Hamming Index for Some Classes of Graphs with Respect to Edge-Vertex Incidence Matrix

Wilda Sitorus (Department of Mathematics, Universitas Sumatera Utara, Medan, Indonesia)

Saib Suwilo (Department of Mathematics, Universitas Sumatera Utara, Medan, Indonesia)

Mardiningsih (Department of Mathematics, Universitas Sumatera Utara, Medan, Indonesia)

Proceedings of MICoMS 2017

eISBN: 978-1-78756-793-1

ISSN: 2516-2853

Publication date: 4 May 2018

pdf (516 KB) ePub (540 KB)

Abstract

Hamming distance of a two bit strings u and v of length n is defined to be the number of positions of u and v with different digit. If G is a simple graph on n vertices and m edges and B is an edge–vertex incidence matrix of G, then every edge e of G can be labeled using a binary digit string of length n from the row of B which corresponds to the edge e. We discuss Hamming distance of two different edges of the graph G. Then, we present formulae for the sum of all Hamming distances between two different edges of G, particularly when G is a path, a cycle, and a wheel, and some composite graphs.

Keywords

Citation

Sitorus, W., Suwilo, S. and Mardiningsih, (2018), "Hamming Index for Some Classes of Graphs with Respect to Edge-Vertex Incidence Matrix", Proceedings of MICoMS 2017 (Emerald Reach Proceedings Series, Vol. 1), Emerald Publishing Limited, Leeds, pp. 621-628. https://doi.org/10.1108/978-1-78756-793-1-00098

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:

Emerald Publishing Limited

License

Published in the Emerald Reach Proceedings Series. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) licence. Anyone may reproduce, distribute, translate and create derivative works of this article (for both commercial and non-commercial purposes), subject to full attribution to the original publication and authors. The full terms of this licence may be seen at http://creativecommons.org/licences/by/4.0/legalcode

1. Introduction

Let Z2={0,1}. Notice that 〈 Z2,+2 〉 is a group with binary operation addition modulo 2. A bit string x of length n can be thought as an element of the set ???Z2×Z2×⋯×Z2︷n. Moreover, the set Z2×Z2×⋯×Z2︷nwith binary operation defined by

x⊕2y=(x1+2y1,x2+2y2,…,xn+2yn)

is a group. For a bit string x∈Z2×Z2×⋯×Z2 the weight of x, denoted by wt(x), is the number of 1 contained in x. For two bit strings x,y∈Z2×Z2×⋯×Z2 the Hamming distance of x and y, denoted by Hd(x,y), is the number of positions in x and y with different digits. Hamming distance can also be defined by Hd(x,y)=wt(x⊕2y) (Pless, 1998).

Let G(V,E) be a simple graph on n vertices and m edges with vertex set V={v1,v2,…,vn} and edge set E={e1,e2,…,em} . If the edge e={u,v} , we say that the edge e incident to vertices u and v. Two edges ei and ej are adjacent if they have one end vertex in common.

An edge–vertex incidence matrix of a graph G is an m×n matrix B=(bij) defined by

bij={ 1,if the edgeeiis incident to the vertexvj, 0,if the edgeeiis not incident to the vertexvj.

Notice that every row of the edge–vertex incidence matrix of G is a bit string of length n and thus the edge–vertex incidence matrix contains m bit strings each of length n. For i=1,2,…,m we define s(ei)=B(i,:) where B(i,:) is the ith row of B. For each two different edges ei and ej the Hamming distance between ei and ej is defined to be

Hd(s(ei),s(ej))=Hd(B(i,:),B(j,:))=wt(B(i,:)⊕2B(j,:)).

The sum of Hamming distances of a graph G, denoted by H(G), is defined as

H(G)=∑1≤i<j≤mHd(s(ei),s(ej)) =∑1≤i<j≤mHd(B(i,:),B(j,:))=∑1≤i<j≤mwt(B(i,:)⊕2B(j,:)).

The sum of Hamming distances is also called the Hamming index.

The notion of Hamming distance and sum of Hamming distances of edges of simple graph G is initiated by Ramane et al. (2015). They discuss a general formula for Hamming index of a graph and apply this formula to get the Hamming index of classes of regular graph. Hamming distance and Hamming index of a graph can also be defined using the adjacency matrix of the graph (Ganagi and Ramane, 2016). In this paper, we discuss Hamming distance and sum of Hamming distances of simple graph with respect to its edge–vertex incidence matrix. We especially discuss a way of finding the Hamming index of a graph from the known Hamming index of its subgraph. We first present formulae for Hamming index of some graphs consist of a path or a cycle as its subgraph. We then present a formula for Hamming index of some composite graphs.

