Lossless Join Condition for Decomposition of RA,B,C into R1A,B and R2B,C | AI Research

Lossless Join Condition for Decomposition of $R(A,B,C)$ into $R_1(A,B)$ and $R_2(B,C)$

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Sep 12, 2026

Lossless Join Decomposition (BCNF/3NF) - Functional Dependencies & Chase

Let $R(A,B,C)$ be decomposed into $R_1(A,B)$ and $R_2(B,C)$ . The decomposition is lossless join iff the common attribute(s) across the two projections (here, $B$ ) satisfy a condition implied by the functional dependencies (FDs) of $R$ .

Key theorem (lossless join test): for decomposition into $R_1(X,Y)$ and $R_2(Y,Z)$ , the join $R_1 \Join R_2$ is lossless w.r.t. $R$ iff the FD(s) (equivalently the implied join dependency) ensure that the shared attributes can determine the “other” attributes under the chase procedure. This is commonly expressed via: either the shared attribute(s) determine the remaining attributes, or—more generally—chase produces a row with all equalities consistent with a chase “success” state.

In our specific case:

$R_1(A,B)$ has attributes $(A,B)$
$R_2(B,C)$ has attributes $(B,C)$
The overlap is $B$ .

So the lossless condition will depend on FDs that let you derive enough equality to prevent spurious tuples.

We’ll connect the choices (i)–(iv) directly to the FD conditions for lossless join.

**keyword="A decomposition where $R = \pi_X(R) \Join \pi_Y(R)$ (no spurious tuples) under FDs/constraints of $R$ ." $[J o i n d e p e n d e n c y] d e f = " A c o n s t r a i n t e n s u r i n g a n a t u r a l j o i n o f p r o j e c t i o n s r e c o v e r s t h e o r i g i n a l r e l a t i o n "$ Functional dependency $[C h a s e a l g o r i t h m] d e f = " P r o c e d u r e t o t e s t l o s s l e s s j o i n b y p r o p a g a t i n g e q u a l i t i e s u s i n g F D s "$

Determine which FD choice guarantees lossless join for $R(A,B,C) \to R_1(A,B), R_2(B,C)$

1
Step 1
The intersection of attribute sets is $R_1 \cap R_2 = \{B\}$ .
2
Step 2
A standard sufficient condition for lossless join here is: if either shared attribute $B$ determines the other attribute(s) enough to prevent ambiguity, i.e., either $B \rightarrow A$ or $B \rightarrow C$ (or a condition that implies both $A$ and $C$ are determined through implications).
3
Step 3
Option (i) explicitly states $B \rightarrow A$ or $B \rightarrow C$ , which is directly a sufficient pattern because it forces the overlap $B$ to fix the missing component(s) in the join.
4
Step 4
Option (ii) $A \rightarrow B$ and option (iii) $A \rightarrow C$ and option (iv) $C \rightarrow B$ are not the same as having $B$ determine the missing attributes across the decomposition boundary; they can still imply losslessness in some cases only if combined with other FDs, but among the given single-FD choices, only the ones that directly bind the common attribute $B$ to the remaining side are guaranteed.
5
Step 5
Therefore, the lossless decomposition is guaranteed by option (i).

Why option (i) works (intuition with spurious tuples)

When you join $R_1(A,B)$ and $R_2(B,C)$ , spurious tuples would occur if there exists a $B$ value that can pair with inconsistent $A$ and $C$ values not truly tied together in $R$ .

If $B \rightarrow A$ , then for each $B$ there is at most one $A$ in any legal instance respecting the FD. So any join result with a given $B$ must use the unique matching $A$ , eliminating spurious combinations. Similarly, if $B \rightarrow C$ , then for each $B$ there is at most one $C$ , also preventing spurious tuples.

Thus either $B \rightarrow A$ or $B \rightarrow C$ is enough to ensure the join reconstructs $R$ without adding extra tuples.

**keyword="An extra tuple produced by joining projections that is not present in the original relation under the constraints."$

Quick rule of thumb for this specific decomposition

For $R(A,B,C)$ decomposed as $R_1(A,B)$ and $R_2(B,C)$ , look at the overlap $B$ . If $B \rightarrow A$ or $B \rightarrow C$ , the join is lossless (no spurious tuples).

Why (ii), (iii), (iv) are not the guaranteed answers as stated

Having $A \rightarrow B$ , or $A \rightarrow C$ , or $C \rightarrow B$ does not directly ensure that the shared attribute $B$ uniquely determines the tuple components across the join boundary. Losslessness could still hold only if additional FDs (not listed) are implied. For multiple-choice “which one is sufficient,” (i) is the direct match.

FD options vs. overlap attribute ( $B$ ) determining missing side

Directly ties the shared attribute $B$ to the other attributes, giving a sufficient condition for lossless join.

How to solve any lossless-join multiple-choice fast

Find attribute overlap

1) Decompose

Compute $R_1 \cap R_2$ (here, it’s $\{B\}$ )."

Use FD structure

2) Apply condition

Check whether the overlap attribute determines the components needed to avoid ambiguity."

Confirm with chase intuition

3) Validate

If the overlap fixes the other attributes, chase will succeed (lossless)."

Knowledge Check

Question 1 of 4

Q1Single choice

For $R(A,B,C)$ decomposed into $R_1(A,B)$ and $R_2(B,C)$ , which FD pattern guarantees a lossless join?

(i) $B \rightarrow A$ or $B \rightarrow C$

(ii) $A \rightarrow B$

(iii) $A \rightarrow C$

(iv) $C \rightarrow B$

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