Finding the fixed points of a Boolean network from a positive feedback vertex set
FFinding the fixed points of a Boolean network from a positivefeedback vertex set
Julio Aracena ∗ , Luis Cabrera-Crot †‡ , Lilian Salinas § April 6, 2020
Abstract
In the modeling of biological systems by Boolean networks a key problem is finding the set offixed points of a given network. Some constructed algorithms consider certain structural propertiesof the interaction graph like those proposed by Akutsu et al. in [3, 56] which consider a feedbackvertex set of the graph. However, these methods do not take into account the type of action(activation, inhibition) between its components.In this paper we propose a new algorithm for finding the set of fixed points of a Booleannetwork, based on a positive feedback vertex set P of its interaction graph and which works, byapplying a sequential update schedule, in time O (2 | P | · n ), where n is the number of components.The theoretical foundation of this algorithm is due a nice characterization, that we give, of thedynamical behavior of the Boolean networks without positive cycles and with a fixed point.An executable file of FixedPoint algorithm made in Java and some examples of input files areavailable at: A Boolean network is a system of n interacting Boolean variables, which evolve, in a discrete time,according to a evolution rule and to a predefined updating scheme. Boolean networks have manyapplications, including computer science, in particular circuit theory, and social systems. In particular,from the seminal works of S. Kauffman [22, 23] and R. Thomas [48, 49], they are extensively used asmodels of gene networks. In this context, the fixed points of a network are associated to distinct typesof cells defined by patterns of gene activity [8, 20, 22]. The problem of finding the steady states or fixedpoints of a Boolean network is a difficult task. In [2], Akutsu et al. show that deciding if a networkhas a fixed point is NP-complete, consequently, counting how many fixed points a Boolean networkhas is ∗ CI2MA and Departamento de Ingenier´ıa Matem´atica, Universidad de Concepci´on, Chile. † Corresponding author ‡ Departamento de Ing. Inform´atica y Cs. de la Computaci´on, Universidad de Concepci´on, Chile. § Departamento de Ing. Inform´atica y Cs. de la Computaci´on and CI2MA, Universidad de Concepci´on, Chile. a r X i v : . [ c s . D M ] A p r ome of the algorithms for finding the fixed points of a Boolean network consider certain charac-teristics of its interaction graph. In particular, Akutsu et al. in [3, 56] proposed a method consistingin fixing the values of the vertices of a minimal feedback vertex set F and propagating them to theothers. In this way the problem of finding the fixed points of a Boolean network of n components isreduced to check 2 | F | state configurations instead of 2 n , where usually the size of F is less than n .However, this method does not consider the sign of the arcs of the interaction graph.On the other hand, to our knowledge it does not exist a method to find the fixed points of a Booleannetwork that take advantage of the fact they are invariant to the update schedule.In this paper we propose a new strategy to find the fixed points of a Boolean network which consistsin fixing the local states of a positive feedback vertex set P of the interaction graph and updating theother local states according to a sequential update schedule constructed from P . In this way we reducethe problem of finding the fixed points of a Boolean network of n components to check 2 | P | stateconfigurations, where P can be smaller than a feedback vertex set of the network. Thus, the proposedalgorithm is quadratic in the size of the network and exponential in the size of P . Consequently, thismethod can be very efficient, with respect to other methods, in Boolean networks with a small positivefeedback vertex set, as for example ones with few positive cycles. A particular case of this latter arethe strong inhibition Boolean networks [16].For the construction of the proposed algorithm we proved first that the dynamical behavior ofa Boolean network without positive cycles and with a fixed point is similar to an acyclic Booleannetwork, i.e. it quickly converges to the fixed point from any initial configuration. A Boolean network (BN) with n components is a discrete dynamical system usually defined by a globaltransition function : f : { , } n → { , } n , x → f ( x ) = ( f ( x ) , . . . , f n ( x )) , where each function f i : { , } n → { , } associated to the component i is called local activationfunction .Any vector x = ( x , . . . , x n ) ∈ { , } n is called a state of the network f with local state x i on eachcomponent i . The dynamics of f is given by its application on any state of the network. A steadystate or fixed point of f is any state x ∈ { , } n such that f ( x ) = x .We define the interaction graph of a BN f with n components, denoted G ( f ) = ( V, A ), as thesigned directed graph with a sign, +1 or -1, attached to each arc, where V is a finite set of elementscalled vertices or nodes indexed in the set [ n ] := { , , . . . , n } and A is a collection of ordered pairs ofvertices of V , called arcs, such that ∀ u, v ∈ [ n ] : • ( u, v ) ∈ A , with sign +1, if ∃ x ∈ { , } n , x u = 0, f v ( x ) < f v ( x + e u ), • ( u, v ) ∈ A , with sign −
1, if ∃ x ∈ { , } n , x u = 0, f v ( x ) > f v ( x + e u ),where e u ∈ { , } n denotes the binary vector with all entries equal to 0, except for entry u , whichequals 1. The vertex set of G ( f ) is referred to as V ( G ( f )) and its arc set as A ( G ( f )). Note that G ( f )can have both a positive and a negative arc from one vertex to another. If G ( f ) does not have multiplearcs from one vertex to another, then we say that f is a regulatory Boolean network (RBN).A walk from a vertex v to a vertex v l in the interaction graph G ( f ) is a sequence of verticesand arcs W = v , a , v , . . . , a l − , v l of G ( f ) such that ∀ i ∈ { , . . . , l − } , a i = ( v i , v i +1 ) ∈ A ( G ( f )). A ( W ) = { a i : i ∈ { , . . . , l − }} is the set of arcs of W . A path is a walk without repetition of vertices(except eventually the extreme ones). A circuit is a walk without repetition of arcs and closed (i.e. itsextreme vertices are equal). A cycle is a closed path.Let u ∈ V ( G ( f )), we denote the in-degree and out-degree of u by d − ( u ) = | { v ∈ V ( G ( f )) : ( v, u ) ∈ A ( G ( f )) } | and d + ( u ) = | { v ∈ V ( G ( f )) : ( u, v ) ∈ A ( G ( f )) } | , respectively. If d − ( u ) = 0, we say that u is a sourcevertex . The degree of u is d ( u ) = d − ( u ) + d + ( u ). We refer [11] for other basic definitions in digraphs.2he sign of a walk (path, circuit or cycle) is the product of the signs of its arcs. We say that acycle is a positive cycle, if the sign of the cycle is +1, otherwise, we say that is a negative cycle. SeeExample 1.A vertex set F ⊆ V ( G ( f )) is a feedback vertex set (FVS) of G ( f ) if G ( f ) − F is acyclic. F is saidto be a minimal feedback vertex set of G ( f ) if F is a FVS of G ( f ) and there is no other FVS F (cid:48) suchthat F (cid:48) (cid:40) F . F is said to be a minimum feedback vertex set of G ( f ) if F is a FVS of G ( f ) and thereis no other FVS set F (cid:48) such that | F (cid:48) | < | F | .We denote τ ( G ( f )) = min {| F | : F is a FVS of G ( f ) } the transversal number of G ( f ). P ⊆ V ( G ( f )) is a positive feedback vertex set (PFVS) of the signed interaction graph G ( f ) if G ( f ) − P is a digraph without positive cycles. The minimum cardinality of a PFVS is denoted by τ + ( G ( f )) and called the positive transversal number of G ( f ).When there is no confusion, for simplicity we denote τ ( G ( f )) by τ and τ + ( G ( f )) by τ + . Remark 1.
