Journal of Mathematical Sciences, Vol. 145, No. 3, 2007

REMOVING VERTICES FROM A k-CONNECTED GRAPH WITHOUT LOSING ITSk-CONNECTIVITY

A. S. Chukhnov∗ UDC 519.172

The paper studies the problem of removing vertices from a k-connected graph without losing its k-connectivity. It isproved that certain inner vertices can be removed from k-blocks, provided that the interior of each block is sufficientlylarge with respect to its boundary, and the degree of every vertex of the graph is no less than 3k−1

2 or 3k2 . Bibliography:

3 titles.

1. Introduction

This work is devoted to studying the structure of k-connected graphs. In this paper, we consider finiteundirected graphs without loops and multiple edges. A graph is said to be k-connected if it has k +1 vertices atleast and remains connected upon the removal of arbitrary k − 1 vertices.

In studying connected graphs, the notions of a block (that is a maximal biconnected subgraph) and of acutpoint are very important. In a k-connected graph, the analogs for cutpoints are k-cutsets, i.e., k-element setswhose removal disconnects the remaining graph. Here, we use the definition of a block in a k-connected graphintroduced in [3].

As is easy to prove, in a connected graph, an arbitrary set of inner vertices (i.e., vertices that are not cutpoints)of some blocks can be removed without losing connectivity. For a k-connected graph, this result is not alwaysvalid. In some cases, the removal of certain sets of inner vertices from different blocks results in losing k-connectivity. Nevertheless, under certain conditions imposed on blocks from which vertices are removed, onecan choose one vertex in each block in such a way that upon removing an arbitrary subset of this set, the graphremains k-connected. In other words, the set chosen does not influence the k-connectivity of the graph.

The best results can be achieved for k-connected graphs all of whose vertices are of degree at least 3k−12 . For

the first time, these graphs were studied in [1], where it was proved that from such a graph any vertex can beremoved without losing the k-connectivity of the graph. In [2], it was demonstrated that in every k-connectedgraph all of whose vertices are of degree at least 2k, in each block whose interior is not smaller than its boundary,one can choose a vertex in such a way that upon removing an arbitrary subset of this set, the graph remainsk-connected.

In this work, similar assertions are proved for graphs all of whose vertices are of degree at least 3k−12 and the

interiors of all blocks are at least twice their boundaries, and also for graphs all of whose vertices are of degreeat least 3k−1

2 and the interiors of all blocks are greater than their boundaries (see Theorems 1 and 2).

2. Basic definitions

For a graph G, as usual, V (G) is the set of its vertices, and E(G) is the set of its edges. The degree of avertex v in G is denoted by deg (v).

For W ⊂ V (G), by G−W we denote the graph obtained by removing from G all vertices in W together withall edges incident with them.

A connected component of a graph G is a maximal (with respect to inclusion) connected subgraph of G.The number of elements in a set M is denoted by |M |. For brevity, the number of vertices in a graph G, i.e.,

the cardinality of V (G), is also denoted by |G|.

Definition 1. A set R of vertices of a graph G is called a cutset if the graph G − R is disconnected. If thenumber of vertices in a cutset should be specified, then the latter set with x vertices is called an x-cutset.

Definition 2. A cutset R in a graph G is said to disconnect a set of vertices X if the removal of R from Gresults in that some of the remaining vertices belong to different connected components. The set of all k-cutsetsof a graph G is denoted by Rk(G).

∗St.Petersburg State University, St.Petersburg, Russia.

Translated from Zapiski Nauchnykh Seminarov POMI, Vol. 340, 2006, pp. 103–116. Original article submitted December19, 2006.

1072-3374/07/1453-4989 c©2007 Springer Science+Business Media, Inc. 4989

Definition 3. Let G be a k-connected graph, let R ∈ Rk(G), and let H be a connected component of the graphG − R. Then H is called a component of G. Also H is said to be separated by R.

A component of a graph G is said to be minimal if it contains no other component separated by a k-cutset.

Definition 4. Let B ⊂ V (G) be a maximal set of vertices in a k-connected graph G such that in order toseparate arbitrary two vertices x, y ∈ B, at least k +1 vertices must be removed. In this case, B is called a block.

