The set D of distinct signed degrees of the vertices in a signed graph G is
called its signed degree set. In this paper, we prove that every non-empty set
of positive (negative) integers is the signed degree set of some connected
signed graph and determine the smallest possible order for such a signed graph.
We also prove that every non-empty set of integers is the signed degree set of
some connected signed graph

Dar, F. A.

Naikoo, T. A.

Pirzada, S.

English

arXiv

summary:The split graph $K_r+\overline {K_s}$ on $r+s$ vertices is denoted by $S_{r,s}$. A non-increasing sequence $\pi =(d_1,d_2,\ldots ,d_n)$ of nonnegative integers is said to be potentially $S_{r,s}$-graphic if there exists a realization of $\pi $ containing $S_{r,s}$ as a subgraph. In this paper, we obtain a Havel-Hakimi type procedure and a simple sufficient condition for $\pi $ to be potentially $S_{r,s}$-graphic. They are extensions of two theorems due to A. R. Rao (The clique number of a graph with given degree sequence, Graph Theory, Proc. Symp., Calcutta 1976, ISI Lect. Notes Series 4 (1979), 251–267 and An Erdős-Gallai type result on the clique number of a realization of a degree sequence, unpublished)

Yin, Jian-Hua

Institute of Mathematics AS CR, v. v. i.

Czechoslovak Mathematical JournalJian Hua YinA Havel-Hakimi type procedure and a sufficient condition for a sequence to bepotentially Sr,s-graphicCzechoslovak Mathematical Journal, Vol. 62 (2012), No. 3, 863–867Persistent URL: http://dml.cz/dmlcz/143030Terms of use:© Institute of Mathematics AS CR, 2012Institute of Mathematics of the Czech Academy of Sciences provides access to digitizeddocuments strictly for personal use. Each copy of any part of this document must contain theseTerms of use.This document has been digitized, optimized for electronic delivery andstamped with digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://dml.czCzechoslovak Mathematical Journal, 62 (137) (2012), 863–867A HAVEL-HAKIMI TYPE PROCEDURE ANDA SUFFICIENT CONDITION FOR A SEQUENCETO BE POTENTIALLY Sr,s-GRAPHICJian-Hua Yin, Haikou(Received August 17, 2011)Abstract. The split graph Kr +Ks on r+ s vertices is denoted by Sr,s. A non-increasingsequence π = (d1, d2, . . . , dn) of nonnegative integers is said to be potentially Sr,s-graphicif there exists a realization of π containing Sr,s as a subgraph. In this paper, we obtaina Havel-Hakimi type procedure and a simple sufficient condition for π to be potentiallySr,s-graphic. They are extensions of two theorems due to A.R.Rao (The clique number ofa graph with given degree sequence, Graph Theory, Proc. Symp., Calcutta 1976, ISI Lect.Notes Series 4 (1979), 251–267 and An Erdős-Gallai type result on the clique number of arealization of a degree sequence, unpublished).Keywords: graph, split graph, degree sequenceMSC 2010 : 05C071. IntroductionA sequence π = (d1, d2, . . . , dn) of nonnegative integers is said to be graphic ifit is the degree sequence of a simple graph G on n vertices, and such a graph G isreferred to as a realization of π. The following well-known result due to Hakimi [1]and Havel [2] gives a necessary and sufficient condition for π to be graphic.Let π = (d1, d2, . . . , dn) be a non-increasing sequence of nonnegative integers.Let d′1 > d′2 > . . . > d′n−1 be the rearrangement in non-increasing order of d2 − 1,d3−1, . . . , dd1+1−1, dd1+2, . . . , dn. Then π′ = (d′1, d′2, . . . , d′n−1) is called the residualsequence of π.This work was supported by National Natural Science Foundation of China (GrantNo. 11161016) and SRF for ROCS, SEM.863Theorem 1.1 (Hakimi [1] and Havel [2]). π is graphic if and only if π′ is graphic.A sequence π = (d1, d2, . . . , dn) is said to be potentially Kr+1-graphic if there isa realization G of π containing Kr+1 as a subgraph.Definition. If π has a realization G containing Kr+1 on those vertices havingdegree d1, . . . , dr+1, then π is potentially Ar+1-graphic.In [4], Rao showed that a non-increasing sequence π = (d1, d2, . . . , dn) is poten-tially Ar+1-graphic if and only if it is potentially Kr+1-graphic. In [4], Rao consid-ered the problem of characterizing potentiallyKr+1-graphic sequences and developeda Havel-Hakimi type procedure to determine the maximum clique number of a graphwith a given degree sequence π. This procedure can also be used to construct a graphwith the degree sequence π and containing Kr+1 on the first r + 1 vertices.Let n > r + 1 and let π = (d1, d2, . . . , dn) be a non-increasing sequence of nonneg-ative integers with dr+1 > r. We construct sequences π1, . . . , πr as follows. We firstconstruct the sequenceπ1 = (d2 − 1, . . . , dr+1 − 1, d(1)r+2, . . . , d(1)n )from π by deleting d1, reducing the first d1 remaining terms of π by one, and thenreordering the last n− r− 1 terms to be non-increasing. For 2 6 i 6 r, we constructπi = (di+1 − i, . . . , dr+1 − i, d(i)r+2, . . . , d(i)n )fromπi−1 = (di − i + 1, . . . , dr+1 − i + 1, d(i−1)r+2 , . . . , d(i−1)n )by deleting di − i + 1, reducing the first di − i + 1 remaining terms of πi−1 by one,and then reordering the last n − r − 1 terms to be non-increasing.Theorem 1.2 (Rao [4]). π is potentially Ar+1-graphic if and only if πr is graphic.In [5], Rao gave a simple