Large induced subgraphs with $k$ vertices of almost maximum degree

TL;DR

This paper proves that large induced subgraphs with many vertices of nearly maximum degree exist in graphs, confirming a conjecture up to a constant factor, and provides bounds based on maximum degree and vertex count.

Contribution

It establishes the existence of large induced subgraphs with specified degree properties, solving a conjecture by Caro and Yuster up to a constant.

Findings

Existence of induced subgraphs with many vertices of almost maximum degree

Bounds depend on maximum degree and number of vertices

Confirms a conjecture of Caro and Yuster up to a constant

Abstract

In this note we prove that for every integer $k$, there exist constants $g_{1}(k)$ and $g_{2}(k)$ such that the following holds. If $G$ is a graph on $n$ vertices with maximum degree $\Delta$ then it contains an induced subgraph $H$ on at least $n - g_{1}(k)\sqrt{\Delta}$ vertices, such that $H$ has $k$ vertices of the same degree of order at least $\Delta(H)-g_{2}(k)$. This solves a conjecture of Caro and Yuster up to the constant $g_{2}(k)$.

Equations

$$\text{if }{d}_G(x_i)>d_G(x_j)\text{ then }{d}_G(x_i)-d_{G^{\prime }}(x_i)>d_G(x_j)-d_{G^{\prime }}(x_j).$$

$$\text{for any }x,y\in A\text{, if }{d}_K(x)<d_K(y)\text{ then }{d}_K(x)-d_{K^{\prime }}(x)<d_K(y)-d_{K^{\prime }}(y).$$

$$\sum _{i=1}^{n^{\prime }-k}\left(e(v_{n^{\prime }-k+j},v_i)-e(v_{n^{\prime }},v_i)\right)$$

$$=(d(v_{n^{\prime }-k+j})-a)-(d(v_{n^{\prime }})-a)=d(v_{n^{\prime }-k+j})-d(v_{n^{\prime }})$$

$$\le M,$$

$$z=\sum _{i=1}^{n^{\prime }-k}{w}_i\in {\left[-M,M\right]}^{k-1}.$$

Full text

1 Introduction

Given a graph $G$, let the repetition number, denoted by $rep(G)$, be the maximum multiplicity of a vertex degree. Trivially, any graph $G$ of order at least two contains at least two vertices of the same degree, i.e. $rep(G)\geq 2$. This parameter has been widely studied by several researchers (e.g., [2, 4, 7, 9, 10]), in particular, by Bollobás and Scott, who showed that for every $k\geq 2$ there exist triangle-free graphs on $n$ vertices with $rep(G)\leq k$ for which $\alpha(G)=(1+o(1))n/k$ ([4]). As there are infinitely many graphs having repetition number two, it is natural to ask what is the smallest number of vertices one needs to delete from a graph in order to increase the repetition number of the remaining induced subgraph. This question was partially answered by Caro, Shapira and Yuster in [6], indeed, they proved that for every $k$ there exists a constant $C(k)$ such that given any graph on $n$ vertices one needs to remove at most $C(k)$ vertices and thus obtain an induced subgraph with at least $\min\{k,n-C(k)\}$ vertices of the same degree. Related to this question, Caro and Yuster ([8]) considered the problem of finding the largest induced subgraph $H$ of a graph $G$ which contains at least $k$ vertices of degree $\Delta(H)$. To do so they defined $f_{k}(G)$ to be the smallest number of vertices one needs to remove from a graph $G$ such that the remaining induced subgraph has its maximum degree attained by at least $k$ vertices. They found examples of graphs on $n$ vertices for which $f_{2}(G)\geq(1-o(1))\sqrt{n}$ and conjectured $f_{k}(G)\leq O(\sqrt{n})$ for every graph $G$ on $n$ vertices. In the same paper they established the conjecture for $k\leq 3$.

The following more general conjecture was posed recently by Caro, Lauri and Zarb in [5].

Conjecture 1**.**

For every $k\geq 2$ there is a constant $g(k)$ such that given a graph $G$ with maximum degree $\Delta$, one can remove at most $g(k)\sqrt{\Delta}$ vertices such that the remaining subgraph $H\subseteq G$ has at least $k$ vertices of degree $\Delta(H)$.

Let us define $g(k,\Delta)=\max\{f_{k}(G):\Delta(G)\leq\Delta\}$. In the same paper, they proved that $g(2,\Delta)=\lceil\frac{−3+\sqrt{8\Delta+1}}{2}\rceil$ and stated that $g(3,\Delta)\leq 42\sqrt{\Delta}$. We should point out that, if true, the conjecture is best possible, as there are graphs on $n$ vertices found in [5] for which any induced subgraph on more than $n-\frac{k}{2}\sqrt{\Delta}$ does not contain $k$ vertices of the same maximum degree. We shall present such constructions in Section 3.

In this note we prove the following approximate version of Conjecture 1.

Theorem 1**.**

For every positive integer $k$, there exist constants $g_{1}(k)$ and $g_{2}(k)$ such that the following holds. If $G$ is a graph on $n$ vertices with maximum degree $\Delta$ then it contains an induced subgraph $H$ on at least $n-g_{1}(k)\sqrt{\Delta}$ vertices, such that $H$ has $k$ vertices of the same degree at least $\Delta(H)-g_{2}(k)$.

2 Proofs

Given a partition of $\{1,2,\ldots,n\}$ into $t$ sets, $A_{1},A_{2}\ldots,A_{t}$, and a strictly decreasing sequence of non-negative integers $r_{1}>r_{2}>r_{3}>\ldots>r_{t}$, we say $\mathcal{A}$ is an $(r_{1},r_{2},\ldots,r_{t})$-uniform cover of $\{1,2\ldots,n\}$ if $\mathcal{A}$ is a multiset of subsets of $\{1,2,\ldots n\}$ such that, whenever $i\in\left\{1,\dots,t\right\}$ and $j\in A_{i}$, we have $|\left\{A\in\mathcal{A}:j\in A\right\}|=r_{i}$. Note that $\mathcal{A}$ is a multiset, hence we allow repetitions.

We call an $(r_{1},r_{2},\ldots,r_{t})$-uniform cover $\mathcal{A}$ of $\{1,2,\ldots,n\}=A_{1}\cup A_{2}\cup\ldots\cup A_{t}$ irreducible if there is no proper $(r^{\prime}_{1},\ldots,r^{\prime}_{t})$-uniform cover $\mathcal{B}\subset\mathcal{A}$, for some strictly decreasing sequence of non-negative integers $r^{\prime}_{1}>r^{\prime}_{2}>\ldots>r^{\prime}_{t}$.

Given a uniform cover $\mathcal{A}$ of $\{1,2,\ldots,n\}$ and a subset $B\subseteq\{1,2,\ldots,n\}$ we define $w_{\mathcal{A}}(B)$ to be the number of times $B$ appears in $\mathcal{A}$.

Lemma 2**.**

For all $n\in\mathbb{N}$, there exists $f(n)$ such that for any $1\leq t\leq n$ and any partition of $\{1,2,\ldots,n\}$ into $t$ sets $A_{1},A_{2},\ldots,A_{t}$, every $(r_{1},r_{2},\ldots,r_{t})$-uniform cover $\mathcal{A}$ of $\{1,2,\ldots,n\}$ contains a proper $(r^{\prime}_{1},r^{\prime}_{2},\ldots,r^{\prime}_{t})$-uniform sub-cover $\mathcal{B}\subset\mathcal{A}$ with $r^{\prime}_{1}\leq f(n)$.

Proof.

We shall prove there are only finitely many irreducible covers. For otherwise, let us assume there exists an infinite sequence $\{B_{i}\}_{i\in\mathbb{N}}$ of irreducible uniform covers. Since there are only finitely many partitions of $\{1,2,\ldots,n\}$, we may pass to an infinite subsequence $\{B_{l_{i}}\}_{i\in\mathbb{N}}$ of uniform covers of the same partition of $\{1,2,\ldots,n\}$. Now, choose $A\subseteq\{1,2,\ldots,n\}$ and consider the sequence of non-negative integers $\{w_{B_{l_{i}}}(A)\}_{i\in\mathbb{N}}$, clearly it must contain an infinite non-decreasing subsequence $w_{B_{l_{i_{1}}}}(A)\leq w_{B_{l_{i_{2}}}}(A)\leq\ldots$. We restrict our attention to this subsequence of the uniform covers $B_{l_{i_{1}}},B_{l_{i_{2}}},\ldots$ and iteratively apply the same argument for the remaining subsets of $\{1,2,\ldots,n\}$, always passing to a subsequence of the previous sequence of uniform covers. After we have done it for every subset of $\{1,2,\ldots,n\}$, we must end up with two distinct irreducible uniform covers (actually an infinite sequence) $\mathcal{A},\mathcal{B}$ for which $w_{\mathcal{A}}(F)\leq w_{\mathcal{B}}(F)$ for every $F\subseteq\{1,2,\ldots,n\}$. This implies $\mathcal{A}\subseteq\mathcal{B}$, which is a contradiction. Take $f(n)$ to be the maximum $r_{1}$ over all irreducible uniform covers of $\{1,2,\ldots,n\}$. ∎

