Proofs and Results - Hyperbolic PolyStep

01

Where this sits

This page settles the claims ledger. The proofs rest on the step as it is defined in Explainer 1. The live checks run the browser core. The results come from the Python reference, under the preregistration and its two amendments.

02

Live checks

Each property is tested here, now, on seeded random instances, by the same hyperbolic_core.js that runs the simulation. Exact claims are checked in exact rational arithmetic. Every check also runs a control: a deliberately wrong input that the check must reject, so that a check which cannot fail would show up.

H1–H7 in this browser

seed

Property	Trials passed	Worst case	Controls rejected	What is measured
Running…

The Python reference checks the same properties with its own 82 tests. Among them, the curvature-0 step is checked to equal polystep’s own step exactly, using the library itself (reference guide).

03

Setting and notation

Minkowski space $\mathbb{R}^{1,d}$ carries the form $\langle a,b\rangle = -a_0b_0 + \sum_{i\ge1} a_ib_i$. Hyperbolic space is the upper sheet $\mathbb{H}^d = \{y : \langle y,y\rangle = -1,\ y_0 > 0\}$, with $\cosh d(y,z) = -\langle y,z\rangle$. The origin is $o = (1,0,\dots,0)$, and $\bar y = (y_1,\dots,y_d)$. The boost $L_x$ carries $o$ to $x$:

$$L_x = \begin{pmatrix} x_0 & \bar x^{\top} \\ \bar x & I + \dfrac{\bar x\,\bar x^{\top}}{1+x_0}\end{pmatrix}.$$

The Klein chart is $\kappa(y) = \bar y / y_0$. One step for one particle $x$ and one factor of dimension $d$ works as follows. A rotation $R\in SO(d)$ turns the orthoplex directions $u_k = \pm e_i$ ($k = 1,\dots,2d$). The corners are

$$V_k = L_x\begin{pmatrix}\cosh r\\ \sinh r\,R u_k\end{pmatrix}.$$

A solver turns the corners’ costs into weights $w$ with $w_k \ge 0$ and $\sum_k w_k = 1$. The new point is

$$x' = \frac{s}{\sqrt{-\langle s,s\rangle}},\qquad s = \sum_k w_k V_k.$$

The weights come from polystep’s softmax, or from its Sinkhorn plan normalised per particle. In a product of factors, a step moves a block of factors. Each corner is cut into chunks $v_{k,i}$ with $\sum_i |v_{k,i}|^2 = 1$, and factor $i$ moves to $L_{x_i}(\cosh(r|v_{k,i}|), \sinh(r|v_{k,i}|)\,\hat v_{k,i})$, at distance $r|v_{k,i}| \le r$.

One fact used throughout

Lemma 0. For $a, b \in \mathbb{H}^d$, $-\langle a,b\rangle \ge 1$, with equality only when $a = b$. Hence if $w \ge 0$ and $\sum w_k = 1$, then $s = \sum w_k V_k$ has $-\langle s,s\rangle \ge 1$ and $s_0 > 0$, so $x'$ is well defined and lies on the sheet.

Since $a_0^2 = 1 + |\bar a|^2$ and $b_0^2 = 1 + |\bar b|^2$, we have $(a_0b_0 - 1)^2 - |\bar a|^2|\bar b|^2 = (a_0 - b_0)^2 \ge 0$. Cauchy–Schwarz then gives $\bar a\cdot\bar b \le |\bar a||\bar b| \le a_0b_0 - 1$, that is $-\langle a,b\rangle = a_0b_0 - \bar a\cdot\bar b \ge 1$. Equality forces $a_0 = b_0$ and $\bar a = \bar b$. For $s$, $-\langle s,s\rangle = \sum_{j,k} w_jw_k(-\langle V_j,V_k\rangle) \ge (\sum w)^2 = 1$, and $s_0 = \sum w_k V_{k,0} > 0$.

04

Lemma S: the step seen from x

One identity carries most of what follows. Seen from the particle, the hyperbolic step moves in exactly the direction polystep would, by a slightly shorter distance.

