Total weight choosability for Halin graphs
Yu-Chang Liang, Tsai-Lien Wong, and Xuding Zhu

TL;DR
This paper proves that generalized Halin graphs satisfy the total weight choosability conjecture, extending the understanding of weight assignments in complex graph classes.
Contribution
The paper introduces new tools and techniques to establish the total weight choosability of generalized Halin graphs, confirming the conjecture for this class.
Findings
Proved the total weight choosability conjecture for generalized Halin graphs.
Developed new methods for analyzing list assignments in complex graphs.
Extended the class of graphs known to satisfy the 1-2-3 total weight conjecture.
Abstract
A proper total weighting of a graph is a mapping which assigns to each vertex and each edge of a real number as its weight so that for any edge of , . A -list assignment of is a mapping which assigns to each vertex a set of permissible weights and to each edge a set of permissible weights. An -total weighting is a total weighting with for each . A graph is called -choosable if for every -list assignment of , there exists a proper -total weighting. As a strenghtening of the well-known 1-2-3 conjecture, it was conjectured in [ Wong and Zhu, Total weight choosability of graphs, J. Graph Theory 66 (2011), 198-212] that every graph without isolated edge isβ¦
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Taxonomy
TopicsGraph Labeling and Dimension Problems Β· Advanced Graph Theory Research Β· Photochromic and Fluorescence Chemistry
Total weight choosability for Halin graphs
Yu-Chang Liang Department of Applied Mathematics, National Pingtung University, Pingtung, Taiwan 90003. Grant number: MOST 105-2811-M-153-001. Email: [email protected] ββ
Tsai-Lien Wong Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung, Taiwan 80424. Grant numbers: 105-2918-I-110-003. Email: [email protected] ββ
Xuding Zhu Department of Mathematics, Zhejiang Normal University, China. Grant numbers: NSF11171310 and ZJNSF Z6110786. Email: [email protected].
Abstract
A proper total weighting of a graph is a mapping which assigns to each vertex and each edge of a real number as its weight so that for any edge of , . A -list assignment of is a mapping which assigns to each vertex a set of permissible weights and to each edge a set of permissible weights. An -total weighting is a total weighting with for each . A graph is called -choosable if for every -list assignment of , there exists a proper -total weighting. As a strenghtening of the well-known 1-2-3 conjecture, it was conjectured in [ Wong and Zhu, Total weight choosability of graphs, J. Graph Theory 66 (2011), 198-212] that every graph without isolated edge is -choosable. It is easy to verified this conjecture for trees, however, to prove it for wheels seemed to be quite non-trivial. In this paper, we develop some tools and techniques which enable us to prove this conjecture for generalized Halin graphs.
**Key words: ** Total weighting, -matrix, Halin graphs
1 Introduction
A total weighting of is a mapping . A total weighting is proper if for any edge of ,
[TABLE]
where is the set of edges incident to . A total weighting with for all vertices is also called an edge weighting.
Proper edge weighting (also called vertex colouring edge weighting) of graphs was introduced in [10]. It was conjectured in [10] that every graph with no isolated edges has a proper edge weighting with for . This conjecture, now called the 1-2-3 Conjecture, has received a lot of attention [2, 3, 8, 10, 11, 12, 14, 18]. It still remains open, and the best partial result on this conjecture was proved in [11]: every graph with no isolated edge has a proper edge weighting with for all .
Proper total weighting was first studied in [14]. It was conjectured in [14] that every graph has a proper total weighting with for all . This conjecture, now called the 1-2 Conjecture has also received a lot of attention and the best partial result was proved in [9]: for any graph , there is a proper total weighting with for each vertex and for each .
A total list assignment of is a mapping which assigns to each element a set of real numbers as permissible weights. An -total weighting is a total weighting with for each . Assume is a mapping which assigns to each vertex or edge of a positive integer. A total list assignment of is called a -total list assignment of if for all . A graph is called -choosable if for every -list assignment of , there exists a proper -total weighting. A graph is called -choosable if is -choosable, where for each vertex and for each edge .
