GATE CS 2015 Set 2 Q32 — Normal Forms

A full natural outer join ($R_1 \bowtie_{outer} R_2$) combines the matching tuples of both relations based on their common attributes, and also preserves the unmatched tuples from both relations by padding the missing attributes with null.

Given relations:

• $R_1(A, B) = \{(1, 5), (3, 7)\}$
• $R_2(A, C) = \{(1, 7), (4, 9)\}$

The common attribute is $A$. Let's compute the join step-by-step:

1. Matching Tuples (Inner Join):
   • For $A = 1$, we have $(1, 5)$ in $R_1$ and $(1, 7)$ in $R_2$. Joining them yields: $$(1, 5, 7) \quad \text{(which is tuple } e\text{)}$$
2. Unmatched Tuples from $R_1$ (Left Outer Join component):
   • For $A = 3$, the tuple $(3, 7)$ in $R_1$ has no matching tuple in $R_2$. We pad the attribute $C$ with null: $$(3, 7, \text{null}) \quad \text{(which is tuple } f\text{)}$$
3. Unmatched Tuples from $R_2$ (Right Outer Join component):
   • For $A = 4$, the tuple $(4, 9)$ in $R_2$ has no matching tuple in $R_1$. We pad the attribute $B$ with null: $$(4, \text{null}, 9) \quad \text{(which is tuple } g\text{)}$$

Thus, the full natural outer join relation $R(A, B, C)$ contains exactly the tuples: $$R = \{e, f, g\}$$

Therefore, $R$ contains $e, f, g$ but does not contain $a, b, c, d$. This matches Option C.