Lemma S PROVEN. Let $m = \sum_k w_k\,R u_k \in \mathbb{R}^d$, the barycentre of the turned orthoplex directions. Then $|m| \le 1$. If $m = 0$, then $x' = x$. Otherwise

$$x' = L_x \begin{pmatrix}\cosh\delta\\ \sinh\delta\ \hat m\end{pmatrix},\qquad \delta = \operatorname{artanh}\big(|m|\tanh r\big),\quad \hat m = m/|m|.$$

So $x' = \exp_x(\delta\,\hat m)$ in the frame carried by $L_x$, and $d(x,x') = \delta$.

$L_x^{-1}$ is linear and preserves the form, so apply it to everything. The corners become $(\cosh r, \sinh r\,Ru_k)$, and $L_x^{-1}s = (\cosh r,\ \sinh r\,m)$. Normalising does not change the Klein point, so $\kappa(L_x^{-1}x') = \tanh r\cdot m$. A point at distance $\delta$ from $o$ in direction $\hat m$ is $(\cosh\delta, \sinh\delta\,\hat m)$, with Klein point $\tanh\delta\,\hat m$. Matching the two gives $\tanh\delta = |m|\tanh r$. Finally, $|m| \le \sum w_k |Ru_k| = 1$.

Consequences.

The direction is polystep’s. polystep moves its particle by $r\,m$, with the same $m$ if it is given the same costs. Pull the objective back to the tangent space at $x$, $f_x = f\circ\exp_x$. There, polystep’s corners are $rRu_k$, exactly the hyperbolic corners seen from $x$.
The length is shorter. $\delta = \operatorname{artanh}(|m|\tanh r) \le r|m|$. The function $a \mapsto \operatorname{artanh}(a\tanh r)$ is convex on $[0,1]$, is 0 at 0 and $r$ at 1, so it lies below the chord $a r$. The difference $r|m| - \delta$ is about $\tfrac{r^3}{3}|m|(1-|m|^2) \le 0.13\,r^3$ for $r \le 1$; the maximum computed over $r \le 1$ is $0.1282\,r^3$.
So: the hyperbolic step is polystep’s step on the pulled-back objective $f_x$, shortened radially by a factor $1 - O(r^2)$, and carried to the sheet by $\exp_x$. Chapter 12 uses this.

The browser core matches Lemma S to $3\times10^{-12}$ relative on 2,000 random steps ($d$ = 2–4, $r$ up to 3).

05

H1 · Everything stays on the sheet PROVEN

For $x\in\mathbb{H}^d$, $L_x$ is a Lorentz matrix that maps the sheet to itself, and every corner, probe and new state lies on the sheet. On the exact track the following points have rational coordinates and satisfy $\langle y,y\rangle = -1$ exactly:

every corner built from a rational $x$, a rational $t = \tanh(r/2)$ and a Cayley rotation;
every snapped state $G(p)$ of a rational $p$ with $|p| < 1$.

Write $c = 1 + x_0$ and use $|\bar x|^2 = x_0^2 - 1 = (x_0-1)c$.

The columns of $L_x$ are orthonormal for the form. The first column $x$ has $\langle x,x\rangle = -1$. The first column against column $j \ge 1$ gives $-x_0\bar x_j + \bar x_j + \bar x_j|\bar x|^2/c = \bar x_j(1 - x_0 + x_0 - 1) = 0$. Columns $i, j \ge 1$ give $\delta_{ij} + \bar x_i\bar x_j\big(-1 + 2/c + |\bar x|^2/c^2\big) = \delta_{ij} + \bar x_i\bar x_j\big(-1 + (2 + x_0 - 1)/c\big) = \delta_{ij}$. So $L_x^{\top}JL_x = J$. Since $(L_x)_{00} = x_0 > 0$ and $L_x o = x$, it keeps the upper sheet.

