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Independent orientations collapse planted labels
Theorem — proved
Fix arbitrary supports , each of size . Independently on every support choose uniformly exactly positive and negative signs, where . Let be the rational span of these signed rows. Set
Then
At , the right side is at most , which tends to zero. The same estimate with permits different support sizes and numbers of positive signs, provided every size is at least three and both signs occur.
Proof
Fix a subspace containing no coordinate difference. On one support, the -sets of positive positions whose signed row lies in cannot differ by swapping one element: subtracting their rows would put in . Their -subsets are therefore disjoint between different members, giving relative density at most . Counting -supersets instead gives density at most . Thus a fresh row belongs to with probability at most .
If the final span avoids coordinate differences, its rank is . Choose sampled rows forming a basis. Conditional on any specified candidate basis whose span has the required avoidance, the other rows all belong to that span with probability at most . Summing over candidate bases and ranks proves (1). For , bound each of the terms by .
Consequence for construction
Any realization satisfying all planted relations has equal labels at indices if . Thus independently orienting or more relations is typically incompatible with distinct labels, including for balanced orientations. A nonzero quotient alone does not guarantee a usable configuration.
This does not exclude rare conditioning or correlated orientations. Those are the remaining possibilities for this planting mechanism, and their full signed closure must still satisfy uniform extraction.