Boundary activity

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

This lesson proves the packing theorem, and with it Theorems A and B [OpenAI-HL, Section 7]. The finite observation is that, while the cutting lines of a short stretch of the sweep turn by a small angle $\theta$, every participating vertex in a fixed region either lies in one of three strips of width about $\theta$, or is accounted for by a token crossing the boundary of the region. Tokens move in a fixed cone of directions, so each crosses the boundary at most twice in all. In the limit, the spatial measure of A spatial measure gives strips little mass, the angles add up to at most $\pi$, and the boundary crossings add up to a bounded amount; Hölder's inequality then contradicts the positive edge mass of the good set.

1. Three strips

Lemma 1.1 (three strips). Let $P$ be a finite set in the plane, $B$ a bounded open set and $D\ge1$ with $|x|\le D$ for $x\in B$. Let $M_0,\dots,M_m\subseteq P$ have the same number of points and be given by cuts $\begin{gathered} M_i=\{x\in P:\nu_i\cdot x<c_i\},\\ |\nu_i|=1,\qquad|\nu_i-\nu_0|\le\theta, \end{gathered}$ with no point of $P$ on any line $\nu_i\cdot x=c_i$. For $i\ge1$ fix a bijection from $M_{i-1}\setminus M_i$ onto $M_i\setminus M_{i-1}$, the transfers of step $i$, and let $C_B$ be the number of transfers with exactly one endpoint in $B$. Let $A=\bigcup_{i\ge1}(M_{i-1}\triangle M_i)$ be the set of participating points. Put $\rho=D\theta$, $c_-=\min c_i$, $c_+=\max c_i$, and for $q\in\{c_0,c_-,c_+\}$ let $S_q=\{x:|\nu_0\cdot x-q|\le3\rho\}$. Then $\#\bigl((A\cap B)\setminus(S_{c_0}\cup S_{c_-}\cup S_{c_+})\bigr)\le2C_B ,$ and hence $\#(A\cap B)\le2C_B+\sum_q\#(P\cap B\cap S_q)$.

Proof. Write $t(x)=\nu_0\cdot x$ and $e_i(x)=(\nu_i-\nu_0)\cdot x$; for $x\in B$, $|e_i(x)|\le\rho$. Let $S$ be the union of the three strips.

Initially marked participants. Let $x\in(A\cap B)\setminus S$ with $x\in M_0$. Then $t(x)<c_0$, and outside $S_{c_0}$, $t(x)<c_0-3\rho$. Since $x$ participates, it is unmarked at some $i$: $t(x)+e_i(x)>c_i\ge c_-$, so $t(x)>c_--\rho$, and outside $S_{c_-}$, $t(x)>c_-+3\rho$. At an index $i_-$ with $c_{i_-}=c_-$, $\nu_{i_-}\cdot x\ge t(x)-\rho>c_-$: $x\notin M_{i_-}$. The existence of such an $x$ gives $c_0-c_->6\rho$. Then every $y\in(P\cap B)\setminus M_0$ has $\nu_{i_-}\cdot y\ge t(y)-\rho>c_0-\rho>c_-$, so $y\notin M_{i_-}$: between the states $0$ and $i_-$, no point of $B$ is gained, and $M_{i_-}\cap B\subseteq M_0\cap B$.

Put a token on each point of $M_0$ and move the tokens along the transfers, one step at a time. At each state the tokens occupy exactly the points of $M_i$. So $\#(M_0\cap B)-\#(M_{i_-}\cap B)$ equals the number of transfers leaving $B$ minus the number entering $B$ during the first $i_-$ steps, which is at most $C_B$. The left side counts the points of $M_0\cap B$ lost by state $i_-$, among them all the $x$ considered.

Initially unmarked participants. For $x\in(A\cap B)\setminus S$ with $x\notin M_0$, symmetrically $t(x)>c_0+3\rho$ and $t(x)<c_+-3\rho$, so $x\in M_{i_+}$ for an index with $c_{i_+}=c_+$; every point of $M_0\cap B$ is still marked at $i_+$, and the gains in $B$ up to state $i_+$ number at most $C_B$. The two classes together have at most $2C_B$ points. $\square$

The lemma includes $\theta=0$, when the strips have width zero, and points that change several times and return to their initial status.

2. Tokens and the boundary of a region

We keep the extremal sequence, the subsequence and the enhanced limiting measures of A spatial measure: the window $K_0$ with closure in $(t_1,t_2)$, the set $G$ with $\lambda_{K_0}(G)>0$, the spatial measure $\beta$, carried by the closed unit disk, with $\beta(B(x,r))\le C_Gr^{d_*}$, $d_*=32/17$, and the normalized plane of each configuration, the image of the original plane under an affine bijection $\Psi_n$ whose linear part maps the lines of slopes $\varsigma(t_1),\varsigma(t_2),\varsigma(t_3)$ to $\mathbb Rv_1,\mathbb Rv_2,\mathbb Rv_3$.

