Upper bound on the minimum degree of minimally 1-tough, bipar-

80 Strengthening some results on toughness of bipartite graphs

so

|S|>

{i∈[k] :d(ui)≥2}

. Therefore,

{i∈[k] :d(u_i) = 1}

=k−

{i∈[k] :d(u_i)≥2}

>2|S| − |S|=|S|, which means that there exists a vertex in S having at least two neighbors in {u₁, . . . , u_k} of degree 1. Then the removal of this vertex leaves at least three components (and note that since G is 3-regular, it cannot leave more than three components), so τ(G)≤1/3.

From this it also follows thatτ(G) = 1/3if and only if there exists a cut-vertex whose removal leaves three components.

To summarize, it can be decided in polynomial time whether a connected 3- regular graph is 2/3-tough, and if it is not, then its toughness is either 1/3 or 1/2.

In both cases the graph contains at least one cut-vertex, and if the removal of any of them leaves (at least) three components, then the toughness of the graph is 1/3, otherwise it is 1/2.

Theorem 5.18 (Katona, Varga, [4]). There is a polynomial time algorithm to recognize 1/2-tough 4-regular graphs.

The proof of this theorem follows directly from the following claim.

Claim 5.19 (Katona, Varga, [4]). The toughness of any connected 4-regular graph is at least 1/2.

Proof. LetGbe a connected 4-regular graph and letS be an arbitrary cutset inG andLbe a component ofG−S. Since every vertex has degree 4 inG, the number of edges between S and L is even (more precisely, it is equal to the sum of the degrees in G of the vertices of L minus two times the number of edges induced by L). Since G is connected, the number of these edges is at least two. On the other hand, since G is 4-regular, there are at most 4|S| edges between S and L. Therefore ω(G−S)≤2|S|, which means that τ(G)≥1/2.

5.5 Upper bound on the minimum degree of min-

Properties of minimally tough graphs 81

Theorem 5.20 (Katona, Varga). Every minimally 1-tough, bipartite graph onn vertices has a vertex of degree at most(n+ 6)/4.

Proof of Theorem 5.20. Suppose to the contrary thatδ(G)>(n+6)/4. Obviously, a 1-tough bipartite graph must be balanced and therefore the number of its vertices must be even, hence δ(G) ≥ n/4 + 2. Consider an arbitrary edge e = uv. By Propostition 2.4, there exists a vertex set S = S(e), whose removal from G−e leaves|S|+ 1connected components, anduandv belong to dierent components.

Let k = |S| and let Lu and Lv denote the components of u and v, respectively.

Obviously, the components of (G−e)−S require ω (G−e)−S

=k+ 1

independent vertices: two of them can beuand v, but the rest of them cannot be adjacent either to u or to v, and the rest of them cannot be either in L_u∪L_v or inS.

Since there are no triangles in the graph, the open neighborhood of u and v contains at least

2·n 4 + 2

−2 = n 2 + 2 vertices and at most k of them belongs to S, so at least

n 2 + 2

−k vertices belong to L_u∪L_v.

Since G is bipartite, the components of G−S are also bipartite and since G is 1-tough, the sizes of the two color classes of these components can dier in at most 1. Therefore,

|L_u|+|L_v| ≥2·n

2 + 2−k

−2 =n−2k+ 2. Hence, for the remaining k−1 components there are only

n−((n−2k+ 2) +k) = k−2 vertices, which is a contradiction.

Summary

The main focus of the thesis is on minimally tough graphs.

Chapter 3is motivated by a conjecture, stating that every minimally 1-tough graph on n vertices has a vertex of degree 2. In this chapter, we give an upper bound on the minimum degree of minimally 1-tough graphs, namely n/3 + 1.

Chapter4investigates the complexity of recognizing minimallyt-tough graphs.

There we prove that this problem is DP-complete for all positive rational number t.

In Chapter 5, we study bipartite graphs. First, we show that recognizing t- tough bipartite graphs is coNP-complete for all positive rational number t ≤ 1 (the case t = 1 was already known). Motivated by two open problems regarding the complexity of recognizing 1-tough 3-connected bipartite graphs and 1-tough 3-regular bipartite graphs, we also prove that recognizing t-tough k-connected bipartite graphs and 1-toughr-regular bipartite graphs is also coNP-complete for any integers k ≥2and r≥6 and for any positive rational number t≤1.

