Bandwidth of graphs resulting from the edge clique covering problem

Let $n,k,b$ be integers with $1 \le k-1 \le b \le n$ and let $G_{n,k,b}$ be the graph whose vertices are the $k$-element subsets $X$ of $\{0,\dots,n\}$ with $\max(X)-\min(X) \le b$ and where two such vertices $X,Y$ are joined by an edge if $\max(X \cup Y) - \min(X \cup Y) \le b$. These graphs are generated by applying a transformation to maximal $k$-uniform hypergraphs of bandwidth $b$ that is used to reduce the (weak) edge clique covering problem to a vertex clique covering problem. The bandwidth of $G_{n,k,b}$ is thus the largest possible bandwidth of any transformed $k$-uniform hypergraph of bandwidth $b$. For $b\geq \frac{n+k-1}{2}$, the exact bandwidth of these graphs is determined. For $b<\frac{n+k-1}{2}$, the bandwidth is asymptotically determined in the case of $b=o(n)$ and in the case of $b$ growing linearly in $n$ with a factor $\beta \in (0,0.5]$, where for one case only bounds could be found. It is conjectured that the upper bound of this open case is the right asymptotic value.


Introduction
The bandwidth problem for graphs is to find a labelling of the vertices with different integers, such that the maximum absolut value of the difference of the labels of two adjacent vertices is minimal. There are many applications such as efficient storage of sparsely populated symmetric matrices, which arise e.g. from discretization of partial differential equations, cf. [20]. Several other applications, including the placement problem for modules of a VLSI design, the binary constraint satisfaction problem and the minimization of effects of noise in the multichannel communication of data are discussed e.g. in [5,3,8]. The bandwidth problem was shown to be NP-hard [18] and even an approximation with a ratio better than 2 is NP-hard [10], so several heuristics such as the Cuthill-McKee-Algorithm [9] or some similar approaches, cf. [12], are very popular in applications. However, for some graph classes the exact bandwidth is known. These include the path, the cycle, the complete graph, the complete bipartite graph [6], the hypercube [13], the grid graph [7], special Hamming graphs [16] and several others, cf. [5]. However, there are still many graphs, where the exact bandwidth is unknown, such as the general Hamming graphs, cf. [15,2]. In this paper, we consider graphs G n,k,b , 1 ≤ k − 1 ≤ b ≤ n, whose vertices are those k-element subsets of {0, . . . , n}, for which the difference of the maximum and the minimum is at most b. There is an edge between two vertices, if the difference of the maximum and the minimum of the union of the corresponding sets is at most b. We start by introducing the necessary notation and a motivation in Section 2 and study some basic properties in Section 3. Based on that, we determine the exact bandwidth for these graphs in the case of b ≥ n+k−1 2 in Section 4. In Section 5, we present some asymptotic results for n → ∞ in the case of b = o(n). The results of Sections 4 and 5 are summarized by the following theorem:

Theorem 1. Let k be a fixed positive integer and
) then for sufficiently large n B(G n,k,b ) = k b k .
Sections 6 to 9 discuss the case b ∼ βn with β ∈ (0, 0.5]. The main result is given by the next theorem: Theorem 2. Let k ≥ 2 be a fixed positive integer, but n → ∞. Let b ∼ βn and let 1 = qβ + r, where q ≥ 2 is a positive integer and 0 ≤ r < β. Let a) If r ≤ q−1 q 2 +q−1 then B(G n,k,b ) ∼ c 1 (β, k)n k .
The part b) gives only bounds instead of an exact asymptotic value. We strongly conjecture that the RHS bound is the right value. The bounds are not too far away from each other because since β > qβ − qr iff r > q−1 q 2 +q−1 and k ≥ 2, q ≥ 2, β ≤ 1 2 . Let U = {β ∈ (0, 0.5] ∶ r > q−1 q 2 +q−1 } be the set of numbers β for which part b) applies and thus the exact asymptotic value is still unknown. Note that r > q−1 q 2 +q−1 iff 1 q+1 < β < q q 2 +q−1 . Thus the Lebesgue measure of U is equal to ∑ ∞ q=2 ( q q 2 +q−1 − 1 q+1 ) = 0.119 . . . , i.e., for the "majority" of numbers β ∈ (0, 0.5] the exact value is known. The proof of Theorem 2 is based on a reduction to a continuous problem on the unit square [0, 1] 2 . Riemann integrals and elementary geometric arguments suffice. The embedding into a more difficult continuous problem on the unit cube was used by Harper [14] to obtain bounds for the bandwidth of Hamming graphs. Also for the edge-bandwidth of multidimensional grids and Hamming graphs (the bandwidth of the line graph of these graphs) Harper's reduction to the unit cube was applied in [1]. Asymptotic bounds for the bandwidth of the d-ary de Bruijn graph were obtained in [19] by an approach based on the use of a continuous domain.

