Oracle-求素数

需求:求200w以内的素数
素数是只能被1和自身整除的数,1不是素数

一.SQL版

先用2w进行测试

--  非1和自身,只要有整除的,通过not exists 剔除 
WITH t AS
 (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= 20000 -1),
t1 AS
 (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= 20000 -1)
SELECT count(*)
  FROM t
 WHERE NOT EXISTS (SELECT t1.rn
          FROM t1
         WHERE t1.rn != t.rn
           AND MOD(t.rn, t1.rn) = 0);


-- 一个数不可能整除比自身大的数
WITH t AS
 (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= 20000 -1),
t1 AS
 (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= 20000 -1)
SELECT count(*)
  FROM t
 WHERE NOT EXISTS (SELECT t1.rn
          FROM t1
         WHERE t1.rn < t.rn
           AND MOD(t.rn, t1.rn) = 0);



-- 其实一个数不能整除比自身平方根更大的数据,所以t1表其实可以缩小限制
-- where后面的谓词条件也可以缩小限制
WITH t AS
 (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= 20000 -1),
t1 AS
 (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= sqrt(20000-1)
SELECT count(*)
  FROM t
 WHERE NOT EXISTS (SELECT t1.rn
          FROM t1
         WHERE t1.rn <= sqrt(t.rn)
           AND MOD(t.rn, t1.rn) = 0);

运行记录:

scoot@orcl>--  非1和自身,只要有整除的,通过not exists 剔除
scoot@orcl>WITH t AS
  2   (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= 20000 -1),
  3  t1 AS
  4   (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= 20000 -1)
  5  SELECT count(*)
  6    FROM t
  7   WHERE NOT EXISTS (SELECT t1.rn
  8            FROM t1
  9           WHERE t1.rn != t.rn
 10             AND MOD(t.rn, t1.rn) = 0);

  COUNT(*)
----------
      2262

已用时间:  00: 00: 32.19
scoot@orcl>
scoot@orcl>
scoot@orcl>-- 一个数不可能整除比自身大的数
scoot@orcl>WITH t AS
  2   (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= 20000 -1),
  3  t1 AS
  4   (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= 20000 -1)
  5  SELECT count(*)
  6    FROM t
  7   WHERE NOT EXISTS (SELECT t1.rn
  8            FROM t1
  9           WHERE t1.rn < t.rn
 10             AND MOD(t.rn, t1.rn) = 0);

  COUNT(*)
----------
      2262

已用时间:  00: 00: 24.42
scoot@orcl>
scoot@orcl>
scoot@orcl>-- 其实一个数不能比自身平方根更大的数据,所以t1表其实可以缩小限制
scoot@orcl>-- where后面的谓词条件也可以缩小限制
scoot@orcl>WITH t AS
  2   (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= 20000 -1),
  3  t1 AS
  4   (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= sqrt(20000-1)
  5  SELECT count(*)
  6    FROM t
  7   WHERE NOT EXISTS (SELECT t1.rn
  8            FROM t1
  9           WHERE t1.rn <= sqrt(t.rn)
 10             AND MOD(t.rn, t1.rn) = 0);

  COUNT(*)
----------
      2262

已用时间:  00: 00: 00.57
scoot@orcl>

一个简单的逻辑上的优化,让CPU少计算了很多不必要的数据
计算时间由32秒提升到0.57秒,性能大幅提升

将2w调整为200w

-- 上例中最后一个由2w调整为200w
WITH t AS
 (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= 2000000 -1),
t1 AS
 (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= sqrt(2000000-1)
SELECT count(*)
  FROM t
 WHERE NOT EXISTS (SELECT t1.rn
          FROM t1
         WHERE t1.rn <= sqrt(t.rn)
           AND MOD(t.rn, t1.rn) = 0);
           
-- 这个是否可以持续优化,当然可以
-- 我们优化算法,通过minus,只要2个数的成绩在20w以内,就进行剔除
-- 然后再把 能整除2的给剔除掉
WITH t AS (
   SELECT ROWNUM+1 rn FROM DUAL CONNECT BY ROWNUM <= (2000000)/2-1
   )
,t_prime AS (
   SELECT 2*ROWNUM+1 rn FROM DUAL CONNECT BY ROWNUM <= (2000000)/2-1
   UNION ALL
   SELECT 2 FROM DUAL
   )
SELECT COUNT(*)
  FROM (SELECT rn from t_prime
        MINUS
        SELECT t1.rn * t2.rn
          FROM t t1, t t2
         WHERE t1.rn <= t2.rn
               AND t1.rn <= (SELECT SQRT(2000000) FROM DUAL)
               AND t1.rn * t2.rn <2000000
       );
 
