FINITE-STATE MACHINES

We now look at mathematical models of sequential machines. These are machines that accept a finite set of inputs in sequential order. At any one time, the machine can be in one of a finite set of internal configurations or states. There may be a finite set of outputs. These outputs and internal states depend not only on the previous input but also on the stored information in the machine, that is, on the previous state of the machine. A pushbutton elevator is an example of such a machine. A digital computer is a very complex finite-state machine. It can be broken down into its component parts, each of which is also a machine. The RS and JK flip-flops, discussed in Exercises 2.69 and 2.70, are examples of two widely used components.

For simplicity, we only consider machines with a finite set of inputs and a finite set of states. We do not mention any outputs explicitly, because the state set can be enlarged, if necessary, to include any outputs. The states can be arranged so that a particular state always gives rise to a certain output.

A finite-state machine, (S, I, m) consists of a set of states S = {s1, s2, . . . , sn}, a set of input values I = {i1, i2, . . . , it }, and a transition function

m: I × S → S,

which describes how each input value changes the states. If the machine is in state sp and an input iq is applied, the machine will change to state m(iq , sp ).

For example, consider a pushbutton elevator that travels between two levels, 1 and 2, and stops at the lower level 1 when not in use. We take the time for the elevator to travel from one level to the other to be the basic time interval, and the controlling machine can change states at the end of each interval. We allow the machine three inputs, so that I = {0, 1, 2}.

0 if no button is pressed in the preceding time interval

input =

1 if button 1 only is pressed in the preceding time interval

2 if button 2 or both buttons are pressed

in the preceding time interval.

Since the elevator is to stop at the bottom when not in use, we only consider states that end with the elevator going down. Let the set of states be

S = {stop, down, up–down, down–up–down}.

For example, in the “up–down” state, the elevator is traveling up, but must remember to come down. If no button is pressed or just button 1 is pressed while it is going up, the machine will revert to the “down” state when the elevator reaches level 2. On the other hand, if someone arrives at level 1 and presses button 2, the machine will change to the “down–up–down” state when the elevator reaches level 2.

monoid of FM(I ) by this relation. It will always be a finite monoid with, at most, nn elements. We first define the notion of a quotient monoid.

Suppose that R is an equivalence relation on a monoid (M, ). Then R is

Proposition 7.5. If R is a congruence relation on the monoid (M, ), the quotient set M/R = {[a]|a M} is a monoid under the operation defined by

[a] [b] = [a b].

This monoid is called the quotient monoid of M by R.

Proof. We first have to verify that the operation is well defined on congruence

[a b] = [a b ]. This shows that is well defined on M/R. The associativity of in M/R follows from the associativity of in M. If e is the identity of M, then [e] is the identity of M/R. Hence (M/R, ) is a monoid.

Let (S, I, m) be a finite-state machine and let the effect of an input sequence

be given by

h: FM(I ) → SS .

Define the relation R on FM(I ) by

αRβ if and only if h(α) = h(β).

This is easily verified to be an equivalence relation. Furthermore, it is a congruence relation on the free monoid (FM(I ), ), because if αRβ, then h(α) = h(β), and h(α γ ) = h(α) Ž h(γ ) = h(β) Ž h(γ ) = h(β γ ); thus (α γ )R(β γ ), and similarly, (γ α)R(γ β).

The quotient monoid (FM(I )/R, ) is called the monoid of the machine

(S, I, m).

We can apply the same construction to the free semigroup of input sequences to obtain the semigroup of the machine.

The monoid of a machine reflects the capability of the machine to respond to the input sequences. There are an infinite number of sequences in FM(I ), whereas the number of elements in the quotient monoid is less than or equal to nn. Two sequences are in the same congruence class if and only if they have the same effect on the machine.

A morphism theorem for monoids can be proved in a similar way to the morphism theorem for groups (Theorem 4.25; see Exercise 7.24). Applying this

to the monoid morphism h: FM(I ) → SS , it follows that the quotient monoid FM(I )/R is isomorphic to Im h. This isomorphism assigns to each congruence class a unique transition between states.

Example 7.6. Draw the state diagram and find the monoid of the following machine (S, I, m). The machine has two states, s0 and s1, and two input symbols, 0 and 1. The effects of the input symbols are given by the functions h(0), h(1): S → S, defined in Table 7.3.

Solution. Let us calculate the effect of inputs of length 2. We have h(ij ) = h(i) Ž h(j ), where j is fed into the machine first. It follows from Tables 7.3 and 7.4 that h(00) = h(01) = h(0) and [00] = [01] = [0] in the monoid of the machine. There are only four functions from {s0, s1} to {s0, s1}, and these are h(0), h(1), h(10), and h(11). Hence the monoid of the machine consists of the four congruence classes [0], [1], [10], and [11]. The table of this quotient monoid is given in Table 7.5, and the state diagram is given in Figure 7.6. For example,

[110] = [0]. Notice that [11] is the identity; thus, in the monoid of the machine, [ ] = [11].

Example 7.7. Describe the monoid of the machine ({start, even, odd}, {0, 1}, m) that determines the parity of the number of 1’s in the input.

Solution. We have already seen that any input sequence with an even number of 1’s has the same effect as 0 and that any sequence with an odd number of 1’s has the same effect as 1. It follows from Table 7.6 that the monoid of the machine contains the three elements [ ], [0], and [1]. The table for this monoid is given in Table 7.7.

Finite-state machines can easily be designed to recognize certain types of input sequences. For example, most numbers inside a computer are in binary form and have a check digit attached to them so that there is always an even number of 1’s in each sequence. This is used to detect any machine errors (see Chapter 14). A finite-state machine like Example 7.7 can be used to perform a parity check on all the sequences of numbers in the computer. The machine can be designed to signal a parity check error whenever it ends in the “odd” state.

Let us now look at a machine that will recognize the pattern 010 in any binary input sequence that is fed into the machine. Figure 7.7 is the state diagram of such a machine. If the machine is initiated in state s1, it will be in state s4 if and only if the preceding inputs were 010, and in this case, the machine sends an output signal.

This machine has four states; thus the total possible number of different functions between states is 44 = 256. Table 7.8 shows that the input sequences of length 0, 1, and 2 all have different effects on the various states. However, seven of the eight sequences of length 3 have the same effect as sequences of length

TABLE 7.6

Figure 7.7. State diagram of a machine that recognizes the sequence 010.

TABLE 7.8. Effects of the Input Sequences on the States of the Machine

Λ

Figure 7.8. Tree diagram of input sequences.

2. The only input sequence with a different effect is 010, the sequence that the machine is designed to recognize. Therefore, the only sequences of length 4 that we check are those whose initial inputs are 010, namely, 0010 and 1010.

We can use the tree diagram in Figure 7.8 to check that we have covered all the possible transition functions obtainable by any input sequence. We label the nodes of the tree by input sequences. At any node α, there will be two upward branches ending in the nodes 0 α and 1 α, corresponding to the two input symbols. We prune the tree at node α, if α gives rise to the same transition function as another