sufficient condition for a graphic sequence to be poten-tially Ar+1-graphic.Theorem 1.3 (Rao [5]). Let n > r + 1 and let π = (d1, d2, . . . , dn) be a non-increasing graphic sequence. If dr+1 > 2r − 1, then π is potentially Ar+1-graphic.Let Sr,s = Kr +Ks, the split graph on r + s vertices, where Ks is the complementof Ks and + denotes the standard join operation. Clearly, Sr,1 = Kr+1. Therefore,the graph Sr,s is an extension of the graph Kr+1. A sequence π = (d1, d2, . . . , dn) issaid to be potentially Sr,s-graphic if there is a realization G of π containing Sr,s as864a subgraph. If π has a realization G containing Sr,s on those vertices having degreesd1, d2, . . . , dr+s such that the vertices of Kr have degrees d1, . . . , dr and the verticesof Ks have degrees dr+1, . . . , dr+s, then π is potentially Ar,s-graphic. Yin [6] showedthat a non-increasing sequence π = (d1, d2, . . . , dn) is potentially Ar,s-graphic if andonly if it is potentially Sr,s-graphic. Related research has been done by Lai et al(see [3]). In the present paper, we develop a Havel-Hakimi type procedure (Theorem1.4) to determine whether a non-increasing sequence π is potentially Ar,s-graphic.This is an extension of Theorem 1.2 (which corresponds to s = 1). This procedurecan also be used to construct a graph with the degree sequence π and containing Sr,son the first r + s vertices.Let n > r + s and let π = (d1, d2, . . . , dn) be a non-increasing sequence of nonneg-ative integers with dr > r + s − 1 and dr+s > r. We construct sequences π1, . . . , πras follows. We first construct the sequenceπ1 = (d2 − 1, . . . , dr − 1, dr+1 − 1, . . . , dr+s − 1, d(1)r+s+1, . . . , d(1)n )from π by deleting d1, reducing the first d1 remaining terms of π by one, and thenreordering the last n− r− s terms to be non-increasing. For 2 6 i 6 r, we constructπi = (di+1 − i, . . . , dr − i, dr+1 − i, . . . , dr+s − i, d(i)r+s+1, . . . , d(i)n )fromπi−1 = (di − i + 1, . . . , dr − i + 1, dr+1 − i + 1, . . . , dr+s − i + 1, d(i−1)r+s+1, . . . , d(i−1)n )by deleting di − i + 1, reducing the first di − i + 1 remaining terms of πi−1 by one,and then reordering the last n − r − s terms to be non-increasing.Theorem 1.4. π is potentially Ar,s-graphic if and only if πr is graphic.Moreover, we also give a simple sufficient condition for a graphic sequence to bepotentially Ar,s-graphic. This is an extension of Theorem 1.3 (which corresponds tos = 1).Theorem 1.5. Let n > r + s and let π = (d1, d2, . . . , dn) be a non-increasinggraphic sequence. If dr+s > 2r + s − 2, then π is potentially Ar,s-graphic.8652. Proofs of Theorems 1.4 and 1.5P r o o f of Theorem 1.4. Assume that π is potentially Ar,s-graphic. Then π hasa realization G with a vertex set V (G) = {v1, v2, . . . , vn} such that dG(vi) = di for1 6 i 6 n andG contains Sr,s on v1, v2, . . . , vr+s so that V (Kr) = {v1, v2, . . . , vr} andV (Ks) = {vr+1, . . . , vr+s}. We now show that π has a realization G such that v1 isadjacent to vertices vr+s+1, . . . , vd1+1. If otherwise, we may choose such a realizationH of π such that the number of vertices adjacent to v1 in {vr+s+1, . . . , vd1+1} is max-imum. Let vi ∈ {vr+s+1, . . . , vd1+1} and v1vi /∈ E(H), and let vj ∈ {vd1+2, . . . , vn}and v1vj ∈ E(H). We may assume di > dj since the order of i and j can be in-terchanged if di = dj . Hence there is a vertex vt, t 6= i, j such that vivt ∈ E(H)and vjvt /∈ E(H). Clearly, G = (H \ {v1vj , vivt}) ∪ {v1vi, vjvt} is a realization ofπ such that dG(vi) = di for 1 6 i 6 n, G contains Sr,s on v1, v2, . . . , vr+s withV (Kr) = {v1, v2, . . . , vr} and V (Ks) = {vr+1, . . . , vr+s}, and G has the number ofvertices adjacent to v1 in {vr+s+1, . . . , vd1+1} larger than that of H . This contradictsthe choice ofH . Clearly, π1 is the degree sequence of G−v1 and is potentially Ar−1,s-graphic. Repeating this procedure, we can see that πi is potentially Ar−i,s-graphicsuccessively for i = 2, . . . , r. In particular, πr is -graphic.Suppose that πr is graphic and is realized by a graph Gr with a vertex set V (Gr) ={vr+1, . . . , vn} such that dGr (vi) = di for r + 1 6 i 6 n. For i = r, . . . , 1, form Gi−1from Gi by adding a new vertex vi that is adjacent to each of vi+1, . . . , vr+s and alsoto the vertices of Gi with degrees d(i−1)r+s+1 −1, . . . , d(i−1)di+1−1. Then, for each i, Gi hasdegrees given by πi, and Gi contains Sr−i,s on r + s− i vertices vi+1, . . . , vr+s whosedegrees are di+1 − i, . . . , dr+s − i so that V (Kr−i) = {vi+1, . . . , vr} and V (Ks) ={vr+1, . . . , vr+s}. In particular, G0 has degrees given by π and contains Sr,s on r + svertices v1, . . . , vr+s whose degrees are d1, . . . , dr+s so that V (Kr) = {v1, . . . , vr} andV (Ks) = {vr+1, . . . , vr+s}. P r o o f of Theorem 1.5. Let n > r + s and let π = (d1, d2, . . . , dn) be a non-increasing graphic sequence with dr+s > 2r+s−2. By Theorem 1.3, π is potentiallyAr-graphic. Therefore, we may assume that G is a realization of π with a vertexset V (G) = {v1, v2, . . . , vn} such that dG(vi) = di for 1 6 i 6 n, G contains Kr onv1, . . . , vr and M = eG({v1, . . . , vr}, {vr+1, . . . , vr+s}) (that is the number of edgesbetween {v1, . . . , vr} and {vr+1, . . . , vr+s}) is maximum. IfM = rs, then G containsSr,s on v1, v2, . . . , vr+s with V (Kr) = {v1, v2, . . . , vr} and V (Ks) = {vr+1, . . . , vr+s}.In other words, π is potentially Ar,s-graphic. Assume thatM < rs. Then there exista vk ∈ {v1, v2, . . . , vr} and a vm ∈ {vr+1, . . . , vr+s} such that vkvm /∈ E(G). LetA = NG\{v1,...,vr+s}(vk) \ NG\{v1,...,vr}(vm),B = NG\{v1,...,vr+s}(vk) ∩ NG\{v1,...,vr}(vm).866Then xy ∈ E(G) for x ∈ NG\{v1,...,vr}(vm) and y ∈ NG\{v1,...,vr+s}(vk). Otherwise, ifxy /∈ E(G), then G′ = (G\{vky, vmx})∪{vkvm, xy} is a realization of π and containsSr,s on v1, v2, . . . , vr+s with V (Kr) = {v1, v2, . . . , vr} and V (Ks) = {vr+1, . . . , vr+s}such thateG′({v1, . . . , vr}, {vr+1, . . . , vr+s}) > M,which contradicts the choice of G. Thus, B is complete. We consider the followingtwo cases.Case 1. A = ∅. Then 2r + s − 2 6 dk = dG(vk) 6 r + s − 2 + |B|, and so|B| > r. Since each vertex in NG\{v1,...,vr}(vm) is adjacent to each vertex in B and|NG\{v1,...,vr}(vm)| > 2r+s−2− (r−1) = r+s−1, it is easy to see that the inducedsubgraph of NG\{v1,...,vr}(vm) ∪ {vm} in G contains Sr,s as a subgraph. Thus, π ispotentially Ar,s-graphic.Case 2. A 6= ∅. Let a ∈ A. If there are x, y ∈ NG\{v1,...,vr}(vm) such thatxy /∈ E(G), thenG′ = (G \ {vmx, vmy, vka}) ∪ {vkvm, avm, xy}is a realization of π and contains Sr,s on v1, v2, . . . , vr+s with V (Kr) = {v1, v2, . . . , vr}and V (Ks) = {vr+1, . . . , vr+s} such thateG′({v1, . . . , vr}, {vr+1, . . . , vr+s}) > M,which contradicts the choice of G. Thus, NG\{v1,...,vr}(vm) is complete. Since|NG\{v1,...,vr}(vm)| > r + s − 1 and vmz ∈ E(G) for any z ∈ NG\{v1,...,vr}(vm),it is easy to see that the induced subgraph of NG\{v1,...,vr}(vm) ∪ {vm} in G is com-plete, and so contains Sr,s as a subgraph. Thus, π is potentially Ar,s-graphic. References[1] S. L. Hakimi: On realizability of a set of integers as degrees of the vertices of a lineargraph. I. J. Soc. Ind. Appl. Math. 10 (1962), 496–506.[2] V.Havel: A remark on the existence of finite graphs. Čas. Mat. 80 (1955), 477–480. (InCzech.)[3] C.H. Lai, L. L. Hu: Potentially Km −G-graphical sequences: a survey. Czech. Math. J.59 (2009), 1059–1075.[4] A.R.Rao: The clique number of a graph with a given degree sequence. Graph theory,Proc. Symp., Calcutta 1976, ISI Lect. Notes 4. 1979, pp. 251–267.[5] A.R.Rao: An Erdős-Gallai type result on the clique number of a realization of a degreesequence. Unpublished.[6] J.H.Yin: A Rao-type characterization for a sequence to have a realization containinga split graph. Discrete Math. 311 (2011), 2485–2489.Author’s address: J i a n - H u a Y i n, Department of Mathematics, College of Infor-mation Science and Technology, Hainan University, Haikou 570228, P. R.China, e-mail:yinjh@ustc.edu.867

