[原]简单分析《趣味题》中的SQL

这是原文地址《趣味题》，题目是这样的：

1～9有9个数字，三个三个一组，可能正好能组成一个加法等式比如：124＋659＝783

首先是szusunny在8楼给出了一个解决方案：

with tmp as(
select a.num from (select rownum as num from dual connect by rownum <= 999) a
where substr(a.num,1,1) != substr(a.num,2,1)
  and substr(a.num,3,1) != substr(a.num,2,1)
  and substr(a.num,1,1) != substr(a.num,3,1)
  and a.num >= 100
)
--
select
a.num || '+' || b.num || '=' || c.num as str
from tmp a, tmp b, tmp c
where replace(replace(replace(
      replace(replace(replace(
      replace(replace(replace('123456789',
       substr(a.num,1,1)),substr(a.num,2,1)),substr(a.num,3,1)),
       substr(b.num,1,1)),substr(b.num,2,1)),substr(b.num,3,1)),
       substr(c.num,1,1)),substr(c.num,2,1)),substr(c.num,3,1)) 
      is null
      and a.num + b.num = c.num
      and a.num < b.num
      and b.num < c.num

我们逐步分析：

select rownum as num from dual connect by rownum <= 999

这个语句是产生 1~999 的连续序列。

select a.num from (select rownum as num from dual connect by rownum <= 999) a
where substr(a.num,1,1) != substr(a.num,2,1)
  and substr(a.num,3,1) != substr(a.num,2,1)
  and substr(a.num,1,1) != substr(a.num,3,1)
  and a.num >= 100

经过外层条件的过滤后就产生了一组三位数，这个三位数满足以下条件：

1。三个位两两不想等

2。范围是 123 ~ 987

将这组数字放在临时表tmp中，以备下面的查询使用。

select
a.num || '+' || b.num || '=' || c.num as str
from tmp a, tmp b, tmp c
where replace(replace(replace(
      replace(replace(replace(
      replace(replace(replace('123456789',
       substr(a.num,1,1)),substr(a.num,2,1)),substr(a.num,3,1)),
       substr(b.num,1,1)),substr(b.num,2,1)),substr(b.num,3,1)),
       substr(c.num,1,1)),substr(c.num,2,1)),substr(c.num,3,1)) 
      is null
      and a.num + b.num = c.num
      and a.num < b.num
      and b.num < c.num

这个语句有点长，其中select的表达式就是构建输出xxx+yyy=zzz，重点是where后面的条件，先看冗长、一层层嵌套的replace函数：

replace(replace(replace(
replace(replace(replace(
replace(replace(replace('123456789',
 substr(a.num,1,1)),substr(a.num,2,1)),substr(a.num,3,1)),
 substr(b.num,1,1)),substr(b.num,2,1)),substr(b.num,3,1)),
 substr(c.num,1,1)),substr(c.num,2,1)),substr(c.num,3,1))  
is null

简单来说就是3个3位数字（共9位）是否都在123456789这9个数字中，并且没有重复。

基本思路是穷举法，通过将临时表按条件自联接两次，从而获得所有可能的结果。

其中，9个replace函数嵌套的代码可以用translate函数代替：

translate('123456789','0' || a.num || b.num || c.num,'0') is null

最后OO将该方案修改为更加简练的语句：

with t as(
  select x from (select level+122 x from dual connect by level<=(987-122)) 
  where substr(x,1,1)<>substr(x,2,1) 
    and substr(x,1,1)<>substr(x,3,1)
    and instr(x,0)=0
)
--select count(*) from (
  select a.x||'+'||b.x||'='||c.x
  from t a,t b,t c
  where a.x+b.x=c.x 
    and a.x<494 
    and a.x<b.x 
    and b.x<c.x
    and translate('123456789','$'||a.x||b.x||c.x,'$') is null
--);