The cone of transfers. In late configurations every batch of $K_0$ lies strictly between the batches containing the times $t_1$ and $t_2$, so every exchange edge of $K_0$, from a lost point to a gained point, points to the right with slope strictly between $\varsigma(t_1)$ and $\varsigma(t_2)$ (Lemma 3.1 of Chains, crossings and Dey's bound). These vectors lie in the cone spanned by $(1,\varsigma(t_1))$ and $(1,\varsigma(t_2))$, which $\Psi_n$ maps to a closed quadrant spanned by $\pm v_1$ and $\pm v_2$. Passing to a subsequence, it is one fixed closed quadrant $\mathcal Q$. Choose linearly independent linear functionals $\ell_1,\ell_2$ positive on $\mathcal Q\setminus\{0\}$; for $\mathcal Q=\{\epsilon_1x\ge0,\ \epsilon_2y\ge0\}$, $\ell_1=2\epsilon_1x+\epsilon_2y$ and $\ell_2=\epsilon_1x+2\epsilon_2y$ will do. Choose the open parallelogram $\begin{aligned} B=\{x:{}&a_1<\ell_1(x)<b_1,\\ &a_2<\ell_2(x)<b_2\} \end{aligned}$ containing the closed unit disk, and $D\ge\max(1,\sup_{x\in B}|x|)$.

Tokens. In each batch of $K_0$ fix one perfect matching from the lost points to the gained points, and move tokens, starting on the marked points at the beginning of $K_0$, along these matchings; there are at most $N$ tokens. Each move of a token is a transfer vector in $\mathcal Q\setminus\{0\}$ after normalization, so $\ell_1$ and $\ell_2$ increase strictly along every token's path. The set of times at which a token lies in $B$ is therefore an interval of consecutive times: if the token is in $B$ at two times, both functionals lie between their values at those times in between. So each token enters $B$ at most once and leaves it at most once, and the number of transfers of $K_0$ with exactly one endpoint in $B$ is at most $2N$.

Budgets. For a window $J\subseteq K_0$ in configuration $n$, let $C_n(J)$ be the number of these transfers in the batches of $J$, and $\theta_n(J)$ the angle spanned, in the normalized plane, by the lines of the states bounding the batches of $J$. The slopes of the states increase, so their lines turn monotonically in the original plane, and also after $\Psi_n$, within half a turn; windows of a partition span angles with disjoint interiors. So for every partition of $K_0$ into windows, $\sum_J\frac{C_n(J)}N\le2,\qquad\sum_J\theta_n(J)\le\pi .$ Along a diagonal subsequence, $C_n(J)/N\to b_J$ and $\theta_n(J)\to\theta_J$ for all the countably many windows $J\subseteq K_0$, and for every partition $\sum_Jb_J\le2$ and $\sum_J\theta_J\le\pi$.

3. Limiting boundary activity

Lemma 3.1. For every window $J\subseteq K_0$, with $\mu_J(G)$ the limiting participating-vertex measure of $J$ of the labels in $G$, $\mu_J(G)\le2b_J+C_G'\theta_J^{\,d_*-1},$ where $C_G'$ depends only on $C_G$ and $D$.

Proof. In configuration $n$ apply Lemma 1.1 in the normalized plane to the marked sets at the states bounding the batches of $J$, with the chosen matchings as transfers; the participating points are the endpoints of the exchange edges of $J$. Each marked set is the set of points below a line of the state's slope, strictly, so after $\Psi_n$ it is given by a cut $\nu_i\cdot x<c_i$ with unit normals $\nu_i$ chosen continuously; they turn within the angle $\theta_n(J)$, so $|\nu_i-\nu_0|\le\theta_n(J)$. Passing to a further subsequence, the normal $\nu_0$ and the three thresholds $c_0,c_-,c_+$ converge, possibly to $\pm\infty$, simultaneously for all windows. A finite limit gives a closed limiting strip of half-width $3D\theta_J$; a threshold tending to infinity gives strips that eventually miss every bounded set. Let $\mathcal S_J$ be the union of the at most three limiting strips.

Let $\alpha_n$ be the image on the position coordinate of the participating-vertex measure of $J$, with mass $1/N$ per vertex, and $\alpha$ its limit. A continuous $f$ with $0\le f\le1$ and compact support in the open set $B\setminus\mathcal S_J$ is, for late $n$, zero on the three strips of configuration $n$, since its support has positive distance from the limiting strips. By Lemma 1.1, $\int f\,d\alpha_n\le2C_n(J)/N$, so $\int f\,d\alpha\le2b_J$, and exhausting the open set, $\alpha(B\setminus\mathcal S_J)\le2b_J$. The positions of the labels in $G$ lie in the closed unit disk inside $B$, and $\mu_J\le\mu$, so $\begin{aligned} \mu_J(G)&\le\alpha(B\setminus\mathcal S_J)+\beta(\mathcal S_J)\\ &\le2b_J+\beta(\mathcal S_J). \end{aligned}$ By Proposition 1.1(c) of A spatial measure, a strip of half-width $r$ has $\beta$-mass at most $C r^{d_*-1}$ for $r<1$, and at most $1$ in any case; so $\beta(\mathcal S_J)\le C'_G\theta_J^{\,d_*-1}$, including $\theta_J=0$, when the strips are lines. $\square$

