A lower bound on the independence number of a graph in terms of degrees and local clique sizes

Discrete Applied Mathematics (

)

–

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A lower bound on the independence number of a graph in terms of degrees and local clique sizes C. Brause a,∗ , B. Randerath b , D. Rautenbach c , I. Schiermeyer a a

Institute of Discrete Mathematics and Algebra, Technical University Bergakademie Freiberg, Freiberg, Germany

b

Institute of Communications Engineering, Cologne University of Applied Sciences, Köln, Germany

c

Institute of Optimization and Operations Research, Ulm University, Ulm, Germany

article

abstract

info

Caro and Wei independently showed that the independence number α(G) of a graph G is 1 at least u∈V (G) d (u)+1 . In the present paper we conjecture the stronger bound α(G) ≥

Article history: Received 21 October 2014 Received in revised form 1 June 2015 Accepted 15 June 2015 Available online xxxx

G

where ωG (u) is the maximum order of a clique of G that contains the vertex u. We discuss the relation of our conjecture to recent conjectures and results concerning the independence number and the chromatic number. Furthermore, we prove our conjecture for perfect graphs and for graphs of maximum degree at most 4. © 2015 Elsevier B.V. All rights reserved. 2 u∈V (G) dG (u)+ωG (u)+1



Keywords: Independent set Chromatic number Degree Clique

1. Introduction We consider finite, simple, and undirected graphs, and use standard terminology and notation. Independence in graphs is one of the most well studied topics of graph theory. Since even the approximation of the independence number α(G) of a graph G is algorithmically very hard [10], lower bounds on the independence number have received much attention. Caro [5] and Wei [16] independently proved the following classical lower bound. Theorem 1.1 (Caro [5], Wei [16]). If G is a graph, then

α(G) ≥



1

d (u) + 1 u∈V (G) G

. n(G)

For a graph G of order n(G) and maximum degree ∆(G), this result immediately implies α(G) ≥ ∆(G)+1 , which, by Brooks’ theorem [4], is best-possible for cliques and odd cycles. Since the clique number ω(G) of G satisfies ω(G) ≤ ∆(G) + 1, the following result due to Fajtlowicz [6] is a strengthening of the above observation. Theorem 1.2 (Fajtlowicz [6]). If G is a graph, then

α(G) ≥

2n(G) . ∆(G) + ω(G) + 1

Recently, Henning et al. [11] showed a more localized version of Fajtlowicz’s bound.

∗

Corresponding author. E-mail addresses: [email protected] (C. Brause), [email protected] (B. Randerath), [email protected] (D. Rautenbach), [email protected] (I. Schiermeyer). http://dx.doi.org/10.1016/j.dam.2015.06.009 0166-218X/© 2015 Elsevier B.V. All rights reserved.

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Theorem 1.3 (Henning et al. [11]). Let G be a graph. If p is an integer such that for every clique C of G, there is a vertex u in C with dG (u) + |C | + 1 ≤ p, then 2n(G)

α(G) ≥

p

.

The minimum value of p satisfying the condition in Theorem 1.3 can be smaller than ∆(G) + ω(G) + 1 because of its more local definition. The following alternative local refinement of Fajtlowicz’s bound was conjectured by Bertram and Horak in [1]. Conjecture 1.4 (Bertram and Horak [1]). If G is a graph, then

α(G) ≥

2



d (u) + ω(G) + 1 u∈V (G) G

.

In [1] Conjecture 1.4 was established for regular graphs as well as for triangle-free graphs. Further strengthenings and versions of Caro’s and Wei’s bound were considered in [2,3,8,9]. In the present paper we propose the following strengthening of Conjecture 1.4, where ωG (u) denotes the maximum size of a clique of a graph G that contains a specified vertex u of G. Conjecture 1.5. If G is a graph, then

α(G) ≥

2



d (u) + ωG (u) + 1 u∈V (G) G

.

As our main contributions we show that Conjecture 1.5 holds for graphs of maximum degree at most 4 and for perfect graphs. Note that the bounds in Conjectures 1.4 and 1.5 coincide if every vertex of the considered graph lies in a maximum clique. Using this observation it follows easily from the result in [1] that Conjecture 1.5 holds for triangle-free graphs. Before we proceed to our results, we explain some connections to the chromatic number χ (G) and the fractional chron(G) n(G) matic number χf (G) of a graph G. Since α(G) ≥ χ (G) ≥ χ (G) for every graph G, upper bounds on the (fractional) chrof

matic number always imply lower bounds on the independence number as their immediate corollaries. In fact, Fajtlowicz’s Theorem 1.2 is an immediate consequence of the following result. Theorem 1.6 (Molloy and Reed [14]). If G is a graph, then

χf (G) ≤

1 2

(∆(G) + ω(G) + 1) .

