The relation that does not have a lossless join dependency preserving BCNF decomposition is AB-C, C-AD.

Here's why:

1. First, we need to find the candidate keys for the relation. In this case, the candidate keys are AB and AD.

2. Next, we check if the dependencies violate BCNF. A dependency X - Y violates BCNF if and only if X is not a superkey. In this case, AB - C and C - AD both violate BCNF because neither AB nor C are superkeys.

3. Finally, we check if the decomposition is dependency preserving. A decomposition is dependency preserving if and only if the closure of the dependencies of the decomposed relations is equal to the closure of the dependencies of the original relation. In this case, the closure of the dependencies of the decomposed relations (AB - C, C - A, and A - D) is not equal to the closure of the dependencies of the original relation (AB - C, C - AD), so the decomposition is not dependency preserving.

Therefore, the relation AB-C, C-AD does not have a lossless join dependency preserving BCNF decomposition.

Question

The relation that does not have a lossless join dependency preserving BCNF decomposition is AB->C, C->AD.

Here's why:

1. First, we need to find the candidate keys for the relation. In this case, the candidate keys are AB and AD.

2. Next, we check if the dependencies violate BCNF. A dependency X -> Y violates BCNF if and only if X is not a superkey. In this case, AB -> C and C -> AD both violate BCNF because neither AB nor C are superkeys.

3. Finally, we check if the decomposition is dependency preserving. A decomposition is dependency preserving if and only if the closure of the dependencies of the decomposed relations is equal to the closure of the dependencies of the original relation. In this case, the closure of the dependencies of the decomposed relations (AB -> C, C -> A, and A -> D) is not equal to the closure of the dependencies of the original relation (AB -> C, C -> AD), so the decomposition is not dependency preserving.

Therefore, the relation AB->C, C->AD does not have a lossless join dependency preserving BCNF decomposition.

Knowee AI · Accepted Answer

The relation that does not have a lossless join dependency preserving BCNF decomposition is AB->C, C->AD.

Here's why:

1. First, we need to find the candidate keys for the relation. In this case, the candidate keys are AB and AD.

2. Next, we check if the dependencies violate BCNF. A dependency X -> Y violates BCNF if and only if X is not a superkey. In this case, AB -> C and C -> AD both violate BCNF because neither AB nor C are superkeys.

3. Finally, we check if the decomposition is dependency preserving. A decomposition is dependency preserving if and only if the closure of the dependencies of the decomposed relations is equal to the closure of the dependencies of the original relation. In this case, the closure of the dependencies of the decomposed relations (AB -> C, C -> A, and A -> D) is not equal to the closure of the dependencies of the original relation (AB -> C, C -> AD), so the decomposition is not dependency preserving.

Therefore, the relation AB->C, C->AD does not have a lossless join dependency preserving BCNF decomposition.

R (A, B, C, D) is a relation. Which of the following does not have a lossless join dependency preserving BCNF decomposition?A->B, B->C, C->DAB->C, C->ADA->B, B->CDA->BCD

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