A reductor with 6 gates and 13 leads proposed by G. SpencerBrown is an example of automaton with an input z to the system. For z=0 it has two balanced states for (a,b,c,d,e,f): A=(1,1,0,1,0,1) and C=(1,1,1,0,1,0), and two balanced states B=(1,0,1,0,1,1) and D=(0,1,0,1,1,1) for z=1. The alternating sequence of input values z=0,1,0,1¼ produces a sequence of states A,B,C,D,A,B,C,D¼ of the period 4. In the extended set [0,1] for z=0 the system has an infinite number of stable points determined by the parametric equation (a,b,c,d,e,f)=(1,1,c,1c,c,1c), and for z=1 we have one additional fixed point (a,b,c,d,e,f)=(j,j,j,j,j,j). In a general case, we can take any z Î [0,1] as an input z and obtain a system of equations with two parameters z and c. This system has a continuum of solutions. L. Kauffman (1994) conjectured that it is not possible to make a determinate (asynchronous) reductor with less than six inverters. Every circuit that has more then one stable state in the discrete set of values {0,1}, has a continuum of stable states (i.e., fixed points) in [0,1], defined by resulting parametric equations. On the other hand, a circuit that has no stable states in the set {0,1}, always has a stable state in the extended set [0,1], i.e., in polyvalent logic. The principle of duality holds for automata with NAND and NOR gates: a system remains unchanged with regard to stable states if all NAND gates are replaced by NOR gates or vice versa. This holds also for automata that contain NAND and AND, or NOR and OR gates. The LinKnot function fAutoSigInp (webMathematica fAutoSigInp) calculates stable states of an automaton given by a list of outgoing edges, signs of vertices, and inputs in vertices. For vertices with the sign 1 the operation NAND is used, while for vertices with sign 1 the operation AND. The result is a list of edge colorings corresponding to stable states and a list of stable states according to gates (see Kauffman, 1994), with values taken from the discrete set {0,1}. If the list of signs is empty, it is treated as (1,...,1), and computations are made with NAND gates in all vertices of a graph. Inputs in vertices can be included by giving the corresponding list of vertices. Because only an input 0 produces a change in NAND or AND gates, all inputs are treated as 0. The function fAutoKL (webMathematica fAutoKL) calculates stable states of an automaton obtained from a KL given in Conway notation, followed by a list of signs of vertices, and inputs in vertices. In vertices with sign 1 the operation NAND is used, while for vertices with sign 1 the logical operation AND. The result is the oriented graph corresponding to a given KL, the list of edge colorings corresponding to its stable states, and the list of stable states according to gates (see Kauffman, 1994). If the list of signs of a given KL is empty, the original list of the signs of a given KL is used for the computation. Inputs in crossings can be included in the same way as for the function fAutoSigInp. By using the function fAutoKL it is possible to examine a behavior of actual KLs with regard to stable states, this means, their edge colorings with two colors 0 and 1 compatible with the requirements of NAND and AND logical gates. For example, an automaton obtained from a figureeight knot 2 2 (or 4_{1}) with the signs 1 in all crossings (i.e., with NAND gates in all crossings), defined by equations has no stable states in the discrete set {0,1}. The corresponding operator generates a periodic sequence (0,0,0,0),(1,1,1,1)¼ for (x,y,z,u) Î {(1,1,1,1),(1,1,0,0),(0,0,1,1)} and for every other initial state (x,y,z,u) with more than two zeros. In all the other cases, it generates a sequence of the period eight (0,1,1,0),(1,1,1,1),(0,1,0,1),(1,1,0,1), (1,0,0,1),(1,0,1,1),(1,0,1,0),(1,1,1,0)... By using the original signs of that knot (this means, NAND gates in the crossings with the sign 1, and AND gates in the crossings with the sign 1), we obtain a stable state of that knot (x,y,z,u)=(1,1,0,0). The mirror image 2 2 of the same knot produces the same result, showing that the figureeight knot is achiral. In the extended set [0,1] for a figureeight knot with all signs equal to 1 we have the stable state (x,y,z,u)=(j,j,j,j) (b). A figureeight knot with the signs 1 in all vertices (this means, with AND gates in all vertices) has two stable states, where all edges are labelled by 0, or all edges are labelled by 1. On the other hand, a trefoil 3 has no stable states with NAND gates in all crossings, but its mirror image 3 with AND gates in all crossings has two stable states from the discrete set {0,1}, where all edges are colored by 0, or by 1. This shows that the trefoil is chiral.