This last result is the fractional version of Reed’s [15] famous conjecture. Conjecture 1.7 (Reed [15]). If G is a graph, then

χ (G) ≤



1 2

 (∆(G) + ω(G) + 1) .

In his thesis, King [12] considered a local strengthening of Reed’s conjecture. Conjecture 1.8 (King [12]). If G is a graph, then

χ (G) ≤ max



u∈V (G)

1 2

 (dG (u) + ωG (u) + 1) .

As observed by McDiarmid, the fractional version of King’s conjecture is true. Theorem 1.9 (McDiarmid, cf. Theorem 2.10 in [12]). If G is a graph, then

χf (G) ≤ max

u∈V (G)

1 2

(dG (u) + ωG (u) + 1) .

This implies the following, which yields further support for our Conjecture 1.5.

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Corollary 1.10. If G is a graph, then

α(G) ≥

2n(G) max (dG (u) + ωG (u) + 1)

.

u∈V (G)

Clearly, the minimum value of p satisfying the condition in Theorem 1.3 is such that p ≤ maxu∈V (G) (dG (u) + ωG (u) + 1), and thus Henning et al.’s Theorem 1.3 is actually stronger than this last corollary. It might be interesting to relate the (fractional) chromatic number to the right hand side of the inequality in our Conjecture 1.5. 2. Results Before we proceed to results and proofs, we introduce some useful notation. Let G be a graph. For a vertex u of G, let fG (u) =

2 dG (u) + ωG (u) + 1

,

and let f (G) = u∈V (G) fG (u). Using this notation, Conjecture 1.5 states α(G) ≥ f (G). For a vertex u of a graph G and a set S of vertices of G, let dG (u, S ) denote |NG [u] ∩ S |, that is, the number of elements of  the closed neighborhood of u that belong to S. For a function g : V (G) → R and a set S of vertices of G, let g (S ) = u∈S g (u). Our first lemma is a quantitative version of the intuitively obvious observation that for every independent set T in the complement of a maximum independent set S of a graph G, the graph G must contain many edges between S and T .



Lemma 2.1. If S is a maximum independent set of a graph G, and T is an independent set of the graph G − S, then EG (S , T ) = {uv ∈ E (G) : u ∈ S and v ∈ T } contains a matching of size |T |. Proof. If there is no such matching, then Hall’s theorem [7] implies the existence of a subset T ′ of T with |NG (T ′ ) ∩ S | < |T ′ |. Now (S \NG (T ′ )) ∪ T ′ is an independent set of G that is larger than S, which is a contradiction.  At several points during later proofs we remove a set U of vertices from a given graph G, and compare f (G − U ) to f (G). Clearly, going from G to G − U, we lose the contribution fG (U ) of the vertices in U to f (G). On the other hand, for a vertex v of G − U that has neighbors in U, its degree dG (v) decreases and its local clique number ωG (v) might decrease. Therefore, the contribution of these vertices to f (G − U ) is larger than their contribution to f (G). The following two lemmas allow to estimate these effects. In the next lemma, the value of X is a lower bound on the total weight with respect to f of the vertices of G − U that have 2 − 2x . The a neighbor in U. If some such vertex v satisfies fG (v) = 2x and has i neighbors in U, then fG−U (v) − fG (v) ≥ x− i number of such vertices will be ni . Note that x − i is of the form dG−U (v) + ωG−U (v) + 1 and hence is at least 2. Lemma 2.2. If the optimization problem min

ni  d   i=1 j=1

2 xi,j − i

−

2



xi,j

ni d   2 i=1 j=1

0< is feasible and 0 < X < 2

d

ni i=1 i ,

Proof. Using the variables yi,j = min

ni d   i=1 j=1 d

xi,j 2 xi,j

≥

X

<

2 i

then the optimum value of (1) is d X 2

2 , xi,j

the problem (1) reads as follows

iy2i,j 2 − iyi,j

ni 

(1)

yi,j

≥

X

0 < yi,j

<

2

i=1 j=1

i

.

We assume that the variables yi,j form an optimal solution.

2

ni i=1 i

−X

.

