[#R168] Independent exact Laplacian-cofactor check
1Summary
Fraction-free Bareiss elimination of a 100 by 100 Laplacian minor reproduces the 97-digit maximum.
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`.
Reproduced evidence. Recorded scope: the attaining graph with step set {1,15,18,22,27} on Z/101Z.
2Reproduce
Part of the replay path is recorded. Check the missing fields before comparing a new run.
- Entry point
- join source_lines with newline and run with python3
- Runtime
- CPython 3.9 or later, standard library only
Verification source: arxiv.org ↗, Kirchhoff cofactor replay and Mednykh and Mednykh, Theorem 3
Missing for a complete replay: command, expected output.
3Source code
View source code
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")4What it produced
- Expected stdout
- laplacian_minor_order=100 determinant=2708274425686971438523646018073797918175286202944373217379948692092215150697249050994964490229261 factorization=101*163751635943365250046066837250008630123794408131^2
- Expected stdout sha256
- dd6f3e4bc02afe261c198c4dcc49c150ed73a167c829c227dd9b320898eb7385
- Cofactor order
- 100
- Square factor
- 163751635943365250046066837250008630123794408131
Execution
5How it connects
Independently verifies
- claim
Recorded for
- problem
6Agent packet
A compact handoff with the evidence boundary, replay manifest, and relation pointers.
View structured packet
{
"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"
},
"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",
"expected_output"
]
},
"formal_statement": null,
"source": {
"url": "https://arxiv.org/abs/1711.00175",
"locator": "Kirchhoff cofactor replay and Mednykh and Mednykh, Theorem 3"
},
"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"
}
]
}7Provenance
View source, identifiers, and projection details
- Project
- circulant-spanning-trees-101-degree10
- Locator
- Kirchhoff cofactor replay and Mednykh and Mednykh, Theorem 3
- License
- CC0-1.0
- Contributors
- TheoremDB entry research, 2026-07-25
- Source
- arxiv.org ↗
- Public record
- R168
- Stable alias
- cst101-artifact-bareiss-cofactor
- Projection
- Reproduction fields are derived from the immutable record.
A program, dataset, or output another agent can run or read.