The set of all vertices in B belonging to no k-cutset of G is referred to as the interior of the block B and isdenoted by I(B). The set of all vertices of B belonging to some k-cutsets is called the boundary of the block Band is denoted by R(B). Vertices from I(B) and R(B) are said to be inner and boundary vertices, respectively.

If I(B) is empty, then the block B is said to be empty. Otherwise B is said to be nonempty.

Lemma 1. For every two intersecting blocks B1 and B2 of a k-connected graph G, there is a set S ∈ Rk(G)that contains B1 ∩ B2.

For the proof of this lemma, see [3].Lemma 1 means that our definitions of the interior and boundary of a block satisfy the natural condition that

blocks can intersect by boundary vertices only.

3. Removal of vertices

In what follows, a graph G is always assumed to be k-connected, and the degrees of its vertices are assumedto be no less than 3k−1

2 .From the definition of a block it follows that on removing an arbitrary inner vertex from a nonempty block of

a k-connected graph G, this graph remains k-connected. We will provide conditions on a graph and on a subsetof its blocks under which one can remove one vertex from each block of this set without losing k-connectivity.

It is easy to prove that the interiors of two distinct blocks always are disjoint. Thus, we can speak about setsof inner vertices of different blocks of a graph G.

Lemma 2. Consider a set M = {v1, v2, . . . , vn} (n > 1) of vertices of a graph G such that no two vertices liein the same minimal component and possessing the following properties:

(1) there is S1 ∈ Rk(G) that disconnects M ;

(2) on removing an arbitrary proper subset of M , the graph G remains k-connected.Let S1, S2, . . . , Sm be all the k-cutsets of a graph G disconnecting M . Set S = ∩n

i=1Si.Then, upon removing arbitrary k − 1 vertices from the graph G − M , all the vertices not belonging to S are

located in one connected component. Moreover, the number of vertices from S not belonging to this connectedcomponent does not exceed k−1

2.

This lemma was proved in [2] for a somewhat wider class of graphs.

Definition 5. A set M consisting of inner vertices of different blocks of a graph G and containing at most onevertex from each block is said to be irremovable if there is a subset of M (possibly coinciding with M) whoseremoval results in that G is no longer k-connected. If, upon removing an arbitrary subset of M , the graph Gremains k-connected, then M is said to be removable.

Lemma 3. If M is a minimal irremovable set in a graph G, then it satisfies conditions (1) and (2) of Lemma 2.

Proof. In [3], it was proved that any minimal component is the interior of a block. Therefore, two inner verticesof different blocks cannot lie in one minimal component. As every two nonempty blocks are separated by a setfrom Rk(G), condition (1) of Lemma 2 is fulfilled.

Condition (2) is fulfilled because M is a minimal irremovable set. �

Theorem 1. Let G be a k-connected graph all of whose vertices are of degree no less than 3k−12 . Let {Bi}n

i=1

be a set of blocks in G such that |I(Bi)| > 2|R(Bi)|. Then in each block Bi one can choose an inner vertex insuch a way that these vertices form a removable set.

Proof. Let R(Bi) = {d1, . . . , dq}. Consider the vertex d1 and mark, in I(Bi), one vertex adjacent to d1 (if suchvertices exist). Then consider the vertex d2 and, in I(Bi), mark one unmarked vertex adjacent to d2 (if suchvertices exist), and so on. At step j, we mark, in I(Bi), one vertex adjacent to dj and unmarked so far (if suchvertices exist). The vertices marked during this process are referred to as the marked vertices of the first kind.Their number does not exceed |R(Bi)|.

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Then we repeat this process once again, marking at step j an unmarked vertex in I(Bi) adjacent to dj (if suchvertices exist). The vertices marked during this process are referred to as the marked vertices of the second kind.Their number does not exceed |R(Bi)| as well. Thus, the total number of marked vertices is at most 2|R(Bi)|,and in I(Bi) there is at least one unmarked vertex. Choose an arbitrary unmarked vertex vi in Bi.