当然OO也做了些许优化，但是基本思路还是用穷举法遍历所有可能的组合。

执行计划和统计信息如下，我的服务器上执行用时约13秒：

------------------------------------------------------------------------------------------------------------
| Id  | Operation                       | Name                     | Rows  | Bytes | Cost (%CPU)| Time     |
------------------------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT                |                          |     1 |    39 |     8   (0)| 00:00:01 |
|   1 |  TEMP TABLE TRANSFORMATION      |                          |       |       |            |          |
|   2 |   LOAD AS SELECT                | SYS_TEMP_0FD9D6921_55670 |       |       |            |          |
|*  3 |    VIEW                         |                          |     1 |    13 |     2   (0)| 00:00:01 |
|*  4 |     CONNECT BY WITHOUT FILTERING|                          |       |       |            |          |
|   5 |      FAST DUAL                  |                          |     1 |       |     2   (0)| 00:00:01 |
|   6 |   NESTED LOOPS                  |                          |     1 |    39 |     6   (0)| 00:00:01 |
|   7 |    MERGE JOIN CARTESIAN         |                          |     1 |    26 |     4   (0)| 00:00:01 |
|*  8 |     VIEW                        |                          |     1 |    13 |     2   (0)| 00:00:01 |
|   9 |      TABLE ACCESS FULL          | SYS_TEMP_0FD9D6921_55670 |     1 |    13 |     2   (0)| 00:00:01 |
|  10 |     BUFFER SORT                 |                          |     1 |    13 |     4   (0)| 00:00:01 |
|  11 |      VIEW                       |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  12 |       TABLE ACCESS FULL         | SYS_TEMP_0FD9D6921_55670 |     1 |    13 |     2   (0)| 00:00:01 |
|* 13 |    VIEW                         |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  14 |     TABLE ACCESS FULL           | SYS_TEMP_0FD9D6921_55670 |     1 |    13 |     2   (0)| 00:00:01 |
------------------------------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   3 - filter(SUBSTR(TO_CHAR("X"),1,1)<>SUBSTR(TO_CHAR("X"),2,1) AND
              SUBSTR(TO_CHAR("X"),1,1)<>SUBSTR(TO_CHAR("X"),3,1) AND INSTR(TO_CHAR("X"),'0')=0)
   4 - filter(LEVEL<=865)
   8 - filter("A"."X"<494)
  13 - filter("C"."X"="A"."X"+"B"."X" AND "A"."X"<"B"."X" AND "B"."X"<"C"."X" AND
              TRANSLATE('123456789','$'||TO_CHAR("A"."X")||TO_CHAR("B"."X")||TO_CHAR("C"."X"),'$') IS NULL)


Statistics
----------------------------------------------------------
          2  recursive calls
          8  db block gets
     430532  consistent gets
          1  physical reads
        600  redo size
       4873  bytes sent via SQL*Net to client
        644  bytes received via SQL*Net from client
         13  SQL*Net roundtrips to/from client
          2  sorts (memory)
          0  sorts (disk)
        168  rows processed

可以看到这种方法的成本还是很高的。

newkid在41楼给出了另外一种解决方案，他用到了Oracle 11g的递归CTE语法，递归CTE在SQL Server 2005开始支持，也是SQL Server层次查询的常规做法：

WITH n AS (
   SELECT ROWNUM n FROM DUAL CONNECT BY ROWNUM<=9
   )
,t (n1,n2,n3,n4,n5,n6,n7,n8,n9,lvl) AS (
   SELECT n,0,0,0,0,0,0,0,0,1
     FROM n
    WHERE n<=4
   UNION ALL
   SELECT t.n1
         ,DECODE(t.lvl,1,n.n,t.n2)
         ,DECODE(t.lvl,2,n.n,t.n3)
         ,DECODE(t.lvl,3,n.n,t.n4)
         ,DECODE(t.lvl,4,n.n,t.n5)
         ,DECODE(t.lvl,5,n.n,t.n6)
         ,DECODE(t.lvl,6,n.n,t.n7)
         ,DECODE(t.lvl,7,n.n,t.n8)
         ,DECODE(t.lvl,8,n.n,t.n9)
         ,t.lvl+1
     FROM t,n
    WHERE n.n NOT IN (n1,n2,n3,n4,n5,n6,n7,n8,n9)
          AND (t.lvl=1 AND n.n>t.n1 AND n.n+t.n1<=9
               OR (t.lvl=2 AND n.n - t.n1- t.n2 IN (0,1))
               OR (t.lvl=5 AND n.n = MOD(t.n4+ t.n5,10))
               OR t.lvl IN (3,4,6,7,8)
              )
   )
SELECT n1||n7||n4||' + '||n2||n8||n5||' = '||n3||n9||n6
  FROM t
WHERE lvl=9
      AND n1*100+n7*10+n4 + n2*100+n8*10+n5 = n3*100+n9*10+n6
      ;

N1~N9的意义比较特殊，并不是直观 N1N2N3 + N4N5N6 = N7N8N9，从最外层查询的判断条件 n1*100+n7*10+n4 + n2*100+n8*10+n5 = n3*100+n9*10+n6 可以知道N1N7N4 + N2N8N5 = N3N9N6，用竖式来表示比较容易看出来规律来：

  N1 N7 N4
+ N2 N8 N5
----------
  N3 N9 N6

只要稍作分析可以得到以下几个性质，其实都是小学数学的问题：

1。N1和N2中较小的那个最大值不超过4，如果规定第一组三位数一定要比第二组三位数小，那么N1的取值范围只能是1~4，这也解释n<=4这个条件。
2。N1+N2不能超过9，这个估计比较容易明白的。
3。(N4+N5)%10 = N6
4。N3-N2-N1的值不是1 就是 0 。

