Monomial Dynamical Systems over Finite Fields

Transcription

1 Monomial Dynamical Systems over Finite Fields Omar Colón-Reyes Dissertation submitted to the Faculty of the Virginia Polytechnic Institute and State University in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Mathematics Reinhard Laubenbacher, Chair Christopher Beattie Ezra Brown Edward Green Peter Linnell April 28, 2005 Blacksburg, Virginia Keywords: polynomial dynamical system, finite field, state space structure Copyright 2005, Omar Colón-Reyes

2 Monomial Dynamical Systems over Finite Fields Omar Colón-Reyes (ABSTRACT) Linking the structure of a system with its dynamics is an important problem in the theory of finite dynamical systems. For monomial dynamical systems, that is, a system that can be described by monomials, information about the limit cycles can be obtained from the monomials themselves. In particular, this work contains sufficient and necessary conditions for a monomial dynamical system to have only fixed points as limit cycles.

3 Dedication To: God To my family: Margarita Reyes-Rosario, Anibal Colón-Negron and Anibal Jr. Colón-Reyes. To the ones that kept me sane: Jose Osvaldo Rodriguez-Rosado, Reinhard Laubenbacher, Trisha Townsend, Kalota Stewart, Brandy Stigler, Dustin Potter, Olgamary Rivera Marrero, Hannah Swiger, Jose Maria Menendez, Juan Ariel Ortiz-Navarro and Marilyn Arzuaga. For the ones that made me stronger: Rebecca E. Pablo, Richard Byrd and John W. Massey. iii

4 Acknowledgments I would like to thank Reinhard Laubenbacher, my advisor, and Abdul Jarrah, for all their help, guidance, time and patience. It was all greatly appreciated. I would also thank Bodo Pareigis, for teaching me so many things and Bernd Sturmfels for his suggestions. Finally I would like to thank the faculty and the staff in the Math Department at Virginia Tech, especially Hannah Swiger. iv

5 Contents 1 Introduction 1 2 Boolean Dynamical Systems Preliminaries on Systems and Dependency Graphs Thestructureofthedependencygraph Thestatespaceofaconnectedcomponent Loopnumbers Fixed points and cycles of strongly connected components Glueing Booleanmonomialfixedpointsystems Monomial Dynamical Systems over Finite Fields Thelogmapandthesupportmap v

6 3.2 Linearsystemsovercommutativerings Thestatespaceoflinearmapsoverrings Linear maps over Z p r Conclusions and Summary 59 vi

7 List of Figures 1.1 State space of f =(x 1 + x 2,x 1 + x 2 ):F 2 2 F State space of f =(x 2,x 3,x 1 ):F 2 2 F Dependency graph of f =(x 1,x 1 x 2 x 3,x 1 ):F 3 2 F Dependency graph of f =(x 3,x 1 x 4,x 4,x 1 ):F 4 2 F Dependency graph of f =(x 5,x 1,x 2,x 6,x 4,x 7,x 8,x 9,x 10,x 11,x 12,x 13,x 5 ) Dependency graph of f =(x 4,x 5,x 4,x 7,x 6,x 3,x 1 x 2 ):F 7 2 F State space of f =(x 2,x 3,x 1 ):F 3 2 F State space of f =(x 4,x 5,x 4,x 7,x 6,x 3,x 1 x 2 ):F 7 2 F Dependency graph of f =(x 2,x 4,x 1 x 2,x 3,x 5 x 4 ):F 5 2 F Dependency graph of f = x 1 : F 2 F Dependency graph of g =(x 2,x 4,x 1 x 2,x 3 ):F 4 2 F State space of h =(x 2,x 4,x 1 x 2,x 3,x 4 ):F 5 2 F vii

8 2.11 Dependency graph of f =(x 2,x 3 x 1,x 1 x 4, 0) : F 4 2 F State space of f =(x 2,x 3 x 1,x 1 x 4, 0) : F 4 2 F Dependency graph of f =(x 1,x 1 x 2 ):F 2 2 F State space of f =(x 1,x 1 x 2 ):F 2 2 F State space of f =(x 1,x 2 1x 2 ):F 2 3 F State space of log(f), f =(x 1, 2x 1 + x 2 ):F 2 2 F State space of the support of f, S f =(x 1,x 1 x 2 ):F 2 2 F State space of f =4x 1 : Z 6 Z State space of g =(0,x 2 ):Z 2 Z 3 Z 2 Z State space of f =2x 1 : Z 6 Z State space of Π 1 =2x 1 : Z 2 Z 2 and state space of Π 2 =2x 1 : Z 3 Z State space of f(x 1,x 2 )=(x 2 1x 2,x 1 ):Z 2 6 Z State space of Π 1 =(x 1 x 2,x 1 ):Z 2 2 Z State space of Π 1 =(x 2 1x 2,x 1 ):Z 2 3 Z State space of g 1 =2x 1 : Z 3 Z State space of g 2 =2x 1 : Z 9 Z Dependency graph of the support of f =(x 4 1x 2,x 2 x 4 3,x 4 1):F 3 5 F viii

9 4.2 State space of the support of f =(x 2 1x 2,x 2 x 2 3,x 3 1):F 3 5 F ix

10 Chapter 1 Introduction Linking the structure of a system with its dynamics is an important problem in the theory of finite dynamical systems. Let X be a finite set and let f : X X be a set map. Such a map is called a finite dynamical system, see [13]. In this study, due to applications, we restrict ourselves to the case X = k n = Fq n,wheref q is a finite field with q elements. We are interested in the study of the dynamics of the map f : X X. The dynamics is given by iterations of f, with the variables being updated simultaneously. In this work we present a family of non-linear systems from which we can obtain information about the dynamics of the family from the structure of the function. The dynamics of f is represented by the state space of f, denoted S(f), which consists of a directed graph. The vertex set of S(f) isf n q. We define a directed edge a b if f(a) =b. We say that the state space of f, S(f) contains a limit cycle of length t>0ifthereisa F n q 1

