可以在没有主键的3NF中使用表吗？ [英] Can a table be in 3NF with no primary keys?
问题描述
1。
如果满足以下条件的一个，则表将自动以3NF表示：
（i）如果一个关系包含两个属性。
（ii）如果2NF表仅包含一个非关键属性。
2。
如果X→A是一个依赖项，则该表位于3NF中，如果存在以下条件之一：
（i）如果X是超键
（ii）如果A是超键的一部分
我从此网站获得了上述声明。 / p>
我认为在两个声明中，第二个子点都是错误的。
第一个是在如果我们拥有所有非键属性，并且表位于2NF中，则2NF将位于3NF中。
考虑具有依赖项A-的示例R（A，B，C） > B。
此处我们没有候选键，因此所有属性均为非素属性，并且该关系不在3NF中，而是在2NF中。
第二个说，对于形式为X-> A的依赖项，如果A是超键的一部分，则它位于3NF中。
考虑示例R（A，B，C）的依赖关系A-> B，B-> C。这里的CK是{A}。现在超级键之一可以是AC，并且FD B-> C的RHS包含AC的一部分，但是上述关系R不在3NF中。
它应该是A应该是候选键的一部分，而不是超级键。
我正确吗？
如果存在<是否存在强功能依赖性？
CK（候选键）是不包含更小的超键的超键。超级键是一组唯一的属性。关系是一组元组。因此，每个关系都有一个超键，即所有属性的集合。因此，它至少具有一个CK。
当定义的一组确定属性的每个值始终显示相同的值时，FD（功能依赖项）按定义成立组。每个关系值或变量都满足琐碎的FD，其中确定的集合是确定的集合的子集。每组属性确定{}。因此，每个关系至少满足一个FD。但是，正确的定义形式通常专门讨论非平凡的 FD。不要使用网络，不要使用教科书，尽管并非全部都写得很好，但其中有几十种是免费在线提供的。许多教科书还忘记了行列式和/或确定集为{}的FD。
您的第一点不是3NF的正确定义。由于其用 if ...代替 if and only if，因此它可能不是要作为定义。但是，这仍然是错误的。 （i）是错误的，因为如果其中一个是CK，而另一个在每个元组中具有相同的值（即由{}确定），则具有两个属性的关系不在3NF中。
类似地，第二点不是正确的定义，即使您仅将其视为3NF的结果（如果...），它也是错误的。如果它使用 if并且仅当并讨论拥有FD的 并说它是不平凡的 FD，那将是一个定义。
由于这些既不是正确的定义也不是正确的含义，因此有无数种方法可以反驳它们。读一本书（或我的帖子）并获得正确的定义。
一些评论是您的推理：
第一个说，如果我们具有所有非键属性，并且表位于2NF中，则2NF中的表将位于3NF中。
我不知道你为什么这么认为。
没有候选键
总是有一个或多个CK。您需要阅读CK的定义。也有非蛮力算法可以找到它们。
第二个说法是，对于形式X-> A的依赖性如果A是超级键的一部分，则它在3NF中。
我不知道您为什么会这样。
A应该是候选关键字的一部分，而不是超级关键字。
正确的定义（如第二点）通常会说 ...或（ii）AX是CK的一部分。但是我听不懂您的推理。
合理的推理涉及从假设开始，并写出我们认为是正确的新语句，因为我们应用了定义，即先前证明的语句（定理）或合理的推理规则，例如，根据 A表示B和 A，我们可以得出 B。您似乎需要阅读有关如何操作的信息。
1. A table is automatically in 3NF if one of the following holds:
(i) If a relation consists of two attributes.
(ii) If 2NF table consists of only one non key attribute.2. If X → A is a dependency, then the table is in 3NF, if one of the following conditions exists:
(i) If X is a superkey
(ii) If A is a part of superkey
I got the above claims from this site.
I think that in both the claims, 2nd subpoint is wrong.
The first one says that a table in 2NF will be in 3NF if we have all non-key attributes and the table is in 2NF.
Consider the example R(A,B,C) with dependency A->B.
Here we have no candidate key, so all attributes are non-prime attributes and the relation is not in 3NF but in 2NF.
The second one says that for a dependency of the form X->A if A is part of a super key then it's in 3NF.
Consider the example R(A,B,C) with dependencies A->B, B->C . Here a CK is {A}. Now one of the super keys can be AC and the RHS of FD B->C contains part of AC but still the above relation R is not in 3NF.
I think it should be A should be part of a candidate key and not super key.
Am I correct?
Also can a particular relation be in 1NF, 3NF or 2NF if there are no functional dependencies present?
A CK (candidate key) is a superkey that contains no smaller superkey. A superkey is a unique set of attributes. A relation is a set of tuples. So every relation has a superkey, the set of all attributes. So it has at least one CK.
A FD (functional dependency) holds by definition when each value of a determining set of attributes appears always with the same value for its determined set. Every relation value or variable satisfies "trivial" FDs, the ones where the determined set is a subset of the determining set. Every set of attributes determines {}. So every relation satisfies at least one FD. However, the correct forms of definitions typically specifically talk about non-trivial FDs. Don't use the web, use textbooks, of which dozens are free online, although not all are well-written. Many textbooks also forget about FDs where the determinant and/or determined set is {}.
Your first point is not a correct definition of 3NF. Since its phrased "if..." instead of "if and only if", maybe it's not trying to be a definition. However, it is still wrong. (i) is wrong because a relation with two attributes is not in 3NF if one is a CK and the other has the same value in every tuple, ie it is determined by {}.
Similarly the second point is not a proper definition and also even if you treat it as only a consequence of 3NF (if...) it's false. It would be a definition if it used if and only if and talked about an FD that holds and it said it was a non-trivial FD and some other things were fixed.
Since those are neither correct definitions nor correct implications, there's a unlimited number of ways to disprove them. Read a book (or my posts) and get correct definitions.
Some comments re your reasoning:
First one says that, a table in 2NF will be in 3NF if we have all non key attributes and table is in 2NF.
I have no idea why you think that.
Here we have no candidate key
There's always one or more CKs. You need to read a definition of CK. There are also non-brute-force algorithms for finding them all.
Second one says that, for the dependency of form X->A if A is part of super key then it's in 3NF.
I have no idea why you think that.
A should be part of candidate key and not super key.
A correct defintion like the second point does normally say "... or (ii) A-X is part of a CK". But I can't follow your reasoning.
Sound reasoning involves starting from assumptions and writing new statements that we know are true because we applied a definition, a previously proved statement (theorem) or a sound rule of reasoning, eg from 'A implies B' and 'A' we can derive 'B'. You seem to need to read about how to do that.
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