gpt pro 5.2 attempt on Math-6124

FORMAL RESTATEMENT

Given $n$ points in the plane, consider the graph on these $n$ vertices in which two points are adjacent iff their Euclidean distance is exactly $1$. Let $f(n)$ be the largest integer such that every set of $n$ points contains an independent set of size at least $f(n)$ (equivalently, the minimum possible independence number among all unit-distance graphs on $n$ vertices). The problem asks for estimates on $f(n)$, and in particular whether $f(n)\ge n/4$.

QUICK LITERATURE/CONTEXT CHECK

The problem file relates this to the chromatic number $\chi$ of the unit distance graph of $\mathbb{R}^2$ (Hadwiger--Nelson). It states the general bounds $n/\chi\le f(n)\le 2n/7$, with the upper bound via the Moser spindle. It also states a lower bound $f(n)\ge n^2/(2m_1+n)$ in terms of the maximal number $m_1$ of unit distances among $n$ points.

ATTACK PLAN

Prove the general bounds from scratch: $1$ $f(n)\ge n/\chi$ by pigeonholing a coloring of the plane; $2$ $f(n)\ge n^2/(2m+n)$ for any graph with $n$ vertices and $m$ edges, then specialize to $m\le m_1(n)$; $3$ Build an explicit 7-point unit-distance configuration (Moser spindle) with independence number $2$ and replicate it far apart to obtain $f(n)\le 2n/7+O(1)$. Include a computational "FAST REALITY CHECK" verifying the Moser-spindle adjacency and independence number.

WORK

Lemma 1070.1 (chromatic number lower bound). Let $\chi$ be the chromatic number of the (infinite) unit distance graph on $\mathbb{R}^2$. Then for every $n$, $f(n)\ge \lceil n/\chi\rceil$.

Proof. By definition of $\chi$, there exists a coloring of all points of $\mathbb{R}^2$ with $\chi$ colors such that no two points at distance $1$ share a color. Given any set $P$ of $n$ points, one color class in $P$ has size at least $\lceil n/\chi\rceil$ (pigeonhole principle). Any two points of the same color are not at distance $1$, so that color class is an independent set in the unit-distance graph on $P$. Thus every $P$ has an independent set of size at least $\lceil n/\chi\rceil$, i.e. $f(n)\ge \lceil n/\chi\rceil$. $\square$

Lemma 1070.2 (edge-count lower bound on independence). Let $G$ be a finite graph with $n$ vertices and $m$ edges. Then \[ \alpha(G)\ \ge\ \frac{n^2}{2m+n}. \] Consequently, if $m_1(n)$ denotes the maximum possible number of unit-distance edges among $n$ planar points, then \[ f(n)\ \ge\ \frac{n^2}{2m_1(n)+n}. \]

Proof. Let the vertex degrees be $d(v)$, $v\in V(G)$, so that $\sum_{v} d(v)=2m$. Choose a uniformly random permutation $\pi$ of $V(G)$ and define \[ I(\pi):=\{v\in V(G): v\text{ appears before all its neighbors in }\pi\}. \] Then $I(\pi)$ is an independent set: if $uv$ is an edge, it is impossible that both $u$ and $v$ appear before each other. Fix a vertex $v$ of degree $d(v)$. Among the set consisting of $v$ and its $d(v)$ neighbors (a set of size $d(v)+1$), each vertex is equally likely to be the earliest in the random permutation $\pi$, hence \[ \mathbb{P}(v\in I(\pi))=\frac{1}{d(v)+1}. \] By linearity of expectation, \[ \mathbb{E}|I(\pi)|=\sum_{v\in V(G)} \frac{1}{d(v)+1}. \] Therefore there exists at least one permutation $\pi$ with \[ \alpha(G)\ge |I(\pi)|\ge \sum_{v\in V(G)} \frac{1}{d(v)+1}. \] Apply Cauchy--Schwarz to the positive numbers $d(v)+1$: \[ \sum_{v} \frac{1}{d(v)+1} \ \ge\ \frac{n^2}{\sum_v (d(v)+1)} \ =\ \frac{n^2}{2m+n}. \] This proves $\alpha(G)\ge n^2/(2m+n)$. For unit-distance graphs on $n$ planar points, the number of edges is at most $m_1(n)$ by definition, hence $f(n)=\min \alpha\ge n^2/(2m_1(n)+n)$. $\square$

Lemma 1070.3 (a 7-vertex unit-distance graph with $\alpha=2$). There exists a set of $7$ points in the plane whose unit-distance graph has independence number $2$. Hence, for every $t\ge 1$, \[ f(7t)\le 2t, \] and in particular $f(n)\le \frac{2}{7}n + O(1)$.

