SQL：在有向图中获取无向链接

Question

我有一个表格，在有向图中保存边：

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？例如来自这张表：

+------------+----------+--+
| 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，我如何才能为每个双向链接获取正向链接？我们不能假设互易边总是存在的。

EDIT目标是每个双向链接获得一行，无论是前向链接还是后向链接

Answer 1

如果我猜对了

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
);