{"id":493,"job_id":1141,"problem_id":1,"lane_id":3,"type":"explore","user_id":36,"model":"gpt-5.6-sol","provider":"openai","report_md":"# Job1141: local interval graphs do not make the phase cover integral\n\nThis rejects a local-integrality shortcut on a sparse arithmetic support. It does not reject a route on complete short-window supports or prove a uniform gap bound. The combinatorial obstruction is known; the contribution here is an explicit realization with the literal twin-phase rule and every old-prime admissibility condition paid.\n\nI reused the current closed-routes register, questions and route list after finding all three byte-identical to1139. I read current routes13/18 and reports420/472. Those sources distinguish fractional cover, actual phase cover and the unresolved full-window arithmetic bridge. Their earlier counts and experiments were not rerun. I searched the changed perfect-graph/rainbow-matching question before choosing this falsifier; details and access limits are in prior-art1141.md.\n\nWrite B_q(s)={-s,-s-2} modulo an odd prime q>3. These are edges of the cycle on phase vertices with step2. The graph G_q joining slots whose B_q sets intersect is an edge-intersection graph. If q>|D|, some cycle edge is absent. Cut there and represent the remaining edges by unit intervals on a path, with repeated edges represented by identical intervals. This constructs an interval representation of G_q. In particular, all19 owner graphs on the frozen51-slot p97 support have this property, since their primes start at101. This local representation does not establish integrality after the separate prime factors are coupled by slot-cover inequalities.\n\nHere is the smallest check I chose, with p=7, the full band Q={11,13}, and only four selected old-admissible slots:\n\n| slot | s mod11 | s mod13 | s mod210 |\n|---|---:|---:|---:|\n| u00=5291 |0|0|41|\n| u01=12221 |0|1|41|\n| u10=21671 |1|0|41|\n| u11=28601 |1|1|41|\n\nThe displayed quotients give a direct check:5291=11*481=13*407;12221=11*1111=13*940+1;21671=11*1970+1=13*1667;28601=11*2600+1=13*2200+1. Each is41 modulo210, so each pair s,s+2 is coprime to2*3*5*7. These are not claimed to be prime pairs.\n\nFor either band prime, the represented residues are0 and1. Their killing phases are respectively{0,q-2} and{q-1,q-3}, which are disjoint for q=11,13. Every nonempty kill set therefore consists of exactly one complete row at11 or one complete column at13. All other phases kill nothing. Both owner graphs are2K2, with two disjoint cliques and an explicit interval representation.\n\nA fractional phase mixture putting probability1/2 on phase0 and1/2 on phase q-1 at each prime has normalization1 and covers every slot with total weight1. Actual phases select at most one row and one column. Their union contains at most3 of the4 slots, so no integral phase cover exists.\n\nThe fractional cover is forced: expected total coverage must be at least4, while each prime can contribute at most2. Thus neither may put mass on empty phases. If a is the probability of the first row and b of the first column, the four cover inequalities are a+b>=1, a>=b, b>=a and a+b<=1. They force a=b=1/2. There is no integral feasible point. For any nonnegative slot weights of total W, the maximum row weight and maximum column weight are each at least W/2, so their sum is at least W. Consequently no strict singleton weighted noncover certificate can detect this example. The counting slack is also exactly4-2-2=0.\n\nThe four two-slot kill masks form the known alternating bichromatic C4. Kelk-Stamoulis, [Integrality Gaps for Colorful Matchings, arXiv1801.07937v3](https://arxiv.org/pdf/1801.07937v3), sections2/4, studies its packing relaxation; Aharoni et al., [Large Rainbow Matchings, Example1.2](https://web.math.princeton.edu/~pds/papers/rainbowmatching/paper.pdf), gives the two perfect-matchings cycle obstruction. At this capacity-tight four-slot target a successful cover would have to be a disjoint rainbow matching, explaining the correspondence. I do not transfer their global approximation or hierarchy theorems.\n\nIn particular, the matching constraint sum(edge variables)<=1 on this bichromatic cycle is not a valid general phase-selection inequality: a row phase and an intersecting column phase may both be selected and cover3 slots. Packing forbids that overlap; phase cover permits it. Applying that cut directly to general phase variables would discard legitimate choices. A useful coupling rule still needs its own sound cover derivation, and may already be supplied by existing conditioning/coherence routes.\n\nThe support has diameter23310, is sparse, and omits many old-admissible starts in its enclosing interval. It does not satisfy route18's diameter+2