Asymmetric list sizes in bipartite graphs

Let the vertices of $B$ be assigned $k$ disjoint lists of length $\delta$, and let the vertices of $A$ be assigned all possible $k$-tuples drawn from these $k$ disjoint lists. ∎

It suffices to check condition $(i)$ of Theorem 4 with $k_{A}=\Delta_{A}^{\varepsilon}$ and $k_{B}=\Delta_{B}^{\varepsilon}$. Indeed, as $\Delta_{B}^{2}>e\Delta_{A}^{\varepsilon}\Delta_{B}$, $\varepsilon/4>1/\Delta_{A}^{\varepsilon}$ and $\Delta_{B}^{\varepsilon/2}>\Delta_{A}$, we have $\Delta_{B}^{\varepsilon}>\left(e\Delta_{A}^{\varepsilon}\Delta_{B}\right)^{1/\Delta_{A}^{\varepsilon}}\Delta_{A}$.

Assume condition $(i)$ of Conjecture 7, and moreover assume the truth of the conjecture only for its most symmetric form $\Delta_{A}=\Delta_{B}=\Delta$. We may assume that $\Delta_{0}>(4/\varepsilon)^{1/\varepsilon}$, and so, without loss of generality, we may also assume by the first part that $\Delta_{A}\leq\Delta_{B}\leq\Delta_{A}^{2/\varepsilon}$. For any bipartite graph $G=(V{=}A\cup B,E)$ with parts $A$ and $B$ having maximum degrees at most $\Delta_{A}$ and $\Delta_{B}$, we have from the assumption that $G$ is $k$-choosable for, say, $k=\Delta_{B}^{\varepsilon^{3}}\leq\Delta_{A}^{2\varepsilon^{2}}$, provided $\Delta_{B}$, and thus $\Delta_{A}$, is large enough as a function of $\varepsilon$. This is smaller than $\Delta_{A}^{\varepsilon}$ for $\Delta_{A}$ sufficiently large as a function of $\varepsilon$, as required. ∎

This follows for $\Delta_{B}$ large enough from condition $(i)$ of Theorem 4 with either $k_{A}=\delta\log\Delta_{B}$ and $k_{B}=(1+\varepsilon)\Delta_{A}$ or with $k_{A}=\log\Delta_{B}$ and $k_{B}=(e+\varepsilon)\Delta_{A}$. ∎

This follows for $\Delta$ large enough from condition $(ii)$ in Theorem 4 with $\Delta_{A}=\Delta_{B}=\Delta$ and either $k_{A}=(1+\varepsilon)\Delta/\log_{4}\Delta$ and $k_{B}=2$ or $k_{A}=(1+\varepsilon)\Delta/\log\Delta$ and $k_{B}=\log\Delta$. ∎

Let $k_{A},k_{B},\Delta_{A},\Delta_{B}$ satisfy condition $(i)$. Let $L$ be a $(k_{A},k_{B})$-list-assignment of $G$. We would like to show that there is a proper $L$-colouring of $G$. We do so by defining a suitable hypergraph $H=(V_{H},E_{H})$.

Let $(w,c)$ be a vertex of $V_{H}$ if $w\in B$ and $c\in L(w)$.

Let $((w_{1},c_{1}),\dots,(w_{k_{A}},c_{k_{A}}))$ be an edge of $E_{H}$ whenever there is some $v\in A$ such that $N(v)\supseteq\{w_{1},\dots,w_{k_{A}}\}$ and $L(v)=\{c_{1},\dots,c_{k_{A}}\}$.

Note that $H$ is a $k_{A}$-uniform vertex-partitioned hypergraph, where the parts are naturally induced by each list in $B$ and so are each of size $k_{B}$. We have defined $H$ and its partition so that any independent transversal corresponds to a special partial $L$-colouring of $G$. In particular, it is an $L$-colouring of the vertices in $B$ for which there is guaranteed to be a colour in $L(v)$ available for every $v\in A$, and so it can be extended to a proper $L$-colouring of all $G$.

Every part in $H$ has degree sum at most $\Delta_{B}\binom{\Delta_{A}-1}{k_{A}-1}k_{A}!\leq\Delta_{B}\Delta_{A}^{k_{A}},$ so the result follows from Lemma 11 with $\ell=k_{B}$ and $\Delta=\Delta_{B}\Delta_{A}^{k_{A}}$. ∎

Write $H=(V,E)$ and suppose $\ell^{k}\geq e(k(\Delta-1)+1)$. Consider the random transversal $\mathbf{T}$ formed by choosing one vertex from each part independently and uniformly. For each edge $f\in E$, let $A_{f}$ denote the event that $\mathbf{T}\supseteq f$. Note that $\,\mathbb{P}(A_{f})\leq 1/\ell^{k}$. Moreover $A_{f}$ is mutually independent of a set of all but at most $k(\Delta-1)$ of the other events $A_{f^{\prime}}$. The transversal $\mathbf{T}$ is independent if none of the events $A_{f}$ occur. Since by assumption $e(1/\ell^{k})(k(\Delta-1)+1)\leq 1$, there is a positive probability that $\mathbf{T}$ is independent by the Lovász Local Lemma. ∎

Let $k_{A},k_{B},\Delta_{A},\Delta_{B}$ satisfy condition $(ii)$. Let $L$ be a $(k_{A},k_{B})$-list-assignment of $G$. We would like to show that there is a proper $L$-colouring of $G$. To this end, colour each vertex $w\in B$, randomly and independently, by a colour chosen uniformly from its list $L(w)$. Let $T_{v,c}$ be the event that $v\in A$ has a neighbour coloured with colour $c$. Let $T_{v}$ be the event that $T_{v,c}$ happens for all $c\in L(v)$. The proof hinges on the following claim, which is related to the coupon collector problem, cf. e.g. [6].

