1. Introduction
Tait conjectured in 1884 [21] that all cubic polyhedral graphs, i.e., all triconnected cubic planar graphs, have a Hamiltonian cycle; this was disproved by Tutte in 1946 [23], and the study of Hamiltonian cubic planar graphs has been a very active area of research ever since, see for instance [1, 10, 16, 18]. Barnette formulated two conjectures that have been at the centre of much of the effort: (1) that bipartite triconnected cubic planar graphs are Hamiltonian (the case of Tait’s conjecture where all face sizes are even) [4], and (2) that triconnected cubic planar graphs with all face sizes or are Hamiltonian, cf. [3, 19]. Goodey [11, 12] proved that the conjectures hold on the intersection of the two classes, i.e., that triconnected cubic planar graphs with all face sizes or are Hamiltonian. When all faces have sizes or , this was a longstanding open problem, especially since these graphs (triconnected cubic planar graphs with all face sizes or ) are the popular fullerene graphs [8]. The second conjecture has now been affirmatively resolved in full [17]. For the first conjecture, two of the present authors have shown in [9] that if the conjecture is false, then the Hamiltonicity problem for triconnected cubic planar graphs is NPcomplete. In view of these results and conjectures, in this paper we call bipartite triconnected cubic planar graphs typeone Barnette graphs; we call cubic plane graphs with all face sizes or typetwo Barnette graphs; and finally we call cubic plane graphs with all face sizes or Goodey graphs. Note that it would be more logical, and historically accurate, to assume triconnectivity also for typetwo Barnette graphs and for Goodey graphs. However, we prove our positive results without needing triconnectivity, and hence we do not assume it.
Cubic planar graphs have been also of interest from the point of view of colorings [6, 14]. In particular, they are interesting for distancetwo colourings. Let be a graph with degrees at most . A distancetwo coloring of is an assignment of colors from to the vertices of such that if a vertex has degree then the colors of and of all the neighbors of are all distinct. (Thus a distancetwo coloring of is a classical coloring of .) Clearly a graph with maximum degree needs at least colors in any distancetwo coloring, since a vertex of degree and its neighbours must all receive distinct colors. It was conjectured by Wegner [24] that a planar graph with maximum degree has a distancetwo colouring where for , for , and for all larger . The case has been settled in the positive by Hartke, Jahanbekam and Thomas [13], cf. also [22].
For cubic planar graphs in general it was conjectured in [13] that if a cubic planar graph is triconnected, or has no faces of size five, then it has a distancetwo sixcoloring. We propose a weaker version of the second case of the conjecture, namely, we conjecture that a bipartite cubic planar graph can be distancetwo sixcolored. We prove this in one special case (Theorem 2.5), which of course also confirms the conjecture of Hartke, Jahanbekam and Thomas for that case. Heggerness and Telle [15] have shown that the problem of distancetwo fourcoloring cubic planar graphs is NPcomplete. On the other hand, Borodin and Ivanova [5] have shown that subcubic planar graphs of girth at least can be distancetwo fourcolored. In fact, there has been much attention focused on the relation of distancetwo colorings and the girth, especially in the planar context [5, 14].
Our results focus on distancetwo colorings of cubic planar graphs, with particular attention on Barnette graphs, of both types. We prove that a cubic plane graph with all face sizes divisible by four can always be distancetwo fourcolored, and a give a simple condition for when a biconnected cubic plane graph with all face sizes divisible by three can be distancetwo fourcolored using only three colors per face. It turns out that the distancetwo fourcoloring problem for typeone Barnette graphs is NPcomplete, while for typetwo Barnette graphs it is not only polynomial, but the positive instances can be explicitly described. They include one infinite family of Goodey graphs (cubic plane graphs with all faces of size or ), and all typetwo Barnette graphs which have all faces of size or . Interestingly, there is an analogous result for quartic (fourregular) graphs: all quartic planar graphs with faces of only sizes or that have a distancetwo five coloring can be explicitly described; there are only two such graphs.
Note that we use the term “plane” graph when the actual embedding is used, e.g., by discussing the faces; when the embedding is unique, as in triconnected graphs, we stick with writing “planar”.
