Question

Finite state machines & Regular Expressions

Please select the best option

1.

For the following questions Let r, s, t be regular expressions for the same alphabet "á" (left column). Get the property on the right side that produces equality for each regular expression.

TrФ Trt+st= ( T+= тф+ =                     Ar B. (r+s) C.

2.

From the diagram of the solution M = (Σ, Q, s,, F) is respectively:


a 92 a b 90 b b 91 a (b)..(90.91.92). 391. F=190) I (b) 0.90.91.92) 5 =90. F-190) CE (b) 0.90.91.921.5-90. F=191) dI (4.b) 0
e would be NONE.

3.

The following graph corresponds to a diagram of:

a 92 a b 90 b b o a

A. Transition machine and states
b. Transition from a deterministic finite automata
c. Transition from a finite non-deterministic automata
d. No response
E. Language of transition and states

4. From the diagram of the solution M = (Σ, Q, s, ,F,) is respectively:
a Diagrama de Transición de M 91 92 b a 90 93 94 a b b I ab) = (01.02.441.500 Ft In(ab) =(2001.01 F01 NONE (a) 0 (0.01.2... F

0 0
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Answer #1

1. Answer:

rφ = φ [By annihilator for concatenation]

rt+st = (r+s)t  [By the reverse distribution rule]

r+r = r  [By idempotent law]

φ+r = r [By identity for union]

φ+φ = φ [By simplification]

2. Answer: option b is correct.

  • The input symbols in machine is a and b. so Σ={a, b}.
  • The states are in machine q0, q1 and q2. So Q={q0, q1, q2}.
  • Initial state is q0 because q0 has --> without coming from any state. So s=q0
  • final state is q0. Because final state is represented with two circles. In machine final state represented with double circle.
  • So these condition matching with option b.

3. Ansswer: option b is correct.

  • Because machine is said to be deterministic automato if the machine is read an input string one symbol at a time. In DFA, there is only one path for specific input from the current state to the next state.
  • So this machine following these rules so it is a deterministic finite automato.

4. Answer: option d is correct.

  • The input symbols in machine is a and b. so Σ={a, b}.
  • The states are in machine is q0, q1, q2, q3 and q4. So Q={q0, q1, q2, q3, q4}.
  • Initial state is q0 because q0 has --> without coming from any state. So s=q0.
  • final states are q1, q2 and q4. Because final state is represented with two circles. In machine final state represented with double circle. F={q1, q2, q4}
  • So these condition matching with option d.

Note: my friend if you have any questions or queries comment below. I will sort out your queries. Thank you my friend.

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