Solved

DES password encryption

Posted on 1997-03-28
2
480 Views
Last Modified: 2013-12-26
Hi there!  Can anyone tell me the exact DES algorithm (1976 standard) so that I may produce WORKING(!) code.  If you have it on hand, I'd appreciate working code as well.
0
Comment
Question by:white wolf
  • 2
2 Comments
 
LVL 2

Accepted Solution

by:
pxh earned 170 total points
ID: 1292476
From http://www.program.com/source/crypto/des-how-to.txt what you are looking for.

Peter (pxh@mpe-garching.mpg.de)



This is a tutorial designed to be clear and compact, and to provide a
newcomer to the DES with all the necessary information to implement it
himself, without having to track down printed works or wade through C
source code. I welcome any comments.
 Matthew Fischer <mfischer@heinous.isca.uiowa.edu>

Here's how to do it, step by step:

 1  Process the key.

 1.1  Get a 64-bit key from the user. (Every 8th bit is considered a
parity bit. For a key to have correct parity, each byte should contain
an odd number of "1" bits.)

 1.2  Calculate the key schedule.

 1.2.1  Perform the following permutation on the 64-bit key. (The parity
bits are discarded, reducing the key to 56 bits. Bit 1 of the permuted
block is bit 57 of the original key, bit 2 is bit 49, and so on with bit
56 being bit 4 of the original key.)

                        Permuted Choice 1 (PC-1)

                          57 49 41 33 25 17  9
                           1 58 50 42 34 26 18
                          10  2 59 51 43 35 27
                          19 11  3 60 52 44 36
                          63 55 47 39 31 23 15
                           7 62 54 46 38 30 22
                          14  6 61 53 45 37 29
                          21 13  5 28 20 12  4

 1.2.2  Split the permuted key into two halves. The first 28 bits are
called C[0] and the last 28 bits are called D[0].

 1.2.3  Calculate the 16 subkeys. Start with i = 1.

 1.2.3.1  Perform one or two circular left shifts on both C[i-1] and
D[i-1] to get C[i] and D[i], respectively. The number of shifts per
iteration are given in the table below.

    Iteration #   1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16
    Left Shifts   1  1  2  2  2  2  2  2  1  2  2  2  2  2  2  1

 1.2.3.2  Permute the concatenation C[i]D[i] as indicated below. This
will yield K[i], which is 48 bits long.

                        Permuted Choice 2 (PC-2)

                           14 17 11 24  1  5
                            3 28 15  6 21 10
                           23 19 12  4 26  8
                           16  7 27 20 13  2
                           41 52 31 37 47 55
                           30 40 51 45 33 48
                           44 49 39 56 34 53
                           46 42 50 36 29 32

 1.2.3.3  Loop back to 1.2.3.1 until K[16] has been calculated.

 2  Process a 64-bit data block.

 2.1  Get a 64-bit data block. If the block is shorter than 64 bits, it
should be padded as appropriate for the application.

 2.2  Perform the following permutation on the data block.

                        Initial Permutation (IP)

                        58 50 42 34 26 18 10  2
                        60 52 44 36 28 20 12  4
                        62 54 46 38 30 22 14  6
                        64 56 48 40 32 24 16  8
                        57 49 41 33 25 17  9  1
                        59 51 43 35 27 19 11  3
                        61 53 45 37 29 21 13  5
                        63 55 47 39 31 23 15  7

 2.3  Split the block into two halves. The first 32 bits are called L[0],
and the last 32 bits are called R[0].

 2.4  Apply the 16 subkeys to the data block. Start with i = 1.

 2.4.1  Expand the 32-bit R[i-1] into 48 bits according to the
bit-selection function below.

                             Expansion (E)

                           32  1  2  3  4  5
                            4  5  6  7  8  9
                            8  9 10 11 12 13
                           12 13 14 15 16 17
                           16 17 18 19 20 21
                           20 21 22 23 24 25
                           24 25 26 27 28 29
                           28 29 30 31 32  1

 2.4.2  Exclusive-or E(R[i-1]) with K[i].

 2.4.3  Break E(R[i-1]) xor K[i] into eight 6-bit blocks. Bits 1-6 are
B[1], bits 7-12 are B[2], and so on with bits 43-48 being B[8].

