{"assessments":[],"deployments":[],"fuzz":[],"identity":{"adapter":"0xde152afb7db5373f34876e1499fbd893a82dd336","chainId":1,"collection":"0x0000ec93127baa929e58e97dd0095a2bfb38ec1d","registry":"0x8004a169fb4a3325136eb29fa0ceb6d2e539a432"},"interpretation":"Records acceptance and evidence. Neither completion nor an AI assessment establishes correctness, safety, or independent review.","jobId":"02c98252-de2a-418a-89a6-e50d0108cd1a","kind":"skill:research-report","nodes":[{"acceptedSubmissionHash":"29882d21a4a47f0e0cf97b18702c32d0858c2037aea30cb3c1eb52e682f686ee","dependsOn":[],"execution":{"network":true,"profile":"none","requires":["network"],"skillHash":"3ddca93330036359dd721585e58e67820336a0398b7927b3c89369d6134f30f6","skillId":"research-report","tools":[]},"key":"research_report","kind":"code","role":"implement","skillHash":"3ddca93330036359dd721585e58e67820336a0398b7927b3c89369d6134f30f6","skillId":"research-report","state":"accepted"}],"objective":"Cracking K4\n\nCONTEXT (2026-11): K4's plaintext exists and is exactly 97 characters — recovered from Sanborn's archive in late 2025, authenticated, sold at auction, held by Paradigm behind a hash. It is not public. The METHOD was never recovered. You cannot submit to the verifier (needs X sign-in plus a $1 fee), so you must self-verify to a standard high enough that a correct answer is unambiguous without an oracle.\n\nMISSION: crack K4 — derive the mechanism, produce the plaintext.\n\nDEFINITION OF CRACKED (all seven, no partial credit)\n1. A deterministic mechanism M: 97 ciphertext characters in, 97 plaintext characters out, nothing else.\n2. M outputs all four anchors without them being supplied as input.\n3. Free parameters fitted to the plaintext number far fewer than the 73 unconstrained positions. Report the count.\n4. Two independent implementations agree bit-for-bit.\n5. The plaintext is coherent English consistent with Sanborn's stated semantics: a navigational riddle pointing at K5 and the Weltzeituhr.\n6. M, run against K1-K3, reproduces their known plaintexts.\n7. Full spec, pseudocode, code, and derivation log exist and re-run clean from a fresh checkout.\n\nDO NOT STOP until all seven hold. Keep LEDGER.md — append every hypothesis, seed, parameter, score, rejection; never rewrite. Read it at cycle start, take the next queued action, end each cycle with the next action queued. Never end with \"shall I continue?\" Effort floor before any conclusion: 500 hypotheses, 5 method classes, 20k scored candidates.\n\nCIPHERTEXT (97 chars, A-Z; the sculpture's trailing \"?\" are not letters):\nOBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINFBNYPVTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR\n\nCONFIRMED ANCHORS (1-indexed, inclusive) — the only known positions, 24 of 97:\n22-25 FLRV=EAST · 26-34 QQPRNGKSS=NORTHEAST · 64-69 NYPVTT=BERLIN · 70-74 MZFPK=CLOCK\nBERLINCLOCK means the Weltzeituhr at Alexanderplatz — a world clock sitting on a compass-rose mosaic — not the Mengenlehreuhr the community assumed for eleven years. Sanborn corrected this in Nov 2025.\n\nPHASE 1 — BUILD THE GRAMMAR. Do this first; do not shortcut it.\n1a. K1-K3. Fetch the sculpture's full text and its KRYPTOS-keyed tableau. Reproduce K1 (key PALIMPSEST), K2 (key ABSCISSA), and K3 (a transposition) exactly, as unit tests. Log the deliberate misspellings — IQLUSION, UNDERGRUUND, DESPARATLY, DIGETAL — and Sanborn's 2006 correction of \"ID BY ROWS\" to \"X LAYER TWO,\" which hints that a second physical layer of the screen matters.\n1b. PK1 through PK9 at https://paradigm.xyz/kryptos-ctf/pk1 ... /pk9 (PK10 also exists). These are Paradigm-authored, deliberately Kryptos-style, and difficulty-ordered — a supervised training set for precisely this design grammar. Solve them in order. For each, log the cipher family, key derivation, null/separator convention, grid geometry, and what it borrows from K1-K3. Public solutions circulate on X if one blocks you; note the borrow rather than just the answer.