Corners. $\langle(\cosh r, \sinh r\,Ru), (\cosh r, \sinh r\,Ru)\rangle = -\cosh^2 r + \sinh^2 r\,|Ru|^2 = -1$, because $R$ is orthogonal and $|u| = 1$. With $t$ rational, $\cosh r = (1+t^2)/(1-t^2)$ and $\sinh r = 2t/(1-t^2)$ are rational and satisfy $\cosh^2 - \sinh^2 = 1$ identically. The Cayley map $R = (I-A)(I+A)^{-1}$ of a rational skew $A$ is defined, since $A$ has imaginary eigenvalues so $I+A$ is invertible. It is rational. And it is orthogonal: $I-A$ and $I+A$ commute, so $R^{\top}R = (I-A)^{-1}(I+A)(I-A)(I+A)^{-1} = I$. The corner $L_x(\cdot)$ is then a product of rational matrices and vectors, and stays on the sheet by the first part.

Snapped states. For $q = |p|^2 < 1$, $G(p) = ((1+q), 2p)/(1-q)$ has $\langle G,G\rangle = (-(1+q)^2 + 4q)/(1-q)^2 = -1$ and $G_0 > 0$.

Float states. $x'$ is on the sheet by Lemma 0. In floating point, that holds up to rounding.

Checked live: exact $\langle y,y\rangle = -1$ for every corner and snapped state. Control: a corner nudged by $2^{-60}$ in one coordinate must fail.

06

H2 · One step moves each factor at most r PROVEN

In every float step, every moved factor satisfies $d(x_i, x_i') \le r$. This includes block moves in products, and holds after the horizon projection. On the exact track the snap adds at most its logged residual.

One factor, full corners. Lemma S gives $d(x,x') = \operatorname{artanh}(|m|\tanh r) \le r$, since $|m| \le 1$.

Block moves. For factor $i$, the corners seen from $x_i$ are at distances $\rho_k = r|v_{k,i}| \le r$, with Klein points $\tanh\rho_k\,\hat v_{k,i}$, each of norm at most $\tanh r$. The Klein point of the centroid is a convex combination of the corners’ Klein points:

$$\kappa(x') = \frac{\sum_k w_kV_{k,\cdot}}{\sum_j w_jV_{j,0}} = \sum_k \lambda_k\,\kappa(V_k),\qquad \lambda_k = \frac{w_kV_{k,0}}{\sum_j w_jV_{j,0}} \ge 0,\quad \sum_k\lambda_k = 1,$$

so its norm is at most $\tanh r$, and $d(x,x') = \operatorname{artanh}|\kappa| \le r$.

The horizon. Let $\Pi$ be the nearest-point projection onto the closed ball of radius 8 about $o$. The ball is convex, and $\mathbb{H}^d$ has non-positive curvature, so $\Pi$ is 1-Lipschitz. The step starts inside, so $\Pi x = x$, and $d(x, \Pi x') = d(\Pi x, \Pi x') \le d(x,x') \le r$. The radial map the code uses (keep the direction from $o$, cap the distance at 8) is this nearest-point projection, because geodesics through $o$ are the rays of the ball.

The snap moves a point by its residual $\eta$, which the code logs on track C. The triangle inequality then gives $d(x, \mathrm{snap}(x')) \le r + \eta$.

Checked live: $d(x,x') \le r$ on every moved factor of every particle, with the horizon on. The check allows for float rounding of the distance itself, about $10^{-13}x_0x'_0/\sinh r$. Control: a move $0.1\%$ longer than $r$ must fail the same check.

07

H3 · The centroid is the best compromise PROVEN

With the squared Lorentzian distance $d_L^2(y,z) = -2 - 2\langle y,z\rangle$, the new point $x'$ is the unique minimiser over $\mathbb{H}^d$ of $F(y) = \sum_k w_k\,d_L^2(y, V_k)$. The minimum value is $2\sqrt{-\langle s,s\rangle} - 2$. This is a known result (Law et al. 2019); here is the short proof.

