Bounding the locality of distributed routing algorithms

Abstract

We examine bounds on the locality of routing. A local routing algorithm makes a sequence of distributed forwarding decisions, each of which is made using only local information. Specifically, in addition to knowing the node for which a message is destined, an intermediate node might also know (1) its local neighbourhood (the subgraph corresponding to all network nodes within $k$ hops of itself, for some fixed $k$), (2) the node from which the message originated, and (3) the incoming port (which of its neighbours last forwarded the message). Our objective is to determine, as $k$ varies, which of these parameters are necessary and/or sufficient to permit local routing on a network modelled by a connected undirected graph. In particular, we establish tight bounds on $k$ for the feasibility of deterministic $k$-local routing for various combinations of these parameters, as well as corresponding bounds on dilation (the worst-case ratio of actual route length to shortest path length).

Appendix: Reducing dilation

We describe a refinement of Rule U2 that reduces the dilation of Algorithm 1 to at most 6. We refer to this new routing strategy as Algorithm 1B.

1.1 Modified routing algorithm

The new rule is defined formally as follows, replacing Rule U2 of Algorithm 1. Let $u$ denote the current node. Recall that Rule U2 applies if $u$ has two active components, $\text{ dist}(u, t) > k$, $u \ne s$, and either $s \not \in G_k^{\prime }(u)$ or $s$ is in an active component of $u$. Let $\text{ dist}^{\prime }(\cdot ,\cdot )$ denote graph distance in $G_k^{\prime }(u)$ ($\text{ dist}(\cdot ,\cdot )$ still denotes graph distance in $G$).

Case U2A. Suppose $s \not \in V(G_k^{\prime }(u))$ or $\text{ dist}^{\prime }(u,s) = k$. Rule U2 remains unchanged; forwarding decisions are illustrated in Fig. 11.

In the remaining cases, U2b–U2f, $u$ has two active components, $\text{ dist}^{\prime }(u,s) < k$, $u \ne s$, $\text{ dist}(u, t) > k$, and $s$ is in an active component of $u$.

Case U2B/C. Suppose the origin node $s$ is a constraint vertex in $G_k^{\prime }(u)$. Suppose furthermore that $s$ has neighbours $c$ and $d$ in $G_k^{\prime }(u)$ that are also constraint vertices in $G_k^{\prime }(u)$. Without loss of generality, assume $\text{ dist}^{\prime }(u,c) = \text{ dist}^{\prime }(u,s)+1$ and $\text{ dist}^{\prime }(u,d) = \text{ dist}^{\prime }(u,s)-1$. Note, if $\text{ dist}^{\prime }(u,s) = 1$ then $d=u$. If $\text{ rank}(c) > \text{ rank}(d)$ (Case U2b), then forwarding decisions are illustrated in Fig. 15b. Otherwise (Case U2c), forwarding decisions are illustrated in Fig. 15c. In Case U2c, $u$ can determine the imminent application of Rule S2 and applies this rule pre-emptively.

Fig. 15

Algorithm 1B, Cases U2b/c: The message is at node $u$. Light blue regions denote the portion of each active component of $u$ that extends to the boundary of $G_k(u)$. Passive components are not illustrated. Blue arrows denote forwarding decisions at $u$ as a function of the neighbour from which the message is received. Green and red arrows denote anticipated forwarding decisions at $s$ (Rule S2) as a function of the neighbour from which $s$ receives the message (color figure online)

Case U2D/E. Suppose the origin node $s$ is in a passive component of some constraint vertex $e$ in $G_k^{\prime }(u)$. Suppose furthermore that $e$ has neighbours $c$ and $d$ in $G_k^{\prime }(u)$ that are also constraint vertices in $G_k^{\prime }(u)$. Without loss of generality, assume $\text{ dist}^{\prime }(u,c) = \text{ dist}^{\prime }(u,e)+1$ and $\text{ dist}^{\prime }(u,d) = \text{ dist}^{\prime }(u,e)-1$. Note, if $\text{ dist}^{\prime }(u,e) = 1$ then $d=u$. If $\text{ rank}(c) > \text{ rank}(d)$ (Case U2d), then forwarding decisions are illustrated in Fig. 16d. Otherwise (Case U2e), forwarding decisions are illustrated in Fig. 16e. In Case U2e, $u$ can determine the imminent application of Rule US2 and applies this rule pre-emptively.

Fig. 16

Algorithm 1B, Cases U2d/e: The message is at node $u$. Light blue regions denote the portion of each active component of $u$ that extends to the boundary of $G_k(u)$. The light red region denotes the passive component of $e$ that contains $s$. Other passive components are not illustrated. Blue arrows denote forwarding decisions at $u$ as a function of the neighbour from which the message is received. Green and red arrows denote anticipated forwarding decisions at $e$ (Rule US2) as a function of the neighbour from which $e$ receives the message (color figure online)

Case U2F. Suppose none of Cases U2b to U2e apply. Rule U2 remains unchanged; forwarding decisions are illustrated in Fig. 11.