Lemma 3**.**

For every $n\in\mathbb{N}$, there exists $f(n)$ such that the following holds. Let $G=(A,B)$ be a bipartite graph with $A=\{x_{1},x_{2},\ldots,x_{n}\}$. Then there exists a subset $W\subseteq V(B)$ of size at most $n\cdot f(n)=f^{\prime}(n)$, such that the induced bipartite graph ${G^{\prime}=G[A,(B\setminus W)]}$ has the property that

[TABLE]

Proof.

Partition $A$ into $A_{1},\dots,A_{t}$, so that two vertices belong to the same part if they have the same degree. Let $r_{i}$ be the degree of the vertices in $A_{i}$. We may assume that $r_{1}>r_{2}>\dots>r_{t}$. The lemma follows as a corollary of Lemma 2. Indeed, for every vertex $w\in B$, let $A_{w}\subseteq\{1,2,\ldots,n\}$ such that $i\in A_{w}$ if $x_{i}$ is a neighbour of $w$ in $G$. Note that $\mathcal{A}=\{A_{w}:w\in B\}$ is an $(r_{1},r_{2},\ldots,r_{t})$-uniform cover of $\{1,2,\ldots,n\}$. Applying now Lemma 2, we can find a $(r^{\prime}_{1},r^{\prime}_{2},\ldots,r^{\prime}_{t})$-uniform sub-cover $\mathcal{B}\subseteq\mathcal{A}$ with $r^{\prime}_{1}\leq f(n)$. Let $W=\{w\in B:A_{w}\in\mathcal{B}\}$ and $G^{\prime}=G[A,(B\setminus W)]$. It is easy to see that $|W|\leq n\cdot f(n)$ and that the property is satisfied by the definition of uniform cover. ∎

Given a positive integer $k$ and a graph $G$ with the vertex set $\left\{x_{1},\dots,x_{n}\right\}$ such that $d(x_{1})\geq\dots\geq d(x_{n})$, let $r_{k}(G):=\Delta(G)-d_{G}(x_{k})$ be the difference between the maximum degree and the degree of vertex $x_{k}$.

Theorem 4**.**

For every positive integer $k$ there exists $h(k)$ such that the following holds. If $G$ is a graph on $n$ vertices with maximum degree $\Delta$ then it contains an induced subgraph $H$ on at least $n-(h(k)+k)\sqrt{\Delta}$ vertices, such that $r_{k}(H)\leq h(k)\cdot k$.

Proof.

The proof consists of two parts. Firstly, we shall show that we can remove at most $k\sqrt{\Delta}$ vertices from $G$ so that in the remaining graph $H^{\prime}$ we have $r_{k}(H^{\prime})\leq\sqrt{\Delta}$. Then we iteratively apply Lemma 3 (at most $\sqrt{\Delta}$ times) in order to obtain an induced subgraph $H$ of $H^{\prime}$ on at least $n-\left(h(k)+k\right)\sqrt{\Delta}$ vertices such that $r_{k}(H)\leq h(k)\cdot k$. We may take $h(k)$ to be $f^{\prime}(k)$ from Lemma 3.

We start by showing there is a large induced subgraph $H^{\prime}\subseteq G$ with $r_{k}(H^{\prime})\leq\sqrt{\Delta}$.

Claim 1**.**

There is an induced subgraph $H^{\prime}$ of $G$ on at least $n-k\sqrt{\Delta}$ vertices such that $r_{k}(H^{\prime})\leq\sqrt{\Delta}$.

Proof of Claim 1.

