In the following grammar X :: = X⊕Y / Y Y :: = Z*Y / Z Z :: = id Which of the…

For multiple '⊕', the derivation is possible only via 'X' which is on left side of '⊕' in the production. Hence it is left associative.

For multiple '∗', the derivation is possible only via 'Y' which is on the right side of '∗' in the production. Hence it is right associative.

Note for more Understanding

1. Showing that ⊕ is Left-Associative

Let's look at the grammar rule: X → X ⊕ Y | Y

We will generate the string: id1 ⊕ id2 ⊕ id3

The Step-by-Step Derivation:

To get two ⊕ signs, we must apply the left-recursive rule X → X ⊕ Y first:

1. Start with the root: X

2. Apply X → X ⊕ Y

3. Expand the leftmost X again using X → X ⊕ Y ⇒ (X ⊕ Y) ⊕ Y

4. Now, safely turn the remaining non-terminals into IDs using the base rules (X → Y → Z → id):

   • First X → id1

   • Middle Y → id2

   • Rightmost Y → id3

The Parse Tree Structure:

Plaintext

         X
       / | \
     X   ⊕  Y
    /|\      |
   X ⊕ Y     Z
   |   |     |
   Y   Z     id₃
   |   |
   Z   id₂
   |
  id₁

Why this means Left-Associative:

In a compiler parse tree, bottom-up execution takes place. The operations deeper in the tree must be calculated before their parent nodes can resolve.

• Look at the tree: ( id1 ⊕ id2) is grouped together at the very bottom left.

• The compiler is forced to calculate ( id1 ⊕ id2) first, and then take that result to compute ⊕ id3

• Execution moves from Left to Right.

2. Showing that * is Right-Associative

Let's look at the grammar rule: Y → Z * Y | Z

We will generate the string: id1 * id2 * id3

The Step-by-Step Derivation:

Because the recursive variable Y is on the right side, the expansion grows to the right:

1. Start with the root variable for this level: Y

2. Apply Y → Z * Y

3. Expand the rightmost Y using Y → Z * Y ⇒ Z * (Z * Y)

4. Turn all non-terminals into IDs via Z → id and the terminating Y → Z → id:

   • Leftmost Z → id1

   • Middle Z → id2

   • Rightmost Y → id3

The Parse Tree Structure:

Plaintext

      Y
    / |  \
   Z  *   Y
   |     /|\
  id₁   Z * Y
        |    |
        id₂  Z
             |
            id₃

Why this means Right-Associative:

Look at where the grouping happens now:

• The sub-tree containing (id2 * id3) is isolated at the very bottom right.

• The compiler cannot evaluate the first * operator until it knows what the right-hand side equals. Therefore, it is forced to evaluate (id2 * id3) first.

• Once that is calculated, it multiplies it with id1

• Execution moves from Right to Left.