$F(y) = -2 - 2\langle y, s\rangle$, so minimising $F$ means minimising $-\langle y,s\rangle$. Write $\sigma = \sqrt{-\langle s,s\rangle}$, so $s/\sigma = x'\in\mathbb{H}^d$. By Lemma 0, $-\langle y, x'\rangle \ge 1$ with equality only at $y = x'$. Hence $-\langle y,s\rangle \ge \sigma$, with equality only at $y = x'$, and $F(x') = -2 + 2\sigma$.

Checked live: $F(y) \ge F(x')$ for points $y$ near $x'$ and far from it. Control: a corner itself must score worse than the centroid.

08

H4 · Any chart gives the same step PROVEN

Let $g$ be an orientation-preserving isometry ($g \in SO^+(1,d)$), and let $f^g = f\circ g^{-1}$ be the objective seen through $g$. Put $Q = (gL_x)^{-1}L_{gx}$. Then $Q = \mathrm{diag}(1, \hat Q)$ with $\hat Q \in SO(d)$, and

$$\mathrm{step}(f^g,\ gx,\ \hat Q^{-1}R) = g\cdot\mathrm{step}(f,\ x,\ R)$$

for any weight rule that depends only on the costs. With a Haar-random $R$, the random step at $gx$ therefore has the same law as $g$ applied to the random step at $x$.

$Qo = (gL_x)^{-1}gx = L_x^{-1}x = o$. A Lorentz matrix fixing $o$ has the block form $\mathrm{diag}(1,\hat Q)$ with $\hat Q \in O(d)$. Its determinant is $\det g\cdot\det L_x^{-1}\cdot\det L_{gx} = 1$, because boosts lie in the identity component; so $\hat Q \in SO(d)$. The corners at $gx$ with rotation $\hat Q^{-1}R$ are

$$L_{gx}\begin{pmatrix}\cosh r\\ \sinh r\,\hat Q^{-1}Ru_k\end{pmatrix} = gL_xQ\begin{pmatrix}\cosh r\\ \sinh r\,\hat Q^{-1}Ru_k\end{pmatrix} = gL_x\begin{pmatrix}\cosh r\\ \sinh r\,Ru_k\end{pmatrix} = gV_k.$$

Their costs are $f^g(gV_k) = f(V_k)$: the same cost matrix, in the same column order, so the same weights. With several particles, each has its own $\hat Q_p$, and Sinkhorn sees the same matrix too. Then $\sum w_k gV_k = gs$, and $-\langle gs,gs\rangle = -\langle s,s\rangle$, so the new point is $gx'$. For the law: Haar measure on $SO(d)$ is invariant under left multiplication, so $\hat Q^{-1}R$ is Haar whenever $R$ is.

Where equivariance stops

The horizon is a ball about a fixed origin, and the snap uses a fixed grid. Neither is equivariant, so H4 is a statement about the step itself.
Cayley rotations are not Haar, so only the deterministic statement holds for them.

Checked live: the core’s step at $gx$, with the rotation carried by $g$, equals $g$ applied to its step at $x$, to $10^{-9}$ in relative coordinates. $\hat Q$ is checked to be a proper rotation. Control: the step at $gx$ with the unadjusted rotation $R$ must differ.

09

H5 · Flat space gives polystep back PROVEN

Fix the spatial point $\xi\in\mathbb{R}^d$, the rotation $R$, the radius $r$ and $\varepsilon$. Let $f$ be continuous in spatial coordinates. Run the step on the hyperboloid of curvature $-c$, $\mathbb{H}_c = \{y : \langle y,y\rangle = -1/c,\ y_0 > 0\}$, at the point with spatial part $\xi$. As $c\to0$, the spatial part of the new point tends to polystep’s step $\sum_k w_k(\xi + rRu_k)$, with polystep’s weights.

