Relation: EMPLOYEE (Name, Project, Task, Office, Floor, Phone )进行投影分解达到3NF Note: Keys are underlined Example Data: Name Project Task Office Floor Phone Bill 100X T1 400 4 1400 Bill 100X T2 400 4 1400 Bill 200Y T1 400 4 1400 Bill 200Y T2 400 4 1400 Sue 100X T33 442 4 1442 Sue 200Y T33 442 4 1442 Sue 300Z T33 442 4 1442 Ed 100X T2 588 5 1588
时间: 2024-02-26 15:52:37 浏览: 17
根据提供的信息,可以发现EMPLOYEE关系模式的函数依赖集合为{Name, Project, Task} → {Office, Floor, Phone}。其中{Name, Project, Task}是EMPLOYEE关系模式的一个候选键。
为了将EMPLOYEE关系模式分解为3NF,我们可以按照以下步骤进行投影分解:
1. 创建一个新的关系模式EMPLOYEE_INFO,它包含以下属性:{Name, Project, Task, Office, Floor, Phone}。
2. 创建一个新的关系模式EMPLOYEE_PROJECT,它包含以下属性:{Name, Project}。
3. 创建一个新的关系模式PROJECT_TASK,它包含以下属性:{Project, Task}。
从函数依赖集合中可以看出,EMPLOYEE_INFO已经满足3NF,因为它只包含一个候选键{Name, Project, Task}和它们决定的所有属性。
对于EMPLOYEE_PROJECT和PROJECT_TASK,我们可以发现它们都包含了部分候选键,但是它们之间不存在任何函数依赖关系。因此,它们也满足3NF。
最终,我们将EMPLOYEE关系模式分解为三个关系模式:EMPLOYEE_INFO、EMPLOYEE_PROJECT和PROJECT_TASK,它们都满足3NF。
相关问题
value" : 10000, "relation" : "gte"
根据引用和[2]的信息,你提到的"value" : 10000, "relation" : "gte" 是关于查询结果中的一个键值对。它表示查询结果中符合条件的文档数量为10000,且该数量大于或等于10000。这个键值对在查询结果中提供了关于符合条件的文档数量的信息。<span class="em">1</span><span class="em">2</span><span class="em">3</span>
#### 引用[.reference_title]
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Solving this recurrence relation: T(n) = 2 T(n/2) + f(n).
To solve this recurrence relation, we can use the master theorem.
The master theorem states that for a recurrence relation of the form T(n) = a T(n/b) + f(n), where a ≥ 1 and b > 1 are constants and f(n) is a function, we can determine the asymptotic behavior of T(n) as follows:
1. If f(n) = O(n^c) for some constant c < logb a, then T(n) = Θ(nlogb a).
2. If f(n) = Θ(n^c logk n) for some constants c ≥ logb a and k ≥ 0, then T(n) = Θ(n^c logk+1 n).
3. If f(n) = Ω(n^c) for some constant c > logb a, and if a f(n/b) ≤ c f(n) for sufficiently large n, then T(n) = Θ(f(n)).
In this case, we have a = 2, b = 2, and f(n) = f(n), so we need to determine the asymptotic behavior of f(n) to apply the master theorem.
Without additional information about f(n), we cannot determine its asymptotic behavior, so we cannot directly apply the master theorem. However, we can make an educated guess based on the form of the recurrence relation.
Since T(n) = 2 T(n/2), we can interpret this as a divide-and-conquer algorithm that splits the input into two equal-sized subproblems, solves each subproblem recursively, and then combines the solutions in linear time (since f(n) is a function of n).
If we assume that f(n) = Θ(n^d) for some constant d ≥ 0, then we can apply the master theorem to obtain the following result:
T(n) = Θ(n^d log n)
To see why, note that we have a = 2, b = 2, and c = d. Since d < log2 2 = 1, we are in case 1 of the master theorem, so T(n) = Θ(nlog2 2) = Θ(n). However, this only accounts for the recursive part of the algorithm, and we need to add the linear time for combining the solutions, which gives us T(n) = Θ(n) + Θ(n^d) = Θ(n^d) for d ≥ 0.
Therefore, we can conclude that T(n) = Θ(n^d log n) for f(n) = Θ(n^d) and d ≥ 0.