On the Stretch Factor of Convex Delaunay Graphs

Let C be a compact and convex set in the plane that contains the origin in its interior, and let S be a finite set of points in the plane. The Delaunay graph DG_C(S) of S is defined to be the dual of the Voronoi diagram of S with respect to the convex distance function defined by C. We prove that DG_C(S) is a t-spanner for S, for some constant t that depends only on the shape of the set C. Thus, for any two points p and q in S, the graph DG_C(S) contains a path between p and q whose Euclidean length is at most t times the Euclidean distance between p and q.


Introduction
Let S be a finite set of points in the plane and let G be a graph with vertex set S, in which each edge (p, q) has a weight equal to the Euclidean distance |pq| between p and q. For a real number t ≥ 1, we say that G is a t-spanner for S, if for any two points p and q of S, there exists a path in G between p and q whose Euclidean length is at most t|pq|. The smallest such t is called the stretch factor of G. The problem of constructing spanners has received much attention; see Narasimhan and Smid [12] for an extensive overview.
Spanners were introduced in computational geometry by Chew [3,4], who proved the following two results. first, the L 1 -Delaunay graph, i.e., the dual of the Voronoi diagram for the Manhattan metric, is a √ 10-spanner. Second, the Delaunay graph based on the convex distance function defined by an equilateral triangle, is a 2-spanner. We remark that in both these results, the stretch factor is measured in the Euclidean metric. Chew also conjectured that the Delaunay graph based on the Euclidean metric, is a t-spanner, for some constant t. (If not all points of S are on a line, and if no four points of S are cocircular, then the Delaunay graph is the well-known Delaunay triangulation.) This conjecture was proved by Dobkin et al. [8], who showed that t ≤ π(1 + √ 5)/2. The analysis was improved by Keil and Gutwin [9], who showed that t ≤ 4π In this paper, we unify these results by showing that the Delaunay graph based on any convex distance function has bounded stretch factor.
Throughout this paper, we fix a compact and convex set C in the plane. We assume that the origin is in the interior of C. A homothet of C is obtained by scaling C with respect to the origin, followed by a translation. Thus, a homothet of C can be written as for some point x in the plane and some real number λ ≥ 0. We call x the center of the homothet x + λC.
For two points x and y in the plane, we define d C (x, y) := min{λ ≥ 0 : y ∈ x + λC}.
If x = y, then this definition is equivalent to the following: Consider the translate x + C and the ray emanating from x that contains y. Let y ′ be the (unique) intersection between this ray and the boundary of x + C. Then d C (x, y) = |xy|/|xy ′ |.
The function d C is called the convex distance function associated with C. Clearly, we have d C (x, x) = 0 and d C (x, y) > 0 for all points x and y with x = y. Chew and Drysdale [5] showed that the triangle inequality d C (x, z) ≤ d C (x, y) + d C (y, z) holds. In general, the function d C is not symmetric, i.e., d C (x, y) is not necessarily equal to d C (y, x). If C is symmetric with respect to the origin, however, then d C is symmetric. Let S be a finite set of points in the plane. For each point p in S, we define If C is not strictly convex, then the set V ′ C (p) may consist of a closed region of positive area with an infinite ray attached to it. For example, in figure 1, the set V ′ C (a) consists of the set of all points that are on or to the left of the leftmost zig-zag line, together with the infinite horizontal ray that is at the same height as the point a. Also, the intersection of two regions V ′ C (p) and V ′ C (q), where p and q are distinct points of S, may have a positive area. As a result, the collection V ′ C (p), where p ranges over all points of S, does not necessarily give a subdivision of the plane in which the interior of each cell is associated with a unique point of S. In order to obtain such a subdivision, we follow the approach of Klein and Wood [10] (see also Ma [11]): first, infinite rays attached to regions of positive area are not considered to be part of the region. Second, a point x in R 2 that is in the interior of more than one region V ′ C (p) is assigned to the region of the lexicographically smallest point p in S for which x ∈ V ′ C (p). To formally define Voronoi cells, let ≺ denote the lexicographical ordering on the set of all points in the plane. Let p 1 ≺ p 2 ≺ . . . ≺ p n be the points of S, sorted according to this order. Then the Voronoi cells V C (p i ) of the points of S are defined as and, for 1 < i ≤ n, where cl (X) and int(X) denote the closure and the interior of the set X ⊆ R 2 , respectively. Thus, in figure 1, the Voronoi cell V C (a) consists only of the set of all points that are on or to the left of the leftmost zig-zag line; the infinite horizontal ray that is at the same height as the point a is not part of this cell.
