THE BALANCED DECOMPOSITION NUMBER AND VERTEX CONNECTIVITY

THE BALANCED DECOMPOSITION NUMBER AND VERTEX CONNECTIVITY SHINYA FUJITA AND HENRY LIU Astract The alanced decomposition numer f(g) of a graph G was introduced y Fujita and Nakamigawa [Discr Appl Math, 156 (008), pp 3339-3344] A alanced colouring of a graph G is a colouring of some of the vertices of G with two colours, such that there is the same numer of vertices in each colour Then, f(g) is the minimum integer s with the following property: For any alanced colouring of G, there is a partition V (G) = V 1 V r such that, for every i, V i induces a connected sugraph, V i s, and V i contains the same numer of coloured vertices in each colour Fujita and Nakamigawa studied the function f(g) for many asic families of graphs, and demonstrated some applications In this paper, we shall continue the study of the function f(g) We give a characterisation for non-complete graphs G of order n which are n -connected, in view of the alanced decomposition numer We shall prove that a necessary and sufficient condition for such n -connected graphs G is f(g) = 3 We shall also determine f(g) when G is a complete multipartite graph, and when G is a generalised Θ-graph (ie, a graph which is a sudivision of a multiple edge) Some applications will also e discussed Further results aout the alanced decomposition numer also appear in two susequent papers of Fujita and Liu Key words Graph decomposition, vertex colouring, k-connected AMS suject classifications 05C15, 05C40, 05C70 1 Introduction In this paper, we only consider finite undirected graphs For such a graph G, its vertex set and edge set are denoted y V (G) and E(G) respectively Our definitions concerning graphs throughout the paper are fairly standard For a graph G and U V (G), the sugraph of G induced y U is denoted y G[U] The graph G U is the sugraph of G induced y V (G) \ U We write G u for G {u} The open neighourhood of U is N(U) = {v V (G U) : vu E(G) for some u U} The set U is a cut-set of G if G U is a disconnected graph For k N, G is a k- connected graph if V (G) k + 1, and G has no cut-set of size at most k 1 Every non-empty graph is 0-connected The maximum k for which G is k-connected is the connectivity of G, and is denoted y κ(g) For U, W V (G) and U W =, we write E(U, W ) for the edges of G which intersect with oth U and W We write E(u, W ) for E({u}, W ) A U W path of G is a su-path where one end-vertex is in U, the other is in W, and no other vertex (if any) is in U W Finally, for u, v V (G), the distance from u to v in G is denoted y d G (u, v) We refer the reader to [1] or [4] for any undefined terms In [9], Fujita and Nakamigawa introduced the alanced decomposition numer of a graph For a graph G with V (G) = n N, a alanced colouring of G is a pair (R, B), where R, B V (G), R B =, and 0 R = B n We shall refer the vertices of R (B) as the red (lue) vertices, those of V (G) \ (R B) Date: Octoer 3, 010 Department of Mathematics, Gunma National College of Technology, 580 Toria, Maeashi 371-8530, Japan (fujita@natgunma-ctacjp) Research supported y JSPS grant (no 0740068) Departamento de Matemática, Faculdade de Ciências e Tecnologia, Universidade Nova de Lisoa, Quinta da Torre, 89-516 Caparica, Portugal (hliu@fctunlpt) Research partially supported y Financiamento Base 008 ISFL-1-97 from FCT/MCTES/PT 1

SHINYA FUJITA AND HENRY LIU the uncoloured vertices, and those of R B the coloured vertices A set U V (G) is a alanced set if U R = U B, and G[U] is connected A alanced decomposition of G, or of (R, B), is a partition V (G) = V 1 V r (for some r 1), such that each V i is a alanced set We may also write the alanced decomposition as P = {V 1,, V r } The size of P is then defined as the maximum of V 1,, V r Now, oserve that, if G is a disconnected graph, then we can take a alanced colouring so that G does not have a alanced decomposition at all, simply y colouring one vertex in one component red, and another vertex in another component lue So, from now on, we shall only consider alanced decompositions for connected graphs Let G e a connected graph on n vertices, and k Z, 0 k n We define f(k, G) = min{s N : every alanced colouring (R, B) of G with R = B = k has a alanced decomposition of size at most s} Note that f(k, G) n, so f(k, G) is well-defined The alanced decomposition numer of G is then defined as { n } f(g) = max f(k, G) : 0 k Fujita and Nakamigawa [9] studied the function f(g) in many directions Among these were the following A graph G with f(g) = if and only if G is a complete graph with at least two vertices f(k m,n ) = n m + 3, if K m,n is the complete ipartite graph with m n The result that f(c n ) = n + 1, if C n is the cycle graph on n vertices, and the conjecture that f(g) n + 1, if G is a -connected graph on n vertices In this paper, we shall prove further results in the direction of each of the aove In Section, we characterise connected graphs G with alanced decomposition numer 3 In Section 3, we determine the alanced decomposition numer of complete multipartite graphs, which extends the complete ipartite graph result In Section 4, we prove an asymptotically tight ound for f(g) when G is a generalised Θ-graph, which is a graph otained y sudividing a multiple edge As we will oserve from our results, the alanced decomposition numer seems to have a deep relationship with vertex connectivity of graphs In Section, we show that a graph G of order n is n -connected if and only if f(g) = or 3 This result will point us to a new direction for the study of vertex connectivity in graphs The prolem of finding non-trivial characterisations for a graph to e k-connected has een well studied, with the cases k = (Whitney [17]) and k = 3 (Tutte [16]) the most well known (see also, eg, Sections 31 and 3 of [4]) On the other hand, if a graph G does not have high connectivity, say, G is -connected, then f(g) is likely to e large (see the aove conjecture for -connected graphs) In view of this, we elieve that the alanced decomposition numer can e a new criterion to measure the connectivity of a graph In Section 5, we propose a prolem aout the relationship etween the alanced decomposition numer and connectivity Also, some applications are discussed We shall see that, we can decide whether a graph satisfies f(g) {, 3} or not with an algorithm in polynomial time We shall also have a discussion aout the relation of the alanced decomposition numer with non-separating sugraphs In two susequent papers [7, 8], more results aout the alanced decomposition numer are proved These include further applications [7], as well as partial results of the aove conjecture, in the cases when the graph is a sudivided K 4, and when it is a -connected series-parallel graph [8] We first recall a trivial remark from [9] Graphs G with f(g) = 3 Proposition 1 (Remark of [9]) Let G e a connected graph with at least vertices Then f(g) = if and only if G is a complete graph