-- 继续优化,把 3 、5、7都给剔除         
WITH t0 AS
 (SELECT 2*rownum+1 rn FROM dual  CONNECT BY rownum <= 1000000 -1),
t AS
 (select rn from t0 where mod(rn,3) != 0 and  mod(rn,5) != 0 and mod(rn,7) != 0 )
SELECT COUNT(*)+1+3 --2,3,5,7
   FROM (SELECT rn from t
         MINUS
         SELECT t1.rn * t2.rn
           FROM t t1, t t2
         WHERE t1.rn <= t2.rn
               AND t1.rn BETWEEN 9 AND (SELECT SQRT(2000000) FROM DUAL)
               AND t1.rn * t2.rn <2000000
       );
-- 剔除越多就越快
WITH t0 AS (
    SELECT 2*ROWNUM+1 rn FROM DUAL CONNECT BY ROWNUM <= (2000000)/2-1-1
    ),
t as(SELECT rn from t0 
      where mod(rn,3)<>0 
            and mod(rn,5)<>0 
            and mod(rn,7)<>0 
            and mod(rn,11)<>0 
            and mod(rn,13)<>0
            and mod(rn,17)<>0
            and mod(rn,19)<>0
            and mod(rn,23)<>0
            and mod(rn,29)<>0
    )
SELECT COUNT(*)+10 --2,3,5,7,11,13,17,19,23,29
   FROM (SELECT rn from t
         MINUS
         SELECT t1.rn * t2.rn
           FROM t t1, t t2
         WHERE t1.rn <= t2.rn
               AND t1.rn BETWEEN 31 AND (SELECT SQRT(2000000) FROM DUAL)
               AND t1.rn * t2.rn <2000000
       );

执行记录

scoot@orcl>-- 上例中最后一个由2w调整为200w
scoot@orcl>WITH t AS
  2   (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= 2000000 -1),
  3  t1 AS
  4   (SELECT rownum+1 rn FROM dual CONNECT BY rownum <= sqrt(2000000-1)
  5  SELECT count(*)
  6    FROM t
  7   WHERE NOT EXISTS (SELECT t1.rn
  8            FROM t1
  9           WHERE t1.rn <= sqrt(t.rn)
 10             AND MOD(t.rn, t1.rn) = 0);

  COUNT(*)
----------
    148933

已用时间:  00: 04: 13.83
scoot@orcl>-- 这个是否可以持续优化   
scoot@orcl>-- 我们优化算法,通过minus,只要2个数的成绩在20w以内,就进行剔除
scoot@orcl>-- 然后再把 能整除2的给剔除掉
scoot@orcl>WITH t AS (
  2     SELECT ROWNUM+1 rn FROM DUAL CONNECT BY ROWNUM <= (2000000)/2-1
  3     )
  4  ,t_prime AS (
  5     SELECT 2*ROWNUM+1 rn FROM DUAL CONNECT BY ROWNUM <= (2000000)/2-1
  6     UNION ALL
  7     SELECT 2 FROM DUAL
  8     )
  9  SELECT COUNT(*)
 10    FROM (SELECT rn from t_prime
 11          MINUS
 12          SELECT t1.rn * t2.rn
 13            FROM t t1, t t2
 14           WHERE t1.rn <= t2.rn
 15                 AND t1.rn <= (SELECT SQRT(2000000) FROM DUAL)
 16                 AND t1.rn * t2.rn <2000000
 17         );

  COUNT(*)
----------
    148933

已用时间:  00: 02: 04.74
scoot@orcl>
scoot@orcl>
scoot@orcl>
scoot@orcl>
scoot@orcl>-- 继续优化,把 3 、5、7都给剔除
scoot@orcl>WITH t0 AS
  2   (SELECT 2*rownum+1 rn FROM dual  CONNECT BY rownum <= 1000000 -1),
  3  t AS
  4   (select rn from t0 where mod(rn,3) != 0 and  mod(rn,5) != 0 and mod(rn,7) != 0 )
  5  SELECT COUNT(*)+1+3 --2,3,5,7
  6     FROM (SELECT rn from t
  7           MINUS
  8           SELECT t1.rn * t2.rn
  9             FROM t t1, t t2
 10           WHERE t1.rn <= t2.rn
 11                 AND t1.rn BETWEEN 9 AND (SELECT SQRT(2000000) FROM DUAL)
 12                 AND t1.rn * t2.rn <2000000
 13         );

COUNT(*)+1+3--2,3,5,7
------------------------
                  148933

已用时间:  00: 00: 14.23

scott@orcl>WITH t0 AS (
  2      SELECT 2*ROWNUM+1 rn FROM DUAL CONNECT BY ROWNUM <= (2000000)/2-1-1
  3      ),
  4  t as(SELECT rn from t0
  5        where mod(rn,3)<>0
  6              and mod(rn,5)<>0
  7              and mod(rn,7)<>0
  8              and mod(rn,11)<>0
  9              and mod(rn,13)<>0
 10              and mod(rn,17)<>0
 11              and mod(rn,19)<>0
 12              and mod(rn,23)<>0
 13              and mod(rn,29)<>0
 14      )
 15  SELECT COUNT(*)+10 --2,3,5,7,11,13,17,19,23,29
 16     FROM (SELECT rn from t
 17           MINUS
 18           SELECT t1.rn * t2.rn
 19             FROM t t1, t t2
 20           WHERE t1.rn <= t2.rn
 21                 AND t1.rn BETWEEN 31 AND (SELECT SQRT(2000000) FROM DUAL)
 22                 AND t1.rn * t2.rn <2000000
 23         )
 24  /

COUNT(*)+10--2,3,5,7,11,13,17,19,23,29
--------------------------------------
                                148933

已用时间:  00: 00: 08.13

运行时间由 4分13秒优化到8.13秒,性能提升明显。

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