Analysing Context-Free Grammar Properties

Let's examine the given context-free grammar:

We need to determine which of the provided statements are true about the strings generated by this grammar. We can analyse the structure of the strings produced by the derivation rules.

Statement (a): Strings produced by the grammar can have consecutive three a's

Let's look at how the terminal symbol 'a' is introduced into a string. The 'a' appears only in the rules for X:

• \(X \rightarrow a\): Introduces a single 'a'.
• \(X \rightarrow YaY\): Introduces an 'a' surrounded by Ys.

Now consider how Y expands: \(Y \rightarrow bbX\). So, the rule \(X \rightarrow YaY\) expands to \(X \rightarrow (bbX) a (bbX)\). This means any 'a' introduced via \(YaY\) will be surrounded by segments starting with 'bb'.

Let's trace some derivations:

• \(S \rightarrow XY\)
• If we use \(X \rightarrow a\), we get \(S \rightarrow aY\). Since \(Y \rightarrow bbX\), this becomes \(S \rightarrow abbX\). The next symbol is 'b'. To get consecutive 'a's, the X needs to start with 'a'. If \(X \rightarrow a\), we get \(S \rightarrow abba\). No consecutive 'a's here.
• If we use \(X \rightarrow YaY\), we get \(S \rightarrow (YaY)Y \rightarrow (bbX a bbX) Y\). The 'a' here is surrounded by 'bbX'. If the X's become 'a', we get \((bb a bb a) Y\). Again, 'a's are separated by 'bb'. The final Y also expands to \(bbX\), adding more 'b's. For example: \((bb a bb a) bb a\). Still no consecutive three 'a's.

In any derivation, an 'a' either comes from \(X \rightarrow a\) or is embedded in \(bbX a bbX\). An 'a' from \(X \rightarrow a\) becomes adjacent to whatever is next in the derivation. If it's the first X in \(S \rightarrow XY\) becoming 'a', we get \(aY \rightarrow abbX\), so 'a' is followed by 'b'. If it's inside \(YaY\), it's surrounded by patterns starting with 'bb'. There is no way to generate three 'a's next to each other.

Therefore, statement (a) is False.

Statement (b): Every string produced by the grammar have alternate a and b

For a string to have alternate 'a' and 'b', it must look like 'ababab...' or 'bababa...'.

Consider the string \(abba\) derived above. It has 'bb', which violates the alternating property. The string \(bbabbabba\) also has 'bb'.

Since we found strings that do not have alternate 'a' and 'b', statement (b) is False.

Statement (c): Every string produced by the grammar have at least two a's

Let's trace the symbols that can generate 'a'. Only X can generate 'a' (either directly or indirectly). The start symbol is S, which generates XY. So, any string generated must come from an X part and a Y part.

• \(S \rightarrow XY\)
• \(Y \rightarrow bbX\). Substituting this into S, we get \(S \rightarrow XbbX\).

Now we have two occurrences of X that must be expanded into terminal strings. Let's analyse what X can generate:

• \(X \rightarrow a\): generates one 'a'.
• \(X \rightarrow YaY \rightarrow bbX a bbX\): generates one 'a' and two more X's.

Every derivation of X into a terminal string must eventually use the rule \(X \rightarrow a\), because it's the only rule for X that doesn't produce non-terminals on the right side (or produces only terminals). Any path starting with \(X \rightarrow YaY \rightarrow bbX a bbX\) introduces an 'a' and two new X's, each of which must eventually generate at least one 'a'. Therefore, any string derived from X must contain at least one 'a'.

Since \(S \rightarrow XbbX\), and each of the two X's must generate at least one 'a', the resulting string must contain at least two 'a's.

Example: The shortest string is \(abba\). It has two 'a's.

Therefore, statement (c) is True.

Statement (d): Every string produced by the grammar have b's in multiple of 2

The terminal symbol 'b' is introduced only by the rule \(Y \rightarrow bbX\), which adds exactly two 'b's to the string being derived.

Let's trace the derivation again starting from S:

• \(S \rightarrow XY\)
• The only rule for Y is \(Y \rightarrow bbX\). So, S must derive \(X(bbX)\). This is the initial introduction of 'b's, specifically two 'b's.

Now we have \(S \rightarrow XbbX\). Further derivation proceeds by expanding the X's.

• \(X \rightarrow a\): This rule introduces no 'b's.
• \(X \rightarrow YaY\): This rule introduces Ys. Since \(Y \rightarrow bbX\), this rule expands to \(X \rightarrow (bbX) a (bbX)\). This step introduces two instances of Y, each adding two 'b's, for a total of \(2 + 2 = 4\) 'b's.

Every time a Y is introduced (which can only happen via \(X \rightarrow YaY\)), exactly two 'b's are added. The initial Y in \(S \rightarrow XY\) also adds two 'b's. The structure of the grammar ensures that 'b's are always introduced in pairs of two via the Y rule.

Let N_b be the total number of 'b's in a derived string. The initial derivation S → XY → XbbX contributes 2 b's. Any subsequent Ys are generated from X → YaY, which expands to bbX a bbX, adding 4 b's (2 from each Y). So, the total number of b's is 2 + 4k, where k is the number of times the production X → YaY is used in the derivation of the string. \(N_b = 2 + 4k = 2(1 + 2k)\). This expression \(2(1+2k)\) is always an even number, meaning the total number of 'b's is always a multiple of 2.

Therefore, statement (d) is True.

Conclusion on Grammar Properties

Based on our analysis:

• Statement (a) is False.
• Statement (b) is False.
• Statement (c) is True.
• Statement (d) is True.

The statements that are true about the grammar are (c) and (d).

Grammar Analysis Revision Table

Revisiting key aspects of the grammar and its properties:

• The grammar is a Context-Free Grammar (CFG).
• The start symbol S immediately leads to S → XY, which implies a structure involving both X and Y components.
• The Y production \(Y \rightarrow bbX\) is the sole source of 'b' terminals. This rule always introduces 'b's in pairs (exactly two).
• The X productions \(X \rightarrow a\) and \(X \rightarrow YaY\) are the only sources of 'a' terminals. Every derivation from X must eventually include \(X \rightarrow a\) to terminate, ensuring at least one 'a' per X derivation.

Additional Information on Language Theory

This problem explores properties of formal languages defined by context-free grammars. Understanding how terminals are introduced and how non-terminals expand is crucial.

• Derivation Tree: Visualizing the derivation using a tree can help understand the structure of the string being generated. Each node represents a non-terminal expanding according to a production rule.
• Language of a Grammar: The set of all terminal strings that can be derived from the start symbol is the language generated by the grammar, denoted L(G).
• Properties of CFGs: CFGs can generate languages with nested structures, but they cannot capture arbitrary dependencies or counts as easily as more powerful grammars (like context-sensitive grammars). Properties like having an equal number of 'a's and 'b's or specific patterns might require careful analysis or might not be possible for a given CFG.
• String Properties: Analysing string properties involves examining all possible derivation paths to see if a property holds for *every* string (like statements c and d) or if it holds for *at least one* string (like statement a).

For this grammar, the mandatory introduction of 'bb' from every Y, and the fact that every X must yield at least one 'a', are key to determining the string properties.