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C. Brause et al. / Discrete Applied Mathematics (

Note that ∂∂y



iy2 2−iy



=

4iy−i2 y2 . (2−iy)2

)

–

Since 4iy − i2 y2 > 0 and 2 − iy > 0 for 0 < y <

d n implies i=1 j=i 1 yi,j = X .  2  2 iy Note that ∂∂y2 2−iy = (2−8iiy)3 . Since 2 − iy > 0 for 0 < y <

we obtain ∂∂y2 2



2 , i

we obtain ∂∂y

iy2 2−iy



iy2 2−iy



> 0, which



> 0, which implies that the  2   2  iy ′ ′ iyi,j iy2 2 ∂ ∂ function y → 2−iy is strictly convex for 0 < y < i . Now the optimality of the yi,j s implies ∂ y = ∂ y ′ ′ 2−iyi ,j′ ′ 2−iyi,j i,j i ,j i ,j  2  y1,1 iy 4 ∂ ′ ′ for all choices of (i, j) and (i , j ). Since ∂ y 2−iy = −1 + (2−iy)2 , we obtain yi,j = i for i ∈ [d] and j ∈ [ni ]. This implies X =

ni d  

ni d   y1,1

yi,j =

i

i=1 j=1

i=1 j=1

and hence y1,1 = d X

ni i=1 i

=

d  ni

i

i =1

2 , i

y1,1

. Note that the condition 0 < X < 2

d

ni i=1 i

implies 0 < yi,j <

2 i

for all i and j. Altogether, the

value of the optimum solution is ni d   i i=1 j=1

 y1,1 2

2−i

i

 y1,1  = i

d  ni

y21,1

2 − y1,1 i=1 i

X2

= 2

d 

ni i

i=1

which completes the proof.

, −X



As noted above, the value of X in the previous lemma is a lower bound on the total weight of the vertices of G − U that have a neighbor in U. If, for every edge uv of G between a vertex u in U and a vertex v of G − U, we have a lower bound on fG (v), it is possible to refine the estimation of f (G − U ) − f (G) as follows. For a set X of vertices of a graph G, let ∂G X denote the set of edges of G between X and V (G)\X . Lemma 2.3. Let G be a graph and let X be a set of vertices of G. If p : ∂G X → N is such that fG (v) ≥ uv ∈ ∂G X with v ̸∈ X , then

 

f (G − X ) ≥ f (G) − fG (X ) +

2

−

p(uv) − 1

uv∈∂G X

2



p(uv)

2 p(uv)

for every edge

.

Proof. Let G′ = G − X . If some vertex v ∈ V (G)\X is connected to X by exactly k edges, say u1 v, . . . , uk v , then 2 dG′ (v) = dG (v) − k. Clearly, ωG′ (v) ≤ ωG (v). Now the monotonicity and convexity of the function x →  for x > 0 x implies fG′ (v) − fG (v) =

≥ =

2 dG′ (v) + ωG′ (v) + 1 2

  i =1

dG (v) + ωG (v) + 1

−

2 dG (v) − 1 + ωG (v) + 1

i =1 k

−

dG (v) − i + ωG (v) + 1

k

≥

2 dG (v) + ωG (v) + 1 2

(dG (v) − k) + ωG (v) + 1 k   2 i =1

≥

−

2 p(ui v) − 1

−

2 p(ui v)



−



2 dG (v) − (i − 1) + ωG (v) + 1



2 dG (v) + ωG (v) + 1

.

Summing these inequalities over all vertices in V (G)\X implies the desired bound.



Our first goal is to show Conjecture 1.5 for perfect graphs. Recall that a vertex u of a graph G is simplicial if the neighborhood NG (u) of u in G is a clique. Lemma 2.4. If G is a counterexample of minimum order for Conjecture 1.5, then no vertex of G is simplicial. Proof. We assume to the contrary that u is a vertex of G whose neighborhood is a clique. Let C = {u} ∪ NG (u). The graph G′ = G − C satisfies α(G) ≥ 1 + α(G′ ). For every vertex v in C , we have dG (v) ≥ dG (u) = |C | − 1 and ωG (v) ≥ ωG (u) = |C |. Since f (G′ ) − f (G) ≥ −

 v∈C

fG (v) = −



2

d (v) + ωG (v) + 1 v∈C G

≥−

 v∈C

2

|C | + | C |

= −1,

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we obtain, by the minimality of G, that α(G) ≥ 1 + α(G′ ) ≥ 1 + f (G′ ) ≥ f (G), which is a contradiction and completes the proof.  Lemma 2.5. If G is a graph whose vertex set is partitioned into Θ cliques, then

Θ≥



2

u∈V (G)

dG (u) + ωG (u) + 1

.

Proof. Let the cliques C1 , . . . , CΘ partition the vertex set V (G) of G. For every vertex u in a clique C of G, we have dG (u) ≥ |C | − 1 and ωG (u) ≥ |C |. This implies

Θ=

Θ 

1=

Θ  

2

|Ci | + |Ci |

i=0 u∈Ci

i=0

which completes the proof.

