# Have characteristics of a 'doubly percolated' lattice been described

Has a system where conducting sites can percolate by hopping over/tunnelling through a non-conducting site been described? If so what are the characteristics, and where can I find more details (such as a paper on the subject)?

In the image below, if the edge of a black square touches the edge of another black it 'conducts' across. That could be described as singularly percolated.

I'm trying to describe a system whereby the 'conduction' can hop over a white square.

Does this have a name? Is it formally described in a paper anywhere?

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I don't know a reference, but perhaps you mean percolation on a percolation cluster? So that you take a large dimensional percolation cluster and try to find a separate nontrivial percolation problem on the cluster? In this case, I would imagine this is difficult, because the fractal dimension of the percolation cluster will usually be too low to support a second nontrivial percolation transition, but I might be wrong. This might be more interesting for the case of k-core percolation, where the percolating cluster has measure. – Ron Maimon Jan 3 '12 at 8:29
@RonMaimon I don't really follow anything you just said....I'm looking for a system that doesn't need to check for 'nearest neighbours' to percolate/conduct to, but can also percolate/conduct to the next nearest neighbour. – Pureferret Jan 3 '12 at 8:37
This might be a better fit for math.SE – Colin K Jan 9 '12 at 21:32
Ah ok, I'm trying to model a physical system. But I can see your point. Can it be migrated? – Pureferret Jan 9 '12 at 21:35
It can be migrated, but based on the topic I believe it's just fine here. Using math to study the behavior of a physical system, even if you're using a model (like a grid) is what physics is all about. Now, if you want to take it to math.SE because you haven't gotten an answer here, that's another story. In that case I'd suggest just making a new question on the Math site and link it to this one, mentioning that you didn't get any answers here. (Alternatively, you could try asking in their chat room to see if anyone there is able to answer this question.) – David Z Jan 10 '12 at 0:13