The auxiliary graph H t,k 00 when t ≥ 1

LetH_t be constructed as follows. Let

V₁⁰ ={v₁⁰, . . . , v_T⁰ }, V₂⁰ ={v⁰_T+1, . . . , v⁰_2T}, V₃⁰ ={v_2T⁰ ₊₁, . . . , v⁰_aT}, V⁰⁰={v⁰⁰₁, . . . , v⁰⁰_T},

U₁⁰ ={u⁰₁, . . . , u⁰_T}, U₂⁰ ={u⁰_T+1, . . . , u⁰_2T}, U₃⁰ ={u⁰_2T+1, . . . , u⁰_bT−1}, U⁰⁰ ={u⁰⁰₁, . . . , u⁰⁰_T0},

and

U₁⁰⁰ ={u⁰⁰₁, . . . , u⁰⁰_T}.

Place a clique on the vertices of V₁⁰, V₂⁰, V₃⁰, andU⁰⁰. For all l ∈ [T] connect v_l⁰⁰ to v⁰_l and to u⁰_l, and connect v_T⁰_+l to u⁰_T+l. Connect all the vertices of V₃⁰ to all the vertices of V₁⁰∪V⁰⁰∪U₁⁰ ∪U₂⁰, and connect all the vertices of V₂⁰ to all the vertices ofU⁰⁰. Finally, add a new vertexx to the graph and connect it to all the vertices of V₁⁰∪U⁰⁰. See Figure 4.5.

x

u⁰⁰₁ u⁰⁰_T0

U⁰⁰

v⁰_T+1 v_2T⁰ V₂⁰

v₁⁰ v⁰_T V₁⁰ v₁⁰⁰ v⁰⁰_T V⁰⁰ u⁰₁ u⁰_T U₁⁰ u⁰_T+1 u⁰_2T U₂⁰ u⁰_2T+1 u⁰_bT−1U₃⁰

v_2T⁰ ₊₁

v_aT⁰ V₃⁰

Figure 4.5: The graph Ht, whent ≥1.

Claim 4.14. For any rational number t ≥ 1, the graph Ht has weighted toughness t with respect to the weight function w that assigns weight 1 to all the vertices of H_t except for the vertex x, to which it assigns weightt.

Properties of minimally tough graphs 45

Proof. LetSbe an arbitrary cutset of H_t. We need to show thatω(H_t−S)≤ w(S)/t.

We can assume that either V₃⁰ ∩S =∅ or V₃⁰ ⊆ S since removing only a proper subset of V₃⁰ does not disconnect anything from the graph. Similarly, we can also assume that either U⁰⁰∩S =∅ orU⁰⁰ ⊆S.

Case 1: V₃⁰∩S =∅ and U⁰⁰∩S =∅.

ThenH_t−S has at most 2 components, and to obtain 2 components, the following must hold:

u⁰_T+l ∈S orv⁰_T+l∈S for all l ∈[T], and x∈S orV₁⁰ ⊆S.

Hence w(S)≥T +t and

ω(H_t−S) = 2≤ T +t

t ≤ w(S) t . Case 2: V₃⁰∩S =∅ and U⁰⁰ ⊆S.

Now we can assume thatx /∈S since after the removal ofU⁰⁰ removing x does not disconnect anything from the graph. Similarly, we can also assume that V₂⁰ * S. Then Ht − S has at most 3 components. To obtain three components, the following must hold:

(i) u⁰_T+l ∈S orv⁰_T+l∈S for all l ∈[T] (but V₂⁰ *S), and (ii) V₁⁰ ⊆S.

Hence w(S)≥T⁰+ 2T =d2te+T and

ω(H_t−S) = 3 ≤ d2te+T

t = w(S) t .

To obtain two components, either (i) or (ii) must hold; in both cases w(S)≥ d2te and

ω(Ht−S) = 2 ≤ d2te

t ≤ w(S) t . Case 3: V₃⁰ ⊆S.

First we show that the following assumptions can be made for S.

46Chapter 4. The complexity of recognizing minimally tough graphs

(1) (U₁⁰ ∪U₂⁰ ∪U₃⁰)∩S =∅.

After the removal of V₃⁰, removing any of the vertices of U₁⁰ ∪U₂⁰ ∪U₃⁰ does not disconnect anything from the graph.

(2) There exists at most one l∈[T]for which v_T⁰_+l ∈/ S, i.e. |V₂⁰\S| ≤1. Suppose that there existl₁, l₂ ∈[T]for which l₁ 6=l₂ and v⁰_T+l

1, v⁰_T+l

2 ∈/ S. By assumption (1), considering the cutset S⁰ =S∪ {v_T⁰_+l

2} instead of S increases both the number of components and the weight of the removed vertex set by 1. Hence it is enough to show that

ω(Ht−S⁰)≤ w(S⁰) t since it implies

ω(H_t−S) =ω(H_t−S⁰)−1≤ w(S⁰)

t −1 = w(S) + 1

t −1≤ w(S) t , where the last inequality is valid sincet ≥1.

