Timeline for On independent sets of graph
Current License: CC BY-SA 3.0
30 events
| when toggle format | what | by | license | comment | |
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| Jan 27, 2015 at 6:15 | comment | added | Włodzimierz Holsztyński | The last portion of the general part was wrong (the two preceding portions, and the small example, were fine). | |
| Jan 27, 2015 at 6:13 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
The last traces of the error removed.
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| Jan 27, 2015 at 6:02 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
An unfortunate Error removed.
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| Jan 27, 2015 at 2:49 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
mth typo (x2)
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| Jan 25, 2015 at 20:43 | comment | added | Włodzimierz Holsztyński | Perhaps. This and variations were on my mind, but I wanted to finish first step to answer properly on your question. Just to establish the $\ \LaTeX\ $ phrases was hard to me (I am slow). Now wider possibilities are open. | |
| Jan 25, 2015 at 20:37 | comment | added | Turbo | I think if you use $|A|=dk-1$, you can get better. | |
| Jan 25, 2015 at 20:33 | comment | added | Włodzimierz Holsztyński | No yet, but there should be more fun ahead of us :-) | |
| Jan 25, 2015 at 20:30 | comment | added | Turbo | So corollary $(N(G))^2 < M(G) < (N(G))^3\implies$ no example to $(2)$ exists? | |
| Jan 25, 2015 at 20:18 | vote | accept | Turbo | ||
| Jan 25, 2015 at 20:38 | |||||
| Jan 25, 2015 at 15:37 | comment | added | Włodzimierz Holsztyński | This general (:-) example is done; it's time for a more dramatic one, hey! | |
| Jan 25, 2015 at 15:31 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
Corollary
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| Jan 25, 2015 at 15:25 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
More complete (THEOREM)
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| Jan 25, 2015 at 15:14 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
smoother
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| Jan 25, 2015 at 14:57 | comment | added | Włodzimierz Holsztyński | I fixed $\ M(G)$--now it's perfect! (every k-subset is used). In particulat the small example got improved as the result. | |
| Jan 25, 2015 at 14:54 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
A typo in Remark 2.
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| Jan 25, 2015 at 14:36 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
REMARK 0 added
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| Jan 25, 2015 at 14:20 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
A TeX typo
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| Jan 25, 2015 at 14:09 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
PROOFS--start
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| Jan 25, 2015 at 12:49 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
Redoing the main part. The small example improved.
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| Jan 25, 2015 at 12:32 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
PROOFS
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| Jan 25, 2015 at 12:31 | comment | added | Turbo | I think your count for $M(G)$ is off. $M(G)$ is a collection of subsets with each subset disjoint having cardinality exactly $\alpha(G)$. | |
| S Jan 25, 2015 at 12:23 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
mth typo fixed
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| Jan 25, 2015 at 12:19 | review | Suggested edits | |||
| S Jan 25, 2015 at 12:23 | |||||
| Jan 25, 2015 at 12:15 | comment | added | Włodzimierz Holsztyński | Could you tell me what does oic mean? | |
| S Jan 25, 2015 at 12:14 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
an extension of the General Construction comments (the final theorem)
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| Jan 25, 2015 at 12:09 | review | Suggested edits | |||
| S Jan 25, 2015 at 12:14 | |||||
| S Jan 25, 2015 at 12:05 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
an extension of the General Construction comments
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| Jan 25, 2015 at 11:58 | review | Suggested edits | |||
| S Jan 25, 2015 at 12:05 | |||||
| Jan 25, 2015 at 11:45 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
A generalization
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| Jan 25, 2015 at 10:38 | history | answered | Włodzimierz Holsztyński | CC BY-SA 3.0 |