slug: krishnan-zero-forcing-generalized-petersen-three
bibkey: krishnan2026correction
doi: 10.48550/arXiv.2607.19412
url: https://arxiv.org/abs/2607.19412v1
triage: theorem
motivation_gids:
  - D5/S3/Combinatorics/GeneralizedPetersen/ZeroForcingThree.result

The Zero Forcing Number of P(n,3) Is Eight for Every n At Least Thirteen

Problem

Arnav Krishnan, A correction to the Zero Forcing Number of the Generalized Petersen Graphs P(n,3), arXiv:2607.19412v1, Conjecture 5:

Z(P(n,3)) = 8 for every n ≥ 13.

The paper proves the upper bound for n ≥ 9 (Proposition 4) and the lower bound for 13 ≤ n ≤ 20 by exhaustive search, and states that a lower-bound proof valid for all n is missing.

Motivation

The frozen theorem D5/S3/Combinatorics/GeneralizedPetersen/ZeroForcingThree.result proves the conjecture for every n ≥ 13. The lower-bound argument transfers a statement about forcing sets into a vertex-isoperimetric inequality for P(n,3), and the isoperimetric inequality is reduced from arbitrary n to finitely many column patterns by deleting empty columns.

Gap

Issue 9685 records the screen made before the work: the arXiv record has only v1, the searched formal-conjecture catalogues have no entry, and nothing under Problems/, D5/ or Library/, and no entry of the three screening records, concerns this paper.

Route

1. Upper bound: the eight consecutive outer vertices u_0, …, u_7 force P(n,3) for every n ≥ 9.
2. First forcers: for any finite graph, if a zero forcing set S performs at least p forces, the set X of its first p forcing vertices satisfies |N(X)| ≤ |S|.
3. Isoperimetry: for every n ≥ 14, every 10-vertex set X of P(n,3) has |N(X)| ≥ 8. Seven consecutive empty columns can be shortened by one column without changing |X| or |N(X)|, so it suffices to treat bounded gap patterns; these are exhausted by kernel-checked interval covers of the remaining layer codes.
4. Steps 2 and 3 exclude 7-vertex forcing sets for n ≥ 14.
5. n = 13: explicit forts, checked in the kernel, meet every 7-vertex candidate up to rotation.

Falsifier

The statement would fail if some P(n,3) with n ≥ 13 had a 7-vertex zero forcing set, equivalently if a 10-vertex set of P(n,3), n ≥ 14, had external boundary at most 7 and could be the first-forcer set of such a forcing process, or if a 7-vertex set of P(13,3) met no fort.

Evidence

SAT checks with CaDiCaL: “|X| = p and |N(X)| ≤ 7” is unsatisfiable for p = 10 and every 14 ≤ n ≤ 40 and satisfiable at n = 13; for p ≤ 9 it is satisfiable for every tested n, so ten first forcers are the least number that works.

Triage

theorem; Tier 1 open problem stated in a 2026 paper, preregistered in issue 9685 before the work. The finite parts (the n = 13 fort certificates and the interval covers of bounded gap patterns) are checker instances consumed by the general proof, recorded in the module headers.

ASSUMED-UNVERIFIED

The literature screen is bounded to the sources named above; no worldwide priority claim is made.