Truszczyński, Mirosław

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Centers and centroids of unicyclic graphs

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Mathematica SlovacaMirosław TruszczyńskiCenters and centroids of unicyclic graphsMathematica Slovaca, Vol. 35 (1985), No. 3, 223--228Persistent URL: http://dml.cz/dmlcz/130902Terms of use:© Mathematical Institute of the Slovak Academy of Sciences, 1985Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access todigitized documents strictly for personal use. Each copy of any part of this document must containthese Terms of use.This paper has been digitized, optimized for electronic delivery and stampedwith digital signature within the project DML-CZ: The Czech Digital MathematicsLibrary http://project.dml.czMath. Slovaca 35,1985, No. 3, 223-228 CENTERS AND CENTROIDS OF UNICYCLIC GRAPHS MIROSLAV TRUSZCZYNSKI The terminology used in this note is standard and follows that of Harary [2]. We consider only simple graphs. The distance between vertices u and v of a graph G is the smallest number of edges in a u-v path in G and is denoted distG(u, v). The eccentricity of a vertex u in G, denoted eG(w), is the distance between u and a vertex in G, farthest from u. The subgraph of G induced by the set of vertices with minimum eccentricity is called the center of G and is denoted by C(G). The distance of a vertex u in G, denoted mG(u), is the sum of distances between u and all vertices of G. The subgraph of G induced by the vertices with minimum distance is called the centroid of G, and is denoted by M(G). For every vertex u of G and every set S of vertices of G we also define mG(u; S) = ^distG(u, v). veS The well-known Jordan's theorem identifies centers and centroids of trees. Theorem 1. (Jordan [4]) IfTisatree then C(T) = Kt orK2 andM(T) = Kt or K2. D Next results of this type, i.e. characterizing centers or centroids of graphs from a specified class appeared only one hundred years after the theorem of Jordan. Proskurowski [5], [6] characterized centers of maximal outerplanar graphs and 2-trees and Hedetniemi et al. [3] determined centers and centroids of C(„>-trees. It is the aim of this paper to find all centers and centroids of unicyclic graphs (a graph is unicyclic if it is connected and has exactly one cycle). Let C be a cycle. By °U(C) we shall denote the class of all unicyclic graphs having C as their cycle, and by $l$(C) the subclass of °U(C) containing graphs whose every vertex not in C is pendant, i.e. has its vertex degree equal to 1. First we shall state two general results. Theorem 2. (Harary and Norman [1]) The center of a connected graph G is contained in a block of G. D Theorem 3. The centroid of a connected graph G is contained in a block of G. 223 Proof. Suppose that the theorem fails and that G is a counterexample to it. Then its centroid M(G) contains two vertices v\ and v2 belonging to different blocks of G, say Bi and B2, respectively. Let u be the cutvertex of G which belongs to Bi and separates Vi and v2. Define Vi to be the set vertices containing u and the vertices of the connected component of G-u which contains vi and put V2 = (V(G)\Vi)u{w}. Finally, put fc, = distG(i),, u), / = 1,2. It is easy to see that mG(t>,; Vi)>mg(u; V ) - fc*| V | , i = 1, 2. Hence m G (u 1 ) = m G (u i ; Vi)+fci| V 2 | + m G (u ; V2) (1) > m G ( u ; Vi)-fci |Vi| + fci|V2|+mG(w; V2) = m G ( u ) - f c i ( | V i | - | V 2 | ) , and analogously m o Ы ^ m o ^ - Ы І V ^ I - І V , ! ) . (2) Since vteM(G), m G (u)^m G (t>,), i = l , 2 , and consequently (1) and (2) imply | Vi | > | V21 and | V21 > | Vi |, respectively. This contradiction completes the proof. D For unicyclic graphs Theorems 2 and 3 imply the following corollaries. Corollary 4. If Ge °U(C) then C(G) = Ki or K2, or C(G)cC. B Corollary S. If Ge °U(C) then M(G) = Ki or K2, or M(G)cC. Q Corollary 6. If Ge °U$(C) then M(G)cC. D We shall now determine all induced subgrahps of a cycle C which are the centers of unicyclic graphs from °ll(C). The