Large induced acyclic and outerplanar subgraphs of 2outerplanar graph ^{†}^{†}thanks: This material is based upon work supported by the National Science Foundation under Grant Nos. CCF1252833 and DMS1359173. Melissa ShermanBennett performed this work while participating in the Summer 2015 REU program in Mathematics and Computer Science at Oregon State University.
Abstract
Albertson and Berman conjectured that every planar graph has an induced forest on half of its vertices. The best known lower bound, due to Borodin, is that every planar graph has an induced forest on two fifths of its vertices. In a related result, Chartran and Kronk, proved that the vertices of every planar graph can be partitioned into three sets, each of which induce a forest.
We show tighter results for 2outerplanar graphs. We show that every 2outerplanar graph has an induced forest on at least half the vertices by showing that its vertices can be partitioned into two sets, each of which induces a forest. We also show that every 2outerplanar graph has an induced outerplanar graph on at least twothirds of its vertices.
1 Introduction
For many optimization problems, finding subgraphs with certain properties is a key to developing algorithms with efficient running times or bounded approximation ratios. For example, balanced separator subgraphs support the design of divideandconquer algorithms for minorclosed graph families [18, 17, 12, 15] and large subgraphs of lowtreewidth^{1}^{1}1Formal definitions of graph theoretic terms will be given at the end of this section. support the design of approximation schemes, also for minorclosed graph families [4, 8, 11]. In the area of graph drawing, one often starts by drawing a subgraph that is somehow easier to draw than the entire graph (such as a planar graph or a tree) and then adds in remaining graph features [5]; the larger the subgraph, the bigger the headstart for drawing and the more structure the subgraph has, the easier the subgraph will be to draw.
In this paper we are concerned with finding large induced subgraphs, in particular large induced forests and large induced outerplanar subgraphs, of input planar graphs. We are motivated both by the intrigue of various conjectures in graph theory but also by the impact that graph theoretic results have on the design of efficient and accurate algorithms. In particular, many algorithms that are specifically designed for planar graphs rely on deep graph theoretic properties of planar graphs [23, 24].
1.1 Large induced forests of planar graphs: known results
Albertson and Berman conjectured that every planar graph has an induced forest on at least half of its vertices [2]; illustrates that this would be the best possible lower bound. A proof of the AlbertsonBerman Conjecture would, among other things, provide an alternative proof, avoiding the 4Color Theorem, that every planar graph has an independent set with at least onequarter of the vertices.
The bestknown lower bound toward the AlbertsonBerman Conjecture has stood for 40 years: Borodin showed that planar graphs are acyclically 5colorable (i.e. have a 5coloring, every two classes of which induce a forest), thus showing that every planar graph has an induced forest on at least twofifths of its vertices [6]. This is the best lower bound achievable toward the AlbertsonBerman Conjecture via acyclic colorings as there are planar graphs which do not have an acyclic 4coloring (for example or the octahedron).
The AlbertsonBerman Conjecture has been proven for certain subclasses of planar graphs. Shi and Xu [21] showed that the AlbertsonBerman Conjecture holds when where and are the number of edges and vertices of the graphs, respectively. Hosono showed that outerplanar graphs have induced forests on at least twothirds of the vertices [14] and Salavatipour showed that every trianglefree planar graph on vertices has an induced forest with at least vertices [20], later improved to by Dross, Montassier and Pinou [10] and to by Le [16]. In bipartite planar graphs, the best bound on the size of the largest induced forest is by Wang, Xie and Yu [22].
One direction toward proving the AlbertsonBerman Conjecture is to partition the vertices of graph into sets such that each set induces a forest; the minimum number, , of such sets is the vertex arboricity of . This implies that has an induced forest with at least of its vertices. Chartran and Kronk first proved that all planar graphs have vertex arboricity at most 3 [7]. Raspaud and Wang proved that if is planar and either has no cycles, any two triangles of are at distance at least , or has at most 20 vertices; they also illustrated a 3outerplanar graph on 21 vertices with vertex arboricity 3 [19]. Yang and Yuan [1] proved that if is planar and has diameter at most .
1.2 Outline of our results
In this paper, we show that 2outerplanar graphs have vertex arboricity at most 2, thus showing that they satisfy the AlbertsonBerman Conjecture and closing the gap for planar graphs with vertex arboricity 2 versus 3 left by Raspaud and Wang’s work (Section 2). We also show that every 2outerplanar graph has an induced outerplanar graph on at least twothirds of its vertices and propose a few related conjectures (Section 3).
1.3 Definitions
We use standard graph theoretic notation [9]. In this paper, all graphs are assumed to be finite and simple (without loops or parallel edges). denotes the induced subgraph of graph on vertex subset : the graph having as its vertices and having as edges every edge in that has both endpoints in . Equivalently, may be constructed from by deleting every vertex and incident edges that is not in . We use to denote degree of vertex in graph and to denote the number of vertices of graph .
BlockCut Tree.
A block of a graph is a maximal twoconnected component of . A blockcut tree of a connected graph is a tree where each vertex of corresponds to a block and there is an edge between two vertices of if two blocks and share a common vertex or are incident to a common edge.
Planar graphs.
A graph is planar if it can be drawn (embedded) in the plane without any edge crossings. Although a planar graph may have many different embeddings, throughout this paper, we will assume that we are given a fixed embedding of the graph. A face of a planar graph is connected region of the complement of the image of the drawing. There is one infinite face, which we denote by . We denote the boundary of , which is the boundary of , by . We say that a vertex is enclosed by a cycle if every curve from the image of to an infinite point must cross the image of .
