Proper 3-orientations of bipartite planar graphs with minimum degree at least 3
aa r X i v : . [ m a t h . C O ] A p r Proper 3-orientations of bipartite planar graphs with minimumdegree at least 3
Kenta Noguchi ∗ Abstract
In this short note, we show that every bipartite planar graph with minimum degree at least3 has proper orientation number at most 3.
Keywords. proper orientation, planar graph, bipartite graph
For basic terminology in graph theory undefined in this paper, refer to [4]. Let G be a simplegraph. An orientation σ of G is a digraph obtained from G by replacing each edge by exactlyone of the two possible arcs with the same end-vertices. Let E σ ( G ) be the resulting arc set. For v ∈ V ( G ), the indegree of v in σ , denoted by d − σ ( v ), is the number of arcs in E σ ( G ) incoming to v . We denote by ∆ − ( σ ) the maximum indegree of σ . An orientation σ with ∆ − ( σ ) at most k is a k -orientation . An orientation σ of G is proper if d − σ ( u ) = d − σ ( v ) for every uv ∈ E ( G ). The proper orientation number of G , denoted by −→ χ ( G ), is the minimum integer k such that G admitsa proper k -orientation. Proper orientation number is defined by Ahadi and Dehghan [1], and seethe related research [2, 3, 6]. Knox et al. [6] showed the following. Theorem 1 ([6], Theorem 3) . Let G be a 3-connected bipartite planar graph. Then −→ χ ( G ) ≤ maximum average degree Mad( G ) of agraph G is defined asMad( G ) = max (cid:26) | E ( H ) || V ( H ) | : H is a subgraph of G (cid:27) . Theorem 2.
Let k be a positive integer and G = G [ X, Y ] be a bipartite graph with Mad( G ) ≤ k .If deg( x ) ≥ k + 1 for every x ∈ X , then −→ χ ( G ) ≤ k + 1.Theorem 2 partially answers Problem 5 in [2], which asks whether −→ χ ( G ) can be bounded bya function of Mad( G ). Using Theorem 2, we can state the following. The minimum degree of G isdenoted by δ ( G ). Theorem 3.
Let G be a bipartite planar graph with δ ( G ) ≥
3. Then −→ χ ( G ) ≤ δ ( G ) and −→ χ ( G ) in Theorem 3 are tight; see Theorems 5 and 6, respec-tively. The following corollary immediately follows from Theorem 3, which is the improvement ofTheorem 1. Corollary 4.
Let G be a 3-connected bipartite planar graph. Then −→ χ ( G ) ≤ ∗ Department of Information Sciences, Tokyo University of Science, 2641 Yamazaki, Noda, Chiba 278-8510, Japan.Email: [email protected] K , .Furthermore, for quadrangulations , which are bipartite planar graphs whose each face is boundedby a 4-cycle, the orientation number is completely determined as follows. Theorem 5.
Let G be a quadrangulation with δ ( G ) ≥
3. Then −→ χ ( G ) = 3 unless G is isomorphicto the 3-cube Q . If G is isomorphic to Q , then −→ χ ( G ) = 2.Finally, we show the tightness of the minimum degree condition in Theorem 3. Note that,when δ ( G ) = 1, Araujo et al. [2, Corollary 2] showed that there exists a tree T with −→ χ ( T ) = 4. Theorem 6.
There exist bipartite planar graphs G with δ ( G ) = 2 and −→ χ ( G ) = 4. The proof of Theorem 2 is very simple.
Proof of Theorem 2.
Let G = G [ X, Y ] be a bipartite graph with Mad( G ) ≤ k satisfying deg( x ) ≥ k +1 for every x ∈ X . By Hakimi’s result [5], every graph G with Mad( G ) ≤ k has a k -orientation σ . Let S σ = { ( u, v ) ∈ E σ ( G ) | u ∈ X } . For every x ∈ X , since deg( x ) ≥ k + 1, we can switchthe orientation of some edges in S σ so that the indegree of x is exactly k + 1. In the resultingorientation σ ′ of G , observe that d − σ ′ ( y ) ≤ k for every y ∈ Y . Thus, σ ′ is a proper ( k +1)-orientationof G .To prove Theorem 3, we use the following well-known lemma. Lemma 7.
