λ)kxk 2
(3.13)
because πorg(x) = kxk 2
. Any linear combination can be written as a scalar multiple of an affine combination, so we can get the weighted distance function from (3.13) applied to the relation given in the Weighted Distance Lemma.
Weighted averages. As proved in [Aurenhammer, 1990], the order-k Voronoi diagram of a finite set of points is the ordinary Voronoi diagram of the k-fold averages. We generalize this result, showing that every poset diagram is an ordinary Voronoi diagram, of course of a different set of balls. To construct this set, we say a function λ : [m] → R is anti-parallel to a cotransitive partial order if P λ(i) = 1 and λ(i) > λ(j) iff ui ≺ uj.
Hence, λ is constant on the nodes in an antichain, and using the ordered partition into maximal antichains (3.2), there are values λ1 > λ2 > . . . > λs, such that λ(i) = λp iff
ui ∈ Cp. Given B and U, the weighted average ball of a permutation γ : [m] → [m] is
Bγ = m
X
i=1
λ(i)Bγ(i), (3.14)
and we write B = B(B, U) for the set of all such weighted averages. Noting that
Bγ = Bγ0 iff γ and γ0 define the same domain, we see that the number of balls in B is
card B = Qs m!
p=1(card Cp)!
. (3.15)
The Voronoi domain generated by a weighted point may or may not be empty, which implies that the left-hand side of (3.15) is an upper bound on the number of non-empty domains in VU(B).
Main result. Importantly, the poset diagram of B and U is equal to the ordinary
Voronoi diagram of B.
Theorem 4 (Poset Diagram Theorem). Let B be a finite set of balls in Rn, let U be a
cotransitive partial order on the same number of nodes, and set B = B(B, U). Then
Proof. Fixing a permutation γ, we prove that a point x belongs to the Voronoi domain of γ, VorU(γ), iff x belongs to the Voronoi domain of the weighted average ball, Vor(B
γ).
To see VorU(γ) ⊆ Vor(B
γ), we recall that the weighted distance of x from B0 = Bγ
satisfies π0(x) =
Pm
i=1λ(i)πγ(i)(x) by Lemma 6. Assuming x ∈ VorU(γ), we have
πγ(i)(x) ≤ πγ(j)(x)as well as λ(i) > λ(j) whenever ui ≺ uj. It follows that the weighted
distance to any other weighted average ball is larger. Indeed, this other weighted distance is obtained by switching some of the λ(i). We thus go away from the global minimum, which we get by sorting the πγ(i)(x)and the λ(i) in anti-parallel fashion.
Since both the VorU(γ)and the Vor(B
γ)domains are interior-disjoint closed convex
polyhedra that cover Rn, we have VorU(γ) = Vor(B
γ)for every γ. Indeed, if VorU(γ)
domains were a strict subset of Vor(Bγ), then there would be points in the interior of
the Voronoi domain that are not covered by any VorU domain.
Geometric dual. Recall that V (B) has a geometric dual, namely the Delaunay triangu- lation of B, D(B). We call this Delaunay triangulation the poset complex of B and U,
DU(B). As shown in Figure 3.3, it has cells that are not necessarily simplicial even
Figure 3.3: The Delaunay complex superimposed on the dotted degree-2 Voronoi dia- gram. Only the edges and faces corresponding to Voronoi cells visible in the window are drawn.
for generic sets B. Specifically, a Delaunay cell is the convex hull of the centers of a maximal collection of average balls whose corresponding Voronoi regions have a given non-empty common intersection. As usual, the dimension of the common intersection of Voronoi domains and of the dual Delaunay cell are complementary, that is: they add to n.