sql - 在 SQL 的有向图中计算不同的无向边

Question

给定一个在有向图中保存边的表格，如下所示：

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

获取特定节点的不同无向链接数量的最佳方法是什么？没有任何重复的有向边，也没有任何节点直接链接到它们自己，我只是想避免计算重复的无向边（例如(1,2)和(2,1)）两次。

这行得通，但NOT IN对我来说味道很糟糕：

SELECT COUNT(*)
FROM edges
WHERE from_here = 1
   OR (to_there = 1 AND from_here NOT IN (
        SELECT to_there 
        FROM edges 
        WHERE from_here = 1
   ))

PostgreSQL 特定的解决方案对此很好。

Answer 1

如果是这样的情况，对于每条边，都有一个倒数（例如，如果(1,2)存在，则(2,1)必须存在），那么你可以简单地缩小你的列表，如下所示：

 Select Count(*)
 From edges
 Where from_here < to_here
    And from_here = 1

如果我们不能假设互易边总是存在，那么您可以使用 except 谓词：

Select Count(*)
From    (
        Select from_here, to_there
        From edges
        Where from_here = 1
            Or to_there = 1
        Except
        Select to_there, from_here
        From edges
        Where from_here = 1
        ) As Z

Answer 2

select count(*) from (
  select to_there from edges where from_here = 1
  union
  select from_here from edges where to_there = 1
) as whatever

Answer 3

SELECT COUNT(DISTINCT CASE to_here WHEN 1 THEN from_here ELSE to_here END)
FROM edges
WHERE from_here = 1
   OR to_here = 1
/* or WHERE 1 IN (from_here, to_here) */