Generalised Majority Colourings of Digraphs

Ant\'onio Gir\~ao; Teeradej Kittipassorn; Kamil Popielarz

arXiv:1701.03780·math.CO·March 26, 2018

Generalised Majority Colourings of Digraphs

Ant\'onio Gir\~ao, Teeradej Kittipassorn, Kamil Popielarz

PDF

TL;DR

This paper investigates the minimum number of colours needed to colour directed graphs so that each vertex shares its colour with at most a 1/k proportion of its out-neighbours, providing bounds and supporting conjectures.

Contribution

It establishes bounds for the minimal colours needed in majority colourings of digraphs and extends results to majority choosability, advancing understanding in graph colouring.

Findings

01

m(k) is either 2k-1 or 2k

02

Supports the conjecture that m(2)=3

03

Provides bounds for majority choosability

Abstract

The purpose of this note is to draw attention to problems related to a concept called majority colouring recently studied by Kreutzer, Oum, Seymour, van der Zypen and Wood. They raised a problem of determining, for a natural number $k$ , the smallest number $m = m (k)$ such that every digraph can be coloured with $m$ colours where each vertex has the same colour as at most $1/ k$ proportion of its out-neighbours. We show that $m (k) \in {2 k - 1, 2 k}$ . We also prove a result supporting the conjecture that $m (2) = 3$ . Moreover, we prove similar results for a more general concept called majority choosability.

Equations14

P (x is bad)

P (x is bad)

\leq exp (- \frac{( 1/2 ) ^{2}}{3} E (B_{x})) = exp (- d^{+} (x) /36) \leq exp (- 2^{i - 1} /36) .

E (X)

E (X)

\leq 205 + i \geq 11 \sum 2^{- i + 8} = 205 + \frac{1}{4} < 206.

f (v_{1}, v_{2}, \dots, v_{1023})

f (v_{1}, v_{2}, \dots, v_{1023})

j = 1 \sum i v_{j} \leq 2 i + 1, for i \in {1, 2, \dots, 1023}

v_{i} \geq 0, for i \in {1, 2, \dots, 1023} .

j \in f^{- 1} (r) \sum (b_{ij} + b_{j i}) \leq j \in f^{- 1} (l) \sum (b_{ij} + b_{j i}) .

j \in f^{- 1} (r) \sum (b_{ij} + b_{j i}) \leq j \in f^{- 1} (l) \sum (b_{ij} + b_{j i}) .

m j \in f^{- 1} (r) \sum (b_{ij} + b_{j i}) \leq j \in f^{- 1} (L_{i}) \sum (b_{ij} + b_{j i}) \leq j = 1 \sum n (b_{ij} + b_{j i}) = 2 u_{i} .

m j \in f^{- 1} (r) \sum (b_{ij} + b_{j i}) \leq j \in f^{- 1} (L_{i}) \sum (b_{ij} + b_{j i}) \leq j = 1 \sum n (b_{ij} + b_{j i}) = 2 u_{i} .

Peer Reviews

No public reviews on file for this paper yet. If you reviewed it on a platform where reviews are public (OpenReview, ICLR, NeurIPS, ICML), you can paste yours below so the community can read it here.

Videos

No videos yet. Explain this paper in a talk, walkthrough, or lecture? Add one.

Full text

Generalised Majority Colourings of Digraphs

António Girão Department of Pure Mathematics and Mathematical Statistics, University of Cambridge, Wilberforce Road, Cambridge CB3 0WB, UK; [email protected].

Teeradej Kittipassorn Departamento de Matemática, Pontifícia Universidade Católica do Rio de Janeiro (PUC-Rio), Rua Marquês de São Vicente 225, Gávea, Rio de Janeiro, RJ 22451-900, Brazil; [email protected].

Kamil Popielarz Department of Mathematics, University Of Memphis, Memphis, TN 38152, USA; [email protected]

Abstract

We almost completely solve a number of problems related to a concept called majority colouring recently studied by Kreutzer, Oum, Seymour, van der Zypen and Wood. They raised the problem of determining, for a natural number $k$ , the smallest number $m=m(k)$ such that every digraph can be coloured with $m$ colours where each vertex has the same colour as at most a $1/k$ proportion of its out-neighbours. We show that $m(k)\in\{2k-1,2k\}$ . We also prove a result supporting the conjecture that $m(2)=3$ . Moreover, we prove similar results for a more general concept called majority choosability.

For a natural number $k\geq 2$ , a $\frac{1}{k}$ -majority colouring of a digraph is a colouring of the vertices such that each vertex receives the same colour as at most a $1/k$ proportion of its out-neighbours. We say that a digraph $D$ is $\frac{1}{k}$ -majority $m$ -colourable if there exists a $\frac{1}{k}$ -majority colouring of $D$ using $m$ colours. The following natural question was recently raised by Kreutzer, Oum, Seymour, van der Zypen and Wood [6].

Question 1.

Given $k\geq 2$ , determine the smallest number $m=m(k)$ such that every digraph is $\frac{1}{k}$ -majority $m$ -colourable.

In particular, they asked whether $m(k)=O(k)$ . Let us first observe that $m(k)\geq 2k-1$ . Consider a tournament on $2k-1$ vertices where every vertex has out-degree $k-1$ . Any $\frac{1}{k}$ -majority colouring of this tournament must be a proper vertex-colouring, and hence it needs at least $2k-1$ colours. Conversely, we prove that $m(k)\leq 2k$ .

Theorem 2.

Every digraph is $\frac{1}{k}$ -majority $2k$ -colourable for all $k\geq 2$ .

This is an immediate consequence of a result of Keith Ball (see [3]) about partitions of matrices. We shall use a slightly more general version proved by Alon [1].

Lemma 3.

Let $A=(a_{ij})$ be an $n\times n$ real matrix where $a_{ii}=0$ for all $i$ , $a_{ij}\geq 0$ for all $i\neq j$ , and $\sum_{j}a_{ij}\leq 1$ for all $i$ . Then, for every $t$ and all positive reals $c_{1},\dots,c_{t}$ whose sum is $1$ , there is a partition of $\{1,2,\dots,n\}$ into pairwise disjoint sets $S_{1},S_{2},\dots,S_{t}$ , such that for every $r$ and every $i\in S_{r}$ , we have $\sum_{j\in S_{r}}a_{ij}\leq 2c_{r}$ .

Proof of Theorem 2.

Let $D$ be a digraph on $n$ vertices with vertex set $\left\{v_{1},v_{2},\dots,v_{n}\right\}$ and write $d^{+}(v_{i})$ for the out-degree of $v_{i}$ . Let $A=(a_{ij})$ be an $n\times n$ matrix where $a_{ij}=\frac{1}{d^{+}(v_{i})}$ if there is a directed edge from $v_{i}$ to $v_{j}$ and $a_{ij}=0$ otherwise. We apply Lemma 3 with $t=2k$ and $c_{i}=\frac{1}{2k}$ for $1\leq i\leq 2k$ obtaining a partition of $\left\{1,2,\dots,n\right\}$ into sets $S_{1},S_{2},\dots,S_{2k}$ , such that for every $r$ and every $i\in S_{r}$ , $\sum_{j\in S_{r}}a_{ij}\leq\frac{1}{k}$ . Equivalently, the number of out-neighbours of $v_{i}$ that have the same colour as $v_{i}$ is at most $\frac{d^{+}(v_{i})}{k}$ where the colouring of $D$ is defined by the partition $S_{1}\cup S_{2}\cup\dots\cup S_{2k}$ . ∎

Question 1 has now been reduced to whether $m(k)$ is $2k-1$ or $2k$ .

Question 4.

Is every digraph $\frac{1}{k}$ -majority $(2k-1)$ -colourable?

Surprisingly, this is open even for $k=2$ . Kreutzer, Oum, Seymour, van der Zypen and Wood [6] gave an elegant argument showing that every digraph is $\frac{1}{2}$ -majority $4$ -colourable and they conjectured that $m(2)=3$ .

Conjecture 5.

Every digraph is $\frac{1}{2}$ -majority $3$ -colourable.

We provide evidence for this conjecture by proving that tournaments are almost $\frac{1}{2}$ -majority $3$ -colourable.

Theorem 6.

Every tournament can be $3$ -coloured in such a way that all but at most $205$ vertices receive the same colour as at most half of their out-neighbours.

Proof.

The proof relies on an observation that in a tournament $T$ , the set $S_{i}=\{x\in V(T):2^{i-1}\leq d^{+}(x)<2^{i}\}$ has size at most $2^{i+1}$ . Indeed, the sum of the out-degrees of the vertices of $S_{i}$ is at least $\binom{|S_{i}|}{2}$ , the number of edges inside $S_{i}$ . On the other hand, this sum is at most $(2^{i}-1)|S_{i}|$ by the definition of $S_{i}$ . Therefore, $\binom{|S_{i}|}{2}\leq(2^{i}-1)|S_{i}|$ and hence, $|S_{i}|\leq 2^{i+1}-1$ .

