SQL:在有向图中获取无向链接

Question

我在有向图中有一张保持边的表格：

CREATE TABLE edges ( 
    from_here int not null, 
    to_there  int not null
)

如何在此表上进行选择而不考虑所有反向链接？例如，从本表中：

+------------+----------+--+
| from_there | to_there |  |
+------------+----------+--+
|          1 |        2 |  |
|          2 |        1 |  |
|          3 |        4 |  |
+------------+----------+--+

我想得到这个结果：

+------------+----------+--+
| from_there | to_there |  |
+------------+----------+--+
|          1 |        2 |  |
|          3 |        4 |  |
+------------+----------+--+

换句话说，如何才能得到每个双向链路--仅仅是前向链路？我们不能假定互惠边总是存在的。

编辑目标是每个双向链接获得一行，不管是前向还是后向链路。

Answer 1

如果我做对了

select from_there, to_there
from edges x
where not exists (
    select 1 from edges y 
    where x.from_there = y.to_there
      and x.to_there = y.from_there
      and x.to_there < y.to_there
);