For the $i^{\text{th}}$ colour $c_{i}$ in $L(v)$, let the number of occurrences of $c_{i}$ in the neighbouring lists of $v$ be $x_{i}$. Note that $\,\mathbb{P}(T_{v,c_{i}})=1-(1-1/k_{B})^{x_{i}}.$ Using $x_{i}\leq\Delta_{A}$ for every $1\leq i\leq k_{A}$ and the claim, we have

Noting that $\sum_{i=1}^{k_{A}}x_{i}\leq k_{B}\Delta_{A}$ and that the function $\log(1-(1-1/k_{B})^{x})$ is concave and increasing, Jensen’s Inequality applied with the claim implies that

Each event $T_{v}$ is mutually independent of all other events $T_{u}$ apart from those corresponding to vertices $u\in A$ that have a common neighbour with $v$ in $G$. As there are at most $\Delta_{A}(\Delta_{B}-1)$ such vertices besides $v$, the Lovász Local Lemma guarantees with positive probability that none of the events $T_{v}$ occur, i.e. there is a proper $L$-colouring, as desired. ∎

We define a $(k_{A},k_{B})$-list-assignment $L$ as follows. Let ${\mathcal{F}}_{1}$ be a family of $\overline{M}(k_{1},k_{A},\ell)$ $k_{A}$-element subsets of $[\ell]$ without Property A$(k_{1},k_{A},\ell)$. Let ${\mathcal{F}}_{2}$ be a family of $\overline{M}(k_{2},k_{B},\ell)$ $k_{B}$-element subsets of $[\ell]$ without Property A$(k_{2},k_{B},\ell)$. Assign the sets of ${\mathcal{F}}_{1}$ as lists to the vertices in $A$ and the sets of ${\mathcal{F}}_{2}$ as lists to the vertices in $B$. Suppose that $c$ is an $L$-colouring. Then $C_{1}=\{c(a)\mid a\in A\}$ intersects every set in ${\mathcal{F}_{1}}$ and so $|C_{1}|\geq k_{1}+1$ by assumption and similarly $C_{2}=\{c(b)\mid b\in B\}$ has cardinality $|C_{2}|\geq k_{2}+1$. So $|C_{1}|+|C_{2}|\geq k_{1}+k_{2}+2>\ell$, implying that $c$ cannot be a proper colouring.∎

Let $L$ be $(k_{A},k_{B})$-list-assignment of $G$. Let $U=\cup_{v\in V}L(v)$, i.e. $U$ is the union of all colour lists. Define a random partition of $U$ into parts $L_{A}$ and $L_{B}$ as follows. For each colour in $U$, randomly and independently assign it to $L_{B}$ with probability $p$ and otherwise assign it to $L_{A}$.

First assume (1). For a given vertex $v\in A$, the probability that $L(v)\cap L_{A}=\emptyset$ is $p^{k_{A}}$. For a given vertex $v\in B$, the probability that $L(v)\cap L_{B}=\emptyset$ is $(1-p)^{k_{B}}$. So the expected number of vertices $v$ which cannot be coloured with the corresponding list $L_{A}$ or $L_{B}$ is equal to $ap^{k_{A}}+b(1-p)^{k_{B}}.$ So the probabilistic method then guarantees some choice of the parts $L_{A}$ and $L_{B}$ such that every vertex in $A$ can be coloured with a colour from $L_{A}$ and every vertex in $B$ with a colour from $L_{B}.$

Otherwise assume (2). For a given vertex $w\in B$, the random variable $|L(w)\cap L_{B}|$ has a binomial distribution of parameters $k_{B}$ and $p$. So by a Chernoff bound the probability that $|L(w)\cap L_{B}|$ is smaller than $(1-\varepsilon)k_{B}p$ is smaller than $\exp{(-\varepsilon^{2}k_{B}p/2)}$. Thus, with probability greater than $1-b\exp{(-\varepsilon^{2}k_{B}p/2)}$, we have $|L(w)\cap L_{B}|\geq(1-\varepsilon)k_{B}p$ for all $w\in B$.

For a given vertex $v\in A$, the probability that $L(v)\cap L_{A}=\emptyset$ is $p^{k_{A}}$. So the expected number of vertices $v$ for which this holds is $p^{k_{A}}|A|$. Thus by Markov’s inequality, the probability that there are at least $(1-\varepsilon)k_{B}p$ such vertices $v$ is smaller than $\frac{p^{k_{A}-1}a}{(1-\varepsilon)k_{B}}$.

Due to (2), the probabilistic method guarantees a fixed choice of partition of $U$ into parts $L_{A}$ and $L_{B}$ such that $|L(w)\cap L_{B}|\geq(1-\varepsilon)k_{B}p$ for every $w\in B$ and the number of vertices $v\in A$ such that $L(v)\cap L_{A}=\emptyset$ is smaller than $(1-\varepsilon)k_{B}p$. Colour any vertex $v\in A$ having $L(v)\cap L_{A}=\emptyset$ with an arbitrary colour from $L(v)$. Colour any other vertex $v^{\prime}\in A$ with a colour from $L(v^{\prime})\cap L_{A}$. Finally colour any vertex $w\in B$ with some colour in $L(w)\cap L_{B}$ unused by any vertex of $A$ — this is possible as there were fewer than $(1-\varepsilon)k_{B}p$ colours of $L_{B}$ used to colour vertices in $A$ and the lists in $B$ all have at least $(1-\varepsilon)k_{B}p$ colours from $L_{B}$.

In either case, we are guaranteed a proper $L$-colouring of $G$, as promised. ∎