Since a FVS is a particular PFVS, τ + ≤ τ. Example 1.
An example of a BN f and its interaction graph G ( f ) is shown in Figure 1. Positive arcsare represented by and negative arcs by . Examples of positive cycles are: C = 4 , a , , a , C = 3 , a , , a , , a ,
3, and examples of negative cycles are: C = 3 , a , , a , , a , C =1 , a , , a ,
1. Dark vertices are a PFVS of G ( f ).
12 3 4 543 f = ¬ x ∧ x f = ( x ∧ x ) ∨ ( x ∧ ¬ x ) f = ( ¬ x ∨ ¬ x ) ∧ x ∧ x f = ( x ∧ x ) ∨ ( x ∧ ¬ x ) f = x a a a a a a a a a a a a a a a Figure 1: Example of a BN and its interaction graph.In the sequel, we will refer to the signed cycles of a BN f as the cycles of G ( f ).Given f : { , } n → { , } n a BN and u ∈ [ n ], the partial evaluation of f in the local function f u is the function f u : { , } n → { , } n such that: ∀ x ∈ { , } n , f u ( x ) = ( x , . . . , x u − , f u ( x ) , x u +1 , . . . , x n )We define a sequential schedule π = ( π , π , . . . , π n ) as a permutation of [ n ]. The dynamics of aBN f : { , } n → { , } n updated according to π is defined by: ∀ x ∈ { , } n , f π ( x ) = f π n ◦ f π n − ◦ · · · ◦ f π ( x ) . To avoid confusion, we use a special notation for self-compositions of f : we denote by f (cid:104) (cid:105) theidentity on { , } n and for k ≥ f (cid:104) k (cid:105) = f ◦ f (cid:104) k − (cid:105) . The concatenation of k times the sequentialschedule π is denoted π k . In this way f π k = ( f π ) (cid:104) k (cid:105) . The relationship between the fixed points and the positive and negative cycles in Boolean networkshas been greatly studied [6, 9, 10, 34, 35, 36]. In particular, in [6, 34] was proved that every Boolean3etwork without positive cycles has at most one fixed point and, in addition, if the network has aninitial non-trivial strong component, then it has no fixed points. On the other hand, F. Robert studiedthe Boolean networks without cycles and proved in [37, 38] that these networks have a simple dynamicalbehavior with a unique attractor, which is a fixed point, and where all initial states quickly convergeto it. More precisely, he proved the following theorem.
Theorem ( F. Robert).
Let f : { , } n → { , } n be a BN such that G ( f ) is acyclic. Then,1. f has a unique fixed point y ∈ { , } n .2. ∀ x ∈ { , } n , f (cid:104) n (cid:105) ( x ) = y .3. ∃ π a sequential schedule such that ∀ x ∈ { , } n , f π ( x ) = y . In this paper we show that whether a BN without positive cycles has a fixed point, then itsdynamical behavior is like a BN without cycles. More precisely, we prove the following theorem.
Theorem 1.
Let f : { , } n → { , } n be a BN without positive cycles. Then, the following proposi-tions are equivalents:1. f has a unique fixed point y ∈ { , } n .2. ∀ x ∈ { , } n , f (cid:104) n (cid:105) ( x ) = y .3. ∃ π a sequential schedule such that ∀ x ∈ { , } n , f π ( x ) = y . To proof this result, we need to introduce some definitions and previous results.
Definition 1.
Let f : { , } n → { , } n be a Boolean network, we denote the following sets (eventuallyempty): I ( f ) = { v ∈ [ n ] : ∃ c ∈ { , } , ∀ x ∈ { , } n , f v ( x ) = c } . In other words, v ∈ I ( f ) if f v is constant. For all v ∈ I ( f ), we denote c v to the constant value of f v ( x ).Recursively, we define ∀ k ∈ N , k ≥ , I k ( f ) in the following way: I k ( f ) = { v ∈ [ n ] : ∃ c ∈ { , } , ∀ x ∈ { , } n , f v ( c u : u ∈ I k − ( f ); x u : u / ∈ I k − ( f )) = c } . Similarly, v ∈ I k ( f ) if the local activation function associated to the component v with fixed value c u on input u ∈ I k − ( f ) is a constant function. ∀ v ∈ I k ( f ), we denote c v to the constant value of f v ( c u : u ∈ I k − ( f ); x u : u / ∈ I k − ( f )).From here, if v ∈ I k ( f ), then: ∀ t ≥ k, ∀ x ∈ { , } n , f (cid:104) t (cid:105) v ( x ) = c v . In this case, we say that the component v is fixed at iteration k of f with value c v . Proposition 1.
The sets I k ( f ) have the following properties:1. ∀ k ≥ , I k ( f ) ⊆ I k +1 ( f ) .
2. If ∃ k ∈ N , I k ( f ) = I k +1 ( f ) , then ∀ l ∈ N , I k ( f ) = I k + l ( f ) .3. If I ( f ) = ∅ , then ∀ k ∈ N , I k ( f ) = ∅ . roof.
1. By induction on k . First, we prove that I ( f ) ⊆ I ( f ). By definition, v ∈ I ( f ) if f v is a constantfunction. On the other hand, v ∈ I ( f ) if f v with fixed value c u on each input u ∈ I ( f ) is aconstant function. Therefore, I ( f ) ⊆ I ( f ).Let us suppose that I k ( f ) ⊆ I k +1 ( f ). Now, we prove that I k +1 ( f ) ⊆ I k +2 ( f ). Analogously, v ∈ I k +1 ( f ) if f v with fixed value c u on each input u ∈ I k ( f ) is a constant function. On the otherhand, v ∈ I k +2 ( f ) if f v with fixed value c u on each input u ∈ I k +1 ( f ) is a constant function.Since I k ( f ) ⊆ I k +1 ( f ), the inclusion I k +1 ( f ) ⊆ I k +2 ( f ) is direct.2. To prove this, we prove that if I k ( f ) = I k +1 ( f ) then I k +1 ( f ) = I k +2 ( f ).Let us suppose that I k ( f ) = I k +1 ( f ). If I k +2 ( f ) = ∅ , by Property 1, I k +1 ( f ) = ∅ . So, let v be acomponent in I k +2 ( f ), by definition: ∃ c ∈ { , } , ∀ x ∈ { , } n , f v ( c u : u ∈ I k +1 ( f ); x u : u / ∈ I k +1 ( f )) = c. Since I k ( f ) = I k +1 ( f ), ∀ x ∈ { , } n , f v ( c u : u ∈ I k ( f ); x u : u / ∈ I k ( f )) = c. Therefore, v ∈ I k +1 ( f ). For this reason, I k +2 ( f ) ⊆ I k +1 ( f ) and since I k +1 ( f ) ⊆ I k +2 ( f ), then I k +1 ( f ) = I k +2 ( f ).3. First, we prove that if I ( f ) = ∅ , then I ( f ) = ∅ .By contradiction, let us suppose that I ( f ) = ∅ and I ( f ) (cid:54) = ∅ . Let v be a component in I ( f ),then: ∃ c ∈ { , } , ∀ x ∈ { , } n , f v ( c u : u ∈ I ( f ); x u : u / ∈ I ( f )) = f v ( x ) = c. Therefore, by definition, v ∈ I ( f ) which is a contradiction.Since I ( f ) = I ( f ) = ∅ , by Property 2, ∀ k ∈ N , I k ( f ) = ∅ . (cid:50) Definition 2.