82

Appendix A

A.1 Proof of Claim 4.6

Here we present the missing details of the proof of Claim4.6. First we handle the case t= 1 and to complete it, we only need to prove Lemma4.7.

Proof of Lemma 4.7. LetS ⊆V(G⁰_1,k)be a cutset in G⁰_1,k. We need to show that ω(G⁰_1,k −S)≤ |S|.

Case 1: W ⊆S.

If S contains a vertex of V, then we can assume that its only neighbor in U does not belong to S and, therefore, is an isolated vertex in the graph G⁰_1,k−S. Hence there are two types of components in G⁰_1,k−S:

(a) isolated vertices from U and

(b) components containing at least one vertex from V.

There are at most α(G)components of type (b) since picking a vertex of V from each such component forms an independent set of G⁰_1,k[V]. On the other hand, there are exactly|V ∩S| ≤ |S\W|=|S| −k components of type (a). Thus

ω(G⁰_1,k−S)≤ |S| −k+α(G)≤ |S|. Case 2: W *S.

First, we make some convenient assumptions for S. (1) U ∩S=∅.

Suppose thatu_i1,j1,1 ∈S for some i₁ ∈[n], j₁ ∈[k]. Ifv_i1,j1,1 ∈S orw_i1,j1,1 ∈ S, then considering S⁰ = S \ {ui1,j1,1} instead of S either increases the

83

84 Appendix

number of components by one or does not change it. Then it is enough to show thatω(G⁰_1,k−S⁰)≤ |S⁰| since it implies

ω(G⁰_1,k−S)≤ω(G⁰_1,k−S⁰)≤ |S⁰| ≤ |S| −1≤ |S|.

If v_i1,j1,1, w_j1,1 ∈/ S, then considering S⁰ = S \ {u_i1,j1,1} instead of S either decreases the number of components by one or does not change it. This means that if S⁰ is a cutset inG⁰_1,k, then it is enough to show thatω(G⁰_1,k− S⁰)≤ |S⁰| since it implies

ω(G⁰_1,k−S)≤ω(G⁰_1,k−S⁰) + 1≤ |S⁰|+ 1 = |S| −1

+ 1 =|S|. If S⁰ is not a cutset in G⁰_1,k, then ω(G⁰_1,k−S) = 2 and |S| ≥ 2 since u_i1,j1,1 has degree 2 and not a cut-vertex inG⁰_1,k; thus

ω(G⁰_1,k−S) = 2≤ |S|. (2) For all i∈[n], either V_i ⊆S or V_i∩S =∅.

Suppose that only a proper subset ofV_i1 is contained inS for some i₁ ∈[n], and let

J =

j ∈[k]

v_i1,j,1 ∈S . Then J 6=∅ and J 6= [k]. Let j₁ ∈J.

If w_j1,1 ∈ S, then considering S⁰ = S\ {v_i1,j1,1} instead of S decreases the number of components by one. This means that ifS⁰ is a cutset inG⁰_1,k, then it is enough to show that ω(G⁰_1,k−S⁰) ≤ |S⁰| since (similarly as earlier) it implies that ω(G⁰_1,k−S)≤ |S|. If, however, S⁰ is not a cutset in G⁰_1,k, then ω(G⁰_1,k−S) = 2 since J 6= [k], and |S| ≥2 since vi1,j1,1 is not a cut-vertex inG⁰_1,k; thus

ω(G⁰_1,k−S) = 2≤ |S|.

Ifw_j1,1 ∈/ S, then by assumption (1), consideringS⁰ = S\{v_i1,j1,1}

∪{w_j1,1} instead ofSdoes not decrease the number of components. Thus it is enough to show that ω(G⁰_1,k−S⁰) ≤ |S⁰| since (similarly as earlier) it implies that ω(G⁰_1,k−S)≤ |S|.

(3) There existsi₀ ∈[n]for which V_i0 ∩S =∅.

