Talk:Graph pebbling

Does this mean...

..."after a series of pebbling moves, it is possible to move a pebble to any vertex in the graph."

Does this mean to make it so that each vertex has at least one pebble?

If not, then I don't understand the 2 nodes 1 edge example. If we start with 1 pebble on each node we can't move any pebble? --anon

The challenge is, more accurately, to reach every vertex. If we start with a pebble on a vertex, we have already reached that vertex. --AaronRosenberg 19:58, 27 April 2007 (UTC)

Insufficient Context Note

Why is this note still there? Is there really still insufficient context for this page? I'll grant that it needs to be linked more thoroughly, but the page itself is sound in my opinion. --AaronRosenberg 19:58, 27 April 2007 (UTC)

--Insufficient Context Note-- I agree, and after having made a few changes (hopefully for the better), I'm just going to count two opinions as a consensus and remove that tag.--Nathan Clement 01:09, 24 September 2007 (UTC)

incorrect formula

In the --γ(G) for families of graphs-- section, the formula is incorrect for wheel graphs of n vertices ($\scriptstyle K_{4}$ is isomorphic to $\scriptstyle W_{4}$, yet the covering numbers are different), although it might be correct for wheel graphs of n+1 vertices. I think the problem lies in the inconsistent numerical notation for wheel graphs. I don't have the knowledge to judge whether the formula is wrong just in the $\scriptstyle n=4$ case or in any general case. —Preceding unsigned comment added by 173.15.69.60 (talk) 17:58, 10 September 2009 (UTC)

This was fixed in https://en.wikipedia.org/w/index.php?title=Graph_pebbling&diff=prev&oldid=646524752. - Patrick (talk) 08:54, 20 December 2017 (UTC)

Needs a Picture

Any kind of visual reference would be nice. 2604:3D08:497F:F430:686E:35E:4941:322B (talk) 21:23, 7 November 2019 (UTC)