Let G be a bipartite planar graph with n ≥ | E ( G ) | ≤ n − G is a quadrangulation. Proof of Theorem 3.
Let G be a bipartite planar graph with δ ( G ) ≥
3. By Lemma 7, G hasmaximum average degree less than 4. Thus, −→ χ ( G ) ≤ Proof of Theorem 5.
Let G [ X, Y ] be a quadrangulation with δ ( G ) ≥
3, where | X | ≥ | Y | . Notethat n = | X | + | Y | ≥ | Y | ≥
3. By Theorem 3, −→ χ ( G ) ≤
3. We show that G does not admit aproper 2-orientation or is isomorphic to Q . Since | E ( G ) | = 2 n − σ of G , then the number of vertices of indegree 2 in σ is at least n −
4. Sincetwo vertices of indegree 2 in σ cannot be adjacent in G , | Y | ≤
4. If | Y | = 3, then three vertices x , x , x in X must be adjacent to all vertices in Y , and hence there exists a complete bipartitegraph K , , contrary to the planarity of G . If | Y | = 4, let Y = { y , y , y , y } .Case 1. Suppose that there exist two vertices x , x ∈ X such that each of them is adjacentto the same three vertices in Y , say y , y , y (see Figure 1). We may assume that y is in theface f = x y x y by symmetry. Then the other vertices in X must be in f to be adjacent to atleast 3 vertices in Y by the planarity of K , . So the degree of y is exactly 2, which contradictsto δ ( G ) ≥
3. 2igure 2: The bipartite planar graph G .Case 2. Suppose that there exists no such pair of vertices in X . In this case one can seethat | X | = 4 and that G is isomorphic to Q . It is easy to show that −→ χ ( Q ) = 2 (see also [2,Proposition 2]). Proof of Theorem 6.
Consider the bipartite planar graph (quadrangulation) G defined as follows(see Figure 2). Let V ( G ) = { a ijk }∪{ b ij , c ij }∪{ d ik }∪{ p i , q i }∪{ s, t } , where 1 ≤ i, j ≤ , ≤ k ≤ E ( G ) = { a ijk b ij , a ijk c ij | ≤ i, j ≤ , ≤ k ≤ } ∪{ b ij p i , b ij q i , c ij p i , c ij q i | ≤ i, j ≤ } ∪{ d ik p i , d ik q i | ≤ i ≤ , ≤ k ≤ } ∪{ p i s, p i t, q i s, q i t | ≤ i ≤ } . Suppose that −→ χ ( G ) ≤ σ be a proper 3-orientation of G . Thoughout the proof, anindegree means one in σ .Step 1. Consider the graph induced by { a ijk } ∪ { b ij , c ij } for fixed i, j . Since d − σ ( b ij ) ≤ d − σ ( c ij ) ≤
3, there exists a vertex of indegree 2 in { a ijk } . So b ij and c ij cannot be indegree 2 forall 1 ≤ i, j ≤ { a ijk } ∪ { b ij , c ij } ∪ { d ik } ∪ { p i , q i } for a fixed i . Ifall vertices in { b ij , c ij } have indegree at most 1, then at least one of the indegrees of p i and q i must be at least 4, a contradiction. By this fact and Step 1, there exists a vertex of indegree 3in { b ij , c ij } . So p i and q i cannot be indegree 3 for all 1 ≤ i ≤
4. Moreover, since there exists avertex of indegree 2 in { d ik } by the same argument in Step 1, p i and q i cannot be indegree 2 forall 1 ≤ i ≤ p , . . . , p , q , . . . , q have indegree at most 1. This leads tothe fact that at least one of the indegrees of s and t must be at least 4, a contradiction.3hus, −→ χ ( G ) ≥
4. It is easy to show that −→ χ ( G ) ≤
4. (An infinite family of such graphs areeasy to construct; add some vertices a , . . . , a k of degree 2 to G for example.) Acknowledgements
The author’s work was partially supported by JSPS KAKENHI Grant Number 17K14239.