We prove that the set {vi}ni=1 is removable. Suppose the contrary. Then this set contains a minimal irremovable

set. Without loss of generality, one may assume that this set is M = {vi}li=1. In the graph G − M , there is a

(k − 1)-cutset R. By applying Lemmas 3 and 2, we obtain that all the components of G−M that are separatedby R, possibly except one of them, contain in total no more than k−1

2 vertices, and all their vertices belong tothe intersection of all the sets from Rk(G) that disconnect M . Among these components, choose one and denoteit by F1. This is feasible because otherwise at most one component would be separated by R. Thus, F1 is acomponent of the graph G − M that is separated by R and belongs to every set from Rk(G) disconnecting M .In particular, among the vertices of F1 there are no inner vertices of the blocks, and |F1| ≤ k−1

2 (see Fig. 1).

Fig. 1

Compute the sum of degrees of the vertices in F1. Consider a block Bi (1 ≤ i ≤ l). Assume that in Bi thereare ti marked vertices of the first kind adjacent to F1 and si marked vertices of the second kind adjacent to F1.The marked vertices belong to neither F1 (as they are inner vertices of some blocks) nor M . Hence all thesevertices belong to R, and we obtain the inequality

l∑i=1

(ti + si) ≤ k − 1.

In marking vertices of the second kind, we considered all vertices in F1 and, at each step, we marked at mostone vertex still unmarked and adjacent to the vertex in F1 under consideration. Since in the block Bi there aresi marked vertices of the second kind adjacent to F1, this means that in considering vertices in Bi, we markedvertices of the second kind at no more than si steps related to vertices from F1. If, in F1, there are more than si

vertices adjacent to vi, then, in considering one of these vertices, we marked no new vertex of the second kind.But this is impossible because, at this moment, there was an unmarked vertex adjacent to the vertex considered(vi). Thus, vi is adjacent to at most si vertices in F1.

Similarly, one can prove that vi is adjacent to at most ti vertices in F1. Therefore, vi is adjacent to at most(ti+si)

2 vertices in F1. Summing up these inequalities over all i from 1 to l, we conclude that there are at most12

∑li=1 (ti + si) edges connecting vertices in M = {vi}l

i=1 with vertices in F1.A vertex from F1 can only be adjacent in G to vertices in M = {vi}l

i=1, to vertices in the (k − 1)-cutset R,and to other vertices in F1. The number of vertices in F1 is denoted by p.

As the degree of every vertex in F1 does not exceed 3k−12 , by using the inequalities for the sums of ti and si,

we obtain the following inequality for the sum of degrees of the vertices in F1:

p

(3k − 1

2

)≤ p(p − 1) + p(k − 1) +

∑li=1 (ti + si)

2≤ p(p − 1) + p(k − 1) +

k − 12

,

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whence

0 ≤ p2 − p

(k + 3

2

)+

k − 12

= (p − 1)(

p − k + 12

)− 1.

On the right-hand side, we have a quadratic trinomial with a positive highest coefficient. At p = 1 andp = k+1

2, this trinomial takes the value −1. But 1 ≤ p ≤ k−1

2. Thus, at every admissible p, the value of this

trinomial is negative. The contradiction obtained completes the proof of the theorem. �Theorem 2. Let G be a k-connected graph all of whose vertices are of degree no less than 3k

2. Let {Bi}n

i=1 bea set of blocks in G such that |I(Bi)| > |R(Bi)| + 1. Then in each block Bi one can choose an inner vertex insuch a way that these vertices form a removable set.

Proof. As the proof of this theorem is quite lengthy, first we describe its idea.At the beginning, we mark some vertices in the interior of every block in such a way that there are at least

two unmarked vertices in each block, and every boundary vertex of each block is adjacent to at least one markedvertex. We will search for a removable set among the vertices unmarked. To this end, we consider a minimalirremovable set and prove that after its removal, for each (k−1)-cutset, one of the components separated by thiscutset contains only one vertex. Then we consider all the vertices separated in this way and prove that each ofthem must be adjacent to quite many of the unmarked vertices (see inequality (1)). Further we construct a setof unmarked vertices such that for every separated vertex this inequality cannot be valid. Hence the removal ofthis set results in that none of these vertices is separated by a (k − 1)-cutset. Thus, the set constructed cannotbe irremovable.