11 Chapter 1. Introduction 2 such that f t (a) =a and there is no smaller t such that f t (a) =a. Equivalently, we say that there is a limit cycle in S(f) oflengthn if there exists a closed loop of minimum length n, based at a in S(f). Figure 1.1 shows a trivial cycle (cycle of length 1), based on the element a =(0, 0). Example Let f =(x 1 + x 2,x 1 + x 2 ):F 2 2 F 2 2. The state space of f is in Figure Figure 1.1: State space of f =(x 1 + x 2,x 1 + x 2 ):F 2 2 F 2 2. Remark- The figures on the Examples in this study were created with the DVD software developed by the Applied Discrete Mathematics Group at the Virginia Bioinformatic Institute. In Figure 1.2 we can see the state space of the map f =(x 2,x 3,x 1 ):F 3 2 F 3 2,which contains two limit cycles of length 2 and two trivial cycles. Every map from a finite field to itself can be written as a polynomial, therefore f can be written as a tuple of n polynomials, f =(f 1,..., f n ), where each f i is in the ring k[x 1,..., x n ]

12 Chapter 1. Introduction Figure 1.2: State space of f =(x 2,x 3,x 1 ):F 2 2 F 3 2. [12]. The problem we are facing then becomes one of establishing conclusions about the structure of the state space S(f) from the structure of the polynomials f i. In this work we are particularly interested in fixed point systems, (FPS), that is finite dynamical systems in which all limit cycles have length 1. The map f on Example is an Example of a FPS. The following is a brief survey of existing results. For all linear systems over an arbitrary finite field, we can represent such a system as a matrix A, then the structure of the state space of f can be determined from the factorization of the characteristic polynomial of A. Moreover, results from Hernández, [10] establish that if f is a linear map, f : F n q F n q then f is a fixed point system if and only if the characteristic polynomial of the matrix that f represents can be written in the form x n 0 (x 1) m and the matrix is annihilated by x n 0 (x 1). Example For f in Example 1.0.1: A = , and the characteristic polynomial of A is x 2. Moreover A vanishes on x 2 (x 1) as well.

13 Chapter 1. Introduction 4 For two-dimensional linear systems, Brown and Vaughan [2] determine the number of limit cycles for systems over more general rings than finite fields. If A is the matrix that represents f, then Brown and Vaughan determine the number of limit cycles when f : Zm 2 Zm 2 and m =gcd(m, det(a)) = 1. When the polynomials f i have a constant c different than zero, the system f =(f 1,..., f n ) is then called an affine system. The structure of the limit cycles of such systems have been determined by Milligan and Wilson, [15]. For non-linear systems an approach was proposed by P. Cull [8]. In Cull s study a general nonlinear system is embedded in a linear one and then, using results similar to the ones published in B. Elpas [9] and R. Hernández [10], we can obtain the number and length of the limit cycles. The difficulty with this method is that if the nonlinear system has dimension n and the field has q elements, then the linear system has dimension q n. The direct effect of the specific non-linear functions on the state space structure of f is also very difficult to see. Results concerning non-linear systems are rare. In Wolfram s work [16], some nonlinear one-dimensional cellular automata results can be found. For sequentially updated systems Barret, et. al. have shown that updated Boolean NOR systems do not have any fixed points [3] and [4]. When we view the univariate polynomials over Q p as a one-dimensional dynamical system, including the case of monomials, the problem of obtaining information about the dynamics from the structure of the function has been studied extensively over the p-adic numbers by

14 Chapter 1. Introduction 5 A. Khrennikov and M. Nilsson [11]. Mullen and Lidl, [14] considered one dimensional dynamical systems. They called such systems permutation polynomials. These dynamicals system are called invertible system, that is dynamical system whose state space contains only cycles. When the function is a monomial, then necessary and sufficient conditions for such monomial to be a permutation polynomial were given, [14]. For a family of binomials, Colón-Reyes and Cáceres, [5], provided necessary and sufficient conditions for such family to be permutation binomials. The general case for binomials is still open. In this work we concentrate on FPS. Moreover, we provide necessary and sufficient conditions for a monomial dynamical system to be a FPS. By a monomial dynamical system we mean a function f =(f 1,..., f n ) in which each f i is a monomial. Chapter 2 of this work deals with this problem over the field with 2 elements. We associate a directed graph to the map f, it its dependency graph D(f). Our main result is that the cycle structure of the state space S(f) can be determined from D(f). The key role is played by a numerical invariant associated with a strongly connected directed graph, that is, a graph in which there exists a walk (a directed path) between any two vertices. For such a graph one can define its loop number as the minimum of the distances of two walks from some vertex to itself. It turns out that the dependency graph of a monomial system can be decomposed into strongly connected components whose loop numbers determine whether the system has only fixed points. If the loop number of every strongly connected component is one, then the state space has only trivial limit cycles, that is, f is a fixed point system.

15 Chapter 1. Introduction 6 In chapter 3 we deal with the same problem but over arbitrary finite fields. Given a monomial dynamical system, f : F n q F n q, we associate two maps with f: thelogmap,log(f) and the support map, S f. These two maps are defined exclusively from the structure of the monomials f i and it is shown that f is a FPS if and only if both maps, log(f) ands f,are FPS. As will be shown later, the log map is a linear map over a ring of size q 1, and the support map is a boolean monomial dynamical system. A criterion to determine when the support map is a FPS can be found in Chapter 2. We dedicate most of Chapter 3 to provide necessary and sufficient conditions for the log map to be a FPS.

16 Chapter 2 Boolean Dynamical Systems The results we are going to present in this chapter have already been published in O. Colón- Reyes et. al. [7]. In this chapter we deal with monomial dynamical systems over the finite field with two elements, F 2. Our main result in this chapter is that we provide a sufficient condition for a monomial dynamical system to be a fixed point system. The key role is played by a numerical invariant, called the loop number, which is associated with the dependency graph of f. We will associate a directed graph called the dependency graph of f and we will focus on the strongly connected components of the dependency graph. By strongly connected components we mean a subgraph in which we can always find a directed path between any two vertices. We will prove that if the dependency graph of f is strongly connected and the loop number is 1, then this is enough to conclude that f is a fixed point system. 7

17 Chapter 2. Boolean Dynamical Systems 8 Because the dependency graph is not always strongly connected, we define an operation among graphs called glueing. One of our main results establishes that if we can glue together two dependency graphs that are strongly connected and each dependency graph corresponds to a fixed point system, then the glueing provides a dependency graph for a fixed point system as well. Therefore, if the dependency graph of a monomial dynamical system consists of strongly connected components that are glued together, each component having the loop number 1, then f is a fixed point system as well. 2.1 Preliminaries on Systems and Dependency Graphs We follow [7] to define a monomial dynamical system. Let f = (f 1,..., f n ) be an n- dimensional Boolean monomial system, where the monomials are functions of the form f i = α i x ɛ 1i 1 x ɛ 2i 2 x ɛ ni n with α i {0, 1} and ɛ ij {0, 1}. If α i =0wesetallɛ ji =0. Letf m be the m-fold composition of the map f with itself. We write f m =(f m 1,f m 2,..., f m n). By definition we have f m i = α i (f m 1 1 ) ɛ i1 (f m 1 n ) ɛ ni. Definition With f we associate a digraph X, called dependency graph, with vertex

