2017
DOI: 10.1137/16m1060868
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Number of Fixed Points and Disjoint Cycles in Monotone Boolean Networks

Abstract: Given a digraph G, a lot of attention has been deserven on the maximum number φ(G) of fixed points in a Boolean network f : {0, 1} n → {0, 1} n with G as interaction graph. In particular, a central problem in network coding consists in studying the optimality of the feedback bound φ(G) ≤ 2 τ , where τ is the minimum size of a feedback vertex set of G. In this paper, we study the maximum number φ m (G) of fixed points in a monotone Boolean network with interaction graph G. We establish new upper and lower bound… Show more

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Cited by 37 publications
(44 citation statements)
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“…If the period of a periodic orbit is one, i.e., x(t 0 ) = x(t 0 + k) for any k ≥ 1, then the state x(t 0 ) is said to be a fixed point. We refer the reader to [46], [47] for studies on the number of fixed points of a BN.…”
Section: Preliminaries and Problem Formulationmentioning
confidence: 99%
“…If the period of a periodic orbit is one, i.e., x(t 0 ) = x(t 0 + k) for any k ≥ 1, then the state x(t 0 ) is said to be a fixed point. We refer the reader to [46], [47] for studies on the number of fixed points of a BN.…”
Section: Preliminaries and Problem Formulationmentioning
confidence: 99%
“…We then say that the packing is principal if the following holds: for every cycle C i and for every vertex v in C i , if there exists a principal path from a cycle C j = C i to v, then either there exists a principal path from C i or a source to v, or there exists a special cycle containing v. We denote by ν ′ (D) the maximum size of a special packing in D. Then fix + [D, 2] ≥ 2 ν ′ . A comparison between ν and ν ′ is also carried out in [4].…”
Section: Number Of Fixed Pointsmentioning
confidence: 99%
“…Secondly, if u is not a source, then f u ≡ φ or f u ≡ ψ for some function g u . We then introduce the function 2], defined as…”
Section: Strict Monotone Stabilitymentioning
confidence: 99%
“…Now let f ∈ F + [K • m,m , 2] with maximum stability andA := |{i : f l i ≡ φ}| + |{j : f r j ≡ ψ}| B := |{i : f l i ≡ ψ}| + |{j : f r j ≡ φ}|.We then have A + B = 2m, hence one of the two, say A, satisfies A ≥ m (the proof is similar if B ≥ m instead). Then let y ∈[2] 2m such that y L = (0, . .…”
mentioning
confidence: 99%