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Construct NPDA's that accept the following regular languages:

a) L1 = L(aaa*bab)

b) L2 = L(aab*aba*)

c) The union of L1 and L2

1 Answer

5 votes

Answer:

Step-by-step explanation:

a) To construct an NPDA that accepts the language L1 = L(aaa*bab), we can follow these steps:

Start with a single initial state, q0.

Create a transition from q0 to a new state, q1, upon reading 'a'.

From q1, create a transition to q2 upon reading 'a' again.

From q2, create a transition to q3 upon reading 'a' again.

From q3, create a transition to q3 upon reading 'a' (looping on 'a').

From q3, create a transition to q4 upon reading 'b'.

From q4, create a transition to q5 upon reading 'a'.

From q5, create a transition to q6 upon reading 'b'.

From q6, create a transition to a new state, q7, upon reading the end of the input symbol ('$').

Additionally, make sure to include empty transitions (epsilon transitions) for transitions that do not consume any input symbols.

b) To construct an NPDA that accepts the language L2 = L(aababa), we can follow these steps:

Start with a single initial state, q0.

Create a transition from q0 to a new state, q1, upon reading 'a'.

From q1, create a transition to q2 upon reading 'a'.

From q2, create a transition to q3 upon reading 'b'.

From q3, create a transition to q4 upon reading 'a'.

From q4, create a transition to q5 upon reading 'a' (looping on 'a').

From q5, create a transition to q6 upon reading 'b'.

From q6, create a transition to a new state, q7, upon reading the end of the input symbol ('$').

c) To construct an NPDA that accepts the union of L1 and L2, we can combine the NPDA for L1 and the NPDA for L2 as follows:

Create a new initial state, q0, and connect it with epsilon transitions to the initial states of the NPDA for L1 and L2.

Connect the accepting states of the NPDA for L1 and L2 with epsilon transitions to a new accepting state.

This combined NPDA will accept strings that belong to either L1 or L2, representing the union of the two languages.

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