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S Sep 27, 2013 at 3:23 history suggested Yili Dong CC BY-SA 3.0
I have changed my problem completely.
Sep 27, 2013 at 3:15 review Suggested edits
S Sep 27, 2013 at 3:23
Sep 11, 2013 at 3:37 comment added David Benson-Putnins Deterministically we can get at most $n^2$ steps on $n$ nodes by simply remembering all vertices/edges that we have seen (not visited, but seen via the 'knows all neighbors'), and at each step picking a random unvisited vertex and walking to it. We can walk to the kth one in at most k steps, so the total number of steps is $\sum_{k=1}^{n} k = n(n+1)/2$
Sep 10, 2013 at 17:32 history closed Suvrit
Tom Leinster
Yemon Choi
Chris Godsil
Dima Pasechnik
Needs details or clarity
S Sep 10, 2013 at 15:20 history suggested Wlodek Kuperberg CC BY-SA 3.0
several grammatical and notational correcions
Sep 10, 2013 at 15:11 answer added Chassaing timeline score: 2
Sep 10, 2013 at 14:53 comment added Wlodek Kuperberg What exactly is your question? A most efficient walk? An algorithm? Complexity?
Sep 10, 2013 at 14:49 review Suggested edits
S Sep 10, 2013 at 15:20
S Sep 10, 2013 at 14:31 review First posts
Sep 10, 2013 at 14:37
S Sep 10, 2013 at 14:31 review Close votes
Sep 10, 2013 at 17:37
Sep 10, 2013 at 14:15 history asked Dong Yili CC BY-SA 3.0