The Voronoi diagram VD C (S) of S with respect to C is defined to be the collection of Voronoi cells V C (p), where p ranges over all points of S. An example is given in figure 1.
As for the Euclidean case, the Voronoi diagram VD C (S) induces Voronoi cells, Voronoi edges, and Voronoi vertices. Each point in the plane is either in the interior of a unique Voronoi cell, in the relative interior of a unique Voronoi edge, or a unique Voronoi vertex. Each Voronoi edge e belongs only to the two Voronoi cells that contain e on their boundaries. Observe that Voronoi cells are closed.
The Delaunay graph is defined to be the dual of the Voronoi diagram:  We consider the Delaunay graph DG C (S) to be a geometric graph, which means that each edge (p, q) is embedded as the closed line segment with endpoints p and q.
Before we can state the main result of this paper, we introduce two parameters whose values depend on the shape of the set C. Let x and y be two distinct points on the boundary ∂C of C. These points partition ∂C into two chains. For each of these chains, there is an isosceles triangle with base xy and whose third vertex is on the chain. Denote the base angles of these two triangles by α xy and α ′ xy ; see figure 2 (left). We define α C := min{max(α xy , α ′ xy ) : x, y ∈ ∂C, x = y}.
Consider again two distinct points x and y on ∂C, but now assume that x, y, and the origin are collinear. As before, x and y partition ∂C into two chains. Let ℓ xy and ℓ ′ xy denote the lengths of these chains; see figure 2 (right). We define κ C,0 := max max(ℓ xy , ℓ ′ xy ) |xy| : x, y ∈ ∂C, x = y, and x, y, and 0 are collinear .
Clearly, the convex distance function d C and, therefore, the Voronoi diagram VD C (S), depends on the location of the origin in the interior of C. Surprisingly, the Delaunay graph DG C (S) does not depend on this location; see Ma [11, Section 2.1.6]. We define κ C := min {κ C,0 : 0 is in the interior of C} .
In this paper, we will prove the following result: Theorem 1 Let C be a compact and convex set in the plane with a non-empty interior, and let S be a finite set of points in the plane. The stretch factor of the Delaunay graph DG C (S) is less than or equal to Thus, for any two points p and q in S, the graph DG C (S) contains a path between p and q whose Euclidean length is at most t C times the Euclidean distance between p and q.
We emphasize that we do not make any "general position" assumption; our proof of Theorem 1 is valid for any finite set of points in the plane.
Throughout the rest of this paper, we assume that the origin is chosen in the interior of C such that κ C = κ C,0 .
The rest of this paper is organized as follows. In Section 2, we prove some basic properties of the Delaunay graph which are needed in the proof of Theorem 1. In particular, we give a formal proof of the fact that this graph is plane. Even though this fact seems to be well known, we have not been able to find a proof in the literature. Section 3 contains a proof of Theorem 1. This proof is obtained by showing that the Delaunay graph satisfies the "diamond property" and a variant of the "good polygon property" of Das and Joseph [6]. The proof of the latter property is obtained by generalizing the analysis of Dobkin et al. [8] for the lengths of so-called one-sided paths.

Some properties of the Delaunay graph
Recall that in the Euclidean Delaunay graph, if two points p and q of S are connected by an edge, then there exists a disk having p and q on its boundary that does not contain any point of S in its interior. The next lemma generalizes this result to the Delaunay graph DG C (S).
Lemma 1 Let p and q be two points of S and assume that (p, q) is an edge in the Delaunay graph DG C (S). Then, the following are true.
1. The line segment between p and q does not contain any point of S \ {p, q}.

For every point
Thus, the Voronoi cells of p and q do not share an edge and, therefore, (p, q) is not an edge in the Delaunay graph. This is a contradiction.
To prove the second claim, let x be an arbitrary point in Thus, if we define λ := d C (x, p), then λ > 0, both p and q are on the boundary of the homothet x + λC, and no point of S is in the interior of this homothet.
As can be seen in figure 1, Voronoi cells are, in general, not convex. They are, however, star-shaped: x is in the interior or on the boundary of this Voronoi cell).