(3) For all l∈[T]if v_l⁰ ∈S, thenv⁰⁰_l ∈/ S.

After the removal of V₃⁰ and v_l⁰, removing v_l⁰⁰ does not disconnect any-thing from the graph.

(4) For all l∈[T]if v_l⁰ ∈/ S, thenv⁰⁰_l ∈S.

Suppose that there exists l ∈ [T] for which v_l⁰, v_l⁰⁰ ∈/ S. By assump-tion (1), considering the cutset S⁰ = S ∪ {v_l⁰⁰} instead of S increases both the number of components and the weight of the removed vertex set by 1. Hence, similarly as in assumption (2), it is enough to show that

ω(H_t−S⁰)≤ w(S⁰) t . (5)

(V₁⁰∪V⁰⁰)∩S =T.

It follows directly from assumptions (3) and (4).

Case 3.1: (V₃⁰ ⊆S and) U⁰⁰ ⊆S.

Properties of minimally tough graphs 47

Now we can assume thatx /∈S since after the removal ofU⁰⁰ removing x does not disconnect anything from the graph. Similarly, by assumption (2), we can also assume that V₂⁰ *S, i.e. |V₂⁰∩S|=T −1. Hence

w(S) = |V₃⁰|+|U⁰⁰|+

(V₁⁰∪V⁰⁰)∩S

+|V₂⁰∩S|

= (aT −2T) +T⁰+T + (T −1) = aT +T⁰−1

and every component of H_t−S contains exactly one of the vertices u⁰₁, . . . , u⁰_bT−1, x, i.e.

ω(H_t−S) = bT = aT

t ≤ aT +T⁰−1

t = w(S) t . Case 3.2: (V₃⁰ ⊆S and) U⁰⁰∩S =∅.

In this case we can make some further assumptions for S. (6) If V₁⁰ ⊆S, then x /∈S.

After the removal of V₁⁰ removing xdoes not disconnect anything from the graph.

(7) If V₁⁰ *S, then x∈S.

Suppose that x /∈S. Then considering the cutsetS⁰ =S∪ {x} instead of S increases the number of components by 1 and the weight of the removed vertex set by t. Hence it is enough to show that ω(H_t−S⁰)≤ w(S⁰)/t since it implies

ω(Ht−S) =ω(Ht−S⁰)−1≤ w(S⁰)

t −1 = w(S) +t

t −1 = w(S) t . (8) V₂⁰ ⊆S.

Suppose that V₂⁰ *S. Then by assumption (2), there existsl ∈[T] for which V₂⁰\S ={v_T⁰ _+l}. But by assumption (1), considering the cutset S⁰ =S∪ {v_T⁰_+l}instead of S increases both the number of components and the weight of the removed vertex set by 1. Then, similarly as in assumption (2), it is enough to show that ω(H_t−S⁰)≤w(S⁰)/t. Case 3.2.1: (V₃⁰ ⊆S,U⁰⁰∩S=∅ and) V₁⁰ ⊆S.

48Chapter 4. The complexity of recognizing minimally tough graphs

Hence

w(S) =|V₃⁰|+|V₂⁰|+|V₁⁰|=aT and

ω(H_t−S) = bT = w(S) t . Case 3.2.2: (V₃⁰ ⊆S, U⁰⁰∩S =∅ and) V₁⁰ *S. Hence

w(S) = |V₃⁰|+|V₂⁰|+|(V₁⁰ ∪V⁰⁰)∩S|+w(x) = aT +t and

ω(H_t−S) = bT + 1 = w(S) t .

Therefore H_t is weighted t-tough with respect to w (meaning that the weighted toughness of H_t is at least t).

Consider the cutset

S =V₁⁰∪V₂⁰ ∪V₃⁰. Sincew(S) = aT and

ω(H_t−S) = bT = w(S) t ,

the weighted toughness ofHt with respect tow is at most t.

Thus the weighted toughness of H_t with respect tow is exactly t.

Deleting an edge may decrease the weighted toughness, and now we delete edges not induced by U⁰⁰ until the weighted toughness with respect to the weight function w remains at least t but the deletion of any other edge not induced by U⁰⁰ would result in a graph with weighted toughness less than t. LetH_t⁰ denote the obtained graph.

According to the following claim we could not delete the edges induced byV₁⁰ or incident to any of the vertices of {x} ∪V₂⁰∪U⁰⁰.