BALANCED DECOMPOSITION NUMBER AND CONNECTIVITY 3 So it is natural to ask: Which connected graphs G have f(g) = 3? With f(g) small, we would expect such graphs G to have many edges So, these graphs may conceivaly e highly connected as well Before we proceed, we need Theorem elow, which is a well known consequence of Menger s theorem (see, for example, [1], Ch III, Corollary 6, which implies Theorem ) Theorem Let k N, and let G e a k-connected graph with V (G) k Then, for any U, W V (G) with U W = and U = W = k, there exist k vertex disjoint U W paths Now, here is the characterisation result Theorem 3 Let G e a connected graph with n 3 vertices and G K n Then f(g) = 3 if and only if G is n -connected Proof Suppose that f(g) = 3, and assume for contradiction that G is not n -connected Then, let C V (G) e a cut-set with C = c n 1, and let V (G C) = P Q, where P = p > 0, Q = q > 0, and E(P, Q) = Assume without loss of generality that p q Then, p 1 (n c) 1 (c+ c) = c +1 If q c + 1, then we take the alanced colouring (R, B) with P R = c + 1, C R = c, Q B = c + 1, and C B = c, so that R = B = c + 1 n If q c, then p = n q c c + q c = c q + We take the alanced colouring (R, B) where P R = c q +, C R = q 1, Q B = q, and C B = c q + 1 So again, R = B = c + 1 n Note that c q +, c q + 1 q + 1 > 0 In either case, it is easy to check that f(g) 4 Indeed, for any such alanced colouring, if we can find a alanced decomposition P of size at most 3, then we cannot have two coloured vertices of the same colour in any memer of P So, the vertices of P R are in distinct memers of P, say, A 1,, A P R Also, for every i, A i cannot contain a vertex of Q B Otherwise, since E(P, Q) =, we would have A i also containing a vertex of C, which is always coloured So, A i must contain a vertex of C B Now, distinct A i and A j must contain distinct vertices of C B, and this is impossile since P R = C B + 1 Hence, G is n -connected Conversely, suppose that G is n -connected Since G K n, y Proposition 1, we have f(g) 3 We shall show, y downward induction on k, that f(k, G) 3 for every k n, which will suffice Firstly, for k = n, suppose that we have a alanced colouring (R, B) with R = B = n Since G is n -connected, applying Theorem with k = n, U = R and W = B gives f( n, G) 3 Now, suppose that the implication holds for l + 1 k n, ut not for k = l That is, l is the maximal integer such that f(l + 1, G) 3 and f(l, G) > 3 Let (R, B) e a alanced colouring of G with R = B = l, and let U = V (G) \ (R B) e the uncoloured vertices of (R, B) Let (R, B ) e a alanced colouring of G, where R = R {y }, B = B {x }, and y, x U In other words, to get the sets R and B, we take vertices y, x U, colour y red and append it to R, and, colour x lue and append it to B By the induction hypothesis, there exists a alanced decomposition P for (R, B ) of size at most 3 We may assume without loss of generality that the structure of P is as follows A set of size induces an edge of G with one end-vertex in R, the other in B ; call these sets P We may assume that P is maximal, so that a set of size 3 induces a path of length in G with one end-vertex in R, the other in B (so there is no edge of G joining these end-vertices); call these sets P 3 A set of size 1 is a vertex of U \ {x, y } If y and x are in the same alanced set of P, then P will also e a alanced decomposition of size at most 3 for (R, B) So, we may assume that x A r and y A, where A r, A P are distinct, and A r R = {x}, A B = {y}, for some x R and y B Define P i = P i \ {A r, A } for i =, 3, and U 1 = U \ {A : A P 3 } We have the following oservation Oservation 1 V (G {x, y}) is a disjoint union of memers of P, P 3 and U 1 For a set C P 3, let {c r } = C R, {c } = C B, and {c u } = C U Now, we construct a sugraph H G containing x and y Moreover, we shall derive a partition V (H) = X Y Z which we will need later in the proof The sugraph H will contain many trees in a specific form which we shall descrie first

4 SHINYA FUJITA AND HENRY LIU Suppose that we have susets Q P, Q 3 P 3 and U U 1 Suppose that we also have a red vertex w R \ {A : A Q Q 3 } We shall grow a tree y successively attaching elements of Q, Q 3 and U to what we already have This will e ased on two operations Start with w If at some stage we have the tree T, then we can form a new tree T T y doing one of the following If u V (T ) R, then join v U \ V (T ) to u if vu E(G) That is, T = T vu If u V (T ) (R U), A Q Q 3, A V (T ), and {v} = A B, then unite G[A] with T y joining vu, if vu E(G) That is, T = T G[A] vu We shall call a tree that can e constructed y successively performing these two operations in some order a red tree, and such a red vertex w is the seed of the red tree Denote this red tree y R w Similarly, switching the roles of red and lue, we call a tree B w that can e constructed from a lue vertex w B \ {A : A Q Q 3 } as a result a lue tree Examples of these trees are shown in Figure 1 Groups of vertices in the oxes are elements of P and P 3 The uncoloured vertices not inside the oxes are elements of U 1 The vertices w and w are the seeds In susequent diagrams, red, lue and uncoloured vertices will e depicted as in Figure 1 w w Red vertex Blue vertex Uncoloured vertex R w B w Figure 1 A red tree and a lue tree So with this, using Q = P, Q 3 = P 3 and U = U 1, we can construct a red tree R x with seed x, and a lue tree B y with seed y Choose R x and B y such that V (R x ) V (B y ) =, and V (R x ) V (B y ) is maximal Let G 0 = R x B y (so, G 0 is a disconnected graph with V (G 0 ) = V (R x ) V (B y )) Now, suppose that we have constructed a sugraph G with G 0 G G satisfying the following properties (i) There are families of red trees and lue trees, R and B, and some C P 3 such that the following hold R x R, B y B, and the memers of R B are vertex disjoint For every T R B and C C, we have C {C P 3 : C V (T )} For all C C, we have V (R x ) C = V (B y ) C = If T (R B) \ {R x, B y } and w is the seed of T, we have V (T w) C = for every C C, and there is a unique C C such that V (T ) C = {w} V ( G) = {V (T ) : T R B} {C : C C } Hence, V ( G {x, y}) is a union of memers of P, P 3 and U 1 Note that, if we have C C and c r (c ) has no red (lue) tree with order at least constructed onto it, then c r (c ) may or may not e a red (lue) tree itself, and we must declare such a status for c r (c )

BALANCED DECOMPOSITION NUMBER AND CONNECTIVITY 5 (ii) (Adjacency conditions) If C C, then in G, one of the following holds (a) c u sends at least one edge of G to a red vertex in some red tree in R \ {R cr } Moreover, no edge of G joins c u to any lue vertex of any lue tree in B, and c is not the seed of a lue tree () c u sends at least one edge of G to a lue vertex in some lue tree in B \ {B c } Moreover, no edge of G joins c u to any red vertex of any red tree in R, and c r is not the seed of a red tree (c) c u sends at least one edge of G to a red vertex in some red tree in R \ {R cr }, and to a lue vertex in some lue tree in B \ {B c }, and c r and c are oth seed vertices, of a red tree in R and a lue tree in B, respectively We partition C = C r C C as follows C r = {C C : c u satisfies (a) aove}, C = {C C : c u satisfies () aove}, C = {C C : c u satisfies (c) aove} In addition, the following hold (d) In G, every C 1 C r C has the following property For some t 1, there is a sequence of distinct sets C 1,, C t C r C such that for every i, we have a red tree R c i r R, and c i u sends an edge of G to V (R c i+1) R, where y convention, R r c t+1 = R r x (e) A similar statement to (d) holds when we switch the roles of red and lue, x and y, R and B, and C r and C (iii) (Maximality condition) Every red tree and lue tree of G cannot e extended in the following sense: There is no element of {C P P 3 U 1 : C V ( G)} that can e appended to any red tree or lue tree, in accordance to the rules of the construction of the red trees and lue trees Note that G = G 0 satisfies (i) and (ii) vacuously, and also (iii) Figure shows a possile structure of G Each large ox represents a red tree or a lue tree, with the seed vertex at the top of each tree For the oxed 3-sets, we have C 1, C 4 C r, C 3, C 7 C, and C, C 5, C 6 C R x x y B y z 1 C 1 C C 3 z z 3 C 4 C 5 C 6 C 7 Figure The structure of G

