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Apr 22, 2010 at 4:17 comment added Hooked Perhaps more simply, two nodes are nearest neighbors on a unit lattice if the Cartesian distance between them is one (with the provision that node $n_i$ is not a nearest neighbor to nodes $n_{i-1}$ or $n_{i+1}$. So no, I don't think this makes it a graph in the usual sense.
Apr 22, 2010 at 2:48 comment added Leandro hi Hooked, I think I am not understanding what you are calling nearest neighbors of a random walk. Suppose that we have a non-intersect walk $\omega=\{v_0,\{v_0,v_1\},v_1,\{v_1,v_2\},\ldots,\{v_{n-1},v_n\},v_n\}$. We can think about it as a graph, I will call it $G$. My question is: what are you calling nearest neighbors pairs of $\omega$ ? Are they the set of pairs $\{v_i,v_j\}\notin E(G)$, satisfying $\|v_i-v_j\|_1=1$ ? ($E(G)$, as usual, is the edge set of $G$ and $\|(z_1,\ldots,z_d)\|_1=\sum_{j=1}^d|z_j|$
Apr 22, 2010 at 0:31 history edited Hooked CC BY-SA 2.5
Correction on example
Apr 21, 2010 at 23:45 answer added Douglas S. Stones timeline score: 1
Apr 21, 2010 at 20:43 history edited Hooked CC BY-SA 2.5
added example
Apr 21, 2010 at 20:01 comment added Yemon Choi For instance, can you give me a concrete example (with d=2, say) where two paths are supposed to have EXACTLY the same set of nearest neighbours?
Apr 21, 2010 at 19:22 comment added Yemon Choi What does it mean for two non-intersecting walks to have "the same set of nearest neighbors"?
Apr 21, 2010 at 15:35 history edited Hooked CC BY-SA 2.5
small grammer fix - added origin to description
Apr 21, 2010 at 15:23 history asked Hooked CC BY-SA 2.5