answer:a. Leftmost Derivation for the sentence S: (1 then # 2 else 3) 4 To perform a leftmost derivation, we start with the start symbol and in each step, we replace the leftmost non-terminal with one of its productions until we reach the terminal symbols that make up the sentence S. 1. Start with the start symbol `expr` 2. Since we have a parenthesis at the beginning of S, we should use the production `expr -> '(' expr ')'`. `expr -> '(' expr ')'` 3. Inside the parentheses, we have another expression that follows the structure of a conditional expression `expr 'then' expr 'else' expr`. We apply this production. `expr -> '(' expr 'then' expr 'else' expr ')'` 4. Now we have to derive the leftmost `expr` which is a plain INT (the number 1 in our sentence). `expr -> '(' INT 'then' expr 'else' expr ')'` 5. Next, we have the `expr` that follows the '#' unary operator. We use the production `expr -> '#' expr`. `expr -> '(' INT 'then' '#' expr 'else' expr ')'` 6. The `expr` after the '#' is another INT (the number 2 in our sentence). `expr -> '(' INT 'then' '#' INT 'else' expr ')'` 7. Now, the only `expr` left inside the parentheses is the INT 3. `expr -> '(' INT 'then' '#' INT 'else' INT ')'` 8. We replace the INTs with their corresponding tokens: `expr -> '(' 1 'then' '#' 2 'else' 3 ')'` 9. Finally, we have to append the `'' expr` part of the sentence S. Since we have a binary operator structure, we use the production `expr -> expr '' expr`. `expr -> expr '' expr` 10. The left side of the '' operator is now fully derived as `'( 1 then # 2 else 3 )'`. So only the remaining right side of the '' operator is an INT (the number 4 in our sentence): `expr -> '(' 1 'then' '#' 2 'else' 3 ')' '' INT` 11. Replacing the INT with the token 4 gives the final derived sentence S: `expr -> '(' 1 'then' '#' 2 'else' 3 ')' '' 4` Putting the steps together, we have the leftmost derivation sequence for the given sentence S: ``` expr -> '(' expr ')' -> '(' expr 'then' expr 'else' expr ')' -> '(' INT 'then' expr 'else' expr ')' -> '(' 1 'then' expr 'else' expr ')' -> '(' 1 'then' '#' expr 'else' expr ')' -> '(' 1 'then' '#' INT 'else' expr ')' -> '(' 1 'then' '#' 2 'else' expr ')' -> '(' 1 'then' '#' 2 'else' INT ')' -> '(' 1 'then' '#' 2 'else' 3 ')' -> expr '' expr -> '(' 1 'then' '#' 2 'else' 3 ')' '' expr -> '(' 1 'then' '#' 2 'else' 3 ')' '' INT -> '(' 1 'then' '#' 2 'else' 3 ')' '' 4 ``` b. Showing Ambiguity in the Grammar To demonstrate that a grammar is ambiguous, we need to show that there is at least one string in the language that the grammar generates that has two different parse trees (or equivalently, two different leftmost or rightmost derivations that result in the same string). Consider the string `1 2 3`. This string can be derived in at least two different ways: First derivation (left associativity of ``): 1. expr -> expr '' expr 2. expr -> expr '' expr '' expr 3. expr -> INT '' expr '' expr 4. expr -> 1 '' expr '' expr 5. expr -> 1 '' INT '' expr 6. expr -> 1 '' 2 '' expr 7. expr -> 1 '' 2 '' INT 8. expr -> 1 '' 2 '' 3 Second derivation (right associativity of ``): 1. expr -> expr '' expr 2. expr -> expr '' (expr '' expr) 3. expr -> expr '' (INT '' expr) 4. expr -> expr '' (2 '' expr) 5. expr -> expr '' (2 '' INT) 6. expr -> expr '' (2 '' 3) 7. expr -> INT '' (2 '' 3) 8. expr -> 1 '' (2 '' 3) Both derivations yield the same string `1 2 3` but associate the operations differently, meaning the grammar allows for two different interpretations (parse trees) of the same string. This demonstrates the grammar is ambiguous since it does not unambiguously dictate the structure of the expression. Specifically, the binary operator `` does not have a defined associativity within the grammar, leading to ambiguity.