2. Necessary background

Let G be a graph with vertex set V={v1,v2,…,vn} and edge set E={e1,e2,…,em}. For each vertex vi the degree of vi, denoted by deg(vi), is the number of edges in G that are incident to vi. It is well known that ∑i=1ndeg(vi)=2m (Korte and Vygen, 2006).

We note that by definition, the Hamming index of a graph can be calculated using the formula H(G)=∑1≤i<j≤mwt(B(i,:)⊕2B(j,:)). Employing this formula, the Hamming index of a certain graph can be determined using the following algorithm (Figure 1).

Figure 1.

Algorithm for Sum of Hamming Distances

The following lemma, due to Ramane et al. (2015), presents a formula for Hamming distance between two edges of a simple graph. We present a simpler proof than in Ramane et al. (2015).

Lemma 2. Let G be a simple graph. Then

Hd(s(ei),s(ej))={ 2,ifeiis adjacent toej,4,ifeiis not adjacent toej.

Proof. Since every edge of G is incident to exactly two vertices, then for each i=1,2,…,m, we have wt(s(ei))=2. Let s(ei)=x1x2x3⋯xn and s(ej)=y1y2y3⋯yn. Ifedge ei is adjacent to ej, then there is a common vertex, say vk, of edges ei and ej. This implies xk=yk=1, and if t≠k then xt=yt=0 or xt≠yt. Therefore, if the edge ei is adjacent to the edge ej,

Hd(s(ei),s(ej))=wt(s(ei))+w(s(ej))−2=2.

If the edge ei is not adjacent to the edge ej, then there is no common vertex vk of ei and ej. This implies for each t=1,2,…,n,xt=yt=0 or xt≠yt. Therefore, if the edge ei is not adjacent to the edge ej, we have

Hd(s(ei),s(ej))=wt(s(ei))+w(s(ej))=4.□

Theorem 3. Let G a simple graph on m edges. If there are n(P) adjacent pairs of edges in G, then H(G)=2m(m−1)−2n(P).

Proof. By Lemma 2 we have H(G)=2n(P)+4n(Q), where n(Q) is the number of pairs of non-adjacent edges in G. Since the graph G contains m edges, then we have n(Q)=m(m−1)/2−n(P). Therefore, H(G)=2m(m−1)−2n(P). □

Proposition 4. Let Pn a path on n≥3 vertices. Then H(Pn)=2(n−2)2.

Proof. Notice that E(Pn)={ei={vi,vi+1}:i=1,2,…,n−1}. Therefore, there are n(P)=n−2 pairs of adjacent edges ei and ei+1. Since Pn has m=n−1 edges, then by Theorem 3 we have

H(Pn)=2m(m−1)−2n(P)=2(n−1)(n−2)−2(n−2)=2(n−2)2.□

Proposition 5. Let Cn be a cycle on n vertices. Then H(Cn)=2n(n−2).

Proof. There are n pairs of adjacent edges ei and ei+1 (we note that en+1=e1), i=1,2,…,n. Hence n(P)=n. By Theorem 3 we have

H(Cn)=2m(m−1)−2n(P)=2n(n−1)−2n=2n(n−2).□

3. Main results

Let G be a graph and F be a subgraph of G. We discuss the Hamming index of the graph G in term of the Hamming index of its subgraph F. We note that

H(G)=∑ei,ej∈E(G)Hd(s(ei),s(ej))=∑ei,ej∈E(F)Hd(s(ei),s(ej))+∑ei∈E(F),ej∉E(F)Hd(s(ei),s(ej))+∑ei,ej∉E(F)Hd(s(ei),s(ej))=H(F)+∑ei∈E(F),ej∉E(F)Hd(s(ei),s(ej))+∑ei,ej∉E(F)Hd(s(ei),s(ej)).

A fan F1,n is a graph on n+1 vertices and 2n−1 edges, where its vertex set is V(F1,n)={v1,v2,…,vn+1} and its edge set is

E(F1,n)={{vi,vi+1}:i=1,2,…,n−1}∪{{vi,vn+1}:i=1,2,…,n}.

Theorem 6. For a fan F1,n, we have H(F1,n)=7n2−17n+12.fo

Proof. Notice that a fan F1,n contains the path Pn as its subgraph where V(Pn)={v1,v2,…,vn} and E(Pn)={{vi,vi+1}:i=1,2,…,n−1}. From Proposition 4, we have H(Pn)=2(n−2)2=2n2−8n+8.

There are n(n−1) pairs of edges ei and ej for which ei∈Pn and ej∉Pn. Among them there are 2(n−1) adjacent edges. This implies

∑ei∈Pn,ej∉PnHd(s(ei),s(ej))=2(2(n−1))+4(n2−3n+2)=4n2−8n+4.

