SQL素数函数

问题描述 投票:0回答:9

如果我有一个数字 X 并且想说

IsPrime(X) = true/false
使用 sql-server 最好的方法是什么?

我是否只导入素数表,或者是否有一种对于较小素数相当有效的算法?

注意:我对大于大约的数字不感兴趣。 1000万。

最终使用以下内容:

CREATE FUNCTION [dbo].[isPrime]
(
    @number INT
)
RETURNS VARCHAR(10)

BEGIN


    DECLARE @retVal VARCHAR(10) = 'TRUE';

    DECLARE @x INT = 1;
    DECLARE @y INT = 0;

    WHILE (@x <= @number )
    BEGIN

            IF (( @number % @x) = 0 )
            BEGIN
                SET @y = @y + 1;
            END

            IF (@y > 2 )
            BEGIN
                SET @retVal = 'FALSE'
                BREAK
            END

            SET @x = @x + 1

    END

    RETURN @retVal
END
sql sql-server sql-server-2012 primes
9个回答
12
投票

正如您所说,您可以拥有一个表来存储最多 1000 万个素数。那么查找一个数是否是素数就很简单了。那么问题是哪种方法会更快。我怀疑该表会快得多(我没有测试过这个说法)。

Prime 表解决方案

SQL 函数解决方案

解决方案0

这是通过 使用 Transact-SQL 函数查找素数的一种解决方案:

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
–- =============================================- Author:        Nicolas Verhaeghe
–- Create date: 12/14/2008- Description:   Determines if a given integer is a prime
/*

      SELECT dbo.IsPrime(1)

      SELECT dbo.IsPrime(9)

      SELECT dbo.IsPrime(7867)

*/- =============================================
CREATE FUNCTION [dbo].[isPrime]
(
      @NumberToTest int
)
RETURNS bit
AS
BEGIN
      -Declare the return variable here
      DECLARE @IsPrime bit,
                  @Divider int- To speed things up, we will only attempt dividing by odd numbers

      –- We first take care of all evens, except 2
      IF (@NumberToTest % 2 = 0 AND @NumberToTest > 2)
            SET @IsPrime = 0
      ELSE
            SET @IsPrime = 1- By default, declare the number a prime

      –- We then use a loop to attempt to disprove the number is a prime

      SET @Divider = 3 -Start with the first odd superior to 1- We loop up through the odds until the square root of the number to test
      –- or until we disprove the number is a prime
      WHILE (@Divider <= floor(sqrt(@NumberToTest))) AND (@IsPrime = 1)
      BEGIN- Simply use a modulo
            IF @NumberToTest % @Divider = 0
                  SET @IsPrime = 0- We only consider odds, therefore the step is 2
            SET @Divider = @Divider + 2
      END- Return the result of the function
      RETURN @IsPrime

END
解决方案1

这是另一种解决方案,通过 how to find if is a prime or non prime with one select statements? 其他评论中也有更多信息。

CREATE FUNCTION isPrime
(
    @number INT
)
RETURNS VARCHAR(10)
BEGIN
    DECLARE @prime_or_notPrime INT
    DECLARE @counter INT
    DECLARE @retVal VARCHAR(10)
    SET @retVal = 'FALSE'

    SET @prime_or_notPrime = 1
    SET @counter = 2

    WHILE (@counter <= @number/2 )
    BEGIN

        IF (( @number % @counter) = 0 )
        BEGIN
            set @prime_or_notPrime = 0
            BREAK
        END

        IF (@prime_or_notPrime = 1 )
        BEGIN
            SET @retVal = 'TRUE'
        END

        SET @counter = @counter + 1
    END
    return @retVal
END

9
投票

我怀疑很多人都没有想到这一点,但你需要检查的是每个新数字是否能被以前的素数整除......

create table prime (primeno bigint)
declare @counter bigint
set @counter = 2
while @counter < 1000000
begin
if not exists(select top 1 primeno from prime where @counter % primeno = 0 )

-- 上面,添加 AND 素数 < @counter / 2 etc will reduce checking overheads further.

insert into prime select @counter
set @counter = @counter + 1
end
select * from prime order by 1

懒惰编码,但即使在像我旁边的一台慢速虚拟电脑上,您也将在几分钟内获得高达一百万的所有内容。我一共做了 78,498 个(如果你不算 1),除非我忽略了一些事情。


5
投票

有一种有趣的方法可以基于

序列生成
来生成素数,无需任何函数或迭代过程(while)。基本上生成了一个
2 .. @max
序列,我们选择序列中没有其他
current%other = 0
:

的所有数字
declare @max INT = 10000

;WITH all_numbers(n) AS
(
    SELECT 2
    UNION ALL
    SELECT n+1 FROM all_numbers WHERE n < @max
)
select all1.n as prime
from all_numbers all1
where not exists (select 1 from all_numbers all2 
                  where all2.n <= all1.n/2 -- Only need to check until half numbers
                       AND all1.n % all2.n = 0)
order by all1.n
-- beware, 0 means no limit. Otherwise 32767 can be the max specified
OPTION (MAXRECURSION 0)

该解决方案的主要缺点是性能(例如,生成 20000 之前的所有素数大约需要 17 秒),但它更 SQLish,因为它不依赖于显式迭代块(即

while


4
投票
CREATE  proc prime  
@no int  
as   
declare @counter int  
set @counter =2  
begin  
while(@counter)<@no  
 begin  
 if(@no%@counter=0)  
  begin  
  select 'Not prime'  
  return  
  end  
  set @counter=@counter+1  
 end  
 select 'prime'  
 return  
end  

--exec prime 10  

1
投票

这是一个运行良好的简单脚本。 根据需要调整@num:

declare @nonprimes table (number int) 
declare @primes table (number int) 
declare @num int = 2
declare @divisor int  = 1
while @num<10000
begin

while @divisor>1
begin
if @num % @divisor=0 
insert @nonprimes
select @num
if @@rowcount>0 goto A

set @divisor=@divisor-1

end
insert @primes
select @num
A: set @num=@num+1 
set @divisor=@num-1

end

select sum(number) from @primes

1
投票

这个怎么样(在 PostgreSQL 上测试):

SELECT Listagg (num, ',') 
         within GROUP (ORDER BY num) 
FROM   (SELECT n1.num   num, 
               SUM(CASE 
                     WHEN MOD(n1.num, n2.num) = 0 THEN 1 
                     ELSE 0 
                   END) AS cnt 
        FROM   (SELECT ROWNUM num 
                FROM   dual 
                CONNECT BY LEVEL <= 1000) n1, 
               (SELECT ROWNUM num 
                FROM   dual 
                CONNECT BY LEVEL <= 1000) n2 
        WHERE  n1.num <> 1 
               AND n2.num <> 1 
               AND n1.num >= n2.num 
        GROUP  BY n1.num) a 
WHERE  cnt = 1; 

1
投票
declare @max INT = 50000
declare @all_numbers table (n int not null primary key, squareRoot decimal(10,1))
;WITH all_odds(n) AS
(
    SELECT 3
    UNION ALL
    SELECT n+2 FROM all_odds WHERE n+2 < @max
)
    INSERT INTO @all_numbers
    SELECT 2 as n, SQRT(2)
    UNION ALL
    SELECT n, SQRT(n) FROM all_odds
    OPTION (MAXRECURSION 0)

select all1.n as prime
from @all_numbers all1
where not exists (select 1 from @all_numbers all2 where all2.n <= all1.squareRoot AND all1.n % all2.n = 0)
order by all1.n

50k 需要不到一秒


1
投票

这是数字生成和素数检查的组合。

这是一个快速运行的素数检查器。

生成器将创建最多 10,000 个数字的列表。
@CheckNumber 可以设置为任何需要的值。

您可以随意扩大最大生成数量以满足您的需求。

DECLARE @CheckNumber INT;
SET @CheckNumber = 29;

with data as (
    SELECT (ones.n + 10*tens.n + 100*hundreds.n + 1000*thousands.n ) + 1 as number
FROM (VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) ones(n),
     (VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) tens(n),
     (VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) hundreds(n),
     (VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) thousands(n)
)

select sub1.number1
from (
  select distinct d1.number as number1, d2.number as number2
  from data as d1 
  cross join data as d2
  where d1.number % d2.number != 0
) as sub1
group by sub1.number1
having count(sub1.number2) = max(sub1.number2)-2
and sub1.number1 = @CheckNumber
order by sub1.number1

-3
投票
Create PROC prime @num int , @flag int OUTPUT
AS
   Declare @i=50
   set

   while (@i<@num)
Begin
     if @i%@num=0
     break
set
     @i=@i+1
  End
  if @i=@num
 set
    @glag=1
else
   set
      @flag=2


Declare @ans int
Exec prime 50,@flag=@ans OUTPUT
if @ans=1
print 'The number is prime'
else
print 'The number is composite'
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