关系R、S如下表所示，R÷(πA1,A2 (σ1大于3(S))的结果为（此空作答），R、S的左外联接、右外联接和完全外联接的元组个数分别为（ ）。
- A.{4}
- B.{3,4}
- C.{3,4,7}
- D.{(1,2),(2,1),(3,4),(4,7)}
正确答案及解析
正确答案
解析
关系代数的除法运算：给定关系R (X，Y) 和S (Y，Z)，其中X，Y，Z为属性组。R中的Y与S中的Y可以有不同的属性名，但必须出自相同的域集。R与S的除运算得到一个新的关系P(X)，P是R中满足下列条件的元组在X属性列上的投影：元组在X上分量值x的象集Yx包含S在Y上投影的集合。具体计算过程如下：（1）找出R中与S中不相同的属性列X，本题是A3。分别求出A3各属性的象集，这里相同属性取一个即可。A3可以取3个值{3，4，7}，其中3的象集为{(1，2)}，4的象集为{(2，1)，(3，4)}，7的象集为{(4，6)}。（2）找出关系R和关系S中相同的属性，即Y属性。在关系S中对Y做投影，并消除重复部分。本题的S为(πA1,A2 (σ1大于3(S))，结果如下。
（3）判断关系R中X各个值的像集Y是否包含关系S中属性Y的所有值。显然，当关系R的属性A3的值为4时，其象集才会包含关系S中属性Y的所有值，即R÷S={4}。两个关系R和S进行自然连接时，选择两个关系R和S公共属性上相等的元组，去掉重复的属性列构成新关系。在这种情况下，关系R中的某些元组有可能在关系S中不存在公共属性值上相等的元组，造成关系R中这些元组的值在运算时舍弃了；同样关系S中的某些元组也可能舍弃。为此，扩充了关系运算左外连接、右外连接和完全外连接。左外连接是指R与S进行自然连接时，只把R中舍弃的元组放到新关系中。右外连接是指R与S进行自然连接时，只把S中舍弃的元组放到新关系中。完全外连接是指R与S进行自然连接时，把R和S中舍弃的元组都放到新关系中。45题R与S的左外联接、右外联接和完全外联接的结果，如下表所示：
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