Graphs with Obstacle Number Greater than One
Abstract.
An obstacle representation of a graph is a straightline drawing of in the plane together with a collection of connected subsets of the plane, called obstacles, that block all nonedges of while not blocking any of the edges of . The obstacle number is the minimum number of obstacles required to represent .
We study the structure of graphs with obstacle number greater than one. We show that the icosahedron has obstacle number , thus answering a question of Alpert, Koch, & Laison asking whether all planar graphs have obstacle number at most . We also show that the skeleton of a related polyhedron, the gyroelongated bipyramid, has obstacle number . The order of this graph is , which is also the order of the smallest known graph with obstacle number .
Some of our methods involve instances of the Satisfiability problem; we make use of various “SAT solvers” in order to produce computerassisted proofs.
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1. Introduction
All graphs will be finite, simple, and undirected. Following Alpert, Koch, & Laison [2], we define an obstacle representation of a graph to be a straightline drawing of in the plane, together with a collection of connected subsets of the plane, called obstacles, such that no obstacle meets the drawing of , while every nonedge of intersects at least one obstacle. By nonedge, we mean a pair of distinct vertices of where is not an edge of . The least number of obstacles required to represent is the obstacle number of , denoted . For clarity, we will sometimes refer to this as the ordinary obstacle number. In an obstacle representation of a graph , an outside obstacle is an obstacle that is contained in the unbounded component of the complement of the drawing of . Any other obstacle is an interior obstacle. Note that by [10], determining whether for a given is not in NP.
Figure 1 shows several obstacle representations for the Petersen graph . The usual drawing of requires five obstacles. This is illustrated in Figure (a)a (edges of are shown as thick line segments, while nonedges are shown as thin line segments).
An obstacle may be any connected plane region; however, it is convenient to expand each obstacle until it is as large as possible. Each of the resulting obstacles is an open plane region forming a connected component of the complement of the drawing of the graph. The boundary of the obstacle—which is not part of the obstacle itself—is formed by appropriate parts of the edges of the graph. Figure (b)b illustrates such obstacles for .
However, can be represented using fewer obstacles; is in fact . Figure (c)c shows an obstacle representation of using only a single obstacle (adapted from Laison [13]).
A related parameter, in which each obstacle is a single point, was studied by Matoušek [14] and by Dumitrescu, Pach, and Tóth [7]. The study of the obstacle number per se was initiated by Alpert, Koch, & Laison [2]. These parameters have since been investigated by others [8, 10, 16, 17, 18, 19].
Alpert, Koch, & Laison [2, Thm. 2] showed that there exist graphs with arbitrarily high obstacle number and asked [2, p. 229] for the smallest order of a graph with obstacle number greater than . They proved [2, Thm. 4] that the graph has obstacle number , where (with ) denotes the graph obtained by removing a matching of size from the complete bipartite graph . Pach & Sariöz [19, Thm. 2.1] found a smaller example of a graph with obstacle number .
Theorem 1.1 (Pach & Sariöz).
.
The PachSariöz example, of order , is apparently the smallest known graph with obstacle number greater than . In section 2 we consider the question of whether any graph of smaller order has obstacle number greater than . We do not answer this question; however, in Proposition 5.3 we will provide an example of a planar graph of order with obstacle number .
We define the outside obstacle number of to be the least number of obstacles required to represent , such that one of the obstacles is an outside obstacle—or zero if has obstacle number zero. We denote the outside obstacle number of by .
Clearly we have for every graph . Alpert, Koch, & Laison [2, p. 229] asked (using different terminology) whether every graph with also has . We ask a more general question.
Question 1.2.
Is it true that for every graph ?
Alpert, Koch, & Laison [2, p. 231] asked whether every planar graph has obstacle number at most (also see a series of questions in the Open Problem Garden [9]). They further asked for the obstacle numbers of the icosahedron and the dodecahedron. In Sections 3, 4, and 5 we develop tools for determining the obstacle numbers of particular graphs, and we use them to address these questions.
In Section 3 we consider the outside obstacle number. We make use of instances of the Satisfiability problem (SAT) to produce computerassisted proofs that the outside obstacle number of certain graphs is at least .