≥

Θ  

2

i=0 u∈Ci

dG (u) + ωG (u) + 1

=



2

d (u) + ωG (u) + 1 u∈V (G) G

,



Theorem 2.6. If G is perfect, then

α(G) ≥

2



d (u) + ωG (u) + 1 u∈V (G) G

.

Proof. By Lovász’s weak perfect graph theorem [13], a perfect graph G can be partitioned into α(G) cliques and Lemma 2.5 implies the desired result.  The next four lemmas collect further structural properties of counterexamples for Conjecture 1.5. Lemma 2.7. If G is a counterexample of minimum order for Conjecture 1.5, then δ(G) ≥ 3. Proof. For a contradiction, we assume that u is a vertex of degree less than 3 in G. Lemma 2.4 implies that δ(G) ≥ 2 and that u has exactly two non-adjacent neighbors v1 and v2 . Let F = fG (u) + fG (v1 ) + fG (v2 ) and let N = (NG (v1 ) ∪ NG (v2 ))\{u}. If fG (N ∪ NG [u]) ≤ 2, then f (G) − f (G − (N ∪ NG [u])) ≤ fG (N ∪ NG [u]) ≤ 2. Since v1 and v2 are not adjacent, we have α(G) ≥ α(G − (N ∪ NG [u])) + 2, and the minimality of G implies

α(G) ≥ α(G − (N ∪ NG [u])) + 2 ≥ f (G − (N ∪ NG [u])) + 2 ≥ f (G), which is a contradiction. Hence fG (N ∪ NG [u]) > 2, which implies fG (N ) > 2 − fG (NG [u]) = 2 − F . Let G′ = G − NG [u]. Clearly, α(G) ≥ α(G′ ) + 1. For i ∈ [2], let N contain ni vertices wi,1 , . . . , wi,ni such that |NG (wi,j )∩{v1 , v2 }| = i for j ∈ [ni ]. Let xi,j = dG (wi,j )+ωG (wi,j )+ 1 for i ∈ [2] and j ∈ [ni ]. Clearly, xi,j ≥ max{2, i}+ 2 + 1 = max{5, i + 3} and

2

ni i =1 i

≤

2

i=1

2 ni i=1

2 j=1 xi,j

= fG (N ) ≥ 2 − F . Since

2 i

>

2 , i +3

we have 0 < 2 − F < 2

2

ni i =1 i .

Note that

ni = dG (v1 ) + dG (v2 ) − 2. Therefore, by Lemma 2.2, we obtain

f (G) − f (G′ ) = fG (NG [u]) + fG (N ) − fG′ (N )

≤F−

ni  2   i=1 j=1

2 xi,j − i

−

2



xi,j

(2 − F )

2

≤F−

2

2

 i=1

≤F−

ni i

− (2 − F )

(2 − F )2 . 2(dG (v1 ) + dG (v2 ) − 2) − (2 − F )

(2−F )2

Claim. F − 2(d (v )+d (v )−1)−(2−F ) ≤ 1. G 1 G 2 Proof of the claim. Let F ∗ = F → F −

2 5

+

2 dG (v1 )+3

+

2 . dG (v2 )+3

Since the function

(2 − F )2 2(dG (v1 ) + dG (v2 ) − 1) − (2 − F ) (2−F ∗ )2

is monotonously decreasing in F and F = fG (u) + fG (v1 ) + fG (v2 ) ≤ F ∗ , it suffices to show F ∗ − 2(d (v )+d (v )−1)−(2−F ∗ ) ≤ 1. G 1 G 2 If dG (v1 ) ≥ 7, then dG (v2 ) ≥ 2 implies F ≤ F ∗ ≤ 1, and the desired bound holds. Therefore, by symmetry, we may assume

6

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2 ≤ dG (v1 ) ≤ dG (v2 ) ≤ 6. Considering the finitely many choices for dG (v1 ) and dG (v2 ) leads to the desired bound, which completes the proof of the claim.  Now the minimality of G implies α(G) ≥ α(G′ )+ 1 ≥ f (G′ )+ 1 ≥ f (G), which is a contradiction and completes the proof.



