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RC4 Encryption - Decryption

Author

Topic

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2006-12-12 : 07:35:35

This function is used to initialize the seed for the RC4 algorithm

CREATE FUNCTION dbo.fnInitRc4
(
	@Pwd VARCHAR(256)
)
RETURNS @Box TABLE (i TINYINT, v TINYINT)
AS

BEGIN
	DECLARE	@Key TABLE (i TINYINT, v TINYINT)

	DECLARE	@Index SMALLINT,
		@PwdLen TINYINT

	SELECT	@Index = 0,
		@PwdLen = LEN(@Pwd)

	WHILE @Index <= 255
		BEGIN
			INSERT	@Key
				(
					i,
					v
				)
			VALUES	(
					@Index,
					 ASCII(SUBSTRING(@Pwd, @Index % @PwdLen + 1, 1))
				)

			INSERT	@Box
				(
					i,
					v
				)
			VALUES	(
					@Index,
					@Index
				)

			SELECT	@Index = @Index + 1
		END


	DECLARE	@t TINYINT,
		@b SMALLINT

	SELECT	@Index = 0,
		@b = 0

	WHILE @Index <= 255
		BEGIN
			SELECT		@b = (@b + b.v + k.v) % 256
			FROM		@Box AS b
			INNER JOIN	@Key AS k ON k.i = b.i
			WHERE		b.i = @Index

			SELECT	@t = v
			FROM	@Box
			WHERE	i = @Index

			UPDATE	b1
			SET	b1.v = (SELECT b2.v FROM @Box b2 WHERE b2.i = @b)
			FROM	@Box b1
			WHERE	b1.i = @Index

			UPDATE	@Box
			SET	v = @t
			WHERE	i = @b

			SELECT	@Index = @Index + 1
		END

	RETURN
END

ANd this function does the encrypt/decrypt part

CREATE FUNCTION dbo.fnEncDecRc4
(
	@Pwd VARCHAR(256),
	@Text VARCHAR(8000)
)
RETURNS	VARCHAR(8000)
AS

BEGIN
	DECLARE	@Box TABLE (i TINYINT, v TINYINT)

	INSERT	@Box
		(
			i,
			v
		)
	SELECT	i,
		v
	FROM	dbo.fnInitRc4(@Pwd)

	DECLARE	@Index SMALLINT,
		@i SMALLINT,
		@j SMALLINT,
		@t TINYINT,
		@k SMALLINT,
      		@CipherBy TINYINT,
      		@Cipher VARCHAR(8000)

	SELECT	@Index = 1,
		@i = 0,
		@j = 0,
		@Cipher = ''

	WHILE @Index <= DATALENGTH(@Text)
		BEGIN
			SELECT	@i = (@i + 1) % 256

			SELECT	@j = (@j + b.v) % 256
			FROM	@Box b
			WHERE	b.i = @i

			SELECT	@t = v
			FROM	@Box
			WHERE	i = @i

			UPDATE	b
			SET	b.v = (SELECT w.v FROM @Box w WHERE w.i = @j)
			FROM	@Box b
			WHERE	b.i = @i

			UPDATE	@Box
			SET	v = @t
			WHERE	i = @j

			SELECT	@k = v
			FROM	@Box
			WHERE	i = @i

			SELECT	@k = (@k + v) % 256
			FROM	@Box
			WHERE	i = @j

			SELECT	@k = v
			FROM	@Box
			WHERE	i = @k

			SELECT	@CipherBy = ASCII(SUBSTRING(@Text, @Index, 1)) ^ @k,
				@Cipher = @Cipher + CHAR(@CipherBy)

			SELECT	@Index = @Index  +1
      		END

	RETURN	@Cipher
END

Peter Larsson
Helsingborg, Sweden

harsh_athalye
Master Smack Fu Yak Hacker

5581 Posts

Posted - 2006-12-12 : 07:56:42

Wonderful implementation, Peter!

Thanks for sharing.

Harsh Athalye
India.
"The IMPOSSIBLE is often UNTRIED"

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2006-12-12 : 07:59:31

Thanks.

I kept it in two parts, because then the EncDec can be rewritten to encrypt/decrypt a column with same password a lot faster, without having to call Init for each row.

As of now, any row could have different password.

Peter Larsson
Helsingborg, Sweden

harsh_athalye
Master Smack Fu Yak Hacker

5581 Posts

Posted - 2006-12-12 : 08:05:11

Have you timed it, Peter?
I think it will be good candidate for CLR-SP in SQL 2005.

Harsh Athalye
India.
"The IMPOSSIBLE is often UNTRIED"

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2006-12-12 : 08:29:17

Haven't got time.