Supposing thatV_i∩S 6=∅for alli∈[n], assumption (2) implies thatV ⊆S and thus, by assumption (1),

|S|=|V|+|W ∩S|=kn+|W ∩S|

Properties of minimally tough graphs 85

and

ω(G−S) = |W ∩S| ·n+|W \S|=|W|+|W ∩S| ·(n−1)

≤k+k(n−1) = kn≤S.

Let i₀ ∈ [n] for which V_i0 ∩S = ∅, and let j₀ ∈ [k] for which w_j0,1 ∈/ S. By assumption (1), the component ofw_j0,1 contains all the verticesu_1,j0,1,. . .,u_i0,j0,1, . . ., u_n,j0,1 and thus all the vertices of V_i0, which, by assumption (1), implies that this component also contains all the remaining vertices of W and thus all the remaining vertices of V. Therefore, in G⁰_1,k −S there are isolated vertices from U and one more component containing the remaining vertices of W and V. By assumption (2) and by the fact that all the verticesu_1,j0,1, . . ., u_i0,j0,1, . . ., u_n,j0,1 are contained in the component of w_j0,1 in G⁰_1,k−S, there are less than |V ∩S|

isolated vertices fromU. Thus,

ω(G⁰_1,k−S)≤ |V ∩S| ≤ |S|.

Hence,G⁰_1,k is 1-tough, i.e.τ(G⁰_1,k)≥1. Clearly,τ(G⁰_1,k)≤1since every vertex of U has degree 2. Thus, τ(G⁰_1,k) = 1.

Now we prove Claim 4.6 in the case when t≥2.

Proof of Claim 4.6, when t ≥2. First we need the following lemma.

Lemma A.1. If α(G)≤k, then G⁰_t,k is t-tough.

Proof. LetS ⊆V(G⁰_t,k)be a cutset. We need to prove that ω(G⁰_t,k−S)≤ |S|/t. Case 1: W ⊆S.

In this case there are two types of components in G⁰_t,k−S: (a) components containing vertices only from U and

(b) components containing at least one vertex from V.

There are at most α(G) components of type (b). To obtain a component of type (a), we need to remove at leastt vertices of V ∪U. So there are at most

(V ∪U)∩S t

86 Appendix

components of type (a). Now

|S|=

(V ∪U)∩S +tk and

ω(G⁰_t,k−S)≤

(V ∪U)∩S

t +α(G)≤ |S|

t . Case 2: W *S.

First we show that the following assumptions can be made for S. (1) For any j ∈[k]either W_j ⊆S orW_j∩S =∅.

Otherwise consideringS⁰ =S\W_j instead ofSdoes not decrease the number of components, but decreases the number of removed vertices.

(2) If Sn

i=1U_i,j1 ⊆S for some j₁ ∈[k], then W_j1 ∩S =∅.

Otherwise considering S⁰ = S \W_j1 instead of S increase the number of components, but decreases the number of removed vertices.

(3) If V_i1,j1 ∪W_j1 ⊆S for some i₁ ∈[n] and j₁ ∈[k], then U_i1,j1 ∩S =∅.

Otherwise considering S⁰ = S\U_i1,j1 instead of S increase the number of components, but decreases the number of removed vertices.

Let w_j0,l0 ∈W \S be xed and let I₀ =

i∈[n]

U_i,j0 ⊆S . Since w_j0,l0 ∈/S, assumption (1) implies W_j0 ∩S =∅. Case 2.1: (W *S and) I₀ = [n], i.e. Sn

i=1U_i,j0 ⊆S.

First note that in this case the vertices of W_j0 are isolated in G⁰_t,k−S. Case 2.1.1: (W *S, I₀ = [n] and) ω(G⁰_t,k−S) =t+ 1.

This means that except for the t isolated vertices of W_j0, there is only one other component in G⁰_t,k−S. Then

|S| ≥

n

[

i=1

U_i,j

=nt,

Properties of minimally tough graphs 87

and since n≥t+ 1,

ω(G⁰_t,k−S) =t+ 1≤n≤ |S|

t . Case 2.1.2: (W *S,I₀ = [n] and) ω(G⁰_t,k−S)> t+ 1. Let

I₁ =

i∈[n]

V_i,j0 ⊆S . Since ω(G⁰_t,k−S)> t+ 1,

S⁰ = S\ [

i∈I1

U_i,j0

!!