Planar duality.
Every planar graph has a corresponding dual planar graph : the vertices of correspond to the faces of and the faces of correspond to the vertices of ; an edge of connects two vertices of if the corresponding faces of share an edge (in this way the edges of the two graphs are in bijection).
Outerplanarity.
A nonempty planar graph with a given embedding is outerplanar (or outerplanar) if all vertices are in . A planar graph is outerplanar for if deleting the vertices in results in a outerplanar graph. A outerplanar graph has a natural partition of the vertices into layers: is the set of vertices in ; is the set of vertices in the boundary of . We denote by if is outerplanar. For a 2outerplanar graph, we define the between degree of a vertex to be the number of adjacent vertices in .
Facial Block.
Let be the set of facial cycles bounding finite faces of . For each , let be the set of vertices enclosed by in . Then we call the graph a facial block of .
2 2outerplanar graphs have vertexarboricity 2
In this section, we prove:
Theorem 1.
If is a 2outerplanar graph, then the vertex arboricity of is at most : .
We call a set of vertexdisjoint induced forests of induced pforests if their vertices partition the vertex set of . We consider a counterexample graph of minimal order. By studying the structure of this minimal counterexample, we will derive a contradiction. Let be an edge that is not in . We observe:
Observation 2.
If , then .
Observation 2 allows us to assume w.l.o.g. that is connected (by adding edges between components while maintaining 2outerplanarity) and that is a disk triangulation, i.e., that every face except the outer face of is a triangle (by adding edges inside nontriangular faces while maintaining 2outerplanarity). Let be the bipartition of the vertices of into layers.
Observation 3.
is twoconnected.
Proof.
Suppose otherwise. Let be a cut vertex of . Then is also a cut vertex of since is the outermost layer. Let , be two induced subgraphs of that share the cut vertex and . Since is minimal, we can partition each into two induced forests and , . W.l.o.g, we assume that . Then, and are two induced pforests of , contradicting that is a counterexample. ∎
Claim 4.
Every vertex in has degree at least 4.
Proof.
Suppose has a vertex of degree at most 3. Since is a minimal order counterexample and is a 2outerplanar graph, . Let and be two induced pforests of . Since has at most 3 neighbors in , one of or , w.l.o.g. say , contains at most one of these neighbors. Therefore is a forest of and are two induced pforests of , contradicting that is a counterexample. ∎
By Observation 3, is a simple cycle. Thus, the graph, say , of obtained from the dual graph of by removing the dual vertex corresponding to the infinite face of is a tree. Let be a facial block of that has the boundary cycle corresponding to a leaf of . Then, either has exactly one edge not in or . In the former case, let be the shared edge; in the later case, let be any edge of . Denote . We have:
Claim 5.
.
Proof.
If , then is a triangle since is a disktriangulation and vertices have degree at least . Then, the vertex of that is not an endpoint of has degree 2 in , contradicting Claim 4. If , by Claim 4, has at least four neighbors in and thus, at least one neighbor of in is not an endpoint of . Then the degree of in is 3, contradicting Claim 4. ∎
Claim 6.
Let be a vertex in that has between degree at least . Then, either is a cut vertex of or is adjacent to both endpoints of .
Proof.
Let be neighbors of in in clockwise order around . Let be the clockwise segment of from to , . We define , which is a cycle of . Assume is not a cut vertex, at most one cycle of encloses a vertex of , say . Thus, is only adjacent to and two other neighbors, say , of . Since and enclose no vertex of , and are faces of . If neither nor , then , contradicting Claim 4. ∎
Observation 7.
If there exists such that , then must be adjacent to both endpoints of .
Let be the endpoints of . Since is a triangulation, there is a vertex such that is a face of . We call the separating vertex of .
Claim 8.
If is a vertex in that is adjacent to both endpoints of , then, is a cut vertex of .
Proof.
We will prove that has at least one neighbor in inside the triangle and at least one neighbor in outside the triangle ; thus is a cut vertex of .
By planarity, the triangle encloses . Let which is a cycle of . Since is a disk triangulation and the edge is embedded outside , there must be an edge or a path inside connecting and . Thus, has at least one neighbor in inside the triangle .
Suppose that the cycle does not enclose any vertex of . Since is a facial block that only has as a possible edge not in , every vertex in must have as a neighbor and has degree 3, contradicting Claim 4. Thus, must enclose at least one vertex of . That implies has at least one neighbor in outside the triangle as desired. ∎
Observation 9.
Only the separating vertex of can have .
If the blockcut tree of has at least two vertices, let be a leaf block of that does not contain the separating vertex of . In this case, by Observation 9, . Otherwise, let . We refer to the cut vertex of in the former case and the separating vertex of in the latter case as the separating vertex of . By Claim 6, we have:
Observation 10.
Nonseparating vertices of have between degree at most .
We call a triangle of a critical triangle with top if and is nonseparating. By Observation 10 and Claim 4, has exactly two neighbors in , that we denote by (see Figure 1). Since is a disk triangulation, two edges and are edges of .
Claim 11.
Vertices and have degree at least 5.
Proof.
Neither nor has degree less than 4 by Claim 4. For contradiction, w.l.o.g, we assume that . Recall are three neighbors of . Let be the only other neighbor of . Since and is a simple cycle (Observation 3), must be in (see Figure 1). Since is a disk triangulation, . Let be the graph obtained from by contracting and and removing parallel edges. Then is a minor of (and so is 2outerplanar) with fewer vertices. Let be two induced pforests of that exist by the minimality of . Without loss of generality, we assume that . We have two cases:

If , then . If , adding to does not destroy the acyclicity of in . Thus, are two induced forests of . If , the cycle separates from so and are in different trees in . Thus, are two induced pforests of .

Otherwise, . We have three subcases:

If are both in , then are two induced pforests of .

If , then, are two induced pforests of .

If are in different induced pforests of , then, are two induced pforests of .

In each case, the resulting pforests contradict that is a minimal order counter example. ∎
Claim 12.
.
Proof.
If , as noted in the definition of , . If , then by Claim 5, and by Observation 9, . Suppose that . Then, is a triangle. Let be two neighbors of the separating vertex in . Then, is a critical triangle with top (or ). By Claim 4 and Observation 10, and both have between degree 2. Thus, and have a common neighbor on which therefore has degree , contradicting Claim 11. ∎
Suppose that and of a critical triangle with top of have a common neighbor in . We have:
Claim 13.
If (resp. ) is in , then (resp. ) must be the separating vertex.
Proof.
For a contradiction (and w.l.o.g), we assume that and is nonseparating. See Figure 2. Let be the graph obtained from by contracting and and removing parallel edges. Then, is a minor of (and so is 2outerplanar) with fewer vertices. Let be two induced pforests of , which are guaranteed to exist by the minimality of . Without loss of generality, we assume that . We consider two cases:

If , then are in . If edge , then, are two induced pforests of . If , cycle separates from so and are in different trees of . Thus, are two induced pforests of .

Otherwise, . We have three subcases:

If are both in , then are two induced pforests of .

If are both in , then are two induced pforests of .

If are in different forests of , then, are two induced pforests of .