2. Relations to edgecolorings and facecolorings
Distancetwo colorings have a natural connection to edgecolorings.
Theorem 2.1.
Let be a graph with degrees at most that admits a distancetwo coloring, with odd. Then admit an edgecoloring with colors.
Proof.
The even complete graph can be edgecolored with colors by the Walecki construction [2]. We fix one such coloring , and then consider a distancetwo coloring of . If an edge of has colors at its endpoints, we color in with the color . It is easy to see that this yields an edgecoloring of with colors. ∎
We call the resulting edgecoloring of the derived edgecoloring of the original distancetwo coloring.
In this paper, we mostly focus on the case (the subcubic case). Thus we use the edgecoloring of by colors red, blue, green. This corresponds to the unique partition of into perfect matchings. Note that for every vertex of and every edgecolor , there is a unique other vertex of adjacent to in edgecolor . Thus if we have the derived edgecoloring we can efficiently recover the original distancetwo coloring. In the subcubic case, in turns out to be sufficient to have just one color class of the edgecoloring of .
Theorem 2.2.
Let be a subcubic graph, and let be a set of red edges in . The question of whether there exists a distancetwo fourcoloring of for which the derived edgecoloring has as one of the three color classes can be solved by a polynomial time algorithm. If the answer is positive, the algorithm will identify such a distancetwo coloring.
Proof.
We may assume in red joins colors blue joins colors and green joins colors . Note that we may also assume that is a matching that covers at least all vertices of degree three, otherwise we answer in the negative. We may further assume that some vertex gets an even color ( or ). The parity of the color of a vertex determines the parity of the color of its neighbors, namely the parity is the same if they are adjacent by an edge in , and they are of different parity otherwise. We may thus extend from the assignment of parities to all the vertices, unless an inconsistency is reached, in which case no coloring exists. Otherwise, at this point all vertices have only two possible colors, namely for odd and for even.
Define an auxiliary graph with vertices , and edges in if is a red edge in or if there is a path without red edges in . Note that these edges join vertices of the same parity, and must have different colors. If has an odd cycle, then no solution exists. Otherwise is bipartite, and we may choose in different sides of a bipartition of for odd vertices, and in different sides for even vertices.
Each vertex will have at most one neighbor of the same parity in , namely the one joined to it by the red edge, and is an edge of . This guarantees different colors for . The at most two other neighbors of have different parity from , and the path in ensures the edge is in . This guarantees different colors for . Thus the colors for are all different at each vertex , and we have a distancetwo coloring of . ∎
There is also a relation to facecolorings. It is a folklore fact that the faces of any bipartite cubic plane graph can be threecolored [20]. This threefacecoloring induces a threeedgecoloring of by coloring each edge by the color not used on its two incident faces. (It is easy to see that this is in fact an edgecoloring, i.e., that incident edges have distict colors.) We call an edgecoloring that arises this way from some facecoloring of a special threeedgecoloring of . We first ask when is a special threeedgecoloring of the derived edgecoloring of a distancetwo fourcoloring of .
Theorem 2.3.
A special threeedgecoloring of is the derived edgecoloring of some distancetwo fourcoloring of if and only if the size of each face is a multiple of .
Proof.
The edges around a face alternate in colors, and the vertices of can be colored consistently with this alternation if and only if the size of is a multiple of 4. This proves the “only if” part. For the “if” part, suppose all faces have size multiple of . If there is an inconsistency, it will appear along a cycle in . If there is only one face inside , there is no inconsistency. Otherwise we can join some two vertices of by a path inside , and the two sides of inside give two regions that are inside two cycles . The consistency of then follows from the consistency of each of by induction on the number of faces inside the cycle. ∎
Corollary 2.4.
Let be a cubic plane graph in which the size of each face is a multiple of four. Then can be distancetwo fourcolored.
We now prove a special case of the conjecture stated in the introduction, that all bipartite cubic plane graphs can be distancetwo sixcolored. Recall that the faces of any bipartite cubic plane graph can be threecolored.
Theorem 2.5.
