The proof of Theorem 4.1 when t ≥ 1

48 The complexity of recognizing minimally tough graphs

Properties of minimally tough graphs 49

For all i∈[n] let

V_i = [

j∈[k], m∈[M]

V_i,j,m

and add a clique on the vertices ofV_i. For alli₁, i₂ ∈[n]if v_i1v_i2 ∈E(G), then add a complete bipartite graph on (V_i1;V_i2). For all i ∈ [n], j ∈ [k], m ∈ [M] glue the graph H_t⁰⁰ to the vertex set V_i,j,m by identifying v_i,j,l,m with the vertex v_l⁰ of H_t⁰⁰ for all l ∈ [T]. For all i ∈ [n], j ∈ [k], m ∈ [M] let H^i,j,m, U_i,j,m⁰⁰ and X_i,j,m denote the (i, j, m)-th copies of H_t⁰⁰, U⁰⁰ and X, respectively. For all i ∈ [n], j ∈ [k], l ∈ [T⁰], m ∈ [M] let u⁰⁰_i,j,l,m denote the (i, j, m)-th copy of u⁰⁰_l, and for all i ∈ [n], j ∈ [k], l ∈ [T], m ∈ [M] let Y_i,j,l,m denote the (i, j, m)-th copy of Y_l. For allj ∈[k], m∈[M]add the vertex set

W_j,m =

w_j,l,m

l ∈[T⁰]

to the graph and for alli∈ [n], j ∈[k], l∈ [T⁰], m ∈[M]connect w_j,l,m tou⁰⁰_i,j,l,m. Let

V = [

i∈[n]

Vi, U⁰⁰ = [

i∈[n], j∈[k], m∈[M]

U_i,j,m⁰⁰ , W = [

j∈[k], m∈[M]

Wj,m,

and let

U =













 [

i∈[n], j∈[k], m∈[M]

V(H^i,j,m)















\V.

Add the vertex set

W⁰ ={w₁⁰, . . . , w_{(M T}⁰ 0/t−1)k}

to the graph and add the bipartite graphH_t,k^∗∗ on(W;W⁰). See Figure4.6. Now k is part of the input of the problemα-Critical, therefore the graphH_t,k^∗∗ must be constructed in polynomial time and by Theorem 2.11, this can be done. On the other hand,tis not part of the input of the problem Min-t-Tough, therefore the graphH_t⁰⁰ can be constructed in advance. Hence, G_t,k can be constructed from G in polynomial time.

50 The complexity of recognizing minimally tough graphs

V_1,1,1

V_n,k,M

H^1,1,1

H^n,k,M U_1,1,1⁰⁰

U_n,k,M⁰⁰ u⁰⁰_1,1,1,1

u⁰⁰_n,k,T0,M

w_1,1,1

w_k,T⁰_,M W

w₁⁰

w_{(M T}⁰ 0/t−1)k

W⁰

H_t,k^∗∗

Figure 4.6: The graph Gt,k, whent ≥1.

To show that G is α-critical with α(G) = k if and only if G_t,k is minimally t-tough, rst we prove the following lemma.

Lemma 4.19. Let G be an arbitrary 3-connected graph on n ≥ t+ 1 vertices with α(G)≤k. Then G_t,k ist-tough.

Proof. Let S ⊆V(Gt,k) be a cutset in Gt,k. We need to show thatω(Gt,k−S)≤

|S|/t.

Case 1: W ⊆S.

After the removal ofW, the vertices ofW⁰are isolated, therefore we can assume that W⁰∩S =∅.

Let

C =C(S) =

(i, j, m)∈[n]×[k]×[M]

V_i,j,m⊆S , c_i,j,m=

V(H^i,j,m)∩S −T for all (i, j, m)∈C, and

d_i,j,m=

V(H^i,j,m)∩S for all (i, j, m)∈ [n]×[k]×[M]

\C. Let D=D(S) =

(i, j, m)∈ [n]×[k]×[M]

\C

d_i,j,m >0 . Using these notations it is clear that

|S|=|C| ·T + X

(i,j,m)∈C

c_i,j,m+ X

(i,j,m)∈D

d_i,j,m+kM T⁰.

Properties of minimally tough graphs 51

By the assumption that W ⊆ S, in G_t,k −S the (M T⁰/t−1)k vertices of W⁰ are isolated. Since α Gt,k[V]

= α(G), the removal of V ∩S from Gt,k[V] leaves at most α(G) components. By Claim 4.16, for any (i, j, m) ∈ C the removal of V(H^i,j,m)∩S fromH^i,j,m leaves at most

max

|V(H^i,j,m)∩S|

t ,1

= max

c_i,j,m+T t ,1

= c_i,j,m+T t

components. By Claim 4.16, for any (i, j, m) ∈ D the removal of V(H^i,j,m)∩S fromH^i,j,m leaves at most

max

|V(H^i,j,m)∩S|

t + 1,1

= max

d_i,j,m t + 1,1

= d_i,j,m t + 1

components, but the component of the remaining vertices ofV_i,j,mhas been already counted. Hence

ω(G_t,k−S)≤

M T⁰ t −1

k+α(G) + X

(i,j,m)∈C

c_i,j,m+T

t + X

(i,j,m)∈D

d_i,j,m t

≤ kM T⁰+|C| ·T +P

(i,j,m)∈Cc_i,j,m+P

(i,j,m)∈Dd_i,j,m

t = |S|

t . 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 of W but no vertices of U ∪V, (d) isolated vertices of W⁰.