In each case, the resulting pforests contradicts that is a minimal order counter example. ∎
If the edge is shared with another critical triangle with top , then we call a pair of critical triangles. See Figure 3. Note that we are assuming that is a common neighbor of and in .
Claim 14.
If there exists a pair of critical triangles and in , then must be the separating vertex of .
Proof.
Note that neither nor can be the separating vertex by definition of critical triangles. Suppose for contradiction that is nonseparating. Let be two neighbors of as defined above and and be the neighbors of in . We first argue that . Suppose otherwise. Since is a disk triangulation, are edges of . Let be the subpath of between and that does not contain and . Note that could simply be edge . Since is a facial block that shares at most one edge with other facial blocks and is nonseparating, has exactly one neighbor on . That implies would have degree , contradicting Claim 11.
We also note that (for otherwise, would not be in ) and . See Figure 3. Let be the graph obtained from by contracting and and removing parallel edges. Thus, is a minor of with fewer vertices. By minimality, has two induced pforests . Without loss of generality, we assume that . We will reconstruct two induced pforests of by considering two cases:

If , then . If edge , then, by planarity, and are in different trees of . Thus, has no cycle which implies are two induced pforests in . Otherwise, has no cycle. Thus, are two induced pforests in .

Otherwise, . We have four subcases:

If are both in , then are two induced pforests of .

If are both in , then are two induced pforests of .

If , then are two induced pforests of .

If , then are two induced pforests of .

Thus, in all cases, the resulting induced pforests contradict that is a counterexample. ∎
We are now ready to complete the proof of Theorem 1 by considering a triangle of of , say , containing the separating vertex of and has the most edges in common with . Since is separating, contains at least one edge in . We note that where is the dual vertex of the infinite face of , is a tree that we denote by . Recall that is a block of . We root at the vertex corresponding to the triangle . Consider the deepest leaf and its parent . Let be the triangle corresponding to such that the dual edge of is . Then . Since , and thus, it is a critical triangle with top . Let be the triangle that corresponds to . Note here it may be that . We have three cases:

If , then . Thus, two edges are both in but only one of the two vertices can be the separating vertex of . This contradicts Claim 13.

If , then exactly one of two edges ; w.l.o.g, we assume that . Then, by Claim 13, must be the separating vertex of . Thus, only two triangles and contain the separating vertex. Since is the triangle containing the separating vertex with most edges in , , contradicting our choice of triangle .