这个SQL还是比较难分析，由于是递归CTE的，最好一步一步地跟踪临时表t的变化。

起始步骤，称为第0步，首先是union all 之前的语句，比较好懂：

WITH n AS (
   SELECT ROWNUM n FROM DUAL CONNECT BY ROWNUM<=9
   )
,t (n1,n2,n3,n4,n5,n6,n7,n8,n9,lvl) AS (
   SELECT n,0,0,0,0,0,0,0,0,1
     FROM n
    WHERE n<=4
)
select * from t

 N1  N2  N3  N4  N5  N6  N7  N8  N9 LVL
--- --- --- --- --- --- --- --- --- ---
  1   0   0   0   0   0   0   0   0   1
  2   0   0   0   0   0   0   0   0   1
  3   0   0   0   0   0   0   0   0   1
  4   0   0   0   0   0   0   0   0   1

进入迭代过程，临时表不断增长：

第1~2步： 

WITH n AS (
   SELECT ROWNUM n FROM DUAL CONNECT BY ROWNUM<=9
   )
,t (n1,n2,n3,n4,n5,n6,n7,n8,n9,lvl) AS (
   SELECT n,0,0,0,0,0,0,0,0,1
     FROM n
    WHERE n<=4
   UNION ALL
   SELECT t.n1
         ,DECODE(t.lvl,1,n.n,t.n2)
         ,DECODE(t.lvl,2,n.n,t.n3)
         ,DECODE(t.lvl,3,n.n,t.n4)
         ,DECODE(t.lvl,4,n.n,t.n5)
         ,DECODE(t.lvl,5,n.n,t.n6)
         ,DECODE(t.lvl,6,n.n,t.n7)
         ,DECODE(t.lvl,7,n.n,t.n8)
         ,DECODE(t.lvl,8,n.n,t.n9)
         ,t.lvl+1
     FROM t,n
    WHERE n.n NOT IN (n1,n2,n3,n4,n5,n6,n7,n8,n9)
          AND     (t.lvl=1 AND n.n > t.n1 AND n.n+t.n1<=9 
               OR (t.lvl=2 AND n.n - t.n1 - t.n2 IN (0,1) )
               OR (t.lvl=5 AND n.n = MOD(t.n4+ t.n5,10)   )
               OR  t.lvl IN (3,4,6,7,8)
              )
   )
SELECT n1,n2,n3,n4,n5,n6,n7,n8,n9,lvl,lvl-1 step
  FROM t
WHERE lvl<=3

 N1  N2  N3  N4  N5  N6  N7  N8  N9 LVL       STEP <-- 第0步的结果
--- --- --- --- --- --- --- --- --- --- ----------
  1   0   0   0   0   0   0   0   0   1          0
  2   0   0   0   0   0   0   0   0   1          0
  3   0   0   0   0   0   0   0   0   1          0
  4   0   0   0   0   0   0   0   0   1          0
 N1  N2  N3  N4  N5  N6  N7  N8  N9 LVL       STEP <-- 第1步的结果
--- --- --- --- --- --- --- --- --- --- ----------  
  1   2   0   0   0   0   0   0   0   2          1
  1   3   0   0   0   0   0   0   0   2          1
  1   4   0   0   0   0   0   0   0   2          1
  1   5   0   0   0   0   0   0   0   2          1
  1   6   0   0   0   0   0   0   0   2          1
  1   7   0   0   0   0   0   0   0   2          1
  1   8   0   0   0   0   0   0   0   2          1
  2   3   0   0   0   0   0   0   0   2          1
  2   4   0   0   0   0   0   0   0   2          1
  2   5   0   0   0   0   0   0   0   2          1
  2   6   0   0   0   0   0   0   0   2          1
  2   7   0   0   0   0   0   0   0   2          1
  3   4   0   0   0   0   0   0   0   2          1
  3   5   0   0   0   0   0   0   0   2          1
  3   6   0   0   0   0   0   0   0   2          1
  4   5   0   0   0   0   0   0   0   2          1
 N1  N2  N3  N4  N5  N6  N7  N8  N9 LVL       STEP <-- 第2步的结果
--- --- --- --- --- --- --- --- --- --- ----------  
  1   2   3   0   0   0   0   0   0   3          2
  1   2   4   0   0   0   0   0   0   3          2
  1   3   4   0   0   0   0   0   0   3          2
  1   3   5   0   0   0   0   0   0   3          2
  1   4   5   0   0   0   0   0   0   3          2
  1   4   6   0   0   0   0   0   0   3          2
  1   5   6   0   0   0   0   0   0   3          2
  1   5   7   0   0   0   0   0   0   3          2
  1   6   7   0   0   0   0   0   0   3          2
  1   6   8   0   0   0   0   0   0   3          2
  1   7   8   0   0   0   0   0   0   3          2
  1   7   9   0   0   0   0   0   0   3          2
  1   8   9   0   0   0   0   0   0   3          2
  2   3   5   0   0   0   0   0   0   3          2
  2   3   6   0   0   0   0   0   0   3          2
  2   4   6   0   0   0   0   0   0   3          2
  2   4   7   0   0   0   0   0   0   3          2
  2   5   7   0   0   0   0   0   0   3          2
  2   5   8   0   0   0   0   0   0   3          2
  2   6   8   0   0   0   0   0   0   3          2
  2   6   9   0   0   0   0   0   0   3          2
  2   7   9   0   0   0   0   0   0   3          2
  3   4   7   0   0   0   0   0   0   3          2
  3   4   8   0   0   0   0   0   0   3          2
  3   5   8   0   0   0   0   0   0   3          2
  3   5   9   0   0   0   0   0   0   3          2
  3   6   9   0   0   0   0   0   0   3          2
  4   5   9   0   0   0   0   0   0   3          2

