Timeline for binary code with constant hamming distance
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| May 20, 2015 at 7:23 | comment | added | kodlu | Actually, if you don't require constant Hamming distance, you can double the size of the Walsh-Hadamard code by allowing the binary complement of each codeword to be also a codeword. | |
| Sep 12, 2011 at 15:52 | comment | added | Emil Jeřábek | @Gil: Thanks for editing it in the answer. | |
| Sep 9, 2011 at 22:12 | history | edited | Gil Kalai | CC BY-SA 3.0 |
Added additional material from the comments.; deleted 1 characters in body
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| Sep 9, 2011 at 18:16 | comment | added | Emil Jeřábek | That’s a good point. Moreover, the conjecture holds for Jean-Michel’s $n=80$: one can take the tensor product of an order 4 Hadamard matrix as above with an order $2(9+1)$ Paley construction II as explained in en.wikipedia.org/wiki/Paley_construction . So the answer to the original question is 80. | |
| Sep 9, 2011 at 17:44 | comment | added | Gil Kalai | Indeed it is a famous conjecture that such codes exist whenever n =0 (mod 4). It is easy to see that this is necessary. See en.wikipedia.org/wiki/Hadamard_matrix | |
| Sep 9, 2011 at 15:53 | comment | added | Emil Jeřábek | I should also mention that the construction for $n=2^k$ is know as the Walsh–Hadamard code en.wikipedia.org/wiki/Walsh%E2%80%93Hadamard_code . | |
| Sep 9, 2011 at 15:51 | vote | accept | Jean-Michel | ||
| Sep 9, 2011 at 15:34 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |