How many productions will be there, after constructing the reduced grammar for the given grammar below?
\[ \text{1. } X \rightarrow aYa \text{2. } Y \rightarrow Xb \text{3. } Y \rightarrow bCC \text{4. } C \rightarrow ab \text{5. } E \rightarrow aC \text{6. } Z \rightarrow aZY \]
Step 1: Reducing the grammar.
The given grammar contains several rules that can be reduced. We will look at the productions involving non-terminal \( Y \) and attempt to eliminate unnecessary rules.
1. \( X \rightarrow aYa \): This production can be kept as is.
2. \( Y \rightarrow Xb \): Substitute \( X \) from the first rule, so \( Y \rightarrow (aYa)b \), which can be reduced.
3. \( Y \rightarrow bCC \): Can be reduced by eliminating \( C \) in subsequent steps.
4. \( C \rightarrow ab \): Reduced to a single rule.
5. \( E \rightarrow aC \): A single rule for \( E \).
6. \( Z \rightarrow aZY \): This can be reduced to a simpler production.
After simplifying the above rules, we are left with 5 production rules in total.
Step 2: Conclusion.
Thus, the correct number of reduced productions is 5, so the correct answer is (3) Five.
Find the least upper bound and greatest lower bound of \( S = \{X, Y, Z\} \) if they exist, of the poset whose Hasse diagram is shown below:
Suppose \( D_1 = (S_1, \Sigma, q_1, F_1, \delta_1) \) and \( D_2 = (S_2, \Sigma, q_2, F_2, \delta_2) \) are finite automata accepting languages \( L_1 \) and \( L_2 \), respectively. Then, which of the following languages will also be accepted by the finite automata:
(A) \( L_1 \cup L_2 \)
(B) \( L_1 \cap L_2 \)
(C) \( L_1 - L_2 \)
(D) \( L_2 - L_1 \)
Choose the correct answer from the options given below: