EXERCISES

EXERCISES

Find the number of different types of circular necklaces that could be made from the sets of beads described in Exercises 6.1 to 6.4, assuming that all the beads are used on one necklace.

6.1.Three black and three white beads.

6.2.Four black, three white, and one red bead.

6.3.Seven black and five white beads.

6.4.Five black, six white, and three red beads.

6.5.How many different circular necklaces containing ten beads can be made using beads of at most two colors?

6.6.Five neutral members and two members from each of two warring factions are to be seated around a circular armistice table. In how many nonequivalent ways, under the action of D9, can they be seated if no two members of opposing factions sit next to each other?

6.7.How many different chemical compounds can be made by attaching H, CH3, C2H5, or Cl radicals to the four bonds of a carbon atom? The radicals lie at the vertices of a regular tetrahedron, and the group is the tetrahedral group A4.

6.8.How many different chemical compounds can be made by attaching H, CH3, or OH radicals to each of the carbon atoms in the benzene ring of Figure 6.3? (Assume that all of the C–C bonds in the ring are equivalent.)

6.9.How many ways can the vertices of a cube be colored using, at most, three colors?

6.10.How many ways can the vertices of a regular tetrahedron be colored using, at most, n colors?

6.11.How many different tetrahedra can be made from n types of resistors when each edge contains one resistor?

6.12.How many ways can the faces of a regular dodecahedron be colored using, at most, n colors?

Find the number of different colorings of the faces of the solids described in Exercises 6.13 to 6.16.

6.13.A regular tetrahedron with two white faces and two black faces.

6.14.A cube with two white, one black, and three red faces.

6.15.A regular icosahedron with four black faces and 16 white faces.

6.16.A regular dodecahedron with five black faces, two white faces, and five green faces.

6.17.How many ways can the faces of a cube be colored with six different colors, if all the faces are to be a different color?

6.18.(a) Find the number of binary relations, on a set with four elements, that are not equivalent under permutations of the four elements.

(b)Find the number of equivalence relations, on a set with four elements, that are not equivalent under permutations of the four elements.

6.19.How many different patchwork quilts, four patches long and three patches wide, can be made from five red and seven blue squares, assuming that the quilts cannot be turned over?

6.20.If the quilts in Exercise 6.19 could be turned over, how many different patterns are possible?

6.21.Find the number of ways of distributing three blue balls, two red balls, and four green balls into three piles.

6.22.If the cyclic group Cn, generated by g, operates on a set S, show that the number of orbits is

1 |Fix gn/d | · φ (d), n

d/n

where the Euler φ-function, φ (d), is the number of integers from 1 to d that are relatively prime to d. (See Exercises 4.55 and 4.56.)

6.23.Some transistor switching devices are sealed in a can with three input sockets at the vertices of an equilateral triangle. The three input wires are connected to a plug that will fit into the input sockets as shown in Figure 6.11. How many different cans are needed to produce any boolean function of three input variables?

6.24.How many different ways can the elements of the poset in Figure 6.12 be colored using, at most, n colors?

6.25.Verify that the number of nonequivalent switching functions of four variables, under permutation of the inputs, is 3984.

(S, ) is just a set S together with an associative binary operation, . For example, (P, +) is a semigroup, but not a monoid, because the set of positive integers, P, does not contain zero.

Just as one of the basic examples of a group consists of the permutations of any set, a basic example of a monoid is the set of transformations of any set. A transformation is just a function (not necessarily a bijection) from a set to itself. In fact, the analogue of Cayley’s theorem holds for monoids, and it can be shown that every monoid can be represented as a transformation monoid.

Example 7.2. If X = {0, 1}, write out the table for the transformation monoid

(XX , Ž ).

Solution. XX has four elements, e, f, g, h, defined as follows.

The table for (XX , Ž ) is shown in Table 7.1. For example, g Ž f (0) = g(f (0)) = g(0) = 1, and g Ž f (1) = g(f (1)) = g(0) = 1. Therefore, g Ž f = h. The other compositions can be calculated in a similar manner.

Example 7.3. Prove that (Z, ) is a commutative monoid, where x y = 6 − 2x − 2y + xy for x, y Z.

TABLE 7.1. Transformation Monoid of {0, 1}

Solution. For any x, y Z, x y Z, and x y = y x, so that is a commutative binary operation on Z. Now

x (y z) = x (6 − 2y − 2z + yz) = 6 − 2x + (−2 + x)(6 − 2y − 2z + yz)

= −6 + 4x + 4y + 4z − 2xy − 2xz − 2yz + xyz.

