bcnf

How to BCNF decompose when an attribute has no relation with any others

由 二次信任 提交于 2019-12-12 02:31:47
问题 Assignment: Consider a relation 𝑅(𝐴,𝐵,𝐶,𝐷,𝐸,𝐺,𝐻) and its FD set 𝐹 = {𝐴𝐵 → 𝐶𝐷, 𝐸 → 𝐷, 𝐴𝐵𝐶 → 𝐷𝐸, 𝐸 → 𝐴𝐵, 𝐷 → 𝐴𝐺, 𝐴𝐶𝐷 → 𝐵𝐸}. Decompose it into a collection of BCNF relations if it is not in BCNF. Make sure your decomposition is lossless-join. Explanation Hi, I'm working on my database homework (relation design chapter). I think I've commanded the basic process based on in-class examples. However, the tricky part here is we have an attribute 'H' which has no relation with others, which confuses

Is a candidate key determinant good enough for BCNF?

由 半城伤御伤魂 提交于 2019-12-11 20:06:43
问题 A question that came up in my homework is as follows: If a determinant is part of a candidate key, is that good enough for BCNF? I don't think so, because a relation is in BCNF if all non-key attributes depend on the whole primary key, and nothing else. This says the determinant is part of a candidate key, so does this imply a partial functional dependency? However, I began to second guess myself because there's a possibility that the candidate key is a super key, but it doesn't seem like

BCNF Decomposition?

由 北慕城南 提交于 2019-12-11 11:18:30
问题 Given relation R(A,B,C,D,E) and FDs = {A->BC, CD->E, B->D, E->A} While converting R into BCNF: ABCDE -> BD & ABCE will be first decomposition BD is in BCNF Regarding ABCE: Opinion 1: ABCE has FDs: A->BCE, E->ABC, BC->AE Therefore its already in BCNF Opinion 2 ABCE has FDs: A->BC, E->ABC Therefore needs to be decomposed into ABC & AE I believe that first one should be correct because in second we are assuming that A->E and BC->E are not possible because D is not one of the attributes of the

BCNF Decomposition algorithm not working

由 和自甴很熟 提交于 2019-12-02 21:17:42
问题 I have the following problem: R(ABCDEFG) and F ={ AB->CD, C->EF, G->A, G->F, CE ->F}. Clearly, B & G should be part of the key as they are not part of the dependent set. Further, BG+ = ABCDEFG, hence candidate key. Clearly, AB->CD violates BCNF. But when I follow the algorithm, I do not end up in any answer. Probably doing something wrong. Can anyone show me how to apply the algorithm correctly to reach the decomposition? Thanks in advance. 回答1: First, you should calculate a canonical cover

What is a good KISS description of Boyce-Codd normal form?

由 a 夏天 提交于 2019-12-02 16:42:07
What is a KISS (Keep it Simple, Stupid) way to remember what Boyce-Codd normal form is and how to take a unnormalized table and BCNF it? Wikipedia 's info: not terribly helpful for me. Chris Date's definition is actually quite good, so long as you understand what he means: Each attribute Your data must be broken into separate, distinct attributes/columns/values which do not depend on any other attributes. Your full name is an attribute. Your birthdate is an attribute. Your age is not an attribute, it depends on the current date which is not part of your birthdate. must represent a fact Each

BCNF: Looking for example that actually uses superkey instead of candidate key

由 左心房为你撑大大i 提交于 2019-12-02 11:19:31
问题 The definition of the Boyce–Codd normal form states that the determinants of all non-trivial functional dependencies have to be superkeys. All the examples for relations in BCNF I found make use of candidate keys. I am looking for an example that actually has a superkey as determinant which is not a candidate key. I fail to come up with a relation that only uses superkeys which can't be transformed to use candidate keys. Let's say we have a relation with an candidate key and an additional

BCNF decomposition process

由 a 夏天 提交于 2019-12-02 08:50:08
What is the BCNF decomposition for these dependencies? A->BCD BC->DE B->D D->A What is the process to get to the answer? We can first convert the relation R to 3NF and then to BCNF. To convert a relation R and a set of functional dependencies( FD's ) into 3NF you can use Bernstein's Synthesis . To apply Bernstein's Synthesis - First we make sure the given set of FD's is a minimal cover Second we take each FD and make it its own sub-schema. Third we try to combine those sub-schemas For example in your case: R = {A,B,C,D,E} FD's = {A->BCD,BC->DE,B->D,D->A} First we check whether the FD's is a

BCNF: Looking for example that actually uses superkey instead of candidate key

由 落花浮王杯 提交于 2019-12-02 04:14:39
The definition of the Boyce–Codd normal form states that the determinants of all non-trivial functional dependencies have to be superkeys. All the examples for relations in BCNF I found make use of candidate keys. I am looking for an example that actually has a superkey as determinant which is not a candidate key. I fail to come up with a relation that only uses superkeys which can't be transformed to use candidate keys. Let's say we have a relation with an candidate key and an additional functional dependency with a superkey as determinant. R1(A,B,C) {A} A,B -> C This additional FD is

What is the difference between 3NF and BCNF?

由 守給你的承諾、 提交于 2019-11-28 15:59:41
Can someone please explain the difference between 3NF and BCNF to me? It would be great if you could also provide some examples. Thanks. Mosty Mostacho The difference between 3NF and BCNF is subtle. 3NF Definition A relation is in 3NF if it is in 2NF and no non-prime attribute transitively depends on the primary key. In other words, a relation R is in 3NF if for each functional dependency X ⟶ A in R, at least one of the following conditions are met: X is a key or superkey in R A is a prime attribute in R Example Given the following relation: EMP_DEPT(firstName, employeeNumber, dateOfBirth,

BCNF decomposition algorithm explanation

由 半腔热情 提交于 2019-11-28 00:36:02
I looked in Decomposing a relation into BCNF answers and tried it on my homework, but i don't get the correct answers, so i ask for help in BCNF decomposition Consider R=(ABCDEG) & F={BG->CD, G->A, CD->AE, C->AG, A->D} . I start pick A->D . Now i got S=(AD), R'=(ABCEG). I pick G->A . Now i got S=(AD,AG) R'=(BCEG) . I pick C->G . Now i think i need to get S=(AD,AG,CG) and R'=(BCE) , But the answer in the end is (AD,AG,CGE,BC) .what went wrong? or perhaps, a better algorithm? To convert a relation R and a set of functional dependencies( FD's ) into 3NF you can use Bernstein's Synthesis . To
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