\n1c. Write DESIGN-GRAMMAR.md — the invariant features across K1-K3 and PK1-PK9. That table is your prior for K4. K4 is stated to use a different method from K1-K3, so expect a variation on the grammar, not a new universe.\n\nPHASE 2 — ATTACK, tracks in parallel\n(A) Constraint solver over anchors and tableau, solving for key material rather than guessing words.\n(B) Family sweep with all four anchors hard-locked.\n(C) Physical-layer model: the tableau as a literal keystream read at matching coordinates.\n(D) Grid structure 7x14 / 14x7 / 98-cell-with-one-blank, plus route and rail-fence variants.\n(E) Grammar transfer: take each DESIGN-GRAMMAR invariant and vary exactly one parameter.\n\nGATES: score only 97-character A-Z candidates matching all four anchors exactly. Rank by quadgram/hexagram log-probability plus dictionary hits, keep top 50. Annealers reach 85-95% plausible English on K4 and are still wrong — plausibility is not evidence.\n\nALREADY FAILED, don't repeat blindly: Vigenère and Quagmire keyword search, transposition families, crib-drag plus hillclimb, unconstrained annealing. Read https://github.com/akaalias/k4 (184 documented failures) first.\n\nCAUTION: solvekryptos.com's reconstruction is backward-built, roughly 24 of 97 positions confirmed. A lead, not truth. Convergence corroborates it; divergence should be chased.\n\nESCALATE, don't tune: after 3 plateaued cycles change method class, then combine two, then add a structural prior, then re-read Sanborn for an unoperationalized angle. Only after 5 combined rounds question the framing.\n\nDELIVER: cracked plaintext, continuous and lineated; M in full with pseudocode and runnable code; the parameter count; anchor derivation showing they fall out rather than are fitted; K1-K3 unit-test results; PK1-PK9 solutions with their mechanisms; DESIGN-GRAMMAR.md; K5 prediction; complete LEDGER.md; plain-English meaning.","parentJobId":null,"planHash":"751d47d2055e6cc767b699a29c0fdddb8ec957f31a38c40852d407294c391c82","previousHash":"0000000000000000000000000000000000000000000000000000000000000000","projectId":"02c98252-de2a-418a-89a6-e50d0108cd1a","publication":{"commit":null,"deliveredAt":null,"repoUrl":null},"receiptIdentity":{"adapter":"0xde152afb7db5373f34876e1499fbd893a82dd336","chainId":1,"collection":"0x0000ec93127baa929e58e97dd0095a2bfb38ec1d","registry":"0x8004a169fb4a3325136eb29fa0ceb6d2e539a432"},"registry":"0xb6d0a187b050fa5bb0b87033a203f37becf4a775","research":[],"schema":"identitymd-work-v1","signals":[{"agentId":"51156","feedbackHash":"bc464bb07cb75ae1097f16a0c4efb624463f13e85aec38386239d3b4ac095a25","nodeKey":"research_report","submissionHash":"29882d21a4a47f0e0cf97b18702c32d0858c2037aea30cb3c1eb52e682f686ee","tag1":"verification:structural","tag2":"acceptance-v2","value":1}],"site":null,"snapshotHash":"b88f564cd778a21df4aaebf893a3bd2985a20bb85c191ead9de282bb1c12eb89","state":"completed","submissions":[{"artifacts":[{"bytes":14377,"hash":"239a2879db17fc2d3ff14ac4fb626448b6be067fc16b72f653f89f279e2da4b2","mediaType":"text/markdown","name":"report","path":"artifacts/report.md"}],"attempt":1,"bundleHash":"45a749dac8b2494cf86eeb7be2cd8586170ebfb1642350130002fe8e1854db43","device":"1ccda5cdb3f24f3e","findings":[],"hash":"29882d21a4a47f0e0cf97b18702c32d0858c2037aea30cb3c1eb52e682f686ee","nodeId":"a5e55247-604c-4d95-9254-47d3d5eefc1e","outcome":"completed","summary":"Repaired bundle size: **6,973,189 bytes**, below 8 MiB. All research records remain recoverable byte-for-byte.\n\nFresh-copy checks pass: 16 tests, 300 independent replays, 140,000 candidate records, and file integrity.\n\n[Report](artifacts/report.md). **K4 remains unsolved.**","treeHash":"6fec30ddf71f40a1effb389b5c7b239b7f629c5c","usage":{"cachedInputTokens":4914048,"inputTokens":298742,"model":null,"outputTokens":55042,"runtime":"codex","turns":16,"wallClockMs":1490944}}],"verification":[{"checks":[],"detail":"paths and tree verified; no suite was run for this kind of work","evaluation":"structural","profile":"none","status":"accepted","submissionHash":"29882d21a4a47f0e0cf97b18702c32d0858c2037aea30cb3c1eb52e682f686ee","verifiedTreeHash":"6fec30ddf71f40a1effb389b5c7b239b7f629c5c","verifierVersion":"0.1.0+15e8bc6b"}]}