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Apr 13, 2017 at 12:19 history edited CommunityBot
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May 4, 2016 at 20:34 comment added mhsnk This is a nice analogy. Do you know the exact value of the minimum number of edges to be added to make this graph connected.
May 4, 2016 at 18:56 vote accept mhsnk
May 4, 2016 at 14:31 comment added Sam Zbarsky When k=2, this is equivalent to taking a graph on n+1 vertices and adding random edges until it is connected. Here the vector $e_i+e_j$ corresponds to edge $(i,j)$ and the vector $e_i$ corresponds to edge $(0,i)$
May 4, 2016 at 8:59 answer added Sam Zbarsky timeline score: 5
May 3, 2016 at 17:37 comment added mhsnk Thank you Robert! I have seen this post before and also the mentioned papers. My guess is that if $k$ is larger than the XORSAT satisfiability threshold, the average minimum required number of vectors will be equal to $n(1+o(1))$ but I don't know how to prove or disprove it?
May 3, 2016 at 17:21 comment added Robert Israel You might look at this related MO question of mine and its answer by Kevin Costello.
May 3, 2016 at 16:17 history asked mhsnk CC BY-SA 3.0