∪W_j0

is a cutset inG⁰_t,k. It is not dicult to see that

|S⁰|=|S| − |I1| ·t+t and

ω(G⁰_t,k−S⁰) =ω(G⁰_t,k−S) +|I₁| −t. Now there are two cases.

Case 2.1.2.1: (W *S, I₀ = [n], ω(G⁰_t,k−S)> t+ 1 and) |I₁| ≥(t+ 1)/2. Then it is enough to prove that ω(G⁰_t,k−S⁰)≤ |S⁰|/t since it implies

ω(G⁰_t,k−S) =ω(G⁰_t,k−S⁰)− |I₁|+t ≤ |S⁰|

t − |I₁|+t

= |S| − |I₁| ·t+t

t − |I₁|+t = |S|

t −2|I₁|+t+ 1

≤ |S|

t −2·t+ 1

2 +t+ 1 = |S|

t .

Case 2.1.2.2: (W *S, I₀ = [n], ω(G⁰_t,k−S)> t+ 1 and) |I₁|<(t+ 1)/2. In G⁰_t,k−S there are three types of components:

(a) components containing at least one vertex of V, (b) isolated vertices of W,

(c) components containing at least one vertex of U and not containing any vertices ofV.

88 Appendix

Since G is d(t+ 1)/2e-connected and |I₁| < (t+ 1)/2 holds, (G⁰_t,k −S)[V] is connected, i.e. there is only one component of type (a).

Let

J = (

j ∈[k]

n

[

i=1

U_i,j ⊆S )

.

Obviously, j₀ ∈J, hence J 6= ∅. Assumption (2) implies that W_j ∩S =∅ for all j ∈J. Hence there are exactly|J| ·t components of type (b).

Assume that u_i1,j1,l1 belongs to a component of type (c) for somei₁ ∈[n], j₁ ∈ [k], l₁ ∈[t]. To obtain this component, we need to remove at least t vertices from V_i1,j1∪U_i1,j1. Let m denote the number of components of type (c).

Then

ω(G⁰_t,k−S) = 1 +|J| ·t+m and

|S| ≥ |J| ·nt+mt, hence

ω(G⁰_t,k−S) = 1 +|J| ·t+m = |J| ·nt+mt

t − |J| ·n+|J| ·t+ 1

≤ |S|

t − |J| ·(n−t) + 1≤ |S|

t − |J|+ 1≤ |S|

t ,

where the second last inequality is valid since n ≥ t+ 1, and so is the last one since J 6=∅.

Case 2.2: (W *S and) I₀ 6= [n], i.e. Sn

i=1U_i,j0 *S.

Case 2.2.1: (W * S, I₀ 6= [n] and) for any i ∈ [n]\I₀ there exists l ∈ [t] for which v_i,j0,l, u_i,j0,l ∈/S.

Then V_i,j0 *S for all i∈[n]\I₀. We can also assume that V_i,j0 *S holds for alli∈I₀, otherwise sinceI₀ 6= [n], considering S⁰ =S\U_i,j0 instead ofS does not change the number of components.

Therefore,(G⁰_t,k−S)[V]is connected. Thus there are three types of components:

(a) components containing at least one vertex of V, (b) isolated vertices of W,

(c) components containing at least one vertex of U and not containing any vertices ofV.

Properties of minimally tough graphs 89

Let

J = (

j ∈[k]

n

[

i=1

U_i,j ⊆S )

.

IfJ 6=∅, then similarly as in Case 2.1.2.2, we can conclude thatω(G⁰_t,k−S)≤ |S|/t holds.

So assume that J =∅, i.e. there are no components of type (b). Since (G_t,k− S)[V] is connected, there is only one component of type (a).

Assume thatu_i1,j1,l1 belongs to a component of type (c) for some i₁ ∈[n], j₁ ∈ [k], l₁ ∈[t]. Now to obtain this component, we need to remove not only at least t vertices fromVi1,j1∪Ui1,j1, but at least t other vertices: either all thet vertices of W_j1 or at least (n−1)t vertices from

[

i∈[n], i6=i₁

V_i,j1 ∪U_i,j1 .