 2.4.4  Substitute the values found in the S-boxes for all B[j]. Start
with j = 1. All values in the S-boxes should be considered 4 bits wide.

 2.4.4.1  Take the 1st and 6th bits of B[j] together as a 2-bit value  
(call it m) indicating the row in S[j] to look in for the substitution.

 2.4.4.2  Take the 2nd through 5th bits of B[j] together as a 4-bit
value (call it n) indicating the column in S[j] to find the substitution.

 2.4.4.3  Replace B[j] with S[j][m][n].

                       Substitution Box 1 (S[1])

            14  4 13  1  2 15 11  8  3 10  6 12  5  9  0  7
             0 15  7  4 14  2 13  1 10  6 12 11  9  5  3  8
             4  1 14  8 13  6  2 11 15 12  9  7  3 10  5  0
            15 12  8  2  4  9  1  7  5 11  3 14 10  0  6 13

                                  S[2]

            15  1  8 14  6 11  3  4  9  7  2 13 12  0  5 10
             3 13  4  7 15  2  8 14 12  0  1 10  6  9 11  5
             0 14  7 11 10  4 13  1  5  8 12  6  9  3  2 15
            13  8 10  1  3 15  4  2 11  6  7 12  0  5 14  9

                                  S[3]

            10  0  9 14  6  3 15  5  1 13 12  7 11  4  2  8
            13  7  0  9  3  4  6 10  2  8  5 14 12 11 15  1
            13  6  4  9  8 15  3  0 11  1  2 12  5 10 14  7
             1 10 13  0  6  9  8  7  4 15 14  3 11  5  2 12

                                  S[4]

             7 13 14  3  0  6  9 10  1  2  8  5 11 12  4 15
            13  8 11  5  6 15  0  3  4  7  2 12  1 10 14  9
            10  6  9  0 12 11  7 13 15  1  3 14  5  2  8  4
             3 15  0  6 10  1 13  8  9  4  5 11 12  7  2 14

                                  S[5]

             2 12  4  1  7 10 11  6  8  5  3 15 13  0 14  9
            14 11  2 12  4  7 13  1  5  0 15 10  3  9  8  6
             4  2  1 11 10 13  7  8 15  9 12  5  6  3  0 14
            11  8 12  7  1 14  2 13  6 15  0  9 10  4  5  3

                                  S[6]

            12  1 10 15  9  2  6  8  0 13  3  4 14  7  5 11
            10 15  4  2  7 12  9  5  6  1 13 14  0 11  3  8
             9 14 15  5  2  8 12  3  7  0  4 10  1 13 11  6
             4  3  2 12  9  5 15 10 11 14  1  7  6  0  8 13

                                  S[7]

             4 11  2 14 15  0  8 13  3 12  9  7  5 10  6  1
            13  0 11  7  4  9  1 10 14  3  5 12  2 15  8  6
             1  4 11 13 12  3  7 14 10 15  6  8  0  5  9  2
             6 11 13  8  1  4 10  7  9  5  0 15 14  2  3 12

                                  S[8]

            13  2  8  4  6 15 11  1 10  9  3 14  5  0 12  7
             1 15 13  8 10  3  7  4 12  5  6 11  0 14  9  2
             7 11  4  1  9 12 14  2  0  6 10 13 15  3  5  8
             2  1 14  7  4 10  8 13 15 12  9  0  3  5  6 11

 2.4.4.4  Loop back to 2.4.4.1 until all 8 blocks have been replaced.

 2.4.5  Permute the concatenation of B[1] through B[8] as indicated below.

                             Permutation P

                              16  7 20 21
                              29 12 28 17
                               1 15 23 26
                               5 18 31 10
                               2  8 24 14
                              32 27  3  9
                              19 13 30  6
                              22 11  4 25

 2.4.6  Exclusive-or the resulting value with L[i-1]. Thus, all together,
your R[i] = L[i-1] xor P(S[1](B[1])...S[8](B[8])), where B[j] is a 6-bit  
block of E(R[i-1]) xor K[i]. (The function for R[i] is written as, R[i] =
L[i-1] xor f(R[i-1], K[i]).)