Now we present the detailed proof.A vertex w in R(Bi) is said to have degree m with respect to the block Bi if it is adjacent to exactly m inner

vertices of this block. The degree of a vertex w with respect to a block Bi is denoted by deg (w, Bi). The set ofall vertices of degree m with respect to a block Bi is denoted by R(Bi, m). Recall that R(Bi, 2) ⊂ R(Bi).

For every block Bi, we construct a supplementary multigraph (i.e., a graph, possibly, with multiple edges) Dsuch that V (D) = I(Bi), E(D) = R(Bi, 2), and vertices u and w are connected in D by an edge w if and only ifthe vertex w is adjacent in G to both u and v.

Consider all the connected components of this multigraph that have more than one vertex, i.e., the connectedcomponents with one edge at least. Denote them by D1, . . . , Dz. Denote the spanning tree of the graph Dj byT (Dj) and the total number of edges in the multigraphs Ds with subsripts s < j by Nj . For generality, the totalnumber of edges in the multigraph D is denoted by Nz+1.

Consider an arbitrary leaf of the tree T (Dj) and denote it by uj. At step r, we arbitrarily choose an edge ofT (Dj) one of whose ends is already denoted, whereas the other one is not. This edge is denoted by dNj+r . Thenondenoted end of this edge is denoted by vNj+r. This operations is repeated until all the vertices of the treeT (Dj) are denoted (and each edge of T (Dj) is denoted by ds for a certain s). If, in the multigraph Dj , a vertexuj is not a leaf, then we consider an edge d ∈ E(Dj) incident with this vertex and not belonging to E(T (Dj )).This edge is denoted by dNj+|T (Dj)|, i.e., it receives the number following that of the edge numbered last; thevertex uj is redenoted by vNj+|T (Dj)|. The remaining edges are arbitrarily denoted by ds, where s varies fromeither Nj + |T (Dj)| (if uj is a leaf in Dj) or Nj + |T (Dj)| + 1 (if uj is not a leaf in Dj) to Nj+1.

Thus, d1, . . . , dNz+1 are all the edges of the multigraph D. Since, in constructing this multigraph, we haveassumed that the set of its edges is R(Bi, 2), all the vertices of the original graph that are of degree two withrespect to the block Bi are denoted in this way.

Continue vertex enumeration in the graph G. Vertices in R(Bi) whose degrees with respect to Bi differ fromtwo are arbitrarily denoted by ds, s varying from Nz+1 + 1 to |R(Bi)|. Thus, all the vertices of degree two withrespect to Bi are put at the beginning of the vertex list in a chosen order, and vertices of other degrees are putat the end of this list in an arbitrary order.

Consider the vertex d1 and, in Bi, mark the inner vertex v1, adjacent to d1 (which was chosen during theenumeration process described above), if deg (d1, Bi) = 2 or an arbitrary vertex adjacent to d1 (if such a vertexexists) otherwise.

Next, at step s, we consider the vertex ds and mark a vertex in Bi adjacent to ds in the following way.If deg (ds, Bi) = 2 and, in the multigraph D, there is a vertex denoted by vs, then this vertex is marked. Ifdeg (ds, Bi) = 2 and in D there is no vertex vs, then, by construction, both ends of ds have subscripts less thans. Hence they are already marked, i.e., there are no more unmarked vertices adjacent to ds. If deg (ds, Bi) �= 2,then we mark an arbitrary vertex adjacent to ds (if such a vertex exists).

Thus, in the interior of Bi, at most |R(Bi)| vertices are marked.

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Remark 1. If, at step s, no vertex is marked, then no unmarked vertex is adjacent to ds.

Since, by assumption, |I(Bi)| > |R(Bi)| + 1, in the block Bi there are at least two unmarked inner vertices.Choose two such vertices arbitrarily and denote them by wi1 and wi2.

Since for r = 1, 2 the vertex wir was not marked, we conclude that in considering the multigraph D, thisvertex was denoted by vs for no s such that deg (ds, Bi) = 2 (otherwise it would be marked in considering thevertex ds or previously). Therefore, wir (as a vertex of the multigraph D) is either a vertex of zero degreebelonging to no graph Dj or a leaf uj for a certain j.