We proceed by randomly assigning one of three colours to each vertex independently with probability $1/3$ . Given a vertex $x$ , let $B_{x}$ be the number of out-neighbours of $x$ which receive the same colour as $x$ . We say that $x$ is bad if $B_{x}>d^{+}(x)/2$ . Trivially $\mathbb{E}(B_{x})=d^{+}(x)/3$ , and hence by a Chernoff-type bound, it follows that, for $x\in S_{i}$ ,

[TABLE]

Notice that if $i\geq 11$ then $\mathbb{P}(x\text{ is bad})\leq 2^{-(2i-7)}$ . Let $X$ denote the total number of bad vertices. Since the vertices of out-degree [math] cannot be bad,

[TABLE]

Hence, there is a $3$ -colouring such that all but at most $205$ vertices receive the same colour as at most half of their out-neighbours. ∎

Observe also that the same argument proves a special case of Conjecture 5.

Theorem 7.

Every tournament with minimum out-degree at least $2^{10}$ is $\frac{1}{2}$ -majority $3$ -colourable.

We remark that Theorem 6 can be strengthened ( $205$ can be replaced by $7$ ) by solving a linear programming problem. Recall that the expected number of bad vertices of out-degree at least $1024$ is at most $1/4$ . We shall use linear programming to show that the expected number of bad vertices of out-degree less than $1024$ is less than $7.75$ . Let $V_{i}$ be the set of vertices of out-degree $i$ for $i\in\left\{1,2,\dots,1023\right\}$ and note that the expected number of bad vertices of out-degree at most $1023$ is $f(v_{1},v_{2},\dots,v_{1023})=\sum_{i=1}^{1023}v_{i}p_{i}$ where $v_{i}=|V_{i}|$ and $p_{i}=\sum_{j=\lceil\frac{i+1}{2}\rceil}^{i}\binom{i}{j}(1/3)^{j}(2/3)^{i-j}$ . As before, observe that the number of vertices of degree at most $i$ is at most $2i+1$ , and therefore, $\sum_{j=1}^{i}v_{i}\leq 2i+1$ , leading to the following linear program.

[TABLE]

See Appendix A for the source code. Similarly, we can replace $2^{10}$ in Theorem 7 by $55$ , by using the same linear program to show that the expected number of bad vertices of out-degree in $[55,1023]$ is less than $3/4$ .

Let us now change direction to a more general concept of majority choosability. A digraph is $\frac{1}{k}$ -majority $m$ -choosable if for any assignment of lists of $m$ colours to the vertices, there exists a $\frac{1}{k}$ -majority colouring where each vertex gets a colour from its list. In particular, a $\frac{1}{k}$ -majority $m$ -choosable digraph is $\frac{1}{k}$ -majority $m$ -colourable. Kreutzer, Oum, Seymour, van der Zypen and Wood [6] asked whether there exists a finite number $m$ such that every digraph is $\frac{1}{2}$ -majority $m$ -choosable. Anholcer, Bosek and Grytczuk [2] showed that the statement holds with $m=4$ . We generalise their result as follows.

Theorem 8.

Every digraph is $\frac{1}{k}$ -majority $2k$ -choosable for all $k\geq 2$ .

Theorem 8 was independently proved by Fiachra Knox and Robert Šámal [5]. We prove Theorem 8 using a slight modification of Lemma 3 whose proof is very similar to that of Lemma 3.

Lemma 9.

Let $A=(a_{ij})$ be an $n\times n$ real matrix where $a_{ii}=0$ for all $i$ , $a_{ij}\geq 0$ for all $i\neq j$ , and $\sum_{j}a_{ij}\leq 1$ for all $i$ . Then, for every $m$ and subsets $L_{1},L_{2},\dots,L_{n}\subset\mathbb{N}$ of size $m$ , there is a function $f:\left\{1,2,\dots,n\right\}\rightarrow\mathbb{N}$ such that, for every $i$ , $f(i)\in L_{i}$ and $\sum_{j\in f^{-1}(r)}a_{ij}\leq\frac{2}{m}$ where $r=f(i)$ .

Proof.

