Finite Automata Exercises

---

---

Question 1

Design a two-input, two-output sequence detector that generates an output '1' every time the sequence 1011 is detected. Overlapping sequences should also be counted. Draw the state table and the state diagram.

Solution

We want to detect sequence: 1011

States and Transitions

• S1: Initial state. If input is 0, stay in S1 with output 0. If input is 1, move to S2 with output 0.
• S2: If input is 0, move to S3 with output 0. If input is 1, stay in S2 with output 0.
• S3: If input is 0, return to S1 with output 0. If input is 1, move to S4 with output 0.
• S4: If input is 0, return to S1 with output 0. If input is 1, move to S2 with output 1 (sequence detected).

The sequence 1011 is a sequence that requires a starting point from S1. If input 0 is not received, the sequence is confined in S1, producing output 0. If input 1 is received, the sequence is confined in S2, producing output 0.

If input 1 is received, the sequence is confined in S3, producing output 0. If input 0 is received, the sequence is confined in S1, producing output 0. If input 011 is received, the sequence detector can produce 1.

Here is its State Table −

Question 2

Design a two-input, two-output sequence detector that generates an output '1' every time the sequence 10101 is detected. Overlapping sequences should also be counted. Draw the state table and the state diagram.

Solution

We want to detect sequence: 10101

States and Transitions

• S1: Initial state. If input is 0, stay in S1 with output 0. If input is 1, move to S2 with output 0.
• S2: If input is 0, move to S3 with output 0. If input is 1, stay in S2 with output 0.
• S3: If input is 0, move to S4 with output 0. If input is 1, move to S1 with output 0.
• S4: If input is 0, move to S5 with output 0. If input is 1, move to S2 with output 0.
• S5: If input is 0, return to S1 with output 0. If input is 1, move to S2 with output 1 (sequence detected).

Here is its State Table −

Question 3

Find the equivalent partition for the following machine −

Solution

For a string with no input, all states are 0-equivalent, with P0 = (ABCDE). For string length 1, there are two input types: 0 and 1, with states A and B giving output 0 and states C, D, and E giving output 1. Thus, states in set P0 are divided into (AB) and (C, D, E), with P1 = ((AB) (CDE).

For input string length 1, A and E are distinct because they produce different outputs. For input string length 2, the distinguishability is determined by the next state combination. States A and B produce next states B and A, and next states C and E for input 1. B and A belong to the same set, and C and E also belong to the same set. AB cannot be partitioned for input string length 2. The set (CDE) is partitioned into (C) and (DE), and the new partition becomes P2= ((AB) ((C) (DE)).

For input string length 3, the same process is performed, with states A and B producing next states B and A, and next states C and E for input 1. C is a single state and cannot be partitioned further. States D and E produce next states E and D, respectively, for input 0, and B and E belong to the same set, so D and E cannot be divided. The new partition becomes P3= (A)(B)(C)(DE).

To check for input string length 4, three subsets contain single states A, B, and C, which cannot be partitioned further. States D and E produce the next states E and D for input 0 and B for input 1, respectively. D and E belong to the same set, and their partition is the same as P3, making P3 the equivalent partition of the machine.

After minimization

Question 4

Simplify the following incompletely specified machine −

Solution

The partitions are −

• P0 = (ABCDEF)
• P1 = (ADE)(BCF) (Depending on output for input 0)
• P2 = (ADE)(B)(CF) (For input 0, the next state of B goes to another set)
• P3 = (A)(DE)(B)(CF) (For input 0, the next state of A goes to another set)
• P4 = (A)(DE)(B)(CF)

As P3 and P4 are the same, P3 is the equivalent partition.

The unique equivalent partition, P4, represents the minimized machine's state. To simplify, we can rename the partitions (A, B, DE, and CF) as S1, S2, S3, and S4, respectively, to simplify the process.

Question 5

Test whether the following machine is finite or not −

Solution

The partitions are −

• P0 = (ABCDEF)
• P1 = (AD)(BCE) (Depending on output for input 1)
• P2 = (AD)(BE)(C) (For input 0, the next state of C goes to another set)
• P3 = (A)(D)(BE)(C) (For input 0, the next state of A and D goes to a different set)
• P4 = (A)(D)(BE)(C)

From here, we can see P3 and P4 are same, so P3 is equivalent partition.

The unique equivalent partition, P4, represents the minimized machine's state. To simplify, we can rename the partitions (A, BE, C, and D) as S1, S2, S3, and S4, respectively, to simplify the process.

Question 6

Minimize the following incompletely specified machine −

Solution

We can introduce temporary state T that is placed in the next state where the next states are not specified, and no output is required. As a temporary state T is considered, T is put in the present state with the next state T for all inputs with no output.

Question 7

Construct a compatible graph for the following incompletely specified machine −

Solution

Incompletely specified machines lack all next states or outputs. To minimize these machines, we use the merger graph and compatible graph method. From the merger graph, we find compatible pairs and construct a compatible graph. From the compatible graph, we find the closed partition, which are the minimized states of the machine.

The machine has five states, and its merger graph consists of five nodes: A, B, C, D, and E. The outputs of A and B do not differ, creating an arc between them. The next state conflicts, creating an interrupted arc.

The next state pairs (CE) are placed in the interrupted portion. A and C have no arc between them, and C and D have an uninterrupted arc between them. The merger graph is constructed, crossing the arcs between A, E, and (AD), BE, BD, and BC.

Next, we need to consider the compatibility graph, where the graph consists of four vertices (AB, CD, DE), and (CE), with a closed covering for the machine if all outgoing edges and terminal vertices belong to the subgraph. The minimized machine states are (AB) and (CDE).

To create a closed covering, rename (AB) as S1 and (CDE) as S2, and draw a directed arc from (AB) to (CE). This ensures that every state of the machine is covered by at least one vertex.

So, the minimized machine is −

Question 8

Check whether the following machine is finite or not −

Solution

The Testing Table-Testing Graph Method involves creating a divided table with two halves for a machine with two inputs and two outputs.

The upper half contains single state input-output combinations, while the lower half contains two state combinations. Six combination pairs are formed, resulting in a total of four states.

The testing table contains six present state combinations and six nodes, each labelled with an input output combination. The testing graph for finite memory is a directed arc from S_iS_j [i≠ j] to S_pS_q [p≠ q], where S_p and S_q is the implied pair of S_iS_j.

The testing graph is loop-free, indicating a finite machine. The longest path is 1, and the order of finiteness is 1+1 = 2. This can be proven using the vanishing connection matrix method.

The connection matrix consists of six rows and columns, labelled with present state combinations. The (i, j)th entry in the matrix is 1 if there is an entry in the corresponding testing table, otherwise, it is 0.

Consider the red marked are crossed. The matrix contains all '0' rows AC, AD, BC, and BD, resulting in the corresponding columns and rows labelled AC, AD, BC, and BD disappearing.

The matrix contains '0' rows AB and CD, causing the columns to vanish. The number of steps required to do so is 2, indicating the machine's order of finiteness is 2.

Question 9

Check whether the following machine is information lossless or not. If lossless, find its order.

Solution

To determine if a machine is lossless, construct a testing table, which should be divided into two halves.

The testing table does not contain any repeated entry, so the machine is an information lossless machine.

The testing graph for information loss lessness is loop-free. The order of lossless-ness is m = 1 + 2 = 3. The length of the longest path of the graph is 1.

Question 10

Consider the following DFA, then find the number of strings of length 8 will be accepted by the DFA.

Solution

With all possible strings of length three on set {0, 1}, we can achieve the end state. The ultimate state can be reached in eight different ways. After that, there will be a five-length string with two possible outcomes for each of the five positions: 0 or 1.

Thus, 8 * 2⁵ = 2⁸ = 256 will be the total number of strings approved.