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List Colourings of Complete Multipartite Graphs with 2 Big Parts

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Statement

Definitions and mathematical statement

Fix integers $a,b\ge2$. The complete bipartite graph $K_{a,b}$ has disjoint independent vertex sets of sizes $a$ and $b$, with every edge between the two sets. The join $G+H$ of disjoint graphs adds every edge between $V(G)$ and $V(H)$. Write $K_t$ for the complete graph on $t$ vertices, including $K_0$ with no vertices. Thus $K_{a,b}+K_t$ is a complete multipartite graph with part sizes $a,b,1,\ldots,1$.

A proper colouring assigns different colours to adjacent vertices. The chromatic number of this join is $t+2$: the two large parts require two colours, and each of the $t$ joined clique vertices requires its own additional colour. Its list chromatic number $\chi_\ell$ is the least integer $q$ such that every assignment of $q$ permitted colours to each vertex admits a proper colouring from those lists.

Determine, for every $a,b\ge2$, the parameter \[ \phi(a,b)=\min\{t\in\mathbb N_0: \chi_\ell(K_{a,b}+K_t)=t+2\}. \] The source discusses a theorem ensuring the defining set is nonempty. A complete determination must supply both a universal list-colouring argument at the proposed $t$ and, for every smaller $t$, a list assignment that cannot be coloured with $t+2$ available colours per vertex. Ordinary chromatic number alone does not decide the threshold. Interchanging $a$ and $b$ leaves the graph unchanged, so the sought function is symmetric.

Short English statement

How many mutually adjacent universal vertices must be added to a complete bipartite graph before its choice number equals its chromatic number?

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Math-1112-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-1112

List Colourings of Complete Multipartite Graphs with 2 Big Parts

Definitions and mathematical statement

Fix integers $a,b\ge2$. The complete bipartite graph $K_{a,b}$ has disjoint independent vertex sets of sizes $a$ and $b$, with every edge between the two sets. The join $G+H$ of disjoint graphs adds every edge between $V(G)$ and $V(H)$. Write $K_t$ for the complete graph on $t$ vertices, including $K_0$ with no vertices. Thus $K_{a,b}+K_t$ is a complete multipartite graph with part sizes $a,b,1,\ldots,1$.

A proper colouring assigns different colours to adjacent vertices. The chromatic number of this join is $t+2$: the two large parts require two colours, and each of the $t$ joined clique vertices requires its own additional colour. Its list chromatic number $\chi_\ell$ is the least integer $q$ such that every assignment of $q$ permitted colours to each vertex admits a proper colouring from those lists.

Determine, for every $a,b\ge2$, the parameter \[ \phi(a,b)=\min\{t\in\mathbb N_0: \chi_\ell(K_{a,b}+K_t)=t+2\}. \] The source discusses a theorem ensuring the defining set is nonempty. A complete determination must supply both a universal list-colouring argument at the proposed $t$ and, for every smaller $t$, a list assignment that cannot be coloured with $t+2$ available colours per vertex. Ordinary chromatic number alone does not decide the threshold. Interchanging $a$ and $b$ leaves the graph unchanged, so the sought function is symmetric.

Short English statement

How many mutually adjacent universal vertices must be added to a complete bipartite graph before its choice number equals its chromatic number?

Research attempt

Scope and process.
This is a GPT-6 Astra Ultra research attempt dated 2 October 2026, using primary-source browsing and elementary mathematical checks. The canonical definition above is retained. Results below are restricted results or reductions, not a claim of a new solution to the entire catalogue question. No formal proof assistant or external mathematical referee checked this text.
Literature and approach.
The original question [S2] attributes this threshold problem to Allagan and records bounds stronger than the elementary ones below for some unbalanced parameters. Noel, Reed and Wu [R1] prove Ohba's theorem: a graph with at most $2\chi+1$ vertices is chromatic-choosable. We use that theorem explicitly as an external input for upper bounds, while independently proving the small obstructions, the zero-threshold classification and monotonicity facts. These deductions do not determine $\phi(a,b)$ for every pair.
The parameter really is a threshold.
Suppose $K_{a,b}+K_t$ is $(t+2)$-choosable. Given lists of size $t+3$ on $K_{a,b}+K_{t+1}$, colour one universal clique vertex arbitrarily. Remove its colour from all other lists. Every remaining list has at least $t+2$ colours, so the assumed property colours the remaining graph. Hence the defining property, once true, stays true for every larger $t$.