Also,

(x y) z = (6 − 2x − 2y + xy) z = 6 + (−2 + z)(6 − 2x − 2y + xy) − 2z

=−6 + 4x + 4y + 4z − 2xy − 2xz − 2yz + xyz

=x (y z).

Hence is associative.

Suppose that e x = x. Then 6 − 2e − 2x + ex = x, and 6 − 2e − 3x + ex = 0. This implies that (x − 2)(e − 3) = 0. Hence e x = x for all x Z if and only if e = 3. Therefore, (Z, ) is a commutative monoid with 3 as the identity.

Since the operation in a monoid, (M, ), is associative, we can omit the parentheses when writing down a string of symbols combined by . We write the element x1 (x2 x3) = (x1 x2) x3 simply as x1 x2 x3.

In any monoid (M, ) with identity e, the powers of any element a M are defined by

a0 = e, a1 = a, a2 = a a, . . . , an = a an−1 for n N.

The monoid (M, ) is said to be generated by the subset A if every element of M can be written as a finite combination of the powers of elements of A. That is, each element m M can be written as

m = a1r1 a2r2 · · · anrn for some a1, a2, . . . , an A.

For example, the monoid (P, ·) is generated by all the prime numbers. The monoid (N, +) is generated by the single element 1, since each element can be written as the sum of n copies of 1, where n N. A monoid generated by one element is called a cyclic monoid.

A finite cyclic group is also a cyclic monoid. However, the infinite cyclic group (Z, +) is not a cyclic monoid; it needs at least two elements to gener-

the elements of a cyclic monoid, and the arrows correspond to multiplication by the element c.

Figure 7.1. Finite cyclic monoid.

A computer receives its information from an input terminal that feeds in a sequence of symbols, usually binary digits consisting of 0’s and 1’s. If one sequence is fed in after another, the computer receives one long sequence that is the concatenation (or juxtaposition) of the two sequences. These input sequences together with the binary operation of concatenation form a monoid that is called the free monoid generated by the input symbols.

Let A be any set (sometimes called the alphabet), and let An be the set of n-tuples of elements in A. In this chapter, we write an n-tuple as a string of elements of A without any symbols between them. The elements of An are called words of length n from A. A word of length 0 is an empty string; this

Let FM(A) denote the set of all words from A, more formally

∞

FM(A) = A0 A A2 A3 · · · = An.

n=0

Then (FM(A), ) is called the free monoid generated by A, where the operationis concatenation, and the identity is the empty word . Another common notation for FM(A) is A .

If we do not include the empty word, , we obtain the free semigroup generated by A; this is often denoted by A+.

If α and β are words of length m and n, then α β is the word of length m + n obtained by placing α to the left of β.

If A consists of a single element, a, then the monoid FM(A) = { , a, aa, aaa, aaaa, . . .} and, for example, aaa aa = aaaaa. This free monoid, generated by one element, is commutative.

If A = {0, 1}, then FM(A) consists of all the finite sequences of 0’s and 1’s, FM(A) = { , 0, 1, 00, 01, 10, 11, 000, 001, . . .}.

We have 010 1110 = 0101110 and 1110 010 = 1110010, so FM(A) is not commutative.

If A = {a, b, c, d, . . . , y, z, , .}, the letters of the alphabet together with a space, , and a period, then

the sky FM(A) and the sky is b lue. = the sky is blue.

Of course, any nonsense string of letters is also in FM(A); for example, pqb.a .. xxu FM(A).

There is an important theorem that characterizes free monoids in terms of monoid morphisms. If (M, ) and (N, ·) are two monoids, with identities eM and eN , respectively, then the function f : M → N is a monoid morphism from

(M, ) to (N, ·) if

(i)f (x y) = f (x) · f (y) for all x, y M.

(ii)f (eM ) = eN .

y . Hence f (1 + 1) = 4, whereas f (1) + f (1) = 2.

Theorem 7.4. Let (FM(A), ) be the free monoid generated by A and let i: A → FM(A) be the function that maps each element a of A into the corresponding word of length 1, so that i(a) = a.

Then if l: A → M is any function into the underlying set of any monoid (M, ·), there is a unique monoid morphism h: (FM(A), ) → (M, ·) such that h Ž i = l. This is illustrated in Figure 7.2.

Proof. If h satisfies h Ž i = l, then h must be defined on words of length 1 by h(a) = l(a). Once a morphism has been defined on its generators, it is determined completely as follows. Let α be a word of length n 2 in FM(A). Write α as β c, where β is of length n − 1 and c is of length 1. Then we have

h(α) = h(β c) = h(β) · h(c) = h(β) · l(c). Hence h can be determined by using

Figure 7.2. The function l factors through the free monoid FM(A).