In Section 4 we consider the ordinary obstacle number. We show that a lower bound on the outside obstacle number of a graph implies a lower bound on the obstacle number of a different graph. We use this to answer the first of the abovementioned questions by constructing a planar graph with obstacle number .
In Section 5 we develop methods similar to those of Section 3 for producing computerassisted proofs that the ordinary obstacle number of certain graphs is at least . We answer another of the abovementioned questions by showing that the obstacle number of the icosahedron is . We also show that the obstacle number of the dodecahedron is . We further describe a graph of order that has obstacle number . This graph, which we call , has the same order as the PachSariöz example, but, unlike that graph, it is planar.
We conclude this section with some easy observations, which we will make use of throughout the remainder of this paper.
Observation 1.3.
Given an obstacle representation of a graph , we can perturb all vertices an arbitrarily small distance to obtain an essentially equivalent obstacle representation in which no three vertices are collinear.
Because of the above observation, we will generally assume that our obstacle representations have the property that no three vertices are collinear. Lest there be any confusion, only obstacles block edges, vertices do not, so this has no impact on the definition of obstacle number.
Observation 1.4.
Obstacles are not required to be simply connected, but adding this requirement does not change the obstacle number.
To see why this is true we note two facts: (1) whether an edge is blocked or not by an obstacle is determined only by whether the edge somewhere crosses a portion of the boundary of the obstacle; (2) if an obstacle is not simply connected, then it may be replaced with a simply connected obstacle that blocks those same edges by removing sufficiently small open sets about curves connecting portions of the complement that do not intersect the points on the boundary where edges of the graph cross the obstacle (the simplest case is cutting a small wedge out of an annulus blocking an edge connecting a vertex inside the annulus to a vertex outside the annulus).
Observation 1.5.
Let be a graph, and let be an induced subgraph of . Given an obstacle representation of , removing all vertices of that are not in , and keeping the same obstacles, results in an obstacle representation of .
The next result follows easily from the above observation.
Proposition 1.6.
Let be a nonnegative integer. The class of all graphs such that is closed under taking induced subgraphs, and similarly for .
Proposition 1.6 does not hold if the word “induced” is removed, as the classes in question are generally not closed under taking arbitrary subgraphs. For example, Alpert, Koch, & Laison [2, Thm. 2] showed that there are graphs with arbitrarily large obstacle number. But every graph is a subgraph of a complete graph, and for every .
2. Small Graphs
As noted in the previous section, Pach & Sariöz showed that (see Theorem 1.1). What can we say about graphs of small order with obstacle number greater than ?
Proposition 2.1.
Let be a graph. Let be a graph obtained by starting with and adding a new vertex of degree at most . If is complete and is not complete, then . Otherwise, . Furthermore, these continue to hold if is replaced by .
Proof.
We give the proof for ; the proof for is essentially the same.
We have if and only if is complete; the cases when is complete follow easily. Suppose that is not complete, and let be the new vertex in .
Begin with an obstacle representation of . First suppose that the degree of in is . In this case, Let be any obstacle of . Let be a point in the interior of , replace by , and place vertex at the point . Then since the edges between and the vertices of are blocked by .
If instead has degree , then let be the neighbor of in . Choose a point in the interior of an obstacle so that the line segment intersects no other obstacle of ; note that may cross edges of . (To find such a choose any obstacle and a point on its interior, and then let be the first obstacle crossed by the ray .) Now replace by , and place vertex at the point . (See Figure 2 for an illustration.) This results in an obstacle representation of using the same number of obstacles as that for , and also using an outside obstacle if the representation of used one.∎
We restate [2, Thm. 5] as follows.
Theorem 2.2.
The obstacle number of an outerplanar graph is at most 1.
Corollary 2.3.
Let be a graph with the property that every cycle in has length . Then .
Proof.
Graph can have no minor isomorphic to or , since these graphs both have cycles. Hence is outerplanar, so by Theorem 2.2, the obstacle number is at most . In their proof, Alpert, Koch, & Laison construct an obstacle representation using a single outside obstacle, thus showing that is at most . The result follows.∎
Proposition 2.4.
If a graph has order at most , then .
Proof.
By Proposition 2.1, we may assume that has minimum degree at least , since vertices of degree at most do not affect the obstacle number. There are only four graphs of order at most whose minimum degree is at least : , , , and . All of these are easily checked. So assume that has order .