Notation and motivation
. This definition can be easily generalized to hypergraphs H = (V, E). There we have Now we formally define the subject of our study. Let k and b be positive integers and edge set These graphs arise in the following context: Let G = (V, E) be a graph. A clique is a subset of V that induces a complete subgraph of G. Consider the following transformation, which was used in the NP-completeness proof of the edge clique covering problem in [17] by reducing it to the vertex clique covering ) be the vertex clique covering number (resp. edge clique covering number) of the graph G = (V, E), i.e., the minimal number of cliques whose induced subgraphs cover all vertices (resp. edges) of G. It can be shown, Thus, from an algorithmic point of view, it is enough to study the vertex clique covering problem. For bounded bandwidth, and more generally for bounded treewidth, there is a linear time dynamic programming algorithm for the solution [4]. In an application, which will be described below, we were lead to the weak edge covering problem on a hypergraph whose bandwidth is small (and thus, theoretically, considered as bounded). This implies the following question: Given a hypergraph H of bandwidth b, how large can be the bandwidth of the weak edge clique graphG H of H? Here we discuss only k-uniform hypergraphs though many results can be simply generalized to hypergraphs whose edges have size at most k. For later computations, it is more suitable to take [0, n] as the vertex set of H instead of [1, n]. If, without loss of generality, f (i) = i is the bandwidth numbering of H then, obviously,G H has maximal bandwidth if H contains all k-element subsets X of [0, n] with X − X ≤ b. In this case,G H is exactly the graph G n,k,b , which motivates the study of G n,k,b . We came to these questions in the study of multielectrode recordings of neuronal signals, so-called spikes. Such recordings are carried out on multielectrode arrays, which can be used in-vivo or in-vitro. The denser the electrodes are placed the more likely it is for the neurons to be simultaneously recorded at different electrodes. The resulting similarities in the recordings of the electrodes can provide useful information. In [11] we developed an algorithm to estimate the (unknown) neighborhood of a neuron, i.e., the set of electrodes which record the signals of this neuron. Such neighborhood information is also used as an additional tool in [21] for the so called spike sorting, which is an estimated assignment of the recorded signals to the neurons. Fix a short time interval in which several electrodes record signals. We consider these electrodes as vertices of a graph, which we call similarity-graph for the fixed time interval. First we mention that some neurons may always spike simultaneously. We combine such a set of neurons to one (artificial) new neuron. It might be an accident that two electrodes record a signal at almost the same time, but the simultaneous recording can also be caused by the fact that one spiking neuron has contact to both electrodes. Thus we do not test only one short time interval but several such intervals. If there are sufficiently many simultaneous recordings of two (or k) fixed electrodes, one may expect that these recordings are indeed caused by only one neuron and thus we draw an edge (hyperedge) between the corresponding vertices in the similarity-graph. By algorithmic reasons, it is easier to check only pairs of electrodes, see [11]. But, with some more effort, also k-element subsets of electrodes could be checked for similarities if k is small. This leads to edges and hyperedges of electrodes. If a spiking neuron has contact to an unknown set S of electrodes, all edges between any two vertices of S (all hyperedges of any k vertices of S) are drawn in the similarity-graph. Though these edges may also be caused by different simultaneously spiking neurons having contact in each case to two (or k) neurons, it is more likely that only one neuron is the source. Such a neuron yields the edges of a weak clique in the similarity-graph. Once the similarity-graph is constructed, it remains the question what is the basic cause for this graph. A reasonable answer is that as few as possible neurons yield the graph. Consequently, a minimum weak edge clique covering has to be determined. Because of the bounded length of the axons, only nearby electrodes, which are placed in form of a twodimensional bounded grid (or some similar variants), may have contact to the same neuron. Hence the similarity-graph is a relatively sparse graph and edges are only drawn between electrodes which have a small Euclidean distance. Thus it is reasonable to expect that also this graph has a small bandwidth.