Claim 4.15. Let t ≥ 1 be a rational number. For any edge e ∈ E(Ht) induced byV₁⁰ or incident to any of the vertices of{x} ∪V₂⁰∪U⁰⁰, there exists a cutsetS =S(e)⊆V(H_t) inH_t−e for which

ω (Ht−e)−S

> w(S) t .

Properties of minimally tough graphs 49

Proof. Let e∈E(H_t)be an arbitrary edge induced by V₁⁰ or incident to any of the vertices of {x} ∪V₂⁰∪U⁰⁰.

Case 1: e is incident to a vertex of {x} ∪V₂⁰.

Let y ∈ {x} ∪V₂⁰ denote one of the endpoints of e, and let z denote the other one. Let S be the neighborhood of the vertex y except for z. Since y has degree d2te and all of its neighbors have weight 1,

w(S) = d2te −1.

Since the removal of S fromH_t−e leaves the vertexy isolated, ω (H_t−e)−S

≥2 = 2t

t > d2te −1

t = w(S) t . Case 2: e is incident to a vertex of U⁰⁰.

If e is incident to a vertex of U⁰⁰, then either it is incident to a vertex of {x} ∪V₂⁰ and this case was already settled in Case 1, or it is induced by U⁰⁰ and therefore it was not allowed to be deleted.

Case 3: e is induced byV₁⁰, i.e. e=v_l⁰

1v_l⁰

2 for some l₁, l₂ ∈[T],l₁ 6=l₂. Then

S = V₁⁰\ {v_l⁰₁, v_l⁰₂}

∪ {v⁰⁰_l1, v⁰⁰_l2} ∪V₂⁰∪V₃⁰∪ {x}

is a cutset in H_t−e such that

w(S) = (T −2) + 2 +T + (aT −2T) +t=aT +t and

ω (H_t−e)−S

=bT + 2 = aT +t

t + 1 = w(S)

t + 1> w(S) t .

Claim 4.16. Let t≥1be a rational number and H_t⁰⁰=H_t⁰− {x}. Then the following hold.

(i) The graph H_t⁰⁰ is connected.

(ii) For any cutset S of H_t⁰⁰,

ω(H_t⁰⁰−S)≤ |S|

t + 1.

50Chapter 4. The complexity of recognizing minimally tough graphs

(iii) If V₁⁰ ⊆S holds for a cutset S of H_t⁰⁰, then ω(H_t⁰⁰−S)≤ |S|

t .

(iv) For any edge e ∈ E(H_t⁰⁰) not induced by U⁰⁰ there exists a vertex set S=S(e) whose removal fromH_t⁰⁰−e disconnects the graph and

ω (H_t⁰⁰−e)−S

> |S|

t . Proof.

(i) Suppose to the contrary that H_t⁰⁰ is not connected. Then x is a cut-vertex in H_t⁰. Since the weighted toughness of H_t⁰ with respect to w ist,

2≤ω H_t⁰− {x}

≤ w(x) t = t

t = 1, which is a contradiction.

(ii) LetSbe an arbitrary cutset ofH_t⁰⁰. SinceSis a cutset inH_t⁰⁰, the vertex setS∪ {x}is a cutset in H_t⁰, and

ω(H_t⁰⁰−S) = ω H_t⁰−(S∪ {x})

≤ w(S∪ {x})

t = |S|+t t = |S|

t + 1.

(iii) LetSbe a cutset ofH_t⁰⁰for whichV₁⁰ ⊆S. We can assume thatU⁰⁰∩S =

∅since removing any of the vertices ofU⁰⁰ fromH_t⁰⁰does not disconnect anything from the graph. Then all the neighbors of the vertexxbelong to the same component inH_t⁰⁰−S, so S is a cutset inH_t⁰ as well and

ω(H_t⁰⁰−S) = ω(H_t⁰−S)≤ w(S) t = |S|

t , where the last equality is valid since x /∈S.

Properties of minimally tough graphs 51

(iv) Let e ∈ E(H_t⁰⁰) be an arbitrary edge not induced by U⁰⁰. Then by the properties of H_t⁰, there exists a vertex set S ⊆ V(H_t⁰) whose removal from H_t⁰−e disconnects the graph and

ω (H_t⁰−e)−S

> w(S) t ≥ |S|

t ,

where the last inequality is valid since t≥1. Let S⁰ =S\ {x}. Then ω (H_t⁰⁰−e)−S⁰

≥ω (H_t⁰ −e)−S

> |S|

t ≥ |S⁰| t .

In document KittiVarga Propertiesofminimallytoughgraphs Ph.D.Thesis (Pldal 46-53)