On $\mathbb{H}_c$, $x_0 = \sqrt{1/c + |\xi|^2}$, and the boost from $o_c = (1/\sqrt c, 0)$ to $x$ is

$$L^{(c)}_x = \begin{pmatrix}\sqrt c\,x_0 & \sqrt c\,\xi^{\top}\\ \sqrt c\,\xi & I + \dfrac{c\,\xi\xi^{\top}}{1+\sqrt c\,x_0}\end{pmatrix}.$$

The corner in direction $u = Ru_k$ at distance $r$ is $L^{(c)}_x(\cosh(\sqrt c\,r)/\sqrt c,\ b)$, with $b = \tfrac{\sinh(\sqrt c\,r)}{\sqrt c}\,u$. Its spatial part is

$$\bar V_k = \xi\cosh(\sqrt c\,r) + b + \frac{c\,\xi\,(\xi\cdot b)}{1+\sqrt c\,x_0}.$$

As $c\to0$: $\cosh(\sqrt c\,r)\to1$, $b\to ru$, and $\sqrt c\,x_0 = \sqrt{1 + c|\xi|^2}\to1$, so the last term tends to 0. Hence $\bar V_k \to \xi + rRu_k$, and $\sqrt c\,V_{k,0}\to1$. The costs $f(\bar V_k)$ converge by continuity, and the weights with them (softmax is continuous; so is Sinkhorn at fixed $\varepsilon$). The centroid’s spatial part is $\bar s/(\sqrt c\,\sigma)$, with $\sqrt c\,\sigma = \sqrt{(\sqrt c\,s_0)^2 - c|\bar s|^2} \to 1$, since $\sqrt c\,s_0 \to \sum w_k = 1$ and $c|\bar s|^2 \to 0$. So it tends to $\sum_k w_k(\xi + rRu_k)$.

Separately, the code’s curvature-0 geometry is polystep’s step: the corners are $x + rRu_k$ and the move is the weighted barycentre. The Python tests check this against the library itself, and the live check against the formula.

Checked live: the gap to polystep’s step falls as $c$ goes from $10^{-1}$ to $10^{-3}$ to $10^{-5}$, and is below $10^{-3}$ at the last. The core’s flat step equals the formula to $10^{-12}$. Control: at $c = 1$ the step must differ from polystep’s.

10

H6 · Small ε picks the best corner PROVEN

With softmax weights, let the least cost be attained only at corner $k^*$, with gap $\Delta > 0$ to the next. Then $\sum_{k\ne k^*}w_k \le (2d-1)e^{-\Delta/\varepsilon}$, and $x'\to V_{k^*}$ as $\varepsilon\to0$. With ties, $x'$ tends to the centroid of the tied corners with equal weights. The elitist incumbent (the best cost scored so far) never increases.

$w_k = e^{-(C_k - C_{\min})/\varepsilon}/Z$ with $Z \ge 1$, so each $k\ne k^*$ has $w_k \le e^{-\Delta/\varepsilon}$. The map $w\mapsto x' = s/\sqrt{-\langle s,s\rangle}$ is continuous on the simplex, because the denominator is at least 1 by Lemma 0. At $w = e_{k^*}$ it gives $V_{k^*}$. The tie case is the same argument at the uniform weights on the tied set. The incumbent is the minimum of all costs scored so far, and adding terms to a minimum cannot raise it.

Not claimed: the particle itself may move to a worse cost when $\varepsilon > 0$. With Sinkhorn, as $\varepsilon\to0$ the plan tends to an optimal transport plan, which need not send each row to its own cheapest corner.

Checked live: at $\varepsilon = 10^{-3}$ with a gap of 0.1, the centroid sits on the best corner to rounding, and incumbents never rise over 15 steps. Control: at $\varepsilon = 1$ the centroid must be off the corner.

11

H7 · The register’s holonomy is exact PROVEN

Let $x_0, \dots, x_{720}$ be one factor’s exact states at the frames of a register, rational points of the sheet. Let $\Delta_n = L_{x_n}L_{x_{n-1}}^{-1}$, the identity when the factor did not move. Every $\Delta_n$ is a rational Lorentz matrix. Put:

$P = \Delta_{720}\cdots\Delta_1$;
$B_m$, the ordered product over the $m$-th run of ten frames, and $\delta_m = B_m^{-1}$;
$Q_{10} = \delta_1\cdots\delta_{72}$;
$Q_8 = A_0\cdots A_9$, where $A_\ell$ is the ordered product of the $\delta_m$ with $\lfloor 10m/72\rfloor = \ell$;
$H = Q_8Q_{10}P$.

