The proof of Theorem 4.1 when t ≥ 1

Theorem 4.18 (Katona, Kovács, Varga, [2]). For any rational numbert≥1, the problem Min-t-Tough is DP-complete.

Proof. Lett≥1be a rational number. In Proposition 4.5we already proved that the problem Min-t-Tough is in DP. To show that it is DP-hard, we reduce the variant of α-Critical dened in Proposition 2.21 to it.

Let T = dte, and T⁰ = d2te − dte, and M =

2dte/d2te

as before. Let a, bbe the smallest positive integers such thatb ≥3and t=a/b, let Gbe an arbitrary 3-connected graph on the vertices v₁, . . . , v_n with n ≥ t+ 1, let k be a positive integer that is divisible by a and letGt,k be dened as follows.

For alli∈[n], j ∈[k], m∈[M] let then place a complete bipartite graph on(Vi1;Vi2). (This subgraph is denoted by G˜ in Figure 4.6.) For all i ∈ [n], j ∈ [k], m ∈ [M] glue the graph H_t⁰⁰

Properties of minimally tough graphs 53

Now k is part of the input of the problem α-Critical, therefore the graph H_t,k^∗∗ must be constructed in polynomial time and by Theorem 2.11, this can be done. On the other hand, t is not part of the input of the problem Min-t-Tough, therefore the graphH_t⁰⁰can be constructed in advance. Hence, G_t,k can be constructed from Gin polynomial time.

G˜ t-tough, rst we prove the following lemma.

Lemma 4.19. LetGbe an arbitrary3-connected graph onn ≥t+ 1vertices with α(G)≤k. ThenG_t,k is t-tough.

54Chapter 4. The complexity of recognizing minimally tough graphs

Proof. LetS ⊆V(G_t,k) be a cutset in G_t,k. We need to show that ω(G_t,k − S)≤ |S|/t.

Case 1: W ⊆S.

After the removal of W, the vertices of W⁰ are isolated, therefore we can assume thatW⁰ ∩S =∅. Using these notations it is clear that

|S|=|C| ·T + X fromH^i,j,m leaves at most

max components, but the component of the remaining vertices of V_i,j,m has been already counted. Hence

Properties of minimally tough graphs 55

Case 2: W *S.

There are four types of components in G_t,k−S: (a) components containing at least one vertex of V,

(b) components containing at least one vertex of U but no vertices of V, (c) components containing at least one vertex ofW but no vertices ofU∪V, (d) isolated vertices of W⁰.

Letw_j0,l0,m0 ∈W \S be xed. First we show that the following assump-tions can be made for S.

(1) S∩U⁰⁰ =∅.

Obviously, the number of vertices of W that belong to a component of type (c) is at most |S ∩U⁰⁰|/n. Since the neighborhood of any vertex of U⁰⁰ spans a clique in G_t,k, considering the cutset

S⁰ = (S\U⁰⁰)∪

w∈W

w belongs to a component of type (c) instead of S can only increase the number of components of types (a), (b) and (d), while it decreases the number of components of type (c) to 0, i.e., by at most |S∩U⁰⁰|/n. Hence,

|S⁰| ≤ |S| − |S∩U⁰⁰|+|S∩U⁰⁰| n and

ω(G_t,k−S⁰)≥ω(G_t,k−S)−|S∩U⁰⁰| n .

Then it is enough to prove that ω(G_t,k−S⁰)≤ |S⁰|/t since it implies ω(G_t,k−S)≤ω(G_t,k−S⁰) + |S∩U⁰⁰|

n ≤ |S⁰|

t + |S∩U⁰⁰| n

≤ |S| − |S∩U⁰⁰|+|S∩U⁰⁰|/n

t + |S∩U⁰⁰| n

= |S|

t − |S∩U⁰⁰| · n−t−1 nt ≤ |S|

t , where the last inequality is valid since n≥t+ 1.

56Chapter 4. The complexity of recognizing minimally tough graphs

(2) There are no components of type (c) in G_t,k−S. It follows directly from assumption (1).

(3) remaining vertices ofW_j0,m0. Now considering the cutset

S⁰ =S∪ {W_j0,m0}

instead of S increases the number of removed vertices by at most T⁰, and it increases the number of components by at least dT⁰/te since it disconnects the vertex sets U_i,j0,m0, i ∈ I from each other. Then it is

Proceeding further, we can obtain a cutsetS^∗ for which W ⊆S^∗ holds;

and such sets were already handled in Case 1.

(4) (G_t,k−S)[V]is connected, i.e. there is only one component of type (a).

Since G is 3-connected, assumption (2) implies that (G_t,k −S)[V] is connected.

