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

Let H_t be constructed as follows. Let

V₁⁰ ={v⁰₁, . . . , 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},

Properties of minimally tough graphs 41

U⁰⁰ ={u⁰⁰₁, . . . , u⁰⁰_T0}, and

U₁⁰⁰ ={u⁰⁰₁, . . . , u⁰⁰_T}.

Add cliques on the vertices of V₁⁰, V₂⁰, V₃⁰, and U⁰⁰. For alll ∈[T] connect v_l⁰⁰ to v_l⁰ and to u⁰_l, and connect v⁰_T+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 ofV₂⁰ to all the vertices of U⁰⁰. Finally, add a new vertex x to the graph and connect it to all the vertices of V₁⁰∪U⁰⁰. See Figure 4.5.

x

u⁰⁰₁ u⁰⁰_T0

U⁰⁰

v_T⁰ ₊₁ v_2T⁰ V₂⁰

v₁⁰ v_T⁰ V₁⁰ v⁰⁰₁ v_T⁰⁰ 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 H_t, whent ≥1.

Claim 4.14. For any rational number t ≥ 1, the graph H_t has weighted tough- nesstwith respect to the weight functionwthat assigns weight 1 to all the vertices of H_t except for the vertex x, to which it assigns weightt.

Proof. Consider the cutset

S =V₁⁰∪V₂⁰∪V₃⁰. Since w(S) = aT and

ω(H_t−S) = bT = w(S) t ,

the weighted toughness ofHt with respect to wis at most t.

42 The complexity of recognizing minimally tough graphs

Now we show that the weighted toughness ofH_twith respect towis at leastt. Let S be an arbitrary cutset of Ht. We need to prove thatω(Ht−S)≤w(S)/t.

We can assume that either V₃⁰ ∩S =∅ orV₃⁰ ⊆S, otherwise considering S⁰ = S\V₃⁰ instead ofS does not change the number of components, but decreases the number of removed vertices. Similarly, we can also assume that eitherU⁰⁰∩S =∅ or U⁰⁰ ⊆S.

Case 1: V₃⁰∩S =∅ and U⁰⁰∩S=∅.

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

u⁰_T+l ∈S or v_T⁰_+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 that x /∈ S since after the removal of U⁰⁰ removing x does not disconnect anything from the graph. Similarly, we can also assume that V₂⁰ * S. Then H_t−S has at most 3 components. To obtain three components, the following must hold:

(i) u⁰_T+l ∈S or v_T⁰_+l ∈S for all l∈[T] (but V₂⁰ *S), and (ii) V₁⁰ ⊆S.

Hence w(S)≥T⁰ + 2T =d2te+T and

ω(Ht−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(H_t−S) = 2≤ d2te

t ≤ w(S) t . Case 3: V₃⁰ ⊆S.

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

Properties of minimally tough graphs 43

(1) (U₁⁰ ∪U₂⁰ ∪U₃⁰)∩S =∅.

Otherwise, since V₃⁰ ⊆ S, considering S⁰ = S\(U₁⁰ ∪U₂⁰ ∪U₃⁰) instead of S does not decrease the number of components, but decreases the number of removed vertices.

(2) There exists at most one l∈[T] for which v_T⁰_+l ∈/ S, i.e. |V₂⁰\S| ≤1.

Suppose that there exist l₁, l₂ ∈ [T] for which l₁ 6=l₂ and v_T⁰_+l1, v_T⁰ _+l2 ∈/ 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(H_t−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, thenv_l⁰⁰∈/ S.

After the removal of V₃⁰ and v⁰_l, removing v⁰⁰_l does not disconnect anything from the graph.

(4) For all l∈[T] if v_l⁰ ∈/ S, thenv_l⁰⁰∈S.

Suppose that there exists l ∈ [T] for which v_l⁰, v⁰⁰_l ∈/ S. By assumption (1), considering the cutset S⁰ =S ∪ {v_l⁰⁰} instead of S increases both the num- ber 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.

