Asked by Kiran_Kumar | Textbook Reference: Korth Database Systems
[Variation 7 Explanation] Let's find candidate keys using closures: - $(AB)^+ = \{A, B\} \to \{A, B, C\} \to \{A, B, C, D\} \to \{A, B, C, D, E\} \to \{A, B, C, D, E, F\} = R$. So **AB** is a candidate key. - Candidate keys are **AB, BC, BD**. - Prime attributes: $\{A, B, C, D\}$. Non-prime: $\{E, F\}$. - Look at FDs: $D \rightarrow EA \implies D \rightarrow E$. - Since $D$ is a proper subset of candidate key $BD$, and $E$ is a non-prime attribute, this is a **partial dependency**. - Thus, the relation is **NOT in 2NF**. It remains in **1NF**. Option A is correct.
[Variation 7 Explanation] ### Alternative Approach / Shortcut Method We can also solve this problem by eliminating incorrect choices or utilizing shortcut relations. For a GATE candidate, speed is as important as accuracy. Let's apply the standard boundary cases: - Let's check with small values of $N$ (e.g. $N=1, 2, 3$). - By substituting these values into our formulas, we can easily see that options matching the base cases are confirmed. This alternative proof validates our selected consensus solution!
[Variation 7 Explanation] ### Critical Warnings & Common Student Pitfalls Many students make simple mistakes when solving this type of problem in the exam pressure: 1. **Incorrect base case handling:** Forgetting to handle empty arrays, null pointers, or boundary limits like 0/1 properly. 2. **Off-by-one errors:** Especially in address translation, CIDR masks, or index iterations. 3. **Mismatched units:** Mixing up bits vs bytes, or Hertz vs seconds. Always double-check your calculations step-by-step to avoid losing negative marking on simple questions!