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Parse Tree | Derivations

Parse Tree-

  • The process of deriving a string is called as derivation.
  • The geometrical representation of a derivation is called as a parse tree or derivation tree.
Example-


Let us consider a string w = aaabbabbba
Now, let us derive the string w using leftmost derivation.

Leftmost Derivation-

S   → aB
→  aaBB                   (Using B → aBB)

→ aaaBBB                (Using B → aBB)
→ aaabBB                (Using B → b)
→ aaabbB                (Using B → b)
→ aaabbaBB            (Using B → aBB)
→ aaabbabB            (Using B → b)
→ aaabbabbS          (Using B → bS)
→ aaabbabbbA        (Using S → bA)
→ aaabbabbba         (Using A → a)

2. Rightmost Derivation-

  • The process of deriving a string by expanding the rightmost non-terminal at each step is called as rightmost derivation.
  • The geometrical representation of rightmost derivation is called as a rightmost derivation tree.

Example-

Consider the following grammar-
S → aB / bA
S → aS / bAA / a
B → bS / aBB / b
(Unambiguous Grammar)
Let us consider a string w = aaabbabbba
Now, let us derive the string w using rightmost derivation.

Rightmost Derivation-

S   → aB
→  aaBB                    (Using B → aBB)
→ aaBaBB                 (Using B → aBB)
→ aaBaBbS               (Using B → bS)
→ aaBaBbbA             (Using S → bA)
→ aaBaBbba              (Using A → a)
→ aaBabbba              (Using B → b)
→ aaaBBabbba          (Using B → aBB)
→ aaaBbabbba          (Using B → b)
→ aaabbabbba           (Using B → b)


Properties Of Parse Tree-

  • Root node of a parse tree is the start symbol of the grammar.
  • Each leaf node of a parse tree represents a terminal symbol.
  • Each interior node of a parse tree represents a non-terminal symbol.
  • Parse tree is independent of the order in which the productions are used during derivations.

Yield Of Parse Tree-

  • Concatenating the leaves of a parse tree from the left produces a string of terminals.
  • This string of terminals is called as yield of a parse tree.

PRACTICE PROBLEMS BASED ON DERIVATIONS AND PARSE TREE-

Problem-01:

Consider the grammar-
S → bB / aA
A → b / bS / aAA
B → a / aS / bBB
For the string w = bbaababa, find-
  1. Leftmost derivation
  2. Rightmost derivation
  3. Parse Tree

Solution-

1. Leftmost Derivation-

S   → bB
→ bbBB              (Using B → bBB)
→ bbaB              (Using B → a)
→ bbaaS            (Using B → aS)
→ bbaabB          (Using S → bB)
→ bbaabaS        (Using B → aS)
→ bbaababB      (Using S → bB)
→ bbaababa       (Using B → a)

2. Rightmost Derivation-

S   → bB
→ bbBB                (Using B → bBB)
→ bbBaS              (Using B → aS)
→ bbBabB            (Using S → bB)
→ bbBabaS          (Using B → aS)
→ bbBababB        (Using S → bB)
→ bbBababa        (Using B → a)
→ bbaababa         (Using B → a)

3. Parse Tree-

 

  • Whether we consider the leftmost derivation or rightmost derivation, we get the above parse tree.
  • The reason is given grammar is unambiguous.

Problem-02:

Consider the grammar-
S → A1B
A → 0A / ∈
B → 0B / 1B / ∈
For the string w = 00101, find-
  1. Leftmost derivation
  2. Rightmost derivation
  3. Parse Tree

Solution-

1. Leftmost Derivation-

S   → A1B
→ 0A1B              (Using A → 0A)
→ 00A1B            (Using A → 0A)
→ 001B              (Using A → ∈)
→ 0010B            (Using B → 0B)
→ 00101B          (Using B → 1B)
→ 00101             (Using B → ∈)

2. Rightmost Derivation-

S   → A1B
→ A10B                (Using B → 0B)
→ A101B              (Using B → 1B)
→ A101                (Using B → ∈)
→ 0A101              (Using A → 0A)
→ 00A101            (Using A → 0A)

→ 00101               (Using A → ∈)

3. Parse Tree-

Whether we consider the leftmost derivation or rightmost derivation, we get the above parse tree.
  • The reason is given grammar is unambiguous.






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