Parsing Practice Questions

Both grammars, G1 and G2, can correctly generate the specified C-like array declaration. For a declaration like int a[10][3];, both grammars can derive the string using their production rules.

Grammar G1 uses recursion on the right (E -> num] [E) to generate multiple dimensions.
Grammar G2 uses left recursion (E -> E[num]) to achieve the same result.

Since both grammars can produce valid multi-dimensional array declarations of at least one dimension, both are correct.

A top-down parser begins the parsing process from the start symbol $S$ of a given grammar. It systematically expands the leftmost non-terminal symbol in the current sentential form at each step to match the input string. This strategy of always selecting the leftmost non-terminal to replace ensures that the sequence of production rules applied corresponds precisely to a leftmost derivation. By contrast, bottom-up parsers typically construct the rightmost derivation in reverse. Therefore, a top-down parser generates a leftmost derivation.

To parse the input aab using a bottom-up parser, we perform shift-reduce operations corresponding to the SDTS rules:

1. Shift a. (Stack: a)
2. Shift a. (Stack: aa)
3. Reduce a (the second a) to $S$ using $S \rightarrow a$ $\{print\ 2\}$. Output: 2. (Stack: aS)
4. Shift b. (Stack: aSb)
5. Reduce Sb to $A$ using $A \rightarrow Sb$ $\{print\ 3\}$. Output: 23. (Stack: aA)
6. Reduce aA to $S$ using $S \rightarrow aA$ $\{print\ 1\}$. Output: 231. (Stack: S)

The sequence of printed actions is $2, 3, 1$.

Recursive descent parsing constructs a parse tree starting from the start symbol (the root) and expands non-terminals to match the input tokens (the leaves). This approach of working from the root down to the leaves is the fundamental definition of a top-down parsing strategy. In contrast, bottom-up parsers (such as $LR$ parsers) begin with the input tokens and perform reductions back to the start symbol. Since recursive descent mimics a derivation process (e.g., $S \Rightarrow \alpha$), it is categorized as a top-down parser.

An ambiguous grammar is one that can produce more than one leftmost derivation (or parse tree) for the same string. For the terminal string "c", this grammar has two distinct leftmost derivations:

1. $S \Rightarrow F \Rightarrow c$
2. $S \Rightarrow H \Rightarrow c$
   Since the grammar is ambiguous, it cannot be parsed by any deterministic parsing algorithm. Both LL(1) and LR(1) parsers are deterministic. Therefore, the grammar is neither LL(1) nor LR(1).

In shift-reduce parsing, the stack holds a sequence of grammar symbols. The prefixes of right-sentential forms that can appear on the stack of a shift-reduce parser are called "viable prefixes". By definition, at any point during the parsing process, the contents of the stack form a viable prefix. Therefore, the stack contains only a set of viable prefixes.

LR parsing methods, including $SLR$, are deterministic and strictly require the underlying context-free grammar to be unambiguous, as ambiguity creates unavoidable shift/reduce conflicts. The relationship between these classes is hierarchical, represented as $SLR \subset LALR \subset LR(1)$. While every grammar parsed by $SLR$ must be unambiguous, $SLR$ relies on restricted $FOLLOW$ sets which are insufficient to resolve all possible shift/reduce conflicts in complex structures. Consequently, there exist unambiguous grammars that do not satisfy the requirements of an $SLR$ parser, making the statement in option C correct.

Top-down parsers construct the parse tree starting from the start symbol down to the leaves (input tokens). A Recursive descent parser uses a set of recursive procedures to implement this top-down approach, directly mimicking the grammar structure. Conversely, SLR($k$) and LR($k$) are bottom-up parsers that reduce input symbols toward the start symbol using shift-reduce actions. Since Option B is not a standard parsing method, Recursive descent is the only correct choice.

SLR (Simple LR) parsers are the easiest to implement among the LR parsing methods because they use the smallest parsing tables. They construct FOLLOW sets to determine reductions, which simplifies table generation.

Canonical LR parsers are the most powerful among the three. They construct item sets that keep track of lookahead symbols directly in the states, allowing them to make more precise parsing decisions and handle a wider range of grammars compared to SLR and LALR.

LALR parsers offer a good balance, being more powerful than SLR but less powerful than canonical LR, while having tables the same size as SLR.

Therefore, SLR is the easiest to implement, and canonical LR is the most powerful.

An operator grammar forbids two specific conditions: (i) the existence of $\epsilon$-productions ($A \rightarrow \epsilon$), and (ii) the presence of two adjacent non-terminals on the right-hand side (RHS) of any production rule.

1. $A \rightarrow BC$ contains adjacent variables $B$ and $C$, violating the operator grammar constraint.
2. $A \rightarrow CcBb$ and $3. A \rightarrow BaC$ contain alternating variables and terminals, which is permitted.
3. $A \rightarrow \epsilon$ violates the requirement that no production has an empty RHS.

Thus, rules 1 and 4 violate the requirements of an operator grammar.

The grammar defines $E$ in terms of $F$, placing $F$ at a lower level in the derivation tree. Specifically, the operators $*$ and $+$ are defined within the rules for $E$, while the $-$ operator is defined within the rules for $F$. In formal grammars, operations defined by non-terminals further from the start symbol are evaluated before those closer to the start symbol. Since $F$ is nested within $E$, the subtraction operation ($-$) has higher precedence than both multiplication ($*$) and addition ($+$). Thus, $-$ has higher precedence than $*$.

To determine the associativity of the operators $*$ and $+$, we examine their production rules for left or right recursion.
For operator $*$, the rule is $T \rightarrow T*U$. Since $T$ appears on the left side of itself in the production, it indicates left recursion. Therefore, $*$ is left associative.
For operator $+$, the rule is $P \rightarrow Q+P$. Since $P$ appears on the right side of itself in the production, it indicates right recursion. Therefore, $+$ is right associative.
Combining these observations, $+$ is right associative, while $*$ is left associative.

To evaluate the given grammar, we analyze the properties of the strings generated:

1. Parity of lengths: $|A|$ is always odd because $A \rightarrow a$ (length 1) and $A \rightarrow BaB$ (length $2|B|+1$). Since $|B| = 2 + |A|$, $|B|$ is also odd. Thus, $|S| = |A| + |B|$ is always even (Option A is true).
2. Length of B: As derived, $|B| = 2 + |A|$. Since $|A|$ is odd, $|B|$ is always odd (Option C is true).
3. Consecutive a's: Since $A \rightarrow BaB$ and $B \rightarrow bbA$, every $A$ is enclosed by $bb$ sequences, preventing three consecutive $a$'s (Option B is true).
4. Consecutive b's: We can derive a string containing four $b$'s: $B \rightarrow bbA \rightarrow bb(BaB) \rightarrow bb(bbA)a(bbA) = bbbbAa bbA$. Replacing $A$ with $a$, we get $bbbbaabba$, which contains $bbbb$. Therefore, statement D is false.

The given canonical set of items are $S \rightarrow L. \lt R$ and $Q \rightarrow R$.
When the input symbol is '$<$', we look at the item $S \rightarrow L. \lt R$. Since there is a lookahead symbol '$<$' immediately after the dot, this indicates a shift action is possible (shifting '$<$').
For the item $Q \rightarrow R.$ (where the dot is at the end), it implies a reduce action is possible using the production rule $Q \rightarrow R$.
In LR parsing, if a state has both a shift item (dot before a terminal) and a reduce item (dot at the end), and the terminal for the shift item is also a lookahead for the reduce item, it leads to a shift-reduce conflict. Here, the lookahead for $Q \rightarrow R.$ is not specified, but the shift is on the terminal '$<$'.
If a state has two or more reduce items, it leads to a reduce-reduce conflict. However, we only have one reduce item $Q \rightarrow R.$.
Therefore, there is no shift-reduce conflict as the input symbol '$<$' leads to a shift for $S \rightarrow L. \lt R$ and is not a lookahead for reduction $Q \rightarrow R.$. Also, there is no reduce-reduce conflict because there is only one reduce item.

To derive the string $((()())())$ using the given grammar $S \rightarrow SS$, $S \rightarrow (S)$, and $S \rightarrow \varepsilon$, we can use the recursive property of the number of steps $f(S)$ required to derive a string $S$. The number of steps is defined by the grammar rules: $f(\varepsilon) = 1$, $f(S_1 S_2) = 1 + f(S_1) + f(S_2)$, and $f((S)) = 1 + f(S)$.