Some basic properties
we have l ≥ q + 1 and thus the distance is at least q + 1.
Consequently, X and Y are adjacent by Lemma 1 and their distance is 1. Indeed, from the conditions on X and Y it follows that The graph G n,k,b has the following number of vertices: Proof. The proof follows directly from the partition In the following, we often write the elements of V n,k,b as k-tuples in ascending order, i.e., We collect all vertices that are adjacent to all other vertices in the set We denote the set of remaining vertices by R = V n,k,b ∖ C, and split it into two parts: We define a partition (3) we obtain the assertion. Recall the definition of the lexicographic ordering < lex on the set of all k-tuples of integers: We define a proper numbering of V n,k,b in the form of a total order ≤. The minimal element gets label 1, the next elements get labels 2, 3, . . . and the maximal element gets label V n,k,b . Each total order will be given in the form of an ordinal sum of suborders: We define a total order ≤ spo , which we call the simple palindrom ordering (SPO), as follows: with the following suborders: , respectively. Thus it remains the case that X ∈ R 0 and Y ∈ R 1 . To reach a maximal f spo -distance, the form X = (X, X +1, . . . , X +k −1) To prove the assertion, it is sufficient to show the following: We note that b) follows from a) because of the symmetry of the ordering. To show a) we define This case is easy, because C ⊆ I j for all j.
Now the bandwidth of f spo can be determined: As they both differ from each other by at most 1 it follows that Now we are able to prove the first part of Theorem 1.
Proof of Theorem 1.a). We know from Lemma 6 that B (G n,k, ⌉.
Let f be an arbitrary proper numbering of G n,k,b . Let X V be the vertex with number 1 and X V the vertex with number V n,k,b . Further let X C be the vertex of C with smallest number, denoted α, and X C be the veretx of C with largest number, denoted β. Then β − α ≥ C − 1. Further X V and X C as well as X C and The sum of them is The maximum of both f -distances is therefore at least ⌈ Lemma 3 and (2) we obtain

Asymptotic bandwidth for b = o(n)
In this section, we consider the case, where b grows sublinearly with respect to n. First we take a simple proper numbering, which provides an upper bound for the bandwidth.

Lemma 7. Let n, k, b be arbitrary integers with
Proof. We order the vertices of G n,k,b in a lexicographic way, see (4). Let f lex (X) be the label of X ∈ V n,k,b with respect to this ordering. Now let X and Y be two adjacent vertices with X < lex Y and let X ′ = [X, X + k − 1] and Moreover, for j ∈ [0, n − b], and, for j ∈ [n − b + 1, n], Since Y ′ is the lexicographically smallest vertex with minimum element Now (5) and (6) imply which proves the assertion. Chvátal observed in [6] that a lower bound for the bandwidth is given by Here the diameter diam(G) of the graph G = (V, E) is the maximal distance of any two vertices of G. By Corollary 1, the distance of any two vertices of G n,k,b is at most Now we have all preparations to prove Theorem 1 b).

Proof of Theorem 1 b). From Lemma 7 we know that
For the lower bound, we use the fact that b We have by Lemma 3, (7) and (8) for n → ∞: .
Thus, B(G n,k,b ) ≥ k b k for sufficiently large n, which proves the first part of the assertion.
which shows that B(G n,k,b ) ≳ k b k as n → ∞, which proves the second part of the assertion.

Further basic properties for the asymptotics
where n → ∞ and i is a positive integer. If n is sufficiently large, then X and Y have distance at most i.
Proof. Without loss of generality, let (X, X) ≤ lex (Y , Y ), i.e., X < Y or X = Y as well as X < Y . By Corollary 1, X and Y have distance at most (1) β+o (1) Let P be a polygon in Ω = {(x, y) ∈ R 2 ∶ 0 ≤ x ≤ y ≤ 1} and let int(P ) be the interior of P . Let Accordingly, The RHS is a Riemann sum for the integral ∬ P (y − x) k−2 dx dy, which shows that For δ > 0 let P δ = {(x, y) ∈ P ∶ y − x ≥ δ}. Clearly, P δ ⊆ P . Obviously, for any ε > 0 there is some δ > 0 such that Moreover, for any z ≥ δ there is some n 0 such that for all n > n 0 This implies The reasoning for V o n,k (P ) is the same. For the sake of brevity, we define the measure of the polygon P ⊆ Ω by

Corollary 2. If S ⊆ [0,n]
k is a family of sets that contains all X ∈ [0,n] k with 1 n (X, X) ∈ int(P ) and some X ∈ [0,n] k with 1 n (X, X) on the boundary of P , then S ∼ µ(P )n k as n → ∞.

