Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Comp-B3, Data Bases & File Systems — National Exams, December 2014. 3 hours, closed book (calculators permitted). Candidates were instructed to answer five questions: one of Questions 1/2, one of Questions 3/4, and three of Questions 5–8 — only those five are marked. All 8 questions are answered below for completeness (this is a study resource covering the full syllabus).
Check: the source page header prints “98-Comp-B3/ December 2014” while the title block prints “National Exams May 2014” — a date inconsistency on the printed cover page. This is treated as the December 2014 exam period, matching every subsequent page header. Also, the page-1 marking scheme lists Question 8 as having two part-(c) entries (“(c) 4 marks; (c) 6 marks”); read as a mislabelled (c)/(d), matching the body text's actual four sub-parts (a)(b)(c)(d).
Reference texts: Silberschatz, Korth & Sudarshan, Database System Concepts (7th ed.) — RAID, indexing, ER modelling, normal forms, transactions and serializability; Ramakrishnan & Gehrke, Database Management Systems (3rd ed.) — B+-tree indexing, SQL, relational algebra, and concurrency control.
Find. (a) a minimal cover of F; (b) whether $XZA\rightarrow YB \in F^+$; (c)/(d) whether the given binary decomposition is lossless-join and dependency-preserving.
Approach. Compute everything from ATTRIBUTE CLOSURE ($X^+$ under F): singleton-ize right-hand sides, drop the resulting trivial FD, test each remaining FD for redundancy and each multi-attribute left-hand side for an extraneous attribute to get the minimal cover; test implication and superkey-hood by direct closure; test dependency preservation with the standard closure-restricted-to-each-piece algorithm.
(a) Minimal cover. Split right-hand sides to singletons: $XZ\to Z,\ XZ\to Y,\ XZ\to B,\ YA\to C,\ YA\to G,\ C\to W,\ B\to G,\ XZ\to G$. Drop the trivial $XZ\to Z$ (Z is already in its own left side). Test each remaining FD for redundancy by recomputing $\text{LHS}^+$ using only the OTHER FDs: $XZ\to G$ is redundant, since $\{X,Z\}^+$ via $XZ\to Y,\ XZ\to B,\ B\to G$ already reaches $\{X,Z,Y,B,G\}$ — drop it. The remaining five ($XZ\to Y,\ XZ\to B,\ YA\to C,\ YA\to G,\ C\to W,\ B\to G$, six total after re-listing) are each independently non-redundant, and neither $X$ nor $Z$ (nor $Y$ nor $A$) is extraneous in the two-attribute left sides (dropping either one, alone, derives nothing). $$\boxed{F_{min} = \{XZ\to Y,\ XZ\to B,\ YA\to C,\ YA\to G,\ C\to W,\ B\to G\}}$$ (equivalently grouped: $XZ\to BY,\ YA\to CG,\ C\to W,\ B\to G$).
(b) Is $XZA\to YB$ implied by F? Compute $\{X,Z,A\}^+$: $XZ\to Y$ adds $Y$; now $YA\to C$ fires (add $C$), $YA\to G$ fires (add $G$); $C\to W$ adds $W$; $XZ\to B$ adds $B$. Closure reaches $\{X,Z,A,Y,C,G,W,B\}$ — every attribute of $R$. $$\boxed{\{X,Z,A\}^+ = R \Rightarrow XZA \text{ is a superkey of } R \Rightarrow XZA\to YB \text{ IS implied}}$$ (in fact $XZA$ functionally determines every attribute, so it implies far more than just $YB$).
(c) Is $\{XZYAB,\ YABCGW\}$ lossless-join? The standard binary test: lossless iff $(R_1\cap R_2)^+$ is a superkey of $R_1$ or $R_2$. Here $R_1\cap R_2 = \{Y,A,B\}$. Compute $\{Y,A,B\}^+$: $YA\to C$ adds $C$; $YA\to G$ (also $B\to G$) adds $G$; $C\to W$ adds $W$. Closure $=\{Y,A,B,C,G,W\} = R_2$ exactly. $$\boxed{\{Y,A,B\}^+ = R_2 \Rightarrow YAB \text{ is a superkey of } R_2 \Rightarrow \text{LOSSLESS-JOIN}}$$
(d) Is the decomposition dependency preserving? Test every FD in the minimal cover with the closure-restricted-to-$R_1,R_2$ algorithm (grow a result set by repeatedly intersecting with each $R_i$, closing within F, and re-intersecting): each of the six FDs' right-hand side is recovered this way (e.g. for $XZ\to Y$: restricting $\{X,Z\}$ to $R_1$ stays $\{X,Z\}$, its F-closure $\{X,Z,Y,B,G\}$ intersected back with $R_1$ gives $\{X,Z,Y,B\}$, which already contains the target $Y$). All six pass. $$\boxed{\text{DEPENDENCY-PRESERVING}}$$