If contains no  or cycle, then we may apply Corollary 2.3. If has a cycle, then arrange the vertices on a circle in the plane so that each edge in the cycle joins vertices that are not consecutive on the circle (so the cycle is a pentagram). No matter what other edges lie in , an outside obstacle can block all nonedges (see Figure 3 for an illustration).
So we may assume that contains no cycle, but does contain a cycle. Choose a cycle in , and let be the vertex of that does not lie on this cycle. Since has minimum degree at least , must be adjacent to at least vertices of the cycle. The neighbors of cannot include two consecutive vertices on the cycle, since then would have a cycle. Consequently, has degree exactly , and its neighbors are nonconsecutive vertices on the cycle. It remains to consider adjacencies among vertices in the cycle. The nonneighbors of cannot be adjacent, as this would form a cycle in .
The only question left is whether the neighbors of are adjacent. Thus there are exactly graphs that satisfy our assumptions: and . Both are easily shown to have outside obstacle number .∎
Corollary 2.5.
Let be the minimum order of a graph with . Then . Similarly, the minimum order of a graph with satisfies .
Proof.
The lower bound follows from Proposition 2.4. For the ordinary obstacle number, the upper bound follows from the result of Pach & Sariöz (Theorem 1.1). The obstacle representation constructed by Pach & Sariöz [19, Fig. 1] uses an outside obstacle, and so as well. The upper bound for the outside obstacle number follows.∎
We now consider what properties a small graph with must have. We show that such a graph cannot be a subgraph of .
Proposition 2.6.
Let be a subgraph of . Then .
Proof.
Let , be the smallest values such that is a subgraph of . Let be the subgraph of formed by those edges that do not lie in .
If has a vertex of degree or , then we may apply induction, by Proposition 2.1. So assume has minimum degree at least . Then has maximum degree at most .
We observe that, if there is a set of vertices containing at most vertices from each partite set, such that contains an endpoint of each edge in , then we can draw so that every nonedge is blocked by an outside obstacle, by placing the two partite sets in (near) lines, with the vertices in at the ends of their respective lines. (See Figure 4 for an illustration.) We will use this observation repeatedly in the remainder of this proof.
If does not contain a matching of size , then has a vertex cover containing at most vertices, by the KönigEgerváry Theorem (see, e.g., Bondy & Murty [5, p. 74]). If includes members of both partite sets, then let . Otherwise, replace one of the vertices in with the other endpoints of the (at most ) edges of it is incident with; again, we have our set .
So we may assume that has a matching of size . Thus . Since has minimum degree at least , we have two cases: in the first case contains two vertexdisjoint cycles; in the second contains an cycle. In the former case, the vertices of one of the cycles form our set . In the latter case, if contains at least one edge that is not in the cycle, then and edges of the cycle form a cycle in . Let consist of the vertices of that do not lie on this cycle.
It remains only to handle the case when is an cycle with no additional edges. In this case there is no set with the properties we are looking for. However, we can draw with every edge blocked by an outside obstacle as in Figure 5. ∎
3. Outside Obstacle Number
We now begin our development of tools for explicitly determining the (outside) obstacle number of a particular graph. We will apply these tools to various planar graphs. In this section, we cover results related to outside obstacle number; we will consider the ordinary obstacle number in the following sections.
Our ideas are based on the Satisfiability Problem (SAT). An instance of this problem is a particular kind of Boolean formula. In this context, a variable is a Boolean variable; i.e., its value is either true or false. A literal is either a variable or the negation of a variable. A clause is the inclusiveOR of one or more literals. For example, given variables , , and , the following is an example of a clause:
An instance of SAT consists of a number of clauses. The instance is said to be satisfiable if there exists a truth assignment for the variables such that every clause in the instance is true; i.e., a truth assignment in which each clause contains at least one true literal.
We will show how to construct, for each graph , a SAT instance encoding necessary conditions for the existence of an obstacle representation of using no interior obstacles. Thus, if we can show that the instance is not satisfiable, then we know that . There are a number of freely available, highquality implementations of algorithms to determine satisfiability of a SAT instance. Using these, we will construct computeraided proofs that for various planar graphs.