18 Chapter 2. Boolean Dynamical Systems 9 set {a 1,..., a n,ɛ}. There is a directed edge from a i to a j if α i =1and x j is a factor of f i. There is a directed edge from a i to ɛ if α i =0. Self loops a i a i are allowed. Such loops occur if f i has the factor x i. If there is an edge a i ɛ (f i = 0) then there is no edge a i a j for any j. It is evident that the monomial system f is completely described by the dependency graph X. Example Consider f(x 1,x 2,x 3 )=(x 1,x 1 x 2 x 3,x 1 ):F 3 2 F 3 2. Figure 2.1: Dependency graph of f =(x 1,x 1 x 2 x 3,x 1 ):F 3 2 F 3 2. Example The map f(x 3,x 1 x 4,x 4,x 1 ) has the dependency graph shown in Figure 2.2 Figure 2.2: Dependency graph of f =(x 3,x 1 x 4,x 4,x 1 ):F 4 2 F 4 2. The following proposition establishes a relationship between the factors of the monomials f m i and paths of certain lengths in the dependency graph of f.

19 Chapter 2. Boolean Dynamical Systems 10 Proposition Let X be the dependency graph of f and assume f m i 0. There exists a walk p : a i a j of length m in X if and only if f m i contains the factor x j. Proof. Assume that there are directed edges from a i to a j1,...,a jt. This means that: f i = x j1 x j2 x jt. Any walk leaving a i goes through one of the a j1,...,a jt as a first step. Hence f 2 i = γ i f j1 f j2 f jt so f m i = α i f m 1 j 1...f m 1 j t. The claim now follows by induction on m and the observation that x 2 j = x j. For the converse observe that from f m i = α i f m 1 j 1...f m 1 j t it follows that, if x j is a factor of f m i, it must also be a factor of some f m 1 j k.wecanthen again proceed by induction to get a walk of length m from a i to a j.qed Corollary f r i is the product of all functions f r s j for all walks p : a i a j of length s r. Proof. This follows by induction, as in Proposition QED. Corollary If there is a walk p : a i a j of length less than r, andiff 1 j = f j =0, then the function f r i is zero. Example In Figure 2.1 we can see a path of length 2 from x 2 and x 1. Notice that f 2 2 = x 2 x 3 x 1 is divided by x 2.

20 Chapter 2. Boolean Dynamical Systems 11 As Example shows, there is a relationship between walks in the dependency graph and the composition of the map f with itself. 2.2 The structure of the dependency graph Since we are interested in identifying cycles in the state space of f, and cycles are closed loops, we are then interested in closed walks in the dependency graph of f. Therefore, we are going to assign an equivalence relation to the dependency graph in which two vertices a and b are equivalent if a and b are in a strongly connected component of X f. Definition Let X be the dependency graph of f. 1. We write a X for a V X \{ε}. 2. For vertices a, b X,wecalla and b strongly connected, andwritea b, if and only if there is a walk p : a b and a walk q : b a. Observe that there is always a walk of length zero (the empty walk) from a to a. Thena b is an equivalence relation on V X \{ε}, called strong equivalence. 3. The equivalence class of a Xis called a (strongly) connected component and is denoted by ā. LetE(X ) be the set of equivalence classes. 4. A vertex a with an edge a ε is called a zero. 5. For a, b X,letp : a b be a walk. We denote the length of the walk p by p.

21 Chapter 2. Boolean Dynamical Systems 12 Example The dependency graph in Figure 2.1 partitions into the equivalence classes ā 1, ā 2, ā 3. The dependency graph in Figure 2.2 has equivalence classes ā 2 and ā 3. Notice that we can enumerate the smallest strongly connected components: 1. If f i = 1, then the vertex a i defines a one element strongly connected component which only contains the empty walk. 2. If f i = 0, then there exists an edge a i ɛ, anda i defines a one element strongly connected component. 3. If f i = x i, then we again have only one strongly connected component with loops of any length based on a i. Lemma Define ā b if and only if there is a walk from a to b. Then E(X ) is a poset. Proof. Observe that, given a, a ā and b, b b, there is a walk from a to b if and only if there is a walk from a to b. Ifthereisawalkfroma to b and a walk from b to a, thena and b are strongly connected and thus ā = b. QED. Observe that a i defines a one point connected component ā i if and only if f i =0, 1, or f i = x i. This partial order will come up again in the discussion of glueing. It covers all the edges that do not occur in strongly connected components.

22 Chapter 2. Boolean Dynamical Systems The state space of a connected component In this section we will study strongly connected components. The study of the relations between connected components that lead to the poset structure of E(X ) will be studied in the next section. First we examine trivial strongly connected components, so we do not have to worry about them later on. If X = {a}, witha a zero, then f m = 0 for every m 1. If the dependency graph X consists of only one element, X = {a} and f =1,thenf m = 1 for every m 1. If X is the dependency graph of the dynamical system f : F r 2 F r 2 and if a i is a vertex in X, such that there exists a walk from a i to zero, a j ɛ of length n 0, then by Corollary we have f n+m i = 0 for every m 1. Consider a strongly connected component in X, called ā. If there exists a walk of length n from any element in ā to a zero a j ɛ, then for all a i equivalent to a we have that f n+m i =0 for all m 1. In particular, if X is strongly connected, then we have only one equivalence class and f isafixedpointsystemwiththeonlyfixedpoint(0,..., 0). Therefore, for the rest of Section 2.3 we can assume that X is strongly connected and we can exclude the cases f(x 1 )=1andf(x 1 )= Loop numbers Definition The loop number of a X is the minimum of all numbers t 1 with t = p q for all closed walks p, q : a a. If there is no closed walk from a to a then we

23 Chapter 2. Boolean Dynamical Systems 14 set the loop number to be zero. This last case occurs only if ā = {a} and there is no edge from a to a (i.e. f =0, 1, but we have excluded f =0.). Example The dependency graph illustrated in Figure 2.1 has loop number 1. The one in Figure 2.2 has two strongly connected components. One component has loop number 0 and the other has loop number 3. If there is a loop of length 1, p : a a then the loop number of a is p 2 = pp p =1. If we choose a, b X we will show that the loop number is invariant under the choice of a or b, i.e. the loop numbers of a and of b are equal, if X is strongly connected. Lemma The loop number is constant on any strongly connected X, so the loop number of a strongly connected X is a well defined number L(X ). Proof. Let p : a b and q : b a be walks. Then p pq,p qq : b b are closed walks with p pq p qq = t, so the loop number of b is less than or equal to the loop number of a. By symmetry the loop numbers of a and b are equal. Hence the loop number is constant on X. QED. The following lemma establishes a relationship between the loop number and the length of the walks on X. Lemma Let the loop number of X be t. Let p : a i a j and q : a i a j be walks. Then p q (t) Z.