Thus, each of these vertices is adjacent to at most one vertex of degree two with respect to the block Bi.Let W = {wir | i = 1, . . . , n; r = 1, 2}.If there is a number set {ri | ri ∈ {1, 2}}n

i=1 such that the vertex set {wiri}ni=1 ⊂ W is irremovable, then, in

the latter set, there is a minimal irremovable subset. Without loss of generality, we assume that this subset isM = {wiri}l

i=1. In the graph G− M , there is a (k − 1)-cutset R. By Lemma 3, in the graph G −M , there is acomponent F1 separated by R that is contained in every set from Rk(G) disconnecting M , and |F1| ≤ k−1

2.

Compute the sum of degrees of the vertices in F1. Consider a block Bi (1 ≤ i ≤ l). Let Bi contain ti markedvertices adjacent to F1. All these vertices belong to R. Thus, we obtain the inequality

l∑i=1

ti ≤ k − 1.

Since in the block Bi there are ti marked vertices adjacent to F1, we marked vertices at at most ti stepsperformed for vertices in F1. By Remark 1, if at step s no vertex was marked, then no unmarked vertex adjacentto ds existed. In particular, the vertex wiri is not adjacent to ds. Therefore, the vertex wiri , which was removedfrom Bi, was adjacent to at most ti vertices in F1.

In the graph G, a vertex from F1 can be adjacent to vertices in M = {wiri}li=1, to vertices in R ∈ Rk−1(G−M),

and to other vertices in F1. The number of vertices in F1 is denoted by p.Since the degree of every vertex in F1 must be at least 3k

2 , by using the inequalities for the sum of ti, weobtain the following inequality for the total sum of degrees of the vertices in F1:

p

(3k

2

)≤ p(p − 1) + p(k − 1) +

l∑i=1

ti ≤ p(p − 1) + p(k − 1) + k − 1,

whence

0 ≤ p2 − p

(k

2+ 2

)+ k − 1 = (p − 2)

(p − k

2

)− 1.

On the right-hand side, we have a quadratic trinomial with a positive highest coefficient. At p = 2 and p = k2 ,

this trinomial takes the value −1. But 1 ≤ p ≤ k−12 . Hence, at every admissible p, except for p = 1, the value

of this trinomial is negative. Therefore, p = 1, i.e., F1 consists of a single vertex. This vertex is denoted by x.The set of all vertices x for which one can find a minimal irremovable set M ⊂ W consisting of inner vertices ofdifferent blocks and R ∈ Rk−1(G − M) separating x is denoted by X.

As M is minimal and R separates x in G−M , x must be adjacent in G to k− 1 vertices in R, to all l verticesin M , and only to them (see Fig. 2). Thus, l + k − 1 = deg v ≥ 3k

2 , implying that l ≥ k2 + 1. For all i = 1, . . . , l,

the vertices wiri are adjacent to x, and the number of vertices in F1 = {x} adjacent to wiri does not exceed ti.Therefore, for all such i, ti ≥ 1. As x is adjacent to ti marked vertices of the block Bi and to the vertex wiri ,we have the inequality deg (x, Bi) ≥ ti + 1 ≥ 2. Thus, for all i = 1, . . . , l, the numbers hi = deg (x, Bi) − 2 arenonnegative.Therefore, we have

l∑i=1

hi =l∑

i=1

(deg (x, Bi) − 2) ≤ deg (x) − 2l = l + k − 1 − 2l = 2(

k

2+ 1

)− l − 3 ≤ l − 3.

Since all the numbers his are nonnegative, we conclude that at least three of them are zero, i.e., the degrees ofx with respect to at least three blocks are equal to two.

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Fig. 2

The vertices wiri adjacent to x and such that hi = 0 are said to be distinguished for x. A vertex distinguishedfor x does not necessarily belong to M . Since, as was shown above, every vertex wir is adjacent to at most onevertex of degree two with respect to the block Bi, it can be distinguished for a unique vertex x ∈ X.