 2.4.7  L[i] = R[i-1].

 2.4.8  Loop back to 2.4.1 until K[16] has been applied.

 2.5  Perform the following permutation on the block R[16]L[16].

                       Final Permutation (IP**-1)

                        40  8 48 16 56 24 64 32
                        39  7 47 15 55 23 63 31
                        38  6 46 14 54 22 62 30
                        37  5 45 13 53 21 61 29
                        36  4 44 12 52 20 60 28
                        35  3 43 11 51 19 59 27
                        34  2 42 10 50 18 58 26
                        33  1 41  9 49 17 57 25


This has been a description of how to use the DES algorithm to encrypt
one 64-bit block. To decrypt, use the same process, but just use the keys
K[i] in reverse order. That is, instead of applying K[1] for the first
iteration, apply K[16], and then K[15] for the second, on down to K[1].

Summaries:

 Key schedule:
  C[0]D[0] = PC1(key)
  for 1 <= i <= 16
   C[i] = LS[i](C[i-1])
   D[i] = LS[i](D[i-1])
   K[i] = PC2(C[i]D[i])

 Encipherment:
  L[0]R[0] = IP(plain block)
  for 1 <= i <= 16
   L[i] = R[i-1]
   R[i] = L[i-1] xor f(R[i-1], K[i])
  cipher block = FP(R[16]L[16])

 Decipherment:
  R[16]L[16] = IP(cipher block)
  for 1 <= i <= 16
   R[i-1] = L[i]
   L[i-1] = R[i] xor f(L[i], K[i])
  plain block = FP(L[0]R[0])


To encrypt or decrypt more than 64 bits there are four official modes
(defined in FIPS PUB 81). One is to go through the above-described
process for each block in succession. This is called Electronic Codebook
(ECB) mode. A stronger method is to exclusive-or each plaintext block
with the preceding ciphertext block prior to encryption. (The first
block is exclusive-or'ed with a secret 64-bit initialization vector
(IV).) This is called Cipher Block Chaining (CBC) mode. The other two
modes are Output Feedback (OFB) and Cipher Feedback (CFB).

When it comes to padding the data block, there are several options. One
is to simply append zeros. Two suggested by FIPS PUB 81 are, if the data
is binary data, fill up the block with bits that are the opposite of the
last bit of data, or, if the data is ASCII data, fill up the block with
random bytes and put the ASCII character for the number of pad bytes in
the last byte of the block. Another technique is to pad the block with
random bytes and in the last 3 bits store the original number of data bytes.

The DES algorithm can also be used to calculate checksums up to 64 bits
long (see FIPS PUB 113). If the number of data bits to be checksummed is
not a multiple of 64, the last data block should be padded with zeros. If
the data is ASCII data, the first bit of each byte should be set to 0.
The data is then encrypted in CBC mode with IV = 0. The leftmost n bits
(where 16 <= n <= 64, and n is a multiple of 8) of the final ciphertext
block are an n-bit checksum.


0
 
LVL 2

Expert Comment

by:pxh
ID: 1292477
Also look on  http://www.program.com/source/crypto/des/

I think the link is selfexplanatory ;-)

Peter

0

Featured Post

How to run any project with ease

Manage projects of all sizes how you want. Great for personal to-do lists, project milestones, team priorities and launch plans.
- Combine task lists, docs, spreadsheets, and chat in one
- View and edit from mobile/offline
- Cut down on emails

Join & Write a Comment

In this article, I'll describe -- and show pictures of -- some of the significant additions that have been made available to programmers in the MFC Feature Pack for Visual C++ 2008.  These same feature are in the MFC libraries that come with Visual …
Introduction: Dialogs (2) modeless dialog and a worker thread.  Handling data shared between threads.  Recursive functions. Continuing from the tenth article about sudoku.   Last article we worked with a modal dialog to help maintain informat…
This video will show you how to get GIT to work in Eclipse.   It will walk you through how to install the EGit plugin in eclipse and how to checkout an existing repository.
This demo shows you how to set up the containerized NetScaler CPX with NetScaler Management and Analytics System in a non-routable Mesos/Marathon environment for use with Micro-Services applications.

708 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

10 Experts available now in Live!

Get 1:1 Help Now