如何用关系代数找到MAX ?

Working with databases, how can I find MAX using relational algebra?

使用数据库，如何使用关系代数找到MAX ?

7 个解决方案

#1

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.

假设你有一个关系，a，有一个单独的属性a(减少一个更复杂的关系是关系代数中的一个简单的任务，我相信你们已经学过了)，所以现在你想要找到a的最大值。

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:

一种方法是找到A和它自己的叉乘，一定要重命名“A”，这样你的新关系就有具有不同名称的属性。例如:

(rename 'a' as 'a1') X (rename 'a' as 'a2')

(将'a'重命名为'a1') X(将'a'重命名为'a2')

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:

现在选择'a1' < 'a2'，结果关系将具有除最大值以外的所有值。要得到最大值，只需找出你最初的关系之间的差异:

(A x A) - (select 'a1' < 'a2') ((rename 'a' as 'a1')(A) x (rename 'a' as 'a2')(A))

Then use the project operator to reduce down to a single column as Tobi Lehman suggests in the comment below.

然后，按照雷曼兄弟在下面评论中建议的那样，使用项目运营商将项目缩减为一栏。

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:

如果我没记错的话，用关系代数学符号来写这个。注意最后一重命名(即ρ)是最后一个属性名称相同的原始关系:

ρ_a/a1(π_a1((A x A) - σ_{a1 < a2} (ρ_a1/a(A) x ρ_a2/a(A))))

ρa / a1(πa1((x)——σa1 < a2(ρa1 /(A)xρa2 /(A))))

#2

Just my two cents as I was trying to solve this today myself.

我今天就想解决这个问题。

Lets say we have A = 1,2,3

假设A = 1 2 3

If you use

如果你使用

A x A - (select 'a1' < 'a2') ((rename 'a' as 'a1')(A) x (rename 'a' as 'a2')(A))

you will not get the single max value rather two columns like 1|1, 2|1,3|2,3|1,3|2,3|3

你不会得到唯一的最大值而是两列，比如1|1 2|1 3|2 3|1 3|2 3|3

the way to get just 3 is

得到3的方法是。

project(a)A - project(a1)((select 'a1' < 'a2') ((rename 'a' as 'a1')(A) x (rename 'a' as 'a2')(A)))

At least that is what I had to do in a similar situation.

至少在类似的情况下，我必须这么做。

Hope it helps someone

希望它能帮助一些人

#3

I've forgotten most of the relational algebra syntax now. A query just using SELECT, PROJECT, MINUS and RENAME would be

现在我已经忘记了大部分的关系代数语法。查询只需使用SELECT、PROJECT、MINUS和RENAME即可

SELECT v1.number
FROM values v1
MINUS
SELECT v1.number
FROM values v1 JOIN values v2 ON v2.number > v1.number

Hopefully you can translate!

希望你可以翻译!

#4

lets think we have a relation with an attribute A and values 1,2,3

假设我们有一个与属性a和值1,2,3的关系

A

1
2
3

so now..

project A values and rename with A1

所以现在. .项目A值并使用A1重命名

A1
1
2
3

again project A values and rename with A2

再次投影一个值并使用A2重命名

A2
1
2
3

join this with A2<A1 i.e \join_{A2<A1}
so the - Output schema: (A2 integer, A1 integer)

加上A2

A2<A1

1|2
1|3
2|3

hear always A2 values will be less than A1 because we join like that(a2<a1)

总是A2的值小于A1因为我们像这样连接(A2 < A1)

now project A2 the output is like below

现在项目A2的输出如下所示

A2
1
2

now diff with original attribute

现在，diff有了原来的属性。

A diff A2

A
1
2
3

 diff

A2
1
2

Output is 3 which is maximum value

输出是3，这是最大值

Hi, i know some one have to help in editing, for better look

嗨，我知道有人需要帮忙编辑，为了更好的看。

#5

I know this is old, but here is a hand-written formula which might be handy!

我知道这是旧的，但这是一个手写的公式，可能方便!

Relation A: 1,2,3,4

关系:1、2、3、4

1. First we want to PROJECT and RENAME relation A
2. We then to a THETA JOIN with the test a1<a2
3. We then PROJECT the result of the relation to give us a single set of values 
   a1: 1,2,3 (not max value since a1<a2)

4. We then apply the difference operator with the original relation so: 
   1,2,3,4 --- 1,2,3 returns 4

   4 is the Max value.

#6

Find the MAX:

找到马克斯:

• Strategy:

  策略:

  1. Find those x that are not the MAX.

     找出那些不是最大值的x。

     • Rename A relation as d so that we can compare each A x with all others.
     • 将一个关系重命名为d，以便我们可以将每个A x与其他所有的x进行比较。

  2. Use set difference to find those A x that were not found in the earlier step.

     使用set difference查找在前面步骤中没有找到的A x。

• The query is:

  这个查询的方法是:

#7

 Project x(A) - Project A.x
(Select A.x < d.x (A x Rename d(A)))