24 Chapter 2. Boolean Dynamical Systems 15 Proof. Assume p > q and let p q = rt + s with 0 s<t.wewanttoshows =0. Let p, q : a i a i be such that q p = t. Wehaver 0. Then p p q q = p + p q q = rt + s t =(r 1)t + s. Hence there are walks p,q : a i a j with p q = s. Letp : a j a i be a walk. Then p p p q = s = 0 because of the minimality of the loop number t. So p q (t). QED. Corollary Let the loop number of X be t. Let p : a a be a closed walk. Then p (t). Proof. In Lemma take p and pp. QED. Proposition Let the loop number of X be t 1. For each a, b X there exists an m N, wheren denotes the natural numbers, and walks p i : a b of length p i = m + i t for all i N. Proof. Let p, q : a a be closed walks with q p = t. By Corollary p is divisible by t. Let r := p /t. Let p : a b beawalkoflengths := p. Let m := s +(r 2 r)t. Write i N as i = jr + k with 0 k<r. Then the composition of walks p q k p r 1+j k : a b

25 Chapter 2. Boolean Dynamical Systems 16 has length p q k p r 1+j k = s + k(r +1)t +(r 1+j k)rt = s +(kr + k + r 2 r + jr kr)t = s +(r 2 r)t +(jr + k)t = m + i t QED Fixed points and cycles of strongly connected components We are interested in cycles in the state space of f, moreover we want to link the loop number with the length of such cycles. Lemma For a i,a j X we define a i a j : walk p : a i a j with p (t). This is an equivalence relation, called loop equivalence. Proof. We have a i a i with a walk of length zero and thus reflexivity. Transitivity is trivial. For symmetry let a i a j, then there is a walk p : a i a j with p (t). Let q : a j a i be any walk. Then qp : a i a i is a walk with qp = q + p (t) by Corollary 2.3.5, hence q (t) and thus a j a i. QED. Lemma There are exactly t loop equivalence classes in X.

26 Chapter 2. Boolean Dynamical Systems 17 Proof. Since there are walks starting at a i for all lengths 0, take a walk a i a i+1... a i+t of length t. Thea i+j, j =0,...,t 1 are in different equivalence classes ā i, ā i+1,...,ā i+t = ā i by Lemma Every a k is in one of these equivalence classes, since there is a walk q : a i+t a k of length q = rt+ s, sothereisawalkq : a i+s a k of length q (t), hence a i+s a k. QED. We may now label and enumerate the vertices of X in the following way (a 1,...,a i1 ), (a i1 +1,...,a i2 ),...,(a it 1 +1,...,a it ) where each group (a ij +1,...,a ij+1 ) is a loop equivalence class and there is an edge a ij a ij +1 for all j =1,...,t 1. If s j is the number of elements in each loop class, then t i=1 s i = n. Example Consider the connectivity graph in Figure 2.3 for the function f =(x 5,x 1,x 2,x 6,x 4,x 7,x 8, The loop number is 5, and then we can enumerate the vertices of X in the following way: Figure 2.3: Dependency graph of f =(x 5,x 1,x 2,x 6,x 4,x 7,x 8,x 9,x 10,x 11,x 12,x 13,x 5 ). (x 1,x 8,x 13 ), (x 5,x 9 ), (x 4,x 10 ), (x 6,x 11,x 3 ), (x 2,x 12,x 7 )

27 Chapter 2. Boolean Dynamical Systems 18 Corollary Let a X and let {a 1,...,a u } be the loop class of a (with a 1 = a). Then there is an m N such that for all j =1,...,u and all i N there is a walk a a j of length (m + i)t. Proof. Use Proposition to obtain m j and walks p ij : a a j of lengths m j + it for all i N and all j = 1,...,u. By Lemma each m j is divisible by t. Take m := max{m 1,...,m u }/t. QED. Before we study the fixed points and cycles of an arbitrary connected dependency graph we look at a simple Example. Proposition The state space of a directed t-gon is isomorphic to the set of orbits of the action of the cyclic group of order t acting on F t 2,thet-dimensional hypercube, by cyclically exchanging the canonical basis vectors. Proof. The dynamical system f of a directed t-gon is f =(x 2,x 3,...,x 1 ):F2 t F2.Every t element v of F2 t can be written as t 2 v = λ ij e i i=1 j=1 where e i are the canonical vectors, and λ ij F 2. Notice then that for t>0: t 2 t 2 f t (v) = λ ij f t (e i )= λ ij e i+1 i=1 j=1 i=1 j=1 Therefore if there exists a cycle of length t 0 > 0 based on v, we have the following cycle on the state space of f: t 0 2 v = λ ij e i i=1 j=1 f t 0 2 λ ij e i+1 i=1 j=1 f f t 0 2 f λ ij e i+t0 1 i=1 j=1 t 0 2 λ ij e i = v i=1 j=1

28 Chapter 2. Boolean Dynamical Systems 19 Definenowtheactionφ : C t F t 2 F t 2 of C t, a cyclic group of order t, on the hypercube F t 2 as: t 2 φ(z k,w)= λ ij e σ k (i) i=1 j=1 where σ =(1, 2,..., t) is a permutation in the symmetric group S t. Notice that φ(z k,w)= f k (w), and there is a one-to-one correspondence between orbits and cycles. Therefore the state space of f is isomorphic to the orbits of the action φ. QED. Example Consider the function f =(x 2,x 3,x 1 ):F 3 2 F 3 2. The state space of f is illustrated in Figure 2.5. The orbits under the action φ are: {(0, 0, 0}, {(0, 0, 1), (0, 1, 0), (1, 0, 0)}, {(0, 1, 1), (1, 1, 0), (1, 0, 1)}, {(1, 1, 1)}. Theorem Let X be strongly connected with loop number t 1 and n vertices. Let (a 1,...,a i1 ), (a i1 +1,...,a i2 ),...,(a it 1 +1,...,a it ) be the enumeration of vertices of X as described above. Then there is an m such that f mt =(y 1,...,y }{{ 1,y } 2,...,y 2,...,y }{{} t,...,y }{{} t s 1 times s 2 times s t times ) where y 1 = x 1... x i1, y 2 = x i x i2,..., y t = x it x it. Furthermore f mt+1 =(y 2,...,y }{{ 2,y } 3,...,y 3,...,y }{{} 1,...,y 1 ). }{{} s 1 times s 2 times s t times.

