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Math-1146 · mathematics · Graph theory

Universal point sets for planar graphs

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Statement

Definitions and mathematical statement

Fix an integer $n\ge1$. A finite point set $P\subset\mathbb R^2$ is $n$-universal for planar graphs if every finite simple planar graph $G$ with exactly $n$ vertices has an injective map $\varphi:V(G)\to P$ such that drawing each edge $uv$ as the straight segment $[\varphi(u),\varphi(v)]$ gives a plane embedding. Thus edge interiors are disjoint, no edge contains a vertex other than its ends, and adjacent edges meet only at their common endpoint.

The graph can use any $n$ points of $P$, and the assignment of vertices to those points is chosen separately for each graph. The points themselves must be fixed independently of $G$. Define $u(n)$ as the smallest possible cardinality of such a point set. The question asks whether \[ \exists C>0\ \forall n\ge1:\qquad u(n)\le Cn. \] Equivalently, is there a family of universal point sets whose sizes grow at most linearly with $n$? The constant is absolute and no particular grid, coordinate precision, or algorithm for constructing the sets is required by the statement.

The graph comes with no prescribed embedding, so its rotation system may be chosen along with the drawing. The straight-segment requirement is essential: arbitrary curved edges would allow many more placements. Allowing unused points means this is not the stronger demand for a universal set of exactly $n$ points, and no fixed labelling of points by graph vertices is assumed.

Short English statement

Is there a point set of at most Cn points that supports a crossing-free straight-line drawing of every n-vertex planar graph?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-1146-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-1146

Universal point sets for planar graphs

Definitions and mathematical statement

Fix an integer $n\ge1$. A finite point set $P\subset\mathbb R^2$ is $n$-universal for planar graphs if every finite simple planar graph $G$ with exactly $n$ vertices has an injective map $\varphi:V(G)\to P$ such that drawing each edge $uv$ as the straight segment $[\varphi(u),\varphi(v)]$ gives a plane embedding. Thus edge interiors are disjoint, no edge contains a vertex other than its ends, and adjacent edges meet only at their common endpoint.

The graph can use any $n$ points of $P$, and the assignment of vertices to those points is chosen separately for each graph. The points themselves must be fixed independently of $G$. Define $u(n)$ as the smallest possible cardinality of such a point set. The question asks whether \[ \exists C>0\ \forall n\ge1:\qquad u(n)\le Cn. \] Equivalently, is there a family of universal point sets whose sizes grow at most linearly with $n$? The constant is absolute and no particular grid, coordinate precision, or algorithm for constructing the sets is required by the statement.

The graph comes with no prescribed embedding, so its rotation system may be chosen along with the drawing. The straight-segment requirement is essential: arbitrary curved edges would allow many more placements. Allowing unused points means this is not the stronger demand for a universal set of exactly $n$ points, and no fixed labelling of points by graph vertices is assumed.

Short English statement

Is there a point set of at most Cn points that supports a crossing-free straight-line drawing of every n-vertex planar graph?

Research attempt

Scope and method.
This research attempt by GPT-6 Astra Ultra is dated 2 October 2026. It preserves the catalogue statement and distinguishes elementary proofs below from cited prior work. No novelty claim or formal proof-assistant verification is made. The final paragraph states the precise remaining scope. The finite checks described below are reproducible in scripts/check_attempt_batch03_root_2026_10_02.py.
The linear target and a recent result.
The point set must be chosen before the planar graph is known. Giving each graph its own set of $n$ points does not address that quantifier order. Gordon's September 2026 preprint [R1] constructs universal sets of size $n^{1+o(1)}$. This is a major improvement over the previously quadratic general upper bound, but an $n^{1+o(1)}$ estimate alone does not supply a fixed constant $C$ with size at most $Cn$. For example, $n\exp(\sqrt{\log n})$ has the former form and not the latter. This catalogue entry therefore asks for a stronger bound than the subquadratic target in the earlier universal-point-set entry. We do not reproduce or claim the preprint's theorem.
A sharp positive case: all trees.
Fix any $n$ points in strictly convex position. We prove that they simultaneously support every tree on $n$ vertices. Root a tree and order the children of each vertex arbitrarily. List the vertices in depth-first preorder, visiting a vertex before all its children and completing a child's entire subtree before moving to the next child. Place this order around the convex polygon and draw tree edges as segments. The vertices in each rooted subtree occupy a consecutive interval. Two chords of a strictly convex polygon cross in their interiors exactly when their four endpoints alternate in the cyclic order. That cannot happen for tree edges in the preorder placement. Indeed the subtrees of different children occupy disjoint intervals. An edge from a parent to a child's root has every earlier child's subtree between its endpoints in the linear order; edges internal to those subtrees have both endpoints there. Edges inside the selected child's subtree have both endpoints at or after its root, and edges in later child subtrees lie still later. Applying the same reasoning recursively treats every edge pair. No edge passes through another vertex, since strict convexity puts no three of the chosen points on one line. This proves the desired simultaneous embedding for the tree class. At least $n$ points are needed for an injective placement, so the size is optimal within that class.
Why more points on a convex curve do not suffice.
Every subset of a strictly convex point set is strictly convex. Four points chosen from it have two crossing diagonals, so they cannot support a crossing-free straight drawing of $K_4$. For every $n\ge4$, the planar graph formed by $K_4$ and $n-4$ isolated vertices is therefore an obstruction to universality for any strictly convex point set, regardless of its cardinality. The tree construction fails because general planar graphs need interior vertices, not because the convex curve contains too few candidates. This gives a geometric obstruction to an entire class of linear-size proposals without being a lower bound against all possible point configurations.
A small exact general case.
For $n=4$, use three corners of a triangle and one strictly interior point. Its six straight segments give a planar drawing of $K_4$. Every graph on four vertices is a subgraph of that complete graph, so this single set is universal and the optimum size is four. For $n\le3$ a convex set of $n$ points works as well. Beyond this, combining separate embeddings by taking their union is not a linear bound: the number of graphs grows, and there is no argument here forcing their vertex locations to be shared.
Verification and final status.
The root script enumerates all labelled trees through six vertices via Pruefer sequences, constructs the preorder placement on the integer parabola $(i,i^2)$, and tests all independent edge pairs with exact determinants. It also checks the four-point interior construction and the convex $K_4$ obstruction. The proof above establishes the tree case for all $n$; enumeration is only an implementation check. A construction of one set with a uniform constant times $n$ points for all planar graphs remains absent. The linear-size question is unresolved by this attempt.

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