A Havel-Hakimi type procedure and a sufficient condition for a sequence to be potentially $S_{r,s}$-graphic

summary:We give a necessary and sufficient condition for the existence of a tree of order $n$ with a given degree set. We relate this to a well-known linear Diophantine problem of Frobenius

Gupta, Gautam

Joshi, Puneet

Tripathi, Amitabha

Czechoslovak Mathematical JournalGautam Gupta; Puneet Joshi; Amitabha TripathiGraphic sequences of trees and a problem of FrobeniusCzechoslovak Mathematical Journal, Vol. 57 (2007), No. 1, 49–52Persistent URL: http://dml.cz/dmlcz/128153Terms of use:© Institute of Mathematics AS CR, 2007Institute of Mathematics of the Czech Academy of Sciences provides access to digitized documentsstrictly for personal use. Each copy of any part of this document must contain these Terms of use.This document has been digitized, optimized for electronic delivery andstamped with digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://dml.czCzechoslovak Mathematical Journal, 57 (132) (2007), 49–52GRAPHIC SEQUENCES OF TREES ANDA PROBLEM OF FROBENIUSGautam Gupta, Puneet Joshi, andAmitabha Tripathi, New Delhi(Received August 23, 2004)Abstract. We give a necessary and sufficient condition for the existence of a tree oforder n with a given degree set. We relate this to a well-known linear Diophantine problemof Frobenius.Keywords: graphic, tree-graphicMSC 2000 : 05C07Graphic sequences are sequences of positive integers which can be realized as thesequence of degrees of vertices of some simple graph. Graphic sequences have beenmuch studied in the literature, and some of the most well-known characterizationsof such sequences are given in [2], [4], [5]. Graphic sets are sets of positive integerswhich are formed by the degrees of vertices of some simple graph. Each set of positiveintegers is graphic, as is easy to see. For a given set S of positive integers, the leastorder of a graph for which S is its degree set was determined in [6]. This result hasbeen generalized in [1] to the determination of all n for which there exist n-vertexgraphs with degree set S.We call a sequence of positive integers tree-graphic provided it can be realizedas the degree sequence of a tree. The following result characterizes tree-graphicsequences. Although it is well-known, we give a proof for the sake of completeness.Theorem 1. Let a1, . . . , an be positive integers, with n > 2. Then there exists atree with degree sequence a1, . . . , an if and only ifn∑i=1ai = 2n− 2. . Suppose s := a1, . . . , an is a sequence of positive integers with sum2n − 2. We show, by induction on the number of terms in the sequence, that there49exists a tree with degree sequence s. The only 2-term sequence is 1, 1 and correspondsto the tree K2. Assuming the result for all such sequences with fewer than n terms,let s := a1, . . . , an be a sequence, arranged in decreasing order. Then an−1 = an = 1and a1 > 1. Consider the (n− 1)-term sequence s′ := a1 − 1, a2, . . . , an−1 of positiveintegers with sum 2n − 4. By induction hypothesis, let T ′ be a tree with degreesequence s′. The graph constructed by adding a new vertex to T ′ and adjoining itto the vertex of degree a1 − 1 is a tree with degree sequence s. This completes theinductive proof. The converse is straightforward. An immediate and easy consequence of Theorem 1 is that every nontrivial treemust have at least two vertices of degree 1 since the average degree of a vertex inany tree is less than 2 and since there cannot be an isolated vertex. Therefore, inorder to realize a set of positive integers as the degree set of a tree it is necessarythat the set have least element 1. It is easy to determine the least order of a treewith a given degree set as a consequence of Theorem 1. Later on, we will also obtainTheorem 2 as a special case of Theorem 3.Theorem 2. Let S be a set of k positive integers with least element 1 whoseelements sum to σ. Let ltree(S) be the least order of a tree with degree set S. Thenltree(S) = σ − k + 2. . Let any tree of least order have n vertices. There must be a uniquevertex corresponding to each degree > 1, and hence n − (k − 1) end-vertices. ByTheorem 1, this implies (σ − 1) + (n − k + 1) = 2n− 2, whence n = σ − k + 2. Theorem 2 can be extended to the problem of characterizing n for which there isa tree with n vertices with a given degree set S.Theorem 3. Let S = {1, a1, a2, . . . , ak−1}, with 1 < a1 < a2 < . . . < ak−1. Thenthere exists a tree of order n with degree set S if and only if the equation(1) (a1 − 1)x1 + (a2 − 1)x2 + . . . + (ak−1 − 1)xk−1 = n − 2is solvable in integers xi > 1. If S = {1}, the only tree with degree set S is K2. . The case S = {1} is easy. We henceforth assume that |S| > 1. Let T bea tree of order n, with degree set S. Let there be xi vertices of degree ai for 1 6 i 6k−1, and letm denote the number of end-vertices. Thenm = n−(x1+x2+. . .+xk−1),and by Theorem 1,a1x1 + a2x2 + . . . + ak−1xk−1 = (2n − 2)−m = n − 2 + (x1 + x2 + . . . + xk−1).Hencek−1∑i=1(ai − 1)xi = n − 2, with each xi > 1.50Conversely, suppose (1) has a solution in positive integers xi. We construct atree of order n with degree set S = {1, a1, . . . , ak−1}. Observe that K2 is the onlypossibility when S = {1}. If |S| > 1, beginning with T1 = K1,a1 , we adjoin (a1 − 1)new vertices to any one of the end-vertices to form T2, and continue this process ofadjoining (a1−1) new vertices to any one of the end-vertices in Ti to obtain Ti+1. Foreach i > 1, each vertex in Ti+1 not in Ti has degree 1, while the vertex in Ti with newvertices adjoined has degree a1 in Ti+1. Moreover, Tx1 has (a1+1)+(a1−1)(x1−1) =2 + (a1 − 1)x1 vertices. We continue this process to obtain larger and larger treesadjoining (a2−1) new vertices x2 times, then adjoining (a3−1) new vertices x3 times,and so on, until we have adjoined (ak−1 − 1) new vertices xk−1 times. The tree thusformed has 2+k−1∑i=1(ai−1)xi vertices and degree set S. This completes the constructiveproof. It is easy to see that Theorem 2 follows immediately from Theorem 3 by choosingeach xi = 1. Theorem 3 shows that the problem of characterizing n for which thereexist an n-vertex tree with degree set S is dependent on solving a linear Diophantineequation. A necessary and sufficient condition for the solvability of (1) in integers isthatd := gcd(a1 − 1, a2 − 1, . . . , ak−1 − 1) | (n− 2).Since we seek solutions with each xi > 1, the divisibility condition is only necessary.The case k = 2 is trivial; trees of order n exist if and only if n ∈ (a1 − 1)  + 2. Fork > 2, we reduce (1) to the equivalent(2) b1x1 + b2x2 + . . . + bk−1xk−1 =n− 2d,where bi = (ai−1)/d. There is no nice sufficient condition for the solvability of (2) inpositive integers for k > 3, but it is not difficult to show that there is a least positiveinteger f = f(b1, b2, . . . , bk−1) such that (2) has a solution whenever the right-handside of (2) exceeds f . Moreover, the least such positive integer in the case k = 3is given by f(b1, b2) = b1b2. This is the much studied problem of Frobenius; for anextensive study, see [3].References[1] T.S. Ahuja, A. Tripathi: On the order of a graph with a given degree set. The Journalof Combinatorial Mathematics and Combinatorial Computing 57 (2006), 157–162.[2] P. Erdös, T. Gallai: Graphs with prescribed degrees of vertices. Mat. Lapok 11 (1960),264–274. (In Hungarian.) Zbl 0103.39701[3] R.K. Guy: Unsolved Problems in Number Theory. Unsolved Problems in Intuitive Math-ematics, Volume I, Third Edition. Springer-Verlag, New York, 2004.51[4] S.L. Hakimi: On the realizability of a set of integers as degrees of the vertices of agraph. J. SIAM Appl. Math. 10 (1962), 496–506. Zbl 0168.44705[5] V. Havel: A remark on the existence of finite graphs. Čas. Pěst. Mat. 80 (1955), 477–480.(In Czech.) Zbl 0068.37202[6] S.F. Kapoor, A.D. Polimeni, and C.E. Wall: Degree sets for graphs. Fund. Math. 95(1977), 189–194. Zbl 0351.05129Authors’ address: 	   	     ,          ,              ,Department of Mathematics, Indian Institute of Technology, Hauz Khas, New Delhi-110016,India, e-mails: mau01469@ccsun50.iitd.ernet.in, mau01477@ccsun50.iitd.ernet.in,atripath@maths.iitd.ac.in.52