6 SHINYA FUJITA AND HENRY LIU Now, we want to extend G to some sugraph Ĝ with G Ĝ G We shall do this with the following algorithm Step 1 Let G G 0 satisfy properties (i) to (iii), with R, B, C r, C and C defined as efore Now, suppose that we have C P 3, C V ( G) with c u sending an edge of G to a red vertex in some red tree in R, or to a lue vertex in some lue tree in B, or oth For the first case, append the edges {c u w E(G) : w V (T ) R for some T R} to G Using the memers of P P 3 U 1 availale to us; that is, the family {A P P 3 U 1 : A V ( G) C}, construct a red tree R cr, while c is left alone and is not considered as a lue tree Choose R cr so that V (R cr ) is maximal Set R 0 = R {R cr }, B 0 = B, C r 0 = C r {C}, C 0 = C and C 0 = C Similarly, for the second case, append the edges {c u w E(G) : w V (T ) B for some T B} to G Using the memers of {A P P 3 U 1 : A V ( G) C}, construct a lue tree B c, while c r is left alone and is not considered as a red tree Choose B c so that V (B c ) is maximal Set R 0 = R, B 0 = B {B c }, C r 0 = C r, C 0 = C {C} and C 0 = C For the last case, append the edges {c u w E(G) : w V (T ) R for some T R} and {c u w E(G) : w V (T ) B for some T B} to G, and note that there is at least one edge of each type We then construct a red tree R cr and a lue tree B c, again using memers of {A P P 3 U 1 : A V ( G) C}, so that V (R cr ) V (B c ) = and V (R cr ) V (B c ) is maximal Set R 0 = R {R cr }, B 0 = B {B c }, C r 0 = C r, C 0 = C and C 0 = C {C} In all three cases, let G 0 e the new graph otained If we cannot perform Step 1, then we set Ĝ = G Step Now, suppose that, for some t 0, we have found the families R i, B i, C r, i C i, C i, with C r i C i C i = C r C C {C}, and the graphs G i, for 0 i t, ut not for i = t+1, with G 0 G G 0 G t Suppose that G t partially satisfies properties (i) to (iii) as follows Instead of properties (ii)(a-c), assume that the following is satisfied instead (ii ) In G t, one of the following holds (a ) If C C r, t then c u sends at least one edge of G to some red vertex of some red tree in R t \ {R c r }, and c is not the seed of a lue tree in B t ( ) If C C t, then c u sends at least one edge of G to some lue vertex of some lue tree in B t \ {B c }, and c r is not the seed of a red tree in R t (c ) If C C t, then c u sends at least one edge of G to a red vertex in some red tree in R t \ {R c r }, and to a lue vertex in some lue tree in B t \ {B c }, and c r and c are oth seed vertices, of a red tree in R t and a lue tree in B t, respectively As for properties (i) and (iii), they are all satisfied with G t, R t, B t, C r, t C t and C t in place of G, R, B, Cr, C and C Let (i ) and (iii ) e the modified properties Note that of course G 0 satisfies the properties (i ) to (iii ) Now, suppose that we have E(V (R dr ) R, d u) for some D C r t C t, some red tree R dr R t, and some D C t (If not, then set t = t and go to Step 3) Add all edges of E(V (R dr ) R, d u) to G t t+1 Now, set C r = C r, t Ct+1 = C t \ {D } and C t+1 = C t {D } Next, construct a new red tree R d r with elements from {A P P 3 U 1 : A V ( G t )}, and with V (R d r ) maximal Set R t+1 = R t {R d r } and B t+1 = B t Let G t+1 G t e the new graph otained Note that indeed, we have C r 0 t+1 = = C r It is easy to check that G t+1 also satisfies the properties (i ) to (iii ) Repeat this procedure successively, starting with G 0 Note that the construction of new red trees enhances this possiility For some t 0, we must have E(V (S) R, d u ) = for all red trees S R t t (including S = R x ) and D C This is ecause in each step, we are moving a set from some C t to some C t These sets are not moved again, so this procedure cannot last forever

BALANCED DECOMPOSITION NUMBER AND CONNECTIVITY 7 Step 3 Next, we do a similar thing as Step, switching the roles of red and lue We now have a graph G t Suppose that for some t t 0, we have found R i, B i, Ci r, Ci, Ci and G i for 0 i t, ut not for i = t + 1, with G 0 G G 0 t G G t Moreover, suppose that G t satisfies the properties (i ) to (iii ) Now, suppose that E(V (B d ) B, d u) for some lue tree B d B t, D C t C t, and D C r t Carry out Step in an analogous manner, switching the roles of red and lue, and C r t and C t We otain Rt+1, B t+1, Ct+1 r, Ct+1, Ct+1 and G t+1 G t, again satisfying the properties (i ) to (iii ) Note that at this stage, it is not possile to apply Step, since we do not t t+1 construct any new red trees in Step 3, and indeed, we have C = = C So, starting with G t, repeat Step 3 successively until it stops For some t t 0, we otain R t, B t t t, C r, C, C t and G t such that, E(V (S) R, d u ) = for all red trees S R t t and D C ; E(V (T ) B, d u) = for all lue trees T B t and D t C r So now, the graph G t satisfies the properties (i) to (iii) that the graph G at the eginning satisfied Now, we set Ĝ = G t Roughly speaking, in Step 1 we attempt to extend a graph G that we already have to a new graph G 0, y adding a permissile element of P 3 and constructing new red and lue trees In Steps and 3, we tidy up the new graph G 0 y moving vertices and constructing more red and lue trees, so that we end up with Ĝ having a similar structure to the previous graph G Now, starting with G 0, with C r = C = C =, run the aove algorithm successively, replacing Ĝ y G each time we move from Step 3 ack to Step 1 This process must terminate at some stage, ecause we are using more and more elements of P 3 every time we apply Step 1 When we cannot apply Step 1, let H e the final state of the graph G, and let C r, C and C e the final states of C r, C and C We have now constructed a graph H with G 0 H G Moreover, H satisfies properties (i) to (iii) (with H in place of G) Now, let X (Y ) e the union of the vertex sets of the red (lue) trees in H with {C : C C r } ( {C : C C }), and Z = {{c u } : C C} Note that we have V (H) = X Y Z Let W = U \ V (H), and W = (R B) \ V (H) If (X W ) U Y U, set X = X W and Y = Y W Otherwise, set X = X W W and Y = Y Note that V (G) = X Y Z Our aim now will e to delete a cut-set of G of size at most n 1, which will e a final contradiction This cut-set will e (X (B U)) (Y R) Z if X U Y U, and (X B) (Y (R U)) Z if X U > Y U We must therefore prove certain non-adjacencies in G In order to tackle this, we shall digress and descrie a special type of tree which will e crucial to our discussion For this, we shall forget aout G for the moment, as well as R, B and U Let F e a family of edges and F 3 e a family of paths of length, where each memer has one end-vertex coloured red, the other coloured lue, and in the case of a memer of F 3, the middle vertex uncoloured Let F 1 e a set of uncoloured vertices Also, the memers of F, F 3 and F 1 are mutually vertex disjoint, and we may think of each of F, F 3 and F 1 having infinitely many memers Let w (w ) e another red (lue) vertex Let R, B and Ũ denote the red, lue and uncoloured vertices We say that a tree T with at least two vertices is alternating if T can e constructed as follows Start with the vertex w R We will now successively append memers of F F 3 F 1 Suppose that at some stage, we have constructed a tree S, and A F F 3 F 1 {w} is the last sugraph appended to S Then, we can otain a new tree S S y doing either one of the following If we have u V (A) R, we may join v F 1 \ V (S) to u That is, S = S vu If we have u V (A) ( R Ũ), C F F 3, C S, and {v} = V (C) B, we may unite C with S y joining vu That is, S = S C vu We do this successively, and stop at any point we wish Let T r e the final tree otained, and C r e the final sugraph appended, where C r F F 3 F 1 {w} Now, repeat the aove two steps, starting with the vertex w B, switching the roles of red and lue, and that any sugraph that we append is not involved in the construction of T r We otain a similar tree T, vertex disjoint from T r Let C e the final sugraph appended, where C F F 3 F 1 {w } Furthermore, assume that we do not have oth C r F 1 and C F 1

