# Reducing the maximum degree of a graph: comparisons of bounds

Let λ(G) be the smallest number of vertices that can be removed from a non-empty graph G so that the resulting graph has a smaller maximum degree. Let λ_ e(G) be the smallest number of edges that can be removed from G for the same purpose. Let k be the maximum degree of G, let t be the number of vertices of degree k, let M(G) be the set of vertices of degree k, let n be the number of vertices in the closed neighbourhood of M(G), and let m be the number of edges incident to vertices in M(G). Fenech and the author showed that λ(G) ≤n+(k-1)t/2k, and they essentially showed that λ (G) ≤ n ( 1- k/k+1( n/(k+1)t) ^1/k ). They also showed that λ_ e(G) ≤m + (k-1)t/2k-1 and λ_ e (G) ≤ m ( 1- k-1/k( m/kt) ^1/(k-1) ). These bounds are attained if k ≥ 2 and G is the union of t pairwise vertex-disjoint (k+1)-vertex stars. For each of λ(G) and λ_ e(G), the two bounds on the parameter are compared for the purpose of determining, for each bound, the cases in which the bound is better than the other. This work is also motivated by the likelihood that similar pairs of bounds will be discovered for other graph parameters and the same analysis can be applied.

## Authors

• 3 publications
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## 1 Introduction

For basic terminology and notation in graph theory, we refer the reader to [2, 6]; the definitions of terms and notations used here are given in the papers [3, 5], which are the basis of the work presented here.

The set of positive integers is denoted by . For any , the set is denoted by . For a set , the set of all -element subsets of is denoted by . Arbitrary sets are taken to be finite.

Every graph is taken to be simple, that is, its vertex and edge set satisfy . We may represent an edge by . For , denotes , denotes , denotes , and denotes () and is called the degree of . For , is denoted by and called the closed neighbourhood of . The maximum degree of is and is denoted by . The set of vertices of of degree is denoted by . For , denotes the subgraph of induced by , that is, . For , denotes the subgraph of obtained by removing the vertices in from , that is, . For , denotes the subgraph of obtained by removing the edges in from , that is, .

We call a subset of a -reducing set of if or (note that is the smallest -reducing set of if and only if ). We call a subset of a -reducing edge set of if or . We denote the size of a smallest -reducing set of by , and we denote the size of a smallest -reducing edge set of by .

Let denote the subgraph of induced by , and let denote the subgraph of with vertex set and edge set . As explained in [3, 5], we clearly have

 Δ(Gv)=Δ(G),M(Gv)=M(G),λ(Gv)=λ(G), (1) Δ(Ge)=Δ(G),M(Ge)=M(G),λe(Ge)=λe(G), (2) |V(Gv)|≤∑v∈M(G)|NG[v]|=(Δ(G)+1)|M(G)|, (3) |E(Ge)|≤∑v∈M(G)|EG(v)|=Δ(G)|M(G)|. (4)

By the handshaking lemma and (2), , so

 |E(Ge)|≥Δ(G)|M(G)|/2. (5)

The graph parameters and were investigated in [3] and [5], respectively. For each of them, two main general bounds were obtained, and the bounds are sharp.

The following is the first main general bound proved in [3].

###### Theorem 1.1 ([3])

If is a graph, , , and , then

 λ(G)≤n+(k−1)t2k.

In [3], the result is actually stated with , but the improvement given by is immediately deduced from (1). The extremal structures are determined in [4]. Using a probabilistic argument similar to that used by Alon in [1], it was also shown in [3, Proof of Theorem 2.7] that

 λ(G)≤u(p)=np+t(1−p)k+1 for any real number p such that 0≤p≤1, (6)

and that this yields the bound

 λ(G)≤nln(k+1)+tk+1.

However, by differentiating with respect to , we find that the minimum value of occurs at (note that this satisfies by (3)), and hence it is . Thus, by (6), the following was essentially established in [3].

###### Theorem 1.2

If is a graph, , , and , then

 λ(G)≤n⎛⎝1−kk+1(n(k+1)t)1/k⎞⎠.

The following are the two main general bounds proved in [5].

###### Theorem 1.3 ([5])

If is a graph, , , and , then

 λe(G)≤m+(k−1)t2k−1.
###### Theorem 1.4 ([5])

If is a graph, , , and , then

 λe(G)≤m(1−k−1k(mkt)1/(k−1)).