There are n(n−1)/2 pairs of edges ei and ej for which ei∉Pn and ej∉Pn. All of them are adjacent pairs. This implies

∑ei∉Pn,ej∉PnHd(s(ei),s(ej))=2(n(n−1)/2)=n2−n.

We now conclude that

H(F1,n)=H(Pn)+∑ei∈Pn,ej∉PnHd(s(ei),s(ej))+∑ei,ej∉PnHd(s(ei),s(ej)) =(2n2−8n+8)+(4n2−8n+4)+(n2−n)=7n2−17n+12.

An n-wheel Wn is a graph on n+1 vertices and 2n edges with vertex set V(Wn)={v1,v2,…,vn+1} and edge set E(Wn)={{vi,vi+1}:i=1,2,…,n−1}∪{{vn,v1}}∪{{vi,vi+n}:i=1,2,…,n}. □

Theorem 7. If n≥2, then H(Wn)=7n2−9n.

Proof. We note that the cycle Cn is a subgraph of Wn. Proposition 5 guarantees that H(Cn)=2n2−4n. There are n2 pairs of edges ei and ej for which ei∈Cn and ej∉Cn. Among them there are 2n pairs of adjacent edges. This implies

∑ei∈Cn,ej∉CnHd(s(ei),s(ej))=2(2n)+4(n2−2n)=4n2−4n.

There are n(n−1)/2 pairs of edges ei and ej for which ei,ej∉Wn. All of them are adjacent pairs. This implies

∑ei,ej∉WnHd(s(ei),s(ej))=2(n(n−1)/2)=n2−n.

We now conclude that

H(Wn)=H(Cn)+∑ei∈Cn,ej∉CnHd(s(ei),s(ej))+∑ei,ej∉CnHd(s(ei),s(ej)) =(2n2−4n)+(4n2−4n+(n2−n)=7n2−9n.

An n-Sheep Steering Wheel, denoted by SSWn, is a graph on 2n+1 vertices and 3n edges such that its vertex set is V(SSWn)=V(Wn)∪{vn+2,vn+3,…,v2n+1} and its edge set is E(SSWn)=E(Wn)∪{{vi,vn+1+i}:i=1,2,…,n}. □

Theorem 8. If n≥3, then H(SSWn)=17n(n−1).

Proof. Notice that an n-wheel Wn is a subgraph of SSWn. If the pair of edges ei and ej are both in Wn, then by Theorem 7

∑ei,ej∈E(Wn)Hd(s(ei),s(ej))=7n2−9n.

Let the pair of edges ei and ej are such that ei is in Wn and ej is not in Wn. There are 3n adjacent pairs of such edges. There are n(2n−3) non-adjacent pairs of such edges. Therefore

∑ei∈E(Wn),ej∉E(Wn)Hd(s(ei),s(ej))=2(3n)+4(n(2n−3))=8n2−6n.

There are n(n−1)/2 pairs of non-adjacent edgesei and ej such that both of them are not in Wn. There are no pairs of adjacent edges ei and ej such that both of them are not in Wn. Therefore

∑ei,ej∉E(Wn)Hd(s(ei),s(ej))=2n(n−1).

We now conclude that

H(SSWn)=∑ei,ej∈E(SSWn)Hd(s(ei),s(ej))=7n2−9n+8n2−6n+2n2−2n=17n(n−1).

An (n,m)-Jahangir graph, denoted by Jn,m, is a graph on nm+1 vertices and m(n+1) edges such that its vertex set is V(Jn,m)=V(Cmn)∪{vmn+1} and its edge set is

E(Jn,m)=E(Cmn)∪{{v(j−1)n+1,vmn+1}:j=1,2,…,m}.□

Theorem 9. If Jn,m is a (n,m)-Jahangir graph, then H(Jn,m)=(2nm+4m)(nm−2)+m(m+3).

Proof. Notice that the cycle Cnm is a subgraph of the Jahangir graph Jn,m. If the pair of edges ei and ej are both in Cnm, then by Proposition 5 we have

∑ei,ej∈CnmHd(s(ei),s(ej))=HB(Cnm)=2nm(nm−2).

If the pair of edges ei and ej is such that ei is in Cnm and ej is not in Cnm, then there are 2m pairs of edges that are adjacent and there are m(nm−2) pairs of edges which are not adjacent. Therefore

∑vi∈V(Cnm),vj∉V(Cnm)Hd(s(ei),s(ej)=2(2m)+4m(nm−2).

If both edges ei and ej are not in Cnm, then there are m(m−1)/2 pairs of edges which are adjacent and there are no pairs of edges which are not adjacent. Therefore

∑vi,vj∉V(Cnm)Hd(s(ei),s(ej))=m(m−1).