The list version of total weighting are studied in a few papers [7, 13, 15, 20, 19, 21] It is known [21] that is -choosable if and only if is (vertex) -choosable. So the concept of -choosability is a common generalization of vertex choosability, edge weighting and total weighting of graphs. As strengthening of the 1-2-3 conjecture and the 1-2 conjecture, it was conjectured in [7, 21] that every graph with no isolated edges is -choosable and conjectured in [21] that every graph is -choosable. These two conjectures are called the -choosability conjecture and the -choosability conjecture, respectively.
In the study of total weighting of graphs, one main algebraic tool is Combinatorial Nullstellensatz.
For each , let be a variable associated to . Fix an arbitrary orientation of . Consider the polynomial
[TABLE]
Assign a real number to the variable , and view as the weight of . Let be the evaluation of the polynomial at . Then is a proper total weighting of if and only if . The question is under what condition one can find an assignment for which .
An index function of is a mapping which assigns to each vertex or edge of a non-negative integer . An index function of is valid if . Note that is the degree of the polynomial . For a valid index function , let be the coefficient of the monomial in the expansion of . It follows from the Combinatorial Nullstellensatz [4, 6] that if , and is a list assignment which assigns to each a set of real numbers, then there exists a mapping with such that
[TABLE]
Therefore, to prove that a graph is -choosable, it suffices to show that there exists an index function with for each vertex and for each edge and .
The coefficient is related to the permanent of the martix below (see Equation (1)).
We write the polynomial as
[TABLE]
Then for and , if (oriented from to ), then
[TABLE]
Now is a matrix, whose rows are indexed by the edges of and the columns are indexed by edges and vertices of . Given a vertex or edge of , let be the column of indexed by . For an index function of , let be the matrix, each of its column is a column of , and each column of occurs times as a column of . It is known [5] and easy to verify that for a valid index function of ,
[TABLE]
where denotes the permanent of the square matrix . Recall that if is an matrix, then
[TABLE]
where is the symmetric group of order .
A square matrix is permanent-non-singular if . A square matrix of the form is called an -matrix if for each vertex and for each edge . Motivated by an edge weighting and an total weighting problem of graphs, the following two conjectures were proposed in [7] and [21], respectively.
Conjecture 1
Every graph has a permanent-non-singular -matrix.
Conjecture 2
Every graph without isolated edges has a permanent-non-singular -matrix.
Conjecture 1 and Conjecture 2 have been studied in many papers (see [17] for a survey of partial results on these two conjectures), and both conjectures remain largely open. It is easy to verify both conjectures for trees. However, proving these two conjectures for wheels seem to be quite non-trivial. It was proved in [15] that Conjecture 1 is true for wheel, and in [22] for Halin graphs. Quite surprising, Conjecture 2 remained open for wheels for a long time. In this paper, we develop some tools and techniques and settle Conjecture 2 for generalized Halin graphs.
2 Main theorem and some observations
A Halin graph is a planar graph obtained by taking a plane tree (an embedding of a tree on the plane) without degree vertices by adding a cycle connecting the leaves of the tree cyclically. If the tree is allowed to have degree vertices, then the resulting graph is called a generalized Halin graph.
Theorem 1
Every generalized Halin graph has a permanent-non-singular -matrix.
We will prove this theorem in the next two sections. In the proof, we shall frequently use the following observations:
Observation 1
If is a matrix whose columns are integral liner combinations of columns of and and each olumn occurs in at most times in the combinations, then there is an index function with and . Moreover, if for some prime , then .
This can be derived directly from the multilinear property of permanent.
Observation 2
( [21] ) For an edge of ,
[TABLE]
The above follows easily from the definition of the matrix (cf. [21]):
A balloon is a graph obtained by attaching a path to a cycle (i.e., identify one end vertex of a path with a vertex of a cycle). If the cycle is of odd length, then the ballon is called an odd balloon. The path could be a single vertex, in which case the balloon is simply a cycle. If the path is not a single vertex, then the unique vertex of degree is called the root of the ballon. Otherwise, the root of the ballon (which is a cycle) is an arbtriary vertex of the cycle.
Observation 3
If is an odd balloon with root , then is an integral linear combination of for .
Indeed, if and and the balloon is obtained by identifying with , let (for ) and (for and ), then
[TABLE]
3 Non-bipartite generalized Halin graphs
In this section, we consider non-bipartite generalized Halin graphs.
Lemma 1
Let be a connected non-bipartite graph. If there is a matrix whose columns consists of vertex columns of and for some odd prime , then has a permanent-non-singular -matrix.
Proof. Since is an odd prime, by replacing each column with in , the resulting matrix has . Since is connected and non-bipartite, for any vertex , there is an odd balloon of with root . By Observation 3, can be written as an integral linear combination of edge columns of . By Observation 1, there is an index function with for each vertex such that . It is obvious that if is an index function for which for some edge , then . Therefore for each . I.e., is a permanent-non-singular -matrix of .
Corollary 1
If is a prime, is a connected non-bipartite -degenerate graph, then has a permanent-non-singular -matrix.
Proof. Order the vertices in such a way that each vertex has back-degree , i.e., has at most neighbours with . Let be the matrix consisting copies of the column of indexed by for . It is easy to verify that . Hence . It follows from Lemma 1 that has a permanent-non-singular -matrix.
Lemma 2
Assume is an odd prime, is a connected non-bipartite graph, is a vertex of of degree and has a permanent-non-singular -matrix . If there are edge disjoint odd balloons with root such that for any and , and , then has a permanent-non-singular -matrix.
Proof. Let except that . Let be obtained from by replacing each copy of by . Then . By Observation 3, each copy of can be written as integral linear combination of edge columns for , i.e., for each , where are integers. As , we can replace the copies of with integral linear combinations , so that each is used at most times. Therefore we can write each colummn of as linear combination of edge columns of , and each edge column is used at most times. So has a permanent-non-singular -matrix.
By Lemma 1, to prove that a non-bipartite generalized Halin graph has a permanent-non-singular -matrix, it suffices to show that there is a matrix consisting of vertex columns of and . By Equation (1), this is equivalent to the existence of a valid index function of such that for each vertex , for each edge and .
Recall that the graph polynomial of is defined as , where is an orientation of . So is obtained from by letting for each edge of . Therefore, if for all edges , is indeed the coefficient of the monomial in the expansion of the graph polynomial of . For the purpose of calculating for such an index function , we use a result of Alon and Tarsi [6].
A sub-digraph (not necessarily connected) of a directed graph is called Eulerian if the in-degree of every vertex of is equal to its out-degree . An Eulerian sub-digraph is even if it has an even number of edges, otherwise, it is odd. Let and denote the sets of even and odd Eulerian subgraphs of , respectively. The following result was proved in [6].
Lemma 3
[6]** Let be an orientation of an undirected graph , and is the out-degree of in . Then the coefficient of in the graph polynomial of is .
Lemma 4
Let be a non-bipartite generalized Halin graph. Then has a permanent-non-singular -matrix.
Proof. Assume is obtained from a tree plane by adding edges connecting its leaves into a cycle . We choose non-leave vertex of as the root of . If has an even number of leaves, then we orient the edges of in such a way that the edges in the tree are all oriented towards to the root vertex, and orient the edges of so that it becomes a directed cycle. In such an orientation of , by repeated deleting sink vertices (that must isolated vertices in any Eulerian subgraph), the resulting graph is a directed even cycle . As has no odd Eulerain sub-digraph, and has even Eulerian sub-digraph (the empty digraph and ). As each vertex has out-degree at most . The conlcusion follows from Lemmas 1, 3 and Observation 2.
Assume has an odd number of leaves. Hence is an odd cycle.
Assume first that is not a wheel. Let be a non-leaf vertex of all its sons are leaves. Assume has leaf sons .
