[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。
