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Math-604 · mathematics · Graph theory

Albertson Conjecture

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Statement

Definitions and mathematical statement

Shared notation.
Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

The crossing number $\operatorname{cr}(G)$ is the least number of proper edge crossings over plane drawings with vertices distinct and edges drawn as arcs, in general position. The conjecture is $\operatorname{cr}(G)\geq\operatorname{cr}(K_{\chi(G)})$ for every finite graph $G$. Straight-line drawings are not required. [S1] [S2]

Statement recorded in the supplied source.
Can the crossing number of a graph be lower-bounded by the crossing number of a complete graph with the same chromatic number? [S1]
Residual task reported by the archive.

The embedded review (2026-08-17) records: Establish the crossing-number lower bound for every chromatic number and graph order. [S1]

Short English statement

Must every graph incur at least as many unavoidable plane crossings as a complete graph requiring the same number of vertex colours?

Sources

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Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-604-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-604

Albertson’s Crossing-Number Conjecture

Definitions and mathematical statement

Shared notation.
Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

The crossing number $\operatorname{cr}(G)$ is the least number of proper edge crossings over plane drawings with vertices distinct and edges drawn as arcs, in general position. The conjecture is $\operatorname{cr}(G)\geq\operatorname{cr}(K_{\chi(G)})$ for every finite graph $G$. Straight-line drawings are not required. [S1] [S2]

Statement recorded in the supplied source.
Can the crossing number of a graph be lower-bounded by the crossing number of a complete graph with the same chromatic number? [S1]
Residual task reported by the archive.

The embedded review $2026-08-17$ records: Establish the crossing-number lower bound for every chromatic number and graph order. [S1]

Short English statement

Must every graph incur at least as many unavoidable plane crossings as a complete graph requiring the same number of vertex colours?

Research attempt

Status: UNRESOLVED. We reduce a possible counterexample to a vertex-critical graph, derive an elementary crossing lemma, and prove the conjectured comparison when that critical graph has sufficiently many vertices relative to its chromatic number. This leaves a concrete finite range of orders for each fixed chromatic number, but does not eliminate it.
Literature check.
Cranston [S2], primary abstract consulted 1 October 2026, reports the conjecture through chromatic number 24 and stronger restrictions on the order of a minimum counterexample. We do not reproduce those results. The calculation below is a weaker self-contained reduction illustrating the role of criticality and crossing density. No improvement of the published constants is claimed.
Passing to a critical subgraph.
Let $r=\chi(G)$. Repeatedly delete a vertex whose deletion leaves chromatic number $r$, stopping at a graph $H$ for which every vertex deletion lowers it. Then $\chi(H)=r$, and $\operatorname{cr}(H)\le\operatorname{cr}(G)$ by restricting a drawing. Every vertex of $H$ has degree at least $r-1$: otherwise an $(r-1)$-colouring of $H-v$ would leave a colour available for $v$. Writing $n=|H|$ and $m=|E(H)|$ gives $m\ge(r-1)n/2$.

If $n=r$, the graph must be complete. Indeed any missing edge permits its endpoints to share a colour while every other vertex receives its own, giving at most $r-1$ colours. Thus order $r$ itself satisfies the desired comparison exactly. The possible difficulty starts at larger critical graphs.

An elementary crossing-number inequality.
We may take a drawing realizing the crossing number in which no two adjacent edges cross and every crossing is between two edges with four distinct endpoints. A crossing between adjacent edges can be removed by exchanging their initial arcs and making a local perturbation, without increasing other intersections. Degenerate multiple crossings can also be separated into pairwise crossings.

In any such drawing, removing at most one edge for each crossing leaves a planar simple graph. The planar edge bound, weakened to $3n$ so that all small orders are included, therefore yields $c\ge m-3n$, where $c$ is the number of crossings. Now retain each vertex independently with probability $p$ and restrict the drawing to the induced subgraph. Expected vertex, edge, and crossing counts are $pn,p^2m,p^4c$. Applying the same weak planar inequality to each sampled drawing and averaging gives \[ p^4c\ge p^2m-3pn. \] For $m\ge4n$, choose $p=4n/m\le1$. Substitution gives \[ \operatorname{cr}(H)\ge\frac{m^3}{64n^2}. \] The fourth power of $p$ is why the four-distinct-endpoints convention had to be justified. Using $p^4$ for a crossing of adjacent edges would be incorrect.

A range where the target follows.
For $r\ge9$, criticality ensures $m\ge4n$, hence \[ \operatorname{cr}(H)\ge\frac{(r-1)^3n}{512}. \] Place $r$ points in strictly convex position and draw all straight chords. Every four-set determines exactly one crossing between its diagonals, so this gives a drawing of $K_r$ with $\binom r4$ crossings. Thus \[ \operatorname{cr}(K_r)\le\binom r4. \] It follows that the desired inequality holds whenever \[ n\ge\frac{512\binom r4}{(r-1)^3}. \] This sufficient threshold is approximately $(64/3)r$ for large $r$. The coefficient is intentionally weak; the source reports much sharper order reductions. The deduction does not require knowing the exact crossing number of $K_r$, whose general formula is itself a separate unresolved target.
A second, weaker inequality for moderate order.
The unsampled planar inequality combined with criticality gives \[ \operatorname{cr}(H)\ge \frac{(r-7)n}{2}. \] For $r>7$ this is positive, but grows only linearly with $r$ at fixed $n/r$. It generally cannot match a quartic complete-graph crossing scale. This makes precise why simply deleting one edge per crossing and applying a colouring degree bound is insufficient in the critical order range.

For $r\le4$, $K_r$ has a plane drawing, so $\operatorname{cr}(K_r)=0$ and the target follows trivially. No appeal to the converse assertion that every planar graph is four-colourable is needed for this direction.

Verification and remaining gap.
The random sampling is used only as an averaging proof; no simulation is claimed. The parameter choice has been restricted to $m\ge4n$ so that it is a probability. Monotonicity is used for taking subgraphs, where it is immediate by deleting edges and vertices from a drawing; no unsupported claim about arbitrary graph contractions is used.

For each fixed $r\ge9$, the remaining candidates after this elementary argument have critical order strictly between $r$ and the displayed threshold. Finiteness at fixed $r$ does not give one finite computation for all $r$. The missing step is either a stronger structural lower bound for crossings in that range or a verified counterexample drawing. Our work reduces the search but does not settle Albertson's conjecture.

Sources

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