By increasing some of the numbers $a_{ij}$ , if needed, we may assume that $\sum_{j}a_{ij}=1$ for all $i$ . We may also assume, by an obvious continuity argument, that $a_{ij}>0$ for all $i\neq j$ . Thus, by the Perron–Frobenius Theorem, $1$ is the largest eigenvalue of $A$ with right eigenvector $\left(1,1,\dots,1\right)$ and left eigenvector $\left(u_{1},u_{2},\dots,u_{n}\right)$ in which all entries are positive. It follows that $\sum_{i}u_{i}a_{ij}=u_{j}$ . Define $b_{ij}=u_{i}a_{ij}$ , then $\sum_{i}b_{ij}=u_{j}$ and $\sum_{j}b_{ij}=u_{i}\left(\sum_{j}a_{ij}\right)=u_{i}$ .

Let $f:\left\{1,2,\dots,n\right\}\rightarrow\mathbb{N}$ be a function such that $f(i)\in L_{i}$ and $f$ minimises the sum $\sum_{r\in\mathbb{N}}\sum_{i,j\in f^{-1}(r)}b_{ij}$ . By minimality, the value of the sum will not decrease if we change $f(i)$ from $r$ to $l$ where $l\in L_{i}$ . Therefore, for any $i\in f^{-1}(r)$ and $l\in L_{i}$ , we have

[TABLE]

Summing over all $l\in L_{i}$ , we conclude that

[TABLE]

Hence, $\sum_{j\in f^{-1}(r)}u_{i}a_{ij}=\sum_{j\in f^{-1}(r)}b_{ij}\leq\sum_{j\in f^{-1}(r)}(b_{ij}+b_{ji})\leq\frac{2u_{i}}{m}$ . Dividing by $u_{i}$ , the desired result follows. ∎

Proof of Theorem 8.

The proof is the same as that of Theorem 2, using Lemma 9 instead of Lemma 3. ∎

In fact, the same statement also holds when the size of the lists is odd.

Corollary 10.

Every digraph is $\frac{2}{m}$ -majority $m$ -choosable for all $m\geq 2$ .

This statement generalises a result of Anholcer, Bosek and Grytczuk [2] where they prove the case $m=3$ which says that, given a digraph with colour lists of size three assigned to the vertices, there is a colouring from these lists such that each vertex has the same colour as at most two thirds of its out-neighbours.

We have established that the $\frac{1}{k}$ -majority choosability number is either $2k-1$ or $2k$ . Let us end this note with an analogue of Question 4.

Question 11.

Is every digraph $\frac{1}{k}$ -majority $(2k-1)$ -choosable?

Appendix A Linear program

We use the toolkit [4] to solve the linear program with the following source code:

Bibliography6

The reference list from the paper itself. Each links out to its DOI / PubMed record.

1[1] N. Alon , Splitting digraphs , Combin. Probab. Comput., 15 (2006), pp. 933–937.
2[2] M. Anholcer, B. Bosek, and J. Grytczuk , Majority choosability of digraphs , ar Xiv: 1608.06912, (2016).
3[3] J. Bourgain and L. Tzafriri , Restricted invertibility of matrices and applications , in Analysis at Urbana, Vol. II (Urbana, IL, 1986–1987), vol. 138 of London Math. Soc. Lecture Note Ser., Cambridge Univ. Press, Cambridge, 1989, pp. 61–107.
4[4] GLPK , GNU Linear Programming Kit . https://www.gnu.org/software/glpk/ .
5[5] F. Knox and R. Šámal , Linear Bound for Majority Colourings of Digraphs , ar Xiv: 1701.05715, (2017).
6[6] S. Kreutzer, S. Oum, P. D. Seymour, D. van der Zypen, and D. R. Wood , Majority colourings of digraphs , Electr. J. Comb., 24 (2017), p. P 2.25.

TL;DR

Contribution

Findings

Abstract

Peer Reviews

Videos

Generalised Majority Colourings of Digraphs

Abstract

Question 1**.**

Theorem 2**.**

Lemma 3**.**

Proof of Theorem 2.

Question 4**.**

Conjecture 5**.**

Theorem 6**.**

Proof.

Theorem 7**.**

Theorem 8**.**

Lemma 9**.**

Proof.

Proof of Theorem 8.

Corollary 10**.**

Question 11**.**

Appendix A Linear program

Question 1.

Theorem 2.

Lemma 3.

Question 4.

Conjecture 5.

Theorem 6.

Theorem 7.

Theorem 8.

Lemma 9.

Corollary 10.

Question 11.