I am in a debate right now woth senile developer about requirements.
He wrote all requirements down on a paper and I did exactly what it said, but now he is so mad and angry with me I think he will burst any minute. Now he says he never wrote the requirements.

Peter Larsson
Helsingborg, Sweden

Michael Valentine Jones
Yak DBA Kernel (pronounced Colonel)

7020 Posts

Posted - 2006-12-12 : 08:48:27

I see you linked to this thread from Wikipedia
http://en.wikipedia.org/wiki/RC4

CODO ERGO SUM

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2006-12-12 : 09:32:06

Yes, I have learnt how to do it.
Thanks again Michael.

Peter Larsson
Helsingborg, Sweden

afrika
Master Smack Fu Yak Hacker

2706 Posts

Posted - 2006-12-12 : 09:39:03

Must say,

"Peso on fire"

Brilliant work

Michael Valentine Jones
Yak DBA Kernel (pronounced Colonel)

7020 Posts

Posted - 2006-12-12 : 09:43:55

quote:
Originally posted by Peso

Yes, I have learnt how to do it.
Thanks again Michael.

Peter Larsson
Helsingborg, Sweden

Are you going to start doing 50 posts/day in Wikipedia now?

CODO ERGO SUM

madhivanan
Premature Yak Congratulator

22864 Posts

Posted - 2006-12-12 : 11:47:52

<<
Are you going to start doing 50 posts/day in Wikipedia now?
>>

He will score a Century

Madhivanan

Failing to plan is Planning to fail

avmreddy17
Posting Yak Master

180 Posts

Posted - 2007-10-04 : 13:06:44

Peso,

How to you decrypt using this function
fnEncDecRc4

Thanks

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2007-10-04 : 15:33:14

Same function, hence the name fnEncDecRc4.

E 12°55'05.25"
N 56°04'39.16"

khautinh
Starting Member

10 Posts

Posted - 2007-10-19 : 19:32:18

Peso,
I am new in these type of encrypt and decrypt. How can I use the same function to decrypt the text please? Would you more specific please.

I tried

declare @test varchar(50)
select @test = 'password'
select dbo.fnEncDecRc4(@test, '2345')
print @test = ÍÆ;

How can I use the same function to decrypt (ÍÆ;)?

Thanks again.

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2007-10-20 : 03:23:12

select dbo.fnEncDecRc4(@test, 'ÍÆ ;')

E 12°55'05.25"
N 56°04'39.16"

tkizer
Almighty SQL Goddess

38200 Posts

Posted - 2007-10-31 : 18:21:00

Peso, according to this thread, your function has an error: http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=78552. Check out Helly's post. Not sure if it's valid or not, just wanted to give you a heads up since he posted it in the other thread rather than this one.

Tara Kizer
Microsoft MVP for Windows Server System - SQL Server
http://weblogs.sqlteam.com/tarad/

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2007-11-01 : 12:33:55

Thanks for the link. However I am not able to reproduce the alleged bug.

SELECT	dbo.fnEncDecRc4('Yek', (SELECT dbo.fnEncDecRc4('Yik', 'This is the secret message.')))	-- Fails
SELECT	dbo.fnEncDecRc4('Yek', (SELECT dbo.fnEncDecRc4('Yek', 'This is the secret message.')))	-- Works
SELECT	dbo.fnEncDecRc4('Yek', (SELECT dbo.fnEncDecRc4('Yak', 'This is the secret message.')))	-- Fails

See last example. I think that poster used two different passwords, and when decrypthing the encrypted text, he used wrong password and thus produced a string containing an ascii zero character, which seesm to be a string terminator in SQL Server.

E 12°55'05.25"
N 56°04'39.16"

Ahmad Farid
Starting Member

7 Posts

Posted - 2008-07-22 : 11:11:55

Can anyone please show with a full example how to decrypt some text?
Thank you :)

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2008-07-22 : 12:56:13

The post just before your's show to to both encrypt and decrypt.

E 12°55'05.25"
N 56°04'39.16"

Ahmad Farid
Starting Member

7 Posts

Posted - 2008-07-27 : 05:36:52

yeah I wasn't able to understand it well. Thank you Peso. But I have another question. Is it possible that I do the encryption through the .NET and the decryption through this SQL function?

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2008-07-28 : 03:32:24

Yes, there are a number of .NET RC4 encryption and decryption libraries.

E 12°55'05.25"
N 56°04'39.16"

Ahmad Farid
Starting Member

7 Posts

Posted - 2008-07-28 : 05:13:28

I only found RC2, DES, TripleDES and some else but never found RC4. Can you please mention its full path of namespaces? Thank you.

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