Graphic sequences of trees and a problem of Frobenius

Czech digital mathematics library

We give a necessary and sufficient condition for the existence of a tree of order n with a given degree set. We relate this to a well-known linear Diophantine problem of Frobenius

Gautam Gupta

Puneet Joshi

Amitabha Tripathi

CiteSeerX

GRAPHIC SEQUENCES OF TREES AND A PROBLEM OF FROBENIUS

summary:The set $D$ of distinct signed degrees of the vertices in a signed graph $G$ is called its signed degree set. In this paper, we prove that every non-empty set of positive (negative) integers is the signed degree set of some connected signed graph and determine the smallest possible order for such a signed graph. We also prove that every non-empty set of integers is the signed degree set of some connected signed graph

Czechoslovak Mathematical JournalShariefuddin Pirzada; T. A. Naikoo; F. A. DarSigned degree sets in signed graphsCzechoslovak Mathematical Journal, Vol. 57 (2007), No. 3, 843–848Persistent URL: http://dml.cz/dmlcz/128210Terms of use:© Institute of Mathematics AS CR, 2007Institute of Mathematics of the Czech Academy of Sciences provides access to digitized documentsstrictly for personal use. Each copy of any part of this document must contain these Terms of use.This document has been digitized, optimized for electronic delivery andstamped with digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://dml.czCzechoslovak Mathematical Journal, 57 (132) (2007), 843–848SIGNED DEGREE SETS IN SIGNED GRAPHSS. Pirzada, T. A. Naikoo and F. A. Dar, Srinagar(Received June 1, 2005)Abstract. The set D of distinct signed degrees of the vertices in a signed graph G iscalled its signed degree set. In this paper, we prove that every non-empty set of positive(negative) integers is the signed degree set of some connected signed graph and determinethe smallest possible order for such a signed graph. We also prove that every non-emptyset of integers is the signed degree set of some connected signed graph.Keywords: signed graphsMSC 2000 : 05C201. IntroductionAll graphs in this paper are finite, undirected, without loops and multiple edges.A signed graph G is a graph in which each edge is assigned a positive or a negativesign. These were first introduced by Harary [3]. The signed degree of a vertex viin a signed graph G is denoted by sdeg(vi) (or simply by di) and is defined as thenumber of positive edges incident with vi less the number of negative edges incidentwith vi. So, if vi is incident with d+i positive edges and d−i negative edges, thensdeg(vi) = d+i − d−i . A signed degree sequence σ = [d1, d2, . . . , dn] of a signed graphG is formed by listing the vertex signed degrees in non-increasing order. A sequenceσ = [d1, d2, . . . , dn] of integers is graphical if σ is a signed degree sequence of somesigned graph. Also, a non-zero sequence σ = [d1, d2, . . . , dn] is a standard sequenceif σ is non-increasing,n∑i=1di is even, d1 > 0, each |di| < n, and |d1| > |dn|.The following result, due to Chartrand et al. [1], gives a necessary and sufficientcondition for a sequence of integers to be graphical, which is similar to Hakimi’sresult for degree sequences [2].843Theorem 1.1. Let σ = [d1, d2, . . . , dn] be a standard sequence. Then, σ isgraphical if and only if there exist integers r and s with d1 = r − s and 0 6 s 612(n − 1 − d1) such thatσ′ = [d2 − 1, d3 − 1, . . . , dr+1 − 1, dr+2, dr+3, . . . , dn−s, dn−s+1 + 1, . . . , dn + 1]is graphical.The next characterization for signed degrees in signed graphs is given by Yan etal. [5].Theorem 1.2. A standard integral sequence σ = [d1, d2, . . . , dn] is graphical ifand only ifσ′m = [d2 − 1, . . . , dd1+m+1 − 1, dd1+m+2, . . . , dn−m, dn−m+1 + 1, . . . , dn + 1]is graphical, where m is the maximum non-negative integer such that dd1+m+1 >dn−m+1.In [4], Kapoor et al. proved that every non-empty set of distinct positive integersis the degree set of a connected graph and determined the smallest order for such agraph.2. Main resultsFirst we have the following definition.Definition. The set D of distinct signed degrees of the vertices in a signed graphG is called