GPT 6 Astra Ultra attempt on Math-5905

Erdős Problem 799: list colouring a typical graph

Statement and claim.

For $G\sim G(n,1/2)$, with probability tending to one, \[ \chi_L(G)=O\left(\frac{n\log\log n}{\log n}\right)=o(n). \] We prove this using a random allocation of list colours and a reserve palette. The external graph-theoretic input is the classical ordinary-colouring bound $\chi(G(n,1/2))=O(n/\log n)$ with probability tending to one. This bound, and its use in the known solution, are recorded in Section 4 of Alon's primary paper Choice numbers of graphs: a probabilistic approach, https://www.cs.tau.ac.il/~nogaa/PDFS/choice.pdf. See https://www.erdosproblems.com/799.

A typical graph has small colour classes.

Let $s=\lceil3\log_2 n\rceil$. By the first-moment bound, \[ \Pr(\alpha(G)\geq s) \leq\binom ns2^{-\binom s2}\longrightarrow0. \] Together with the stated ordinary-colouring theorem, we may therefore fix a graph having a proper partition into $r\leq K n/\log n$ independent classes, each of size at most $s$. Necessarily $r\geq n/s$. We will prove a list-colouring bound for every such graph, so there will be no union bound over assignments of lists.

Allocating colours, including a reserve.

Fix arbitrary lists of $k=\lceil16r\log s\rceil$ distinct colours at every vertex. Independently for each colour appearing anywhere in the lists, assign it to a reserve palette with probability $1/2$, and to each of the $r$ class palettes with probability $1/(2r)$. These are disjoint palettes of colour names; the same colour has the same allocation at every vertex. Call a vertex bad if its list misses the palette belonging to its proper colour class. For every fixed vertex, \[ \Pr(\text{bad})=(1-1/(2r))^k\leq e^{-k/(2r)} \leq s^{-8}. \] Thus the expected number of bad vertices is at most $ns^{-8}\leq rs^{-7}$, and Markov's inequality gives $\Pr(\text{at least }r\text{ bad vertices})\leq s^{-7}$.

The number of reserve colours in any one list is $\operatorname{Bin}(k,1/2)$. The elementary Chernoff bound gives $\Pr(\text{fewer than }k/4\text{ reserve colours})\leq e^{-k/16}$. By the union bound, the probability of any such deficient list is at most $ne^{-k/16}=o(1)$, since $k\geq16(n/s)\log s$ and $s=O(\log n)$. For sufficiently large $n$, both desirable properties therefore hold simultaneously for some allocation: fewer than $r$ bad vertices, and at least $k/4\geq r$ reserve colours in every list.

Completing the list colouring.

Every good vertex chooses a colour from its own class palette. Adjacent good vertices lie in different classes and use disjoint palettes, so this is proper. Colour the fewer than $r$ bad vertices one at a time, using different reserve colours. Each has at least $r$ available reserve colours in its list, and fewer than $r$ have previously been used. These colours cannot conflict with good vertices because the reserve is disjoint from all class palettes. This gives a proper colouring from the originally arbitrary lists, proving $\chi_L(G)\leq k$.

Conclusion and scope.

Substituting $r=O(n/\log n)$ and $s=O(\log n)$ gives the claimed $o(n)$ bound. The probability concerns the random graph; for every graph in the typical class, the deterministic conclusion holds for every assignment of lists. The sharper known order $\Theta(n/\log n)$ is due to Alon--Krivelevich--Sudakov; it is not needed for the complete affirmative answer proved here. The only substantial external input in this reconstruction is the ordinary-colouring theorem, not a list-colouring theorem.

Completion estimate.

COMPLETION: 100%