Consider the following procedure. Let $G_{0}=G$ and suppose that ${G_{0}\supset\dots\supset G_{i}}$ have been defined. If $G_{i}$ does not have the required property then let $G_{i+1}$ be obtained from $G_{i}$ by removing $k$ vertices with largest degrees in $G_{i}$. Notice that $\Delta(G_{i+1})\leq\Delta(G_{i})-\sqrt{\Delta}$ and $|G_{i+1}|=|G_{i}|-k$. Observe that the procedure will stop after at most $\sqrt{\Delta}$ steps, as otherwise the obtained graph would have maximum degree [math]. Since $|G_{i}|\geq n-i\cdot k$ we have that $|H^{\prime}|\geq n-k\sqrt{\Delta}$. ∎

We now proceed to the second part of the proof and iteratively apply Lemma 3. In each step we remove at most $h(k)$ vertices from $H^{\prime}$ while decreasing the value of $r_{k}$ and we stop when $r_{k}$ is at most $k\cdot h(k)$.

Let $H_{0}=H^{\prime}$ and suppose that $H_{0},\dots,H_{i}$ have already been defined. If $r_{k}(H_{i})\leq k\cdot h(k)$ then we are done, so we may assume that $r_{k}(H_{i})>k\cdot h(k)$. Let $A=\left\{x_{1},\dots,x_{k}\right\}$ be a set of $k$ vertices with the largest degrees in $H_{i}$ and write $B$ for $H_{i}\setminus A$. Without loss of generality we may assume that $d_{H_{i}}(x_{1})\geq\dots\geq d_{H_{i}}(x_{k})$. Since $r_{k}(H_{i})\geq k\cdot h(k)$ there must exist $l\in\left\{2,\dots,k\right\}$ such that $d_{H_{i}}(x_{l})>d_{H_{i}}(x_{l-1})+h(k)$. Now consider the bipartite subgraph $K=H_{i}[A,B]$. By Lemma 3, with $G=K$ and $n=k$, we can remove a set $W\subset B$ of at most $f^{\prime}(k)=h(k)$ vertices from $B$, and obtain $K^{\prime}=H_{i}[A,(B\setminus W)]$ such that

[TABLE]

Let $H_{i+1}=H_{i}\setminus W$ (hence $|H_{i+1}|\geq|H_{i}|-|W|\geq|H_{i}|-h(k)$). The following claim asserts that the above procedure will stop after at most $\sqrt{\Delta}$ steps.

Claim 2**.**

$r_{k}(H_{i+1})<r_{k}(H_{i})$.

Proof of Claim 2.

Let $z$ be a vertex with the maximum degree and $w$ a vertex with the $k$’th largest degree in $H_{i+1}$. Observe that $z=x_{t}$ for some $t\geq l$ and $d_{H_{i+1}}(w)\geq d_{H_{i+1}}(x_{s})$ for some $s<l$. First, notice that $d_{H_{i}}(x_{t})-d_{H_{i}}(x_{s})\leq d_{H_{i}}(x_{1})-d_{H_{i}}(x_{k})=r_{k}(H_{i})$. Hence, $r_{k}(H_{i+1})=d_{H_{i+1}}(z)-d_{H_{i+1}}(w)\leq d_{H_{i+1}}(x_{t})-d_{H_{i+1}}(x_{s})<d_{H_{i}}(x_{t})-d_{H_{i}}(x_{s})\leq r_{k}(H_{i})$, where the strict inequality follows from (1) since $d_{K}(x_{t})>d_{K}(x_{s})$. ∎

As in each iteration the value of $r_{k}$ decreases, we must stop after at most $r_{k}(H^{\prime})=\sqrt{\Delta}$ steps thus getting an induced subgraph $H\subset H^{\prime}$ with $r_{k}(H)\leq k\cdot h(k)$ and $|H|\geq|H^{\prime}|-h(k)\sqrt{\Delta}\geq n-(h(k)+k)\sqrt{\Delta}$. ∎

In order to prove Theorem 1 we need the following theorem of Caro, Shapira and Yuster, appearing in [6], whose proof is inspired by the one used by Alon and Berman in [1].

Theorem 5**.**

For positive integers $r,d,q,$ the following holds. Any sequence of $n\geq\left(\lceil q/r\rceil+2\right)\left(2rd+1\right)^{d}$ elements of $\left[-r,r\right]^{d}$ whose sum, denoted by $z$, is in $\left[-q,q\right]^{d}$ contains a subsequence of length at most $\left(\lceil q/r\rceil+2\right)\left(2rd+1\right)^{d}$ whose sum is $z$.

