3 ms·

I don't think mesh makes sense from an economics perspective. The number of edges scales O(n^2), where n is the number of people in the network. That's not very

I don't think mesh makes sense from an economics perspective. The number of edges scales O(n^2), where n is the number of people in the network. That's not very efficient. Even if you think wireless internet is feasible in dense urban areas (I think this is still an open question) it seems like a star topology connected by wireless backhaul to fiber makes more sense. This scales O(n).

teraflop 9y ago
> The number of edges scales O(n^2), where n is the number of people in the network. I don't see how this is true for any reasonable definition of "edge". It's certainly not the case that every node directly communicates with every other node; if they did, you wouldn't need a mesh.
- jeffreyrogers 9y ago
  A mesh network is defined as a network where the nodes connect to as many other nodes as possible. In the limit this is a complete graph with O(n^2) edges. It's true that in practice you won't have a complete graph but it's still less efficient than a more hierarchical, tree-like network structure.
  - mpfundstein 9y ago
    the fact that a wifi network node only covers a very limited area is sufficient to show that your theoretical limit is never going to be reached -> by a large margin. You can probably assume that in a dense network, every node will be reachable by every other node through at most 7 hops. But thats it [1] https://en.wikipedia.org/wiki/Small-world_network https://en.wikipedia.org/wiki/Small-world_network
  - MR4D 9y ago
    Am I missing something here? Let me approach your math with a reasonable argument: Assume a 1 sq mile area (1mi x 1mi). Assume each node has a 50 yard diameter reach. If you space a node every 50 yards, you will have 35 nodes on a side, for a total of 1,225 total nodes. Each node will only be able to physically connect to 3 other nodes on the edges, and 4 other for interior nodes. That means the node connectivity is only .00326 for an interior node. The longest path is diagonal across the square. Given the 35 modes on a side, and A^2 +B^2 = C^2, that means 35^2 = 35^2 = C^2. And c^2 = 2,450, meaning C = 50 (rounding up). So I’m not sure how O^2 even applies in this example. Yes, it’s theoretically a max, but theory >< the real world. Is this not correct?
nycmesh 9y ago
That's why we build a hybrid network with p2p and p2mp backhaul.

teraflop 9y ago

> The number of edges scales O(n^2), where n is the number of people in the network. I don't see how this is true for any reasonable definition of "edge". It's certainly not the case that every node directly communicates with every other node; if they did, you wouldn't need a mesh.

jeffreyrogers 9y ago
A mesh network is defined as a network where the nodes connect to as many other nodes as possible. In the limit this is a complete graph with O(n^2) edges. It's true that in practice you won't have a complete graph but it's still less efficient than a more hierarchical, tree-like network structure.
- mpfundstein 9y ago
  the fact that a wifi network node only covers a very limited area is sufficient to show that your theoretical limit is never going to be reached -> by a large margin. You can probably assume that in a dense network, every node will be reachable by every other node through at most 7 hops. But thats it [1] https://en.wikipedia.org/wiki/Small-world_network https://en.wikipedia.org/wiki/Small-world_network
- MR4D 9y ago
  Am I missing something here? Let me approach your math with a reasonable argument: Assume a 1 sq mile area (1mi x 1mi). Assume each node has a 50 yard diameter reach. If you space a node every 50 yards, you will have 35 nodes on a side, for a total of 1,225 total nodes. Each node will only be able to physically connect to 3 other nodes on the edges, and 4 other for interior nodes. That means the node connectivity is only .00326 for an interior node. The longest path is diagonal across the square. Given the 35 modes on a side, and A^2 +B^2 = C^2, that means 35^2 = 35^2 = C^2. And c^2 = 2,450, meaning C = 50 (rounding up). So I’m not sure how O^2 even applies in this example. Yes, it’s theoretically a max, but theory >< the real world. Is this not correct?

nycmesh 9y ago

That's why we build a hybrid network with p2p and p2mp backhaul.