The excluded minors for isometric realizability in the plane
Abstract.
Let be a graph and . The parameter is the least integer such that for all and all vectors , there exist vectors satisfying
It is easy to check that is always finite and that it is minor monotone. By the graph minor theorem of Robertson and Seymour [RS04], there are a finite number of excluded minors for the property .
In this paper, we determine the complete set of excluded minors for . The two excluded minors are the wheel on vertices and the graph obtained by gluing two copies of along an edge and then deleting that edge. We also show that the same two graphs are the complete set of excluded minors for . In addition, we give a family of examples that show that is unbounded on the class of planar graphs and is not bounded as a function of treewidth.
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1. Introduction
Let be a finite set and . We say that is a metric space if satisfies the following properties: for all , if and only if , and for all . For define and . Recall that is a norm for all . Throughout this article we denote by the metric space where .
A natural way for comparing two metric spaces and is through the use of distance preserving maps from one space to the other. Formally, an isometric embedding of into is a function such that for all .
Typically, the requirement that all pairwise distances are preserved exactly is too restrictive to be useful in practice. To cope with this, a successful theory of embeddings with distortion has been developed, where the requirement that distances are preserved exactly is relaxed to the requirement that no distance shrinks or stretches excessively. In this direction, the celebrated theorem of Bourgain [Bourgain85] asserts that every point metric space can be embedded into an space with distortion. Moreover, this is best possible up to a constant factor.
Another popular approach is to only require a subset of the distances to be preserved exactly. This viewpoint is very graph theoretical, and is the approach that we take in this paper.
All graphs in this paper are finite and do not contain loops or parallel edges. A graph is a minor of a graph , if can be obtained from a subgraph of by contracting some edges. When taking minors, we always suppress parallel edges and loops.
Let be a graph and . We define to be the least integer such that for all and all vectors , there exist vectors satisfying
It is not obvious that this parameter is always finite, but from the conic version of Carathéodory’s Theorem, it follows that for all and all vertex graphs (see [B90] and [DL97, Proposition 11.2.3]). For , Barvinok [Barvinok95] showed the better bound for graphs with edges.
Let denote the complete graph on vertices. The study of for varying values of is a fundamental problem in the theory of metric embeddings. For the case , Holsztynski [H78] (and subsequently Witsenhausen [W86]) showed that
Furthermore, Witsenhausen [W86] showed that for , which was later improved to
by Ball [B90]. Lastly, Ball [B90] also showed that
and that there is a constant such that
The lower bound of uses the biclique covering number, which is the minimum number of complete bipartite subgraphs needed to cover the edges of a graph. Rödl and Ruciński [RR97] have since shown that there is a constant such that for every there exists an vertex graph that cannot be covered with complete bipartite subgraphs. This implies that there is a constant such that
The parameters are also widely studied in rigidity theory. We refer the interested reader to Kitson [Kitson15] and Sitharam and Gao [SG10] and the references therein.
It is easy to show that for all , the parameter is minor monotone. By the graph minor theorem of Robertson and Seymour [RS04], there are a finite number of minorminimal graphs with . We call these graphs the excluded minors for .
The excluded minors for , , and were determined by Belk and Connelly [Belk, BC07].
Theorem 1 ([Belk, Bc07]).
For every graph ,

iff has no minor;

iff has no minor;

iff has no minor and no minor.
In this article we mainly focus on the case . The spaces are particularly interesting due to their “universal” nature in terms of isometric embeddings, as illustrated by the following theorem of Fréchet.
Theorem 2 ([F10]).
Every point metric space can be isometrically embedded in .
Theorem 2 allows us to rephrase the condition as follows. Let be a graph and . The length of a path in is defined as . Throughout this work we call a distance function on if for all edges , every path from to has length at least (in other words, the path consisting of the edge is a shortest path). We remark that is allowed in this definition, and that defines a corresponding metric space on at most points as follows. First contract all edges with , and then consider the shortest path lengths between pairs of vertices. Hence, by Theorem 2, if and only if for all distance functions on , there exist vectors satisfying
Note that for all , . Thus, by Theorem 1, if and only if has no minor. In this paper we determine the complete set of excluded minors for . Let denote the wheel on vertices and be the graph obtained by gluing two copies of along an edge and then deleting , see Figure 1. Using techniques from rigidity matroids, Sitharam and Willoughby [SW15] determined for all graphs with at most 5 vertices, except for . They conjectured that is an excluded minor for , and that is the only excluded minor for . We verify their first conjecture, but disprove the second by showing that is also an excluded minor for .