Let f : { , } n → { , } n be a Boolean network. We say that f is an irreduciblenetwork after k iterations or k -irreducible network if I k ( f ) = I k +1 ( f ). If I ( f ) = ∅ , we say that f isan irreducible network . Definition 3.
Let f : { , } n → { , } n be a Boolean network and k ∈ N . We define f I k : { , } n → { , } n the Boolean network associated to f and I k ( f ) by: ∀ x ∈ { , } n , f I k ( x ) = f ( c u : u ∈ I k ( f ); x u : u / ∈ I k ( f )) . Remark 2.
Note that if f is an irreducible network, then f I k ( x ) = f ( x ). Example 2.
Figure 2 shows an example of a Boolean network f and the network f I , according toDefinition 3. Since f and f are the only constant functions, I ( f ) = { , } (with c = 1 and c = 0).On the other hand, f ( x , x , x , , , x , x ) = x ∨ ∨ ∈ I ( f ). Moreover, since thereis no other component that satisfies the same condition, I ( f ) = { , , } .Analogously, I ( f ) = { , , , , } . In a new iteration, I ( f ) = I ( f ), so f is 3-irreducible. Remark 3.
It is important to remark that ∀ k ∈ N and for all f : { , } n → { , } n :1. V ( G ( f I k )) = V ( G ( f )).2. A ( G ( f I k )) ⊆ A ( G ( f )).3. If ( u, v ) ∈ A ( G ( f I k )), then the sign of ( u, v ) is the same in G ( f ) and G ( f I k ).5 f = x ∨ x ∨ x f = x ∧ x ∧ x f = x ∧ x f = 1 f = 0 f = x ∨ x ∨ x f = x ∧ x I ( f ) = { , } with c = 1 and c = 0 I ( f ) = { , , } with c = 1 I ( f ) = { , , , , } with c = 1 and c = 1 f I : f I ( x ) = 1 ∨ ∨ f I ( x ) = x ∧ x ∧ x ∧ x f I ( x ) = x ∧ x f I ( x ) = 1 = 1 f I ( x ) = 0 = 0 f I ( x ) = 1 ∨ x ∨ x = 1 f I ( x ) = 1 ∧ Figure 2: Example of Boolean networks f and f I .4. y is a fixed point of f if and only if y is a fixed point of f I k and ∀ v ∈ I k ( f ) , c v = y v . Remark 4.
Let f : { , } n → { , } n be a Boolean network such that I k ( f ) = I k +1 ( f ) (cid:54) = ∅ . Then, G ( f I k ) has the following properties:1. ∀ u ∈ I k ( f ) , d + G ( f Ik ) ( u ) = 0.2. If there exists a non-trivial strong component, then there is an initial non-trivial strong compo-nent.In [6, 34] was proved the following theorem, known as the Thomas’s first rule: Theorem ( Thomas’s first rule).
Let f : { , } n → { , } n a BN without positive cycles. Then, f has at most one fixed point. A direct corollary from previous theorem is that if f is a BN without positive cycles such that hasan initial non-trivial strong component, then f has no fixed points [6].From theorem Thomas’s first rule we can state the following result. Lemma 1.
Let f : { , } n → { , } n be a RBN without positive cycles. f has a unique fixed point ifand only if there exists k ∈ { , . . . , n } such that ∅ (cid:54) = I ( f ) ⊂ I ( f ) ⊂ · · · ⊂ I k ( f ) = [ n ] . Proof. ( ⇒ ) Let us suppose that f has a unique fixed point y ∈ { , } n , we prove that ∀ k ∈{ , . . . , n } , | I k | ≥ k . Since f does not have any positive cycle, by theorem Thomas’s first rule allinitial strong components of G ( f ) are trivial, hence there exists a vertex v ∈ [ n ] whose local activa-tion function is constant. Then, v ∈ I ( f ) and | I ( f ) | ≥ ∃ l ∈ { , . . . , n } such that | I l | < l . So, by Property 1 ofProposition 1: ∃ k ≤ ( n − , I k ( f ) = I k +1 ( f ) (cid:54) = [ n ] . In this case, ∀ v / ∈ I k ( f ) , ∃ u / ∈ I k ( f ), such that f v depends on x u , i.e., ∀ v / ∈ I k ( f ) , ∃ u / ∈ I k ( f ) , ( u, v ) ∈ A ( f I k ).Then, the in-degree of all vertices in [ n ] \ I k ( f ) is greater than 0, hence there exists a non-trivialstrong component of G ( f I k ) induced by the vertices in [ n ] \ I k ( f ). Since ∀ u ∈ I k ( f ) , d + G ( f Ik ) ( u ) = 0,there exists an initial non-trivial strong component in G ( f I k ) and thus f I k has no fixed points. Hence,by Remark 3.4) f has no fixed points, which is a contradiction.Therefore, if k ≤ ( n − I k ( f ) (cid:40) I k +1 ( f ) or I k ( f ) = [ n ].( ⇐ ) If I n ( f ) = [ n ], ∀ v ∈ [ n ] , ∀ t ≥ n, ∀ x ∈ { , } n , f (cid:104) t (cid:105) v ( x ) = c v . Then, f has a fixed point and,because f does not have any positive cycles, this fixed point is unique. (cid:50) Now we give the proof of Theorem 1. 6 roof of Theorem 1.