It is well known that the Euclidean Delaunay graph is a plane graph; see, for example, de Berg et al. [7, page 189]. The following lemma states that this is true for the Delaunay graph DG C (S) as well.

Lemma 3 The Delaunay graph DG C (S) is a plane graph.
Proof. By the first claim in Lemma 1, DG C (S) does not contain two distinct edges (p, q) and (p, r) that are collinear and overlap in a line segment of positive length. Again by the first claim in Lemma 1, DG C (S) does not contain two distinct edges (p, q) and (r, s) such that r is on the open line segment joining p and q.
It remains to show that DG C (S) does not contain two edges (p, q) and (r, s) that cross properly. The proof is by contradiction. Thus, let p, q, r, and s be four pairwise distinct points of S, no three of which are collinear, and assume that the line segments (p, q) and (r, s) are edges of DG C (S) that have exactly one point in common.
Since (p, q) is an edge of DG C (S), there exists a point x in the relative interior of V C (p) ∩ V C (q). Thus, by the second claim in Lemma 1, there exists a real number λ > 0, such that the homothet x + λC contains p and q on its boundary and no point of S is in the interior of this homothet. Observe that x is in the interior of x + λC. Let D be a Euclidean disk centered at x that is contained in the interior of x+λC and that is contained in V C (p)∪V C (q). We define B to be the set of all 2-link polygonal chains (p, z, q), with z ∈ D; see figure 3.
Observe that B has a positive area. Since V C (p) and V c (q) are star-shaped (by Lemma 2), Thus, neither r nor s is in the interior of the convex hull of B. Since pq and rs intersect in a point, the line segment rs crosses the set B.
By a symmetric argument, since (r, s) is an edge of DG C (S), there exist a point y in the relative interior of V C (r) ∩ V C (s) and a real number µ > 0, such that y + µC contains r and s on its boundary and no point of S is in the interior of this homothet. Let D ′ be a Euclidean disk centered at y that is contained in the interior of y + µC and that is contained in V C (r)∪V C (s). We define B ′ to be the set of all 2-link polygonal chains (r, z, s), with z ∈ D ′ .
x + λC The set B ′ has a positive area, the line segment pq crosses this set, B ′ ⊆ V C (r) ∪ V C (s), and neither p nor q is in the interior of the convex hull of B ′ . It follows that B and B ′ overlap in a region of positive area. Since B ⊆ V C (p) ∪ V C (q) and B ′ ⊆ V C (r) ∪ V C (s), however, the area of the intersection B ∩ B ′ is equal to zero. This is a contradiction. It follows that the edges (p, q) and (r, s) do not cross.

The stretch factor of Delaunay graphs
In this section, we will prove Theorem 1. first, we show that the Delaunay graph DG C (S) satisfies the diamond property and a variant of the good polygon property of Das and Joseph [6]. According to the results of Das and Joseph, this immediately implies that the stretch factor of DG C (S) is bounded. In fact, we will obtain an upper bound on the stretch factor which is better than the one that is implied by Das and Joseph's result.

The diamond property
Let G be a plane graph with vertex set S and let α be a real number with 0 < α < π/2. For any edge e of G, let ∆ 1 and ∆ 2 be the two isosceles triangles with base e and base angle α; see figure 4. We say that e satisfies the α-diamond property, if at least one of the triangles ∆ 1 and ∆ 2 does not contain any point of S in its interior. The graph G is said to satisfy the α-diamond property, if every edge e of G satisfies this property.
Lemma 4 Consider the value α C that was defined in Section 1. The Delaunay graph DG C (S) satisfies the α C -diamond property. Proof. Let (p, q) be an arbitrary edge of DG C (S) and let x be any point in the relative interior of V C (p) ∩ V C (q). By Lemma 1, there exists a real number λ > 0 such that p and q are on the boundary of the homothet x + λC and no point of S is in the interior of x + λC. The points p and q partition ∂(x + λC) into two chains. For each of these chains, there is an isosceles triangle with base pq and whose third vertex is on the chain. We denote the base angles of these two triangles by β and γ; see figure 5. We may assume without loss of generality that β ≥ γ. Let a denote the third vertex of the triangle with base angle β. If we translate x + λC so that x coincides with the origin and scale the translated homothet by a factor of 1/λ, then we obtain the set C. This translation and scaling does not change the angles β and γ. Thus, using the notation of Section 1 (see also figure 2), we have {β, γ} = {α pq , α ′ pq }. The definition of α C then implies that α C ≤ max(α pq , α ′ pq ) = β. Let ∆ be the isosceles triangle with base pq and base angle α C such that a and the third vertex of ∆ are on the same side of pq. Then ∆ is contained in the triangle with vertices p, q, and a. Since the latter triangle is contained in x + λC, it does not contain any point of S in its interior. Thus, ∆ does not contain any point of S in its interior. This proves that the edge (p, q) satisfies the α C -diamond property.