Properties of minimally tough graphs 57

By assumption (2), there are no components of type (c), and by assump-tion (4), there is only one component of type (a). By the properties of H_t,k^∗∗, the removal of S∩(W ∪W⁰)from H_t,k^∗ leaves at most

components of type (d). Similarly as before, for any (i, j, m)∈Cthe removal of V(H^i,j,m)∩S from H^i,j,m leaves at most

components; the component of the remaining vertices of V_i,j,m is of type (a), all the others can be of type (b). By assumption (1), all the vertices of Sn

i=1U_i,j0,m0 belong to the component of w_j0,l0,m0, hence the component of w_j0,l0,m0 has been counted multiple times (more than once). Therefore,

ω(G_t,k−S)≤

Now we return to the proof of Theorem 4.18 and we show that G is α-critical withα(G) =k if and only if Gt,k is minimally t-tough.

Let us assume that G is α-critical with α(G) = k. By Lemma 4.19, the graph G_t,k ist-tough, i.e.τ(G_t,k)≥t.

58Chapter 4. The complexity of recognizing minimally tough graphs

Let I be an independent vertex set of size α(G) in G, and recall the denition of the setsX and Y₁, . . . , Y_T constructed in Subsection 4.4.3. Let

and after the removal ofS from G_t,k, the vertices of W⁰ are isolated and the other components of G_t,k−S are exactly the components of H^i,j,m− S ∩ V(H^i,j,m)

Properties of minimally tough graphs 59

Lete∈E(G_t,k)be an arbitrary edge. We need to show thatτ(G_t,k−e)< t. Now we have four cases.

Case 1: e has an endpoint in U⁰⁰.

Then this endpoint has degree d2te −1in Gt,k−e, so τ(G_t,k−e)≤ d2te −1

2 < 2t 2 =t. Case 2: e has an endpoint in W⁰.

By the properties of H_t,k^∗ , there exists a cutset S ⊆ W in H_t,k^∗ −e for which

ω (H_t,k^∗ −e)−S

> |S|

t . Note that S is also a cutset inG_t,k−e and

ω (Gt,k−e)−S

=ω (H_t,k^∗ −e)−S

> |S|

t , so τ(G_t,k−e)< t.

Case 3: e is induced byH^i0,j0,m0 for some i₀ ∈[n], j₀ ∈[k], m₀ ∈[M]. The case whene is induced byU_i⁰⁰_0,j0,m0 was already covered in Case 1. So assume thate is not induced by U_i⁰⁰_0,j0,m0. Then by Claim4.16, there exists a vertex set S ⊆V(H_t⁰⁰) for which

ω (H_t⁰⁰−e)−S

> |S|

t .

Consider the (i₀, j₀, m₀)-th copy of the vertex set S inG_t,k−e; let us denote it withS_i0,j0,m0. IfV_i0,j0,m0 ⊆S_i0,j0,m0, thenS_i0,j0,m0 is a cutset in G_t,k−eand

ω (G_t,k−e)−S_i0,j0,m0

=ω (H_t⁰⁰−e)−S

> |S|

t ,

so τ(Gt,k−e)< t. Assume that Vi0,j0,m0 *Si0,j0,m0. LetI be an independent vertex set of size α(G) in G that contains v_i0 (by Proposition 2.18, such an independent vertex set exists). Let

J =

i∈[n]

v_i ∈I

60Chapter 4. The complexity of recognizing minimally tough

Case 4: e connects two vertices of V.

Since the case when e is induced by H^i0,j0,m0 for some i₀ ∈ [n], j₀ ∈

Properties of minimally tough graphs 61

62Chapter 4. The complexity of recognizing minimally tough graphs

and similarly as before, ω (G_t,k−e)−S

=nkM bT +α(G) +

M T⁰ t −1

k > nkM bT +kM T⁰ t

= nkM aT +kM T⁰

t = |S|

t ,

soτ(G_t,k)< t, which means that G_t,k is not minimally t-tough.

Case II: α(G)≤k.

Since G is not α-critical with α(G) = k there exists an edge e ∈ E(G) such thatα(G−e)≤k. By Lemma4.19, the graph(G−e)_t,k ist-tough, but it can be obtained fromGt,k by edge-deletion, which means that Gt,k is not minimallyt-tough.

Therefore the problem Min-1-Tough is DP-complete, so by Claim 2.6, we can conclude the following.

Corollary 4.20 (Katona, Kovács, Varga, [2]). Recognizing almost minimally 1-tough graphs is DP-complete.

Let Almost-Min-1-Tough denote the problem of determining whether a given graph is almost minimally 1-tough.

In document KittiVarga Propertiesofminimallytoughgraphs Ph.D.Thesis (Pldal 54-64)