The following is our main result.
Theorem 3 (Main Theorem).
The excluded minors for are and .
The proof of Theorem 3 is given in Section 6. Note that unlike the case, given points with there does not necessarily exist an isometry of which maps to and to . For example, take and in . Indeed, the isometries of correspond to signed permutation matrices. Therefore, our proof technique for the case is quite different from the case. For example, we will show that the property is not closed under taking sums.
We also prove the following result, which follows from Theorem 3 with a little extra work.
Corollary 4.
The excluded minors for are and .
Robertson and Seymour [RS95] proved that testing for a fixed minor can be done in cubic time. Therefore, our results give an explicit cubictime algorithm to test if (equivalently . We simply have to test if our input graph contains a minor or a minor.
In a previous version of this paper, we asked whether is bounded on the class of planar graphs. We also asked whether is bounded as a function of treewidth. We now have found an example that shows that the answer to both of these questions is negative.
Theorem 5.
For every there exists a planar graph with treewidth such that .
Paper Organization.
In Section 2 we present a few equivalent ways to think about and prove some upper and lower bounds. In Section 3, we show . In Section 4 we show that we can suppress degree vertices when computing . In Section 5 we show that and are excluded minors for . In Section 6 we show that and are the only excluded minors for , and explain how to deduce Corollary 4 from the main theorem. We conclude the paper in Section 7 by proving Theorem 5 and discussing some open problems.
2. Potentials and Implicit Realizations
In this section we present several equivalent ways to think about the parameter .
Consider an vertex graph , a distance function on , and a realization of in ; that is, a collection of points such that for all . We can write a matrix whose columns are the vectors for . In this section we analyze this matrix by looking at its rows, which turn out to be potentials of a natural directed graph associated to .
Let be an edgeweighted directed graph and let be the length function on the arcs of . Note that negative lengths are allowed. A function is called a potential on if for all arcs . We recall the following wellknown result characterizing the existence of a potential.
Theorem 6.
A weighted directed graph admits a potential if and only if it does not contain any negative length directed cycle.
Now let be the weighted directed graph obtained from as follows. First, we bidirect all edges of . For every edge , we define the length of both and to be . That is, the length function on is given by
(1) 
Note that is a potential on if and only if . An edge is tight for a potential on if .
Let be a realization of in . Clearly, if we define for and , we have that is a potential for all . Moreover, every edge of is tight in some . It is easy to see that the converse also holds.
Lemma 7.
Let be a graph. A distance function on admits a realization in if and only if the directed graph with lengths as in (1) admits a collection of potentials such that every edge is tight in some . Moreover, in this equivalence we can take for all and .
In view of Lemma 7, we get a combinatorial approach for constructing and analyzing realizations. For , let denote some orientation of . We say that is a feasible orientation (with respect to ) if there exists a potential on such that for all . See Figure 2 for an illustration. We say that is feasible if it admits a feasible orientation. If a set of edges is not feasible, we say that it is infeasible. Notice that is a feasible orientation if and only if the opposite orientation is a feasible orientation. Furthermore, note that a subset of a feasible set is also feasible.
The notion of feasible sets allows to reformulate Lemma 7 as follows.
Lemma 8.
Let be a graph and be a distance function on . The pair admits a realization in if and only if there exist feasible sets such that .
Given an orientation , we define a modification of the length function as follows.
(2) 
We denote this length function by . Note that is a feasible orientation if and only if admits a potential. By Theorem 6, this happens if and only if the weighted digraph does not contain a directed cycle of negative length.
We demonstrate the usefulness of Lemma 8 by quickly deriving some nontrivial upper and lower bounds for .
Note that for every distance function on and every vertex of , the star centered at is always feasible with respect to , as can be seen by orienting all the edges of the star outwards (as in Figure 2). From this we obtain the following upper bound.
Lemma 9.
For every graph ,
where denotes the minimum size of a vertex cover of .
We say that a distance function is generic with respect to if for every cycle in and , we have . Every distance function on can be perturbed to a nearby generic distance function . Furthermore, we have if and only if can be realized in for every generic distance function .
Observe that if is generic, every feasible set is acyclic. Therefore, we immediately obtain the following lemma.
Lemma 10.
For every graph ,
where denotes the minimum number of forests required to partition .
Our next result implies that, if is generic, every maximal feasible set is a spanning forest.
Lemma 11.
Let be a graph and be a distance function on . Then every maximal feasible set contains a spanning forest.
Proof.