Then $P = L_{x_{720}}L_{x_0}^{-1}$, $Px_0 = x_{720}$, and $Q_{10} = Q_8 = H = P^{-1}$.

$L_x$ has rational entries for rational $x$, since $1 + x_0 \ne 0$. Its inverse $JL_x^{\top}J$ is rational too, so every $\Delta_n$ is a rational Lorentz matrix. The product telescopes to $L_{x_{720}}L_{x_0}^{-1}$, and applying it to $x_0 = L_{x_0}o$ gives $L_{x_{720}}o = x_{720}$. By associativity $P = B_{72}\cdots B_1$, so $Q_{10} = B_1^{-1}\cdots B_{72}^{-1} = (B_{72}\cdots B_1)^{-1} = P^{-1}$. $Q_8$ is the same word $\delta_1\cdots\delta_{72}$, bracketed into ten groups, so it equals $Q_{10}$. Finally $H = P^{-1}P^{-1}P = P^{-1}$.

What the check actually certifies

The identities $Q_{10} = Q_8 = H = P^{-1}$ are group algebra; they hold for any invertible increments. The checks have teeth for two reasons:

The inverse. The code inverts with $JL^{\top}J$, which is the true inverse only for Lorentz matrices. A corrupted increment then breaks the identities.
The end point. $Px_0 = x_{720}$ holds only if the recorded increments compose to the motion actually made. Swapping two increments breaks it.

So H7 certifies the exact bookkeeping of every trajectory. It says nothing about the quality of the answer.

Checked live: random exact registers close, and every increment is Lorentz. Controls: two swapped increments, or a wrong end point, must fail.

12

Convergence (T1) OPEN

polystep’s paper proves convergence in its theory mode: a radius that decays like a power, $r_t = r_0(t+1)^{-(1/2+\gamma)}$ (square-summable but not summable), a fixed $\varepsilon$, and jittered probes. Does the theorem carry over to the hyperboloid? Lemma S reduces the question to a precise one, without settling it.

What carries over

Each step is polystep’s step in a chart. By Lemma S, step $t$ is: polystep’s step on $f_{x_t} = f\circ\exp_{x_t}$ at the origin of the tangent space, shortened radially by a factor $1 - O(r_t^2)$, then mapped by $\exp_{x_t}$. The square-summable radius makes the total shortening $\sum_t O(r_t^3)$ finite.
The iterates stay in a compact set. The horizon keeps them in the ball of radius 8. There the pulled-back objectives $f_x$ have derivatives bounded uniformly in $x$, if $f$ is smooth, because the curvature terms of $\exp_x$ are bounded on that ball.
The space is a Hadamard manifold. It has curvature $-1$ and is complete and simply connected. This is the setting of Riemannian stochastic approximation (for example Bonnabel 2013), where descent arguments with step sizes $\sum\gamma_t = \infty$, $\sum\gamma_t^2 < \infty$ are carried out in exactly this way.

What is missing

Locality and robustness. The transfer needs polystep’s proof to have two properties. It must be local: each step’s analysis uses only that step’s geometry. It must be robust to a radial shrinkage of $1 - O(r_t^2)$ and to an $x$-dependent objective. Neither has been checked line by line against the paper’s proof.
Constants. The constants of the pulled-back objectives depend on $x$ through the curvature. A bound uniform over the horizon ball is plausible but not written.

So T1 stays OPEN, and the reduction is recorded as proved (Lemma S). Below is evidence, not proof: theory mode run live on a smooth objective, $f(y) = d(y, y^*)^2$.