If is even, then we orient the cycle as a directed cycle, orient the tree with all edges towards the root, except that the edge is oriented from to . Straightforward counting shows that among Eulerian sub-digraphs of containing the directed edge , are odd and are even. The empty Eulerian subgraph is even, and the directed cycle is odd. Hence . As each vertex has out-degree at most , we are done.
If is odd, then we oriented the edges of as in the case that is even, except that the edge in oriented towards is reversed as an edge oriented away from (so becomes a source vertex in the cycle ). Straightforward counting shows that among Eulerian sub-digraphs of containing the directed edge , are even and are odd.
There is one even Eulerian sub-digraph not using the edge (the empty sub-digraph) and no odd Eulerian sub-digraph not using the edge . Again each vertex has out-degree at most , we are done.
Assume is an odd wheel with , and is the center of the wheel. If , then it can be checked directly that has a permanent-non-singular -matrix. Assume . Consider the graph . We order the vertices of as . Then each vertex has back-degree at most . As in the proof of Corollary 1, for the index function with for , . It is easy to check that each vertex is the root of an odd balloon in that does not contain any edge incident to , and does not contain the edges and . By Observation 3 (cf. the proof of Lemma 1), we know that there is an index function of with such that for all , for any edge of , and for . Now the vertex is the root of two edge disjoint odd balloons with and with . As , by Lemma 2, has a permanent-non-singular -matrix.
4 Bipartite generalized Halin graphs
Lemma 5
*If is a prime, is a connected bipartite -degenerate graph, is a vertex of degree , then has a *permanent-non-singular matrix in which each edge column occurs at most times, the vertex column indexed by occurs once and there are no other vertex column.
Proof. Order the vertices in such a way that each vertex has back-degree , i.e., has at most neighbours with , and . Let be the matrix consisting copies of the column of indexed by for . Similarly, .
Assume is a column in indexed by and . Let be the matrix obtained from by replacing by . Since has two copies of the column , which has only one nonzero entry, we know that . Therefore, if we replace by , the resulting matrix has the same permanent as .
For each vertex column in of the form for , we replace it by , where the sign is determined by the parity of the distance between the two vertices and : if the distance is odd, then choose , and otherwise choose . Denote the resulting matrix by . Then .
Similarly as in the proof of Corollary 1, each column of other than the column indexed by can be written as an integral linear combination of edge columns of . As in the proof of Lemma 1, there is a matrix consisting of edges columns of plus one column indexed by , such that , where each edge column occurs at most in .
Assume is a graph and are subsets of . We denote by the set of edges with one end vertex in and one end vertex with . Let .
Lemma 6
Assume is a connected graph, and is a partition of . If the subgraph induced by edges in has a permanent-non-singular matrix which contains no columns indexed by edges in and is -degenerate, then has a permanent-non-singular -matrix.
Proof. Assume has connected components . For each , let be an edge connecting to . Let . By Lemma 5, has a permanent-non-singular matrix in which each edge column occurs at most twice, the vertex column index by occurs once and there are no other vertex column.
Let be the index function of so that .
Let be the matrix obtained from by deleting the column indexed by and the row index by . Since the column indexed by has only one nonzero entry, we conclude that .
Let be a permanent-non-singular matrix of which contains no columns indexed by edges in . Let be the index function of so that .
For each edge of , let
[TABLE]
Let . Note that for each edge of . Now is of the form
[TABLE]
Therefore , and hence is a permanent-non-singular -matrix of .
Observe that any proper subgraph of a generalized Halin graph is -degenerate. Therefore to prove that a generalized Halin graph has a permanent-non-singular -matrix, by Lemma 6, it suffices to find a partition so that has a permanent-non-singular matrix which contains no columns indexed by edges in .
Let be an oriented graph and be an edge in . We call a sink edge if all edges adjacent to are oriented towards (i.e, towards the common end vertex of and ) and a source edge if all edges adjacent to are oriented away from .
Lemma 7
Assume is a connected graph and is a partition of . If there is an orientation of edges in and a mapping such that for each , is a source or a sink edge incident to , and for each , , then the subgraph has a permanent-non-singular matrix which contains no columns indexed by edges in .
Proof. Let be the subgraph of induced by edges in . Let be an orientation of edges in , and be a mapping from such that for each , is a source or a sink edge incident to , and for each , .
Let for each edge . We shall show that has non-zero permanant. Note that the column vector is non-negative if is a sink edge and non-positive if is a source edge. Thus to prove that has non-zero permanant, it suffices to find a one-to-one mapping between the rows and columns of such that for each row , the entry . The rows of are indexed by edges in and columns are indexed by a multiset of edges in , with each occurs times. Since edges in are incident to , the mapping is such a one-to-one mapping.
Lemma 8
Let be a bipartite generalized Halin graph. Then has a permanent-non-singular -matrix.
Proof. By Lemma 6, it suffices to choose a set such that the subgraph has a a permanent-non-singular matrix which contains no columns indexed by edges in . In all the figures below, vertices of are indicated by open dots, and vertices of are indicated by solid dots.
Recall that is obtained from a plane tree by adding a cycle connecting its leaves in order. If is a path, by choosing with three consecutive vertices and using twice of edges in as column vectors, then it can easily be verified that this is a permanent-non-singular -matrix of . Assume is not a path. We choose a vertex of degree at least as the root of . Let be a leaf with maximum depth. Since is bipartite, the father of has only one son (i.e. ). Let be the father of .
Case 1: has two or three sons.
Let be a leaf son of and choose . We orient the edges in so that is a sink vertex and is a source vertex.
If has two sons, as depicted in Figure 1, then let be the matrix consisting two copies of columns of indexed by and one copy of the column of indexed by . I.e.,
[TABLE]
Then , and we are done.
If has three sons, as depicted in Figure 2, we choose twice the columns of indexed by .
Then
[TABLE]
Then and we are done.
Case 2: has at least four sons or .
Let be the set consisting and all its descendants. Let . Again, orient the edges of so that is a sink vertex, and all vertices at distance from are source vertices as in Figure 3.
Then all the edges in are source or sink edges.
As has at least four sons, there is at least one son, say , such that is not a leaf of . Let be the son of (note that since is bipartite, if a son of is not a leaf of , then it has exactly one son), and be the father of . Let , , , and .
Each edge in is either incident to or is of the form , where is a son of and is the son of . Let be the mapping defined as follows: Cyclically order the edges of incident to as , and . Let , where the indices are modulo . Let . For each son of which is not a leaf of , let be the son of , and let . In particular, and .
Assume . If , then . Otherwise , if is to the left of , then is the tree edge incident to and to the right of ; if is to the right of , then is the tree edge incident to and to the left of . (In particular, both cycle edges incident to are mapped to ).
It is easy to verify that for each , is a source or a sink edge incident to , and for each , . So it follows from Lemma 7 that the subgraph has a matrix which contains no columns indexed by edges in .
Case 3: has only one son.
Let be the father of . If there is a son of with , then the depth of is the same as . We choose to play the role of , and we are in Cases 1 and 2. Hence, we may assume that each son of has degree at most two in . Let be the set consisting of and all the descendants of that have distance at most to . Let . Orient the edges in so that is a sink vertex, and all vertices at distance from are source vertices as in Figure 4.
In this orientation, all the edges in are source or sink edges. Similarly as in the previous case, it is easy to find a mapping such that for each , is a source or a sink edge incident to , and for each , . By Lemma 7, the subgraph has a matrix which contains no columns indexed by edges in .
This completes the proof of Lemma 8.
It was proved in [23] that every graph has a permanent-non-singular -matrix. However, the following two conjectures which are weaker than Conjectures 1 and 2, respectively, remain open.
Conjecture 3
There is a constant such that every graph has a permanent-non-singular -matrix.
Conjecture 4
There is a constant such that every graph without isolated edges has a permanent-non-singular -matrix.
Acknowledgement: This paper is finished while the 2nd author is visiting Professor Shinya Fujita at Yokohama City University. She thanks the hospitality of Professor Fujita and Yokohama City University.
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