Suppose the faces of a bipartite cubic plane graph are threecolored red, blue and green, so that the red faces are of arbitrary even size, while the size of each blue and green face is a multiple of . Then can be distancetwo sixcolored.
Proof.
Let be the multigraph obtained from by shrinking each of the red faces. Clearly is planar, and since the sizes of blue and green faces in are half of what they were in , they will be even, so is also bipartite. Let us label the two sides of the bipartition as and respectively. Now consider the special threeedge coloring of associated with the face coloring of . Each red edge in this special edgecoloring joins a vertex of with a vertex of ; we orient all red edges from to . Now traversing each red edge in in the indicated orientation either has a blue face on the left and green face on the right, or a green face on the left and blue face on the right. In the former case we call the edge class one in the latter case we call it class two. Each vertex of is incident with exactly one red edge; the vertex inherits the class of its red edge. The vertices around each red face in are alternatingly in class 1 and class 2. We assign colors to vertices of class one and colors to vertices of class two. It remains to decide how to choose from the three colors available for each vertex. A vertex adjacent to red edges in class has only three vertices within distance two in the same class, namely the vertex across the red edge, and the two vertices at distance two along the red face in either direction. Therefore distancetwo coloring for class corresponds to threecoloring a cubic graph. Since neither class can yield a , such a threecoloring exists by Brooks’ theorem [7]. This yields a distancetwo sixcoloring of . ∎
3. Distancetwo fourcoloring of typeone Barnette graphs is NPcomplete
We now state our main intractability result.
Theorem 3.1.
The distancetwo fourcoloring problem for triconnected bipartite cubic planar graphs is NPcomplete.
We will begin by deriving a weaker version of our claim.
Theorem 3.2.
The distancetwo fourcoloring problem for bipartite planar subcubic graphs is NPcomplete.
Proof.
Consider the graph in Figure 1.
We will reduce the problem of coloring planar graphs to the distancetwo fourcoloring problem for bipartite planar subcubic graphs. In the coloring problem we are given a planar graph and and the question is whether we can color the vertices of with colors that are vertices of so that adjacent vertices of obtain adjacent colors. This can be done if and only if is threecolorable, since the graph both contains a triangle and is threecolorable itself. (Thus any threecoloring of is an coloring of , and any coloring of composed with a threecoloring of is a threecoloring of .) It is known that the threecoloring problem for planar graphs is NPcomplete, hence so is the coloring problem.
Thus suppose is an instance of the coloring problem. We form a new graph obtained from by replacing each vertex of by a ring gadget depicted in Figure 2. If has degree , the ring gadget has squares. A link in the ring is a square followed by the edge . A link is even if is even, and odd otherwise. Every even link in the ring will be used for a connection to the rest of the graph , thus vertex has available links. For each edge of we add a new vertex that is adjacent to a vertex in one available link of the ring for and a vertex in one available link of the ring for . (We use primed letters for the corresponding vertices in the ring of to distinguish them from those in the ring of .) The actual choice of (the even) subscripts does not matter, as long as each available link is only used once. The resulting graph is clearly subcubic and planar. It is also bipartite, since we can bipartition all its vertices into one independent set consisting of all the vertices with odd in all the rings, and another independent set consisting of the vertices with even in all the rings. Moreover, we place all vertices into the set . Note that in any distancetwo fourcoloring of the ring, each link must have four different colors for vertices , and the same color for and . Thus all have the same color and all have the same color. The pair of colors in is also the same for all ; we will call it the characteristic pair of the ring for . For any pair of colors from , there is a distancetwo coloring of the ring that has the characteristic pair .
We prove that is colorable if and only if is distancetwo fourcolorable. In an coloring of , the vertices of are actually assigned unordered pairs from , since the vertices of are labeled by pairs. (Note that two vertices of are adjacent if and only if the pairs they are labeled with intersect in exactly one element.) Thus suppose that we have an coloring of . If (i.e., the vertex of is assigned the vertex of labeled by the pair ), then we colour the ring of so that its characteristic pair is . This still leaves a choice of which of the colors is in which , in each of the links . Since is an coloring, adjacent vertices are assigned pairs that intersect if exactly one element. This makes it possible to color each so that all colors at distance at most two are distinct. For instance if vertices and are adjacent in and colored by by , and if is adjacent to the vertices in the ring for and in the ring for , then both in the ring for and in the ring for are colored , as is , while in the ring for and in the ring for are colored and respectively. It is easy to see that this is a distancetwo fourcoloring of .