Lemma 2.8. Let G be a counterexample of minimum order for Conjecture 1.5. Let a maximum independent set S of G be chosen such that (i) fG (S ) is maximum,  and (ii) subject to (i), x∈S fG (NG [x]) is minimum. If u ∈ S is a vertex of degree 3, then

fG (v) v∈NG [u] dG (v,S )



≤ 1.

f (v)

2 G Proof. Let F = v∈NG [u] dG (v,S ) . By Lemma 2.7, we have δ(G) ≥ 3, which implies that fG (v) ≤ 6 for every vertex v of G. Let H = G[NG (u)]. By Lemma 2.4, the vertex u is not simplicial, and hence H is isomorphic to either 3K1 , or P3 , or K2 ∪ K1 . If H is 3K1 , then Lemma 2.1 implies the existence of two neighbors, say v1 and v2 , of u with dG (v1 , S ), dG (v2 , S ) ≥ 2, and hence F ≤ 26 + 16 + 16 + 26 = 1. If H is P3 , then, for 1 ≤ i ≤ 3, we have ωG (vi ) ≥ 3, which implies fG (u), fG (vi ) ≤ 27 . Furthermore,



Lemma 2.1 implies the existence of a neighbor, say v1 , of u with dG (v1 , S ) ≥ 2, and hence F ≤ 72 + 17 + 27 + 27 = 1. Hence, we may assume that H is K2 ∪ K1 . Let v1 and v2 be the two adjacent neighbors of u, and let v3 be the third neighbor of u. Note that fG (u), fG (v1 ), fG (v2 ) ≤ 27 . If dG (v3 , S ) = 1, then Lemma 2.1 implies dG (v1 , S ), dG (v2 , S ) ≥ 2, and hence F ≤ 27 + 17 + 71 + 26 < 1. Hence we may assume that dG (v3 , S ) ≥ 2. If fG (v3 ) ≤

+ 27 + 27 + 17 = 1. Hence fG (v3 ) = 26 , which implies dG (v3 ) = 3 and ωG (v3 ) = 2. If dG (v1 , S ) ≥ 2, then F ≤ + + + 16 < 1. Hence dG (v1 , S ) = 1, and, by symmetry, 2 2 2 1 dG (v2 , S ) = 1. If dG (v1 ) ≥ 4, then F ≤ 7 + 8 + 7 + 6 < 1. Hence dG (v1 ) = 3, and, by symmetry, dG (v2 ) = 3, which implies fG (v1 ) = fG (v2 ) = 27 . If v1 and v2 have a common neighbor w that is distinct from u, then fG (w) ≤ 72 , and S ′ = (S \{u}) ∪ {v1 } is a maximum independent set of G with fG (S ′ ) = fG (S ) − fG (u) + fG (v1 ) = fG (S ), and     2 2 fG (NG [x]) − + < fG (NG [x]) − fG (v3 ) + fG (w) ≤ fG (NG [x]) = fG (NG [x]), x∈S ′

2 , 7

then F ≤ 2 7

1 7

2 7 2 7

6

x∈S

x∈S

7

x∈S

which contradicts the choice of S. Hence ωG−u (v1 ) = ωG−u (v2 ) = 2, which implies f (G) − f (G − u) =

2 6

−

 2 5

2 7

2 7

−

2 5



+ ≤ 0. Now the minimality of G implies α(G) ≥ α(G − u) ≥ f (G − u) ≥ f (G), which is a contradiction and completes

the proof.

+2



Lemma 2.9. If G is a counterexample of minimum order and maximum degree at most 4 for Conjecture 1.5, and u is a vertex of degree 4 in G, then ωG (u) ≤ 3. Proof. By Lemma 2.4, no vertex of G is simplicial, which implies that ωG (x) ≤ dG (x) for every vertex x of G. Furthermore, by n(G) Brooks’ theorem, this implies that G is 4-colorable, and hence α(G) ≥ 4 . Since G is a counterexample for Conjecture 1.5,

it contains at least one vertex y with fG (y) > 82 . For a contradiction, we assume that u is a vertex with dG (u) = ωG (u) = 4, that is, fG (u) ≤ 29 . If fG (v) ≤ 92 for every neighbor v of u, then dG (v) = ωG (v) = 4 for every neighbor v of u. Since G is connected and contains at least one vertex y with fG (y) ≥ 28 , this implies that we may assume that some neighbor, say v4 , of u satisfies fG (v4 ) ≥ 82 . Let v1 , v2 , and

v3 denote the neighbors of u that are distinct from v4 . Since fG (v4 ) ≥ 28 , the vertex v4 does not belong to the clique of order 4 that contains u. Hence v1 , v2 , and v3 form a triangle, and thus, by Lemma 2.4, we have dG (vi ) = ωG (vi ) = 4 for i ∈ [4]. If ωG−u (v1 ) < 4, then v1 , v2 , and v3 do not have a common neighbor that is distinct from u, which implies ωG−u (v2 ), ωG−u (v3 ) < 4, and thus 4  (fG (vi ) − fG−u (vi )) i =1         2 2 2 2 2 2 2 2 2 ≤ + − + − + − + −

f (G) − f (G − u) = fG (u) +

9

9

< 0.