Theory mode, live

r₀ 0.80 γ 0.10 ε 0.10 steps

Each line is one seed: the closest particle’s distance to the optimum against the step, on log scales. Distances are computed as $2\operatorname{asinh}(\tfrac12\sqrt{\langle a-b,a-b\rangle})$, which stays accurate for close points, where $\operatorname{acosh}$ loses half its digits. Solid lines are the hyperbolic step on $\mathbb{H}^2$. Dashed lines are the flat step on $|z - z^*|^2$, with $z^*$ the optimum’s tangent coordinates. Both get 4 particles, a fixed $\varepsilon$, probe jitter 0.1 and the same power-decay radius. On this smooth bowl both steps converge quickly: the weighted barycentre of a symmetric orthoplex acts like a smoothed gradient step. Lines that keep falling are what the theorem would predict; they are not a proof that it holds.

13

Benchmark results

The preregistered benchmarks, run by the Python reference as registered and amended. Every cell is shown. Each decision is the mean paired difference with a 95% bootstrap interval (10,000 resamples, seed 7). A claim is supported when the interval lies above 0, refuted when it lies at or below 0, and inconclusive otherwise.

Loading reference/results/decisions.json…

Reading the results

7,740 measured runs, about 29 CPU-hours, all as registered and amended, plus the 750 reruns of hyper3 that Amendment 3 required. In short:

the hyperbolic step did not beat the same step at curvature 0 in any cell;
its steps do shrink in Klein coordinates as particles move out, in every cell where that was measured;
three channels did not beat one channel with three times the particles.

C1 is refuted: 0 cells supported, 12 refuted and 1 inconclusive (B4).
- B1 sits at a floor. Only 2 of B1’s 3,240 runs found a proper colouring, one for flat3 and one for hyper1, both at n = 60. So most B1 cells are refuted with the interval [0, 0]: no evidence of a benefit, rather than evidence of harm. The preregistered secondary metric, final conflicts, says the same more finely. hyper3 ends within 0.2–3.3 conflicts of flat3. It is significantly behind in 5 of the 9 cells, and no cell shows a difference in its favour.
- B3 graded. Curvature 0 finds the target in 73%, 100% and 100% of runs at depths 4, 6 and 8, against 24%, 60% and 41% for hyper3.
- B4. Test accuracy is 32.5% for hyper3 and 33.4% for flat3: a difference of −0.9 points, with interval [−2.2, +0.4]. Chance is 12.5%.
- B2 (descriptive, since C1 is not tested there). flat3 satisfies 2.3–2.7 points more of the clauses.
C2 is supported in all 12 of its cells. The mean Spearman correlation between a block’s Klein radius and its Klein step length is about −0.90 in every B1 cell, and −0.95, −0.92 and −0.88 in B2. A fixed hyperbolic radius gives ever shorter Klein steps as a particle commits, as conjectured.
C3 is refuted in all 9 B1 cells, 8 of them floors. On final conflicts, one channel of 12 particles and three channels of 4 are indistinguishable in 8 cells. In the ninth, one channel is ahead by 1.7.
The control passes. Under the membership objective no method beats the bound. Random search lands at about (N + 1)/2, as Prop 5.13 says it must. The particle methods almost never find the target: they get no signal, so their weights stay uniform and they barely move.

What Amendment 3 changed

A defect in the float membrane let hyper3’s states drift off the sheet near the horizon and turn to NaN (Amendment 3). With the defect:

hyper3 ended B1 at n = 300, c = 4.4 with 82.2 conflicts; corrected, it ends with 34.9;
C2 was refuted in B1’s n = 300 cells (Spearman +0.05 to +0.08); corrected, it is supported there at −0.90.

So C2 moved from mixed to supported, and C3 from 8 refuted and 1 inconclusive to 9 refuted. C1 stayed refuted, but its one inconclusive cell moved from B1 to B4. Both sets of decisions are on this page; the original files are in reference/results/pre_amendment3/.