answer:a. Leftmost derivation for the sentence S containing tokens (1 then # 2 else 3) 4: 1. Start with the non-terminal `expr`. 2. Apply the rule `expr → expr '' expr` to introduce ``. ``` expr → expr '' expr ``` 3. Apply the rule `expr → '(' expr ')'` to handle the left expression of ``. ``` expr '' expr → '(' expr ')' '' expr ``` 4. Inside the parentheses, apply the rule `expr → expr 'then' expr 'else' expr` to introduce `then` and `else`. ``` '(' expr ')' '' expr → '(' expr 'then' expr 'else' expr ')' '' expr ``` 5. Resolve the first expression within parentheses to an `INT`. ``` '(' expr 'then' expr 'else' expr ')' '' expr → '(' INT 'then' expr 'else' expr ')' '' expr → '(' 1 'then' expr 'else' expr ')' '' expr ``` 6. Apply the rule `expr → '#' expr` for next part of the `then` expression. ``` '(' 1 'then' expr 'else' expr ')' '' expr → '(' 1 'then' '#' expr 'else' expr ')' '' expr ``` 7. Resolve the expression after `#` to an `INT`. ``` '(' 1 'then' '#' expr 'else' expr ')' '' expr → '(' 1 'then' '#' INT 'else' expr ')' '' expr → '(' 1 'then' '#' 2 'else' expr ')' '' expr ``` 8. Resolve the last expression within parentheses, following the `else`, to an `INT`. ``` '(' 1 'then' '#' 2 'else' expr ')' '' expr → '(' 1 'then' '#' 2 'else' INT ')' '' expr → '(' 1 'then' '#' 2 'else' 3 ')' '' expr ``` 9. Finally, resolve the rightmost `expr` to an `INT` to complete the sentence. ``` '(' 1 'then' '#' 2 'else' 3 ')' '' expr → '(' 1 'then' '#' 2 'else' 3 ')' '' INT → '(' 1 'then' '#' 2 'else' 3 ')' '' 4 ``` So, the leftmost derivation for the sentence S is: ``` expr → expr '' expr → '(' expr ')' '' expr → '(' expr 'then' expr 'else' expr ')' '' expr → '(' INT 'then' expr 'else' expr ')' '' expr → '(' 1 'then' expr 'else' expr ')' '' expr → '(' 1 'then' '#' expr 'else' expr ')' '' expr → '(' 1 'then' '#' INT 'else' expr ')' '' expr → '(' 1 'then' '#' 2 'else' expr ')' '' expr → '(' 1 'then' '#' 2 'else' INT ')' '' expr → '(' 1 'then' '#' 2 'else' 3 ')' '' expr → '(' 1 'then' '#' 2 'else' 3 ')' '' INT → '(' 1 'then' '#' 2 'else' 3 ')' '' 4 ``` b. To show that the grammar is ambiguous, we need to find a string for which there are two different parse trees (or equivalently, two different leftmost or rightmost derivations). Consider the string `1 2 : 3`. There are two possible interpretations (parse trees) for this string: Interpretation 1: ``` expr → expr '' expr → INT '' expr → 1 '' expr → 1 '' expr ':' expr → 1 '' INT ':' expr → 1 '' 2 ':' expr → 1 '' 2 ':' INT → 1 '' 2 ':' 3 ``` Here, the expression is grouped as `(1 2) : 3`. Interpretation 2: ``` expr → expr ':' expr → expr '' expr ':' expr → INT '' expr ':' expr → 1 '' expr ':' expr → 1 '' expr ':' INT → 1 '' INT ':' INT → 1 '' 2 ':' INT → 1 '' 2 ':' 3 ``` Here, the expression is grouped as `1 (2 : 3)`. Each interpretation provides a different parse tree, which means the grammar is ambiguous because the string `1 2 : 3` can be parsed in two different ways, leading to two different meanings without additional context or precedence rules. This ambiguity arises because the grammar does not specify operator precedence or associativity, allowing such expressions to group in various ways. To resolve the ambiguity, one would typically introduce different levels of non-terminals for different precedence levels, or use additional rules that enforce associativity.

answer:a. Leftmost Derivation for Sentence S: (1 then # 2 else 3) 4 1. Start with the non-terminal 'expr'. 2. Apply the rule `expr → expr '' expr` because we have a '' operator in the sentence. `expr => expr '' expr` 3. Apply the rule `expr → '(' expr ')'` to the leftmost 'expr' to handle the parentheses. `expr '' expr => '(' expr ')' '' expr` 4. Apply the rule `expr → expr 'then' expr 'else' expr` to the 'expr' inside the parentheses. `'(' expr ')' '' expr => '(' expr 'then' expr 'else' expr ')' '' expr` 5. Replace the first 'expr' after 'then' with `'#' expr` since we see a use of the unary operator '#' before '2'. `'(' expr 'then' expr 'else' expr ')' '' expr => '(' expr 'then' '#' expr 'else' expr ')' '' expr` 6. Apply the rule `expr → INT` to all 'expr' that represent integer values to obtain the numeral '1', '2', and '3'. `'(' expr 'then' '#' expr 'else' expr ')' '' expr => '(' INT 'then' '#' expr 'else' INT ')' '' expr` `'(' INT 'then' '#' INT 'else' INT ')' '' expr => '(' INT 'then' '#' INT 'else' INT ')' '' INT` 7. Apply the rule `expr → INT` to the 'expr' representing the number '4'. `'(' INT 'then' '#' INT 'else' INT ')' '' INT => '(' INT 'then' '#' INT 'else' INT ')' '' INT` 8. Plug in the actual INT values into our derivation. `'(' INT 'then' '#' INT 'else' INT ')' '' INT => '(1 then # 2 else 3) 4` And this completes the leftmost derivation. b. Proof of Ambiguity for the Given Grammar A