29 Chapter 2. Boolean Dynamical Systems 20 f mt+j =(y j+1,...,y j+1,y j+2,...,y j+2,...,y j,...,y t ). }{{}}{{}}{{} s 1 times s 2 times s t times Proof. For each loop class c we use some m c from Corollary and compute the values f mt i for all i. By Proposition we get f mt i = x 1...x u for all i =1,...,u. If we use the maximum m of all such m c, then we have the structure of f mt. To determine f mt+1 i observe that for i =1,...,i 1 there are walks a i a j of length mt +1 for all a j {a i1 +1,...,a i2 }. By Lemma no walk of length mt + 1 starting in a i can end in a vertex different from these a j. Hence f mt+1 i = y 2 for all i =1,...,i 1. The rest of the proof follows by induction on m. QED. This theorem allows us to give a complete description of the cycles in the state space of X. Example Consider the monomial dynamical system whose dependency graph is illustrated in Figure 2.4 Figure 2.4: Dependency graph of f =(x 4,x 5,x 4,x 7,x 6,x 3,x 1 x 2 ):F 7 2 F 7 2. Then using the notation of Theorem we can enumerate the vertices as (x 1,x 2,x 3 ), (x 4,x 5 ), (x 7,x 6 ).

30 Chapter 2. Boolean Dynamical Systems 21 It is clear that s 1 =3, s 2 =2and s 3 =2, moreover we have that, if m =3,then f 9 (x 1,x 2,x 3,x 4,x 5,x 6,x 7 )=(x 1 x 2 x 3,x 1 x 2 x 3,x 1 x 2 x }{{ 3,x } 4 x 5,x 4 x 5,x }{{} 7 x 6,x 7 x 6 ) }{{} 3 times 2times 2times Corollary Let X be as in Theorem Then the subgraph of cycles in the state space of f is isomorphic to the state space of a directed t-gon, hence the set of orbits in the hypercube F t 2 under the action of the cyclic group C t. Proof. For every choice of arguments for the f i weendupintheformoff mt.thenfacts on these points by a cyclic permutation, which defines the various cycles in the state space. By reducing the number of y s with the same subscript to one (see proof of Theorem , we get the system: g =(y 1,...,y t ):F t 2 F t 2 with the same cyclic action g(y 1,...,y n )=(y 2,y 3,...,y 1 )andg j (y 1,...,y n )=(y j+1,y j+2,...,y j ). This system arises from a directed t-gon. We are left to prove that S(g) is embedded in the state space of S(f). Consider the map I from S(g) tos(f), such that: I(a 1,..., a t )=(a 1,..., a }{{ 1,a } 2,..., a 2,..., a }{{} t,..., a t ) }{{} s 1 times s 2 times s t times The map I is a well-defined map and is clearly an injection. However, we have to ensure its edges are preserved under the map I. Leta b an edge on S(g). Notice that f(i(a)) = f(a 1,..., a }{{ 1,a } 2,..., a 2,..., a }{{} t,..., a t ) }{{} s 1 times s 2 times s t times = (a 2,..., a }{{ 2,..., a } t,..., a t,a }{{} 1,..., a 1 ) }{{} s 2 times s t times s 1 times

31 Chapter 2. Boolean Dynamical Systems 22 Now b = g(a), so b =(a 2,..., a t,a 1 )andi(b) =f(i(a)). Hence I(a) I(b) isanedgein S(f), and I preserves edges. QED Example We continue with Example, The state space of a dynamical system whose dependency graph is a 3-gon is illustrated in Figure Figure 2.5: State space of f =(x 2,x 3,x 1 ):F 3 2 F 3 2. In Figure 2.6 we can observe the state space whose dependency graph is in Figure 2.4. We can observe two 3-cycles and two 2-cycles, that is precisely what occurs in the state space illustrated in Figure 2.5. Corollary Let X be as in the Theorem. 1. The system f has fixed points (0,...,0) and (1,...,1). 2. The system f has limit cycles of all lengths dividing t. 3. The system f is a fixed point system if and only if the loop number of X is 1. Example The map f =(x 5,x 1,x 2,x 3 x 6,x 4,x 7,x 8,x 9,x 10,x 11,x 12,x 13,x 5 ) has dependency graph in Figure 2.3, and has loop number 5. By Corollary the state space of f contains cycles of all lengths dividing 5.

32 Chapter 2. Boolean Dynamical Systems 23 Figure 2.6: State space of f =(x 4,x 5,x 4,x 7,x 6,x 3,x 1 x 2 ):F 7 2 F 7 2.

33 Chapter 2. Boolean Dynamical Systems 24 So far Corollary is a criterion to determine if f is a fixed point system only if the dependency graph of f is a strongly connected graph. When the dependency graph is not a strongly connected graph, it can still be decomposed into strongly connected components. Such components are glued together by edges among them. Example Consider the dependency graph X shown in Figure 2.7. Clearly X is not Figure 2.7: Dependency graph of f =(x 2,x 4,x 1 x 2,x 3,x 5 x 4 ):F 5 2 F 5 2. a strongly connected graph, but is the glueing of two strongly connected component graphs. 2.4 Glueing Definition Let X and Y be dependency graphs of functions f : F r 2 F r 2 and g : F s 2 F s 2, respectively. A glueing X #Y of Y to X consists of a digraph with vertices V X V Y and edges E X E Y (disjoint union), together with a set of additional directed edges from vertices in Y to vertices in X.