4. The contradiction

Let $\alpha_*=d_*-1=15/17$. Fix $M\ge1$ and divide $K_0$ into the windows $J$ of length $\delta=2^{-m}$, at most $\delta^{-1}$ of them. By Lemma 1.1(d) of Limiting measures, at most $C_1M^{-4/3}\delta^{-1}$ of them have $s_J>M\delta^{3/4}$; each has $\lambda_J$-mass $\delta$, so together they carry at most $C_1M^{-4/3}$ of $\lambda_{K_0}(G)$. For the other windows, Lemma 2.1 of Limiting measures, for the enhanced labels, gives $\lambda_J(G)\le C^{1/3}M\delta^{3/4}\mu_J(G)^{1/3}$. By Hölder's inequality, Lemma 3.1, the budgets, and concavity of $\theta\mapsto\theta^{\alpha_*}$, $\begin{aligned} &\sum_J\mu_J(G)^{1/3}\le\delta^{-2/3}\Bigl(\sum_J\mu_J(G)\Bigr)^{1/3}\\ &\quad\le\delta^{-2/3}\Bigl(4+C_G'\sum_J\theta_J^{\alpha_*}\Bigr)^{1/3},\\ &\sum_J\theta_J^{\alpha_*}\le\delta^{\alpha_*-1}\Bigl(\sum_J\theta_J\Bigr)^{\alpha_*}\le\pi\,\delta^{\alpha_*-1}. \end{aligned}$ Since $\delta^{\alpha_*-1}\ge1$, the bracket is at most $(4+\pi C_G')\delta^{\alpha_*-1}$, and $\begin{aligned} &\lambda_{K_0}(G)=\sum_J\lambda_J(G)\\ &\quad\le C_1M^{-4/3}+C''M\,\delta^{3/4-2/3+(\alpha_*-1)/3}\\ &\quad=C_1M^{-4/3}+C''M\,\delta^{3/68}, \end{aligned}$ because $\frac34-\frac23-\frac2{51}=\frac{153-136-8}{204}=\frac{9}{204}=\frac3{68}$. Let $\delta\to0$ with $M$ fixed, and then $M\to\infty$: $\lambda_{K_0}(G)=0$. This contradicts $\lambda_{K_0}(G)>0$.

Conclusion. No extremal sequence exists, so Theorem 4.1 of From packing to a power saving holds. By Lemma 4.2 there, the packing theorem (Theorem 1.1) holds, and by Proposition 3.1 there, Theorems A and B are proved. The exponent $3/68$ only serves this contradiction; the saving $\varepsilon$ of Theorem B comes from the induction and has no explicit value.

5. Exercises

5.1 (easy). In Lemma 1.1 take $\theta=0$: all cuts are parallel. Show directly that every participating point of $B$ lies on a line $\nu_0\cdot x=c$ with $c\in\{c_0,c_-,c_+\}$ or is counted by $C_B$, and that in fact a participating point lies strictly between $c_-$ and $c_+$.

5.2 (medium). Show that the bound $2N$ on transfers crossing the boundary of $B$ fails for a disk $B$ if the transfers may point in all directions.

5.3 (easy). Verify $\frac34-\frac23+\frac13\bigl(\frac{15}{17}-1\bigr)=\frac3{68}$, and find the exponent obtained if the spatial measure only had disk exponent $d_*=7/4$.

6. Solutions

5.1. With parallel cuts, $M_i=\{\nu_0\cdot x<c_i\}\cap P$, nested by threshold. A point $x$ with $t(x)<c_-$ is marked at every state, and one with $t(x)>c_+$ is never marked; so participating points have $c_-<t(x)<c_+$. The strips of width zero exclude only the three lines, and the lemma bounds the remaining participants in $B$ by $2C_B$.

5.2. If transfers may point in all directions, a token can enter and leave $B$ many times; with $N$ tokens moving back and forth across the boundary, the number of crossing transfers is unbounded in terms of $N$. The cone condition makes $\ell_1$ and $\ell_2$ monotone along tokens, and the parallelogram is defined by these functionals.

5.3. $\frac34-\frac23=\frac1{12}$ and $\frac13\cdot(-\frac2{17})=-\frac2{51}$; $\frac1{12}-\frac2{51}=\frac{17-8}{204}=\frac3{68}$. With $d_*=7/4$, $\alpha_*=3/4$, and the exponent is $\frac1{12}-\frac1{12}=0$: the argument would just fail. This is why a disk exponent strictly greater than $7/4$ is needed.

References

[OpenAI-HL] OpenAI, A power saving for planar halving lines, OpenAI Math Release preprint, 25 September 2026, Section 7. https://github.com/openai/math/tree/main/preprints/A-power-saving-for-planar-halving-lines-September-25-2026