SQL素数函数 [英] SQL Prime number function

问题描述

如果我有一个数字 X 并且想说 IsPrime(X) = true/false 使用 sql-server 什么是最好的方法?

If I have a number X and want to say IsPrime(X) = true/false using sql-server what is the best approach?

我是只导入素数表还是有一种算法对较小的素数相当有效?

Do I just import a table of primes or is there an algorithm that is fairly efficient for the smaller primes?

注意:我对大于约的数字不感兴趣.1000 万.

Note: I'm not interested in numbers greater than approx. 10 million.

最终使用了以下内容:

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

推荐答案

你可以，如你所说，有一个 存储最多 1000 万个质数.那么查找一个数是否为素数就变得微不足道了.那么问题是哪种方法会更快.我怀疑这个表会快得多(我没有测试过这个说法).

You could, as you said, have a table that stores all the primes up to 10 million. Then it would be trivial to look up whether a number was prime or not. The question then is which method would be faster. I suspect the table would be much faster (I have not tested this claim).

• https://www.simple-talk.com/sql/t-sql-programming/celkos-summer-sql-stumpers-prime-numbers/
  提供了一些解决方案，还有更多评论.

• https://www.simple-talk.com/sql/t-sql-programming/celkos-summer-sql-stumpers-prime-numbers/
  Provides a few solutions and there are more in the comments.

https://sqlserverfast.com/blog/hugo/2006/09/the-prime-number-challenge-great-waste-of-time/
同样在这里.提供了一些解决方案.

https://sqlserverfast.com/blog/hugo/2006/09/the-prime-number-challenge-great-waste-of-time/
Same here. A few solutions provided.

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

这是另一个解决方案，通过 如何用一个 select 语句确定是素数还是非素数? 其他评论中也有更多信息.

Solution 1

Here's another solution via how to find whether is a prime or non prime with one select statement? There's more information in other comments as well.

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

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