As usual, we write $R(k)$ (see e.g. [3]) for the two coloured Ramsey number, the least integer $n$ such that in any two colouring of the edges of the complete graph on $n$ vertices, there is a monochromatic $K_{k}$.

Proof of Theorem 1.

Firstly, we apply Theorem 4 with $k=R(k)$ to find a large induced subgraph $G^{\prime}\subset G$ of order at least $n^{\prime}\geq n-(h(R(k))+R(k))\sqrt{\Delta}$ and with vertex set $\left\{x_{1},\dots,x_{n^{\prime}}\right\}$ where $d(x_{1})\geq d(x_{2})\geq\dots\geq d(x_{n^{\prime}})$ and $d(x_{1})-d(x_{R(k)})\leq{h(R(k))\cdot R(k)=M}$. Now we follow the proof of Theorem 1.1 in [6].

By the definition of $R(k)$ we can find a set $S$ of $k$ vertices in $\left\{x_{1},\dots,x_{R(k)}\right\}$ that induces either a complete graph or an independent set.

Without loss of generality, assume that $S=\left\{v_{n^{\prime}-k+1},\dots,v_{n^{\prime}}\right\}$ and $V(G)\setminus S=\left\{v_{1},\dots,v_{n^{\prime}-k}\right\}$. Let $e(v_{i},v_{j})$ be equal to $1$ if there is an edge between $v_{i}$ and $v_{j}$, and [math] otherwise. We construct a sequence $X$ of $n^{\prime}-k$ vectors $w_{1},\dots,w_{n^{\prime}-k}$ in $\left[-1,1\right]^{k-1}$ as follows. The coordinate $j$ of $w_{i}$ is $e(v_{n^{\prime}-k+j},v_{i})-e(v_{n^{\prime}},v_{i})$ for $i=1,\dots,n^{\prime}-k$ and $j=1,\dots,k-1$. It is clear that $e(v_{n^{\prime}-k+j},v_{i})-e(v_{n^{\prime}},v_{i})\in[-1,1]$ as required. Consider the sum of all the $j$’th coordinates,

[TABLE]

where $a=k-1$ if $G^{\prime}[S]$ is complete, and $a=0$ otherwise. Hence,

[TABLE]

By Theorem 5, with $d=k-1$ and $q=M$, there is a subsequence of $X$ of size at most $(M+2)(2k-1)^{k-1}$ whose sum is $z$. Deleting the vertices of $G^{\prime}$ corresponding to the elements of this subsequence results in an induced subgraph $H\subset G^{\prime}$ in which all the $k$ vertices of $S$ have the same degree of order at least $\Delta(H)-\left(M+(M+2)(2k-1)^{k-1}\right)$. Choosing $g_{1}(k)=g_{2}(k)=h(R(k))(4k)^{k}$ we conclude the theorem. ∎

3 Remarks

In the previous section, we proved that every graph contains a large induced subgraph with at least $k$ vertices having the same degree of order almost the maximum degree. Note that Theorem 1 is sharp up to the size of the functions $g_{1}(k)$ and $g_{2}(k)$. Indeed, there are graphs for which one needs to remove ”roughly” $\frac{k}{2}\sqrt{\Delta}$ vertices to force the remaining subgraph to have $k$ vertices with the same degree ”near” the maximum degree. For any $k$ and $\Delta$, let $G^{\Delta}$ be the disjoint union of the stars $K_{1,n_{1}},\dots,K_{1,n_{t}}$, where $n_{i}=i\cdot\sqrt{\Delta}$, for $i\in\left\{1,\dots,t=\sqrt{\Delta}\right\}$ and let $G_{k}^{\Delta}$ be the disjoint union of $k/2$ copies of $G^{\Delta}$. It is easy to see that, for any constant $D$, one needs to remove at least $\frac{k}{2}\sqrt{\Delta}-\frac{k}{2}D$ vertices from $G_{k}^{\Delta}$ in order to obtain an induced graph $H$ with $k$ vertices of the same degree of order at least $\Delta(H)-D$.

Whether removing $C(k)\sqrt{\Delta}$ vertices is enough to force the remaining induced subgraph to have at least $k$ vertices attaining the maximum degree remains an interesting open question.

4 Acknowledgments

This project was carried through during our stay in IMT School for Advanced Studies Lucca. We would like to thank the Institute for their kind hospitality.