Conversely, in any distancetwo fourcoloring of , the color of a vertex determines the same color in the ’s of its adjacent links of the rings for and , whence the characteristic pairs of these two rings intersect in exactly one element. Thus we may define a mapping of to by assigning to each vertex the characteristic pair of the ring for . Then is an coloring of , since adjacent vertices of are assigned pairs that are adjacent in . ∎
To prove the full Theorem 3.1, the construction of the graph is modified as suggested in Figure 3. Recall that in the construction of , for each edge of a separate vertex was made adjacent to in the ring of and in the ring of . Recall that both and are even, and the vertices (with both subscripts odd) remained available for connection. We now make a new edgegadget around the vertex , making it directly adjacent to , and connected to by a path, as depicted in Figure 3. In both rings, the two “” type vertices in the two consecutive links are joined together by an additional edge; specifically, we add the edges and . (Note that this forces the corresponding “” type vertices and to be colored differently in any distancetwo fourcoloring, and similarly for and ). Moreover, further vertices and edges are added, as depicted in Figure 3. The shaded tensided region is identified with the tensided exterior face of the graph depicted in Figure 5, which has a unique distancetwo fourcoloring, shown there. (The heavy edges correspond to the tensided shaded figure.) (Note that the graph in Figure 5 was obtained from the graph in Figure 7 by the deletion of two edges.) Note that the construction is not symmetric, as it depends on which ring is viewed as the “bottom” ring for the vertex . (The depicted figure has the ring of on the bottom, but the conclusions are the same if it were the ring of .) We can choose either way, independently for each edge of . It can be seen that the resulting graph, which we denote by , is bipartite, planar, and cubic. We may assume that is biconnected (the threecoloring problem for biconnected planar graphs is still NPcomplete), and therefore is also triconnected (as no two faces share more than one edge). Using the unique distancetwo fourcolouring of the graph in Figure 5, it also follows that in any distancetwo fourcoloring of the vertices and have different colors, while both vertices and have the same color (the color of ), in any distancetwo fourcoloring of . To facilitate checking this, we show in Figure 4 a partial distancetwo fourcoloring, by circles, squares, up triangles, and down triangles; this coloring is forced by arbitrarily coloring and its three neighbours by four distinct colors. Since the colors of the pair and the pair have exactly one color in common, the previous NPcompleteness proof applies, i.e., is colorable if and only if is distancetwo fourcolorable.
We remark that (with some additional effort) we can prove that the problem is still NPcomplete for the class of triconnected bipartite cubic planar graphs with no faces of sizes larger than .
4. Distancetwo fourcoloring of Goodey graphs
Recall that Goodey graphs are typetwo Barnette graph with all faces of size and [11, 12]. In other words, a Goodey graph is a cubic plane graph with all faces having size or . By Euler’s formula, a Goodey graph has exactly six square faces, while the number of hexagonal faces is arbitrary.
A cyclic prism is the graph consisting of two disjoint even cycles and with the additional edges , . It is easy to see that cyclic prisms have either no distancetwo fourcoloring (if is odd), or a unique distancetwo fourcoloring (if is even). Only the cyclic prisms with are Goodey graphs, and thus from Goodey cyclic graphs only the cube (the case of ) has a distancetwo coloring, which is moreover unique.
In fact, all Goodey graphs that admit distancetwo fourcoloring can be constructed from the cube as follows. The Goodey graph is the cube, i.e., the cyclic prism with . The Goodey graph is depicted in Figure 7. It is obtained from the cube by separating the six square faces and joining them together by a pattern of hexagons, with three hexagons meeting at a vertex tying together the three faces that used to meet in one vertex. The higher numbered Goodey graphs are obtained by making the connecting pattern of hexagons higher and higher. The next Goodey graph has two hexagons between any two of the six squares, with a central hexagon in the centre of any three of the squares, the following Goodey graph has three hexagons between any two of the squares and three hexagons in the middle of any three of the squares, and so on. Thus in general we replace every vertex of the cube by a triangular pattern of hexagons whose borders are replacing the edges of the cube. We illustrate the vertex replacement graphs in Figure 6, without giving a formal description. The entire Goodey graph is depicted in Figure 7.