When interpreting the effect of the satisfiability or nonsatisfiability of the SAT instances on the obstacle number problem, it is important to note that if the SAT instance is not satisfiable, then we are guaranteed that it is impossible to draw the graph using a single outside obstacle; i.e., . However, if the SAT instance is satisfiable, then it does not follow that ; satisfiability is a necessary but not sufficient condition for concluding that .
For and , we say that is a clockwise triple if , and appear in clockwise order. More formally, identifying each point in with the corresponding point in the plane in via the map , we denote the determinant of the matrix with columns , and by . Then is a clockwise triple if . We similarly define counterclockwise triple. Note that for a triple , exactly one of the following is true: is clockwise, is counterclockwise, or , and are collinear (in which case ).
The following two lemmas give properties that hold for all point arrangements in the plane. We will use these properties to construct clauses for our SAT instances.
Lemma 3.1 (Point Rule).
Let , , , be distinct points in . If , , and are clockwise triples, then must also be a clockwise triple.
Proof.
The Point Rule is equivalent to what D. Knuth called the interiority property of triples of points; see Knuth [12, p. 4, Axiom 4]. In addition, the Point Rule is a consequence of the fact that a point configuration is a rank 3 acyclic oriented matroid. Intuitively, the Point Rule follows from the observation that if is a clockwiseoriented triple, then cannot be “outside” all the lines , , (where “outside the lines” means on the negative side of the closed halfplanes determined by the orientation of ). See Figure 6. ∎
We encode the Point Rule as follows. For each triple we introduce a variable representing the statement that is a clockwise triple. By Observation 1.3 we may assume that no three vertices are collinear, so represents the statement that is a counterclockwise triple. The point rule is represented by the clause
(1) 
Lemma 3.2 (Point Rule).
Let , , , and be distinct points in . If , , , and are clockwise triples, then either

both and are clockwise triples, or

both and are counterclockwise triples.
Proof.
The Point Rule is represented by the following two clauses:
(2)  
(3) 
In constructing our SAT instance, we will make use of the following lemma, which describes a restriction that must hold for an obstacle representation without interior obstacles.
Lemma 3.3.
Suppose we are given an obstacle representation of a graph using no interior obstacles. Let be a nonedge of . Then there exists a halfplane determined by the line such that, for each path in , some internal vertex of lies in .
Proof.
Denote by and the two closed halfplanes determined by the line .
Suppose that there exist two distinct paths and in such that (without loss of generality) and . Perturbing slightly if necessary, we assume that no internal vertex of or lies on the line (see Observation 1.3). We observe then that forms a closed curve in the plane; the open line segment from to lies in a bounded component of the complement of this closed curve. By assumption there is no obstacle lying in such a bounded component, and thus there is nothing to block the line segment , a contradiction. ∎
Given a graph , we use Lemmas 3.1, 3.2, and 3.3 to create a SAT instance whose satisfiability is a necessary condition for the existence of a obstacle representation of using no interior obstacles. If this SAT instance is not satisfiable, then we may conclude that graph requires an interior obstacle, that is, that .
Since we are placing the vertices of in the plane, the Point Rule (Lemma 3.1) must hold for any vertices in the graph, the Point Rule (Lemma 3.2) must hold for any vertices in the graph, and by Observation 1.3 we may assume that no three vertices are collinear. Our SAT instance includes clauses from the Point Rule corresponding to clause (1) for every set of vertices of our graph, and every permutation of these vertices. It also includes clauses from the Point Rule corresponding to clauses (2) and (3) for every set of vertices of our graph, and every permutation of these vertices.
Note that there are six ways to say vertices and lie in clockwise order. Variables corresponding to even permutations of (, , ) represent equivalent statements. Variables corresponding to odd permutations of (, , ) represent their negations.
When we construct our SAT instance, we may choose one of the variables from among as the canonical variable; we represent the other five using either the canonical variable or its negation, as appropriate.
Additionally, the actions of even permutations of on clauses (1 and (2) result in statements equivalent to the original; likewise an even permutation of does not alter clause (3). This reduces the number of clauses required by a factor of . Thus, for an vertex graph, our SAT instance includes clauses based on the Point Rule and clauses based on the Point Rule.