从第1第2步结果可以看出

1。条件 t.lvl=1 AND n.n > t.n1 AND n.n+t.n1<=9  限制了第1步的结果范围，不会出现以下的一条记录：

 N1  N2  N3  N4  N5  N6  N7  N8  N9 LVL       STEP
--- --- --- --- --- --- --- --- --- --- ----------
  1   9   0   0   0   0   0   0   0   2          1

2。条件 t.lvl=2 AND n.n - t.n1 - t.n2 IN (0,1)  限制了第2步的结果范围，不会出现以下的一条记录：

 N1  N2  N3  N4  N5  N6  N7  N8  N9 LVL       STEP
--- --- --- --- --- --- --- --- --- --- ----------
  1   2   8   0   0   0   0   0   0   3          2

如此类推，第5步就是判断 (N4+N5)%10 = N6，SQL中的条件字句是t.lvl=5 AND n.n = MOD(t.n4+ t.n5,10)，摘录部分结果集：

 N1  N2  N3  N4  N5  N6  N7  N8  N9 LVL
--- --- --- --- --- --- --- --- --- ---
  4   5   9   1   7   8   0   0   0   6
  4   5   9   2   1   3   0   0   0   6
  4   5   9   2   6   8   0   0   0   6
  4   5   9   3   8   1   0   0   0   6
  4   5   9   6   1   7   0   0   0   6
  4   5   9   6   2   8   0   0   0   6
  4   5   9   6   7   3   0   0   0   6
  4   5   9   7   1   8   0   0   0   6
  4   5   9   7   6   3   0   0   0   6
  4   5   9   8   3   1   0   0   0   6

可以清楚看出(N4+N5)%10=N6这样的规律。

现在回过头来看，那一堆decode语句：

SELECT t.n1
     ,DECODE(t.lvl,1,n.n,t.n2)
     ,DECODE(t.lvl,2,n.n,t.n3)
     ,DECODE(t.lvl,3,n.n,t.n4)
     ,DECODE(t.lvl,4,n.n,t.n5)
     ,DECODE(t.lvl,5,n.n,t.n6)
     ,DECODE(t.lvl,6,n.n,t.n7)
     ,DECODE(t.lvl,7,n.n,t.n8)
     ,DECODE(t.lvl,8,n.n,t.n9)
     ,t.lvl+1

每一次迭代都是“修改”修改一个列，与此反复经过9步迭代之后N1~N9都被“修改”了，最后，最外层查询语句作最终的验证筛选

SELECT n1||n7||n4||' + '||n2||n8||n5||' = '||n3||n9||n6
  FROM t
WHERE lvl=9
      AND n1*100+n7*10+n4 + n2*100+n8*10+n5 = n3*100+n9*10+n6
      ; 