8 SHINYA FUJITA AND HENRY LIU If we have z V (C r ) ( R Ũ) and z V (C ) B, or z V (C r ) R and z V (C ) Ũ (note that such z and z always exist), then we unite T r and T y joining zz Let T e the tree otained That is, T = T r T zz Also, we call T r a red alternating tree, and the red vertex w is the seed of T r Similarly, T is a lue alternating tree, with seed w See Figure 3 The sugraphs of order and 3 in oxes are elements of F and F 3 The uncoloured vertices not in the oxes are elements of F 1 T r w w C r z C z T Figure 3 An alternating tree We have the following claim Claim ( R, B) is a alanced colouring for T, and ( R, B) has a alanced decomposition of size at most 3 Proof Let F, F 3, F 1, w, w, T r, T, C r and C e defined as efore Oviously, R = B, since V (T r ) R = V (T r ) B + 1 and V (T ) B = V (T ) R + 1 We define a linear ordering r on V (T r ) as follows We have a r a if either d Tr (w, a) < d Tr (w, a ), or d Tr (w, a) = d Tr (w, a ) and a R, a B Note that the only way to have d Tr (w, a) = d Tr (w, a ) is when we join the uncoloured vertex in some C F 3 to a lue vertex in some D F F 3, whence {a} = V (C) R and {a } = V (D) B Now, let s = V (T r ) R 1 = V (T r ) B 0 Suppose that, starting with w, as we move along V (T r ), with respect to r, the coloured vertices that we come across are, in order, c 1,, c s+1 We have c 1 = w It is easy to see, for example, y induction on the numer of elements of F F 3 F 1 {w} used in T r, that c 1, c 3,, c s+1 R and c, c 4,, c s B If s 1, then let K V (T r ) e the vertices coming after c s We otain a alanced decomposition for T r K as follows For 1 i s, let [c i 1, c i ] = { V (ci 1 c i ), if d Tr (w, c i 1 ) < d Tr (w, c i ), {u, c i 1, c i }, if d Tr (w, c i 1 ) = d Tr (w, c i ), where c i 1 c i is the su-path of T r with end-vertices c i 1 and c i (which has order at most 3), and u is the uncoloured vertex preceding c i 1 Note that we have uc i 1, uc i E(T r ) So, we have the following

BALANCED DECOMPOSITION NUMBER AND CONNECTIVITY 9 alanced decomposition with size at most 3 for T r K s V (T r K) = [c i 1, c i ] { {u } : u V (T r K) \ i=1 s i=1 } [c i 1, c i ] Of course, if s = 0, then we just ignore this alanced decomposition We can carry out a similar procedure on T, switching the roles of red and lue Defining a similar linear ordering on V (T ), we have a similar set of coloured vertices d 1,, d t+1, for some t 0, which alternate in colour (starting with lue), and a similar set L V (T ) containing the vertices coming after d t R, with respect to Again, we can find a alanced decomposition of T L with size at most 3 Finally, it is easy to check, y a simple case y case analysis, that T [K L] has a alanced decomposition of size at most 3, using the fact that the edge zz exists (In fact, there are four possile cases for the structure of K (L), each one containing one red (lue) vertex and at most two uncoloured vertices) We shall not go into details here Now we return to the graph G Before we prove the claim regarding non-adjacencies in G, we first consider extracting a red (lue) alternating tree from a red (lue) tree in H G, and introduce a notation Suppose that R u is a red tree in H with seed u, and v V (R u ) Let v C 1 {u} P P 3 U 1 for some C 1 R u If C 1 {u}, then, when C 1 was constructed in R u, it was appended y eing joined to some C {u} P P 3 U 1, C R u Repeat this procedure with C, and successively We otain distinct sets C 1,, C t {u} P P 3 U 1 in R u, for some t 1, with v C 1 and C t = {u} This is indeed the case To see this, consider the linear ordering on the sets of {u} P P 3 U 1 used in R u, in the order that they appeared in the construction of R u, and oserve that C t C 1 Now, define R u, v = R u [C 1 C t ] Similarly, if B u is a lue tree in H with seed u, and v V (B u ), we define B u, v analogously With this definition, we have the following oservation Oservation 3 R u, v ( B u, v ) is a red (lue) alternating tree with seed u (u ) Moreover, V (R u ) \ R u, v (V (B u ) \ B u, v ) is a disjoint union of memers of P, P 3 and U 1 Now, we are ready to prove the following non-adjacencies claim Claim 4 (a) If (X W ) U Y U, so that X = X W and Y = Y W, then E(X R, Y (B U)) = () If (X W ) U > Y U, so that X = X W W and Y = Y, then Proof (a) It suffices to prove that E(X (R U), Y B) = (1) E(X R, (Y (B U)) (W B)) = Let u X R and v (Y (B U)) (W B), and assume that uv E(G) We shall prove that the existence of the edge uv either contradicts the maximality of some red tree or lue tree in H (property (iii)), or it implies that (R, B) has a alanced decomposition of size at most 3; hence, f(l, G) 3 Now, u is a red vertex in some red tree R c 1 r of H where, with a slight ause of notation, either {c 1 r} = C 1 R for some C 1 C r C, or c 1 r = x If v W B, then this contradicts the maximality of R c 1 r So let v Y (B U) Then v is a lue or an uncoloured vertex in some lue tree B d 1 in H, where {d 1 } = D1 B for some D 1 C C, or d 1 = y Our aim now will e to find a family T of vertex disjoint alternating trees such that V (G)\ {V (S) : S T } is a disjoint union of memers of P, P 3 and U 1 Then f(l, G) 3 follows from Claim If c 1 r x, then in H, y property (ii)(d), for some s 1, we have distinct sets C 1,, C s 1 C r C,