Theorem 1.4 was obtained by means of a probabilistic argument similar to that for Theorem 1.2.

The bounds in Theorems 1.11.4 are attained if, for example, and is the union of pairwise vertex-disjoint -vertex stars (see [3, 4, 5]). If , then and are the same union of pairwise vertex-disjoint -vertex stars, and hence the bounds in Theorems 1.1 and 1.3 are attained.

In this paper, we compare the bounds on in Theorems 1.1 and 1.2, and we compare the bounds on in Theorems 1.3 and 1.4. We use several well-known results from real analysis to determine, for each bound, a significant proportion of the cases in which the bound is better than the other bound for the same parameter. Our main contribution is the solution of the problem for for sufficiently large.

###### Theorem 1.5

Let , , and be as in Theorems 1.3 and 1.4, and let be the real number such that .
(a) If , then the bounds in Theorems 1.3 and 1.4 are equal.
(b) If and is sufficiently large, then the bound in Theorem 1.3 is smaller than the bound in Theorem 1.4.
(c) If and is sufficiently large, then the bound in Theorem 1.4 is smaller than the bound in Theorem 1.3.

A stronger version that addresses any is proved in Section 2.

This work is also motivated by the likelihood that similar pairs of bounds will be discovered for other graph parameters and the same analysis can be applied.

## 2 The bounds in Theorems 1.3 and 1.4

Let , , and be as in Theorems 1.3 and 1.4. Let and be the bound in Theorem 1.3 and the bound in Theorem 1.4, respectively; that is,

 b1(k,t,m)=m+(k−1)t2k−1andb2(k,t,m)=m(1−k−1k(mkt)1/(k−1)).

By (4), , and equality holds if is the union of pairwise vertex-disjoint -vertex stars, in which case the two bounds are equal and attained. We now consider .

If , then

 b2(k,t,m)−b1(k,t,m)=m−m24t−m+t3=(2t−m)(3m−2t)12t>0

as and by (5). Thus, if . We now consider .

###### Theorem 2.1

Suppose and . Let be the real number such that .
(a) There exists a unique real number such that and . We have

 b1(k,t,m)

The larger is, the larger is. Moreover, for any real , if is sufficiently large.
(b) There exists a unique real number such that and . We have

 b2(k,t,m)

The larger is, the smaller is. Moreover, for any real , if is sufficiently large.

Since can be at most , this result tells us that the range of values of for which the bound in Theorem 1.3 is better than the bound in Theorem 1.4 is wider than that for which the opposite holds.

We now prove Proposition 2.1. The set of real numbers is denoted by , and the set of positive real numbers is denoted by . We shall make use of standard notation for real intervals. Let be the base of the natural logarithm, that is, .

###### Lemma 2.2

If is the function given by

 f(x)=(1+1x)x+1

for , then decreases as increases, and .

Proof. Let be the function given by

 g(z)=z−ln(1+z)

for . The derivative is , which is negative for , for , and positive for . Thus, increases from as increases from to infinity, and hence

 g(z)>0for z>0. (7)

We have . Using implicit differentiation, we obtain . Thus, by (7) with , , and hence, since , we obtain . Therefore, decreases as increases. Now

###### Lemma 2.3

For any , let be the function given by

 fc(x)=ce(x−1)/2x

for .
(a) As increases from to infinity, increases to infinity.
(b) There exists a unique real number such that , and .
(c) If with , then .
(d) For any real , and for any .

Proof. Using differentiation, we obtain that the minimum value of occurs at , and that has no other turning points. Thus, decreases from to as increases from to , and increases as increases from . Since , increases from to infinity as increases from to infinity. This yields (a) and (b).

Let with . Since , we obtain (c).

Let . Let . Let . By (a) and (b), . We have , so . By (c), (d) follows.

###### Lemma 2.4

Let . Let be the function given by

 f(x,y)=(2y−12y−x)y−1/2−x

for . For any , for some unique , and for any and such that or .

Moreover, let be the real number such that .
(a) If with , then .
(b) For any real , there exists some such that for any .

Proof. Let such that for . We have

 dgdx=(y0−1/2)(2y0−12y0−x)y0−3/22y0−1(2y0−x)2−1=12(2y0−12y0−x)y0+1/2−1.

As increases from to , the value of increases from to , and hence increases from to . Thus, there exists a unique such that is at , and . Thus, decreases from to , and then increases from to . Consequently, there exists a unique such that and for each .