its signed degree set.Now, we obtain the following results.Theorem 2.1. Every non-empty set D of positive integers is the signed degreeset of some connected signed graph and the minimum order of such a signed graphis N + 1, where N is the maximum integer in the set D.P r o o f. Let D be a signed degree set and n0(D) denotes the minimum order ofa signed graph G realizing D. Since N is the maximum integer in D, therefore thereis a vertex in G which is adjacent to at least N other vertices with a positive sign.Then, n0(D) > N + 1. Now, if there exists a signed graph of order N + 1 with Das signed degree set, then n0(D) = N + 1. The existence of such a signed graph isobtained by using induction on the number of elements of D.844Let D = {d1, d2, . . . , dn}, where d1 < d2 < . . . < dn, be a set of positive integers.For n = 1, let G be a complete graph on d1 +1 vertices, that is Kd1+1, in which eachedge is assigned a positive sign. Then,sdeg(v) = (d1 + 1 − 1) − 0 = d1, for all v ∈ V (G).Therefore, G is a signed graph with signed degree set D = {d1}.For n = 2, let G1 be a complete graph on d1 vertices, that is Kd1 , in which eachedge is assigned a positive sign and let G2 be a null graph on d2−d1 +1 > 0 vertices,that is Kd2−d1+1. Join every vertex of G1 to each vertex of G2 with a positive edge,so that we obtain a signed graph G on d1 + d2 − d1 + 1 = d2 + 1 vertices withsdeg(u) = (d1 − 1) + (d2 − d1 + 1) − 0 = d2, for all u ∈ V (G1),andsdeg(v) = (0) + (d1) − 0 = d1, for all v ∈ V (G2).Therefore, the signed degree set of G is D = {d1, d2}.For n = 3, let G1 be a complete graph on d1 vertices, that is Kd1 , in which eachedge is assigned a positive sign, G2 be a complete graph on d2 − d1 + 1 > 0 vertices,that is Kd2−d1+1, in which each edge is assigned a positive sign, and G3 be a nullgraph on d3−d2 > 0 vertices, that is Kd3−d2 . Join every vertex of G1 to each vertexof G2 with a positive edge and join every vertex of G1 to each vertex of G3 with apositive edge, so that we obtain a signed graph G on d1+d2−d1+1+d3−d2 = d3+1vertices withsdeg(u) = (d1 − 1) + (d2 − d1 + 1) + (d3 − d2) − 0 = d3, for all u ∈ V (G1),sdeg(v) = (d2 − d1 + 1 − 1) + (d1) − 0 = d2, for all v ∈ V (G2),andsdeg(w) = (0) + (d1) − 0 = d1, for all w ∈ V (G3).Therefore, the signed degree set of G is D = {d1, d2, d3}.Assume that the result holds for k. We show that the result is true for k + 1.Let D = {d1, d2, . . . , dk, dk+1} be a set of k + 1 positive integers with d1 <d2 < . . . < dk < dk+1. Clearly, 0 < d2 − d1 < d3 − d1 < . . . < dk − d1. Therefore,by induction hypothesis, there is a signed graph G1 realizing the signed degree setD1 = {d2 − d1, d3 − d1, . . . , dk − d1} on dk − d1 + 1 vertices as |V (D1)| < k. LetG2 be a complete graph on d1 vertices, that is Kd1 , in which each edge is assigneda positive sign and G3 be a null graph on dk+1 − dk > 0 vertices, that is Kdk+1−dk .845Join every vertex of G2 to each vertex of G1 with a positive edge and join everyvertex of G2 to each vertex of G3 with a positive edge, so that we obtain a signedgraph G on dk − d1 + 1 + d1 + dk+1 − dk = dk+1 + 1 vertices withsdeg(u) = (di − d1) + (d1) − 0 = di, for all u ∈ V (G1) where 2 6 i 6 k,sdeg(v) = (d1 − 1) + (dk − d1 + 1) + (dk+1 − dk) − 0 = dk+1, for all v ∈ V (G2),andsdeg(w) = (0) + (d1) − 0 = d1, for all w ∈ V (G3).Therefore, the signed degree set of G is D = {d1, d2, . . . , dk, dk+1}. Clearly, byconstruction, all the signed graphs are connected. Hence, the result follows.Theorem 2.2. Every non-empty set D of negative integers is the signed degreeset of some connected signed graph and the minimum order of such a graph is |M |+1,where M is the minimum integer in the set D.P r o o f. Let D be a signed degree set and let m0(D) denote the minimum orderof a signed graph G realizing D. Since |M | is the maximum integer in D, thereforethere is a vertex in G which is adjacent to at least |M | other vertices with a negativesign. Then, m0(D) > |M | + 1. Now, if there exists a signed graph of order |M | + 1with D as signed degree set, then m0(D) = |M | + 1.Let D = {−d1,−d2, . . . ,−dn}, −d1 > −d2 > . . . > −dn, be a set of negativeintegers