We similarly construct SAT clauses based on Lemma 3.3. Since this lemma concerns a line defined by two points, and which side of this line certain other points lie on, the lemma can be stated in terms of clockwise or counterclockwise triples. Specifically, this lemma implies that, for each pair of nonadjacent vertices , of , one of the two halfplanes determined by segment is “special”; that is, this halfplane contains at least one internal vertex from each path in . We create a new variable representing the statement that the special halfplane is the halfplane containing points such that is a clockwise triple.
Let be the sequence of vertices in some path . Then the following clauses represent the statement of Lemma 3.3 for :
(4)  
(5) 
As with the variables, we choose one of and to be the canonical variable, and we represent the other by its negation.
Observation 3.4.
Note that if all that is known is that the SAT instance is satisfiable, then we have no definitive information about , although the satisfiability of the instance may provide us with a starting point for finding—by hand—a useful embedding.
Using these ideas, we can determine the exact value of the outside obstacle number for some special planar graphs.
Following [11], for we define the gyroelongated bipyramid to be a convex polyhedron formed by adding pyramids to the top and bottom base of the antiprism; see Figure 7. The gyroelongated square bipyramid, when constructed using equilateral triangles, is also known as the Johnson solid . The gyroelongated pentagonal bipyramid, again when constructed of equilateral triangles, is the regular icosahedron. We denote the skeleton of the gyroelongated bipyramid by . We also refer to as , since it is the icosahedron.
The graph can be constructed as two disjoint wheels, connected by a cycle that alternates between vertices of the wheel boundaries taken cyclically. Figures 7, 8, 9, and 11 show gyroelongated bipyramids for various values of , while Figure 10 illustrates the general case. In each of these figures, the wheel boundaries are are labeled with consecutive numbers and are shown in green, the spokes of the wheel in blue, and the connecting cycle in black; the cycle connecting the two wheel boundaries correspond to the sequence of labeled vertices . Nonedges are shown with thin gray or pink lines.
Proposition 3.5.
All of the following hold.

.

.

.
Proof.
The lower bounds were found using a computer. Two of the authors independently created SAT instances as described above, using clauses derived from Lemmas 3.1 (the Point Rule), 3.2 (the Point Rule), and 3.3, as discussed above. See [6] for software to generate the SAT instances. For each graph, standard SAT solvers (we used MiniSat [15], PicoSAT [3], and zChaff [20]) indicate that the SAT instance is not satisfiable. Thus, by Observation 3.4, we have for each graph.
Note that Figure (a)a exhibits fold dihedral symmetry, while the other representations have only one vertical mirror of symmetry.
Figure 10 shows how the embeddings in Figures (b)b and 9 can be generalized to a obstacle embedding of for arbitrarily large . Thus we observe the following.
Theorem 3.6.
Let . Then .
For the gyroelongated bipyramid we have (see Figure 11), while for we have by Proposition 3.5. It seems likely that, for all larger values of , the graph cannot be represented without an interior obstacle; however the corresponding SAT instances are too large for our computational methods to be feasible.
Conjecture 3.7.
If , then .
4. Obstacle Number, Part I
We have shown that there exist planar graphs with outside obstacle number greater than . We now turn our attention to the ordinary obstacle number.
Knowing that certain graphs require an interior obstacle allows us to reason about the obstacle number of a larger graph; in particular this will allow us to argue that certain planar graphs have obstacle number greater than or equal to .
We denote the disjoint union of graphs , by .
Lemma 4.1.
Let and be graphs such that and . Then .
Proof.
Let , be as in the statement of the result, and let . Suppose for a contradiction that we have an obstacle representation of using only one obstacle .
Based on this representation of , construct an obstacle representation of by eliminating those vertices and edges that do not lie in (see Observation 1.5). Let be the largest connected open set that contains but does not meet the drawing of . We may replace by in our obstacle representation of . Since , this obstacle must be an interior obstacle. We similarly construct an obstacle representation of using a single interior obstacle . Note that each of , is the interior of a bounded polygonal region, and that .
Denote the topological closure of a subset of the plane by . Let be a nonzero vector, and consider the family of parallel closed halfplanes determined by . As we sweep this family across , beginning outside of , there will be some that is the first member of with the property that meets both and .