Towards a contradiction, suppose that is a maximal feasible set that does not contain a spanning forest of . Let be the vertex set of a component of such that contains at least one edge with exactly one end in . Let be any potential that makes all the edges of tight but no other edges. Let be as large as possible with the property that is a potential, where denotes the characteristic vector for the vertex . Then the set of edges that are tight with respect to is a proper superset of , a contradiction. ∎
3.
Since for , it follows that , and . Thus, is the smallest value for which is unknown. In this section we show that . This result is not needed for our main theorem but may be of independent interest.
Proposition 12.
.
Proof.
We already know that , let us prove that . To this aim, enumerate the vertices of as , and define a linear ordering on its edges by letting, for and ,
if , or and . Let be the number of edges. Define a distance function on the graph by letting for each edge , where is the rank of in the ordering . (Thus has rank and has rank .) It is easy to check that is a generic distance function.
We claim that cannot be realized in . Arguing by contradiction, assume it can. Consider a partition of the edges into four feasible forests . Before analyzing these, let us note a few properties of a feasible forest (the easy proofs are left to the reader).

a feasible orientation of cannot contain a length directed path, hence is uniquely determined (up to reversing all arcs);

if then at most one of the two edges and is in ;

if then at most two of the three edges , , are in .
Now color each edge of the graph with the index of the forest it is included in. By (2) we may assume without loss of generality that , , and are colored , , and respectively. By the same property, none of the two edges , are colored or , and they cannot both be colored (otherwise would be a triangle in ), thus there exists such that is colored .
Next consider the four edges between the set and . None of these is colored by (2) (because of the edge ) or (because of the edge ), so each of them is colored or . Moreover, in order to avoid monochromatic triangles, the four edges are split into two matchings and of size , colored and respectively.
Let be the set of edges with that are distinct from . (Thus .) No edge in is colored (because of ). We claim that no edge in is colored either. This is clear for those not incident to , thanks to the edge of that is incident to . Now, suppose for a contradiction that is incident to and is colored . Then letting be the edge of incident to , we see that the edges are all in , contradicting property (3).
All edges in are colored or but has size and spans only vertices. Therefore, there is a monochromatic cycle in . This final contradiction concludes the proof. ∎
4. Degree Vertices
In this section we show that we can essentially ignore degree vertices when computing .
Let and be graphs that each contain a clique of size . A sum of and along is a graph obtained by gluing and along and then deleting some of the edges of . In the special case of sums, we use the notation if we keep the edge , and if we delete the edge .
Lemma 13.
Let be a graph and let . If , then .
Proof.
Set , let and let be the newly added vertex in . Clearly so it suffices to show that . Let be any distance function on . The restriction of to is also a distance function. Let be a collection of feasible sets of such that .
First, note that each is feasible in . Indeed, since is a distance function, and in particular , we can extend any potential on to a potential on by carefully choosing the potential value at between the value at and that at . Without loss of generality, we may assume that . Now extend to a maximal feasible set . By Lemma 11, contains a spanning forest. Hence, contains either or . Without loss of generality, assume that .
Now let be a feasible orientation of . By reversing all the arcs of if necessary, we may assume that . We claim that is a feasible orientation. Indeed, let be a negative directed cycle in with respect to . Since is a feasible orientation, we may assume that . Now , which means that the length of does not increase if we shortcut it from to . Since is a feasible orientation, the length of the shortcut cycle is nonnegative, which contradicts our assumption that has negative length. Hence, is a feasible orientation and the corresponding edge set is feasible.
We have found feasible sets , , , …, that cover each edge of . Thus can be realized in . The lemma follows. ∎
We note that the assumption that in Lemma 13 is necessary. This can easily be seen by taking and .
We say that is obtained from by subdividing an edge if .
Lemma 14.
Let and be graphs such that is obtained from by subdividing an edge. Then .
Proof.
Clearly since is a minor of . It remains to prove . If then is a forest, and so is , implying . Hence we may assume that . Say that is obtained from by subdividing an edge with a new vertex . Let . Since is obtained from by adding a new vertex adjacent to the ends of the edge , we have that by Lemma 13. The graph being a minor of , it follows that . ∎
5. The graphs and
In this section we show that and are excluded minors for .
Lemma 15.
We have that .
Proof.
By Lemma 9, . Towards a contradiction suppose . Let be the distance function on given in Figure 3 and let be an isometric embedding of in . Note that all lie on two consecutive sides of a square centered at with side length . By symmetry we may assume that , that where and that or for . We say that is directly right of if (in this case is directly left of ), is directly below if , and that and are diagonal.