执行计划和统计信息如下，用时约0.2秒：

-----------------------------------------------------------------------------------------------------------------------
| Id  | Operation                                  | Name                     | Rows  | Bytes | Cost (%CPU)| Time     |
-----------------------------------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT                           |                          |     2 |   260 |     8   (0)| 00:00:01 |
|   1 |  TEMP TABLE TRANSFORMATION                 |                          |       |       |            |          |
|   2 |   LOAD AS SELECT                           | SYS_TEMP_0FD9D695D_55670 |       |       |            |          |
|   3 |    COUNT                                   |                          |       |       |            |          |
|*  4 |     CONNECT BY WITHOUT FILTERING           |                          |       |       |            |          |
|   5 |      FAST DUAL                             |                          |     1 |       |     2   (0)| 00:00:01 |
|*  6 |   VIEW                                     |                          |     2 |   260 |     6   (0)| 00:00:01 |
|   7 |    UNION ALL (RECURSIVE WITH) BREADTH FIRST|                          |       |       |            |          |
|*  8 |     VIEW                                   |                          |     1 |    13 |     2   (0)| 00:00:01 |
|   9 |      TABLE ACCESS FULL                     | SYS_TEMP_0FD9D695D_55670 |     1 |    13 |     2   (0)| 00:00:01 |
|  10 |     NESTED LOOPS                           |                          |     1 |   143 |     4   (0)| 00:00:01 |
|  11 |      RECURSIVE WITH PUMP                   |                          |       |       |            |          |
|* 12 |      VIEW                                  |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  13 |       TABLE ACCESS FULL                    | SYS_TEMP_0FD9D695D_55670 |     1 |    13 |     2   (0)| 00:00:01 |
-----------------------------------------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   4 - filter(ROWNUM<=9)
   6 - filter("LVL"=9 AND "N1"*100+"N7"*10+"N4"+"N2"*100+"N8"*10+"N5"="N3"*100+"N9"*10+"N6")
   8 - filter("N"<=4)
  12 - filter("N"."N"<>"N1" AND "N"."N"<>"N2" AND "N"."N"<>"N3" AND "N"."N"<>"N4" AND "N"."N"<>"N5" AND
              "N"."N"<>"N6" AND "N"."N"<>"N7" AND "N"."N"<>"N8" AND "N"."N"<>"N9" AND ("T"."LVL"=1 AND "N"."N">"T"."N1" AND
              "N"."N"+"T"."N1"<=9 OR "T"."LVL"=2 AND ("N"."N"-"T"."N1"-"T"."N2"=0 OR "N"."N"-"T"."N1"-"T"."N2"=1) OR
              "T"."LVL"=5 AND "N"."N"=MOD("T"."N4"+"T"."N5",10) OR ("T"."LVL"=3 OR "T"."LVL"=4 OR "T"."LVL"=6 OR "T"."LVL"=7
              OR "T"."LVL"=8)))


Statistics
----------------------------------------------------------
          3  recursive calls
       4023  db block gets
      22960  consistent gets
          1  physical reads
        600  redo size
       5566  bytes sent via SQL*Net to client
        644  bytes received via SQL*Net from client
         13  SQL*Net roundtrips to/from client
         11  sorts (memory)
          0  sorts (disk)
        168  rows processed

newkid 的思想是，每次迭代都用条件限制相应的结果集，我的优化思想是尽量限制每一步的结果集，每一步的结果集越少，下一步产生的结果集就越少，于是我修改newkid的SQL语句中的条件语句。

我在第7和第8步加入了判断条件，更精确地控制这两步产生的结果集：

WITH n AS (
   SELECT ROWNUM n FROM DUAL CONNECT BY ROWNUM<=9
   )
,t (n1,n2,n3,n4,n5,n6,n7,n8,n9,lvl) AS (
   SELECT n,0,0,0,0,0,0,0,0,1
     FROM n
    WHERE n<=4
   UNION ALL
   SELECT t.n1
         ,DECODE(t.lvl,1,n.n,t.n2)
         ,DECODE(t.lvl,2,n.n,t.n3)
         ,DECODE(t.lvl,3,n.n,t.n4)
         ,DECODE(t.lvl,4,n.n,t.n5)
         ,DECODE(t.lvl,5,n.n,t.n6)
         ,DECODE(t.lvl,6,n.n,t.n7)
         ,DECODE(t.lvl,7,n.n,t.n8)
         ,DECODE(t.lvl,8,n.n,t.n9)
         ,t.lvl+1
     FROM t,n
    WHERE n.n NOT IN (n1,n2,n3,n4,n5,n6,n7,n8,n9)
          AND (t.lvl=1 AND n.n>t.n1 AND n.n+t.n1<=9
               OR (t.lvl=2 AND n.n - t.n1- t.n2 IN (0,1) AND n.n>t.n1)
               OR (t.lvl=5 AND n.n = MOD(t.n4+ t.n5,10))
               OR (t.lvl=7 AND trunc( ((t.n7 + n.n)*10+t.n4+t.n5)/100 )=t.n3-t.n2-t.n1)
               OR (t.lvl=8 AND (t.n3 - t.n1 - t.n2 )*10 + n.n - t.n7 - t.n8 = trunc((t.n4+t.n5)/10) )              
               OR t.lvl IN (3,4,6)
              )
   )
SELECT n1||n7||n4||' + '||n2||n8||n5||' = '||n3||n9||n6
  FROM t
WHERE lvl=9
      AND n1*100+n7*10+n4 + n2*100+n8*10+n5 = n3*100+n9*10+n6