Otherwise, we have . Then, none of is in , so . Let and be the other two neighbors of in with as the parent of . Then, and have the same depth. By our choice of , must also be a leaf. Thus, the triangle, say , corresponding to is critical. Thus is a pair of critical triangles. Since is the parent of , cannot be the separating vertex of , contradicting Claim 14.
This completes the proof of Theorem 1. ∎
3 2outerplanar graphs have large induced outerplanar graphs
In this section, we prove:
Theorem 15.
Let be a 2outerplanar graph on n vertices. has an induced outerplanar subgraph on at least vertices whose outerplanar embedding is induced from .
Let be the partition of into layers. Note that is a cactus graph (every edge is in at most 1 cycle). As in Section 2, we assume w.l.o.g that is connected and a disk triangulation. This gives us:
Observation 16.
If is an edge in , then there exists such that is a face.
Observation 17.
The between degree of every vertex in is at least 1.
Lemma 18.
If has between degree 1, then it is incident to exactly two edges in .
Proof.
Let be ’s neighbor in . Since is a disk triangulation, there exist two triangular faces, say and , containing the edge . As the between degree of is 1, and are in , and the edges and are in . Therefore, is incident to at least two edges in .
Suppose for the sake of contradiction that is incident to more than two edges in . Let be a neighbor of such that . Then by Observation 16, there exists such that is a face. Since and is simple, . This implies has between degree at least ; contradicting that ’s between degree is . ∎
Lemma 19.
If a facial block contains a vertex in , then endpoints of any edge are adjacent to a common vertex in .
Proof.
Since is a triangulation, there is a vertex such that is a triangular face; thus contains no vertex of . Suppose that , then is an induced cycle of . Thus, ; contradicting that contains a vertex in . ∎
Lemma 20.
There exists a matching with the following property:
(1) 
Proof.
Let be the set of vertices of between degree 1 in . We proceed by strong induction on . If , any matching has property (1).
Let and be vertices such that ; exists by Lemma 19. By Lemma 18, has exactly one other neighbor such that . Contract and to and delete parallel edges and loops; let the resulting graph be . Since now has between degree at least , the number of vertices of between degree in is strictly less than . Therefore, by the inductive hypothesis, there exists a matching with property (1). Now, consider as a matching in .
If is not covered by in , then are not covered by in . Then, is a matching and has property (1), since has between degree at least 2 as argued above.
If for some , then either or . In the first case, let ; in the second, let . In both cases, is a matching of with property (1). ∎
We are now ready to prove Theorem 15. To find the vertices inducing a large outerplanar graph in , we delete vertices in until all vertices in are “exposed” to the external face. To ensure that the resulting outerplanar graph is sufficiently large, we delete vertices in that expose 2 vertices in or otherwise ensure 2 vertices will be included in the outerplanar graph.
Let be a matching as guaranteed by Lemma 20. We create a list of triples such that each vertex in occurs in exactly one triple. For each not covered by , has between degree at least , and we add to , where are neighbors of in . For each edge , by Observation 16, there exists such that is a face, and we add to .
We then delete vertices from as follows: 1. While there exists such that delete from and delete all triples containing from ; 2. While there exists such that is in two or more distinct triples of delete from and delete all triples containing from ; 3. While delete from and delete from .
Note that if is deleted from , all vertices in a triple with are exposed. Therefore, the undeleted vertices induce an outerplanar subgraph of .
In the first two steps, at least two vertices were exposed for every deleted vertex. In the final step, all triples are disjoint, so each deletion of an vertex exposes one vertex and ensures that one vertex will not be deleted; again, 2 vertices are included in the induced outerplanar subgraph for every deleted vertex. This means that the subgraph contains at least two thirds of the vertices of . This complete the proof of Theorem 15. ∎
This result is tight, as the disjoint union of multiple octahedrons (see Figure 4) is 2outerplanar, and its largest induced outerplanar subgraph is on of its vertices. The result is also tight for arbitrarily large connected 2outerplanar graphs, as the same property holds for graphs constructed by connecting disjoint octahedrons as shown in Figure 5.
Theorem 15 has an immediate corollary for outerplanar graphs.
Corollary 21.
Let be a outerplanar graph on vertices. Then has an induced outerplanar subgraph on at least vertices.
Proof.
We apply Theorem 15 to pairs of successive layers in , finding large induced outerplanar subgraphs for (if is even; if is odd, we end at ). Let . is outerplanar, as , and , as .∎∎
4 Future directions
We define an induced outerplane graph of a planar graph is an induced subgraph of whose embedding inherited from is an outerplanar embedding. We point out that our last result implies an improvement to a graph drawing result of Angelini, Evans, Frati, and Gudmundsson [3] for the class of 2outerplanar graphs. A simultaneous embedding with fixed edges and without mapping (SEFENoMap) of two planar graphs and of the same size is a pair of planar drawings of and that maps any vertex of into any vertex of such that: (i) vertices of both graphs are mapped to the same point set in a plane and (ii) every edge that belongs to both and must be represented by the same curve in the drawing of two graphs. The OptSEFENoMap problem asks for the maximum such that: given any two planar graphs and of size and , respectively, there exists an induced subgraph of such that and have a SEFENoMap where the drawing of is inherited from a planar drawing of . The result of Gritzmann et al. [13], implies that can be as large as the size of any induced outerplane graph of . Angelini, Evans, Frati, and Gudmundsson (Theorem 1 [3]) showed that any planar graph of size has an induced outerplane graph of size at least which implies by the result of Gritzmann et al. [13]. Our Theorem 15 implies the following corollary, which is an improvement of the result of Angelini, Evans, Frati, and Gudmundsson for the class of 2outerplanar graphs.
Corollary 22.
Every vertex 2outerplanar graph and every vertex planar graph have a SEFENoMap.
Based on Theorem 15, we conjecture that:
Conjecture 23.
Any 3outerplanar graph on vertices contains an induced outerplane graph of size at least .
If this conjecture is true, it would, by Hosono’s result [14], imply that the largest induced forest of outerplanar graphs on vertices has size at least , that is an improvement over Borodin’s result. It also improves the result of Angelini, Evans, Frati, and Gudmundsson for 3outerplanar graphs. We note that in the proof of Theorem 15, we only need to delete vertices in , and leave untouched, to get a large induced outerplane graph of 2outerplanar graphs. For outerplanar graph, one may need to delete vertices in as shown by Figure 6.
We also believe that following conjecture, which is also mentioned in in [3], is true:
Conjecture 24.
A planar graph on vertices contains an induced outerplane graph of size at least .
If this conjecture is true, it would imply an improvement of Borodin’s result and Angelini, Evans, Frati, and Gudmundsson’ result for general planar graphs.
Acknowledgments:
The authors thank conversations with Bigong Zheng and Kai Lei. We thank anonymous reviewers for comments that help improving the presentation of this paper.
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