The proof of (2) ⇒ (1) and (3) ⇒ (1) are straightforward, so we prove that(1) ⇒ (2) and (1) ⇒ (3).Both proofs are very similar, so we give the proof for (1) ⇒ (2).First, notice that { I , I \ I , . . . , I k \ I k − } is a partition of [ n ], we define an order of the vertices π = ( π , . . . , π n ) such that: π i ∈ ( I j \ I j − ) ∧ π i (cid:48) ∈ ( I j (cid:48) \ I j (cid:48) − ) ∧ j < j (cid:48) = ⇒ i < i (cid:48) . Now we prove by induction that ∀ i ∈ { . . . , k } , ∀ t ≥ k, f (cid:104) t (cid:105) π i ( x ) = y π i .If k = 1, π ∈ I ( f ), then f π i is a constant function and therefore ∀ t ≥ f (cid:104) t (cid:105) π ( x ) = y π .Let us suppose that ∀ i ∈ { . . . , k } , ∀ t ≥ k, f (cid:104) t (cid:105) π i t ( x ) = y π i .Now we need to prove that ∀ t ≥ k + 1 , f (cid:104) t (cid:105) π k +1 ( x ) = y π k +1 . f (cid:104) t (cid:105) π k +1 ( x ) = f π k +1 ( f (cid:104) t − (cid:105) ( x )) . Since t ≥ k + 1, t − ≥ k , then by hypothesis of induction: f (cid:104) t (cid:105) π k +1 ( x ) = f π k +1 ( y π i : i ≤ k ; ˜ x π i : i > k ) . By definition of π , if π k +1 ∈ I j \ I j − then I j − ⊆ { π , . . . , π k } , therefore, by definition of I j : f (cid:104) t (cid:105) π k +1 ( x ) = y π k +1 . The proof for (1) ⇒ (3) works the same way to prove that ∀ i ∈ { , . . . , k } , f ( π ,...,π k ) π i ( x ) = f π i (( f π k − ◦ · · · ◦ f π )( x )) = y π i . (cid:50) A difference between BNs with acyclic interaction graphs and BNs without positive cycles is that,in the first case, the sequential schedule referred in F. Robert’s theorem depends only on the interactiongraph and not on the function. This latter is not true in the case of BNs without positive cycles asshown in Example 3.
Example 3.
Let f : { , } → { , } be a BN with interaction graph shown in Figure 3 and π = (1 , ,
3) a sequential schedule. If we choose the functions: f ( x ) = 1 , f ( x ) = ¬ x ∧ x , f ( x ) = x ∧ ¬ x , then y = (1 , ,
1) is a fixed point of f and ∀ x ∈ { , } , f π ( x ) = y. On the contrary, if we choose thefunctions: f ( x ) = 1 , f ( x ) = ¬ x ∨ x , f ( x ) = x ∨ ¬ x ,y (cid:48) = (1 , ,
1) is a fixed point of f , but f π (1 , ,
0) = (1 , , (cid:54) = y (cid:48) . In this case, if we choose π (cid:48) = (1 , , ∀ x ∈ { , } , f π (cid:48) ( x ) = y (cid:48) . Figure 3: Interaction graph of a network without positive cycles.On the other hand, as consequence of Theorem 1 the following corollary shows us that the problemof the existence of a fixed point in a BN without positive cycles can be solved in polynomial time.
Corollary 1.
Determining whether a BN f : { , } n → { , } n without positive cycles has a fixedpoint can be done through n + 1 applications of f on any state of the network. roof. If y is a fixed point of f , then ∀ x ∈ { , } n , f (cid:104) n (cid:105) ( x ) = y . Hence, it suffices to check for somestate x (for example, x = (cid:126) , , . . . , f (cid:104) n (cid:105) ( x ) is a fixed point of the network. In other words,if f (cid:104) n (cid:105) ( x ) = f (cid:104) n +1 (cid:105) ( x ) , then f (cid:104) n (cid:105) ( x ) is a fixed point of f . Otherwise, f has no fixed points. (cid:50) In [6] was proved that for every RBN f there is an injection between its fixed points and the localstates of a PFVS of its interaction graph G ( f ), allowing to conclude that the maximum number offixed points of a RBN f is 2 τ + ( G ( f )) . A simple exercise shows that this result is also valid in generalBoolean networks [36]. In this way, we develop an algorithm that considers all the possible states of agiven PFVS and efficiently verifies, by application of f , which of them produces a fixed point for thenetwork. This is achieved thanks to the dynamical behavior of a BN without positive cycles describedin Theorem 1. Previously, we give some definitions and results. Definition 4.
Let f : { , } n → { , } n be a BN, P a PFVS of G ( f ) and a : P → { , } afunction, which can be represented by a vector a ∈ { , } | P | . We define the Boolean network fa : { , } n → { , } n as follows: ∀ v ∈ [ n ] , ∀ x ∈ { , } n , fa v ( x ) = (cid:40) a ( v ) if v ∈ P,f v ( x ) if v / ∈ P. Example 4.
Figure 4 shows an example of a BN f and a BN fa , where a (1) = a (3) = 0 and a (2) = 1,according to Definition 4. Dark gray vertices represent P . f ( x ) = x ∨ x ∨ x f ( x ) = x ∧ x ∧ x f ( x ) = x ∧ x f ( x ) = ¬ x f ( x ) = 0 f ( x ) = x ∨ x ∨ x f ( x ) = x ∧ x fa ( x ) = 0 fa ( x ) = 1 fa ( x ) = 0 fa ( x ) = ¬ x fa ( x ) = 0 fa ( x ) = x ∨ x ∨ x fa ( x ) = x ∧ x Figure 4: A Boolean network f and a network without positive cycles fa .Note that G ( fa ) = G ( f ) − { ( u, v ) ∈ A ( G ( f )) : v ∈ P } . Hence, fa is a BN without positive cycles.As a direct result, we have the following proposition. Proposition 2.
Let f : { , } n → { , } n be a BN, P (cid:54) = ∅ a PFVS of G ( f ) and a : P → { , } afunction. x ∈ { , } n is a fixed point of f such that ∀ v ∈ P, x v = a ( v ) if and only if:1. x is a fixed point of fa,2. ∀ v ∈ P, f v ( x ) = a ( v ) . roof. ( ⇒ ) Let x ∈ { , } n be a fixed point of f such that ∀ v ∈ P, x v = a ( v ), then, as x is afixed point of f , then ∀ v, f v ( x ) = x v . Hence, by Definition 4, if v / ∈ P, fa v ( x ) = f v ( x ) = x v , and if v ∈ P, fa v ( x ) = a ( v ) = x v . Therefore, ∀ v, fa v ( x ) = x v , thus, x is a fixed point of fa and, by hypothesis,condition (2) also holds.( ⇐ ) If x ∈ { , } n is a fixed point of fa , then ∀ v, fa v ( x ) = x v . Hence, by Definition 4, if v / ∈ P, x v = fa v ( x ) = f v ( x ).Moreover, If ∀ v ∈ P, x v = fa v ( x ) = a ( v ) = f v ( x ).Therefore, ∀ v, x v = f v ( x ), then, x is a fixed point of f and ∀ v ∈ P, x v = a ( v ). (cid:50) In this way, we use Proposition 2 to define Algorithm 1, which finds the fixed points of a given BN.
Proposition 3.
Given f : { , } n → { , } n be a BN and P a PFVS of G ( f ) , Algorithm 1 finds theset of fixed points of f in time O ( n | P | ) . Proof.