collection of all such subgraphs will be denoted by <g(C). Theorem 7. (a) If \C\ is even then ^ ( C ) consists of all induced subgraphs of C. (b) If | C\ is odd and H cz C then H e ^ ( C ) if and only if for every path uvw in C, v is in H whenever u and w are in H. Proof, (a) Let H be an arbitrary induced subgraph of C. Put A = {x eC: the farthest vertex from JC in C is not in H} and define a unicyclic graph G by adding | A | new vertices xa, a e A, to C and joining each xa to a. It is readily verified that H=C(G) (see Figure 1). Figure 1. V(H) = {x0, x3y JC5}, A = {xi, x4, x 5 }; the center is shown in dark vertices and bold edges. 224 (b) Suppose first that He ^ ( C ) and let G be a unicyclic graph with a cycle C such that H= C(G). Let uvw be a path of C and let u and w be in H. We shall prove that v is in H, too. Let z be a vertex in G, farthest from v and let P be a shortest path between v and z. There are three possibilities. 1. u $ P, w £ P. In this case distG(w, z) = distG(t>, z) + 1 and, consequently, eG(u)^eG(v) + 1. Hence u£C(G) = H contrary to the assumption. 2. u e F , w^P. In this case distG(w, z)-^distG(f, z) = eG(u), which in turn implies that eG(w)^eG(v). Since w e C ( G ) , it implies that eG(w) = eG(v) and conse-quently v e C(G). 3. w ^ P , w e P . This case can be dealt with as the previous one. Now suppose that for every path uvw of C, u, w eH implies veH. For every x e C let Fx be the set consisting of the two vertices of C which are farthest from JC, and let A = {xe C: Fx c V(C)\V(H)}. Since for every vertex u e V(C)\V(H) at least one of its neighbours is also in V(C)\V(H)9 it follows that V ( C ) \ V ( H ) = U F , x e A Hence the graph G obtained by adding \A \ new vertices jcfl, a e A, to C and joining each xa to a has clearly H as the center (see Figure 2). El Figure 2. V(H) = {JC2, x5}, A = {JC0, JC3, JC4} ; the center is shown in dark vertices. Now we pass on to centroids. Similarly as in the case of centers we define M(C) to be the collection of all induced subgraphs of C which are centroids of unicyclic graphs from °U(C). Theorem 8. (a) If \ C\ is odd then M(C) consists of all induced subgraphs of C. (b) If \C\ is even and H^C then HeM(C) if and only if either H=C or H contains no pair of antipodic vertices of C, i.e. points at distance | C|/2 from each other. Proof. We begin with some simple observations. Suppose that C = x0xi...xm-ix0 and G e °U(C). For 0 ^ i ^ m — 1 let T.be the set of vertices of the maximal tree in G which contains JC, and no other vertices of C. (i) Obviously, mG(jc,) = mG(jc,; T;) + mG(x,; Tl+i) + . . . + m G ( x , ; Ti+m-i) 225 (throughout the proof additions of indices are modulo m) and m ( v . T \-í p | T / + p | + m0(jc,+ P; T I + P) o t " í i + p j ~ l ( m - p ) | T i + P | + m G (x, + P ; T, These two facts imply that c + |T,+, | + 2 |T l + 2 | + ... + fc|T,+t| + ( f c - l ) | T , + t + I | + ... + |T,+ 2*-i| c + |T,+, | + 2 | T , „ | + ... + fc|T,+*| + k\T,+k+1\ + ... + \Ti+2k\ mG(x ŕ) = p ^ m / 2 +p) p>m/2. m = 2fc m = 2fc + l, where c = m G (x 0 ; T0) + . . . + m G ( x m _ 1 ; Tm_i). Now, let Fe tf/(C) be a unicyclic graph which differs from G only in the number of pendant vertices at each JC,, O^i^m — 1. Denote by bt the difference between the numbers of pendant vertices adjacent to JC, in F and G, respectively. The following three observations are simple consequences of (i). (ii) If m = 2fc + l, then for i = 0, 1, ...,2fc mF(jc,) - mG(jc.) = bi + 2bt+1 + ... + (fc + l)bt+k + (k + l)bl+k+l + ... + 2bl+2k. In particular, if fc^2, b, = 2p, bi+l = — 2p, b]+k = p, b]+k+2= —p and all other fc.'s are equal to 0 then mF(xj) — mG(xj) = — p, m F (c ; + 1 ) — mG(jc;+1) = p and mF(jc l)~mG(jc1) = 0, for i + \, j+1. (iii) If m = 2fc, then for i = 0, 1, ..., 2 f c - l mF(xt) - mG(xt) = bi + 2bt+l + ... + kbt+k-\ + (k + l)bt+k + kbt+k+l + ... + 2bI+2fc_i. In particular, if fc ^ 3, b, = b]+k-\ = p, bi+\ = bi+k = —p and all other b,'s are equal to 0, then m F ( jc ; )- mG(jc;) = - 2 p , mF(jc;+fc) - mG(x ; + f