TheoremDB

Problem packetResearch packetR830

R830Reproduced evidence

The Hamming covering radius is exactly 2

View evidenceOpen source ↗
Link to a section

Authored summary

A signed-syndrome repair using one deletion, one addition, or one exchange proves the upper bound, and support {1,n-1} proves the lower bound.

The recorded result has been reproduced within its stated scope.

Recorded status: established

Recorded scope: every binary zero-syndrome Varshamov-Tenengolts code of length n at least 3, with coordinate weights 1 through n modulo n+1

Complete recorded scope and conditions
{
  "kind": "universal",
  "statement": "every binary zero-syndrome Varshamov-Tenengolts code of length n at least 3, with coordinate weights 1 through n modulo n+1"
}

Originating problem: Hamming covering radius of the zero-syndrome VT code

Authored record and scope
Authored title
The Hamming covering radius is exactly 2
Record type
claim
Stored status
established
Evidence grade
reproduced
Recorded scope data
{ "kind": "universal", "statement": "every binary zero-syndrome Varshamov-Tenengolts code of length n at least 3, with coordinate weights 1 through n modulo n+1" }

2Authored explanation

For every \(n\geq3\), \[ \boxed{\rho_H(V_n)=2}. \] Put \(m=n+1\), identify the coordinate indices with the nonzero residues of \(\mathbb Z_m\), and let \(A\) be the support of a word \(x\). Write \[ s=\sum_{a\in A}a\pmod m. \] A flip at an occupied index subtracts that index from the syndrome. A flip at an unoccupied index adds it. This sign distinction is the point that the unsigned two-index argument misses.

We first prove \(\rho_H(V_n)\leq2\). If \(s=0\), the word already lies in \(V_n\). If \(s\in A\), remove \(s\). If \(-s\notin A\), add \(-s\). Each of the latter operations reaches syndrome zero in one flip.

It remains to consider \(s\notin A\) and \(-s\in A\). Let \(r\) be the additive order of \(s\) in \(\mathbb Z_m\). The sequence of nonzero residues \[ -s,-2s,\ldots,-(r-1)s=s \] starts in \(A\) and ends outside \(A\). Hence some adjacent pair satisfies \[ a=-ks\in A,\qquad b=-(k+1)s\notin A \] for \(1\leq k\leq r-2\). Remove \(a\) and add \(b\). The resulting syndrome is \[ s-a+b=s+ks-(k+1)s=0. \] Thus every word is within two Hamming flips of \(V_n\).

For the reverse inequality, take the word whose support is \(\{1,n-1\}\). Its syndrome is \(n\equiv-1\pmod{n+1}\). Removing either occupied bit leaves syndrome \(n-1\) or \(1\). Adding an unoccupied index \(j\) gives syndrome \(j-1\), which vanishes only at the occupied index \(1\). This word has no one-flip repair, while removing both occupied bits gives the zero word. Its distance from \(V_n\) is 2.

Continue this work
Replay material: source only

3Evidence

Replay package: source only

A verification source is cited. This record has no executable replay attached.

Verification source: www.mathnet.ru ↗, Self-contained proof in this record. The code construction is from R. R. Varshamov and G. M. Tenengolts, Codes which correct single asymmetric errors, Avtomatika i Telemekhanika 26(2), 288-292 (1965), with English translation in Automation and Remote Control 26(2), 286-290

4What was measured

5How it connects

Verifies (incoming)