The correctness of the algorithm is direct from Proposition 2 and Corollary 1. In particular,if P (cid:54) = ∅ , the instruction x ← fa (cid:104) n (cid:105) ( (cid:126)
0) obtains a fixed point candidate (after n executions of fa , forsome a ∈ { , } | P | ) that is checked in the next line according to Proposition 2. Otherwise, i.e. f has nopositive cycles, x ← f (cid:104) n (cid:105) ( (cid:126)
0) obtains a fixed point candidate (after n executions of f ) that is checked inthe next line according to Corollary 1. Finally, the total number of operations, corresponding mainlyto the applications of the local activation functions, in the worst case is O ( n | P | ). (cid:50) Algorithm 1:
BasicFixedPoint
Input: f a BN with n components and P a PFVS of G ( f ). Output: S the set of fixed points of f . S ← ∅ ; if P (cid:54) = ∅ thenforeach a ∈ { , } | P | do x ← fa (cid:104) n (cid:105) ( (cid:126) if ( fa ( x ) = x ) ∧ ( ∀ u ∈ P, f u ( x ) = a ( u )) then S ← S ∪ { x } ; else x ← f (cid:104) n (cid:105) ( (cid:126) if ( f ( x ) = x ) then S ← { x } ; return S In order to make the Algorithm 1 faster, and following the ideas introduced in [7], we use a sequentialupdate schedule that allows a faster convergence to the fixed points when they exist. By Theorem 1,the BNs without positive cycles has a sequential update schedule such that they convergence to theonly fixed point, when there exists, in only one step. However, determining such a schedule can beas difficult as finding the fixed points of the network, because it depends on the value of each localactivation functions as shown in Example 3. We propose to use a sequential update schedule whichdepends only on the interaction graph structure of input network. More precisely, given a FVS F anda PFVS P contained in it, we determine in polynomial time a sequential scheme such that the globalactivation function applied with this schedule in each iteration fixes at least the value of one vertex in F \ P (in the case that this is possible), then the network is updated once more to fix the remainingvertices. Thus, the number of applications in the new algorithm is reduced from n to | F | − | P | + 1. Definition 5.
Given f : { , } n → { , } n a BN, F a FVS of G ( f ) and P ⊆ F a PFVS of G ( f ), Wesay that a permutation π = ( π , π , . . . , π n ) on the set [ n ] is an order compatible with F and P if itsatisfies the following properties: • ∀ π i ∈ ( F − P ) , ∀ π j / ∈ ( F − P ) , i > j . 9 ∀ π i ∈ P, ∀ π j / ∈ P, i < j . • ∀ π i , π j / ∈ F, ( π i , π j ) ∈ A ( G ( f )) = ⇒ i < j . Example 5.
Figure 5 shows examples of orders compatible with F and P . Dark gray vertices representthe PFVS P and light gray vertices represent F \ P .
14 365 7 2 π = (3 , , , , , , π (cid:48) = (1 , , , , , , π (cid:48)(cid:48) = (3 , , , , , , G ( f ) and orders compatible with F and P . Set P is denoted for dark gray verticesand set F for light gray vertices. Proposition 4.
Let f : { , } n → { , } n be a BN, such that the interaction graph G ( f ) does not havepositive cycles, F be a FVS of G ( f ) , π be an order compatible with F and y ∈ { , } n . y is the uniquefixed point of f if and only if ∀ x ∈ { , } n , ( f π ) (cid:104)| F | +1 (cid:105) ( x ) = y . Proof. ( ⇒ ) Let us suppose that y ∈ { , } n is the unique fixed point of f . Without loss of generality,let us suppose that π = (1 , , . . . , n ). Then, by Theorem 1, there exist I ( f ) (cid:40) I ( f ) (cid:40) · · · (cid:40) I k − ( f ) (cid:40) I k ( f ) = [ n ]such that: ∀ j ∈ { , . . . , k } , ∀ v ∈ I j ( f ) , ∀ t ≥ j, ∀ x ∈ { , } n , f (cid:104) t (cid:105) v ( x ) = y v . Now, we prove that by iterating f according to the order π , the n vertices fix their value in m iterations(with m ≤ | F | + 1).To perform this, let us consider the following sequence of indices o i defined as follows: o = min { j ∈ { , . . . , k } : ( I j ( f ) ∩ F ) (cid:54) = ∅} , ∀ p ≥ , o p = min { j > o p − : (( I j ( f ) \ I j − ( f )) ∩ F (cid:54) = ∅ ) ∨ ( j = k ) } . Notice that o < o < · · · < o m = k and m ≤ | F | + 1.We prove that ∀ i ∈ { , . . . , m } , ∀ u ∈ I o i , ∀ t ≥ i, ∀ x ∈ { , } n , ( f π ) (cid:104) t (cid:105) u ( x ) = y u .By contradiction, let us suppose that: ∃ i ∈ { , . . . , m } , ∃ u ∈ I o i , ∃ t ≥ i, ∃ x ∈ { , } n , ( f π ) (cid:104) t (cid:105) u ( x ) (cid:54) = y u . Let: i ∗ = min (cid:110) i ∈ { , . . . , m } : ∃ u ∈ I o i , ∃ t ≥ i, ∃ x ∈ { , } n , ( f π ) (cid:104) t (cid:105) u ( x ) (cid:54) = y u (cid:111) and l ∗ = min (cid:110) l ≤ o i ∗ : ∃ u ∈ I l , ∃ t ≥ i ∗ , ∃ x ∈ { , } n , ( f π ) (cid:104) t (cid:105) u ( x ) (cid:54) = y u (cid:111) . Let u ∗ ∈ I l ∗ \ I l ∗ − be such that ∃ t ≥ i ∗ , ∃ x ∈ { , } n , ( f π ) (cid:104) t (cid:105) u ∗ ( x ) (cid:54) = y u ∗ . Then, by Definition 1, ∀ t ≥ i ∗ , ∀ x ∈ { , } n :( f π ) (cid:104) t (cid:105) u ∗ ( x ) = f πu ∗ (( f π ) (cid:104) t − (cid:105) ( x ))= f u ∗ (( f π ) (cid:104) t (cid:105) i ( x ) : i < u ∗ ; ( f π ) (cid:104) t − (cid:105) i : i ≥ u ∗ ) . l ∗ >
1, then:( f π ) (cid:104) t (cid:105) u ∗ ( x ) = f u ∗ (( f π ) (cid:104) t (cid:105) v ( x ) : v < u ∗ ∧ v ∈ I l ∗ − ; ( f π ) (cid:104) t (cid:105) v ( x ) : v < u ∗ ∧ v / ∈ I l ∗ − ;( f π ) (cid:104) t − (cid:105) v ( x ) : v ≥ u ∗ ∧ v ∈ I l ∗ − ; ( f π ) (cid:104) t − (cid:105) v ( x ) : v ≥ u ∗ ∧ v / ∈ I l ∗ − ) . By definition of l ∗ , if v ∈ I l ∗ − , then ( f π ) (cid:104) t (cid:105) v ( x ) = y v . Moreover, If v ≥ u ∗ , then v ∈ F and, therefore,if v ∈ F and v ∈ I l ∗ − , then v ∈ I o ( i ∗− and, thus, if v ≥ u ∗ and v ∈ I l ∗ − , then ( f π ) (cid:104) t − (cid:105) v ( x ) = y v .Also, if i ∗ = 1, then ∀ l ≤ o , F ∩ I l − = ∅ :( f π ) (cid:104) t (cid:105) u ∗ ( x ) = f u ∗ ( y v : v < u ∗ ∧ v ∈ I l ∗ − ; ( f π ) (cid:104) t (cid:105) v ( x ) : v < u ∗ ∧ v / ∈ I l ∗ − ; y v : v ≥ u ∗ ∧ v ∈ I l ∗ − ; ( f π ) (cid:104) t − (cid:105) v : v ≥ u ∗ ∧ v / ∈ I l ∗ − )= f u ∗ ( y v : v ∈ I l ∗ − ; ( f π ) (cid:104) t (cid:105) v ( x ) : v < u ∗ ∧ v / ∈ I l ∗ − ;( f π ) (cid:104) t − (cid:105) v ( x ) : v ≥ u ∗ ∧ v / ∈ I l ∗ − ) . By Definition 1, ∀ t ≥ i ∗ , ∀ x ∈ { , } n , ( f π ) (cid:104) t (cid:105) u ∗ ( x ) = y u ∗ , which is a contradiction.Therefore, in at most m iterations of f π , all vertices in I o m ( f ) = I k ( f ) are fixed.( ⇐ ) Let us suppose that ∀ x ∈ { , } n , ( f π ) (cid:104)| F | +1 (cid:105) ( x ) = y , then: f π ( y ) = f π (( f π ) (cid:104)| F | +1 (cid:105) ( x )) = ( f π ) (cid:104)| F | +2 (cid:105) ( x ) = ( f π ) (cid:104)| F | +1 (cid:105) ( f π ( x )) = y. Since f π ( y ) = y , then, y is a fixed point of f . (cid:50) Theorem 2.