The visible-pair spanner property
For a real number κ ≥ 1, we say that the plane graph G satisfies the strong visible-pair κ-spanner property, if the following is true: For every face f of G, and for every two vertices p and q on the boundary of f , such that the open line segment joining p and q is completely in the interior of f , the graph G contains a path between p and q having length at most κ|pq|. If for every face f of G and for every two vertices p and q on the boundary of f , such that the line segment pq does not intersect the exterior of f , the graph G contains a path between p and q having length at most κ|pq|, then we say that G satisfies the visible-pair κ-spanner property. Observe that the former property implies the latter one. Also, observe that these properties are variants of the κ-good polygon property of Das and Joseph [6]: The κ-good polygon property requires that G contains a path between p and q that is along the boundary of f and whose length is at most κ|pq|; in the (strong) visible-pair spanner property, the path is not required to be along the boundary of f .
In this subsection, we will prove that the Delaunay graph DG C (S) satisfies the visiblepair κ C -spanner property, where κ C is as defined in Section 1. This claim will be proved by generalizing results of Dobkin et al. [8] on so-called one-sided paths.
Let p and q be two distinct points of S and assume that (p, q) is not an edge of the Delaunay graph DG C (S). Consider the Voronoi diagram VD C (S). We consider the sequence of points in S whose Voronoi cells are visited when the line segment pq is traversed from p to q. If pq does not contain any Voronoi vertex, then this sequence forms a path in DG C (S) between p and q. Since, in general, Voronoi cells are not convex, it may happen that this path contains duplicates. In order to avoid this, we define the sequence in the following way.
In the rest of this section, we will refer to the line through p and q as the X-axis, and we will say that p is to the left of q. This implies a left-to-right order on the X-axis, the notion of a point being above or below the X-axis, as well as the notions horizontal and vertical. (Thus, conceptually, we rotate and translate all points of S , the set C, the Voronoi diagram VD C (S), and the DG C (S), such that p and q are on a horizontal line and p is to the left of q. Observe that VD C (S) is still defined based on the lexicographical order of the points of S before this rotation and translation.) In the following, we consider the (horizontal) line segment pq. If this segment contains a Voronoi vertex, then we imagine moving pq vertically upwards by an infinitesimal amount. Thus, we may assume that pq does not contain any Voronoi vertex of the (rotated and translated) Voronoi diagram VD C (S).
The first point in the sequence is p 0 := p. We define x 1 ∈ R 2 to be the point on the line segment pq such that x 1 ∈ V C (p 0 ) and x 1 is closest to q. Figure 6: Illustrating the proof of Lemma 5.
Let i ≥ 1 and assume that the points p 0 , p 1 , . . . , p i−1 of S and the points x 1 , . . . , x i in R 2 have already been defined, where x i is the point on the line segment pq such that x i ∈ V C (p i−1 ) and x i is closest to q. If p i−1 = q, then the construction is completed. Otherwise, observe that x i is in the relative interior of a Voronoi edge. We define p i to be the point of S \ {p i−1 } whose Voronoi cell contains x i on its boundary, and define x i+1 to be the point on the line segment pq such that x i+1 ∈ V C (p i ) and x i+1 is closest to q.
Let p = p 0 , p 1 , . . . , p k = q be the sequence of points in S obtained in this way. By construction, these k + 1 points are pairwise distinct and for each i with 1 ≤ i ≤ k, the Voronoi cells V C (p i−1 ) and V C (p i ) share an edge. Therefore, by definition, (p i−1 , p i ) is an edge in DG C (S). Thus, p = p 0 , p 1 , . . . , p k = q defines a path in DG C (S) between p and q. We call this path the direct path between p and q. If all points p 1 , p 2 , . . . , p k−1 are strictly on one side of the line through p and q, then we say that the direct path is one-sided.