Proof (explicit construction and combinatorial bound). We give coordinates for a Moser-spindle-type configuration. Let $\theta>0$ satisfy $2\sqrt{3}\,\sin(\theta/2)=1$. Define points \[ A=(0,0),\quad B=(1,0),\quad D=\left(\tfrac12,\tfrac{\sqrt{3}}{2}\right),\quad C=B+D=\left(\tfrac32,\tfrac{\sqrt{3}}{2}\right). \] Let $R_\theta$ denote rotation by $\theta$ about the origin and set \[ E=R_\theta(B),\quad F=R_\theta(D),\quad G=R_\theta(C). \] Now check unit distances:

• In $\{A,B,C,D\}$ we have $|AB|=|AD|=|BD|=|BC|=|CD|=1$ (two equilateral triangles sharing the edge $BD$).
• In $\{A,E,F,G\}$ we have the analogous five unit distances $|AE|=|AF|=|EF|=|EG|=|FG|=1$ by rotation invariance.
• Finally, $|CG|=|C-R_\theta(C)|=1$ by the chord-length equation $2|C|\sin(\theta/2)=1$ and $|C|=\sqrt{3}$.

Thus the unit-distance graph on the seven vertices $\{A,B,C,D,E,F,G\}$ contains the two triangles $B\!C\!D$ and $E\!F\!G$, and $A$ is adjacent to $B,D,E,F$. From this structure we show $\alpha\le 2$:

• If an independent set contains $A$, it cannot contain $B,D,E,F$, and among $\{C,G\}$ it can contain at most one because $C$--$G$ is an edge. So it has size $\le 2$.
• If an independent set does not contain $A$, then it meets the triangle $\{B,C,D\}$ in at most one vertex and meets the triangle $\{E,F,G\}$ in at most one vertex, so it has size $\le 2$.

On the other hand, $\{A,C\}$ is independent (since $A$ is not adjacent to $C$), so $\alpha=2$.

To deduce $f(7t)\le 2t$, place $t$ translated copies of this 7-point configuration so far apart (distance $>2$ between distinct copies) that no unit distances occur between points from different copies. Any independent set in the union contains at most $2$ vertices from each copy, hence at most $2t$ vertices in total. $\square$

FAST REALITY CHECK (computation). Using the above coordinates and $\theta=2\arcsin(1/(2\sqrt{3}))\approx 0.5856855$ radians, a brute-force distance check (within $10^{-9}$ tolerance) finds exactly $11$ unit-distance pairs, namely \[ \{AB,AD,AE,AF,BD,BC,CD,CG,EF,EG,FG\}, \] and exhaustive search over all subsets confirms the maximum independent set size is $2$.

VERIFICATION

Lemma 1070.1 is a direct pigeonhole argument from the definition of $\chi$. Lemma 1070.2 is proved via the Caro--Wei expectation bound and Cauchy--Schwarz; all steps are explicit. Lemma 1070.3 supplies coordinates and checks unit distances by construction; the combinatorial argument for $\alpha=2$ uses only the existence of two triangles and the adjacency of $A$. The replication argument is verified by choosing inter-copy distances $>2$, ensuring cross-copy distances are $>1$. The "FAST REALITY CHECK" was done by direct computation on the stated coordinates.

FINAL

UNRESOLVED.

(i) Strongest proved partial result: We have general lower bounds $f(n)\ge \lceil n/\chi\rceil$ (Lemma 1070.1) and $f(n)\ge n^2/(2m_1(n)+n)$ (Lemma 1070.2), and an explicit construction giving $f(n)\le 2n/7+O(1)$ (Lemma 1070.3). In particular, $\limsup f(n)/n\le 2/7$ via the replicated 7-point configuration.

(ii) First gap (crisp statement): Determine the correct linear growth constant for $f(n)$ (if it exists), and specifically decide whether $f(n)\ge n/4$ holds for all $n$.

(iii) Top 3 next moves (concrete lemmas/computations):

1. Improve upper bounds by finding smaller (or denser) finite unit-distance graphs with smaller independence ratio than $2/7$, and then replicate them.
2. Improve lower bounds by sharpening estimates on $m_1(n)$ (maximum number of unit distances), then plugging into Lemma 1070.2.
3. For moderate $n$ (say $n\le 50$), attempt computational search (with exact algebraic coordinates) for configurations minimizing the independence number, to guess the best constant and candidate extremizers.

(iv) Minimal counterexample structure: A counterexample to $f(n)\ge n/4$ would be a concrete $n$-point set whose unit-distance graph has an independent set of size at most $\lfloor n/4\rfloor$. Any asymptotic counterexample to a conjectured constant $c$ would require a family of finite unit-distance graphs with independence ratio $<c$.