 

 

1.2 Correctness of algorithm 1B

We now prove the correctness of Algorithm 1B and derive a tight bound on the corresponding dilation.

Given any connected graph $G$ on $n$ nodes, any $k \ge n/4$, and any $\{s,t\} \subseteq V(G)$, let $\{\sigma _i\}$ and $\{\sigma ^{\prime }_i\}$ denote the sequences of edges in $E(G)$ visited by Algorithms 1 and 1B, respectively, on a route from $s$ to $t$. For any $x \ge 1$, let $\{a_i\}_x$ denote the first $j$ edges in the sequence $\{a_i\}$, where $j = \min (x, |\{a_i\}|)$.

Lemma 14

For any $x \ge 1$, $\{\sigma ^{\prime }_i\}_x$ is a subsequence of $\{\sigma _i\}$.

 

Proof

We use induction on $x$. When the message is first forwarded from the origin node $s$, both routing algorithms apply the identical Rules S1, S2, or S3. Therefore, $\sigma _1 = \sigma ^{\prime }_1$ and the claim holds when $x=1$. Choose any $y \ge 1$ and assume the claim holds for $x=y$. We show the claim holds for $x=y+1$.

Case 1. Suppose $\{\sigma ^{\prime }_i\}_y = \{\sigma ^{\prime }_i\}_{y+1}$ (i.e., $\{\sigma ^{\prime }_i\}$ contains at most $y$ elements). The claim holds for $x=y+1$ by the inductive hypothesis.

Case 2. Suppose $\{\sigma ^{\prime }_i\}_y \ne \{\sigma ^{\prime }_i\}_{y+1}$. Let $u$, $v$, and $b$ denote nodes in $V(G)$ such that $\sigma ^{\prime }_y = (v,u)$ and¹ $\sigma ^{\prime }_{y+1} = (u,b)$. That is, $u$ denotes the current node, having received the message from node $v$, and $b$ denotes the node to which Algorithm 1B forwards the message after $u$. By the inductive hypothesis, $(v,u) \in \{\sigma _i\}$. That is, Algorithm 1 forwards the message from $v$ to $u$. We show that Algorithm 1 eventually forwards the message from $u$ to $b$.

Consider the forwarding rule applied by each algorithm at node $u$.

Case 2A. Suppose both algorithms make identical forwarding decisions, sending the message from $u$ to $b$. Thus, $(u,b)$ must follow $(v,u)$ in both sequences and the claim holds for $x=y+1$.

The algorithms differ only in Rules U2c and U2e. Furthermore, the forwarding decisions made by each algorithm differ only if the message was received from node $b$ (i.e., $v=b$), where $b$ denotes the active neighbour of $u$ opposite $s$ (see Figs. 15c, 16e).

Case 2B. Suppose Algorithm 1B applies Rule U2c and $v=b$. Let $a$ denote the active neighbour of $s$ opposite $b$. Let node $c$ denote the active neighbour of $s$ opposite $a$. See Fig. 15c. Algorithm 1 applies Rule U2 and forwards the message from $u$ to $a$ whereas Algorithm 1B reverses the direction of the message, sending it from $u$ back to $b$. Since $c$ is a constraint vertex in $G^{\prime }_k(u)$, no node on the path $P$ from $b$ to $c$ can have an active edge outside $P$. In particular, Algorithm 1 can only forward the message toward $s$ or reverse its direction back toward $u$. If Algorithm 1 forwards the message all the way to $s$, then Rules S1 or S2 apply. Otherwise, Algorithm 1 applies Rule U1 before the message reaches $s$. In either case, Algorithm 1 sends the message back to $u$ and then to $b$.

Observe that no rule other than S1, S2, U1, or U2 can be applied by Algorithm 1 when forwarding the message along path $P$. In particular, because the message started at node $s$, $\text{ dist}(p,t) > k$ for all nodes $p \in V(P)$; otherwise, Case 1 of Algorithms 1 and 1B would have applied initially at node $s$, and Rule U2c could not have been applied by Algorithm 1B at node $u$.

Consequently, $(u,b)$ must appear after $(v,u)$ in sequence $\{ \sigma _i\}$. Therefore, by the inductive hypothesis, the claim holds for $x=y+1$.