34 Chapter 2. Boolean Dynamical Systems 25 The function with dependency graph X #Y is denoted by f#g : F r+s 2 F r+s 2. Example Let f =(x) :F 2 F 2, the dependency graph of f is given in Figure 2.8 Figure 2.8: Dependency graph of f = x 1 : F 2 F 2. On the other hand, consider the map g =(x 2,x 4,x 1 x 2,x 3 ). The dependency graph of g is given in Figure 2.9. The map h =(x 2,x 4,x 1 x 2,x 3,x 5 x 4 ):F 5 2 F 5 2 has the dependency graph illustrated in Figure 2.7. Figure 2.9: Dependency graph of g =(x 2,x 4,x 1 x 2,x 3 ):F 4 2 F 4 2. We write elements in F r+s 2 = F r 2 F s 2 as pairs (α, β) withα = (α 1,...,α r )andβ = (β 1,...,β s ). Similarly, we write the variables for the function f#g as (x, y) withx = (x 1,...,x r )andy =(y 1,...,y s ). Thus we can write the function f#g as (f#g)(x 1,...,x r,y 1,...,y s )=(f(x 1,...,x r ),h(x 1,...,x r,y 1,...,y s )), with f =(f 1,...,f r )andh =(h 1,...,h s )=((f#g) r+1,...,(f#g) r+s ).

35 Chapter 2. Boolean Dynamical Systems 26 We will use the poset E(X ) as in Lemma together with the glueing procedure to study fixed point systems. We will discuss elements in E(X )withnoedges(f =0, 1) separately. Elements with no edges are the strongly connected components of loop length 0. Lemma Let X #Y be a fixed point system. Then X is a fixed point system. Proof. Let (α, β) F r+s 2.Letm N be such that (f#g) m (α, β) =(f#g) m+1 (α, β). Then we have f m (α) =f m+1 (α) sothatf is a fixed point system. QED. Theorem Let X be a fixed point system and let Y be strongly connected of loop length 1. LetX #Y be a glueing. The following are equivalent: The glueing X #Y is a fixed point system. 1. There is a vertex a Y that is connected with a walk to a zero in X,or 2. Y is a fixed point system. Proof. =: We use the notation introduced above. Assume that there is a vertex a Y that is connected with a walk to a zero in X. Then all vertices of Y are connected to a zero. By Corollary we get an n N such that (f#g) n+m r+i = 0 for all m 1and for all i =1,...,s. Since f is a fixed point system we have f m = f m+1 =... So for a sufficiently large exponent m the component Y in X #Y can only contribute a fixed point β =(0,...,0). Hence (f#g) m =(f#g) m +1.

36 Chapter 2. Boolean Dynamical Systems 27 Let g be a fixed point system. Note that f is a subsystem of f#g. Iterate f#g until the component on X becomes constant: (f#g) m (x, y) =(f m (x), ([(f#g) m ] r+1 (x, y),...,[(f#g) m ] r+s (x, y)), where (f#g) m+k (x, y) =(f m (x), ((f#g) m+k r+1 (x, y),...,(f#g)m+k r+s (x, y)). Set (z 1,...,z s ):=([(f#g) m ] r+1 (x, y),...,[(f#g) m ] r+s (x, y)). Then apply powers of f#g to (f m (x), (z 1,...,z s )) to get (f#g) m+1 (x, y) =(f m (x), (f#g) r+1 (f m (x),z),...,(f#g) r+s (f m (x),z)), and (f#g) m+k (x, y) =(f m (x), (f#g) k r+1 (f m (x),z),...,(f#g) k r+s (f m (x),z)). So, for a fixed choice of x, the functions (f#g) r+1,...,(f#g) r+s, when viewed only as a function on elements from F s 2, are the functions g 1,...,g s multiplied with certain factors from the fixed point f m (x). This defines a new system h x : F s 2 F s 2,sothat (f#g) k (f m (x),z)=(f m (x),h k x(z)). ( ) If one of the factors taken from f m (x) is zero, then by Corollary and by the fact that Y is strongly connected, we get that h x is a fixed point system with the only fixed point β =(0,...,0). If all of the factors taken from f m (x) are 1, then h x = g and hence is a fixed

37 Chapter 2. Boolean Dynamical Systems 28 point system. So by (*) we see that (f#g) k (f m (x),z) ends in a fixed point for all choices of x and y. = : Before we give the proof in this direction, we prove a Lemma. Let Z be the dependency graph of an arbitrary Boolean monomial system. Decompose Z into two parts, where Z 0 is the glueing of all strongly connected components that allow a walk to a zero. Let Z 1 be the glueing of all the other strongly connected components of Z. Then there are no walks from anyvertexinz 1 to any vertex in Z 0,sothatZ is a glueing of Z 0 to Z 1. Lemma Any system of the form Z 1 has a fixed point (1,...,1). Proof. This is proved by induction on the number of connected components of Z 1. Assume that Z 1 = X 1 #Y where Y is a connected component with no zero. We can assume that X 1 has a fixed point (1,...,1). Observe that Y belongs to the component (X #Y) 1 (not connected to a zero) and Y has a fixed point (1,...,1) by Corollary So we get from the induction hypothesis that Z 1 hasalsoafixedpoint(1,...,1). QED. Proof of Theorem continued. Now assume that X #Y is a fixed point system, and that no vertex of Y is connected with a walk to a zero in X. Now we use as initial state (x, y) for the system f#g, withx =(1,...,1, 0,...,0) (a fixed point of f), where the component (1,...,1) belongs to X 1 and (0,...,0) belongs to X 0,andy arbitrary. Then, as discussed above, the system h x : F2 s F2 s coincides with g : F2 s F2 s. Since f#g is a fixed point system, the system g can also only contain fixed points and no proper limit cycles. Thus g is a fixed point system.qed.