Now suppose and . Let and . Then, . We have

 f(x,y)+x =(1+x−12y−x)y−1/2=(1+1z)(z+1)(x−1)/2 =f(x,y0)+x. (8)

Therefore,

 f(x,y)≤f(x,y0)=g(x)≤g(xy0)=0. (9)

If , then , and hence . If , then , , (by Lemma 2.2), and hence by (8). Thus, if or , then or , and hence by (9).

Let such that for . Using differentiation, we obtain that the minimum value of occurs at , and that has no other turning points. Thus, decreases from to as increases from to , and, since , Lemma 2.3 (a) implies that increases to infinity as increases from . Note that . Let . We have by Lemma 2.2. Thus, , and hence .

Next, suppose . By the same argument for , . Let such that for . By the argument above, for , , and for . Since , we have for any , so . Thus, (a) is proved.

Finally, let . Let . Let . Since , . Let such that for . We have and . For any , let . As increases to infinity, increases to infinity. We have . Thus, by Lemma 2.2, decreases from to as increases from to infinity. Thus, there exists some such that . We have , so . By (a), for any . Thus, (b) is proved.

Proof of Theorem 2.1. Let . Since , . By (5), , so . Let be any of the relations , , and . We have

 b1(k,t,m)∼b2(k,t,m)⇔m+(k−1)tm∼(2k−1)(1−k−1k(mkt)1/(k−1)) ⇔1+(k−1)xk∼2k−1−(2k−1)(k−1)kx1/(k−1) ⇔(2k−1)(k−1)x1/(k−1)∼2k(k−1)−(k−1)x⇔(2k−1)x1/(k−1)∼2k−x ⇔(2k−1)k−1∼(2k−x)k−1x>0(as% 1

Let be as in Lemma 2.4. Let . By Lemma 2.4, for some unique , and the larger is, the larger is. Let . It can be checked that if . By Lemma 2.4 (b), for any real , if is sufficiently large. Suppose . Then, . We have by Lemma 2.4. Since , we have by (10). Thus, (a) is proved.

We now prove (b). Let and . We have

 (2k−12k−x)k−1 =(2k−12k−x)k−1/2(2k−12k−x)−1/2=(1+1z)(z+1)(x−1)/2(2k−x2k−1)1/2 ≥c((1+1z)(z+1))(x−1)/2(as x≤2) >ce(x−1)/2(by Lemma\leavevmode\nobreak\ ???). (11)

Let be as in Lemma 2.3. By Lemma 2.3, for some unique , and the larger is, the smaller is. Thus, the larger is, the smaller is. It can be checked that if . By Lemma 2.3 (d), for any real , if is sufficiently large. Suppose . Then, . By Lemma 2.3, we have , so . By (10) and (11),

## 3 The bounds in Theorems 1.1 and 1.2

Let , , and be as in Theorems 1.1 and 1.2. Let and be the bound in Theorem 1.1 and the bound in Theorem 1.2, respectively; that is,

 b1(k,t,n)=n+(k−1)t2kandb2(k,t,n)=n⎛⎝1−kk+1(n(k+1)t)1/k⎞⎠.

If , then clearly , is attained, and . We now consider . By (3), , and equality holds if is the union of pairwise vertex-disjoint -vertex stars, in which case the two bounds are equal and attained. We now consider .

###### Theorem 3.1

Suppose and .
(a) Let be the real number such that . There exists a unique real number such that and . We have

 b1(k,t,n)

The larger is, the larger is. Moreover, for any real , if is sufficiently large.
(b) Let be the real number such that . There exists a unique real number such that and . We have

 b2(k,t,n)

The smaller is, the smaller is.

Since can be at most , this result tells us that the range of values of for which the bound in Theorem 1.1 is better than the bound in Theorem 1.2 is wider than that for which the opposite holds.

We now prove Proposition 3.1. By slightly modifying the function in Lemma 2.4, we obtain the following lemma by the same argument for Lemma 2.4.

###### Lemma 3.2

Let . Let be the function given by

 f(x,y)=(2y2y+1−x)y−x

for . For any , for some unique , and for any and such that or .

Moreover, let be the real number such that .
(a) If with , then .
(b) For any real , there exists some such that for any .

Proof of Theorem 3.1. Let . Since , . Obviously, , so . Let be any of the relations