使用 SQL 查找给定 x、y 坐标的填充矩形

给定以下填充的 x、y 坐标：

0, 0
0, 1
0, 2
1, 0 
1, 1
1, 2
2, 0
2, 1
2, 2
4, 0
4, 1
5, 0
5, 1

如何编写 SQL 查询来确定所有填充的矩形？矩形由其左上角和右下角定义。

期望的结果

x1 | y1 | x2 | y2 
 0    0   2    2 
 0    4   1    5

Because

+---+---+---+---+
|   | 0 | 1 | 2 |
+---+---+---+---+
| 0 | X | X | X |
| 1 | X | X | X |
| 2 | X | X | X |
| 3 |   |   |   |
| 4 | X | X |   |
| 5 | X | X |   |
+---+---+---+---+

有趣的谜题，用 ypercube 的答案中的想法编辑的解决方案：

declare @t table (x int, y int);
insert @t (x,y) values (0, 0), (0, 1), (0, 2), (1, 0),  (1, 1), (1, 2), 
                       (2, 0), (2, 1), (2, 2), (4, 0), (4, 1), (5, 0), (5, 1);

; with  all_rectangles as
        (
        select  lt.x as x1
        ,       lt.y as y1
        ,       rt.x as x2
        ,       lb.y as y2
        from    @t lt -- Left top
        join    @t rt -- Right top
        on      rt.y = lt.y -- Must share top
                and rt.x > lt.x
        join    @t lb -- Left bottom
        on      lb.x = lt.x -- Must share left
                and lb.y > lt.y
        join    @t rb -- Right bottom (limits resultset)
        on      rb.x = rt.x -- Must share right
                and rb.y = lb.y -- Must share bottom
        )
,       filled_rectangles as
        (
        select  rect.x1
        ,       rect.y1
        ,       rect.x2
        ,       rect.y2
        from    all_rectangles rect
        join    @t crossed
        on      crossed.x between rect.x1 and rect.x2
                and crossed.y between rect.y1 and rect.y2
        group by
                rect.x1
        ,       rect.y1
        ,       rect.x2
        ,       rect.y2
        having  count(*) = 
                (rect.x2 - rect.x1 + 1) * (rect.y2 - rect.y1 + 1)
        )
select  *
from    filled_rectangles rect
where   not exists
        (
        select  *
        from    filled_rectangles bigger
        where   bigger.x1 <= rect.x1 and rect.x2 <= bigger.x2
                and bigger.y1 <= rect.y1 and rect.y2 <= bigger.y2
                and (rect.x1 <> bigger.x1 or rect.x2 <> bigger.x2 
                or rect.y1 <> bigger.y1 or rect.y2 <> bigger.y2)
        );

它首先构建所有可能的矩形的列表。然后，它要求填充位置的数量与位置总数（矩形的面积）相匹配。最后，它要求没有其他矩形完全覆盖该矩形。

您可能必须在 PostgreSQL 中采用它，但这个想法应该可行。