Let w_j0,l0,m0 ∈ W \S be xed. First we show that the following assumptions can be made forS.

(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 ofU⁰⁰ spans a clique in Gt,k, considering the cutset

S⁰ = (S\U⁰⁰)∪

w∈W

w belongs to a component of type (c)

52 The complexity of recognizing minimally tough graphs

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. (2) There are no components of type (c) inG_t,k−S.

It follows directly from assumption (1).

(3)

{i∈[n]|V_i,j0,m0 ⊆S}

≤ dT⁰/te. Let

I =I(w_j0,l0,m0, S) =

i∈[n]

V_i,j0,m0 ⊆S

and suppose that |I| ≥ dT⁰/te+ 1. By assumption (1), the component of w_j0,l0,m0 contains every vertex ofSn

i=1U_i,j0,m0 and therefore all the remaining vertices ofW_j0,m0. Now considering the cutset

S⁰ =S∪ {Wj0,m0}

instead of S increases the number of removed vertices by at mostT⁰, and it increases the number of components by at least dT⁰/te since it disconnects the vertex sets Ui,j0,m0, i ∈ I from each other. Then it is enough to show that ω(G_t,k−S⁰)≤ |S⁰|/t since it implies

ω(G_t,k−S)≤ω(G_t,k−S⁰)− T⁰

t

≤ |S⁰| t −

T⁰ t

≤ |S|+T⁰

t −

T⁰ t

≤ |S|

t .

Properties of minimally tough graphs 53

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

Sincet ≥1,

T⁰ t

≤

t+ 1 t

= 1 + 1

t

≤2.

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

Using the previous notations,

|S|=|C| ·T + X

(i,j,m)∈C

c_i,j,m+ X

(i,j,m)∈D

d_i,j,m+|S∩(W ∪W⁰)|.

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

max

|S∩(W ∪W⁰)|

t ,1

components, one of them is the component of w_j0,l0,m0, hence there are at most max

|S∩(W ∪W⁰)|

t ,1

−1≤ |S∩(W ∪W⁰)|

t

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

ci,j,m+T t

components, all of which can be of type (b). For any (i, j, m)∈D the removal of V(H^i,j,m)∩S fromH^i,j,m leaves at most

d_i,j,m t + 1

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=1Ui,j0,m0

54 The complexity of recognizing minimally tough graphs

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)≤



1 + X

(i,j,m)∈C

c_i,j,m+T

t + X

(i,j,m)∈D

d_i,j,m

t +|S∩(W ∪W⁰)|

t





−1 = |S|

t . Thus, τ(G_t,k)≥t.

Now we return to the proof of Theorem 4.18 and we show thatG isα-critical with α(G) =k if and only if G_t,k is minimally t-tough.

Let us assume that Gis α-critical with α(G) = k. By Lemma4.19, the graph G_t,k ist-tough, i.e.τ(G_t,k)≥t.

LetI be an independent vertex set of sizeα(G)inG, and recall the denition of the sets X and Y₁, . . . , Y_T constructed in Subsection 4.4.3. Let

J =

i∈[n]

vi ∈I and

S= [

i∈J

Y_i,1,1,1

!

∪











 [

i∈J, j∈[k]\{1}, m∈[M]\{1}

X_i,j,m













∪













 [

i /∈J, j∈[k], m∈[M]

X_i,j,m















∪W.

Then S is a cutset inG_t,k with

|S|=nkM aT +kM T⁰

and after the removal of S fromG_t,k, the vertices of W⁰ are isolated and the other components of Gt,k −S are exactly the components of H^i,j,m− S∩V(H^i,j,m) for all (i, j, m)∈[n]×[k]×[M]. By Proposition 4.17,

ω(H^i,j,m−X_i,j,m) =bT and

ω(H^i,j,m−Yi,1,1,1) = bT + 1

Properties of minimally tough graphs 55

for all (i, j, m)∈[n]×[k]×[M]. Since |J|=α(G)and

|W⁰|=

M T⁰ t −1

k, it follows that

ω(G_t,k−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.

Therefore,τ(G_t,k) =t.

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

Case 1: e has an endpoint in U⁰⁰.