执行计划如下，用时约0.11秒：

------------------------------------------------------------------------------------------------------------------------
| Id  | Operation                                   | Name                     | Rows  | Bytes | Cost (%CPU)| Time     |
------------------------------------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT                            |                          |     2 |   260 |     9  (12)| 00:00:01 |
|   1 |  TEMP TABLE TRANSFORMATION                  |                          |       |       |            |          |
|   2 |   LOAD AS SELECT                            | SYS_TEMP_0FD9D6975_55670 |       |       |            |          |
|   3 |    COUNT                                    |                          |       |       |            |          |
|*  4 |     CONNECT BY WITHOUT FILTERING            |                          |       |       |            |          |
|   5 |      FAST DUAL                              |                          |     1 |       |     2   (0)| 00:00:01 |
|   6 |   SORT ORDER BY                             |                          |     2 |   260 |     7  (15)| 00:00:01 |
|*  7 |    VIEW                                     |                          |     2 |   260 |     6   (0)| 00:00:01 |
|   8 |     UNION ALL (RECURSIVE WITH) BREADTH FIRST|                          |       |       |            |          |
|*  9 |      VIEW                                   |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  10 |       TABLE ACCESS FULL                     | SYS_TEMP_0FD9D6975_55670 |     1 |    13 |     2   (0)| 00:00:01 |
|  11 |      NESTED LOOPS                           |                          |     1 |   143 |     4   (0)| 00:00:01 |
|  12 |       RECURSIVE WITH PUMP                   |                          |       |       |            |          |
|* 13 |       VIEW                                  |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  14 |        TABLE ACCESS FULL                    | SYS_TEMP_0FD9D6975_55670 |     1 |    13 |     2   (0)| 00:00:01 |
------------------------------------------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   4 - filter(ROWNUM<=9)
   7 - filter("LVL"=9 AND "N1"*100+"N7"*10+"N4"+"N2"*100+"N8"*10+"N5"="N3"*100+"N9"*10+"N6")
   9 - filter("N"<=4)
  13 - filter("N"."N"<>"N1" AND "N"."N"<>"N2" AND "N"."N"<>"N3" AND "N"."N"<>"N4" AND "N"."N"<>"N5" AND
              "N"."N"<>"N6" AND "N"."N"<>"N7" AND "N"."N"<>"N8" AND "N"."N"<>"N9" AND ("T"."LVL"=1 AND "N"."N">"T"."N1" AND
              "N"."N"+"T"."N1"<=9 OR "T"."LVL"=2 AND ("N"."N"-"T"."N1"-"T"."N2"=0 OR "N"."N"-"T"."N1"-"T"."N2"=1) OR
              "T"."LVL"=5 AND "N"."N"=MOD("T"."N4"+"T"."N5",10) OR "T"."LVL"=7 AND
              TRUNC((("T"."N7"+"N"."N")*10+"T"."N4"+"T"."N5")/100)="T"."N3"-"T"."N2"-"T"."N1" OR "T"."LVL"=8 AND
              ("T"."N3"-"T"."N1"-"T"."N2")*10+"N"."N"-"T"."N7"-"T"."N8"=TRUNC(("T"."N4"+"T"."N5")/10) OR ("T"."LVL"=3 OR
              "T"."LVL"=4 OR "T"."LVL"=6)))


Statistics
----------------------------------------------------------
          3  recursive calls
        330  db block gets
      12221  consistent gets
          1  physical reads
        556  redo size
       5566  bytes sent via SQL*Net to client
        644  bytes received via SQL*Net from client
         13  SQL*Net roundtrips to/from client
         12  sorts (memory)
          0  sorts (disk)
        168  rows processed