7

9

7

9

7

8

7

Now the minimality of G implies α(G) ≥ α(G − u) ≥ f (G − u) ≥ f (G), which is a contradiction. Hence ωG−u (v1 ) = 4, which implies that v1 , v2 , and v3 have a common neighbor w that is distinct from u. Since w is not simplicial, we have dG (w) = ωG (w) = 4, and hence fG (w) = 92 . Now f (G) − f (G − v1 ) = fG (v1 ) + (fG (u) − fG−v1 (u)) + (fG (v2 ) − fG−v1 (v2 )) + (fG (v3 ) − fG−v1 (v3 )) + (fG (w) − fG−v1 (w))         2 2 2 2 2 2 2 2 2

≤

9 < 0,

+

9

−

7

+

9

−

7

+

9

−

7

+

9

−

7

and the minimality of G implies a similar contradiction as above. This completes the proof.



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Lemma 2.10. Let G be a counterexample of minimum order and maximum degree at most 4 for Conjecture 1.5. Let a maximum independent set S of G be chosen such that (i) fG (S ) is maximum,  and (ii) subject to (i), x∈S fG (NG [x]) is minimum. If u ∈ S is a vertex of degree 4, then

fG (v) v∈NG [u] dG (v,S )



≤ 1.

f (v)

G Proof. Let F = v∈NG [u] dG (v,S ) . For a contradiction, we assume that F > 1. Since G and S are fixed, we write s(v) instead of dG (v, S ) for every vertex v of G. By Lemmas 2.7 and 2.9, we have δ(G) ≥ 3 and ωG (u) ≤ 3, that is, the neighborhood of u induces a triangle-free subgraph of G. Since ∆(G) = 4, we have fG (v) ∈ { 26 , 27 , 28 } for every vertex v of G. Let NG (u) = {v1 , v2 , v3 , v4 }. We consider different cases.



Case 1 NG (u) is independent. Note that fG (u) =

+ 4 · 16 < 1, which is a contradiction. Hence s(v1 ) = 1. By condition (i) in the choice of S, this implies f (v1 ) ≤ f (u) = 72 . If s(v2 ) + s(v3 ) + s(v4 ) ≥ 8, then we may assume that either (s(v2 ), s(v3 ), s(v4 )) = (4, 2, 2) or (s(v2 ), s(v3 ), s(v4 )) = (3, 3, 2) componentwise. In the first case, F ≤ 27 + 27 + 62·4 + 62·2 + 62·2 < 1, and in the second case, F ≤ 27 + 72 + 62·3 + 62·3 + 62·2 < 1, which is a contradiction. Hence s(v2 ) + s(v3 ) + s(v4 ) ≤ 7. If fG (v2 ) + fG (v3 ) + fG (v4 ) ≤ 2 57 − fG (v1 ) , then F ≤

2 7

2 . 7

By Lemma 2.1, we may assume that s(vi ) ≥ 2 for i ∈ {2, 3, 4}. If s(v1 ) ≥ 2, then F ≤

2 7

1

+ fG (v1 ) + (fG (v2 ) + fG (v3 ) + fG (v4 )) ≤ 1, 2

which is a contradiction. Hence fG (v2 )+ fG (v3 )+ fG (v4 ) > 2 with fG (vi ) =

5

2 . 6

If s(vi ) = 3 for every i ∈ {2, 3, 4} with fG (vi ) = we may assume that fG (v2 ) =

2 6

2 , 6

and s(v2 ) = 2.

7

 − fG (v1 ) ≥ 76 . This implies the existence of some i ∈ {2, 3, 4}

then F ≤

2 7

+ 72 +

2 6·3

+

2 7·2

+

2 7·2

< 1, which is a contradiction. Hence

If fG (v1 ) = 82 , then fG (v2 ) + fG (v3 ) + fG (v4 ) > 2 57 − fG (v1 ) implies that we may assume that fG (v3 ) = 62 and fG (v4 ) ≥ 27 . Note that ∂G (NG [v2 ]) contains at least 7 edges, three of which are incident with v1 , v3 , and v4 . Therefore, by Lemma 2.3, we have



f (G − NG [v2 ]) ≥ f (G) − 3 ·

2 6

−

2 7

 +

2 5

−



2



 +

6

2 6

−

2



 +5

7

2 7

−

Since α(G) ≥ α(G − NG [v2 ]) + 1, this implies a contradiction. Hence fG (v1 ) =

2



8 2

> f (G) − 1.

.

7  Since fG (v2 ) + fG (v3 ) + fG (v4 ) > 2 7 − fG (v1 ) , we may assume that fG (v3 ) ≥ 72 and fG (v4 ) ≥ 28 . Let u2 be the neighbor of v2 in S that is distinct from u, and let w2 be the neighbor of v2 that is distinct from u and u2 . Since (S \{u, u2 }) ∪ {v1 , v2 } is a maximum independent set of G, condition (i) in the choice of S implies that fG (u2 ) ≥ fG (v2 ), that is, fG (u2 ) = 62 . Since fG (v2 ) = 62 , we have ωG (v2 ) = 2 and hence u2 and w2 are not adjacent. Note that ∂G (NG [v2 ]) contains at least 7 edges, three of which are incident with v1 , v3 , and v4 . If fG (w2 ) ≤ 27 , then, by Lemma 2.3,     2 2 2 2 2 2 f (G − NG [v2 ]) ≥ f (G) − 2 · − 2 · + 2 − +5 − > f (G) − 1,

5

6

7

6

7

which implies a contradiction as above. Hence fG (w2 ) = If fG (v3 ) =

2 , 6

2 6

−

2 7

 +

2 5

−

2

2 . 6



 +

6

which implies a contradiction as above. Hence fG (v3 ) = If fG (v4 ) ≥

8

then, by Lemma 2.3,

f (G − NG [v2 ]) ≥ f (G) − 3 ·

2 , 7

7

2 6

−

2



7

 +5

2 7

−

2 8



> f (G) − 1,

2 . 7

then, by Lemma 2.3,

f (G − NG [v2 ]) ≥ f (G) − 3 ·

2 6

−

2 7

 +3

2 6

−

2 7



 +4

2 7

−

2 8



= f (G) − 1,

which implies a contradiction as above. Hence fG (v4 ) = 82 . Since S ′ = (S \{u}) ∪ {v1 } is a maximum independent set of G with fG (S ′ ) = fG (S ), condition (ii) in the choice of S implies that v1 has degree 4 and one of its neighbors outside of S, say w1 , satisfies fG (w1 ) = 26 . Note that ∂G (NG [u]) contains at least 10 edges, three of which are incident with v1 , two of which are incident with v2 , at least two of are incident with v3 , and

8

C. Brause et al. / Discrete Applied Mathematics (

)

–

three of which are incident with v4 . Furthermore, the vertices u2 , w1 , and w2 are incident with those edges. Therefore, by Lemma 2.3, f (G − NG [u]) ≥ f (G) −

2 6

−2·

2 7

−

2 8

 +3

2 5

−

2



 +7

6

2 7

−

2



8

> f (G) − 1,

which implies a contradiction as above. This completes the proof in this case. Case 2 NG (u) is not independent. Note that ωG (u) = 3, and hence fG (u) = 28 . Considering the different possibilities for the subgraph of G induced by NG (u), it follows easily that its complement ¯ [NG (u)] contains 2K2 . Now Lemma 2.1 implies that s(x) ≥ 2 for at least two neighbors x of u. Let s(v3 ), s(v4 ) ≥ 2. G By condition (i), we have fG (x) = 82 for every neighbor x of u with s(x) = 1. If (s(v1 ), s(v2 ), s(v3 ), s(v4 )) ≥ (1, 2, 2, 2)

+ 28 + 3 · 62·2 = 1, which is a contradiction. Hence, we may assume that s(v1 ) = s(v2 ) = 1, and thus fG (v1 ) = fG (v2 ) = If (s(v3 ), s(v4 )) ≥ (2, 4), then F ≤ 28 + 28 + 28 + 62·2 + 62·4 = 1, which is a contradiction. If (s(v3 ), s(v4 )) ≥ (3, 3), then F ≤ 28 + 28 + 82 + 2 · 62·3 < 1, which is a contradiction. Hence we may assume that (s(v3 ), s(v4 )) ∈ {(2, 2), (2, 3), (3, 2)}. If fG (v3 ) = fG (v4 ) = 28 , then F ≤ 3 · 82 + 2 · 82·2 = 1, which is a contradiction. Hence, we may assume that fG (v4 ) ≥ 27 . We consider two different cases according to the neighborhood of v4 in NG (u). Case 2.1 v4 has a neighbor, say v ′ , in {v1 , v2 , v3 }. First, we assume that s(v ′ ) ≥ 2, that is, v ′ = v3 , and hence fG (v3 ) ≤ 27 and fG (v4 ) = 27 . If s(v3 ) ≥ 3 or s(v4 ) ≥ 3, then F ≤ 3 · 82 + 72·2 + 72·3 < 1, which is a contradiction. Hence s(v3 ) = s(v4 ) = 2. If v3 and v4 have two common neighbors in S, then ∂G (NG [v4 ]) contains at least 3 edges, and hence, by Lemma 2.3,   2 2 2 2 f (G − NG [v4 ]) ≥ f (G) − 3 · − + 3 − = f (G) − 1, componentwise, then F ≤