44 The complexity of recognizing minimally tough graphs

Now we can assume that x /∈S since after the removal ofU⁰⁰ removing xdoes 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 ofH_t−Scontains exactly one of the verticesu⁰₁, . . . , 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, thenx /∈S.

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

(7) If V₁⁰ *S, thenx∈S.

Suppose that x /∈S. Then considering the cutset S⁰ =S∪ {x}instead ofS increases the number of components by 1 and the weight of the removed vertex set byt. 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 exists l ∈ [T] for which V₂⁰ \S={v⁰_T+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. Hence

w(S) =|V₃⁰|+|V₂⁰|+|V₁⁰|=aT

Properties of minimally tough graphs 45

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 .

ThereforeH_tis weightedt-tough with respect tow(meaning that the weighted toughness of H_t is at least 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 functionwremains at leasttbut the deletion of any other edge not induced byU⁰⁰ would result in a graph with weighted toughness less than t. Let H_t⁰ denote an arbitrary graph that can be obtained this way.

According to the following claim we could not delete the edges induced by V₁⁰ 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(H_t) induced by V₁⁰ or incident to any of the vertices of {x} ∪V₂⁰ ∪U⁰⁰, there exists a cutset S =S(e)⊆V(H_t)in H_t−e for which

ω (H_t−e)−S

> w(S) t .

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.

46 The complexity of recognizing minimally tough graphs

Since the removal ofS from H_t−e leaves the vertex y isolated, ω (H_t−e)−S

≥2 = 2t

t > d2te −1

t = w(S) t . Case 2: e is incident to a vertex ofU⁰⁰.

Ifeis incident to a vertex ofU⁰⁰, 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=v⁰_l

1v⁰_l

2 for some l₁, l₂ ∈[T], l₁ 6=l₂. Then

S = V₁⁰\ {v⁰_l

1, v⁰_l

2}

∪ {v_l⁰⁰

1, v_l⁰⁰

2} ∪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 ≥ 1 be a rational number and H_t⁰⁰ = H_t⁰ − {x}. Then the following hold.

(i) The graph H_t⁰⁰ is connected.

(ii) For any cutsetS of H_t⁰⁰,

ω(H_t⁰⁰−S)≤ |S|

t + 1. (iii) If V₁⁰ ⊆S holds for a cutset S of H_t⁰⁰, then

ω(H_t⁰⁰−S)≤ |S|

t .

(iv) For any edgee∈E(H_t⁰⁰)not induced byU⁰⁰there exists a vertex setS =S(e) whose removal from H_t⁰⁰−e disconnects the graph and

ω (H_t⁰⁰−e)−S

> |S|

t .

Properties of minimally tough graphs 47

Proof.

(i) Suppose to the contrary that H_t⁰⁰ is not connected. Then x is a cut-vertex inH_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) Let S be an arbitrary cutset ofH_t⁰⁰. Since S is a cutset inH_t⁰⁰, the vertex set S∪ {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) Let S be a cutset of H_t⁰⁰ for which V₁⁰ ⊆ S. We can assume that U⁰⁰∩S =

∅ since removing any of the vertices of U⁰⁰ from H_t⁰⁰ does not disconnect anything from the graph. Then all the neighbors of the vertex x belong to the same component inH_t⁰⁰−S, so S is a cutset in H_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.

(iv) Let e∈E(H_t⁰⁰)be an arbitrary edge not induced by U⁰⁰. Then by the prop- erties ofH_t⁰, there exists a vertex setS ⊆V(H_t⁰)whose removal fromH_t⁰−e disconnects the graph and

ω (H_t⁰−e)−S

> w(S) t ≥ |S|

t ,

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

≥ω (H_t⁰ −e)−S

> |S|

t ≥ |S⁰| t .

48 The complexity of recognizing minimally tough graphs

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