We will show in Lemma 6 that the length of a one-sided path is at most κ C |pq|. The proof of this lemma uses a geometric property which we prove first.
Let C ′ be a homothet of C whose center is on the X-axis, and let x and y be two points on the boundary of C ′ that are on or above the X-axis. The points x and y partition the boundary of C ′ into two chains. One of these chains is completely on or above the X-axis; we denote this chain by arc(x, y; C ′ ). The length of this chain is denoted by |arc(x, y; C ′ )|.
For two points x and y on the X-axis, we write x < X y if x is strictly to the left of y, and we write x ≤ X y if x = y or x < X y.
We now state the geometric property, which is illustrated in figure 6. Recall the value κ C that was defined in Section 1.
Lemma 5 Let C 1 = y 1 + λ 1 C and C 2 = y 2 + λ 2 C be two homothets of C whose centers y 1 and y 2 are on the X-axis. Assume that λ 1 > 0, λ 2 > 0, and y 1 < X y 2 . For i = 1, 2, let ℓ i and r i be the leftmost and rightmost points of C i on the X-axis, respectively. Assume that r 1 ≤ X r 2 and ℓ 1 ≤ X ℓ 2 < X r 1 . Let x be a point that is on the boundaries of both C 1 and C 2 and on or above the X-axis. Let L 1 = |arc(x, r 1 ; C 1 )| and L 2 = |arc(x, r 2 ; C 2 )|. Then Proof. We define L 3 = |arc(ℓ 2 , x; C 2 )|. Let C ′ be the homothet of C whose center is on the X-axis such that the intersection between C ′ and the X-axis is equal to the line segment ℓ 2 r 1 , and let L ′ = |arc(ℓ 2 , r 1 ; C ′ )|; see figure 6. Observe that, for λ := |ℓ 2 r 1 |/|ℓ 2 r 2 |, C ′ is obtained from C 2 by a scaling by a factor of λ. Thus, since |arc(ℓ 2 , r 2 ; C 2 )| = L 2 + L 3 , we have Let C ′′ be the homothet of C whose center is on the X-axis such that the intersection between C ′′ and the X-axis is equal to the line segment r 1 r 2 , and let L ′′ = |arc(r 1 , r 2 ; C ′′ )|. Since C ′′ is obtained from C 2 by a scaling by a factor of 1 − λ, we have Thus, we have L ′ + L ′′ = L 2 + L 3 .
By convexity, we have C ′ ⊆ C 1 ∩ C 2 . Then it follows, again from convexity (see Benson [1, page 42]), that Thus, we have which implies that Since, by the definition of κ C , L ′′ ≤ κ C |r 1 r 2 |, the proof is complete.
We are now ready to prove an upper bound on the length of a one-sided path.

Lemma 6
If the direct path between p and q is one-sided, then its length is at most κ C |pq|.
Proof. As above, we assume that p and q are on the X-axis and that p is to the left of q.
Consider the direct path p = p 0 , p 1 , . . . , p k = q in DG C (S) and the sequence x 1 , x 2 , . . . , x k , as defined above. Since the direct path is one-sided, we may assume without loss of generality that the points p 1 , p 2 , . . . , p k−1 are strictly above the X-axis. We have to show that Recall that, for each i with 1 ≤ i ≤ k, x i is in the relative interior of V C (p i−1 ) ∩ V C (p i ) and x i is on the line segment pq. Therefore, by Lemma 1, if we define λ i := d C (x i , p i−1 ) (which is equal to d C (x i , p i )), then the homothet C i := x i + λ i C contains p i−1 and p i on its boundary and no point of S is in its interior.
For each i with 1 ≤ i ≤ k, let ℓ i and r i be the leftmost and rightmost points of C i that are on the X-axis, respectively. We will prove that for each j with 1 ≤ j ≤ k, For j = k, inequality (2) implies (1), because r k = p k = q. Before we prove (2), we show that therefore, C i−1 is completely contained in the interior of C i . This is a contradiction, because p i−1 is on the boundary of C i−1 , but no point of S is in the interior of C i . Thus, we have Thus, we have shown that ℓ 1 ≤ X ℓ 2 ≤ X . . . ≤ X ℓ k . By a symmetric argument, it follows that r 1 ≤ X r 2 ≤ X . . . ≤ X r k . Now we are ready to prove (2). The proof is by induction on j. For the base case, i.e., when j = 1, we have to show that |arc(p 0 , r 1 ; C 1 )| ≤ κ C |pr 1 |.