2 8 2 . 8

7

8

7

8

which implies a contradiction as above. Hence v3 and v4 have only u as a common neighbor in S. Now ∂G (NG [v4 ]) contains at least 5 edges, and hence, by Lemma 2.3, f (G − NG [v4 ]) ≥ f (G) −

2 6

−2·

2 7

−

2 8

 +5

2 7

−

2



8

> f (G) − 1,

which implies a contradiction as above. Hence s(v ′ ) = 1. Note that this also implies that we may assume that v3 and v4 are not adjacent. This implies that v ′ ∈ {v1 , v2 }, and hence fG (v ′ ) ≤ 82 . Since s(v ′ ) = 1 and ωG (u) = 3, ∂G (NG [v4 ]) contains at least 4 edges, and hence, by Lemma 2.3, f (G − NG [v4 ]) ≥ f (G) −

2 6

2

−

7

−2·

2 8

 +4

2 7

−

2



8

> f (G) − 1,

which implies a contradiction as above, and completes the proof in this case. Note that, if v3 has a neighbor in {v1 , v2 , v4 }, then, by symmetry between v3 and v4 , we obtain fG (v3 ) < 72 , that is, fG (v3 ) = Case 2.2 v4 has no neighbor in {v1 , v2 , v3 }. First we assume that dG (v4 ) = 3. If ωG (v4 ) = 2, then ∂G (NG [v4 ]) contains at least 7 edges, and hence, by Lemma 2.3, f (G − NG [v4 ]) ≥ f (G) − 3 ·

2 6

−

2 8

 +7

2 7

−

2



8

2 . 8

= f (G) − 1,

which implies a contradiction as above. Hence ωG (v4 ) = 3, that is, the two neighbors of v4 that are distinct from u are adjacent, and exactly one of these two vertices belongs to S. Now ∂G (NG [v4 ]) contains at least 5 edges, and hence, by Lemma 2.3, f (G − NG [v4 ]) ≥ f (G) − 3 ·

2 7

−

2 8

 +5

2 7

−

2



8

> f (G) − 1,

which implies a contradiction as above. Hence we may assume that dG (v4 ) = 4. Note that, if v3 has no neighbor in {v1 , v2 , v4 }, then, by symmetry between v3 and v4 , we have dG (v3 ) = 4. Since fG (v4 ) ≥ 72 , we obtain that ωG (v4 ) = 2, and hence fG (v4 ) = 27 . If v3 has no neighbor in {v1 , v2 , v4 }, then dG (v3 ) = 4, and thus fG (v3 ) = implies

2 . 7

Since dG (v3 ) = dG (v4 ) = 4, we obtain that ∂G (NG [u]) contains at least 10 edges. Now Lemma 2.3

f (G − NG [u]) ≥ f (G) − 2 ·

2 7

−3·

2 8

 + 10

2 7

−

2 8



> f (G) − 1,

C. Brause et al. / Discrete Applied Mathematics (

)

–

9

which implies a contradiction as above. Hence, we may assume that v3 has a neighbor in {v1 , v2 , v4 }. As observed above, this implies that fG (v3 ) = 82 . Since G[{v1 , v2 , v3 }] contains at most 2 edges and fG (v1 ) = fG (v2 ) = fG (v3 ) = 82 , the vertices v1 , v2 , and v3 all have degree 4, and ∂G (NG [u]) contains at least 8 edges. Now Lemma 2.3 implies that f (G − NG [u]) ≥ f (G) −

2 7

−4·

2 8

 +8

2 7

−

2



8

= f (G) − 1,

which implies a contradiction as above. This completes the proof in this case. This completes the proof.  We are now in a position to prove our second main result. Theorem 2.11. If G is a graph of maximum degree at most 4, then

α(G) ≥

2



d (u) + ωG (u) + 1 u∈V (G) G

.

Proof. We assume to the contrary that G is a counterexample of minimum order and maximum degree at most 4 for Conjecture 1.5. By Lemma 2.7, we have δ(G) ≥ 3. If S is a maximum independent set of G chosen as in Lemmas 2.8 and  f (v) 2.10, then these lemmas imply that v∈NG [u] d G(v,S ) ≤ 1 for every vertex u in S. Now double-counting implies G

α(G) =



1

u∈S

≥ =

 



dG (v, S )

v∈V (G)

=

fG (v)

d (v, S ) u∈S v∈NG [u] G



fG (v) dG (v, S )

fG (v),

v∈V (G)

which is a contradiction, and completes the proof.



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