Case 2C. Suppose Algorithm 1B applies Rule U2e and $v=b$. The claim holds for $x=y+1$ by an argument analogous to Case 2b. $\square$

 

Lemma 15

Given any connected graph $G$ on $n$ nodes, any $k \ge n/4$, and any $\{s,t\} \subseteq V(G)$, Algorithm 1B successfully delivers a message from node $s$ to node $t$.

 

Proof

Algorithms 1 and 1B each terminate only once the message reaches $t$. Therefore, it suffices to show that $\{\sigma ^{\prime }_i\}$ is finite. By Lemma 7, Algorithm 1 guarantees delivery. In particular, $\{\sigma _i\}$ is finite. By Lemma 14, therefore $\{\sigma ^{\prime }_i\}$ is also finite. $\square$

 

Lemma 16

Algorithm 1B has dilation at most 6. Furthermore, for any $\epsilon > 0$, there exists a network on which Algorithm 1B has dilation at least $6 - \epsilon$.

 

Proof

Lemma 15 shows that every message eventually reaches its destination. Observe that Lemmas 4–6 and Corollary 3 apply directly to Algorithm 1B. By Corollary 3, the message travels only along consistent edges of $G$.

Case 1. Suppose neither Case 2, 3, nor 4 of Algorithm 1B is applied. Therefore, only Case 1 is applied, the message follows a shortest path to its destination, and the dilation is 1.

Case 2. Suppose Cases 2, 3, or 4 of Algorithm 1B are applied. Therefore,

$$\begin{aligned} \text{ dist}(s,t) \ge k + 1. \end{aligned}$$

(16)

In particular, Case 1 of Algorithm 1B is applied at the last $k + 1$ nodes, each of which is visited exactly once.

These nodes form a path $T$ of length $k$. That is,

$$\begin{aligned} |V(T)| = k+1 \,\text{ and}\, |E(T)| = k . \end{aligned}$$

(17)

Let $Q$ denote the consistent subgraph of $G$ induced by the remaining nodes at which Cases 2, 3, or 4 may be applied. By Observation 1, a message can traverse each edge in $Q$ at most twice, once in each direction. Since $k \ge n/4$ and by (17),

$$\begin{aligned} |V(Q)| = n - |V(T)| \le 4k - (k+1) = 3k - 1. \end{aligned}$$

(18)

By Lemmas 5 and 6, and by (18), it follows that $Q$ contains at most one cycle. Therefore,

$$\begin{aligned} |E(Q)| \le |V(Q)| \le 3k - 1. \end{aligned}$$

(19)

Case 2A. Suppose $Q$ is a tree and the message does not reverse direction. Therefore, each vertex in $V(G)$ is visited at most once. Since $k \ge n/4$ and by (16), the dilation, $S(k)$, is bounded by

$$\begin{aligned} S(k) \le \frac{|V(G)-1|}{\text{ dist}(s,t)} \le \frac{n-1}{k+1} \le \frac{n-1}{n/4 + 1} < 4 . \end{aligned}$$

Case 2B. Suppose $Q$ is a tree and the message reverses direction at least once. If a reversal results from Rules S2, US2, U2c, or U2e, then edge $(u,b)$ will have been traversed twice in the same direction (see Figs. 10, 12, 15, 16); by Observation 1, the message will never reach $t$, contradicting Lemma 15. Therefore, a reversal must result from Rules S1, U1, or US1. By Lemma 4, at least $k-1$ edges in $E(Q)$ are not visited. Let $B$ denote this set of edges. Thus,

$$\begin{aligned} |B| \ge k-1 . \end{aligned}$$

(20)

The remaining edges in $E(Q) {\setminus } B$ are visited at most twice. One additional edge of $G$ is followed once from $Q$ to $T$. Therefore, the dilation, $S(k)$, is bounded by

$$\begin{aligned} S(k) \le \frac{|E(T)| + 2(|E(Q)| - |B|) + 1}{\text{ dist}(s,t)} \le \frac{5k-3}{k+1} < 5 , \end{aligned}$$

by (16), (17), (19), and (20).

Case 2C. Suppose $Q$ contains a cycle $C$ and the message never reverses direction. By Lemma 5,

$$\begin{aligned} |E(C)| \ge 2k+1 . \end{aligned}$$

(21)

Let $B = E(Q) {\setminus } E(C)$. By (19) and (21),

$$\begin{aligned} |B| \le k - 2 . \end{aligned}$$

(22)

Consequently, every edge in $E(C)$ is visited at most once and every edge in $B$ is visited at most twice. One additional edge of $G$ is followed once from $Q$ to $T$. Therefore, the dilation, $S(k)$, is bounded by

$$\begin{aligned} S(k)&\le \frac{|E(T)| + 2|B| + |E(C)| + 1}{\text{ dist}(s,t)}\\&= \frac{|E(T)| + |B| + |E(Q)| + 1}{\text{ dist}(s,t)}\\&\le \frac{5k-2}{k+1}\\&< 5 , \end{aligned}$$

by (16), (17), (19), and (22).

