Problem packetResearch packetR830
The Hamming covering radius is exactly 2
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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.
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3Evidence
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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)
- artifact
Informed by
- attempt
Formalized by
- formalization
Recorded for
- problem
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"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",
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"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"
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"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"
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"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"
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"title": "The focused literature search found deletion covering only",
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{
"slug": "R1280",
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}7Provenance
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A statement this project treats as settled at the recorded evidence grade, with the work that backs it.