Neural Network Dynamics without Minimizing Energy

and Applied Analysis 3 otherwise [H A (x, s(t))] i = x i , where b i ∈ R is the threshold of neuron i and the function 1 is the Heaviside function: 1(u) = 1 for u ≥ 0, otherwise 0, which describes an instantaneous unit pulse. On each subsequent time t = 0, 1, . . ., the network generates a vector of neuronal active states according to (8), resulting in the dynamic flow x(t), t = 0, 1, . . .. Theorem 2. Let Ω = [x0, x1, . . . , xp] be a loop of states in {0, 1} n. If A ∈ M n (R) satisfies ⟨A, C (Ω)⟩ ≥ 0, (10) then for any threshold b ∈ R, any initial neural active state x(0) ∈ {0, 1} n, and any updating s(t) ⊂ {1, 2, . . . , n}, t = 0, 1, . . ., the resulting dynamic flow x(t) of (8) cannot behave in x (T) = x 0 , x (T + 1) = x 1 , . . . , x (T + p) = x p (11) for each T = 0, 1, . . .. Proof. Suppose, by contradiction, that there exist b ∈ R, x(0) ∈ {0, 1} n, s(t) ⊂ {1, 2, . . . , n}, t = 0, 1, . . ., and T ≥ 0 such that x(T) = x0, x(T + 1) = x1, . . . , x(T + p) = x. Let Λ + = {t; 0 (x (t)) ∩ 1 (x (t + 1)) ̸ = 0, T ≤ t < T + p} , Λ − = {t; 1 (x (t)) ∩ 0 (x (t + 1)) ̸ = 0, T ≤ t < T + p} . (12) Then Λ+ ̸ = 0 and Λ− ̸ = 0. Indeed, if Λ+ = 0 or Λ− = 0, then x (T) = x (T + 1) = ⋅ ⋅ ⋅ = x (T + p) , (13) contradicting the loop assumption x(T + i) ̸ = x(T + j) for some i, j ∈ {1, 2, . . . , p}. Since 0(x(t)) ∩ 1(x(t + 1)) ⊂ s(t) and 1(x(t)) ∩ 0(x(t + 1)) ⊂ s(t) for each t = 0, 1, . . ., we conclude from (4) and (8) that