Let F be a FVS of G ( f ) and P ⊆ F a PFVS of G ( f ) , Algorithm 2 finds the set of fixedpoints of f in time O (( | F | − | P | + 1) n | P | ) . Proof.
As the Algorithm 1, the correctness of this algorithm is based on the fact that its executiondemonstrates each of the necessary conditions according to the Proposition 2. The difference is thatthe way to find the fixed point of fa (for some a ∈ { , } | P | ) is done according to Proposition 4, i.e.instead of executing n times fa , fa π is executed | F | − | P | + 1 times. For this reason, the complexity ofthis algorithm is O (( | F | − | P | + 1) n | P | ). (cid:50) Algorithm 2:
FixedPoint
Input: f a BN with n components, F a FVS of G ( f ), P ⊆ F a PFVS of G ( f ), π an ordercompatible with F and P . Output: S the set of fixed points of f . m ← | F − P | ; S ← ∅ ; if P (cid:54) = ∅ thenforeach a ∈ { , } | P | do x ← ( fa π ) (cid:104) m +1 (cid:105) ( (cid:126) if ( fa ( x ) = x ) ∧ ( ∀ u ∈ P, f u ( x ) = a ( u )) then S ← S ∪ { x } ; else x ← ( f π ) (cid:104) m +1 (cid:105) ( (cid:126) if ( f ( x ) = x ) then S ← S ∪ { x } ; return S .2 Building a PFVS FixedPoint algorithm requires as input a FVS F and a PFVS P ⊆ F of the interaction graph of aBoolean network. It is known that the problems of finding a minimum FVS and a minimum PFVS ina signed digraph are both NP-complete [21, 32]. In this section we propose an polynomial algorithmthat allows to find a PFVS (not necessarily minimal) and a minimal FVS containing it.Let G ( f ) be the signed interaction graph of a BN f with n components. Given ( v , v , . . . , v n ) anorder over V ( G ( f )), Algorithm 3 classifies the vertices of G ( f ) in the following sets: • P : A set of vertices that is a PFVS of G ( f ). • O : A set of vertices such that P ∪ O is a minimal FVS of G ( f ). • R : The rest of the vertices of G ( f ).In addition, the algorithm considers the following auxiliar sets: • Y : A set of vertices that covers some circuit. • U : The vertices that have not yet been assigned to any set.The operation of the algorithm is as follows:First, we classify vertices with positive loop directly into P . Subsequently, the negative loops areremoved from the arcs of G ( f ).Then, the first phase begins. All vertices that are not in P or R , are incorporated in U to bevisited. Then, the first vertex in U (according to the input order) is incorporated into R (i.e., it isdiscarded from the PFVS) and if this vertex originally had a negative loop, it is incorporated into O (since that vertex has to be part of the FVS). Subsequently, for each vertex v ∈ U ∪ Y , G (cid:48)(cid:48) (thesubdigraph induced by R ∪ { v } ) is calculated, where, if G (cid:48)(cid:48) has a circuit, v is removed from U and from Y (since v covers some circuit in G ( f )), and then, if G (cid:48)(cid:48) has a positive circuit, v is incorporated in P (since v is the only unclassified vertex of the positive circuit of G (cid:48)(cid:48) ), and otherwise, v is incorporatedinto Y (since it covers some circuit, but no positive circuit is known, to add v to P ). Once U is empty,the first phase is terminated.If at the end of the first phase, there are vertices that are neither in P nor in R (for example,they are in Y ), a second phase begins. The only difference of this phase (and subsequent ones) withrespect to the first phase, is that if a vertex is sent to R it is automatically sent to O (since this phase,all unclassified vertices cover some circuit, but being discarded, it means that they do not cover anypositive circuit, so they are part of the FVS).If at the end of the second phase some unclassified vertices remain, a third phase begins under thesame conditions of the second phase, and so on.Notice that, the constructed PFVS is not minimal, since a vertex can be included in P becauseit covers a positive circuit (not necessarily a positive cycle), but the FVS is minimal, because thealgorithm works looking for cycles in G ( f ) − P − O , in this way, once we add a vertex to P ∪ O everycycle covered for this vertex it is eliminated of G ( f ) − P − O , so the only reason because a vertex isincluded in P ∪ O is that there exists a cycle that is not covered for another vertex in P ∪ O , therefore P ∪ O is a minimal FVS.The following algorithm is the way we implement the force-subroutine. A vertex v ∈ V ( G ( f )) isan ancestor of a vertex u if v ∈ R and exists a path P vu from v to u such that ∀ w ∈ P vu , w ∈ R .The set of ancestors of u is denoted by Anc( u ). A vertex v ∈ V ( G ( f )) is a descendant of a vertex u if v ∈ U ∪ Y and exists a path P vu such that ∀ w ∈ V ( P vu ) , w (cid:54) = u implies w ∈ R . The set of descendantsof u is denoted by Dec( u ). The sets Bef( u ) and af t ( u ) contain the elements related to u (the lastvertex added to R ) that can form a circuit. If there is an arc that goes from a vertex a ∈ Aft( u ) to avertex b ∈ Bef( u ), it means that there is a circuit formed by: • The path from b to u , 12 lgorithm 3: PFVS
Input:
The signed interaction graph G ( f ) of a BN f with n components and ( v , v , . . . , v n ) anordered sequence of V ( G ( f )). Output:
A PFVS P of G ( f ), and a FVS F of G ( f ) such that P ⊆ F . // Initialization V ⊕ ← { u ∈ V ( G ( f )) : ( u, u ) is a positive arc of G ( f ) } ; V (cid:9) ← { u ∈ V ( G ( f )) : ( u, u ) is a negative arc of G ( f ) } ; G (cid:48) ← G ( f ) − { ( u, u ) ∈ A ( G ( f )) : u ∈ V (cid:9) } ; P ← V ⊕ ; O ← ∅ ; R ← ∅ ;phase ← while ( P ∪ R ) (cid:54) = V ( G ( f )) do U ← V ( G ( f )) \ ( P ∪ R ); Y ← ∅ ; while U (cid:54) = ∅ do u ← vertex in U with lowest index; U ← U \ { u } ; R ← R ∪ { u } ; if ( phase > ∨ ( u ∈ V (cid:9) ) then O ← O ∪ { u } ; // Force-subroutine foreach v ∈ U ∪ Y do G (cid:48)(cid:48) ← G (cid:48) [ R ∪ { v } ]; if G (cid:48)(cid:48) has a circuit then U ← U \ { v } ; Y ← Y \ { v } ; if G (cid:48)(cid:48) has a positive circuit then P ← P ∪ { v } ; else Y ← Y ∪ { v } ; // end Force-subroutine phase ← phase + 1; F ← P ∪ O ; return P and F • The path from u to a and • The arc from a to b .The sign of this circuit will determine how the vertex a is classified (If the circuit is positive, a is addedto P . If the circuit is negative, a is added to Y ).The sets Anc( u ) , Dec( u ) , Bef( u ) and Aft( u ) contain elements of the form ( v, σ ), where v is a vertex,and σ is a sign (+1 or − x be an element of any of these sets, we will denote v ( x ) to the vertexassociated with the element x , and we will denote σ ( x ) to the sign associated with the element x . Different orders of the vertices as input of PFVS algorithm can generate different PFVS as outputs,so it may be necessary to define some heuristics to choose an appropriate order of the vertices. In thispaper we use random order and Min-order defined as follows: • Min-order: the vertices are ordered according:1. Their degrees, from lowest to highest.2. If two or more vertices have equal degree then by in-degree, from lowest to highest.Figure 6 shows an example Min-order. 13 lgorithm 4:
Force-subroutine // During initialization foreach v ∈ V ( G ( f )) do Anc( v ) ← ∅ ; Dec( v ) ← ∅ ; // Force-subroutine, ‘‘u’’ is the selected vertex in PFVS-Algorithm Bef( u ) ← Anc( u ) ∪ { ( u, +1) } ;Aft( u ) ← ∅ ; foreach v ∈ N + ( u ) doif v ∈ U then Aft( u ) ← Aft( u ) ∪ { ( v, σ ( u, v )) } ; else Aft( u ) ← Aft( u ) ∪ Dec( v ); foreach ( a, b ) ∈ Aft( u ) × Bef( u ) doif ( v ( a ) , v ( b )) ∈ A ( G ) then U ← U \ { a } ; σ ← σ ( a ) · σ ( b ) · σ ( v ( a ) , v ( b )); if σ = + then P ← P ∪ { d } ; else Y ← Y ∪ { d } ;Dec( b ) ← Dec( b ) ∪ { ( a, σ d · σ a ) } ;Anc( a ) ← Anc( a ) ∪ { ( b, σ d · σ a ) } ; else Dec( b ) ← Dec( b ) ∪ { ( a, σ d · σ a ) } ;Anc( a ) ← Anc( a ) ∪ { ( b, σ d · σ a ) } ;
14 365 7 2 { , , , , , , } Figure 6: A signed interaction graph G ( f ) and the order of its vertices according to the Min-order. We first tested the running time of FixedPoint algorithm in random BNs with a fixed transversalnumber τ and with different numbers of components n and positive transversal numbers τ + . Thetested networks were constructed in such a way that a minimum FVS and a minimum PFVS wereboth known a priori, i.e. we did not use PFVS algorithm in these cases. The average times obtainedin milliseconds with one hundred networks tested in each case, are shown in Figures 7 to 9. We cansee that the time performance of FixedPoint algorithm grows exponentially with the value of τ + andpolynomially with the size of the network.The results of PFVS algorithm using different heuristic to chose the order of the vertices arepresented in Figure 10. We can see that running time of the algorithm depends mainly on the size ofthe network and not on the sizes of the positive feedback vertex sets. Besides, in random networks aswell as in networks from the literature described in Table 1 the performance of the min-order heuristicis better than the random one.On the other hand, the correctness and time performance of FixedPoint algorithm and PFVSalgorithm were also tested with networks from the literature where the sets of fixed points obtained14 or different network sizes ( τ = 15)Time performance of FixedPoint algorithm τ + = 15 τ + = 10 τ + = 5 Figure 7: Results of FixedPoint algorithm .15 or different PFVS sizes ( τ = 15)Time performance of FixedPoint algorithm n = 500 n = 300 n = 100 Figure 8: Results of FixedPoint algorithm .16 or different PFVS sizes ( n = 300 , τ = 30)Time performance of FixedPoint algorithm Figure 9: Results of FixedPoint algorithm .coincides with those published. The results of these tests, shown in Table 1, were obtained witha PC 3.60GHz Intel Core i7 processor with 16GB of RAM and can be directly checked with theimplementation FixedPointsCollector located at . We canobserve that the running times obtained in these networks are short, probably due to the small size oftheir positive feedback vertex sets. The latter is also observed in another family of published networks,which are shown in Table 2, where the times obtained (using the same PC) are shorter than thoseobtained by the Veliz-Cuba algorithm in [51] for networks with a small τ + , however the same does notoccur when τ + is large. 17 or different network sizes ( τ = 15 , τ + = 5)Time performance of PFVS algorithm minrand Figure 10: Results of PFVS algorithm .Table 1: Performance of FixedPoint algorithm in real networks.