For fixed $t$, deleting vertices from either large part preserves $(t+2)$-choosability, and the ordinary chromatic number remains $t+2$ as long as both parts are nonempty. Thus $\phi(a,b)$ is symmetric and is nondecreasing in each of its two arguments. These facts follow from actual list-colouring extensions and restrictions, rather than from ordinary chromatic-number arithmetic alone.

A complete classification of when $t=0$ works.
We first show that $K_{2,b}$ is two-choosable for $b\le3$. Give the two vertices in the small part lists $L_1,L_2$ of size two. If their lists intersect, colour both by a common colour; each vertex on the other side retains another permitted colour. If the lists are disjoint, there are four choices of an unordered pair consisting of one colour from each list. A two-element list at a vertex on the other side forbids at most one such pair: it is unusable only if it equals precisely that chosen pair. At most three vertices forbid at most three pairs, so one of the four choices works. Every other vertex then has a colour outside the chosen pair, completing the proper colouring.

For $K_{2,4}$, give the small side lists $\{1,2\},\{3,4\}$, and give the other side the four lists $\{1,3\},\{1,4\},\{2,3\}, \{2,4\}$. Whatever pair is chosen on the small side, the vertex with precisely that pair as its list cannot be coloured. For $K_{3,3}$, assign the three lists $\{1,2\},\{1,3\},\{2,3\}$ once on each side. Each side must use at least two of the three colours, so the used-colour sets intersect, impossible across a complete bipartite join.

By taking these graphs as induced subgraphs, the obstructions extend to every pair $a,b\ge2$ except $(2,2),(2,3),(3,2)$. Therefore \[ \phi(a,b)=0\quad\Longleftrightarrow\quad \{a,b\}=\{2,2\}\text{ or }\{2,3\}. \] The companion script [R2] checks the two explicit bad list assignments by exhaustive enumeration of their permitted vertex colours.

Upper bounds and two exact positive thresholds.
For $H=K_{a,b}+K_t$, its order is $a+b+t$ and its chromatic number is $t+2$. Ohba's theorem applies whenever \[ a+b+t\le2(t+2)+1, \quad\text{equivalently }t\ge a+b-5. \] Hence $\phi(a,b)\le\max\{0,a+b-5\}$. At $(a,b)=(2,4)$ and $(3,3)$ this gives an upper bound one, and the preceding bad lists give the matching lower bound. Thus $\phi(2,4)=\phi(4,2)=\phi(3,3)=1$. The upper direction here is credited to [R1]; it is not independently proved by the finite tests.

More generally, induced-subgraph monotonicity gives $\phi(a,b)\ge\chi_\ell(K_{a,b})-2$. To see a concrete family, put $n=\binom{2q-1}{q}$ and assign all $q$-subsets of a $(2q-1)$-colour universe to each side of $K_{n,n}$. Each side must use at least $q$ colours, giving a conflict. Thus $\chi_\ell(K_{n,n})>q$ and $\phi(n,n)\ge q-1$. This shows that the required number of universal vertices cannot stay bounded as both parts grow, although it leaves a large gap to the general upper bound.

Final status and first remaining gap.
UNRESOLVED. The zero threshold and the three displayed threshold-one pairs are established, with exact obstructions and a clearly identified external upper theorem. For general $a,b$, the missing work is a tight universal list-colouring construction together with uncolourable assignments at every smaller $t$; neither monotonicity nor the order theorem supplies that sharp pair-dependent analysis.

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