38 Chapter 2. Boolean Dynamical Systems 29 Example Let X be the dependency graph in Figure 2.9. It is a strongly connected graph with loop number 1, therefore by Corollary , the monomial dynamical system f with such a dependency graph is a FPS. Let Y be the dependency graph shown in Figure 2.8. Its loop number is 1, therefore the monomial dynamical system g with dependency graph Y is a FPS as well. Now a dependency graph X #Y is illustrated in Figure 2.7, and by Theorem the dynamical system h with dependency graph X #Y must be a FPS. Nevertheless, the state space of h can be observed in Figure Figure 2.10: State space of h =(x 2,x 4,x 1 x 2,x 3,x 4 ):F 5 2 F 5 2. Theorem Let f be a fixed point system and let Y be strongly connected of loop length 0. Let X #Y be a glueing. Then the system f#g corresponding to the glueing X #Y is a

39 Chapter 2. Boolean Dynamical Systems 30 fixed point system. Proof. There are two cases: Y is a one point component with f a = 1 and the case that Y is a one point component with f a =0. Iff a = 0, then there is no additional edge from Y = {a} to X and thus f#g is a fixed point system with fixed points of the form (α, 0), where α is a fixed point of f. If f a = 1 then any additional edge from Y = {a} to a i X adds a factor to the last component, so that f#g =(f 1,...,f n,x i1...x ir ). If α =(α 1,...,α n )isafixed point for f, i.e. f(α) =α) then(f#g)(α, β) =(α, α i1 α ir ) which is again a fixed point. So f#g is a fixed point system. QED. Corollary Let f be a system with dependency graph X.Thenf is a fixed point system if and only if every strongly connected component of X either corresponds to a fixed point system or is connected by a walk to a zero in X. Proof. This is an immediate consequence of Theorem QED. 2.5 Boolean monomial fixed point systems In this section X will be the dependency graph of an arbitrary Boolean monomial dynamical system. Theorem Let X be the dependency graph of f : F n 2 F n 2. The following are equivalent:

40 Chapter 2. Boolean Dynamical Systems The system f : F n 2 F n 2 is a fixed point system. 2. For every vertex a X one of the following holds (a) a allows two closed walks p, q : a a of length q = p +1, (b) a is connected with a walk to a zero, or (c) there is no walk of length 1 from a to a. Proof. This is an immediate consequence of Theorem 2.4.4, Corollary and the remarks at the beginning of Section 2.3. QED. Example Consider the map f =(x 2,x 4,x 1 x 2,x 3,x 4 ) with dependency graph illustrated in Figure 2.9. Condition 2-a of Theorem holds, therefore f is a fixed point system. Example Consider the system f =(x 2,x 3 x 1,x 1 x 4, 0). The dependency graph of f is illustrated in Condition 2-b holds in Theorem 2.5.1, therefore f is a fixed point system. The state space of f can be observed in Figure Example Consider the system f =(x 1,x 2 x 1 ). The dependency graph of f is illustrated in Figure 2.13 Condition 2-c in Theorem holds, therefore f is a fixed point system, moreover Figure 2.14 illustrates the state space of f. Definition A system f =(f 1,f 2,..., f n ):F n 2 F n 2

41 Chapter 2. Boolean Dynamical Systems 32 Figure 2.11: Dependency graph of f =(x 2,x 3 x 1,x 1 x 4, 0) : F 4 2 F Figure 2.12: State space of f =(x 2,x 3 x 1,x 1 x 4, 0) : F 4 2 F 4 2. Figure 2.13: Dependency graph of f =(x 1,x 1 x 2 ):F 2 2 F Figure 2.14: State space of f =(x 1,x 1 x 2 ):F 2 2 F 2 2.

42 Chapter 2. Boolean Dynamical Systems 33 is a triangular system, ifeachf i is of the form f i = α i x ɛ i1 1 x ɛ i2 2 x ɛ ii i, where α i,ɛ ij {0, 1}. Corollary Every triangular system is a fixed point system. Proof. Let f be a triangular system with dependency graph X. ThenX consists of the glueing of components with just one element. Therefore each component corresponds to a fixed point system, and by Theorem we conclude that f is a fixed point system as well. QED. Example The map f in Example is a triangular system. Remark. The order in which we enumerate the variables is negotiable. We will still get the same state space up to isomorphism. Namely, if f =(f 1 (x 1,...,x n ),...,f n (x 1,...,x n )) is a parallel update system and σ S n is a permutation, then σf =(f σ 1 (1)(x σ(1),...,x σ(n) ),...,f σ 1 (n)(x σ(1),...,x σ(n) )) has a state space isomorphic to the state space of f. In particular, this defines a group action of S n on the fixed point system of n variables.

43 Chapter 2. Boolean Dynamical Systems 34 Theorem Let X be the dependency graph of a fixed point system f =(f 1,...,f n ).If there is no walk from a i to a j or if a i or a j have a walk of length greater than or equal to 1 to themselves, then g := (f 1,...,x i f j,...,f n ) is a fixed point system. Proof. By Lemma and the definition of glueing, X is an iterated glueing of connected components: X =(...(X 1 #X 2 )#...)#X r. By Corollary the connected components X i are fixed point systems or they are connected by a directed path to a zero. The connected components that are fixed point systems then have loop number 0 or 1. The multiplication of f j with x i introduces an extra edge a j a i into the digraph X, unless there is already a factor x i in f j, in which case the graph does not change. Let Y be the digraph of g. We distinguish a number of cases. Case 1: Suppose that a i and a j lie in the same component X s. If i j, thenx s has loop number 1 or is connected to a zero. So all components retain their properties responsible for Y being a fixed point system. If i = j, then the loop number of X s becomes 1, so Y is again a fixed point system. Case 2: Let a i lie in X s and a j X t, and assume that there is no walk from X s to X t. If there is a walk from X t to X s then X s X t in the partial order of E(X )). The existence of an edge a j a i does not change this property, so Y is a fixed point system. IfthereisnowalkfromX t to X s, then we can extend the partial order of E(X ) to a total order such that X s X t. Then the edge a j a i does not change the order of the glueing, so Y is a fixed point system.

44 Chapter 2. Boolean Dynamical Systems 35 Case 3: Let a i X s and a j X t. Assume that there is a walk from X s to X t and thus a walk from a i to a j. By hypothesis we have two cases. Either there is a closed walk a i a i of length greater than or equal to 1. Then the component X s has loop number 1. After inserting the edge a j a i the two components X s and X t are joined into one connected component of loop number 1. All other components remain unchanged. So g is a fixed point system. Or there is a closed walk a j a j of length greater than or equal to 1. Then the component X t has loop number 1. After inserting the edge a j a i the two components X s and X t are joined again into one connected component of loop number 1. All other components remain unchanged. So g is again a fixed point system. QED. Corollary Let f =(f 1,...,f n ) be a fixed point system and m a monomial. Then mf =(mf 1,...,mf n ) is a fixed point system. Proof. It is sufficient to prove the corollary for the case where m = x i is a single variable. Consider first the system (f 1,...,x i f i,...,f n ). As in the proof of the previous theorem we get an edge from a i to a i which can only change the loop number of the component of a i to 1. So this system is again a fixed point system. Now we can apply Theorem n 1 times and get that x i f =(x i f 1,...,x i f n )isafixedpointsystem.qed. This chapter covered monomial dynamical systems over the field with two elements. chapter 3 is devoted to monomial dynamical systems over an arbitrary finite field.

