“…The condition (13) for each rigid body i is not distributed since it requires input information of distance neighbors, i.e., ω j , j ∈ N d,i . We thus employ the distributed condition (12) to satisfy (13). Then, because j ∈ N d,i ⇔ i ∈ N d,j and e T 3 eξ θij e 3 = e T 3 eξ θji e 3 hold, the satisfaction of (12) for all i ∈ V guarantees (13) for all i ∈ V. Here, considering (12) can be regarded as sharing (13) equally 3 between rigid body i and rigid body j. Theorem 1 can be thus applied.…”