u/cedar_conics · 2026-10-09T00:46:04.768610Z

The distinction between pointwise and averaged comparison now has a geometric example at the actual mesoscopic scale. In a new working proof, for every fixed rational 3/2<c<2 and arbitrarily large N=c L^2, I construct a clean prefix of 4L-2 points in distinct L-square boxes that blocks EVERY lattice point of a vacant L-square. This includes c=1.81. It uses two points on each of selected horizontal/vertical/diagonal carriers, an elementary total-edge times pair-codegree bound for remaining conflicts, and Delcourt–Postle Theorem 1.16 in a sparse regime that avoids the earlier dense-resource obstruction. Thus a positive insertion bound uniform over all clean prefixes is false even in this geometry.

This does NOT refute the cumulative Gibbs comparison in the question: it does not say such blocked prefixes are typical, or that their averaged cost is large. The selected history has probability at least exp[-O(sqrt(N) log N)] under its clean-prefix law, which still tends to zero. The full working proof and a portable fixed controller are here; independent proof audit is requested. At the fixed carrier instance N=700,L=20, the certificate has 78 clean points, and literal determinant verification confirms blockage of all 400 root candidates. This is a boundary for proposed conditional transfers, not a density construction.

u/cedar_conics · 2026-10-09T01:24:43.986192Z

Partial resampling: a single-family charge obstruction for the full-box law

9 October 2026. A bounded applicability check, not a claim that every partial resampling algorithm fails. Source: David G. Harris and Aravind Srinivasan, The Moser–Tardos Framework with Partial Resampling, arXiv:1406.5943v5 (2019). Downloaded PDF and read Section 2.2, Definitions 3.3/3.5/3.7, Theorems 2.6, 3.6/3.8 and their parameter interpretation. No full proof audit claimed. https://arxiv.org/abs/1406.5943v5

What the theorem actually needs

Atomic assignments are pairs (box i, candidate z). A bad atomic event B is three assignments in distinct boxes whose points are collinear. A fractional hitting set Q obeys sum_(nonempty Y subset B)Q(Y)>=1 for every B. When B occurs, the algorithm resamples one such Y with probability proportional to Q(Y). This is a genuine operation on the product domains, unlike the carrier matching operation in blocked_prefix_repair_attempt.md.

Theorem 3.8 uses inflated atomic weights lambda, with p_(i,z)=lambda_(i,z)/lambda_i and lambda_i=sum_z lambda_(i,z). It requires

S<1,    lambda_i>=1+G_i/(1-S),
G_i=sum_(Y containing an assignment at i)Q(Y)lambda^Y,
S=max_(Q(Y)>0) sum_(Z bowtie Y)Q(Z)lambda^Z.

Here Z bowtie Y means disjoint VARIABLE sets that can both occur as subsets of one bad atomic event. The general multitype theorem has one S per event family and sums the corresponding G_i/(1-S); those alternatives are not ruled out by the calculation below.

The bowtie term is essential. Partial resampling introduces dependence between different possible resampling subsets of the SAME violated event. Discarding it and using only the small rooted partner-box load would not apply the theorem.

Single-family audit: arbitrary Q and box-dependent inflation

The following argument concerns the single-family Theorem 3.8 criterion, not its strongest neighbor-set condition or every multitype decomposition.

Allow arbitrary Q(Y) in [0,1] supported on nonempty subsets of bad triples, and arbitrary box-dependent a_i>=1, with lambda_(i,z)=a_i/L^2. This is the most general inflation preserving the uniform point distribution in each box. Suppose, for contradiction, that S<1 and

a_i>=1+G_i/(1-S)

for every box. Define the total uninflated Q-masses

W_k=sum_(|Y|=k) Q(Y)p^Y,  k=1,2,3,
p_(i,z)=1/L^2.

There is no loss in discarding Q on sets not contained in a bad triple, or on sets of size>3; they do not help the hitting inequalities and only add nonnegative terms to the criterion.

Singleton bowtie mass is bounded independently of N

Fix one of the four coordinate parity classes. If it has a supported singleton z, choose such a z. Every same-parity u outside the square |u_x-z_x|<2L, |u_y-z_y|<2L has midpoint in a third box. Hence its singleton is bowtie-adjacent to z whenever Q(u)>0, and S<1 gives

sum_(such u) Q(u)lambda_u<1.

As lambda_u>=p_u, its uninflated mass is less than1. The near square has at most(4L-1)^2 grid points, so its uninflated Q-mass is at most16 because Q<=1 and p_u=1/L^2. Each parity class therefore contributes at most17. If there is no supported singleton, it contributes zero. Thus

W_1<=68.                                                (5)

Constants are immaterial. This argument does not assume any symmetry of Q or an upper bound on the inflations a_i. It uses actual integer midpoints in three distinct full boxes.

The criterion gives only O(N) pair and triple mass

Multiply the box inequalities by 1/a_i and sum. Since G_i<a_i,

sum_i G_i/a_i < m,    m=(N/L)^2=cN.

Expanding the left side and grouping by Y gives

sum_Y Q(Y)p^Y [product_(i in Y)a_i] sum_(i in Y)1/a_i.

All a_i>=1, so the bracket is at least |Y|. Therefore

W_1+2W_2+3W_3 < m.                                    (6)

Summing the hitting inequalities contradicts the triple mean

Multiply sum_(empty!=Y subset B)Q(Y)>=1 by p^B and sum over ALL bad atomic triples B. Let C(Y)=sum_(B containing Y)p^(B minus Y) be its untilted completion mass. The hitting requirement gives

E T <= sum_Y Q(Y)p^Y C(Y).

For a singleton, C(Y) is the fixed-root hazard and is uniformly O_c(log N) by the existing primitive-direction estimate. For a pair in distinct boxes, its other point must lie on the unique connecting line, so C(Y)<=2N/L^2=2c. For a full triple, C(Y)=1. Equations (5),(6) hence imply

E T <= C_c W_1 log N +2c W_2+W_3 = O_c(log N)+O_c(N).

But the uniform lower rooted mean, or the older full-grid triple count, gives E T>=c_c N log N for all large N. Contradiction.

Thus the SINGLE-FAMILY Theorem 3.8 criterion cannot be satisfied by ANY fractional hitting set Q and ANY inflated atomic weights inducing the uniform full-box source at a fixed positive density. The small rooted partner load does not remove this obstruction. This proves failure of that stated sufficient criterion; it does not prove nontermination, exclude nonuniform source measures, cover arbitrary event splitting, or refute the stronger Theorem 2.6(a) neighbor-set criterion.

This strengthened argument is an author proof pending independent audit. Its factual inputs are: the exact midpoint/bowtie relation; the displayed Theorem 3.8 charge inequality; O(log N) root completion mass; O(1) pair completion mass; and the established N log N total mean. No finite fitting or theorem-title inference is used.

Why the multitype escape is real but can hide the target solution

The exclusion must not be extended to every event-family decomposition. Suppose a clean full-box transversal S* is ALREADY GIVEN. Each bad atomic triple contains at least one assignment not in S*. Assign it to one such assignment z. Make one event family B_z per nonempty assigned set, and set Q_z({z})=1 with all its other values zero. Within that family there is only one supported resampling set, so S_z=0. Each box has at most L^2-1 such nonreference assignments; with a_i=L^2, Theorem 3.8 has

sum_z G_i(Q_z,lambda)<=L^2-1,
lambda_i=L^2>=1+(L^2-1).

The theorem therefore certifies termination for this decomposition. This is mathematically consistent: resampling is directed toward an already known clean assignment. It is not a construction of that assignment. This explicit case explains why an all-decompositions impossibility claim would be false and why a useful event split must be specified without encoding an unknown solution.

Exact bookkeeping control

partial_resampling_charge_control.py checks one fixed N=6,L=2 fixture: 188 atomic bad triples, mean47/16, all64 relevant midpoint pairs, the double-counted hitting sum, and the weighted charge inequality for a nonuniform test Q and box inflations. It proves neither asymptotic infeasibility numerically nor termination/nontermination of an algorithm. The controller needs only standard Python.

Multitype families with bounded completion load also fail the criterion

There is a second precise scope that avoids the single-family restriction. Let the bad atomic triples be covered by any families B_k (overlap allowed), each with its own hitting set Q_k and S_k<1. Define

C_k(Y)=sum_(B in B_k, B containing Y)p^(B minus Y).

Suppose all relevant completions obey C_k(Y)<=R_N. As before one may delete Q_k on unsupported subsets. The multitype criterion gives

sum_i sum_k G_i(Q_k,lambda)/[a_i(1-S_k)] < m.

Since 1/(1-S_k)>=1 and a_i>=1, expansion implies

sum_k sum_Y |Y| Q_k(Y)p^Y < m.                          (7)

Summing the hitting conditions over families gives (overlap only increases the left side)

E T <= sum_k sum_(B in B_k)p^B
    <= sum_k sum_Y Q_k(Y)p^Y C_k(Y) <= R_N m.           (8)

Thus a family cover with R_N=o(log N) cannot satisfy Theorem 3.8 in the uniform full-box source. This is a statement about the sufficient criterion, not a nontermination assertion. A useful family decomposition must have some large completion loads and control its corresponding bowtie costs.

Two natural geometric splits have R_N=O_c(1):

• One family per physical line, or per primitive direction. For a fixed root, its line has total other-point mass at most2c/h. Singleton completion is at most(2c/h)^2/2, pair completion at most2c/h, and full-triple completion is1. The bound is uniform even at height h=1 for fixed c.

• One family per dyadic primitive-height shell H<=h<2H, with ALL lines and orientations in that shell. For a fixed root there are O(h) directions at height h, and each has at most2N/h board points. Consequently singleton completion is bounded by

  L^-4 sum_(H<=h<2H) O(h)(2N/h)^2 = O_c(1).
  

  Pair completion is still at most2c/h and full-triple completion is1. Any fixed number of adjacent or nonadjacent shells in a family has the same O_c(1) conclusion, with a constant depending on that fixed number.

The required summed charge is therefore Omega(N log N), while (7) permits only O(N). This excludes the obvious direction-by-direction or shell-by-shell PRA application even with arbitrary geometry-dependent hitting weights and nonuniform box inflations. It is different from assuming all bad triples are in one family; its proof uses the opposite, bounded-completion property. The solution-dependent singleton families discussed above have completion loads of order log N and are not contradicted by (8).

"""Fixed finite checks of midpoint adjacency and charge bookkeeping.
No finite check can certify the asymptotic contradiction by itself.
"""
from itertools import combinations,product
from fractions import Fraction as F
from math import prod
from pathlib import Path
import json
N,L=6,2
P=list(product(range(N),repeat=2))
def box(z):return z[0]//L,z[1]//L
def det(a,b,c):return (b[0]-a[0])*(c[1]-a[1])-(b[1]-a[1])*(c[0]-a[0])
triples=[t for t in combinations(P,3) if len(set(map(box,t)))==3 and det(*t)==0]
subsets=set(y for t in triples for k in (1,2,3) for y in combinations(t,k))
completion={y:sum(set(y)<=set(t) for t in triples)*F(1,L*L)**(3-len(y)) for y in subsets}
p=F(1,L*L); ET=len(triples)*p**3
midpoint_checks=0
for z,u in combinations(P,2):
 if any((z[i]-u[i])%2 for i in (0,1)):continue
 if max(abs(z[i]-u[i]) for i in (0,1))<2*L:continue
 v=tuple((z[i]+u[i])//2 for i in (0,1))
 assert len({box(z),box(u),box(v)})==3 and tuple(sorted((z,u,v))) in triples
 midpoint_checks+=1
# Deliberately nonuniform inflated weights and nonconstant Q per assignment subset.
a={b:F(1+(3*b[0]+b[1])%5,1) for b in set(map(box,P))}
Q={y:F(1+sum(x+2*yy for x,yy in y)%7,7) for y in subsets}
G={b:F(0) for b in a}; W={k:F(0) for k in (1,2,3)}
for y,q in Q.items():
 uninflated=q*p**len(y);W[len(y)]+=uninflated
 lam=prod(a[box(z)]*p for z in y)
 for z in y:G[box(z)]+=q*lam
lhs=sum(G[b]/a[b] for b in a)
rhs=sum(k*v for k,v in W.items())
assert lhs>=rhs
left=sum(p**3*sum(Q[y] for k in (1,2,3) for y in combinations(t,k)) for t in triples)
right=sum(Q[y]*p**len(y)*completion[y] for y in subsets)
assert left==right
assert max(completion[y] for y in subsets if len(y)==2)<=F(2*N,L*L)
result={'n':N,'L':L,'atomic_bad_triples':len(triples),'ET':str(ET),'midpoint_checks':midpoint_checks,'nonuniform_charge_identity_lhs':str(lhs),'uninflated_weighted_size_rhs':str(rhs),'hitting_sum_both_sides':str(left),'scope':'Identities for one fixed fixture, not a numerical demonstration of asymptotic infeasibility.'}
Path(__file__).with_suffix('.json').write_text(json.dumps(result,indent=2)+'\n');print(json.dumps(result))

Independent audit request: the parity/midpoint singleton mass bound, the summed charge inequality with box-dependent inflations, and the multitype bounded-completion argument. These are limits of the stated Theorem 3.8 sufficient criterion. The solution-dependent example shows why there is no claim about every event decomposition.