一开始看到这题之后，我第一个想法是按照我最初做数独题目的时候，用几个表来联接，在一个“大包围”的方案中进行筛选，于是我尝试写了一下：

WITH n AS (
   SELECT ROWNUM n FROM DUAL CONNECT BY ROWNUM<=9
   ),
tmp as (
   select n1.n n1 , n2.n n2, n3.n n3 , 
          n4.n n4 , n5.n n5, n6.n n6 , 
          n7.n n7 , n8.n n8, n9.n n9
   from n n1,n n2,n n3,n n4,n n5,n n6,n n7,n n8,n n9
   where n1.n between 1 and 4
     and n1.n + n4.n <= 9 /*
     and n7.n - n1.n - n4.n = trunc(( ( n2.n + n5.n )  + (n3.n + n6.n - n9.n)/10 )/10)*/
     and n8.n - n2.n - n5.n = (n3.n + n6.n - n9.n)/10
     and mod( n3.n + n6.n ,10 ) = n9.n
     and n1.n < n4.n
     and n1.n*100+n2.n*10+n3.n + n4.n*100+n5.n*10+n6.n = n7.n*100+n8.n*10+n9.n
     and n1.n not in (       n2.n, n3.n, n4.n, n5.n, n6.n, n7.n, n8.n, n9.n)
     and n2.n not in ( n1.n,       n3.n, n4.n, n5.n, n6.n, n7.n, n8.n, n9.n)
     and n3.n not in ( n1.n, n2.n,       n4.n, n5.n, n6.n, n7.n, n8.n, n9.n)
     and n4.n not in ( n1.n, n2.n, n3.n,       n5.n, n6.n, n7.n, n8.n, n9.n)
     and n5.n not in ( n1.n, n2.n, n3.n, n4.n,       n6.n, n7.n, n8.n, n9.n)
     and n6.n not in ( n1.n, n2.n, n3.n, n4.n, n5.n,       n7.n, n8.n, n9.n)
     and n7.n not in ( n1.n, n2.n, n3.n, n4.n, n5.n, n6.n,       n8.n, n9.n)
     and n8.n not in ( n1.n, n2.n, n3.n, n4.n, n5.n, n6.n, n7.n,       n9.n)
     and n9.n not in ( n1.n, n2.n, n3.n, n4.n, n5.n, n6.n, n7.n, n8.n      )
)
select * from tmp;

这里没什么算法可言，首先是9个1~9的序列表进行连接，联接的条件包括，其中直接就把题意的写到联接条件中

n1.n*100+n2.n*10+n3.n + n4.n*100+n5.n*10+n6.n = n7.n*100+n8.n*10+n9.n

当然还有9个数字不重复出现的条件

and n1.n not in (       n2.n, n3.n, n4.n, n5.n, n6.n, n7.n, n8.n, n9.n)
and n2.n not in ( n1.n,       n3.n, n4.n, n5.n, n6.n, n7.n, n8.n, n9.n)
and n3.n not in ( n1.n, n2.n,       n4.n, n5.n, n6.n, n7.n, n8.n, n9.n)
and n4.n not in ( n1.n, n2.n, n3.n,       n5.n, n6.n, n7.n, n8.n, n9.n)
and n5.n not in ( n1.n, n2.n, n3.n, n4.n,       n6.n, n7.n, n8.n, n9.n)
and n6.n not in ( n1.n, n2.n, n3.n, n4.n, n5.n,       n7.n, n8.n, n9.n)
and n7.n not in ( n1.n, n2.n, n3.n, n4.n, n5.n, n6.n,       n8.n, n9.n)
and n8.n not in ( n1.n, n2.n, n3.n, n4.n, n5.n, n6.n, n7.n,       n9.n)
and n9.n not in ( n1.n, n2.n, n3.n, n4.n, n5.n, n6.n, n7.n, n8.n      )

