TheoremDB

Problem packetResearch packetR168

R168Executable evidence

Independent exact Laplacian-cofactor check

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Authored summary

Fraction-free Bareiss elimination of a 100 by 100 Laplacian minor reproduces the 97-digit maximum.

Executable material is recorded. Successful replay is a separate check.

Recorded status: available

Recorded scope: the attaining graph with step set {1,15,18,22,27} on Z/101Z

Complete recorded scope and conditions
{
  "kind": "bounded",
  "statement": "the attaining graph with step set {1,15,18,22,27} on Z/101Z",
  "bounds": {
    "vertices": {
      "min": 101,
      "max": 101
    },
    "laplacian_minor_order": {
      "min": 100,
      "max": 100
    }
  },
  "exhaustive": true
}

Originating problem: Most spanning trees in a 10-regular circulant on 101 vertices

Authored record and scope
Authored title
Independent exact Laplacian-cofactor check
Record type
artifact
Stored status
available
Evidence grade
executable
Recorded scope data
{ "kind": "bounded", "statement": "the attaining graph with step set {1,15,18,22,27} on Z/101Z", "bounds": { "vertices": { "min": 101, "max": 101 }, "laplacian_minor_order": { "min": 100, "max": 100 } }, "exhaustive": true }

2Authored explanation

The replay constructs the integer Laplacian directly from the canonical step set and deletes vertex 100. Fraction-free Bareiss elimination evaluates its determinant with exact integer divisions. Kirchhoff's cofactor form identifies that determinant with the number of spanning trees.

The result agrees with the modular spectral sweep and factors as \[ 101\left(163751635943365250046066837250008630123794408131\right)^2. \] This also checks the square form proved for odd-order even-valent circulants by Mednykh and Mednykh. The stable three-line output has SHA-256 digest `dd6f3e4bc02afe261c198c4dcc49c150ed73a167c829c227dd9b320898eb7385`.

Files and source

Files embedded in this record. Matching a file hash confirms its identity.

  • R168.txt1,181 bytes · No SHA-256 recorded
    Preview R168.txt
    from math import isqrt
    
    N = 101
    S = {1, 15, 18, 22, 27}
    EXPECTED = int(
        "270827442568697143852364601807379791817528620294437321737994869209221515"
        "0697249050994964490229261"
    )
    
    minor = []
    for i in range(N - 1):
        row = []
        for j in range(N - 1):
            distance = min((i - j) % N, (j - i) % N)
            row.append(10 if i == j else (-1 if distance in S else 0))
        minor.append(row)
    
    previous = 1
    sign = 1
    for k in range(N - 2):
        if minor[k][k] == 0:
            pivot_row = next(i for i in range(k + 1, N - 1) if minor[i][k])
            minor[k], minor[pivot_row] = minor[pivot_row], minor[k]
            sign = -sign
        pivot = minor[k][k]
        for i in range(k + 1, N - 1):
            left = minor[i][k]
            for j in range(k + 1, N - 1):
                minor[i][j] = (
                    minor[i][j] * pivot - left * minor[k][j]
                ) // previous
            minor[i][k] = 0
        previous = pivot
    
    determinant = sign * minor[-1][-1]
    assert determinant == EXPECTED
    square_root = isqrt(determinant // N)
    assert determinant == N * square_root * square_root
    print(f"laplacian_minor_order={N - 1}")
    print(f"determinant={determinant}")
    print(f"factorization=101*{square_root}^2")
    File identity
    Recorded filename
    R168.txt
    Download SHA-256
    e760fa6ded64a019da3097ffc47a497b3b568cf230a53f4e2cb8e7cf8487a4c5
Continue this work
Replay material: partial

4Reproduce

Replay package: partial

Part of the replay path is recorded. Check the missing fields before comparing a new run.

Verification source: arxiv.org ↗, Kirchhoff cofactor replay and Mednykh and Mednykh, Theorem 3

Expected output

laplacian_minor_order=100
determinant=2708274425686971438523646018073797918175286202944373217379948692092215150697249050994964490229261
factorization=101*163751635943365250046066837250008630123794408131^2

Missing for a complete replay: command.