Definition and measure of crucial polygons
Recall that we consider the case b ∼ βn with β ∈ (0, 0.5] and q ∈ N and r ∈ R such that 1 = qβ + r. We define in Ω several sets of points. First let for i = 1, . . . , q In the following we denote lines given by y = ax + b (or, more generally, by ax + by = c) by g y=ax+b (or g ax+by=c ). Note that the points B i and C i lie on g y=x . Furthermore, the points A i lie above or on the line g y=x+β iff r ≤ q−1 q 2 +q−1 . This is the reason for the distinction between a) and b) in Theorem 2.
For the Case a), i.e., for r ≤ q−1 q 2 +q−1 , we define points D i and E i as the intersection points of the segments A i B i resp. A i C i with the line g y=x+β . An easy computation yields with γ = β(1 − 1 q) For the Case b), i.e., for r > q−1 q 2 +q−1 , we define points F i as those points that provide an equipartition into q + 1 parts of the segment between (0, 0) and (1, 1) and points G i as the intersection points of the segments between F i and (0, 1) with the line g y=x+β . It is straightforward that for i = 0, . . . , q + 1 As for the Case a), let E q be the intersection point of the lines g y=x+β and g y=qβ . Finally, we define an auxiliary point H 1 = (r, β). For the Case a) we also use the points In Figures 1 and 2 the points are illustrated for β = 9 20 (i.e., Case a) with q = 2, r = 1 10 < 1 5) as well as for β = 7 20 (i.e., Case b) with q = 2, r = 3 10 > 1 5).
(1, 0)  It is easy to check that the points A i lie on the segments and that the points F 0 , B 1 , F 1 , C 1 , . . . , B q , F q , C q , F q+1 lie in this order on the line g y=x . It is also easy to see that H 1 lies on the segment A 1 B 1 .

Lemma 10.
Let 0 ≤ s < t ≤ 1, let P 1 = (ξ 1 , ξ 1 + s), P 2 = (ξ 2 , ξ 2 + s) be points on the line g y=x+s with 0 ≤ ξ 1 ≤ ξ 2 ≤ 1 − s and let P 3 = (ξ 3 , ξ 3 + t), P 4 = (ξ 4 , ξ 4 + t) be points on the line g y=x+t with 0 ≤ ξ 3 ≤ ξ 4 ≤ 1 − t. Let u = ξ 2 − ξ 1 and v = ξ 4 − ξ 3 . Then Proof. The proof follows directly by computing the integrals, using the coordinate transformation x ′ = x + y, y ′ = y − x. This leads to a domain of integration in form of a trapezoid whose basis is parallel to the x ′ -axis. Since only the difference of the ξ-values has influence we have: Corollary 3. Let 0 ≤ s < t ≤ 1, let P 1 , P 2 and Q 1 , Q 2 be points of Ω on the line g y=x+s with → P 1 P 2 = → Q 1 Q 2 showing to north east and let P 3 , P 4 and Q 3 , Q 4 be points on the line g y=x+t with → P 3 P 4 = → Q 3 Q 4 showing to north east. Then By inserting the corresponding values and using the definition of the c-functions in Theorem 2, we obtain: Corollary 4. We have for all possible i:

Proof of the lower bounds for the bandwidth in Theorem 2
We fix some ε > 0, where ε is sufficiently small, in particular ε < 1. Let P be a polygon in Ω. In the following we work with dilations of P around a given centerpoint by factors of the form (1 − ε) and (1 + ε), respectively. We denote the new polygons by P and P , respectively. But we emphasize that P and P depend on ε and on the centerpoint of dilation. Note that Though the following result follows also directly from Lemma 3, we prove it as an example for our polygon-method: We choose F 0 as the centerpoint of dilation so that e.g. the vertices of T are given by If the dilation factors are (1 − ε) and (1 + ε), respectively, then T ⊆ {(x, y) ∈ Ω ∶ y ≤ x + (1 − ε)β} and T ∩ Ω ⊆ {(x, y) ∈ Ω ∶ y ≤ x + (1 + ε)β}. It is easy to check that for sufficiently large n and any X ∈ [0,n] k the following implications are true: and hence by Corollary 2 With ε → 0 we obtain (using also Corollary 4) Now we prove the first asymptotic lower bound: For the dilation, we choose again F 0 as the centerpoint. With the factor (1 − ε) we obtain Q. Let G ′ be the subgraph of G n,k,b induced by Note that 1 n X ≤ qβ(1 − ε) for all X ∈ V ′ . Now we show that for sufficiently large n Let X and Y be any two distinct vertices of G ′ . Then 0 ≤ X ≤ X ≤ qβ(1 − ε)n and 0 ≤ Y ≤ Y ≤ qβ(1 − ε)n. By Lemma 8, X and Y have distance at most q if n is sufficiently large, which proves (9). From the Chvátal bound (7) it follows that By Corollary 4, Now we prove the second asymptotic lower bound, which applies only for the second case: Proof. Let briefly R = B 1 C 1 H 1 . We choose 1 2 (B 1 + C 1 ) = 1 2 (r + β, r + β) as the centerpoint and (1 − 2 β−r ε) as the factor of dilation and thus obtain R from R. The vertices of R are B 1 = (r + ε, r + ε), C 1 = (β − ε, β − ε) and H 1 = (r + ε, β − ε). Let f be a bandwidth numbering of G n,k,b . Let X V and X V be those vertices for which f (X V ) = 1 and f (X V ) = V n,k,b . Case 1. X V ≤ n(r + ε) and X V ≤ n(r + ε). Then, for sufficiently small ε and sufficiently large n, X V , X V ≤ n(r + ε + β + ε) ≤ 2β(1 − ε)n. Lemma 8 implies that X V and X V have distance at most 2 and by Lemma 11, we have Case 2. X V > n(r + ε) and X V > n(r + ε). Then, for sufficiently large n, X Lemma 8 implies that X V and X V have distance at most q and by Lemma 11, we have In view of n(r + ε) ≤ X R ≤ n(β − ε), n(r + ε) ≤ X V ≤ n(r + ε + β + ε) and r < β we have if ε is sufficiently small and n is sufficiently large. Thus, by Lemma 1, X V and X R are adjacent and hence In view of X R , X V ≥ n(r + ε) and analogously to Case 2, X R and X V have distance at most q and hence From (10), (11) and (12) we obtain and thus by Corollary 4 and Lemma 11 and with ε → 0 the assertion follows.

Proof of the upper bounds for the bandwidth in Theorem 2
In the following, we present two proper numberings f of G n,k,b whose bandwidth is asymptotically equal to the asserted upper bounds. As in Section 4, we define a total order V n,k,b = S 1 ⊕ ⋅ ⋅ ⋅ ⊕ S l with suborders given by means of polygons.
In order to avoid intersections on the boundaries we explicitly describe which part of the boundary is deleted, though the ordering can be given on the whole polygon. For example, a notation of the form C 0 B 1 D 1 E 0 ∖ B 1 D 1 means that the segment B 1 D 1 is deleted from the closed quadrangle C 0 B 1 D 1 E 0 . Case a) r ≤ q−1 q 2 +q+1 . We define the total order ≤ as follows: We still have to define the ordering of the elements of V n,k (C i B i+1 D i+1 E i ), i = 0, . . . , q, and of V n,k (B i C i E i D i ), i = 1, . . . , q (here we may allow the complete boundary).
We use a new coordinate system with the same origin and with transformation matrix and inverse transformation matrix Thus the new coordinate axes have direction of → C i B i+1 and → C i E i . The ordering is a lexicographic ordering of the points 1 n (X, X) with respect to the new coordinate system, i.e., for X, see Figure 3. Figure 3: Schematic illustration of the ordering for Case a).
Here we work with polar coordinates in the coordinate system with origin A i and x-axis in the direction of → A i B i and arbitrary, but fixed unit length. For X ∈ V n,k (B i C i E i D i ) let ϕ i (X) and r i (X) be the angular and radial coordinates of 1 n (X, X) in this coordinate system. The ordering is a lexicographic ordering with respect to the reflected polar coordinates, i.e., for X, Figure 3. Note that for simpler numerical computations ϕ i (X) may be enlarged to an angle such that one leg is parallel to the y-axis, the size of the angle may be replaced by tan(ϕ i (X)) and the Euclidean norm for r i (X) may be replaced by some other norm, e.g. the L 1 -norm. It is easy to check that Lemma 14. Let f be the numbering for Case a). Then, for n → ∞, Proof. Let P = (ξ, ξ) be any point on the segment C 0 B q+1 , i.e., 0 ≤ ξ ≤ 1. With P we associate a new pointP as follows: If P ∈ C i B i+1 for some i then letP be the intersection point of the line g y=x+β with the line through P that is parallel to C i E i . If P ∈ B i C i for some i then letP be the intersection point of the line g y=x+β with the line through P and A i , see Figure 4. Figure 4: Important points for the upper bound for Case a).
Moreover, for X ∈ V n,k,b let Let X, Y ∈ V n,k,b with X ≤ Y . By the definition of the ordering and in view of Thus we have to prove that Note that → Since X and Y are adjacent we have by Lemma 1, Y − X ≤ b ∼ βn. Let ε > 0. Then 1 n (Y − X) ≤ β + ε for sufficiently large n. Let which is clearly also true if 1 n (Y − X) ≤ β. Using Corollary 4, it is easy to check that for all possible i, An illustration of this fact can be found in Figures 5 and 6. Figure 5: Illustration of (18), where P X ∈ C i B i+1 for some i. Both red quadrangles have the same measure by Corollary 3. Figure 6: Illustration of (18), where P X ∈ B i C i for some i. Both red quadrangles have the same measure by Corollary 3 and the intercept theorem.
By Corollary 2, V n,k (P X P YPYP X ) ≲ (c 1 (β, k) + O(ε)) n k and with ε → 0 we get (15). Case b) r > q−1 q 2 +q+1 . For this case, we use an ordering similar to Case a), but with different polygons, due to the different location of their defining points. Let A 0 = (0, q(β − r)), A q+1 = ((q + 1)r, 1)) and I = (0, 1). Note that the points A i , i = 0, . . . , q + 1, lie on the line g y=x+q(β−r) . We define the total order ≤ as follows: We still have to define the ordering of the elements of V n,k ( . . , q (again, we may allow the complete boundary).
, in the coordinate system with origin I and x-axis in the direction of → IA i and arbitrary, but fixed unit length. Similarly to V n,k (B i C i E i D i ) in Case a), the ordering is a lexicographic ordering with respect to the reflected polar coordinates, i.e., for For a point P ∈ C i B i+1 A i+1 A i letP be the intersection point of the line g y=x+q(β−r) with the line through P that is parallel to C i A i . If, in particular, P = 1 n (X, X) with X ∈ V n,k (C i B i+1 A i+1 A i ), then letφ(X) be the angular coordinate ofP in the coordinate system introduced for V n,k (A i A i+1 G i+1 G i ). The combination of the two orderings is as follows: Finally, we discuss V n,k (A i B i C i ). Here the ordering is a lexicographic ordering of the reflected polar coordinates in the coordinate system with origin A i and x-axis in the direction of → A i B i and arbitrary, but fixed unit length. The whole ordering is illustrated in Figure 7. Proof. Let P = (ξ, ξ) be any point on the segment C 0 B q+1 , i.e., 0 ≤ ξ ≤ 1. With P we associate a new pointP as follows: If P ∈ C i B i+1 for some i, i.e., P belongs to the quadrangle C i B i+1 A i+1 A i , then we already definedP . The pointP is the intersection point of the line g y=x+β with the line throughP and I. If P ∈ B i C i for some i then letP be the intersection point of the line g y=x+β with the line through I and A i , see Figure 8. For X ∈ V n,k,b we define P X and P X as in (14) and, analogously to Case a), have to prove that for n → ∞ V n,k (P X P YPYP X ) ≲ (c 2 (β, k) + c 3 (β, k))n k .
We define P ′ as in (17). With the same arguments as for Case a) it is sufficient to prove that µ(P X P ′P ′P X ) = c 2 (β, k) + c 3 (β, k).
Using Lemma 4, one can verify that for all possible i, An illustration of this fact can be found in Figures 9 and 10. Figure 9: Illustration of (20), where P X ∈ C i B i+1 for some i. Both red and both blue quadrangles have the same measure by Corollary 3 and the intercept theorem.  Figure 10: Illustration of (20), where P X ∈ B i C i for some i. Both red quadrangles have the same measure by Corollary 3.

Open problems
We formulate the following conjecture in form of a problem because we are rather convinced that it is correct.

Problem 1.
Prove that the ordering for Case b) presented in Section 9 and illustrated in Figure 7 defines an asymptotically optimal bandwidth numbering.
A larger program is formulated in the second problem: Problem 2. Find and study other interesting graph classes that allow a reduction to the unit square for the asymptotics and lead to interesting and non-trivial orderings on the unit square.