Let x be adjacent to a vertices distinguished for x and not belonging to M . If, for some i, both wi1 and wi2

are adjacent to x, then none of them is distinguished for x. Indeed, if one of them were distinguished for x,then, by the definition of a distinguished vertex, there would be exactly two vertices adjacent to x in Bi. On theother hand, by the construction of marked vertices, there must be a marked vertex adjacent to x, and neitherwi1 nor wi2 is marked. Therefore, as all the vertices in M are adjacent to x and each block Bi (1 ≤ i ≤ l)has a vertex from M , all the a vertices distinguished for x and not belonging to M are located in blocks withsubscripts greater than l. For each such block Bi, we have deg (x, Bi) = 2. Consequently, together all the blockswith subscripts greater than l contain at least 2a vertices adjacent to x. Therefore,

l∑i=1

hi =l∑

i=1

(deg (x, Bi) − 2) ≤ deg (x) − 2a − 2l = l + k − 1 − 2l − 2a ≤ 2(

k

2+ 1

)− l − 3 − 2a ≤ l − 3 − 2a.

This means that at least 2a+3 vertices in M are distinguished for x. Therefore, if there are b vertices distinguishedfor x, then b ≥ 3a + 3, whence a ≤ b

3 − 1. Since a is an integer, we have

a ≤[

b

3

]− 1. (1)

Thus, if we want to construct a removable vertex set Y = {yi | yi ∈ I(Bi)}ni=1, then, for each vertex x ∈ X,

suffice it to ensure that Y does not contain at least[

b3

]vertices distinguished for x. In this case, in Y there is

no minimal irremovable subset M whose removal results in that the vertex x is separated by a (k− 1)-cutset R.Thus, if we arbitrarily partition the vertices distinguished for x into

[b2

]pairs and, possibly, an exceptional

set consisting of a single vertex, then, suffice it to exclude from Y one vertex of each pair. Then Y does notcontain even [ b

2 ] vertices distinguished for x.All the pairs obtained in this way for all vertices x ∈ X are denoted by Ps, s = 1, . . . , m. Recall that each

vertex in W can be a distinguished one for one vertex x ∈ X at most, and therefore can occur in at most onepair Ps.

In W = {wir | i = 1, . . . , n; r = 1, 2}, choose a removable subset Y ={yi | yi ∈ I(Bi)

}n

i=1in the following way.

As y1 take w11, which is an inner vertex of the block B1. If y1 belongs to no pair Ps, then as y2 take the vertexw21. If y1 is contained in a pair Ps, then consider the other vertex of this pair. Without loss of generality, wemay assume it to be w22. Then as y2 also take the vertex w21.

If a vertex yj−1 chosen at step j − 1 and belonging to I(Bj−1) is contained in no pair Ps, then as yj take thevertex wj1. If yj−1 is contained in a pair Ps, then determine which block contains the other vertex of this pair,wtr. If t > j − 1, then, without loss of generality, assume it to be wj2. In this case, as yj also take the vertexwj1. As was shown above, the vertices wt1 and wt2 cannot be distinguished for the same vertex x ∈ X. Hencethey cannot be contained in the same pair, and therefore the case t = j − 1 is impossible. If t < j − 1, then, atstep t + 1, the vertex yj−1 must occur in W\Y . Hence the vertex wtr cannot be in Y .

Therefore, two vertices of a pair cannot simultaneously belong to Y . Thus, we have constructed a set ofinner vertices belonging to different blocks that has no minimal irremovable subset, i.e., we have constructed aremovable set. �

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The author expresses his gratitude to D. V. Karpov and A. V. Pastor for attracting his attention to the topicand for invaluable advice.

Translated by A. S. Chukhnov.

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1. G. Chartrand, A. Kaugars, and D. R. Lick, “Critically n-connected graphs,” Proc. Amer. Math. Soc., 32,No. 1, 63–68 (1972).

2. D. V. Karpov and A. V. Pastor, “On the structure of a k-connected graph,” Zap. Nauchn. Semin. POMI,266, 76–106 (2000).

3. D. V. Karpov, “Blocks in k-connected graphs,” Zap. Nauchn. Semin. POMI, 293, 59–93 (2002).

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