We first consider the case that is directly right of . This implies that must be directly left of as would be too far from (if is directly right of ) or would be too far from (if and are diagonal). Now, cannot be directly right of as would be too far from , and cannot be directly left of as would be too close to . Thus, and are diagonal. But now , which is a contradiction.
We next consider the case that is directly left of . Again, cannot be directly left of . Suppose that is directly right of . Again, cannot be directly right of or left of . Thus, and are diagonal. But now , which is a contradiction. Thus, and must be diagonal. If is directly above or directly below , then , which is a contradiction. Thus, and are diagonal. Since , we must have or . In the first case, and in the second case , both of which are contradictions.
The remaining case is if and are diagonal. Thus, or . Suppose . If and are diagonal, then , a contradiction. If is directly right of , then is too far away from . Thus, . Evidently, cannot be directly left of . If is directly right of we have , a contradiction. If and are diagonal, then and are too close. We finish with the subcase that . Again, we must have . If is directly below , then , a contradiction. If and are diagonal, then and is too close to . This completes the subcase and the proof. ∎
Lemma 16.
The graph is an excluded minor for . Moreover, is the only excluded minor for among all graphs with at most 5 vertices.
Proof.
By the previous lemma, , so to prove that is an excluded minor it suffices to show that every proper minor of satisfies . If , then since . Now, say is obtained from by only deleting edges. Deleting an edge yields a degree vertex, which we can suppress by either Lemma 14 or Lemma 13. Again, we get a graph with at most 4 vertices, so we are done.
For the second part, let be an excluded minor for with . If has a minor, then . So we may assume that has no minor. Let and be edges of . By Lemma 13 we have that . Since has no minor, this implies that is a minor of . But then, , which is a contradiction. ∎
Lemma 17.
We have that .
Proof.
To simplify notation, throughout this proof we set . Furthermore, we use the labeling of the nodes of given in Figure 4.
We first show that . Let be an arbitrary distance function on . Note that and are feasible sets because they are stars. Thus, if is feasible, then can be realized in by Lemma 8. To conclude the proof assume that is not feasible. Note that and are feasible because they are stars. Let and be maximal feasible sets containing and respectively. By Lemma 11, and each span all the vertices of . Therefore, since is not feasible, we must have and . Let . Thus, is a subset of , or . In the first two cases, is feasible since it is a subset of a star. In the last two cases, note and are feasible orientations of and , respectively. Hence, is also feasible in the last two cases. Since , and are feasible sets covering all the edges of , Lemma 8 yields .
To show that it remains to exhibit a distance function on such that is not realizable in . We exhibit such a distance function in Figure 4. Towards a contradiction, suppose that can be partitioned into two feasible sets and . It is easy to check that is a generic distance function, and so and are both forests.^{1}^{1}1If one does not want to check genericity, simply perturb to a nearby generic distance function. Thus, edges. Since , we conclude that and are both spanning trees. Let and be the subgraphs of induced by and , respectively. By interchanging and , we may assume that . Therefore, there are six possibilities for each of and , and these are shown in Figure 6. The six possibilities for are shown along the first column of the table, and the six possibilities for are shown along the first row.
We rule out each of the possibilities for by showing that at least one of or is infeasible. To do this, we show that for all orientations and of and , at least one of or contains an infeasible orientation.
If forms a triangle in , note that is an infeasible orientation. Indeed, the triangle inequality combined with the fact that is generic imply that the directed cycle is negative. We denote this infeasible orientation as . In Figure 5, we list more infeasible orientations that do not come from triangles. These infeasible orientations consist only of the oriented arcs in each picture. However, for the benefit of the reader, we have included dashed edges to indicate the negative cycle in .
The remainder of the proof is summarized in Figure 6. Each entry in the table gives the infeasible orientations to apply in order to obtain a contradiction. For example, consider the fourth row of the table. For this entire row, it suffices to only consider the edges in . By symmetry, we may assume that . Next, implies that . Then, implies . Since , we contradict . Thus, , and are sufficient to derive a contradiction. Sometimes the infeasible orientations need to be applied to instead of to , in which case we have specified so. ∎
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Lemma 18.
The graph is an excluded minor for .
Proof.
By the previous lemma, , so it suffices to show that every proper minor of satisfies . Contracting an edge of yields a vertex graph which is not connected. In particular, the latter graph does not have as a minor. We are done in this case, since by Lemma 16, is the only excluded minor for among graphs on at most vertices.