Abstract
We are given a connected mesh network of $n$ nodes, where each node has a unique name from the range of $[0,2^l)$. We assume that $n \leq 2^l$. Every node is directly connected to a few other nodes.
We present an iterative distributed algorithm: MIterF, that allows each node $x$ to maintain remote contact with a few key nodes. The remote contact with those key nodes allows efficient greedy routing of messages between any two nodes: At each step the message is sent to a remote contact with a name that is closest to the destination.
Given that the mesh network is connected, We prove that the presented iterative distributed algorithm MIterF always converges to a solution that allows efficient greedy routing.
Previous attempts
In Experimenting with Virtual DHT Routing we have presented an iterative distributed algorithm that allows each node $x$ to maintain contact with a few key nodes that are close to $x \pm 2^t$. We used experiments to show that this algorithm can work in some cases, but we didn't provide any rigorous proof for those claims.
In About cycles in stationary states of VDHT routing algorithms we presented a slightly different iterative distributed algorithm, IterF, that also allows each node $x$ to maintain contact to a few key nodes that are close to $x \pm 2^t$.
We have managed to prove that after IterF converges, we end up with possibly a few disjoint sets of nodes (OneRound Theorem). In each such set $S$, for every two nodes $a,b \in S$ it is possible to route messages efficiently between $a$ and $b$. However, we do not know how to efficiently route messages between nodes from different disjoint sets.
We present here a modification to the IterF algorithm, MIterF (Multi named IterF), which is proven to converge to just one set of nodes $S$, such that for every two nodes $a,b \in S$ it is possible to route messages efficiently between $a$ and $b$. This improvement comes at the cost of having every node maintain paths to more nodes.
Nodes that must share a cycle
Recall that the IterF algorithm eventually converges to a stationary state. Also recall that in a stationary state of the IterF algorithm the nodes are divided to disjoint cycles (of one round each). Our analysis didn't show which node will end up at which cycle.
We will now discuss a few cases of pairs of nodes who we know for sure will end up on the same cycle at a stationary state of IterF.
Lemma (SameCycleA): If $x$ is connected directly to a node $y$ of the form $y = \pm 2^k \pm 2^t$ where $0 \leq k,t < l$, then $x$ and $y$ must be on the same cycle on any stationary state of the IterF algorithm.
An example for a case of SameCycleA:
In this example, we know that the two nodes $x$ and $x + 2^3  2^5$ will end up on the same cycle.
Proof: For readability reasons we will only deal with the case of $y = x + 2^k + 2^t$. The other cases can be proved in the same way, but with inverted plus and minus symbols.
Assume a stationary state of the IterF algorithm. We first show that $\overleftarrow{(x + 2^k + 2^t){2^k}} = \overleftarrow{x{2^t}}$. Suppose that $\overleftarrow{(x + 2^k + 2^t){2^k}} \neq \overleftarrow{x{2^t}}$.
Then we can distinct between two cases:
Case 1: $\overleftarrow{(x + 2^k + 2^t){2^k}} \in (\overleftarrow{x{2^t}},x+2^t]$. In that case in the next iteration of the algorithm $x$ will update $\overleftarrow{x_{2^t}}$ to be $\overleftarrow{(x + 2^k + 2^t)_{2^k}}$, because it is closer to $x+2^t$ from the left (clockwise). This is a contradiction to the stationary state of IterF.
Case 2: $\overleftarrow{(x + 2^k + 2^t){2^k}} \in (x+2^t,\overleftarrow{x{2^t}})$. In that case in the next iteration of the algorithm $y = x + 2^k + 2^t$ will update $\overleftarrow{y_{2^k}}$ to be $\overleftarrow{x_{2^t}}$, because it is closer to $x+2^t$ from the left (clockwise). This is a contradiction to the stationary state of IterF.
Thus it must be true that $\overleftarrow{y_{2^k}} = \overleftarrow{x_{2^t}}$.
By the Locality lemma (see About cycles in stationary states of VDHT routing algorithms) there is a direct path between $x$ and $\overleftarrow{x_{2^t}}$, and so $x$ and $\overleftarrow{x_{2^t}}$ must be on the same cycle. In the same way, there is a direct path between $\overleftarrow{y_{2^k}}$ to $y$, therefore $\overleftarrow{y_{2^k}}$ and $y$ must be on the same cycle.
As $\overleftarrow{y_{2^k}} = \overleftarrow{x_{2^t}}$, we conclude that $x$ and $y$ must be on the same cycle. QED
Lemma (SameCycleB): If $x$ is connected directly to a node $y$, which is connected directly to a node $z$ of the form $z = x \pm 2^k$, where $0 \leq k < l$, then $x$ and $z$ must be on the same cycle on any stationary state of the IterF algorithm.