45 Chapter 3 Monomial Dynamical Systems over Finite Fields Given a finite field F q with q elements and a map f =(f 1,..., f n ):F n q F n q, such that each f i is a monic non-constant monomial, in F q [x 1,..., x n ], i.e. f i = x ɛ i1 1 x ɛ i2 2 x ɛ in. We will establish necessary and sufficient conditions for f to be a fixed point system. n Each element of F n q can be described as a vector a =(a 1,..., a n ) in which each entrace a i F q has index i. Therefore, the elements a and b of F n q can be grouped together if they agree in the indices i s such that a i = b i =0. Example Consider F 2 3, that can be written as the union of the sets: {(0, 0)}, {(1, 0), (2, 0)}, {(0, 1), (0, 2)}, {(1, 1), (2, 2)}. 36

46 Chapter 3. Monomial Dynamical Systems over Finite Fields 37 We are going to do define an equivalence relation on F n q. This equivalence relation will partition F n q into equivalence classes in which two elements are equivalent if they share the same indices that make their respective entries non-zero. Definition Define the support, v a, of a vector a =(a 1,..., a n ) as v ai =0if a i =0and v ai =1otherwise. Define the zero pattern, Z(a), of a vector a, tobethesetofcoordinatesi such that a i =0. For a and b in F n q we say that a is equivalent to b, and denoted by, a b if v a = v b. It is clear that is an equivalence relation.the support of the elements of F n q play a vital role in determining if f is a fixed point system. Moreover, we are going to associate two maps to f, called the support of f and the log of f. The behavior of these maps on the elements v a will provide the necessary criteria to determine if f is a fixed point system. 3.1 The log map and the support map Let us consider the elements of F n q that are equivalent to 1 =(1,..., 1). These elements have non-zero entries. Since F q is a field, such entries can be written as a power of a non-zero element of F q. Let the size of F q be greater than two. Then there exists a generator z not equal to 1 in F q such that F q = F q {0} =< z>= {z 0,z,z 2,..., z q 2 }.Fora =(a 1,..., a n ) F n q,leta i F q be written as a i = z α i for α i Z q 1,whereZ q 1 denotes the ring of integers modulo q 1.

47 Chapter 3. Monomial Dynamical Systems over Finite Fields 38 Hence f(a) =f(a 1,..., a n )=f(z α 1,..., z αn )=(z β 1,..., z βn ). We are going to associate a linear transformation with f, called the Log of f and it will be denoted by L. This transformation depends on the f i s. Definition Given a finite field F n q, and a function f =(f 1,..., f n ):F n q F n q,where f i Fq n [x 1,..., x n ], f i = x ɛ i1 1 x ɛ i2 2 x ɛ in. Define the formal sums g i = ɛ i1 x 1 +ɛ i2 x ɛ in x n n in Z q 1 [x 1,...,x n ], and define L : Z n q 1 Z n q 1 as L(x) =(g 1 (x),...,g n (x)). WecallL the log of f. For a equivalent to the vector 1 F n q, f(a) =f(a 1,..., a n )=f(z α 1,..., z αn )=(z β 1,..., z βn ). Since f i = x ɛ i1 1 x ɛ i2 2 x ɛ in n, f i (a) =f i (z α 1,..., z αn )=z α 1ɛ i1 +α 2 ɛ i α nɛ in. Therefore β i = α 1 ɛ i1 + α 2 ɛ i α n ɛ in and L is a linear transformation such that: L(α 1,..., α n ) = (α 1 ɛ α n ɛ 1n,..., α 1 ɛ n α n ɛ nn ) = (β 1,..., β n ) Notice that the matrix representation of L can be written as: ɛ 11 ɛ 1n A L =..... ɛ n1 ɛ nn

48 Chapter 3. Monomial Dynamical Systems over Finite Fields Figure 3.1: State space of f =(x 1,x 2 1x 2 ):F 2 3 F 2 3. Example Consider f =(x 1,x 2 1x 2 ):F 2 3 F 2 3 whose state space is in Figure 3.1. The map L is given by the matrix: Once we consider L as a matrix, then the key observation is that for A p L,withp>1, the (i, j)-entry of A p L is the exponent of x j,wherex j divides f i p,whithf p =(f 1 p,..., f n p ). Example Consider f as in Example Then f 2 =(x 1,x 4 1x 2 ) and 1 0 A 2 L =. 4 1 Hence the entry (2, 1) in A 2 L is 4 and that is the exponent of the variable x 1 in f 2 2. The following result establishes that f being a fixed point system forces L to be a fixed point system as well, i.e., if f is a fixed point system, then there are no non-trivial cycles among elements equivalent to 1 F n q. Lemma Let f and L be as above. If f is a FPS then L is a FPS.

49 Chapter 3. Monomial Dynamical Systems over Finite Fields 40 Proof. For α Zq 1, n suppose L t (α) =α, thena t L (α) =α. Letz to be a generator for F q, for F q > 2, then for a Fq n,wehavea =(z α 1,..., z αn )andf t (a) =(z β 1,..., z βn ). We conclude that A t L (α) =β. Since At L (α) =α, f t (a) =a and since f is a FPS, we can conclude that t =1. QED Notice that if L is a fixed point system, we can guarantee no nontrivial cycles only among elements equivalent to 1. Example Consider L and f as in Example L is a linear map over F 2 2.ThenL is a fixed point system as we can see in Figure 3.2, and there are no cycles in the state space of f among elements equivalent to 1, as we can observe in Figure Figure 3.2: State space of log(f), f =(x 1, 2x 1 + x 2 ):F 2 2 F 2 2. Lemma If L is a FPS then if f t (a) =a for a eqivalent to the vector 1 F n q,thent must equal one. Proof. Using the notation of last lemma, suppose a =(z α 1,..., z αn )andf t (a) =a. This will imply A t L (α) =α, sincel is a FPS, t =1. QED Using the log, we know how the elements of F n q, that are equivalent to 1, behave in the state space of f. The next question is how to determine if there are cycles in the state space of f among elements not equivalent to 1.