For at least one of , , halfplane meets only the boundary of the obstacle. Say this holds for ; then , and so .
The set is bounded by line segments contained in the edges of . Thus some vertex of lies in . Similarly, some vertex of lies in . Since is not an edge of , the line segment must be blocked by obstacle . However, this line segment lies entirely in , and so it does not meet , a contradiction. ∎
As a consequence, we see that each of the graphs , , and has obstacle number at least and at most . The lower bounds are a consequence of Lemma 4.1, using the previously determined outside obstacle numbers for , (see Proposition 3.5). The upper bounds are established by using the previously determined outside obstacle embeddings for and embedding two copies of using disjoint interior obstacles (e.g., by placing the disjoint copies far apart).
However, we can show that each of these graphs actually has obstacle number . Figure 12 shows a twoobstacle embedding of , found by grouping certain related vertices in the obstacle embedding of . The other two graphs can be drawn similarly, and we have the following.
Corollary 4.2.
All of the following hold.

.

.

.
We note that each of , , and is the skeleton of a polyhedron, and thus is planar. Further, the disjoint union of planar graphs is planar; each of the graphs in Corollary 4.2 is a planar graph. Thus we have answered (negatively!) a question of Alpert, Koch, & Laison [2, p. 231], asking whether every planar graph has obstacle number at most . Note that , , and .
Corollary 4.3.
There exists a planar graph with obstacle number greater than .
5. Obstacle Number, Part II
In this section we develop a variation on the SAT instance constructed in Section 3. This new SAT instance will encode statements about arbitrary obstacles, instead of only outside obstacles. It will enable us to show directly that certain graphs have obstacle number greater than .
In addition to the Point Rule (Lemma 3.1) and the Point Rule (Lemma 3.2), we use a new lemma, Lemma 5.1, which plays role similar to that of Lemma 3.3 in Section 3. For clarity in the next lemma, given distinct points and , we denote the two closed halfplanes determined by line as and , where contains all points such that either is on line or is oriented clockwise. An key path with respect to , denoted , is a path from to that does not cross the line ; that is, a path in from to that is entirely contained in one of the closed halfplanes or ; see Figure 13.
Lemma 5.1.
Suppose we are given an obstacle representation of a graph that uses at most one obstacle. Then, for each nonedge , and for each nonedge ( and may share one vertex), there exists a halfplane such that, for every key path with respect to , , some internal vertex of lies in the interior of ).
Proof.
Choose any nonedge , and then choose a second nonedge distinct from . Perturbing slightly if necessary, we assume that as well (see Observation 1.3). Suppose for a contradiction that there exist two distinct paths and in , both key paths with respect to , that lie in different closed halfplanes determined by . Without loss of generality, we may assume that and .
Now, each of , is contained in one of the two halfplanes , , because they are key paths with respect to . Since and have common endpoints, namely and , we see that and , must be contained in the same halfplane determined by .
Therefore forms a closed path in the plane; the open line segment lies in one component of the complement of this closed path; while the open line segment lies in a different component. Since has only one obstacle, it is impossible for both segments to be blocked, a contradiction.
∎
Given a graph , we can use Lemmas 5.1, 3.1 (the Point rule), and 3.2 (the Point Rule) to create a SAT instance encoding necessary conditions for the existence of an obstacle representation of using at most obstacle. Note that we are saying nothing here about outside obstacles. Thus, if this SAT instance is not satisfiable, then we may conclude that graph requires at least two obstacles, that is, that .
We represent triples of vertices as we did before, in Section 3, and we encode the Point Rule and the Point Rule using the same clauses as in our previous SAT instance. We construct new SAT clauses based on Lemma 5.1, much as, in Section 3, we constructed clauses based on Lemma 3.3. We now look at how this is done.
Choose any nonedge and choose any nonedge distinct from . We illustrate the encoding of Lemma 5.1 in terms of SAT clauses by showing how to encode statements about a particular path. Let be the sequence of vertices in some path (not necessarily a key path), where vertices , are not adjacent (so that is a nonedge). Let , be nonadjacent vertices not lying on .
