Timeline for A different criterion for equivalence of codes?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 18, 2012 at 22:45 | answer | added | Gerry Myerson | timeline score: 2 | |
| Jul 18, 2012 at 20:10 | comment | added | Jyrki Lahtonen | Here's the link dl.acm.org/citation.cfm?id=255876.255880 It is a bit embarassing that I didn't remember this right away. My firstborn graduate student worked on their decoding (among other things). | |
| Jul 18, 2012 at 8:49 | comment | added | Jyrki Lahtonen | By "simple" I mean that the number of zeros of a cyclic code was either two or three. The lengths of these are powers of two $\ge 64$ (IIRC) or one less. | |
| Jul 17, 2012 at 19:14 | answer | added | Dilip Sarwate | timeline score: 4 | |
| Jul 16, 2012 at 12:22 | answer | added | Brendan McKay | timeline score: 5 | |
| Jul 16, 2012 at 12:05 | comment | added | Jyrki Lahtonen | Yes. This is the case. There are known families of relatively simple extended cyclic codes that are known to be non-equivalent, but nevertheless have identical weight enumerators. I will look for a reference, when I am back in my office. | |
| Jul 16, 2012 at 10:15 | history | asked | Rob | CC BY-SA 3.0 |