Example
Consider the case where Y is the graph with vertex set {1,2,3} and undirected edges {1,2}, {1,3} and {2,3} (a triangle or 3-circle) with vertex states from K = {0,1}. For vertex functions use the symmetric, boolean function nor : K3 → K defined by nor(x,y,z) = (1+x)(1+y)(1+z) with boolean arithmetic. Thus, the only case in which the function nor returns the value 1 is when all the arguments are 0. Pick w = (1,2,3) as update sequence. Starting from the initial system state (0,0,0) at time t = 0 one computes the state of vertex 1 at time t=1 as nor(0,0,0) = 1. The state of vertex 2 at time t=1 is nor(1,0,0) = 0. Note that the state of vertex 1 at time t=1 is used immediately. Next one obtains the state of vertex 3 at time t=1 as nor(1,0,0) = 0. This completes the update sequence, and one concludes that the Nor-SDS map sends the system state (0,0,0) to (1,0,0). The system state (1,0,0) is in turned mapped to (0,1,0) by an application of the SDS map.
Read more about this topic: Sequential Dynamical System
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