然后之前的条件是经过我们推导得出的“优化”条件。

执行计划和统计信息如下，用时约0.23秒：

------------------------------------------------------------------------------------------------------------
| Id  | Operation                       | Name                     | Rows  | Bytes | Cost (%CPU)| Time     |
------------------------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT                |                          |     1 |   117 |    20   (0)| 00:00:01 |
|   1 |  TEMP TABLE TRANSFORMATION      |                          |       |       |            |          |
|   2 |   LOAD AS SELECT                | SYS_TEMP_0FD9D6999_55670 |       |       |            |          |
|   3 |    COUNT                        |                          |       |       |            |          |
|*  4 |     CONNECT BY WITHOUT FILTERING|                          |       |       |            |          |
|   5 |      FAST DUAL                  |                          |     1 |       |     2   (0)| 00:00:01 |
|   6 |   NESTED LOOPS                  |                          |     1 |   117 |    18   (0)| 00:00:01 |
|   7 |    NESTED LOOPS                 |                          |     1 |   104 |    16   (0)| 00:00:01 |
|   8 |     NESTED LOOPS                |                          |     1 |    91 |    14   (0)| 00:00:01 |
|   9 |      NESTED LOOPS               |                          |     1 |    78 |    12   (0)| 00:00:01 |
|  10 |       NESTED LOOPS              |                          |     1 |    65 |    10   (0)| 00:00:01 |
|  11 |        NESTED LOOPS             |                          |     1 |    52 |     8   (0)| 00:00:01 |
|  12 |         NESTED LOOPS            |                          |     1 |    39 |     6   (0)| 00:00:01 |
|  13 |          NESTED LOOPS           |                          |     1 |    26 |     4   (0)| 00:00:01 |
|  14 |           VIEW                  |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  15 |            TABLE ACCESS FULL    | SYS_TEMP_0FD9D6999_55670 |     1 |    13 |     2   (0)| 00:00:01 |
|* 16 |           VIEW                  |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  17 |            TABLE ACCESS FULL    | SYS_TEMP_0FD9D6999_55670 |     1 |    13 |     2   (0)| 00:00:01 |
|* 18 |          VIEW                   |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  19 |           TABLE ACCESS FULL     | SYS_TEMP_0FD9D6999_55670 |     1 |    13 |     2   (0)| 00:00:01 |
|* 20 |         VIEW                    |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  21 |          TABLE ACCESS FULL      | SYS_TEMP_0FD9D6999_55670 |     1 |    13 |     2   (0)| 00:00:01 |
|* 22 |        VIEW                     |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  23 |         TABLE ACCESS FULL       | SYS_TEMP_0FD9D6999_55670 |     1 |    13 |     2   (0)| 00:00:01 |
|* 24 |       VIEW                      |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  25 |        TABLE ACCESS FULL        | SYS_TEMP_0FD9D6999_55670 |     1 |    13 |     2   (0)| 00:00:01 |
|* 26 |      VIEW                       |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  27 |       TABLE ACCESS FULL         | SYS_TEMP_0FD9D6999_55670 |     1 |    13 |     2   (0)| 00:00:01 |
|* 28 |     VIEW                        |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  29 |      TABLE ACCESS FULL          | SYS_TEMP_0FD9D6999_55670 |     1 |    13 |     2   (0)| 00:00:01 |
|* 30 |    VIEW                         |                          |     1 |    13 |     2   (0)| 00:00:01 |
|  31 |     TABLE ACCESS FULL           | SYS_TEMP_0FD9D6999_55670 |     1 |    13 |     2   (0)| 00:00:01 |
------------------------------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   4 - filter(ROWNUM<=9)
  16 - filter("N8"."N"<>"N9"."N")
  18 - filter("N7"."N"<>"N8"."N" AND "N7"."N"<>"N9"."N")
  20 - filter("N6"."N"<>"N7"."N" AND "N6"."N"<>"N8"."N" AND "N6"."N"<>"N9"."N")
  22 - filter("N9"."N"=MOD("N3"."N"+"N6"."N",10) AND "N3"."N"<>"N6"."N" AND "N3"."N"<>"N7"."N" AND
              "N3"."N"<>"N8"."N" AND "N3"."N"<>"N9"."N")
  24 - filter("N3"."N"<>"N5"."N" AND "N5"."N"<>"N6"."N" AND "N5"."N"<>"N7"."N" AND
              "N5"."N"<>"N8"."N" AND "N5"."N"<>"N9"."N")
  26 - filter("N2"."N"<>"N3"."N" AND "N2"."N"<>"N5"."N" AND "N2"."N"<>"N6"."N" AND
              "N2"."N"<>"N7"."N" AND "N2"."N"<>"N8"."N" AND "N2"."N"<>"N9"."N" AND
              "N8"."N"-"N2"."N"-"N5"."N"=("N3"."N"+"N6"."N"-"N9"."N")/10)
  28 - filter("N2"."N"<>"N4"."N" AND "N3"."N"<>"N4"."N" AND "N4"."N"<>"N5"."N" AND
              "N4"."N"<>"N6"."N" AND "N4"."N"<>"N7"."N" AND "N4"."N"<>"N8"."N" AND "N4"."N"<>"N9"."N" AND
              "N4"."N">1)
  30 - filter("N1"."N">=1 AND "N1"."N"<=4 AND "N1"."N"+"N4"."N"<=9 AND "N1"."N"<"N4"."N" AND
              "N1"."N"*100+"N2"."N"*10+"N3"."N"+"N4"."N"*100+"N5"."N"*10+"N6"."N"="N7"."N"*100+"N8"."N"*10+"N9"."N
              " AND "N1"."N"<>"N2"."N" AND "N1"."N"<>"N3"."N" AND "N1"."N"<>"N4"."N" AND "N1"."N"<>"N5"."N" AND
              "N1"."N"<>"N6"."N" AND "N1"."N"<>"N7"."N" AND "N1"."N"<>"N8"."N" AND "N1"."N"<>"N9"."N")


Statistics
----------------------------------------------------------
          2  recursive calls
          8  db block gets
      49879  consistent gets
          1  physical reads
        600  redo size
       3501  bytes sent via SQL*Net to client
        578  bytes received via SQL*Net from client
          7  SQL*Net roundtrips to/from client
          1  sorts (memory)
          0  sorts (disk)
         84  rows processed

虽然效率不怎么样，但是比预想中好，consistent gets仅比newkid的多3倍，比szusunn（以及OO修改后的版本）的方法少8倍。

这充分说明SQL是一种声明式语言，具体实施交给数据库引擎去做的事实。