Figure 6.5. Types of rotations of a dodecahedron.

TABLE 6.5. Colorings of the Dodecahedron

COUNTING SWITCHING CIRCUITS

The Burnside theorem can still be applied when the sets to be enumerated do not have any geometric symmetry. In this case, the symmetry group is usually the full permutation group Sn.

Consider the different switching circuits obtained by using three switches. We can think of these as black boxes with three binary inputs x1, x2, and x3 and one binary output f (x1, x2, x3), as in Figure 6.6. Two circuits, f and g, are called equivalent if there is a permutation π of the variables so that f (x1, x2, x3) = g(xπ 1, xπ 2, xπ 3). Equivalent circuits can be obtained from each other by just permuting the wires outside the black boxes, as in Figure 6.7.

Figure 6.7. Permutation of the inputs.

Example 6.6. Find the number of switching circuits using three switches that are not equivalent under permutations of the inputs.

Solution. There are eight possible inputs using three binary variables and hence there are 28 = 256 circuits to consider. The symmetric group S3 acts on these 256 circuits, and we wish to find the number of different equivalence classes, that is, the number of orbits.

Table 6.6 lists the number of circuits left fixed by the different types of elements in S3. For example, if the switching function f (x1, x2, x3) is fixed by

By Burnside’s theorem and Table 6.6, the number of nonequivalent circuits is 480/|S3| = 480/6 = 80.

However, this number can be reduced further if we allow permutations and complementation of the three variables. In a circuit consisting of two-state switches, the variable xi can be complemented by simply reversing each of the switches controlled by xi . The resulting circuit is just as simple and the cost is the same as the original one. In transistor networks, we can just permute the input wires and add NOT gates as in Figure 6.8.

The eight input values of a three-variable switching circuit can be considered as the vertices of a three-dimensional cube, as shown in Figure 6.9. The six faces of this cube are defined by the equations x1 = 0, x1 = 1, x2 = 0, x2 = 1, x3 = 0, x3 = 1. The group that permutes and complements the variables takes each face to another face and takes opposite faces to opposite faces. Hence the group is the complete symmetry group, G, of the cube. There is a morphism ψ : G → {1, −1},

TABLE 6.6. Action of S3 on the Inputs of the Switches

Figure 6.8. Permutation and complementation of inputs.

111

000

Figure 6.9. Cube of input values.

which sends proper rotations to 1 and improper rotations to −1; the kernel of ψ is the group of proper rotations of the cube which, by the morphism theorem, must be a normal subgroup of index 2. Therefore, the order of G is 2 · 24 = 48.

Example 6.7. Find the number of switching circuits involving three switches that are nonequivalent under permutation and complementation of the variables.

Solution. Each boolean function in three variables defines a coloring of the vertices of the cube of input values. A vertex is colored black if the function is 1 for the corresponding input value. It is colored white if the function takes the value 0 at that input value.

We can represent the complete symmetry group, G, of the cube by means of permutations of the vertices labeled 0, 1, 2, 3, 4, 5, 6, 7 in Figure 6.10. Since the group of proper rotations of the cube is a normal subgroup of index 2 in G, every element of G can be written as a proper rotation π or as π Ž ρ, where ρ is the reflection of the cube in its center.

There are 28 different switching functions of three variables, and Table 6.7 describes the number of circuits that are fixed by the action of each element of the group G on the eight inputs. For example, consider the element g = (01) Ž (67) Ž (34) Ž (25). If a switching function f is fixed under the action of g, then the images of the input values corresponding to the vertices 0 and 1 must be the same; that is, f (0, 0, 0) = f (0, 0, 1). Similarly, the images of the input values corresponding to the vertices 6 and 7 are the same, and f (1, 1, 0) = f (1, 1, 1).

Figure 6.10. Labeling of the cube.