We have the following results.
Theorem 4.1.
The Goodey graphs have a unique distancetwo fourcoloring, up to permutation of colors.
Proof.
We described as eight triangular regions , each consisting of hexagons, one region for each vertex of the cube. Each has three squares at the corners, which we describe as two squares joined by a chain of hexagons horizontally at the bottom, and a third square on top. (See Figure 6.)
We partition the vertices into horizontal paths , , with each having endpoints of degree 2 and internal vertices of degree 3. The path has length , and the remaining paths , have length . In particular the last has length 4, and is the only that is actually a cycle, pictured as the square at the top. See Figure 8.
We denote . The edges between and are , for , odd, and . We can choose the permutation of colors for the square to be , forcing for neighbors the colors , and completing the adjacent square, or hexagon with the assignment to of colors . This forced process extends similarly through the chain of hexagons until the last square.
We have derived the beginning of as and the beginning of as . After the forced extension, will be an initial segment of and will be an initial segment of .
For , the edges between and are , for , odd, and .
A similar process derives the beginning of for odd as and the beginning of for even as . After the forced extension, for odd will be an initial segment of and for even will be an initial segment of .
This gives a unique coloring for the triangular region after coloring one square , which is uniquely extended to the four triangular regions surrounding , and then uniquely extended to the four triangular regions surrounding opposite to . ∎
Theorem 4.2.
The Goodey graphs are the only bipartite cubic planar graphs having a distancetwo fourcoloring.
Proof.
Consider a Goodey graph with a fixed distancetwo fourcoloring. Recall that Goodey graphs have exactly six squares. Each of the squares is joined by four chains of hexagons to four squares. We consider the dual sixvertex graph whose vertices are squares in , with an edge in if and only if there is a chain of hexagons joining squares and . It can be readily verified that such a chain cannot cross itself or another chain in . Indeed, the colors in the fixed distancetwo fourcoloring are uniquely forced along such chains and they don’t match if the chains should cross. It follows that the graph is planar. A similar argument shows that a chain cannot return to the same square, and two chains from the square cannot end at the same square . Thus has no faces of size one or two, and by Euler’s formula it has edges and faces; therefore all faces of must be triangles, and is the octahedron.
Let be a triangular face in , let be a side of with the smallest number of hexagons in . Then it can again be checked using the coloring that the other two sides of will also have hexagons in . Then corresponds to a triangular region as in Theorem 4.1, and the octahedron yields for . ∎
We can therefore conclude the following.
Corollary 4.3.
The distancetwo fourcoloring problem for Goodey graphs is solvable in polynomial time.
Recognizing whether an input Goodey graph is some can be achieved in polynomial time; in the same time bound can actually be distancetwo fourcolored.
5. Distancetwo fourcoloring of typetwo Barnette graphs is polynomial
We now return to general typetwo Barnette graphs, i.e., cubic plane graphs with face sizes , or . As a first step, we analyze when a general cubic plane graph admits a distancetwo fourcoloring which has three colors on the vertices of every face of .
Theorem 5.1.
A cubic plane graph has a distancetwo fourcoloring with three colors per face if and only if

all faces in have size which is a multiple of ,

is biconnected, and

if two faces share more than one edge, the relative positions of the shared edges must be congruent modulo in the two faces.
The last condition means the following: if faces meet in edges and there are edges between and in (some traversal of) , and edges between and in (some traversal of) , then .
Proof.
Suppose has a distancetwo fourcoloring with three colors in each face. The unique way to distancetwo color a cycle with colors is by repeating them in some order along one of the two traversals of the cycle. Therefore the length is a multiple of so (1) holds. Moreover, there can be no bridge in as that would imply a face that selfintersects and is traversed in opposite directions along any traversal of that face, disagreeing with the order in one of them; thus (2) also holds. Finally, (3) holds because the common edges must have the same colors in both faces.