Since p 0 = p = ℓ 1 , this inequality follows from the definition of κ C .
We distinguish two cases.
Since p j is on the boundary of C j+1 and strictly above the X-axis, we have It follows that Thus, (4) holds.
Case 2: ℓ j+1 < X r j . Since p j is on the boundaries of both C j and C j+1 and strictly above the X-axis, we can apply Lemma 5 with x = p j and obtain |arc(p j , r j+1 ; C j+1 )| ≤ |arc(p j , r j ; C j )| + κ C |r j r j+1 |.
We claim that p j ∈ arc(p j−1 , r j , C j ). Assuming this is true, it follows that |p j−1 p j | + |arc(p j , r j+1 ; C j+1 )| ≤ |arc(p j−1 , p j ; C j )| + |arc(p j , r j ; C j )| + κ C |r j r j+1 | = |arc(p j−1 , r j ; C j )| + κ C |r j r j+1 |, i.e., (4) holds. It remains to prove that p j ∈ arc(p j−1 , r j , C j ). Since p 0 = ℓ 0 and p 1 is strictly above the X-axis, this is true for j = 1. Assume that 2 ≤ j < k and p j ∈ arc(p j−1 , r j , C j ). Then, since p j is strictly above the X-axis, p j−1 is in the relative interior of arc(p j , r j , C j ).
By the definition of the point x j , there is a point y on the X-axis such that y < X x j and the line segment yx j is contained in the Voronoi cell V C (p j−1 ). By Lemma 2, the triangle ∆ with vertices p j−1 , y, and x j is contained in V C (p j−1 ).
Again by the definition of the point x j , there is a point z on the X-axis such that x j < X z and the line segment x j z is contained in the Voronoi cell V C (p j ). By Lemma 2, the triangle ∆ ′ with vertices p j , x j , and z is contained in V C (p j ).
Since p j−1 and p j are strictly above the X-axis and since p j−1 is in the relative interior of arc(p j , r j , C j ), the intersection of ∆ and ∆ ′ has a positive area and is contained in the intersection of V C (p j−1 ) and V C (p j ). This is a contradiction, because the area of the intersection of any two Voronoi cells is zero.
We are now ready to prove that the Delaunay graph satisfies the visible-pair spanner property: Then, G is a t-spanner for t = 2κκ ′ · max 3 sin(α/2) , κ .
We have shown that the Delaunay graph DG C (S) satisfies all these properties: By Lemma 3, DG C (S) is plane. By Lemma 4, DG C (S) satisfies the α C -diamond property. By Lemma 6, the stretch factor of any one-sided path in DG C (S) is at most κ C . By Lemma 7, DG C (S) satisfies the visible-pair κ C -spanner property. If DG C (S) is a triangulation, then obviously, DG C (S) satisfies the visible-pair 1-spanner property. Therefore, we have completed the proof of Theorem 1.

Concluding remarks
We have considered the Delaunay graph DG C (S), where C is a compact and convex set with a non-empty interior and S is a finite set of points in the plane. We have shown that the (Euclidean) stretch factor of DG C (S) is bounded from above by a function of two parameters α C and κ C that are determined only by the shape of C. Roughly speaking, these two parameters give a measure of the "fatness" of the set C.
Our analysis provides the first generic bound valid for any compact and convex set C. In all previous works, only special examples of such sets C were considered. Furthermore, our approach does not make any "general position" assumption about the point set S, while most related works on Delaunay graphs do not consider the case when four points are cocircular.
Note that for the Euclidean Delaunay triangulation (i.e., when the set C is the disk of radius one, and with no four cocircular points), we have α C = π/4 and κ C = π/2, and we derive an upper bound on the stretch factor of 3π sin(π/8) ≈ 24.6. Observe that this is much worse than the currently best known upper bound (as proved by Keil and Gutwin [9]), which is 4π √ 3 9 ≈ 2.42. We leave open the problem of improving our upper bound. In particular, is it possible to generalize the techniques of Dobkin et al. [8] and Keil and Gutwin [9], from the Euclidean metric to an arbitrary convex distance function?