Timeline for Cube-free infinite binary words
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 16, 2011 at 15:34 | vote | accept | JRN | ||
| May 27, 2012 at 15:15 | |||||
| Apr 14, 2011 at 1:56 | comment | added | JRN | Gerry asks a question related to his comment above at mathoverflow.net/questions/61615 | |
| Apr 12, 2011 at 13:15 | comment | added | JRN | @Tony: Wow. I didn't think of that. Thanks! | |
| Apr 12, 2011 at 13:00 | comment | added | Gerry Myerson | This construction gives a countable infinity of cube-free words. I would imagine there would be an uncountable infinity. | |
| Apr 12, 2011 at 11:22 | comment | added | Emil Jeřábek | @Tara: yes, so did I. But then it would be trivial that there are $2^\omega$ $xxx$-free words: anything matching the regular expression $(0?01)*$ will do. | |
| Apr 12, 2011 at 11:20 | comment | added | Tara Brough | So, I agree with you. If $w - j = w - k$ for some $j<k$, then $w = uv^*$ for some words $u$ and $v$. | |
| Apr 12, 2011 at 10:31 | comment | added | Tara Brough | Oh, sorry, I had misunderstood the definition of cube-free. I thought that $x$ was just a single symbol, not a word in $\{0,1\}^*$! | |
| Apr 12, 2011 at 10:27 | comment | added | Tony Huynh | Doesn't your $\omega$ contain the cube of $x=01$? | |
| Apr 12, 2011 at 10:23 | comment | added | Tara Brough | The word $w = 010101\ldots$ is cube-free, and $w - j = w - k$ if $j$ and $k$ are the same modulo $2$. | |
| Apr 12, 2011 at 10:16 | history | edited | Tony Huynh | CC BY-SA 3.0 |
added 4 characters in body
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| Apr 12, 2011 at 10:15 | history | undeleted | Tony Huynh | ||
| Apr 12, 2011 at 10:12 | history | deleted | Tony Huynh | ||
| Apr 12, 2011 at 10:10 | history | edited | Tony Huynh | CC BY-SA 3.0 |
added 54 characters in body; added 5 characters in body
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| Apr 12, 2011 at 10:05 | history | answered | Tony Huynh | CC BY-SA 3.0 |