[SHIN09-git]

Result

How much coordinate mixing is needed when permutations and nonzero scalings are free, and each elementary row addition costs one? This follows the monomial-header question in PR #6710, pinned at 465a0060fbd236a5e202619de4a45a694d44970a.

• The exact price of separating a full affine family's slopes is h − Z, with h nonzero vector charges and Z the maximum number of disjoint zero-sum blocks. An anchored scalar carrier makes this the full proper-minor price as well. For arbitrary binary normalized guard lists the exact full price has the same form after adjoining one dummy charge. Proof and witnesses
• For any ternary full flat, let S and L be its separate slope and scalar-imbalance prices. The true price lies between max(S,L) and 2S+L. Both endpoints are attained at every prescribed pair of nonnegative values S,L across dimensions. Even S+L can underestimate the cost by an arbitrarily large amount in a single pencil. Sharp price bounds
• Two five-element pencils with four active charges need two versus three row additions. All states at smaller cost fail a literal minor check under every final permutation. The variable-prime anchored decision problem is strongly NP-complete via 3-PARTITION. For fixed small alphabets we give constructive algorithms and scalar closed forms. Algorithms · Actual shortest examples
• Over the ternary field, five is the first dimension in which the larger separate bound can be strictly low. Exactly 1,920 labeled five-coordinate pencil tuples have that gap. Orbit classification

The classical monetary debt-partition principle and its complexity are credited to the earlier literature. Here the transfer graph comes from actual inverse dual row operations, with a legal attaining header and checks of the original principal minors. This counts an explicitly scoped header alphabet and does not price the physical caller or change the selected multiplication exponent. Source pins · Overlap and scope

Verification

python3.12 -B scripts/research_package.py --package research/monomial-guard-settlement

Eight registered commands passed on 03caa7f6cb5bd21c834024e9d81f6b43eb325b91. Normal and optimized certificates agree, and 20 private-copy faults are rejected in both modes. The raw searches cover 126,313 state visits plus 148,392 normalized pencil tuples. Three complete matrix graphs include every one of the 181,440 headers in GL3(F4). The separate ternary quotient search retains exact weights; it represents 31,361,929,920 labeled pencils at dimension eight rather than enumerating that many matrices. The official source-history and export checks passed (archive SHA256 83b6606321d8c1a4382525f6296325cc0764caddc94477dba8b4576105a76256). Validation and scope

The optional python3.12 -I -W error -B orbit_oracle.py --extended command from the package directory covers dimensions 9–12 and another 1,034,466 free orbits, with the first gap-three pencil in dimension 12. Optional fixtures

Prepared by SHIN09-git with substantial OpenAI Codex assistance. ToikY and Nick Petalas retain the earlier complete-header credit. The different computational routes share the same assisted authorship.

The read-only Git-free export passed all eight registered commands.

The Ubuntu source-history, executable-mode, full package and onboarding checks passed; the matrix selected this package alone.

Both hosted jobs passed and all eight hosted science receipts match local output. GitHub Actions

Nine higher-dimensional pencils give another 1,020 named subset checks against literal coordinate orders, including non-prime fields. Exact permutation masks

Independent five-coordinate raw-state recount

The published raw graph also visited all 1,568,160 labeled ternary pencil tuples at dimension five. It recovered the same full cost histogram and exactly 1,920 strict gaps, without grouping by free orbits. From the package directory, this short replay prints {0: 1566240, 1: 1920} up to dictionary order.

from collections import Counter
from flat_oracle import distances_to_passing_pencils
from settlement import Field

states, distance = distances_to_passing_pencils(Field(3), 5)
gaps = Counter()
for state in states:
    u, b, w = state[:5], state[5:10], state[10:]
    x = [a*c % 3 for a, c in zip(u, b)]
    y = [a*c % 3 for a, c in zip(u, w)]
    plus, minus = y.count(1), y.count(2)
    S = plus + minus - min(plus, minus) - abs(plus - minus)//3
    D = x.count(1) - x.count(2)
    L = min(abs(D-1), abs(D+2))//3
    gaps[distance[state] - max(S, L)] += 1
print(dict(gaps))