What the runs suggest (exploratory, not preregistered)

The membrane does not help. In B1, corrected, hyper3 and hyper3_free (no membrane) end within about a conflict of each other. In B2 the membrane costs about 1–3 points of clauses satisfied: hyper3 95.6%, 93.5% and 91.1% against hyper3_free’s 98.5%, 96.3% and 92.3%. Pulling the channels together seems to cost more diversity than it gains.
B3 favours the flat tangent-plane step. Its straight steps in the anchor’s tangent plane sweep large hyperbolic arcs at the same radius, which suits a tree whose children sit at distance 3.
Tuning is a weak point of the design. Each setting was tried once, on its own tuning instance, so setting quality is confounded with instance difficulty. On B4 every method chose the same candidate: the one run on instance 9018.
On B2 at n = 2000, hyper3 (91.1%) beats random search (89.2%) and polystep (89.0%). polystep searches all 16,000 coordinates at once, so a single step spends most of the budget.

C1 · hyperbolic against curvature 0

Cell	Comparison	Mean difference	95% interval	Pairs	Decision	Before Amendment 3

C2 · shorter Klein steps further out

Cell	Measure	Mean Spearman	95% interval	Runs	Decision	Before Amendment 3

C3 · three channels against one with three times the particles

Cell	Comparison	Mean difference	95% interval	Pairs	Decision	Before Amendment 3

The B3 control (Prop 5.13)

Under the membership objective, no method can find a uniformly placed target in fewer than $(N+1)/2$ evaluations on average: the first $k$ queries hit it with probability at most $k/N$. So $\mathbb{E}[\min(T, N)] \ge (N+1)/2$, where $T$ is the evaluation that first scores the target and a run that never does counts as $N$. A method whose whole interval lies below the bound would be a bug, with information leaking from the target.

Cell	Method	Mean of min(T, N)	95% interval	(N + 1)/2	Runs	Result

Descriptive comparisons: secondary metrics, the membrane, polystep and the random floor

Cell	Comparison	Mean difference	95% interval	Pairs	Reading

Every cell and method: primary metric, success rate, certified cost

Cell	Method	Runs	Primary metric	Success	Mean certified	Median evals to success

The tuned settings (Amendment 2)

Benchmark	Method	ε	r	μ	Schedule

14

What this decides

Claim	Final status	Basis
H1–H7	PROVEN	Chapters 5–11, checked live above
Lemma S	PROVEN	Chapter 4: the step is polystep’s step in the particle’s frame, shortened
T1 · convergence	OPEN	Reduced by Lemma S; the transfer needs a check of polystep’s proof (chapter 12)
C1 · hyperbolic beats flat	REFUTED	0 supported, 12 refuted, 1 inconclusive (B4), of 13 cells; the secondary metrics agree. Before Amendment 3: refuted
C2 · built-in annealing	SUPPORTED	Supported in all 12 cells (B1, B2), Spearman −0.88 to −0.95. Before Amendment 3: mixed (9 supported, 3 refuted)
C3 · three channels beat one	REFUTED	Refuted in all 9 cells (8 floors); final conflicts show no difference in 8 cells and one channel ahead in 1. Before Amendment 3: refuted (8 refuted, 1 inconclusive)
L2 · the B3 control	PROVEN	Paper Prop 5.13; the implementation passes the check at every depth, for every method

The negative results are as informative as a positive one would have been. On these benchmarks, at these budgets and with this tuning, putting polystep’s step on the hyperboloid and running it as three channels on the zero-point clock did not help, although the step’s predicted geometry (C2) is borne out everywhere it was measured. The exact machinery around it worked as proved: certificates, memory chains and the pump. What the runs point to next would need a new, separately preregistered round: dropping or reworking the membrane, tuning that is not confounded with instance difficulty, and objectives where depth in the tree is what matters. A bug found after the runs (Amendment 3) was fixed, the affected runs redone, and both sets of results shown.

What none of it decides

P versus NP stays open. A benchmark cannot show that a solver of this kind runs in polynomial time on every instance (paper, Thm 5.11).
Other problems, sizes and budgets. The decisions are about these benchmarks, at these sizes and budgets, with this tuning.
Convergence (T1) is open; see chapter 12.