An example for a case of SameCycleB:
In this example, we know that the two nodes $x$ and $x + 2^7$ will end up on the same cycle. (The value of $z$ does not matter).
Proof: For readability reasons we will only deal with the case of $z = x + 2^k$. The other case could be proved in a similar way.
Assume a stationary state of the IterF algorithm. We first show that $\overleftarrow{x_{2^k}} = x + 2^k$. Suppose that $\overleftarrow{x_{2^k}} \neq x + 2^k$. Then in the next iteration $y$ will tell $x$ about $z = x + 2^k$, and $x$ will update $\overleftarrow{x_{2^k}}$ to be $x+2^k$, because it is closer to $x+2^k$ from the left (It has distance 0!). This is a contradiction to the stationary state of IterF. Therefore we conclude that $\overleftarrow{x_{2^k}} = x + 2^k = z$.
By the Locality lemma there is a direct path between $x$ and $\overleftarrow{x_{2^k}} = z$. Therefore there is a direct path between $x$ and $z$, and so $x$ and $z$ must be on the same cycle. QED.
The two lemmas we have proved above tell us something about which nodes should end up on the same cycle, but they are far from complete in describing the structure of the cycles of a stationary state of IterF. For Example, currently we don't know to say much about which nodes will not be on the same cycle.
On the applicability of SameCycleA and SameCycleB
Given the two lemmas above (SameCycleA and SameCycleB), one might wonder how often it is possible to apply those lemmas and conclude that all nodes will end up on the same cycle.
We will perform a very hand wavy calculation to get a feeling for this likelihood.
For a specific node $x$ there are at most $(2l)^2$ possible combinations of $x \pm 2^k \pm 2^t$ for $0 \leq k,t < l$, and at most $2l$ possible combinations of $x \pm 2^k$.
We assume that every node in the network is directly conected to at most $\log(n)$ other nodes.
For the nodes directly connected to $x$: Assume that those $\log(n)$ nodes were chosen randomly. About $\log(n) \cdot \frac{(2l)^2}{n}$ out of these nodes will be of the form $x \pm 2^k \pm 2^t$.
$$\log(n) \cdot \frac{(2l)^2}{n} = \frac{\log(n)\cdot (2l)^2}{n} \leq \frac{l\cdot (2l)^2}{n} = \frac{4l^3}{n}$$
Hence for $n > 4l^3$ we expect that with high probability some node $z$ will not have any direct connections on the network of the form $z \pm 2^k \pm 2^t$.
For nodes that are two hops from $x$: There are about ${\log(n)}^2$ such nodes. Assume that they were chosen randomly. About ${\log(n)}^2 \cdot \frac{2l}{n}$ out of these nodes will be of the form $x \pm 2^k$.
$${\log(n)}^2 \cdot \frac{2l}{n} = \frac{{2l}\cdot{\log(n)}^2}{n} \leq \frac{2l\cdot l^2}{n} = \frac{2l^3}{n}$$
Hence for $n > 2l^3$ we expect that with high probability some node $z$ will not have any node two hops from him of the form $z \pm 2^k$.
As a final rough conclusion, we believe that for large enough values of $n$, for example $n \gg 4l^3$, we expect that for some nodes non of the lemmas SameCycleA or SameCycleB will be applicable.
We might need something else to be able to prove the convergence of IterF to a single cycle.
Multi Naming
For every two nodes $x$ and $y$, we say that $x$ and $y$ are FriendsA if $y$ is of the form $y = x \pm 2^k \pm 2^t$. (A special case would be $y = x \pm 2^k$). Note that this property is symmetric (If $x$ and $y$ are FriendsA, then $y$ and $x$ are FriendsA). We will denote that $x$ and $y$ are FriendsA by $x \sim y$.
Consider some two directly connected nodes $x,y$ on a mesh network. If $x \sim y$ then by the SameCycleA lemma we can show that $x$ and $y$ will end up on the same cycle. However, if $x \nsim y$, we don't have anything meaningful to say. Unfortunately, in most cases we will have $x \nsim y$.
Consider that case of $x \nsim y$. Imagine for a moment that there were some extra nodes between $x$ and $y$: $x = z_1,z_2,\dots,z_r = y$ such that $z_j \sim z_{j+1}$ for $1 \leq j < r$. In that case we conclude that $z_j, z_{j+1}$ are on the same cycle for $1 \leq j < r$, and therefore $x$ and $y$ must end up on the same cycle on every stationary state of the IterF algorithm.