We now conclude that

HB(Jn,m)=2nm(nm−2)+4m+4m(nm−2)+m(m−1) =(2nm+4m)(nm−2)+m(m+3).

We now discuss the hamming index of two classes of composite graphs. Let G1(V1,E1) andG2(V2,E2) are two graphs. The graph G1∪G2 is a graph with vertex set V(G1∪G)2=V(G1)∪V(G2) and edge set E(G1∪G)2=E(G1)∪V(G2). □

Theorem 10. Let G1 be a graph on m1 edges and G2 be a graph on m2 edges. If V(G1)∩V(G2)=∅, then H(G1∪G2)=H(G1)+H(G2)+4m1m2.

Proof. If pairs of edges ei and ej are both G1, then

∑ei,ej∈E(G1)Hd(s(ei),s(ej))=H(G1).

If pairs of edges ei and ej are both on G2, then

∑ei,ej∈E(G2)Hd(s(ei),s(ej))=H(G2).

If ei is on G1 and ej is on G2, then ei and ej are not adjacent. Since there are m1m2 pairs of such edges, by Lemma 3 we have

∑ei∈E(G1),ej∈E(G2)Hd(s(ei),s(ej))=4m1m2.

Therefore, H(G1∪G2)=H(G1)+H(G2)+4m1m2.

Let G be a graph on n vertices V(G)={v1,v2,…,vn} and m edges E(G). Define the thorn graph G⁺ of G to be a graph on 2n vertices and m + n edges such that

V(G+)=V(G)∪{vn+1,vn+2,…,v2n}

and

E(G+)=E(G)∪{{vi,vi+n}:i=1,2,…,n}.□

Theorem 11. Let G be a graph on n vertices and m edges, then H(G+)=H(G)+4mn+2n2−2n−4m.

Proof. If the pair of edges ei and ej are both lie on E(G), then by definition

∑ei,ej∈E(G)Hd(s(ei),s(ej))=H(G).

There are n(n−1)/2 pairs of edges ei and ej where both of them are not on E(G). Since ei and ej are not adjacent, then

∑ei,ej∉E(G)Hd(s(ei),s(ej))=2n(n−1).

We next consider the pairs of edges ei and ej where ei∈E(G) and ej∉E(G). If the edge ej is incident to some vertex vk then ej is adjacent to deg(vk) edges in E(G) and ej is not adjacent to (m−deg(vk)) edges in E(G). This implies there are ∑k=1ndeg(vk)=2m adjacent pairs of edges and there are ∑k=1n(m−deg(vk))=nm−2m pairs of non-adjacent edges. This implies

∑ei∈E(G),ej∉E(G)|Hd(s(ei),s(ej))=2(2m)+4(nm−2m)=4nm−4m.

Thus H(G+)=H(G)+4nm+2n2−2n−4m.

For n≥3, an n-Sun, denoted by Sn, is a graph on 2n vertices and 2n edges withV(Sn)={v1,v2,…,v2n} and E(Sn)={{vi,vi+1}:i=1,2,…,n,vn+1=v1}∪{{vi,vi+n}:i=1,2,…,n}. □

Corollary 12. If n≥3, then H(Sn)=2n(4n−5).

Proof. Notice that Sn=Cn+. Proposition 5 guarantees that H(Cn)=2n(n−2). Since a cycle Cn has m = n edges, by Theorem 11 we have

H(Sn)=H(Cn)+4mn+2n2−2n−4m =2n(n−2)+4n2+2n2−2n−4n=2n(4n−5).□

References

Ganagi, A.B. and Ramane, H.S. (2016). “Hamming Distance between the String Generated by Adjacency Matrix of a Graph and their Sum”. Algebra and Discrete Mathematics, Vol. 22, No. 1, pp. 82–93.

Korte, B and Vigen, J. (2006). Combinatorial Optimization Theory and Algorithms. (3rd ed.). Springer, Berlin.

Pless, V. (1998). Introduction to the Theory of Error-Correcting Codes. John Wiley and Sons, Boston.

Ramane, H.S., Joshi, V.B., Jummannaver, R.B., Manjalapur, V.V., Patil, S.C., Shindhe, S.D., Hadimani, V.S., Kyalkonda, V.K. and Baddi, B.C. (2015). “Hamming Index of a Graph Generated by and Edge-Vertex Incidence Matrix”. International Journal of Mathematics, Science. and Engineeringg Application, Vol. 9, No. I, pp. 93–103.

Corresponding author

Saib Suwilo can be contacted at saib@usu.ac.id

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Hamming Index for Some Classes of Graphs with Respect to Edge-Vertex Incidence Matrix

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