We introduce a new variable to represent the statement that is an key path with respect to . If all vertices of produce triangles having the same orientation, then is a key path. This is encoded by the following two clauses:
(6)  
(7) 
Next we encode the statement that given this and we can find a special side of so that every key path with respect to lies on the special side of . The special side is encoded in another variable, . We canonically choose that represents the statement that the special halfplane is ; thus represents the statement that the special halfplane is . Note that the special side of depends on the choice of nonedge . The following clauses encode the desired statement:
(8)  
(9) 
Observation 5.2.
Using these ideas, we can determine the exact value of the obstacle number for the icosahedron and some similar graphs.
Proposition 5.3.
All of the following hold.

.

.

.
Proof.
The lower bounds were found using a computer. We create a SAT instance as described above, using clauses representing the Point Rule, the Point Rule, and the statement of Lemma 5.1. See [6] for software to generate the SAT instances. For each graph, a standard SAT solver (as before, we used MiniSat [15], PicoSAT [3], and zChaff [20]) indicates that the SAT instance is not satisfiable. Thus, by Observation 5.2, we have for each graph.
For the upper bound, we apply Proposition 3.5, which states that each graph considered here has , and note that for every graph. ∎
Note that Part (2) of Proposition 5.3 answers a question of Alpert, Koch, & Laison [2, p. 231] asking for the obstacle number of the icosahedron.
Note also that the graph , mentioned in part (1) of Proposition 5.3, has order . This is thus the second known example of a graph of order with obstacle number (the first being , shown to have obstacle number by Pach & Sariöz [19, Thm. 2.1]). But unlike the PachSariöz example, graph is planar. We do not know whether there is any planar graph—or, indeed, any graph at all—of smaller order that has obstacle number .
Alpert, Koch, & Laison [2, p. 231] asked for the obstacle number of the dodecahedron. We answer this question as follows.
Proposition 5.4.
Let be the dodecahedron. Then .
Proof.
Figure 16 shows an obstacle representation of the dodecahedron, using a single outside obstacle. ∎
6. Open questions
There are several interesting open questions related to obstacle numbers of graphs with small numbers of vertices.
Question 6.1.
What is the minimum order of a planar graph with obstacle number ?
By Proposition 2.4, the above minimum must be at least .
We have not been able to show that there exists a planar graph with obstacle number greater than . It is natural to ask whether there exists an upper bound on the obstacle numbers of planar graphs, and, if so, what it is. It seems likely that either there is no such upper bound, or else the maximum obstacle number of a planar graph is . We conjecture that the latter option holds.
Conjecture 6.2.
If is a planar graph, then .
In general, little is known about the least order of a graph with any particular (outside) obstacle number.
Question 6.3.
What is the smallest order of a graph with ? With ? With or ?
Again, by Proposition 2.4, in each case the answer must be at least .
We have found the above questions quite resistant to solution. Perhaps this is unsurprising since Johnson and Sarioz showed in [10] that computing the obstacle number of a plane graph is NPhard. Two examples of graphs of order with obstacle number have been found, namely and ; none of smaller order are known. If we knew of a single graph of order or less for which one of the SAT instances we construct is not satisfiable, then we could reduce our current bound of ; however we have found no such graph.
It seems plausible that an approach to answering the above questions would be a bruteforce application of SAT instances to all graphs with order strictly less than . However, this naive approach has two flaws. First, there are a large number of graphs of order at most (for example, there are connected graphs of order and connected graphs of order [1, Sequence A001349]), and there are significant time and computational issues involved in processing the SAT instances for all these graphs.
Second, while nonsatisfiability of the SAT instance for one of these graphs would imply that the corresponding (outside) obstacle number was strictly greater than , satisfiability of an instance does not imply any bound on the corresponding obstacle number. The solution of one of our SAT instances gives only a specification of clockwise/counterclockwise orientation for each triple of points. This might not correspond to any actual point placement in the plane. Or it may correspond to many point placements. And even if one of these gives the desired obstacle representation, others may not; or none of them may. Furthermore, a single SAT instance can have exponentially many solutions, each of which may need to be checked, in order to find an obstacle representation. In any case, satisfiability provides us only with a starting point in the search for an obstacle representation; we know of no efficient, reliable technique for actually finding such a representation without human intervention and invention.
References
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