grammar is considered ambiguous if there exists a string that can be generated by the grammar in more than one way, meaning it has more than one leftmost derivation (or more than one parse tree). The given grammar is ambiguous because there are expressions which can be derived in multiple ways due to the lack of precedence or association rules among the operators 'then', 'else', '', ':', and '#'. To demonstrate the ambiguity, consider the expression "1 2 : 3". This expression can be derived in two different ways: First derivation: 1. `expr` 2. Apply the rule `expr → expr '' expr` `expr => expr '' expr` 3. Apply the rule `expr → expr ':' expr` to the second 'expr' (right association) `expr '' expr => expr '' (expr ':' expr)` 4. Convert all 'expr' to INT `expr '' (expr ':' expr) => INT '' (INT ':' INT)` 5. Substitute INT values `INT '' (INT ':' INT) => 1 2 : 3` Second derivation: 1. `expr` 2. Apply the rule `expr → expr ':' expr` `expr => expr ':' expr` 3. Apply the rule `expr → expr '' expr` to the first 'expr' (left association) `expr ':' expr => (expr '' expr) ':' expr` 4. Convert all 'expr' to INT `(expr '' expr) ':' expr => (INT '' INT) ':' INT` 5. Substitute INT values `(INT '' INT) ':' INT => 1 2 : 3` These two derivations lead to different parse trees, each implying a different precedence between the '' and ':' operators. The left association treats '' as having higher precedence than ':', while right association does the opposite. Since the grammar does not specify operator precedence or associativity, it is ambiguous. Similarly, expressions involving the 'then' and 'else' tokens without clear precedence would also demonstrate the ambiguity, as well as expressions mixing the unary '#' with other operators, since the grammar does not specify how these operations associate in relation to one another.

answer:a. To show a leftmost derivation for the sentence S containing tokens (1 then # 2 else 3) 4, we start with the initial non-terminal `expr` and apply the production rules to derive the sentence S. A leftmost derivation means we always expand the leftmost non-terminal at each step. Here's the leftmost derivation for the sentence: 1. expr 2. expr '' expr (apply the rule expr -> expr '' expr) 3. '(' expr ')' '' expr (apply the rule expr -> '(' expr ')') 4. '(' expr 'then' expr 'else' expr ')' '' expr (apply the rule expr -> expr 'then' expr 'else' expr) 5. '(' INT 'then' expr 'else' expr ')' '' expr (apply the rule expr -> INT, replacing the first expr after '(' with INT) 6. '(' INT 'then' '#' expr 'else' expr ')' '' expr (apply the rule expr -> '#' expr) 7. '(' INT 'then' '#' INT 'else' expr ')' '' expr (apply the rule expr -> INT, replacing the expr after '#') 8. '(' INT 'then' '#' INT 'else' INT ')' '' expr (apply the rule expr -> INT, for the remaining expr within the parentheses) 9. '(' INT 'then' '#' INT 'else' INT ')' '' INT (apply the last rule expr -> INT, to derive the remaining right-hand side expr) 10. (1 then # 2 else 3) 4 (replace INT with the corresponding numbers from the sentence) This is the leftmost derivation for the sentence S: (1 then # 2 else 3) 4. b. To show that the grammar is ambiguous, we need to demonstrate that at least one string has more than one leftmost derivation or parse tree (also called a syntax tree). By definition, a grammar is ambiguous if a string of tokens can be parsed in more than one way without changing the sequence of tokens. Let's use the sentence containing tokens "1 2 3" to show the ambiguity: One possible leftmost derivation could be: 1. expr 2. expr '' expr (First apply the rule for the first '') 3. expr '' expr '' expr 4. INT '' expr '' expr (Replace the leftmost expr with INT) 5. 1 '' expr '' expr (Replace INT with 1) 6. 1 '' INT '' expr (Replace the next expr with INT) 7. 1 '' 2 '' expr (Replace INT with 2) 8. 1 '' 2 '' INT (Finally, replace the last expr with INT) 9. 1 '' 2 '' 3 (Replace INT with 3) Another possible leftmost derivation for the same sentence could be: 1. expr 2. expr '' expr (This time apply the rule for the second '') 3. expr '' expr '' expr 4. INT '' expr '' expr (Replace the leftmost expr with INT) 5. 1 '' expr '' expr (Replace INT with 1) 6. 1 '' expr '' INT (Move to the next expr and apply INT) 7. 1 '' expr '' 3 (Replace INT with 3) 8. 1 '' INT '' 3 (Now go back to the first remaining expr and replace it with INT) 9. 1 '' 2 '' 3 (Replace INT with 2) As we can see, the same string can be derived using different orders of rule application. These different derivation sequences correspond to different parse trees, meaning the language is inherently ambiguous. This ambiguity arises because the grammar does not enforce precedence or associativity rules on the operator '', leading to multiple valid parse trees for certain strings.