Suppose the conditions hold, and consider the dual of . (Note that each face of is a triangle.) We find a distancetwo coloring of as follows. Let be a face in ; according to conditions (12), its vertices can be distancetwo colored with three colors. That takes care of the vertex in . Using condition (3), we can extend the coloring of to any face adjacent to in . Note that we can use the fourth colour, , on the two vertices adjacent in to the two vertices of a common edge. In this way, we can propagate the distancetwo coloring of along the adjacencies in . If this produces a distancetwo coloring of all vertices of , we are done. Thus it remains to show there is no inconsistency in the propagation. If there is an inconsistency, it will appear along a cycle in . If there is only one face inside of , then is a triangle corresponding to a vertex of , and there is no inconsistency. Otherwise we can join some two vertices of by a path inside , and the two sides of inside give two regions that are inside two cycles . The consistency of then follows from the consistency of each of by induction on the number of faces inside the cycle. ∎
It turns out that conditions (1  3) are automatically satisfied for cubic plane graphs with faces of sizes or .
Corollary 5.2.
Typetwo Barnette graphs with faces of sizes or are distancetwo fourcolorable.
Proof.
Such a graph must be biconnected, i.e., cannot have a bridge, since no triangle or hexagon can selfintersect. Moreover, only two hexagons can have two common edges, and it is easy to check that they must indeed be in relative positions congruent modulo on the two faces. (Since all vertices must have degree three.) Thus the result follows from Theorem 5.1. ∎
Theorem 5.3.
Let be typetwo Barnette graph. Then is distancetwo fourcolorable if and only if it is one of the graphs , or all faces of have sizes or .
Proof.
If there are faces of size both and (and possibly size ), then there must be (by Euler’s formula) two triangles and three squares, and as in the proof of Theorem 4.2, the squares must be joined by chains of hexagons, which is not possible with just three squares.
If there is a face of size , then there is no distancetwo fourcoloring since all five vertices of that face would need different colors. ∎
6. Distancetwo coloring of quartic graphs
A quartic graph is a regular graph with all vertices of degree four. Thus any distancetwo coloring of a quartic graph requires at least five colors. A fourgraph is a plane quartic graph whose faces have sizes or . The argument to view these as analogues of typetwo Barnette graphs is as follows. For cubic plane Euler’s formula limits the numbers of faces that are triangles, squares, and pentagons, but does not limit the number of hexagon faces. Similarly, for plane quartic graphs, Euler’s formula implies that such a graph must have triangle faces, but places no limits on the number of square faces.
We say that two faces are adjacent if they share an edge.
Lemma 6.1.
If a fourgraph can be distancetwo fivecolored, then every square face must be adjacent to a triangle face. Thus can have at most square faces.
Proof.
We view the numbers modulo , and number is separate. Let be a square face that has no adjacent triangle face. (This is depicted in Figure 9 as the square in the middle.) Color by . Let the adjacent square faces be . One of must be colored and the other one . Then either all or all are colored , say all are colored , and all are colored . Then cannot be a triangle face, or would be both colored at distance two. Therefore must be a square face. (In the figure, this is indicated by the corner vertices being marked by smaller circles; these must exist to avoid a triangle face.) This means that the original square is surrounded by eight square faces for , and must have color , since have colors .
But then there cannot be a triangle face , since is within distance two of of colors , so each of the adjacent square faces for has adjacent square faces as well. This process of moving to adjacent square faces eventually reaches all faces as square faces, contrary to the fact that there are triangle faces. ∎
It follows that there are only finitely many distancetwo fivecolorable fourgraphs.
Corollary 6.2.
The distancetwo fivecoloring problem for fourgraphs is polynomial.
In fact, we can fully describe all fourgraphs that are distancetwo fivecolorable. Consider the fourgraphs given in Figure 10. The graph has 8 triangle faces and 4 square faces, the graph has 8 triangle faces and 24 square faces. Note that is obtained from the cube by inserting two vertices of degree four in two opposite square faces. Similarly, is obtained from the cube by replacing each vertex with a triangle and inserting into each face of the cube a suitably connected degree four vertex. (In both figures, these inserted vertices are indicated by smaller size circles.)
Theorem 6.3.
The only fourgraphs that can be distancetwo fivecolored are . These two graphs can be so colored uniquely up to permutation of colors.
Proof.
We show that if can be so colored, then either is or every triangle in must be surrounded by six square faces, in which case is .
Suppose has two adjacent triangles and . The vertices adjacent to must be given two colors other than those of . If has two adjacent squares, then it has five adjacent vertices, which must be given the two colors in alternation, a contradiction. Similarly if is adjacent to three triangles then the three vertices adjacent would need three new colors, a contradiction.
We may thus assume a triangle . If there is a square , this square plus the two adjacent triangles would need six colors, a contradiction, so is a triangle, completing adjacent to the fourcycle . Color with color . Then the additional vertex adjacent to for must be given color (modulo 4), so these form a 4cycle, and any additional vertex adjacent to a must get color 5, so there is a single additional with color 5. This gives a uniquely colored , up to permutation of colors.
In the remaining case, each triangle has adjacent squares , with addition modulo . The vertices must be given the two colors different from those of , and in alternation around , so there cannot be a triangle else with the same color would be at distance two. So there are squares , and is surrounded by six squares.
By Lemma 6.1, we must have a triangle adjacent to the square , either or , but not both since six colors would be needed. Let such a triangle be , and we link to the three . These triangles viewed as vertices linked form a cubic graph without triangles , since a triangle face would be three triangles joined in , which would need to have only three squares inside by Lemma 6.1. The graph has 8 vertices for the 8 triangles, so this graph is the cube . Replacing each vertex corresponding to a triangle by the corresponding triangle gives a graph .
Suppose the triangles adjacent to are for . Then going around a face of we notice only one vertex inside this face by Lemma 6.1, giving the construction of . If we assign to the vertex inside this face the color , we notice that the surrounding triangles in must use three colors at most 4, and each must omit a different color of 4. This implies that all vertices in centers of square faces must be 5, and only opposite triangles for use the same 3 out of 4 colors. This proves existence and uniqueness up to permutation of colors of the distancetwo 5coloring of .
Suppose instead the adjacent triangles are , , and . If there is no triangle , then the three squares between and are respectively adjacent to squares , and must be adjacent to a triangle by Lemma 6.1. There must be triangles at both ends of the and these are adjacent to and , a contradiction.
Finally, suppose again the adjacent triangles are , , and , but there is a triangle . This triangle faces , and faces . Triangles facing each other give two diagonals in the square faces of , which implies two opposite faces without such diagonals in , while the four sets of two diagonals form a matching of the vertices of . If the center of a face without diagonals gets assigned , then the adjacent triangles will be assigned a subset of . Then joining the sets of two diagonals assigns a to a vertex of each remaining triangle, which is not possible to the center of the remaining face without diagonals. ∎
We close with a few remarks and open problems.
Wegner’s conjecture [24] that any planar graph with maximum degree can be distancetwo sevencolored has been proved in [13, 22]. That bound is actually achieved by a typetwo Barnette graph, namely the graph obtained from by subdividing three incident edges. Thus the bound of cannot be lowered even for typetwo Barnette graphs.
Wegner’s conjecture for claims that any planar graph with maximum degree four can be distancetwo ninecolored. The fourgraph in Figure 11 actually requires nine colors in any distancetwo coloring. Thus if Wegner’s conjecture for is true, the bound of cannot be lowered, even in the special case of fourgraphs. It would be interesting to prove Wegner’s conjecture for fourgraphs, i.e., to prove that any fourgraph can be distancetwo ninecolored.
Finally, we’ve conjectured that any bipartite cubic planar graph can be distancetwo sixcolored (a special case of a conjecture of Hartke, Jahanbekam and Thomas [13]). The hexagonal prism (a cyclic prism with , which is a Goodey graph), actually requires six colors. Hence if our conjecture is true, the bound of cannot be lowered even for Goodey graphs. It would be interesting to prove our conjecture for Goodey graphs, i.e., to prove that any Goodey graph can be distancetwo sixcolored.
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