Then this endpoint has degree d2te −1 inG_t,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 cutsetS ⊆W inH_t,k^∗ −e for which ω (H_t,k^∗ −e)−S

> |S|

t . Note thatS is also a cutset in G_t,k−e and

ω (G_t,k−e)−S

=ω (H_t,k^∗ −e)−S

> |S|

t , soτ(G_t,k−e)< t.

Case 3: e is induced by H^i0,j0,m0 for somei₀ ∈[n], j₀ ∈[k], m₀ ∈[M]. The case when e is induced by U_i⁰⁰

0,j0,m0 was already covered in Case 1. So assume that e is not induced by U_i⁰⁰_0,j0,m0. Then by Claim 4.16, there exists a vertex setS ⊆V(H_t⁰⁰)for which

ω (H_t⁰⁰−e)−S

> |S|

t .

56 The complexity of recognizing minimally tough graphs

Consider the (i₀, j₀, m₀)-th copy of the vertex set S in G_t,k−e; let us denote it with Si0,j0,m0. If Vi0,j0,m0 ⊆Si0,j0,m0, thenSi0,j0,m0 is a cutset in Gt,k−e and

ω (Gt,k−e)−Si0,j0,m0

=ω (H_t⁰⁰−e)−S

> |S|

t ,

soτ(Gt,k−e)< t. Assume thatVi0,j0,m0 *Si0,j0,m0. LetIbe an independent vertex set of size α(G)in Gthat contains v_i0 (by Proposition2.18, such an independent vertex set exists). Let

J =

i∈[n]

vi ∈I and

S⁰ =S_i0,j0,m0 ∪















[

j∈[k], m∈[M], (j,m)6=(j0,m0)

X_i0,j,m















∪



 [

i∈J\{i₀}

Y_i,1,1,1





∪













 [

i∈J\{i0}, j∈[k]\{1}, m∈[M]\{1}

X_i,j,m















∪













 [

i /∈J, j∈[k], m∈[M]

X_i,j,m















∪W.

Then S⁰ is a cutset inG_t,k−e with

|S⁰|=|S|+ (nkM −1)aT +kM T⁰ and similarly as before,

ω (G_t,k−e)−S⁰

> |S|

t + (nkM −1)bT +α(G) +

M T⁰ t −1

k

= |S|

t + (nkM −1)bT + kM T⁰

t = |S⁰| t , so τ(G_t,k−e)< t.

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 ∈ [k], m₀ ∈ [M] was settled in Case 3, we can assume that there do not exist i ∈ [n], j ∈

Properties of minimally tough graphs 57

[k], m ∈ [M] for which e is induced by V_i,j,m. By Lemma 2.19, the graph G_t,k[V] is α-critical, so in Gt,k[V]− e there exists an independent vertex set I of size α(G) + 1. Let

J =

(i, j, l, m)∈[n]×[k]×[T]×[M]

v_i,j,l,m ∈I , J₁⁰ =

(i, j, m)∈[n]×[k]×[M]

∃!l ∈[T] : v_i,j,l,m ∈I , and

J₂⁰ =

(i, j, m)∈[n]×[k]×[M]

@l∈[T] : v_i,j,l,m ∈I .

By the assumption that there do not exist i∈[n], j ∈ [k], m∈[M] for which e is induced byV_i,j,m,

J₁⁰ ∪J₂⁰ = [n]×[k]×[M], so

S =



 [

(i,j,l,m)∈J

Y_i,j,l,m



∪



 [

(i,j,m)∈J₂⁰

X_i,j,m



∪W is a (well-dened) cutset in G_t,k−e. Then

|S|=nkM aT +kM T⁰ and similarly as before,

ω (G_t,k−e)−S

=nkM bT +α(G) + 1 +

M T⁰ t −1

k

= nkM aT

t +kM T⁰

t + 1> |S|

t , soτ(G_t,k−e)< t.

Now let us assume thatGis notα-critical withα(G) = k, i.e. either α(G)6=k or even though α(G) =k, the graph Gis not α-critical.

Case I: α(G)> k.

Let I be an independent vertex set of sizeα(G) inG. Let J =

i∈[n]

vi ∈I

58 The complexity of recognizing minimally tough graphs

and

S= [

i∈J

Y_i,1,1,1

!

∪











 [

i∈J, j∈[k]\{1}, m∈[M]\{1}

X_i,j,m













∪













 [

i /∈J, j∈[k], m∈[M]

X_i,j,m















∪W.

Then S is a cutset inG_t,k−e with

|S|=nkM aT +kM T⁰ 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 ,

so τ(G_t,k)< t, which means thatG_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 that α(G−e) ≤ k. By Lemma 4.19, the graph (G−e)_t,k is t-tough, but it can be obtained from G_t,k by edge-deletion, which means that G_t,k is not minimally t-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.

No documento Properties of minimally tough graphs (páginas 50-60)