password encryption with 256 bit AES algorithm...???

Hiii leonard,

hiii Rudy,

              Thanks for ur reply but i have already gone through these links. but i didn't understand about SASLprep functionality in pdf password encryption algorithm. Can u give me any example of it, so that i can understand it properly.

                                                                           Thanks,

DevSam

Hii,

     Thanks for ur help, I have already gone through these steps and written a code for this. I want to know that whatever output i have got from this function is correct or not. I am not able to check it out. Can u give me any example of this. So that i could check that am i on right way or not . Is there any software or online tool for this type of algorithm  ??

                                                             Thanks

DevSam

Thanks for your response, let me tell you in brief that actually at which point I am getting the problem. I am giving you an example of a file. In this file:

Open(user) password: abc

Password algorithm: 256 bit AES

/R :   6

/U:F42EC3CF547F226A4D6FE84EB19F83E8E2B9416E603D8C5120E53F6A1A774BE285409545F6DB0FEA34606D1 1D0A012AF

/UE : 18D370D5942D6BB4A2E29EBCB447C904D45BED033F705C5CA9EB1B47E0817ABB95

/V : 5

Now my algorithm which I have to follow is:

To understand the algorithm below, it is necessary to treat the O and U strings in the Encrypt dictionary

as made up of three sections. The first 32 bytes are a hash value (explained below). The next 8 bytes are

called the Validation Salt. The final 8 bytes are called the Key Salt.

1. The password string is generated from Unicode input by processing the input string with the SASLprep

(IETF RFC 4013) profile of stringprep (IETF RFC 3454), and then converting to a UTF-8 representation.

2. Truncate the UTF-8 representation to 127 bytes if it is longer than 127 bytes.

3. Test the password against the user key by computing the SHA-256 hash of the UTF-8 password

concatenated with the 8 bytes of user Validation Salt. If the 32 byte result matches the first 32 bytes of

the U string, this is the user password.

Now what I am doing :

1.       User password(Ascii value): 616263    (abc)

2.       After processing it through SaslPrep output is same:

3.       Concatenating user password with validaition salt the input String : 61626385409545F6DB0FEA

4.       Now I computed SHA256 hash of the above string and output is: f1bd74e34fbacc88b60cb1bbcaf64f88cb6159b29ccc5e0f25602aae48358eb3

5.       But this output does not match  with first 32 bytes of U string as mentioned above.

I have followed steps as defined in algorithm.

Can you tell where I have done mistake?

Hii Rudy,

                              Thanks for your reply, As you said, I have already checked the Char * (charcter values) in the SHA256 function but the problem is same. I think the problem is somewhere else.

                                                                       Thanks,

Hiii Dirk,

                    yes Dirk you are right, this file has revision 6. But same problem also occurs for revision 5. There is not so much difference in both cases. Can u suggest me something else.

                                                                                          Thanks

Hello,

                 As described in Acrobat specification for Encryption Algorithm for 256 Bit AES Encryption I have tried following steps on file having password "abc"

                    1.Prepare User Password string according to SASLPrep Algorithm.

                     2. Concatenate user password string with user validation salt and after cocatenating output is:  6162631A17BFC56DA9C388

                    3. Computed SHA256 hash of this string and ouput was : d78732fd8267a3604234994a8fff8653ece66565594a9430a57ae0818b14d07c

                    4. Now the step of authenticating user password is:

                             Test the password against the user key by computing the SHA-256 hash of the UTF-8 password

                             concatenated with the 8 bytes of owner Validation Salt, concatenated with the 48-byte U string. If the

                             32-byte result matches the first 32 bytes of the O string, this is the owner password.

                   According to this statement how user key is being used in authentication process.

                " If the  32-byte result matches the first 32 bytes of the O string, this is the owner password."

                   I just compared result of SHA256 with 32 bytes of U String but it fails ,SO I Assume user key is playing some role in authenticating password.

                    Can anybody help me understanding this.

Waiting for soon reply as I am really struck and need to implement this .

Thanks