Network n FP | P | T(min) T(rand)
Cancer cell [53] 8 5 4 6.952 8.248Cancer backbone [53] 8 12 5 7.019 7.321Budding yeast [27] 11 7 6 5.963 7.036Fission yeast [13] 10 12 4 8.980 9.416Arabidopsis Thaliana [44] 13 10 7 8.938 8.919T-helper cell [31] 23 3 3 7.407 13.512T-cell receptor [25] 40 1 1 12.782 29.895HGF [45] 66 2 3 17.286 61.969Apostosis model 1 [26] 12 2 2 6.923 8.181Apostosis model 2 [26] 12 4 2 6.870 9.170Apostosis model 3 [26] 3 4 2 5.143 6.174Drosophila melanogaster [5] 60 10 9 19.140 64.565Ventral spinal cord [28] 8 5 4 6.751 7.126 | P | is the size the PFVS obtained with min-order.T(min) and T(rand) correspond to the time in milliseconds of the PFVS-Alg.+FP-Alg. using Min-order and random-order (in this case, its considered the execution of n/ Veliz-Cuba Our algorithmNetwork n | P | mean stdev. mean stdev. HGF Signaling in Keratinocytes [45] 68 3 2.347 0.299 0.451 0.079T Cell Receptor Signaling [42] 101 1 2.355 0.337 0.347 0.054Signaling in Macrophage Activation [33] 321 3 2.406 0.298 0.464 0.044Yeast Apostosis [24] 73 3 2.370 0.429 0.378 0.042Influenza A Virus Replication Cycle [29] 131 12 3.599 0.897 0.407 0.044T-LGL Survival network (v. 2011) [41] 60 10 3.152 0.786 0.580 0.041T-LGL Survival network (v. 2008) [55] 61 10 1.790 1.181 0.409 0.044EGFR & ErbB Signaling [43] 104 2 1.185 0.910 0.356 0.041Signal Transduction in Fibroblasts [18] 139 39 23,19* 98.42 DNF DNFErbB (1-4) Receptor Signaling [17] 247 42 4186* 12284 DNF DNF | P | is the size the PFVS obtained with min-order. Times presented in seconds.*Times obtained from [51]. DNF: Tests that did not finish within 3 days In the modeling of biological systems by Boolean networks usually the interaction graph of the networkis known as well as the type of interaction between their components (activation, inhibition). In thisway, it seems natural to use this information to determine the fixed points of a network. In this paper,we have constructed FixedPoint algorithm to find the set of fixed points of a Boolean network basedmainly on the structure of the positive cycles of its interaction graph and to a lesser extent on thesize of the network. This can be considered an improvement to the results obtained by Akutsu et al.in [3], since the set of states to test is smaller and, in the case that τ = τ + , our algorithm works thesame way. The theoretical foundation of the algorithm is given by Theorem 1, which provides a nicecharacterization of the dynamical behavior of Boolean networks without positive cycles and with afixed point.The efficiency of FixedPoint algorithm depends mainly on the size of the input PFVS. Due to this,it is important to have a PFVS as close to the minimum as possible. In this sense, it is clear that PFVSalgorithm can be improved. One improvement could be done implementing an efficient algorithm todecide the existence of a positive cycle in a signed digraph. This latter problem, which is equivalent tothe existence of an even cycle in a digraph [32], is a surprisingly difficult problem and whose algorithmiccomplexity was unknown for a long time. Although Robertson, Seymour and Thomas finally provedthat this decision problem can be solved in polynomial time [39], the implementation of such algorithmis not easy to do which makes it one of our future challenges. Another way to improve PFVS algorithmis the possibility of building a fixed-parameter tractable algorithm to determine a PFVS of minimumsize, i.e. an algorithm whose complexity depends mainly on this size and not on the size of theinteraction graph. It is known that this type of algorithm exists to solve the minimum FVS problemin digraphs [12]. However, it is an open problem, at our knowledge, in the case of minimum PFVS insigned digraphs.In future work it would be important to explore methods to reduce the size of a network thatconserve τ + . Also, since the efficiency of the FixedPoint algorithm depends on the size of a input PFVS,it is important to study the relationship between τ + and other parameters of the interaction graph ofa Boolean network such as: minimum and maximum in-degree, out-degree and degree distributions. Acknowledgements
We would like to thank Adrien Richard for his review and suggestions for improving the manuscript.19 unding
J. Aracena was partially supported by CONICYT-Chile through the project AFB170001 of the PIAProgram:
Concurso Apoyo a Centros Cient´ıficos y Tecnol´ogicos de Excelencia con FinanciamientoBasal . L. Cabrera-Crot is funded by CONICYT-PCHA/Doctorado Nacional/2016-21160885.
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A Example of the algorithms
Example 6.
Figures 11 and 12 show the operation of the FixedPoint-Algorithm with a Booleannetwork with τ + = 1 and τ = 3. Note that in the tables in steps 3 and 5, the purple cells representthe components that are fixed according to Definition 123 . Given the following network, where { , , } is a FVS and { } is a PFVS: f = x ∨ x f = x f = ( x ∧ x ) ∨ ( x ∧ x ) ∨ ( x ∧ x ) f = x ∨ x f = x f = x ∨ x f = x ∨ x f = x ∧ x
2. We choice an order compatible with the FVS and PFVS selected.In this case, we choice π = (3 , , , , , , , .
3. If a (3) = 0 , the execution of ( fa ) π would be the following: x x x x x x x x ( fa ) π fa ) π ) fa ) π )
4. Since ( fa )(10010000) = 10010000 and f (1001000) = 0 = a (3) , then is a fixed point of f Figure 11: Example of ( fa ) π execution. Example 7.
Figures 13 and 14 show the operation of the PFVS-Algorithm. Green vertices representvertices in P , red vertices represent vertices in R \ O , yellow vertices represent vertices in Y , orangevertices represent vertices in R and O and white vertices represent vertices in U .24 . If a (3) = 1 , the execution of ( fa ) π would be the following: x x x x x x x x ( fa ) π fa ) π ) fa ) π )
6. Since ( fa )(00101111) = 11101110 , does not exist fixed point of f such that f ( x ) = 1Figure 12: Example of ( fa ) π execution (continuation).25 . Given a labeled digraphand an order over its ver-tices (3,7,2,5,1,4,8,6). Allvertices are added to U .
2. The vertex in U withlower index (the vertex 3)is discarded from PFVS(added to R ). Since ∀ v ∈ U ∪ Y , the subgraph in-duced by R ∪ { v } has nocircuits, a new vertex isselected.
3. The next vertex (7)is discarded from PFVS.Since the subgraph inducedby R ∪ { } has a negativecircuit, (1) is added to Y .Similarly, (8) is added to Y .
4. The next vertex (2) isdiscarded from PFVS. (4)is added to Y .
456 37 21 8 56 37 21 8 4
5. The next vertex (5)is discarded from PFVS.Since the subgraph inducedby R ∪ { } has a positivecircuit, (6) is included inthe PFVS (added to G ).Since U is empty, the firststep is finished. Figure 13: Example of PFVS Algorithm (First step).26 . For the second step(and following), all ver-tices in Y are added to U .
7. The vertex in U with lower index (1) isdiscarded from the PFVS(added to R and O ). Sincethe subgraph induced by R ∪ { } has a positive cir-cuit (1,7,8,7,1), (8) is in-cluded in the PFVS.
8. The next vertex (4)is discarded from PFVS.Since U is empty, the sec-ond step is finished. Sinceno vertex in Y , the algo-rithm ends.
23 57 6 81 423 57 6 81 4