如何“提示”递归CTE的基数？

我使用以下递归CTE作为最小示例，但总的来说，优化程序必须对递归CTE使用默认的“猜测”基数：

with recursive w(n) as ( select 1 union all select n+1 from w where n<5 ) select * from w;
/*
 n
---
 1
 2
 3
 4
 5
*/

explain analyze
with recursive w(n) as ( select 1 union all select n+1 from w where n<5 ) select * from w;
/*
                                                    QUERY PLAN
-------------------------------------------------------------------------------------------------------------------
 CTE Scan on w  (cost=2.95..3.57 rows=31 width=4) (actual time=0.005..0.020 rows=5 loops=1)
   CTE w
     ->  Recursive Union  (cost=0.00..2.95 rows=31 width=4) (actual time=0.003..0.017 rows=5 loops=1)
           ->  Result  (cost=0.00..0.01 rows=1 width=0) (actual time=0.001..0.001 rows=1 loops=1)
           ->  WorkTable Scan on w w_1  (cost=0.00..0.23 rows=3 width=4) (actual time=0.002..0.002 rows=1 loops=5)
                 Filter: (n < 5)
                 Rows Removed by Filter: 0
*/

注意上述计划中的rows=31估计rows=5基数和实际基数。在某些情况下，似乎使用100作为估算值​​，我不确定这些猜测背后的确切逻辑。

在我的现实世界问题中，基数估计不佳会阻止选择快速的“嵌套循环”计划。如何“提示”优化器基数，以便递归CTE可以解决此问题？

我已经解决了这样的问题，但我希望有一种更简单的方法：

explain analyze
with recursive w(n) as ( select 1 union all select n+1 from w where n<5 )
select * from w limit (select count(*) from w);
/*
                                                    QUERY PLAN
-------------------------------------------------------------------------------------------------------------------
 Limit  (cost=3.66..3.72 rows=3 width=4) (actual time=0.032..0.034 rows=5 loops=1)
   CTE w
     ->  Recursive Union  (cost=0.00..2.95 rows=31 width=4) (actual time=0.003..0.019 rows=5 loops=1)
           ->  Result  (cost=0.00..0.01 rows=1 width=0) (actual time=0.000..0.000 rows=1 loops=1)
           ->  WorkTable Scan on w w_1  (cost=0.00..0.23 rows=3 width=4) (actual time=0.002..0.002 rows=1 loops=5)
                 Filter: (n < 5)
                 Rows Removed by Filter: 0
   InitPlan 2 (returns $2)
     ->  Aggregate  (cost=0.70..0.71 rows=1 width=0) (actual time=0.029..0.030 rows=1 loops=1)
           ->  CTE Scan on w w_2  (cost=0.00..0.62 rows=31 width=0) (actual time=0.005..0.025 rows=5 loops=1)
   ->  CTE Scan on w  (cost=0.00..0.62 rows=31 width=4) (actual time=0.000..0.002 rows=5 loops=1)
*/