Let’s break down and answer **each part step by step**, as requested! --- ## Step 1: Introduction to the Concept This problem is about the **decidability of the emptiness problem for context-free grammars (CFGs)**. Given a CFG \(G\), the emptiness problem asks whether the language generated by \(G\) is empty, i.e., \(L(G) = \emptyset\). We are to describe how to encode a CFG as a string and describe a Turing Machine (TM) that decides whether a given CFG's language is empty. --- ## Step 2: (a) Encoding **Encoding:** To encode a CFG \(G = (V, \Sigma, R, S)\) as a string \(\langle G \rangle\), we need to represent all components using a finite alphabet (e.g., ASCII symbols): - **Variables (\(V\))**: List each variable, separated by commas (e.g., \(A,B,C\)). - **Terminals (\(\Sigma\))**: List each terminal, separated by commas (e.g., \(a,b\)). - **Start Symbol (\(S\))**: State the start symbol (e.g., \(S\)). - **Rules (\(R\))**: Each rule is encoded as `LHS->RHS`, where LHS is a variable, and RHS is a string of variables and terminals. Rules are separated by semicolons (e.g., `S->aA;A->b`). **Example Encoding:** Suppose \(G = (\{S, A\}, \{a, b\}, \{S \rightarrow aA, A \rightarrow b\}, S)\). The encoding could be: `Variables:S,A#Terminals:a,b#Start:S#Rules:S->aA;A->b` This encoding uses `#` as a separator between sections and `;`/`,` for lists. Every CFG has a unique, well-formed encoding this way. --- ## Step 3: (b) Decider (High-Level Algorithm / Pseudocode) **Goal:** Describe an algorithm that decides if \(L(G) = \emptyset\) for a given CFG encoding \(\langle G \rangle\). **High-Level Algorithm:** 1. **Input:** String \(\langle G \rangle\) encoding a CFG \(G\). 2. **Marking:** - Mark all variables that produce a terminal string directly (i.e., have a rule like \(A \rightarrow w\), where \(w \in \Sigma^*\)). 3. **Propagation:** - Repeatedly mark any variable \(A\) where there is a rule \(A \rightarrow X_1X_2...X_k\), and all \(X_i\) are either terminals or already marked variables. 4. **Check:** - If the start symbol \(S\) is marked, **accept** (i.e., \(L(G) \neq \emptyset\)). - If \(S\) is not marked after no new markings are possible, **reject** (i.e., \(L(G) = \emptyset\)). **Pseudocode:** ``` Input: encoding ⟨G⟩ of CFG G = (V, Σ, R, S) Initialize Marked = {} Repeat For each rule A → X1 X2 ... Xk in R If each Xi is in Σ or in Marked Add A to Marked Until no new variables are added to Marked If S is in Marked Reject (L(G) ≠ ∅) Else Accept (L(G) = ∅) ``` --- ## Step 4: Final Answer (Summary in Few Sentences) To decide if a CFG's language is empty, encode the CFG as a string using clear separators for variables, terminals, start symbol, and rules. Then, use a marking algorithm to see if the start variable can derive any terminal string. If it can, the language is **not** empty; otherwise, it **is** empty. --- ## Step 5: What We Did in Each Step 1. **Introduced** the problem and concept (decidability of CFG emptiness). 2. **Defined** a systematic encoding for CFGs as strings. 3. **Outlined** a high-level algorithm (decider) for the emptiness problem. 4. **Summarized** the core approach in a few sentences. 5. **Explained** each step's purpose and connection.