Non-trivial automata networks do exist that solve the global majority problem with the local majority rule

Abstract

The global majority problem, often referred to as the Density Classification Task, is a classical benchmark in the context of probing the computational capabilities of automata networks. It poses the simple yet challenging problem of determining, by totally local means, whether an arbitrary initial configuration of binary states can evolve to a final, homogeneous global configuration that reflects the initial global majority. Although it is known that in the specific case of cellular automata with periodic boundaries no rule is able to solve the problem, in other formulations solutions are known and, in others, the problem is still open. Aligned with the latter, here we explore the possibility of solving the problem with automata networks, operating only with the local majority rule, with a focus on identifying non-trivial cases where it can be solved and explaining why they do so.

Figures (5)

• Figure 1: Example of transformation of a graph (left) into a graph of an MBAN able to solve DCT (right). The nodes and arcs in blue emphasize the original graph.
• Figure 2: The graph of the complete cycle MBAN of 7 nodes. For all $i$ between $0$ and $5$, the cycle $(0, \cdots, i, 0)$ is the cycle of length $i+1$.
• Figure 3: The complementary-left-right MBAN with $7$ nodes on the left and its complement on the right. According to the definition, $U = (0,2)$, $R = (1,3,4,5,6)$, $S = (0,1)$ and $T = (2,3,4,5,6)$. For example, the node $0$ has for in-neighbours all the nodes except $S_0 = 0$ and $S_1 = 1$; while the node $1$ has for in-neighbours all the nodes except $T_4 = 6$ and $T_3 = 5$.
• Figure 4: On the left, the graph of the complementary-circle-triangle MBAN with $7$ nodes; and on the right, its complement. According to the definition, $U = (3,4,5,6)$ and $R = (0,1,2)$.
• Figure 5: A graph of an MBAN with two intersecting cycles with 7 nodes and node $c = 4$. According to the definition, the cycles $S = (1,2,3,4,0)$ and $T=(4,5,6)$ have node $4$ as intersecting point. Remark that all nodes of $S$, except node $4$, have $1$ predecessor, and all nodes of $T$ have $n-2$ predecessors. For examples, $1$ has for in-neighbor $1-1 = 0$, the node $5$ has for in-neighbors all the nodes except $0$ and $5-3 = 2$ and the nodes $6$ has for in-neighbors all the the nodes of the graph except $0$ and $4$.

Theorems & Definitions (31)

• Definition 2.1
• Definition 2.2
• Definition 2.3
• Definition 2.4
• Proposition 3.1
• Lemma 3.2
• proof
• Lemma 3.3
• proof
• Lemma 3.4