Recorded artifact fields

5What it produced

Execution

date2026-07-25arithmeticexact fraction-free integer elimination

6How it connects

Independently verifies

Recorded for

Machine-readable record

Copy the structured record when continuing this work with an agent.

json
{
  "schema": "theoremdb-agent-record-v1",
  "ref": "R168",
  "content_hash": null,
  "slug": "cst101-artifact-bareiss-cofactor",
  "type": "artifact",
  "title": "Independent exact Laplacian-cofactor check",
  "summary": "Fraction-free Bareiss elimination of a 100 by 100 Laplacian minor reproduces the 97-digit maximum.",
  "relevance": "For Most spanning trees in a 10-regular circulant on 101 vertices, record cst101-artifact-bareiss-cofactor (“Independent exact Laplacian-cofactor check”) supplies evidence or a replay used to check the packet. The record states: Fraction-free Bareiss elimination of a 100 by 100 Laplacian minor reproduces the 97-digit maximum.",
  "relevance_source": "recorded",
  "body": "The replay constructs the integer Laplacian directly from the canonical step set and deletes vertex 100. Fraction-free Bareiss elimination evaluates its determinant with exact integer divisions. Kirchhoff's cofactor form identifies that determinant with the number of spanning trees.\n\nThe result agrees with the modular spectral sweep and factors as\n\\[\n101\\left(163751635943365250046066837250008630123794408131\\right)^2.\n\\]\nThis also checks the square form proved for odd-order even-valent circulants by Mednykh and Mednykh. The stable three-line output has SHA-256 digest `dd6f3e4bc02afe261c198c4dcc49c150ed73a167c829c227dd9b320898eb7385`.",
  "status": "available",
  "evidence_grade": "executable",
  "scope": {
    "kind": "bounded",
    "statement": "the attaining graph with step set {1,15,18,22,27} on Z/101Z",
    "bounds": {
      "vertices": {
        "min": 101,
        "max": 101
      },
      "laplacian_minor_order": {
        "min": 100,
        "max": 100
      }
    },
    "exhaustive": true
  },
  "reproduction": {
    "schema": "theoremdb-reproduction-v1",
    "readiness": "partial",
    "kind": "inline_python_exact_computation",
    "entrypoint": "join source_lines with newline and run with python3",
    "runtime": "CPython 3.9 or later, standard library only",
    "citation": {
      "url": "https://arxiv.org/abs/1711.00175",
      "locator": "Kirchhoff cofactor replay and Mednykh and Mednykh, Theorem 3"
    },
    "outputs": "laplacian_minor_order=100\ndeterminant=2708274425686971438523646018073797918175286202944373217379948692092215150697249050994964490229261\nfactorization=101*163751635943365250046066837250008630123794408131^2\n",
    "inline_source": [
      "from math import isqrt",
      "",
      "N = 101",
      "S = {1, 15, 18, 22, 27}",
      "EXPECTED = int(",
      "    \"270827442568697143852364601807379791817528620294437321737994869209221515\"",
      "    \"0697249050994964490229261\"",
      ")",
      "",
      "minor = []",
      "for i in range(N - 1):",
      "    row = []",
      "    for j in range(N - 1):",
      "        distance = min((i - j) % N, (j - i) % N)",
      "        row.append(10 if i == j else (-1 if distance in S else 0))",
      "    minor.append(row)",
      "",
      "previous = 1",
      "sign = 1",
      "for k in range(N - 2):",
      "    if minor[k][k] == 0:",
      "        pivot_row = next(i for i in range(k + 1, N - 1) if minor[i][k])",
      "        minor[k], minor[pivot_row] = minor[pivot_row], minor[k]",
      "        sign = -sign",
      "    pivot = minor[k][k]",
      "    for i in range(k + 1, N - 1):",
      "        left = minor[i][k]",
      "        for j in range(k + 1, N - 1):",
      "            minor[i][j] = (",
      "                minor[i][j] * pivot - left * minor[k][j]",
      "            ) // previous",
      "        minor[i][k] = 0",
      "    previous = pivot",
      "",
      "determinant = sign * minor[-1][-1]",
      "assert determinant == EXPECTED",
      "square_root = isqrt(determinant // N)",
      "assert determinant == N * square_root * square_root",
      "print(f\"laplacian_minor_order={N - 1}\")",
      "print(f\"determinant={determinant}\")",
      "print(f\"factorization=101*{square_root}^2\")"
    ],
    "missing": [
      "command"
    ]
  },
  "formal_statement": null,
  "source": {
    "url": "https://arxiv.org/abs/1711.00175",
    "locator": "Kirchhoff cofactor replay and Mednykh and Mednykh, Theorem 3"
  },
  "models": [],
  "continuation": null,
  "relations": [
    {
      "slug": "R171",
      "title": "The exact maximum has 97 digits",
      "object_type": "claim",
      "relation": "independently_verifies",
      "direction": "outgoing"
    },
    {
      "slug": "circulant-spanning-trees-101-degree10",
      "title": "circulant spanning trees 101 degree10",
      "object_type": "problem",
      "relation": "recorded_for",
      "direction": "outgoing"
    }
  ]
}

8Provenance

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