10 SHINYA FUJITA AND HENRY LIU and vertices c V (R c r ) R,, c s V (R c s r ) R such that c c 1 u,, c s cu s 1 E(G), where c s r = x and C i = {c i r, c i, ci u} as usual for 1 i s 1 Note that this is all vacuous if s = 1, except that c 1 r = x Similarly, y property (ii)(e), for some t 1, we have distinct D 1,, D t 1 C C, and vertices d V (B d ) B,, d t V (B d t ) B such that d d 1 u,, d t d t 1 u E(G), where d t = y and Dj = {d j r, d j, dj u} for 1 j t 1 This is all vacuous if t = 1, except that d 1 = y Now, it may e the case that we have some of C 1,, C s 1 coinciding with some of D 1,, D t 1 Let D = {C i : C i {D 1,, D t 1 }} Note that we have D C Now, we claim that we can take T = {T } T 1 T T 3, where T = R c 1 r, u B d 1, v uv, T 1 = { R c i r, c i c i c i 1 u c i 1 : i s, C i 1 D }, T = { B d j, d j d j d j 1 u dr j 1 : j t, D j 1 D }, T 3 = { R c p r, c p B d q, dq c p c p 1 u d q : p s, q t, C p 1 = D q 1 D } To prove that this choice of T works, we prove the following claim Suclaim 5 (A) T is a family of vertex disjoint alternating trees (B) We have {V (S) : S T } = {x, y} V (, u Rc1 c1 r r) V ( B d 1, v ) () d1 s V ( R c i r, c i c i r) i= t j= V ( B d j d, j d j ) (C 1 C s 1 D 1 D t 1 ) Hence, V (G) \ ( {V (S) : S T } ) is a disjoint union of memers of P, P 3 and U 1 Proof (A) Clearly, y Oservation 3, T and memers of T 1 and T are alternating trees Also, if S T 3, then S = R c p r, c p B d q, dq c p c p 1 u d q for some p s, q t with C p 1 = D q 1 We have c p 1 u = d q 1 u, so that c p c p 1 u, c p 1 u d q E(H) It is clear from this and Oservation 3 that S is an alternating tree Now, we prove that these alternating trees of T are vertex disjoint Oviously, T is vertex disjoint from memers of T 1 T T 3 Let S 1 = R c i r, c i c i c i 1 u c i 1 T 1 and S = B d j, d j d j du j 1 d j 1 r T, for some i s, j t and C i 1, D j 1 D Oviously, we have V ( R c i r, c i ) V ( B d j, d j ) = Now, c i 1 u d j 1 u follows that c i 1 u since C i 1 D j 1, and c i 1, c i 1 V (S ) Similarly, d j 1 u c i 1 d j, since Ci 1 D j if j t 1, or c i 1, dr j 1 V (S 1 ) Hence, V (S 1 ) V (S ) = y if j = t It Let S 1 = R c i r, c i c i c i 1 u T 1 and S 3 = R c p r, c p B d q, dq c p c p 1 u d q T 3, for some i, p s, q t, C i 1 D, and C p 1 = D q 1 D Since i p, clearly we have V ( R c i r, c i ) V ( R c p r, c p B d q, dq ) = Also, c i 1 u c p 1 u, and c i 1 d q as efore: either C i 1 D q if q t 1, or c i 1 y if q = t It follows that V (S 1 ) V (S 3 ) = Similarly, V (S ) V (S 3 ) = for every S T and S 3 T 3 (B) Note that the last part follows from Oservation 1 and (), since each set on the right hand side of (), apart from {x, y}, is a union of memers of P, P 3 and U 1, and that we have the disjoint union as indicated in () So, it remains to prove () As usual, any meaningless terms, such as C s, will e considered non-existent We prove that the left hand side of () is contained in the right hand side Oviously, V (T ) {x, y} V ( R c 1 r, u c 1 r) V ( Bd 1, v d1 ) C 1 D 1

BALANCED DECOMPOSITION NUMBER AND CONNECTIVITY 11 c i 1 Let S 1 = R c i r, c i c i c i 1 u T 1, where i s and C i 1 D Then, V (S 1 ) {x} V ( R c i r, c i cr) i C i C i 1 A similar argument holds if S T Let S 3 = R c p r, c p B d q, dq c p c p 1 u d q T 3, where p s, q t, and C p 1 = D q 1 Then, V (S 3 ) {x, y} V ( ( R c p r, c p cr) p V Bd q, dq d q ) C i C i 1 Now, we prove the opposite containment {c 1 r, d 1 } V ( ( R c 1 r, u cr) 1 V Bd 1, ) v d1 V (T ) In particular, x V (T ) if s = 1, and y V (T ) if t = 1 If s and i s, then {c i r} V ( R c i r, c i cr) i V (S1 ) for some S 1 T 1 if C i 1 D; or {c i r} V ( R c i r, c i cr) i V (S3 ) for some S 3 T 1 if C i 1 D A similar argument holds for V ( B d j d, j d j ) if t and j t Note that we have considered the set {x, y} here Let s and 1 i s 1 If C i D, then c i u, c i V (S 1) for some S 1 T 1 If C i D, then t, and C i = D j for some 1 j t 1 We have c i u V (S 3 ) for some S 3 T 3, and c i V (T ) if C i = D 1 ; c i V (S ) for some S T if D j 1 D, j and t 3; and c i V (S 3) for some S 3 T 3 if D j 1 D, j and t 3 A similar argument holds for d j u and d j r if 1 j t 1 and t We must also check that the right hand side of () is a disjoint union as stated Indeed, S = { R c 1 r, u c 1 r, B d 1, v d 1 { } Rc i, ci c i r r : i s } { B d j, d j d j : j t} is a family of graphs, each of which is a red or lue alternating tree in a red or lue tree, with the seed deleted So for any F S, none of {x, y}, C 1,, C s 1, D 1,, D t 1 intersects with F The memers of S are also vertex disjoint themselves Finally, x and y do not elong to any of C 1,, C s 1, D 1,, D t 1 So, () holds This proves part (B), and the proof of Suclaim 5 is complete Finally, y Claim and Suclaim 5(A), G [ {V (S) : S T } ] has a alanced decomposition of size at most 3 With the last part of Suclaim 5(B), it follows that f(l, G) 3 This proves part (a) of Claim 4 () It suffices to prove that E((X (R U)) W (W R), Y B) = Switching the roles of X and Y, and of R and B, and with the fact that now W X (instead of W Y in (a)), (1) implies that E((X (R U)) (W R), Y B) = So, it remains to prove that E(W, Y B) = Let u W and v Y B, and assume uv E(G) Then u is an uncoloured vertex not used in H, and v is a lue vertex in some lue tree S in H If u U 1, then this contradicts the maximality of S If u A P 3 for some A V (H), then this contradicts the termination of the algorithm when H was constructed This completes the proof of Claim 4 Now we can easily finish the proof of Theorem 3 If (X W ) U Y U, it follows that X U Y U We delete from G the set (X (B U)) (Y R) Z This leaves the sets X R and Y (B U), which, y Claim 4(a), are disconnected Now, in X, with the exception of x and Z other red vertices, the rest of the set has the same numer of red and lue vertices, and so, X B = 1 ( X Z 1 X U )