c) = 2p and mF(jc.) - mG(xi) = 0 for i±j, j + fc. (iv) If b0= ... = bm-\, then mG(jc l)-mG(jc ;) = mF(jc I)-mF(jc ;) for every O ^ i , j^m — 1. In particular, M(G) = M(F). We are ready to prove the theorem. In the proof we shall use the symbol m(G) to denote the minimum distance of a vertex in G,i.e. m(G) = min{mg(u): ueG}. (a) Let m = 2fc + l.Iffc = l, then suitable unicyclic graphs are shown in Figure 3. V Figure 3. The centroids are indicated by dark vertices and hold lines. So, assume that fc^2. Instead of proving (a) we shall prove the following stronger statement: for every induced subgraph H of C there is G e °U$(C) such 226 that M(G) = H. We apply induction on \H\. If \H\ = 1, say V(H) = {JC,}, then the graph G obtained from C by adding a new vertex y, and joining it to JC, has H as its centroid and, obviously, belongs to 9l$(C). Suppose now, that \H\ = n>l and the assertion holds for subgraphs of order less than n.lf n = 2k-\-1, i.e. if H = C, then C is a graph in $l#(C) having C as its centroid. So, let n < 2k + 1 and JC;- be a vertex of H such that JC ; + I^H. Define H' = H —JC7. Then, by the induction hypothesis, there is a graph G' e °UJ>(C) such that M(G') = H'. Since JC, £ H', mG (jc7)>m(G'). Put r = mG (jc ;)-m(G'). Let G" be a graph obtained from G' by joining to each vertex of C 2r new pendant vertices. By (iv) M(G') = M(G") = H' and mG»(jc7) - m(G ) = r. Let X (resp. Y) be a set consisting of any 2r (resp. r) pendant vertices adjacent to jci+i (resp. to jc7+fc+2) in G . Denote by G the graph obtained from G" by deleting all edges JC,+IJC, JCGX, and jc;+fc+2y, yeY, and joining the vertices of X to JC, and the vertices of Y to jc7+fc. It follows from Corollary 5 and (ii) that M(G) = H. (b) Let m = 2k. Suppose first that HeM(C) and H=£ C. Furthermore, suppose that H contains at least one pair of antipodic vertices of C. Since H+ C, it follows that there is /, 0 ^ / ^ 2 f c - l , such that jc.eH, jc I + f ceH and jc l+fc+i^H. Let us consider a unicyclic graph Ge°U(C) such that M(G) = H. Clearly, mG(jc,)^ mG(jcI+i) and mG(*,+fc)<me(jcI+fc+i). Hence, by (i), and |T,+i| + ... + |T,+ ł,|^ |T,+*+,| + . . . + \TMk\ |т,+ł+,| + ... + |т,+2*|<|т,+,| + ... + |т1+*|, which yields a contradiction (7Vs are defined as at the beginning of the proof). Suppose now, that H cz C contains no pair of antipodic vertices of C (the case H = C is trivial). If k = 2, then suitable unicyclic graphs are shown in Figure 4. \ — / Figure 4. The centroids are indicated by dark vertices and hold lines. So, let us suppose that fc^3. As in (a), we shall prove the following stronger statement: for every H^C which contains no pair of antipodic vertices of C there is G e °U^(C) such that M(G) = H and mG(jc,)-mG(jc;) is even for every O^i, j^m — 1. The assertion holds if |H | = 1, say V ( H ) = { J C , } , since the graph G obtained from C by joining to JC, two new vertices y, and y\ satisfies it, and the induction step follows from Corollary 6, (iii) and (iv) similarly as in (a). • 227 REFERENCES [1] HARARY, F—NORMAN, R. Z.: The dissimilarity characteгisticof Husimi tгees. Ann Math. 58, 1953, 134—141. [2] HARARY, F.: Graph Theory. Addison-Wesley, Reading, MA, 1969. [3] MITCHEL HEDETNIEMI, S. L —HEDETNIEMI, S. T.—SLATER, P. J.: Centers and medians of C(я)-tгees. Utilitas Math. 21C, 1982, 225—235. [4] JORDAN, C : Sur les assemblages des lignes. J. Reine Anew. Math. 70, 1869, 185—190. [5] PROSKUROWSKI, A.: Centeгs of maximal outerplanaг graphs. J. Graph Theoгy. 4, 1980, 75—79. [6] PROSKUROWSKI, A.: Centeгs of 2-trees. Ann. of Discrete Math. 9, 1980, 1—5. Received February 9, 1983 Institute of Mathematics Technical Univeгsity of Waґsaw p. Jedności Robotniczej 1 00-661 Waгsaw, Poland ЦЕНТРЫ И ЦЕНТРОИДЫ УНИЦИКЛИЧЕСКИХ ГРАФОВ Мгго51а\у Тги$2С2уп8к1 Р е з ю м е В настоящей работе характеризуются центры и центроиды унициклических графов. 228 

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Centers and centroids of unicyclic graphs

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