Let m denote the number of components of type (c). Then ω(G⁰_t,k−S) = 1 +m

and

|S| ≥mt+t, hence

ω(G⁰_t,k−S)≤ |S|

t .

Case 2.2.2: (W *S,I₀ 6= [n]and) there existsi₁ ∈[n]\I₀ such thatv_i1,j0,l ∈S orui1,j0,l ∈S for all l∈[t].

IfI₀ 6= [n]\ {i₁}, then S⁰ =S∪W_j0 is a cutset in G⁰_t,k. It is not dicult to see that

ω(G⁰_t,k−S⁰)≥ω(G⁰_t,k−S) + 1 and

|S⁰|=|S|+t.

Then it is enough to show thatω(G⁰_t,k−S⁰)≤ |S⁰|/t since it implies ω(G⁰_t,k−S)≤ω(G⁰_t,k−S⁰)−1≤ |S⁰|

t −1 = |S|+t

t −1 = |S|

t .

90 Appendix

If I₀ = [n]\ {i₁}, then S⁰ =S∪U_i1,j0 is a cutset in G⁰_t,k. It is not dicult to see that

ω(G⁰_t,k−S⁰)≤ω(G⁰_t,k−S) +t−1 and

|S⁰| ≤ |S|+t.

Then it is enough to show that ω(G⁰_t,k−S⁰)≤ |S⁰|/t since it implies ω(G⁰_t,k−S)≤ω(G⁰_t,k−S⁰)−t+ 1≤ |S⁰|

t −t+ 1 = |S|+t

t −t+ 1

= |S|

t −t+ 2≤ |S|

t , where the last inequality is valid since t≥2.

Hence, G⁰_t,k is t-tough, i.e. τ(G⁰_t,k)≥ t. Clearly, τ(G⁰_t,k)≤ t since every vertex of U has degree 2t. Thus, τ(G⁰_t,k) =t.

Now we return to the proof of Claim 4.6. All we have left to show is that ifGisα-critical withα(G) = k, thenτ(G⁰_t,k−e)< tholds for anye∈E(G), if α(G)> k, then τ(G⁰_t,k)< t, and

if either α(G) =k but the graph Gis notα-critical orα(G)< k, then there exists an edge e∈E(G)for which τ(G⁰_t,k−e) = t.

Assume rst thatGisα-critical withα(G) = k. Lete∈E(G⁰_t,k)be an arbitrary edge. Ife is incident to one of the vertices of U, i.e., to a vertex of degree2t, then clearly τ(G⁰_t,k −e) < t. If e is not incident to any of the vertices of U, then it connects two vertices of V. By Lemma 2.19, the subgraphG⁰_t,k[V]isα-critical, so inG⁰_t,k[V]−ethere exists an independent vertex setI of sizeα(G) + 1. LetI⁰ ⊂I be an independent vertex set of sizeα(G) in G⁰_t,k[V] (clearly, such a vertex set I⁰ exists). Let

S = (V \I)∪ {u_i,j |v_i,j ∈I⁰} ∪W. Then

|S|= (|V| −1) +tk and

ω (G⁰_t,k−e)−S

= |V|

t +α(G)> |S|

t ,

Properties of minimally tough graphs 91

soτ(G⁰_t,k−e)< t.

Now assume α(G)> k. Then let I be an independent vertex set of size α(G) inG⁰_t,k[V] and let

S=W ∪(V \I)∪ {u_i,j |v_i,j ∈I}. Then

|S|=|V|+tk and

ω(G⁰_t,k−S) = |V|

t +α(G)> |S|

t , soτ(G_t,k)< t.

Finally, assume that either α(G) = k but the graph G is not α-critical or α(G) < k. Then there exists an edge e ∈ E(G) such that α(G −e) ≤ k. By Lemma A.1, the graph (G−e)⁰_t,k is t-tough, but we can obtain (G−e)⁰_t,k from G⁰_t,k by edge-deletion, which means that G⁰_t,k is not minimally t-tough.

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