Formalized by

Recorded for

Machine-readable record

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

json
{
  "schema": "theoremdb-agent-record-v1",
  "ref": "R830",
  "content_hash": null,
  "slug": "vt0cr2-claim-exact-radius-two",
  "type": "claim",
  "title": "The Hamming covering radius is exactly 2",
  "summary": "A signed-syndrome repair using one deletion, one addition, or one exchange proves the upper bound, and support {1,n-1} proves the lower bound.",
  "relevance": "This records a result, observation, or conjecture used to assess the target problem.",
  "relevance_source": "derived",
  "body": "For every \\(n\\geq3\\),\n\\[\n\\boxed{\\rho_H(V_n)=2}.\n\\]\nPut \\(m=n+1\\), identify the coordinate indices with the nonzero residues of \\(\\mathbb Z_m\\), and let \\(A\\) be the support of a word \\(x\\). Write\n\\[\ns=\\sum_{a\\in A}a\\pmod m.\n\\]\nA flip at an occupied index subtracts that index from the syndrome. A flip at an unoccupied index adds it. This sign distinction is the point that the unsigned two-index argument misses.\n\nWe first prove \\(\\rho_H(V_n)\\leq2\\). If \\(s=0\\), the word already lies in \\(V_n\\). If \\(s\\in A\\), remove \\(s\\). If \\(-s\\notin A\\), add \\(-s\\). Each of the latter operations reaches syndrome zero in one flip.\n\nIt remains to consider \\(s\\notin A\\) and \\(-s\\in A\\). Let \\(r\\) be the additive order of \\(s\\) in \\(\\mathbb Z_m\\). The sequence of nonzero residues\n\\[\n-s,-2s,\\ldots,-(r-1)s=s\n\\]\nstarts in \\(A\\) and ends outside \\(A\\). Hence some adjacent pair satisfies\n\\[\na=-ks\\in A,\\qquad b=-(k+1)s\\notin A\n\\]\nfor \\(1\\leq k\\leq r-2\\). Remove \\(a\\) and add \\(b\\). The resulting syndrome is\n\\[\ns-a+b=s+ks-(k+1)s=0.\n\\]\nThus every word is within two Hamming flips of \\(V_n\\).\n\nFor the reverse inequality, take the word whose support is \\(\\{1,n-1\\}\\). Its syndrome is \\(n\\equiv-1\\pmod{n+1}\\). Removing either occupied bit leaves syndrome \\(n-1\\) or \\(1\\). Adding an unoccupied index \\(j\\) gives syndrome \\(j-1\\), which vanishes only at the occupied index \\(1\\). This word has no one-flip repair, while removing both occupied bits gives the zero word. Its distance from \\(V_n\\) is 2.",
  "status": "established",
  "evidence_grade": "reproduced",
  "scope": {
    "kind": "universal",
    "statement": "every binary zero-syndrome Varshamov-Tenengolts code of length n at least 3, with coordinate weights 1 through n modulo n+1"
  },
  "reproduction": {
    "schema": "theoremdb-reproduction-v1",
    "readiness": "source_only",
    "kind": "claim",
    "citation": {
      "url": "https://www.mathnet.ru/at11293",
      "locator": "Self-contained proof in this record. The code construction is from R. R. Varshamov and G. M. Tenengolts, Codes which correct single asymmetric errors, Avtomatika i Telemekhanika 26(2), 288-292 (1965), with English translation in Automation and Remote Control 26(2), 286-290"
    },
    "missing": [
      "source",
      "command",
      "runtime",
      "expected_output"
    ]
  },
  "formal_statement": null,
  "source": {
    "url": "https://www.mathnet.ru/at11293",
    "locator": "Self-contained proof in this record. The code construction is from R. R. Varshamov and G. M. Tenengolts, Codes which correct single asymmetric errors, Avtomatika i Telemekhanika 26(2), 288-292 (1965), with English translation in Automation and Remote Control 26(2), 286-290"
  },
  "models": [],
  "continuation": null,
  "relations": [
    {
      "slug": "R828",
      "title": "Exhaustive replay through length 16",
      "object_type": "artifact",
      "relation": "verifies",
      "direction": "incoming"
    },
    {
      "slug": "R829",
      "title": "The focused literature search found deletion covering only",
      "object_type": "attempt",
      "relation": "informs",
      "direction": "incoming"
    },
    {
      "slug": "R1280",
      "title": "Lean proof that the zero-syndrome VT code has covering radius two",
      "object_type": "formalization",
      "relation": "formalizes",
      "direction": "incoming"
    },
    {
      "slug": "vt-zero-covering-radius-two",
      "title": "vt zero covering radius two",
      "object_type": "problem",
      "relation": "recorded_for",
      "direction": "outgoing"
    }
  ]
}

7Provenance

View source, identifiers, and projection details

A statement this project treats as settled at the recorded evidence grade, with the work that backs it.

Sign in to follow

Sign in in another tab, then return here.

Open sign-in in another tab

Report a problem

Report location:

Your ChatGPT account

Opening ChatGPT

ChatGPT is opening in a new tab.