In the picture: Before and after adding imaginary extra nodes between $x$ and $y$.
In the real network there are no such extra nodes, but nothing prevents $x$ and $y$ to pretend the existence of those nodes. $x$ will play the role of $z_1,z_2,\dots,z_{\left\lceil r/2 \right\rceil}$ in the network, and $y$ will play the role of $z_{\left\lfloor r/2 \right\rfloor},\dots,z_r$ in the network.
We want to know how large can $r$ be. We know that $0 \leq x,y < 2^l$, hence both $x$ and $y$ could be written using at most $l$ bits. We want to get from $x$ to $y$ using small changes of the form $\pm 2^k \pm 2^t$. Every small change introduces another imaginary extra node.
We start from $z_1 = x$. We can reach $y$ by changing at most two bits at a time. $x$ and $y$ will usually differ at about $l/2$ bits. We assume that worst case, where $x$ and $y$ differ at $l$ bits. Hence we will need at most $\left\lceil l/2 \right\rceil$ imaginary nodes. (At most two bits in every small change. We add the ceiling operator for the case of odd $l$). We conclude that $r \leq \left\lceil l/2 \right\rceil$.
Hence each of $x,y$ will play the role of at most $\left\lceil l/4 \right\rceil$ extra imaginary nodes.
We can add imaginary extra nodes between every two nodes $a,b$ in the network such that $a \nsim b$. A node $z$ that is directly connected to $q$ nodes will have to play the role of at most $q\cdot \left\lceil l/4 \right\rceil$ nodes. This means that every node will maintain remote paths to about $q\cdot(l/4)\cdot(4l) = ql^2$ nodes.
Note that originally, every node had to maintain remote paths to about $2l$ nodes, which is much less.
Adding the extra imaginary nodes should not make message routing less efficient, because routing messages between imaginary nodes does not cost significant time. The modification does require, however, that nodes will maintain more "fingers": More paths to remote nodes. The more direct connections a node has, the more remote connections he will have to maintain.
In the picture: Before and after adding imaginary extra nodes between all pairs of connected nodes. The imaginary nodes are shown with lighter color that matches the color of the real node who plays their role in the algorithm.
Practicality issues
Regarding Network performance: We require every node to maintain remote paths to about $ql^2$ nodes. Assuming that a node $x$ has $5$ direct connections, and we pick $l=128$, this results in $81920$ ($\approx 2^{15}$) which is a lot of connections. I think that this number is on the edge of what is practical these days with communication. I believe that there is a way to slightly improve the MIterF algorithm to acheive more realistic numbers.
Who gets to pick the names for the nodes? This is a major question, as name collisions (two nodes with the same name) could disrupt correct routing. Also, who gets to pick how the path of imaginary FriendsA nodes between two nodes $x,y$ will be constructed? Where will the names come from?
We solve the problem of name collisions by using a large namespace. We pick a large $l$, for example $l=128$. If names are chosen randomly, the probability of a name collision should be very small (Even between the paths of imaginary FriendsA nodes).
Security: Will this system survive a strong adversary? Not in its current form. We need to add some mechanisms to avoid the following possible attacks:

Eclipse attack: An attacker might be able to obtain many names that are close to a node $x$ in the namespace, and thus absorb all the messages that should arrive at $x$. Therefore $x$ will not be able to receive any messages. The question of who gets to pick the names is tightly related to preventing these kinds of attacks.

Denial of Service: A node might be able to send many messages and flood the network. A node might also decide to stop routing certain messages, and thus disrupt routing of messages in the network.
Further questions
We know that the MIterF always converges to a stationary state where all the nodes are in the same cycle (And the cycle has only one round). We still haven't answered the following questions:

How long does it take for MIterF to converge? I think that the answer is something like $O(c\cdot diam(G))$, where $c$ is some constant and $diam(G)$ is the diameter of the network (The longest distance between two nodes), but I don't know of a proof for this.

How long are the paths that we get from each node $x$ to his "fingers"? For example, how long would be the path from $x$ to $\overleftarrow{x_{2^t}}$, as calculated by MIterF? How does it compare to the length of the shortest path in the network between $x$ and $\overleftarrow{x_{2^t}}$? Currently I have no idea about how to estimate the length of paths calculated by MIterF. Note that this question might be related to the shape of the network.
Summary
We presented MIterF, an iterative distributed algorithm that builds a setting for efficient routing of messages between nodes. Assuming that the network is connected, we proved that MIterF always converges to stationary state that allows efficient routing between all pairs of nodes in the network.
During the execution of the algorithm every node $x$ that has direct connections to $q$ nodes must maintain remote paths to at most $ql^2$ nodes.