On Vertex Euclidean Deficiency of One-Point Union and One-Edge Union of Complete Graphs

Zhao-Bin Gao1, Wei Qiu1, Sin-Min Lee2, Tai-Chieh Yang3, Carl Xiaohang Sun4
1College of Math. Sci Harbin Engineering Univ. Harbin, 150001, China
21304N, 1st Avenue Upland, CA 91786 USA
3Dept. of Maths. National Changhua Univ. of Education. Changhua, Taiwan
4Sacramento Waldorf School 3750, Bannister Road, Fair Oaks, CA 95628

Abstract

A (p,g)-graph G is Euclidean if there exists a bijection f:V{1,2,,p} such that for any induced C3-subgraph {v1,v2,v3} in G with f(v1)<f(v2)<f(v3), we have that f(v1)+f(v2)>f(v3). The Euclidean Deficiency of a graph G is the smallest integer k such that GNk is Euclidean. We study the Euclidean Deficiency of one-point union and one-edge union of complete graphs.

Keywords: Euclidean Deficiency, Complete graph

1. Introduction

Definition 1. A (p,g)-graph G is Euclidean if there exists a bijection f:V{1,2,,p} such that for any induced C3-subgraph {v1,v2,v3} in G with f(v1)<f(v2)<f(v3), we have that f(v1)+f(v2)>f(v3). Let Euclid(0) be the set of all Euclidean graphs.

Example 1. An Euclidean graph with 5 vertices:

Figure 1

Example 2. All triangle free graphs are Euclidean.

Example 3. Euclidean (p,g)-graphs with p=1,2,,6

An immediate observation is that C3 is not Euclidean, for the vertices will be labeled with 1,2,3, but 1+23. This observation can also be extended: the label 1 cannot be used on any vertices contained in some C3 subgraph. For the sake of contradiction, assume this C3 subgraph is labeled 1,x,y where 1<x<y, then 1+xy and hence 1+xy for any integer y>x. It follows

Figure 2

That if all vertices are part of some C3 subgraph then, the graph must necessarily not be Euclidean.

Definition 2. The Euclidean deficiency of a (p,g)-graph G is min{k:GNkEuclid(0)}, where Nk is the null graph with k vertices, and we’ll denote this number by μ(G). For a given k>0, let Euclid(k) be the set of all graphs with Euclidean deficiency k.

Example 4. Graphs with Euclidean deficiency 1:

Theorem 1. Let n3 and Kn be the complete graph of order n, then μ(Kn)=n2.

Proof. Observe letting the vertices having labels n1,n,,2n1 works, since n1,n are the smallest labels, it follows that, for label of any two vertices v1,v2, f(v1)+f(v2)n1+n=2n1, greater than all other labels. This establishes that μ(Kn)n2. On the other hand, let x,x+1 be the smallest two labels, then the largest label is x+n and since the graph is complete, the vertices containing these labels form a C3 subgraph, therefore x+x+1>x+n,

Figure 3

or equivalently, x>n1. This establishes that μ(Kn)n2, and therefore μ(Kn)=n2◻

Theorem 2. If HEuclid(k) and HG such that

  1. GH is triangular free,

  2. |V(GH)|k,

then GEuclid(0).

Proof. Instead of NK, we can use the vertices of GH and the result follows. ◻

2. Construction of graphs

Definition 3. Let v1,v2 be vertices of graphs G1,G2, respectively, then the one-point union, OU((G1,{v1}),(G2,{v2})), is the disjoint union G2 to G1 then v2 attaches to v1.

Example 5. We’ll now look at the Euclid deficiency of one-point union of complete graphs. First, looking at OU(K3,Kn) for n3 we see that OU(K3,Kn){Euclid(1)n4,3n5Euclid(n4)n6. By testing out OU(K4,Kn) for n4, we have that OU(K4,Kn){Euclid(2)n5,4n7Euclid(n5)n8. Testing out similar cases, we see that OU(K5,Kn){Euclid(3)n6,5n9Euclid(n6)n10, OU(K6,Kn){Euclid(4)n7,6n11Euclid(n7)n12, OU(K7,Kn){Euclid(5)n8,7n13Euclid(n8)n14.

Figure 4

In general, we have that

Theorem 3. For mn, OU(Km,Kn){Euclid(m2)nm1,n2m1Euclid(nm1)n2m.

Proof. Let G=OU(Km,Kn) and a=μ(G). We’ll label the vertices on Km with a+1,a+2,,a+m with a+m at the common vertex, and a+m,a+m+1,,a+m+n1 on the Kn graph.

If n2m1 then from the subgraph Km consisting of a+1,a+2,a+m, we have that a+1+a+2>a+m or am2. Direct calculation of with a=m2 on the C3 subgraphs of G consisting of the smallest, second smallest and largest labels in Kn and Km respectively: a+1+a+2=2m1>2m2=a+m and a+m+a+m+1=4m3=(2m1)+(2m2)n+a+m>a+m+n1, shows that μ(G)=m2.

If n2m then from the subgraph of Kn consisting of a+m,a+m+1,a+m+n1, we have that a+m+a+m+1>a+m+n1, or equivalently anm1. Direct calculation of with a=nm1 on the C3 subgraphs of G consisting of the smallest, second smallest and largest labels in Kn and Km respectively: a+1+a+2=2n2m+1n+1>n1=a+m and a+m+a+m+1=2n1>2n2=a+m+n1, shows that μ(G)=nm1◻

Figure 5

Definition 4. Let e1,e2 be edges of graphs G1,G2, respectively, then the one-edge union, OE((G1,{v1}),(G2,{v2})), is the disjoint union G2 to G1 then collapse e2 to e1.

Figure 6
Figure 7
Figure 8
Figure 9

This construction is dual of one-point union of graphs.

Example 6. OE((C3,(c1,c2)),(C4,(v1,v2))).

We’ll now look at the Euclid deficiency of one-edge union of complete graphs. First, looking at OE(K3,Kn) for n3 we see that OE(K3,K3)Euclid(1) OE(K3,K5)Euclid(2) or in general, OE(K3,K4)Euclid(1) OE(K3,K6)Euclid(3) OE(K3,Kn){Euclid(1)3n4Euclid(n3)n5. By testing out OE(K4,Kn) for n4, we have that OE(K4,K4)Euclid(2) OE(K4,K6)Euclid(2) or in general, OE(K4,K5)Euclid(2) OE(K4,K7)Euclid(3) OE(K4,Kn){Euclid(2)4n6Euclid(n4)n7. Testing out similar cases, we see that OE(K5,Kn){Euclid(3)5n8Euclid(n5)n9, OE(K6,Kn){Euclid(4)6n10Euclid(n6)n12, OE(K7,Kn){Euclid(5)7n12Euclid(n7)n13.

In general, we have that

Theorem 4. For mn, OU(Km,Kn){Euclid(m2)n2m2Euclid(nm)n2m1.

Proof. The proof is similar to that of Theorem 3. ◻

Acknowledgment

We appreciate Prof. Harris Kwong for his critical and helpful suggestions on this paper.

Conflict of interest

The author declares no conflict of interest.

References:

  1. Gallian, J. A., 2017. A dynamic survey of graph labeling. Electronic Journal of Combinatorics, 20(DS6), pp.1-30.
  2. Harary, F., 1969. Graph theory. Reading, MA: Addison-Wesley.
  3. Lee, S. M., n.d.. On vertex Euclidean graphs. (Manuscript)
  4. Shee, S. C. and Ho, Y. S., 1993. The cordiality of one-point union of copies of graphs. Discrete Mathematics, 117, pp.225-243.