1 SHINYA FUJITA AND HENRY LIU Similarly, Y R = 1 ( Y Z 1 Y U ) So, (X (B U)) (Y R) Z = X B + X U + Y R + Z 1 ( X Z 1 X U ) + X U + 1 ( Y Z 1 X U ) + Z = 1 ( X + Y ) 1 n 1 Similarly, if (X W ) U > Y U, then X U > Y U We delete from G the set (X B) (Y (R U)) Z This leaves the sets X (R U) and Y B, which, y Claim 4(), are disconnected Again, we have X B = 1 ( X Z 1 X U ), and Y R = 1 ( Y Z 1 Y U ) So a similar calculation gives (X B) (Y (R U)) Z < n 1 In either case, we have deleted from G a cut-set of size at most n 1, which is a final contradiction This completes the proof of Theorem 3 3 Complete Multipartite Graphs Our next aim is to determine f(g), where G is a complete multipartite graph with at least 3 classes The case for complete ipartite graphs was solved in [9] We may easily comine Theorems 1 and of [9] to get the following Theorem 4 (Theorems 1 and of [9]) Let 1 m n Then, f(k m,n ) = n m + 3 It turns out that the complete multipartite graphs version is a simple consequence of Proposition 1 and Theorems 3 and 4 In this section, we denote the complete multipartite graph with class sizes k 1 k t y K k1,,k t, where t 3 Also, let V i e the vertex class with order k i, for 1 i t Then, we have the following extension to Theorem 4 Theorem 5 Let k 1 k t 1, where t 3 Then, k1 (3) f(k k1,,k t ) = t i= k + 3 i Proof Let p = V (K k1,,k t ) = k 1 + + k t If k 1 = 1, then K k1,,k t = Kt, so Proposition 1 implies that f(k k1,,k t ) =, and (3) holds Now, assume that k 1, so that K k1,,k t is not a complete graph We shall consider two cases Case 1 k 1 p In this case, we shall apply Theorem 3 For every 1 i t, we have V (K k1,,k t ) \ V i p This implies that K k1,,k t is p -connected Indeed, if C V (K k 1,,k t ) and C p 1, then K k 1,,k t C must contain vertices from at least two different classes But, two vertices from different classes are neighours, and they form a dominating set for K k1,,k t, so that K k1,,k t C is connected Hence y Theorem 3, we have f(k k1,,k t ) = 3, and this is easily seen to e consistent with (3), since we have t i= k i p, so that (k1 )/ ( t i= k i) = 0 Case k 1 p + 1 For this case, we shall use the following simple oservation Oservation 6 If H G are connected graphs and V (H) = V (G), then f(h) f(g) We shall apply Theorem 4 and Oservation 6 with n = k 1 and m = t i= k i Since K k1,,k t K m,n and V (K k1,,k t ) = V (K m,n ), we have f(k k1,,k t ) n m + 3

BALANCED DECOMPOSITION NUMBER AND CONNECTIVITY 13 Also, since V 1 is an independent set, it turns out that we can take alanced colourings for K k1,,k t similar to those for K m,n as descried in [9] (leading to the lower ound) That is, if V = V V t, we take the alanced colouring (R, B) for K k1,,k t where, if (l )m + n (l 1)m + 1 for some l N, then V 1 R = V 1 B = n ; if (l 1)m + n lm + 1 for some l N, then V R = m, V 1 R = n m, and V 1 B = n+m The exact same arguments as in [9] again lead to f(k k1,,k t ) n m + 3 Indeed, for the first alanced colouring, any alanced decomposition has size at least n (l 1)m + 1 n + 1 + 1 = l + 1 = + 3 m m m For the second alanced colouring, any alanced decomposition has size at least ( n m ) (l 1)m + 1 n + 1 + = l + = + 3 m m m So, (3) holds in this case, and we are done 4 Generalised Θ-graphs In this section, we shall study the function f(g), where G is a generalised Θ-graph That is, G is a graph which is a sudivision of a multiple edge More precisely, G is the graph union of t paths, Q 1,, Q t say, with each having the same two end-vertices, x and y say, so that V (Q i ) V (Q j ) = {x, y} for any i j In other words, the Q i are pairwise internally vertex disjoint paths In addition, all ut at most one of the Q i have order at least 3 We egin y recalling a result and a conjecture from [9] Theorem 6 (Theorem 4 of [9]) Let n 3 Then, f(c n ) = n + 1, where C n is the cycle of order n Conjecture 7 (Conjecture 1 of [9]) Let G e a -connected graph of order n Then f(g) n + 1 So, studying f(g) for generalised Θ-graphs G is related to oth Theorem 6 and Conjecture 7, since cycles are generalised Θ-graphs, and generalised Θ-graphs are -connected In Theorem 8, we shall prove that the same upper ound as in Theorem 6 holds for generalised Θ-graphs, and that we have a matching lower ound which is asymptotically tight This result can e considered as a partial solution to Conjecture 7 Theorem 8 Let G e a generalised Θ-graph, formed y uniting the pairwise internally vertex disjoint paths Q 1,, Q t, where t, V (Q 1 ) V (Q t ), and V (Q t 1 ) 3 Let V (G) = n t + 1 Then, we have the following (a) f(g) n, if V (Q 1) n n t+1 + 1 f(g) if V (Q i ) n for every i () f(g) n + 1 In particular, if t is fixed, then f(g) = n + O(1) Proof Let the common end-vertices of the paths Q 1,, Q t e x and y That is, {x, y} = V (Q i ) V (Q j ) for any i j (a) Let V (Q 1 ) n + 1 We take a alanced colouring (R, B) for G where R = {v V (Q 1) : d Q1 (v, x) n 1}, and B = V (G) \ R if n is even, B = V (G z) \ R if n is odd, where z V (Q 1) is the vertex with d Q1 (z, x) = n (so, z is the only uncoloured vertex) We have R = B = n Now, any alanced decomposition P of G can only have at most two sets containing vertices of R Otherwise, if P has at least three such sets, then P must contain a set A such that A R R \ {x, x }, where x V (Q 1 ) is the vertex with d Q1 (x, x) = n 1 But then, G[A R] is disconnected from G[B], a contradiction Now, since there is at most one uncoloured vertex, we have P 3 If P = 3, then n is odd, and {z} is a memer of P But then, it is not possile to have a alanced decomposition of G z into two sets, since a alanced set

