问题描述
使用数据库，如何使用关系代数求 MAX?
Working with databases, how can I find MAX using relational algebra?
推荐答案
假设你有一个关系 A，有一个单一的属性 'a'(减少一个更复杂的关系，这是关系代数中的一个简单任务，我肯定你已经到了这一步)，所以现在你想在 A 中找到最大值.
Assuming you have a relation, A, with a single attribute, 'a' (reducing a more complex relation to this is a simple task in relational algebra, I'm sure you got this far), so now you want to find the maximum value in A.
一种方法是找到 A 与其自身的叉积，确保重命名 'a' 以便您的新关系具有具有不同名称的属性.例如:
One way to do it is to find the cross product of A with itself, be sure to rename 'a' so your new relation has attributes with distinct names. for example:
(将'a'重命名为'a1') X (将'a'重命名为'a2')
(rename 'a' as 'a1') X (rename 'a' as 'a2')
现在选择'a1' <'a2'，结果关系将具有除最大值之外的所有值.要获得最大值，只需找到原始关系之间的差异:
now select 'a1' < 'a2', the resulting relation will have all values except the maximum. To get the max simply find the difference between your original relation:
(A x A) - (select 'a1' < 'a2') ((rename 'a' as 'a1')(A) x (rename 'a' as 'a2')(A))
然后使用 project 运算符将其缩减为单列，正如 Tobi Lehman 在下面的评论中所建议的那样.
Then use the project operator to reduce down to a single column as Tobi Lehman suggests in the comment below.
用关系代数表示法写这个是(如果我没记错的话).注意最后的重命名(即 ρ)只是为了得到一个与原始关系同名的属性:
Writing this in relational algebra notation would be (if I remember correctly). Note the final rename (i.e. ρ) is just to end up with an attribute that has the same name as in the original relation:
ρ(π((A x A) - σ (ρ(A) x ρ(A))))
ρ(π((A x A) - σ (ρ(A) x ρ(A))))
这篇关于如何使用关系代数找到 MAX?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！