For the target string $((()())())$, we decompose it:

1. The string is of the form $S_{total} = (S_A)$, where $S_A = S_B S_C$.
2. $S_B = (()())$, which is $(S_D)$ where $S_D = ()()$.
3. $S_C = ()$, which is $(\varepsilon)$.
4. $f(\varepsilon) = 1$.
5. $f(()) = 1 + f(\varepsilon) = 1 + 1 = 2$.
6. $f(S_D) = f(()()) = 1 + f(()) + f(()) = 1 + 2 + 2 = 5$.
7. $f(S_B) = f((S_D)) = 1 + f(S_D) = 1 + 5 = 6$.
8. $f(S_A) = f(S_B S_C) = 1 + f(S_B) + f(S_C) = 1 + 6 + 2 = 9$.
9. $f(S_{total}) = f((S_A)) = 1 + f(S_A) = 1 + 9 = 10$.

The derivation proceeds as follows (10 steps):

1. $S \Rightarrow (S)$
2. $S \Rightarrow (SS)$
3. $S \Rightarrow ((S)S)$
4. $S \Rightarrow ((SS)S)$
5. $S \Rightarrow (((S)S)S)$
6. $S \Rightarrow (((\varepsilon)S)S)$
7. $S \Rightarrow ((()S)S)$
8. $S \Rightarrow ((()(S))S)$
9. $S \Rightarrow ((()(\varepsilon))S)$
10. $S \Rightarrow ((()())(\varepsilon))$

Shift-reduce parsing constructs a parse tree starting from the input terminals (the leaves) and proceeds toward the root (the start symbol $S$). It utilizes two main operations: "Shift," which pushes input symbols onto a stack, and "Reduce," which replaces a handle with a corresponding non-terminal $A$ based on a production $A \rightarrow \beta$. Because the derivation is constructed from the leaves upwards to the root, this technique is classified as bottom-up parsing, as opposed to top-down methods like predictive or recursive descent. Thus, shift-reduce parsing is definitively bottom-up.

Let's denote the first set of LR(1) items as $I_1$ and the second set as $I_2$.
$I_1 = \{X \rightarrow c.X, \{c, d\}\}, \{X \rightarrow .cX, \{c, d\}\}, \{X \rightarrow .d, \{c, d\}\}$
I_2 = \{X \rightarrow c.X, \{\}}, {X \rightarrow .cX, {$}}, {X \rightarrow .d, {$}}$

In LALR parsing, items are merged if their core (the production rule with the dot position) is identical, regardless of their lookaheads. Both $I_1$ and $I_2$ have identical cores:
$\{X \rightarrow c.X\}, \{X \rightarrow .cX\}, \{X \rightarrow .d\}$
Therefore, these two sets can be merged into a single state in an LALR parser. This contradicts statement 1, so statement 1 is FALSE.

When $I_1$ and $I_2$ are merged, the resulting set will have lookaheads that are the union of the individual lookaheads: \{c, d, \}$. The merged state is:${X \rightarrow c.X, {c, d, $}}, {X \rightarrow .cX, {c, d, $}}, {X \rightarrow .d, {c, d, $}}$
None of the items in this merged state are reduce items (i.e., the dot is not at the end of the production). An S-R conflict occurs if there is a shift item and a reduce item for the same lookahead. An R-R conflict occurs if there are two reduce items for the same lookahead. Since there are no reduce items in the merged state, there cannot be any S-R or R-R conflicts. This makes statements 2 and 3 FALSE.

The 'goto' function on a non-terminal (like X or c) for an LALR state is determined by its core. Since the cores of $I_1$ and $I_2$ are the same, the 'goto' on 'c' (or any other symbol) will lead to the same successor state. This contradicts statement 4, so statement 4 is FALSE.

Since all statements (1, 2, 3, and 4) are false, the

The final answer is $\boxed{D}$

To eliminate left recursion in the grammar $S \rightarrow Aa \mid b$ and $A \rightarrow Ac \mid Sd \mid \epsilon$, we first substitute $S$ into $A$:
$A \rightarrow Ac \mid (Aa \mid b)d \mid \epsilon \implies A \rightarrow Ac \mid Aad \mid bd \mid \epsilon$.
Grouping the terms, we identify the immediate left recursion: $A \rightarrow A(c \mid ad) \mid (bd \mid \epsilon)$.
This follows the form $A \rightarrow A\alpha \mid \beta$, where $\alpha = c \mid ad$ and $\beta = bd \mid \epsilon$.
Applying the standard transformation $A \rightarrow \beta A'$ and $A' \rightarrow \alpha A' \mid \epsilon$:
$A \rightarrow (bd \mid \epsilon)A' \implies A \rightarrow bdA' \mid A'$
$A' \rightarrow (c \mid ad)A' \mid \epsilon \implies A' \rightarrow cA' \mid adA' \mid \epsilon$