14 SHINYA FUJITA AND HENRY LIU containing x must contain all other red vertices Hence, we have P, which implies that f(g) n Now, let V (Q i ) n for every i Note that we have t 3 in this case We take a alanced colouring (R, B) for G as follows For every i, let q i = V (Q i ) e the numer of internal vertices of Q i Let m = n = t i=1 q i e the total numer of internal vertices of all the paths Q i We claim the following Claim 7 There exists a partition [t] = U W, where U, W, such that i U q i m 3 and j W q j m 3 Proof We have q 1 q t 0 and q t 1 1 We are done if q 1 m 3, since then we have m 3 q 1 m 1, and we can take U = {1} and W = {,, t} So, assume q 1 < m 3 Let p e the integer such that p 1 i=1 q i < m 3 and p i=1 q i m 3 We are done if p i=1 q i m 3 (in which case p < t), since then we can take U = {1,, p} and W = {p + 1,, t} So, assume p i=1 q i > m 3 But then, we have q p > m 3 > q 1, a contradiction Now, take a partition [t] = U W as given y Claim 7, and assume that i U q i j W q j Take R = i U V (Q i {x, y}), so that n 3 R n Now, we want to take B j W V (Q j {x, y}) with B = R, spreading the lue vertices as evenly as possile More precisely, note that ( B n ) 3 j W q ) = 1 j, ( (n ) 3 so that we can take B j W V (Q j {x, y}) where no two adjacent vertices in j W V (Q j {x, y}) are uncoloured Note that x and y are uncoloured vertices Let P e a alanced decomposition for (R, B) Firstly, suppose that x and y are in the same memer C of P Then, this implies that C R B Otherwise, if we have a vertex u R \ C (u B \ C), then in G C, u is disconnected from all other lue (red) vertices Now, note that C n W, for otherwise C < n W, then C misses at least two uncoloured vertices in some Q j {x, y} with j W, which is not possile y the choice of B, and G (R B) j W V (Q j {x, y}) Hence, we have f(g) C n W n t + 1 n t+1 Now, suppose that x and y are in different memers of P, say, C x and C y We can apply a similar argument to the previous case with C x C y in place of C, to get C x C y R B, and C x C y n t + 1 Hence, f(g) max{ C x, C y } n t+1 () Let (R, B) e a alanced colouring for G We shall consider two cases: when V (Q i ) n + 1 for every i, and when V (Q 1 ) n + Case 1 V (Q i ) n + 1 for every i Suppose that we cannot find a suitale alanced decomposition for G We egin y proving two claims These will then help us to descrie an algorithm which finds a alanced decomposition of G with size at most n + 1, and with at most three parts, which will prove Case 1 Claim 8 There exists a alanced set A V (G) with A n such that exactly one of x, y is in A Proof If either x or y is uncoloured, then we can simply take A = {x} or A = {y}, whichever is uncoloured So, assume that oth x and y are coloured, and without loss of generality that x R Suppose that such a alanced set A cannot e found Then, for any 1 i t and any x V (Q i {x, y}), since V (Q i y) n, the path xq ix Q i satisfies V (xq i x ) R > V (xq i x ) B So in particular, we have V (Q i y) R V (Q i y) B + 1 for each i So, R = t V (Q i y) R (t 1) i=1 t V (Q i y) B + 1 B i=1 t ( V (Qi y) B + 1 ) (t 1) i=1

BALANCED DECOMPOSITION NUMBER AND CONNECTIVITY 15 Equality holds if and only if y B and V (Q i y) R = V (Q i y) B + 1 for every i But then {V (Q 1 ), V (Q {x, y}),, V (Q t {x, y})} is a alanced decomposition for G, with size at most n + 1, which is a contradiction This proves Claim 8 Claim 9 Let A V (G y) e a alanced set with x A, A n 1, and N(A)\{y} R Then, for some 1 j t, we can find a non-empty alanced set C V (Q j y) \ A, with N(C) A, and C n 1 Proof Define I [t] y I = {j [t] : V (Q j y) \ A } Note that I, otherwise, A n n > n 1, a contradiction If no such alanced set C exists, then for each j I, since the vertex of N(A) V (Q j) is red, and V (Q j {x, y}) n 1, we have (V (Q j y) \ A) R (V (Q j y) \ A) B + 1, y a similar argument as in Claim 8 But then, we have (V (G) \ A) R (V (Q j y) \ A) R j I j I > (V (G) \ A) B We have a contradiction, since V (G) \ A is a alanced set Claim 9 follows (V (Q j y) \ A) B + We remark that with A as in Claim 8 or Claim 9, we have G A is connected We can now descrie the algorithm Step 1 By Claim 8, without loss of generality, we can find a alanced set A 1 V (G y) with x A 1, and A 1 n If A 1 = n, stop; we have a suitale alanced decomposition {A 1, V (G) \ A 1 } for G Otherwise, A 1 n 1; go to Step Step If N(A 1 ) \ {y} R B, go to Step 3 Otherwise, there exists an uncoloured vertex in N(A 1 ) \ {y}; append it to A 1 We have another alanced set A with A 1 A V (G y) and A = A 1 + 1 If A = n, stop; we have a suitale alanced decomposition {A, V (G) \ A } for G Otherwise, A n 1; repeat Step, using A for A 1 Step 3 If N(A 1 ) \ {y} R or N(A 1 ) \ {y} B, go to Step 4 Otherwise, there exist a red vertex and a lue vertex in N(A 1 )\{y}; append them to A 1 We have another alanced set A 3 with A 1 A 3 V (G y) and A 3 = A 1 + If A 3 = n or n + 1, stop; we have a suitale alanced decomposition {A 3, V (G) \ A 3 } for G Otherwise, A 3 n 1; go ack to Step, using A 3 for A 1 Step 4 By Claim 9, for some 1 j t, we can find a non-empty alanced set C V (Q j y) \ A 1, with N(C) A 1, and C n 1 We have a alanced set A 1 C V (G y) with A 1 C > A 1 If A 1 C n, stop; we have a suitale alanced decomposition {A 1, C, V (G) \ (A 1 C)} for G Otherwise, A 1 C n 1; go ack to Step, using A 1 C for A 1 This algorithm must terminate, since as we go through each step, we are creating new alanced sets with strictly increasing orders When the algorithm terminates, we end up with a alanced decomposition of size at most n + 1 for G (and with at most three parts), a final contradiction This completes the proof of Case 1 Case V (Q 1 ) n + We shall use a similar idea to the proof in [9] of the upper ound of Theorem 6, that f(c n ) n + 1, where C n is the cycle with n vertices We numer the vertices of G with 1,, n as follows The vertices of Q 1 are numered with 1,, V (Q 1 ), with vertex v V (Q 1 ) receiving the numer d Q1 (v, x)+1 Next, the vertices of Q {x, y} are numered with the next V (Q ) integers, ordered y distance from y We then repeat this with Q 3 {x, y},, Q t {x, y} successively Now, for i [n], define A(i) V (G) as follows A(i) = { i, i + 1,, i + {i, i + 1,, n} n } 1 { 1,, i n } 1 if 1 i if n + 1, n + i n

16 SHINYA FUJITA AND HENRY LIU In other words, with the numering, A(i) is the n consecutive vertices of V (G), starting at i, modulo n We have the following claim Claim 10 Let A V (G) e a set of consecutive vertices in the numering, modulo n, with A, V (G) \ A n 1 Then, oth G[A] and G[V (G) \ A] are connected Proof Let y e numered with j Since V (Q 1 ) n +, we have {j +1,, n} n ( n +) = n So, neither A nor V (G) \ A can e a suset of {j + 1,, n}, since A, V (G) \ A n 1 It is then easy to check the following If A {x, y} = 0 or, then one of A and V (G) \ A is a suset of V (Q 1 {x, y}) which induces a su-path, and the other induces a sugraph of G which is the union of Q,, Q t with two disjoint su-paths of Q 1, one containing x, the other containing y If A {x, y} = 1, then A and V (G) \ A are oth sudivided stars, one with centre x, the other with centre y In either case, oth G[A] and G[V (G) \ A] are connected So Claim 10 follows We may now complete Case in a similar way to the proof of Theorem 6 in [9] For i [n], define g(i) = A(i) R A(i) B For every i, we have g(i + 1) g(i) (modulo n), and n i=1 g(i) = ( R B ) = 0 We have two possiilities n Either, there exists i with g(i) = 0, whence {A(i), V (G) \ A(i)} is a suitale alanced decomposition for G, since A(i) = n n 1, and V (G) \ A(i) = n n 1, so that Claim 10 applies Or, there exists i with {g(i), g(i + 1)} = { 1, 1}, modulo n If g(i) = 1 and g(i + 1) = 1, then, with respect to the numering, modulo n, the first vertex of A(i) is lue, A(i) has one more lue vertex than red, and the first vertex outside of A(i) is red If w is this red vertex, then {A(i) {w}, V (G) \ (A(i) {w})} is a suitale alanced decomposition for G, since A(i) {w} and V (G) \ (A(i) {w}) are still consecutive vertices with respect to the numering, modulo n, and A(i) {w} = n + 1 n 1 and V (G) \ (A(i) {w}) = n 1, so that Claim 10 applies A similar argument holds if g(i) = 1 and g(i + 1) = 1 for some i; we just switch the roles of red and lue The proof of Case is now complete, and this proves Theorem 8 5 Concluding Remarks Theorem 3 and Conjecture 7 suggest a more general prolem Prolem 9 Let n, t N with n 1 t Let G e a graph of order n, with n sufficiently large Determine a function g(n, t) such that the following holds: If G is g(n, t)-connected, then f(g) t + 1 Looking at Theorem 3, we find that g(n, ) = n is suitale, and moreover, the converse also holds Along with Conjecture 7, we guess that g(n, t) = n t is a good candidate But we remark if g(n, t) = n t, then the converse is not true for t 3 There is a counter-example: Take the graph formed y a K n 1 with another vertex joining to a vertex of the K n 1 We suspect that Prolem 9 is not easy, given the difficulty of Conjecture 7 But a partial result for Prolem 9 may e of interest For example, the case t = 3 Proposition 1 and Theorem 3 also imply that we can determine the time complexity for deciding whether a graph G satisfies f(g) {, 3} or not The prolem of determining a fast algorithm for finding the vertex connectivity κ = κ(g) of a graph G has een considered y many people (Kleitman [13], Hopcroft and Tarjan [1], Even [5], Even and Tarjan [6], Galil [10], Cheriyan and Thurimella [, 3], Nagamochi and Iaraki [14, 15], and Henzinger et al [11], among others) It is known that such an algorithm can e carried out in polynomial time A result of Henzinger et al [11] states that, for a graph G of order n and with connectivity κ, there is an algorithm which determines the connectivity of G in time O(min(κ 3 + n, κn)κn)

BALANCED DECOMPOSITION NUMBER AND CONNECTIVITY 17 Corollary 10 Let G e a connected graph of order n Then, we can decide whether f(g) =, or f(g) = 3, or f(g) 4, using an algorithm with running time O(n 4 ) The alanced decomposition numer will assure us the existence of a good structure, especially when it is small For example, from Theorem 3, if a graph G of order n is n -connected, then we can always find a nearly alanced matching for any aritrary numer and position of red and lue vertices (Here, nearly alanced matching means vertex-disjoint paths of order at most 3 whose end-vertices are coloured y red and lue, respectively) Moreover, these are related to an existence of so called non-separating sugraphs, that is, the sugraphs whose deletion preserve high connectivity For any disjoint susets R, B V (G) with R = B = k in an m-connected graph G with m k, y using Menger s theorem, there are k vertex disjoint paths Q 1,, Q k from R to B However, in general, we can never hope for high connectivity of G k i=1 V (Q i) As for this prolem, our results give the following Corollary 11 Let m, n N with m n, and let G e an m-connected graph of order n Then, for any disjoint susets R, B V (G) with k = R = B < m 3, there are k vertex disjoint paths Q 1,, Q k from R to B, such that G k i=1 V (Q i) is (m 3k)-connected Proof Since G is n -connected, y Proposition 1 and Theorem 3, we have f(g) {, 3} So the alanced colouring (R, B) for G has a alanced decomposition P of size at most 3 This means that there are exactly k vertex disjoint sets of P containing coloured vertices, with each one having exactly one red vertex and one lue vertex Let A 1,, A k P e these k sets For each A i, since A i 3, one can easily check that there exists A i A i such that G[A i ] is an R B su-path of G The corollary follows y letting Q i = G[A i ] for every i, since clearly the Q i are vertex disjoint, and we have k i=1 V (Q i) 3k, so that G k i=1 V (Q i) is (m 3k)-connected Thus, a graph with a small alanced decomposition numer is likely to have a good structure in view of non-separating sugraphs In this sense, further study in this area will e interesting and significant Acknowledgements The second author would like to thank Gunma National College of Technology for their generous hospitality He was ale to carry out part of this research with the first author during his visit there The authors would also like to thank the anonymous referee for valuale comments and suggestions References [1] B Bolloás, Modern graph theory, Springer-Verlag, New York, 1998, xiii+394pp [] J Cheriyan and R Thurimella, On determining vertex connectivity, Technical report UMIACS-TR-90-79 CS-TR-485, Univ of Maryland, College Park, MD, June 1990 [3] J Cheriyan and R Thurimella, Algorithms for parallel k-vertex connectivity and sparse certificates, Proc 3rd ACM Symp on Theory of Computing (1991) [4] R Diestel, Graph Theory (3rd ed), Springer-Verlag, Berlin, 005, xvi+411pp [5] S Even, An algorithm for determining whether the connectivity of a graph is at least k, SIAM J Comput 4(3) (1975), 393-396 [6] S Even and R E Tarjan, Network flow and testing graph connectivity, SIAM J Comput 4 (1975), 507-518 [7] S Fujita and H Liu, Further results on the alanced decomposition numer, Congressus Numerantium, to appear [8] S Fujita and H Liu, The alanced decomposition numer of T K 4 and series-parallel graphs, sumitted [9] S Fujita and T Nakamigawa, Balanced decomposition of a vertex-coloured graph, Discr Appl Math, 156 (008), 3339-3344 [10] Z Galil, Finding the vertex connectivity of graphs, SIAM J Comput 9(1) (1980), 197-199 [11] M R Henzinger, S Rao and H N Gaow, Computing vertex connectivity: new ounds from old techniques, J Algorithms, 34() (000), -50 [1] J E Hopcroft and R E Tarjan, Dividing a graph into triconnected components, SIAM J Comput (1973), 135-158 [13] D J Kleitman, Methods for investigating connectivity of large graphs, IEEE Trans Circuit Theory CT-16 (1969), 3-33