BeRGeR: Byzantine-Robust Geometric Routing

TL;DR

BeRGeR presents an asynchronous Byzantine-robust unicast geometric routing algorithm that tolerates a single Byzantine fault without cryptography or randomness, on planar graphs where removing all edges crossing the source–target segment leaves a $3$-connected subgraph $G - \overline{st}$. It deploys two concurrent cores (left and right) that traverse the green face in opposite directions, augmented with skip-threads (braids) to bypass potential faults; the target delivers only when it observes two matching cores or a core plus a braid carrying the same message. The paper proves correctness (validity, liveness, termination) and derives a fault-free message complexity of $O(N^2)$ for planar graphs, while also outlining a constant-packet-size extension. This work advances Byzantine-tolerant routing in resource-constrained geometric networks without cryptographic primitives, offering practical resilience for planar wireless devices and sensor networks.

Abstract

We present BeRGeR: the first asynchronous geometric routing algorithm that guarantees delivery of a message despite a Byzantine fault without relying on cryptographic primitives or randomization. The communication graph is a planar embedding that remains three-connected if all edges intersecting the source-target line segment are removed. We prove the algorithm correct and estimate its message complexity.

Figures (4)

• Figure 1: A graph, $G$, violating \ref{['G-st_3connected']}. Even though the graph itself is three-connected, $G - \overline{st}$ is disconnected.
• Figure 2: BeRGeR example operation. $\mathsf{L}$ core and $\mathsf{R}$ core traverse the green face, $F$. The second $\mathsf{L}$ thread, the $d$-thread, skips node $d$ and traverses the union of the $d$-blue face, $H$, and the green face $F$.
• Figure 3: Illustration for the proof of \ref{['coreDisjoint']}. If left and right core paths share node $v$, then $v$ can be connected by a continuous curve that separates $s$ from $t$. Hence, every path from $s$ to $t$ contains $v$.
• Figure 4: Illustration for the proof of \ref{['threadDisjoint']}. If a blue face, $H$, bypassing green node $k$ contains a green node, $v$, that lies on the right core path, then $v$ can be connected by a continuous curve inside $F\cup H$ that separates $s$ from $t$ and, hence, every path from $s$ to $t$ either contains $k$ or $v$.

Theorems & Definitions (18)

• lemma 1
• proof
• lemma 2
• proof
• lemma 3: Core validity
• proof
• lemma 4: Thread validity
• proof
• lemma 5
• proof