where d1, d2, . . . , dn are positive integers. Now, D1 = {d1, d2, . . . , dn} is aset of positive integers with d1 < d2 < . . . < dn. By Theorem 2.1, there exists aconnected signed graph G1 on dn + 1 = | − dn| + 1 vertices with signed degree setD1 = {d1, d2, . . . , dn}. Now, construct a signed graph G from G1 by interchangingpositive edges with negative edges. Then, G is a connected signed graph on |−dn|+1vertices with signed degree set D = {−d1,−d2, . . . ,−dn}. This proves the result.Theorem 2.3. Every non-empty set D of integers is the signed degree set of someconnected signed graph.P r o o f. Let D be a set of n integers. We have the following cases.Case I. D is a set of positive (negative) integers. Then, the result follows byTheorem 2.1 (Theorem 2.2).Case II. D = {0}. Then, a null graph G on one vertex, that is K1, has signeddegree set D = {0}.Case III. D is a set of non-negative (non-positive) integers. Let D = D1 ∪ {0},where D1 is a set of positive (negative) integers. Then, by Theorem 2.1 (Theo-rem 2.2), there is a signed graph G1 with signed degree set D1. Let G2 be a null846graph on two vertices, that is K2. Let e = uv be an edge in G1 with positive (neg-ative) sign and let x, y ∈ V (G2). Add the positive (negative) edges ux and vy, andthe negative (positive) edges uy and vx, so that we obtain a connected signed graphG with signed degree set D. We note that addition of such edges do not effect thesigned degrees of the vertices of G1, and the vertices x and y have signed degreeszero each.Case IV. D is a set of non-zero integers. Let D = D1 ∪ D2, where D1 is a setof positive integers and D2 is a set of negative integers. Then, by Theorem 2.1 andTheorem 2.2, there are connected signed graphs G1 and G2 with signed degree setsD1 and D2. Let e1 = uv be an edge in G1 with positive sign and e2 = xy be an edgein G2 with negative sign. Add the positive edges ux and vy, and the negative edgesuy and vx, so that we obtain a connected signed graph G with signed degree set D.We note that addition of such edges do not effect the signed degrees of the verticesof G1 and G2.Case V. D is a set of integers. Let D = D1 ∪ D2 ∪ {0}, where D1 and D2 arethe sets of positive and negative integers respectively. Then, by Theorem 2.1 andTheorem 2.2, there are connected signed graphs G1 and G2 with signed degree setsD1 and D2. Let G3 be a null graph on one vertex, that is K1. Let e1 = uv be anedge in G1 with positive sign, and let x ∈ V (G2) and y ∈ V (G3). Add the positiveedges uy and vx, and the negative edges ux and vy, so that we obtain a connectedsigned graph G with signed degree set D. We note that addition of such edges do noteffect the signed degrees of the vertices of G1 and G2, and the vertex y has signeddegree zero. This completes the proof. Theorem 2.4. If G is a signed graph with vertex set V , where |V | = r, andsigned degree set {d1, d2, . . . , dn}. Then, for each k > 1, there is a signed graph withkr vertices and signed degree set {d1, d2, . . . , dn}.P r o o f. For each i, 1 6 i 6 k, let Gi be a copy of G with vertex set Vi. Define asigned graph H with vertex set W =k⋃i=1Vi where Vi ∩ Vj = ∅ (i 6= j) and the edgesof H are the edges of Gi for all i, where 1 6 i 6 k. Then, H is a signed graph on krvertices with signed degree set {d1, d2, . . . , dn}. References[1] G.Chartrand, H.Gavlas, F. Harary and M. Schultz: On signed degrees in signed graphs.Czech. Math. J. 44 (1994), 677–690. zbl[2] S. L. Hakimi: On the realizability of a set of integers as degrees of the vertices of a graph.SIAM J. Appl. Math. 10 (1962), 496–506. zbl[3] F.Harary: On the notion of balance in a signed graph. Michigan Math. J. 2 (1953),143–146. zbl847[4] S.F.Kapoor, A.O. Polimeni and C.E.Wall: Degree sets for graphs. Fund. Math. 65(1977), 189–194. zbl[5] J.H.Yan, K.W.Lih, D.Kuo and G. J. Chang: Signed degree sequences of signed graphs.J. Graph Theory 26 (1997), 111–117. zblAuthors’ address: S . P i r z a d a, T . A . N a i k o o, F . A . D a r, Department of Math-ematics, University of Kashmir, Srinagar-190006, India, e-mail: sdpirzada@yahoo.co.in,tariqnaikoo@rediffmail.com, sfarooqdar@yahoo.co.in.848