Case 2D. Suppose $Q$ contains a cycle $C$ and the message reverses direction at least once. Let $B = E(Q) {\setminus } E(C)$. Again, (21) and (22) hold as in Case 2c. By (22) and Lemma 4, a reversal cannot result from Rules S1, U1, or US1 of Algorithm 1B. Furthermore, (22) implies that a reversal cannot result from Rules S2 or US2, since Rules U2c or U2e would have been applied earlier. Thus, the reversal must result from Rules U2c or U2e.

By Observation 1, at most one reversal can occur.

Let $u$ denote the node at which Rule U2c or U2e is applied. If Rule U2c is applied, then let nodes be labelled as in Fig. 15c and let $I$ denote the path from $u$ to $s$ in $G^{\prime }_k(u)$. If Rule U2e is applied, then let nodes be labelled as in Fig. 16e and let $I$ denote the path from $u$ to $e$ in $G^{\prime }_k(u)$. The message is never forwarded along any edge in $E(I)$. If $|E(I)| < k - 1$, then some passive component $D$ adjacent to $I$ in $G^{\prime }_k(u)$ must have appeared active in $G^{\prime }_k(b)$ (otherwise, Rules U2c or U2e would have applied at node $b$). Consequently,

$$\begin{aligned} |E(I)| + |E(D)| \ge k - 1 \end{aligned}$$

(23)

Furthermore, since $D \subseteq B$ and by (22),

$$\begin{aligned} |E(D)| \le |E(B)| \le k - 2 . \end{aligned}$$

(24)

Thus, each edge in $Q$ is traversed at most twice. However, not all edges in $Q$ are traversed twice: edges in $I$ are never traversed and edges in $D$ are traversed at most once.

One additional edge of $G$ is followed once from $Q$ to $T$. Therefore, the dilation, $S(k)$, is bounded by

$$\begin{aligned} S(k)&\le \frac{2 ( |E(Q)| - |E(I)| - |E(D)| ) + |E(D)| + |E(T)| + 1}{\text{ dist}(s,t)}\nonumber \\&\le \frac{2[(3k - 1) - (k-1)] + (k-2) + k + 1}{k+1}\nonumber \\&\le \frac{6k - 1}{k+1}\nonumber \\&< 6 , \end{aligned}$$

(25)

by (16), (17), (19), (23), and (24).

In all cases, we get that $S(k) < 6$.

We now show that (25) is tight. Choose any $\epsilon > 0$ and let $n \ge 48/\epsilon$. Let $G$ denote the graph on $n$ vertices illustrated in Fig. 17 along with the corresponding local routing functions for nodes $u$, $e$, and $c$ (Rules U2e, US2 and U3, respectively). Let edge $\{s,d\}$ have minimum rank. Node $s$ lies on a cycle of length $n - 3k + 1$. Since $k \ge n/4$, this cycle is local and edge $\{s,d\}$ is identified as dormant. Rule S1 applies at node $s$. Rule US1 applies at all nodes on the path from $s$ to $a$. Case 1 of Algorithm 1B applies at all nodes on the path from node $d$ to the destination node $t$. Rule U2 applies at all remaining nodes other than $u$, $e$, and $c$. The route followed by Algorithm 1B has length $n + 2k - 6$, whereas the shortest path has length $k+1$. Therefore, when $k = n/4$, Algorithm 1B has dilation

$$\begin{aligned} \frac{n + 2(n/4) - 6}{n/4+1} = 6 - \frac{48}{n+4} > 6 - \epsilon , \end{aligned}$$

showing that (25) is tight. $\square$

Fig. 17

Solid line segments denote routing edges and the dotted line segment denotes a dormant edge. The red arrow denotes the neighbour to which the message is initially forwarded from the passive component containing $s$. Blue arrows denote subsequent forwarding decisions as a function the neighbour from which the message is received. When $k = n/4$, Algorithm 1B forwards the message from the origin node $s$ along the path to node $e$, and then clockwise around the cycle to node $u$. Node $u$ applies Rule U2e, sending the message counter-clockwise around the cycle back to node $c$ before being forwarded along the path from $c$ to the destination node $t$. This route has length $n+2k-6$, whereas the shortest path (via the dormant edge $\{s,d\}$) has length $k+1$ (color figure online)

Theorem 6 follows from Lemmas 15 and 16:

Theorem 6

For every $k \ge n/4$, there exists an origin-aware, predecessor-aware, $k$-local routing algorithm that succeeds on all connected graphs while guaranteeing dilation at most 6.