The DES (Data Encryption Standard) algorithm is the most widely used encryption algorithm in the world. For many years, and among many people, "secret code making" and DES have been synonymous. And despite the recent coup by the Electronic Frontier Foundation in creating a $220,000 machine to crack DES-encrypted messages, DES will live on in government and banking for years to come through a life- extending version called "triple-DES."
How does DES work? This article explains the various steps involved in DES-encryption, illustrating each step by means of a simple example. Since the creation of DES, many other algorithms (recipes for changing data) have emerged which are based on design principles similar to DES. Once you understand the basic transformations that take place in DES, you will find it easy to follow the steps involved in these more recent algorithms.
But first a bit of history of how DES came about is appropriate, as well as a look toward the future.
DES works on bits, or binary numbers--the 0s and 1s common to digital computers. Each group of four bits makes up a hexadecimal, or base 16, number. Binary "0001" is equal to the hexadecimal number "1", binary "1000" is equal to the hexadecimal number "8", "1001" is equal to the hexadecimal number "9", "1010" is equal to the hexadecimal number "A", and "1111" is equal to the hexadecimal number "F".
DES works by encrypting groups of 64 message bits, which is the same as 16 hexadecimal numbers. To do the encryption, DES uses "keys" where are also apparently 16 hexadecimal numbers long, or apparently 64 bits long. However, every 8th key bit is ignored in the DES algorithm, so that the effective key size is 56 bits. But, in any case, 64 bits (16 hexadecimal digits) is the round number upon which DES is organized.
For example, if we take the plaintext message "8787878787878787", and encrypt it with the DES key "0E329232EA6D0D73", we end up with the ciphertext "0000000000000000". If the ciphertext is decrypted with the same secret DES key "0E329232EA6D0D73", the result is the original plaintext "8787878787878787".
This example is neat and orderly because our plaintext was exactly 64 bits long. The same would be true if the plaintext happened to be a multiple of 64 bits. But most messages will not fall into this category. They will not be an exact multiple of 64 bits (that is, an exact multiple of 16 hexadecimal numbers).
For example, take the message "Your lips are smoother than vaseline". This plaintext message is 38 bytes (76 hexadecimal digits) long. So this message must be padded with some extra bytes at the tail end for the encryption. Once the encrypted message has been decrypted, these extra bytes are thrown away. There are, of course, different padding schemes--different ways to add extra bytes. Here we will just add 0s at the end, so that the total message is a multiple of 8 bytes (or 16 hexadecimal digits, or 64 bits).
The plaintext message "Your lips are smoother than vaseline" is, in hexadecimal,
"596F7572206C6970 732061726520736D 6F6F746865722074 68616E2076617365 6C696E650D0A".
(Note here that the first 72 hexadecimal digits represent the English message, while "0D" is hexadecimal for Carriage Return, and "0A" is hexadecimal for Line Feed, showing that the message file has terminated.) We then pad this message with some 0s on the end, to get a total of 80 hexadecimal digits:
"596F7572206C6970 732061726520736D 6F6F746865722074 68616E2076617365 6C696E650D0A0000".
If we then encrypt this plaintext message 64 bits (16 hexadecimal digits) at a time, using the same DES key "0E329232EA6D0D73" as before, we get the ciphertext:
"C0999FDDE378D7ED 727DA00BCA5A84EE 47F269A4D6438190 9DD52F78F5358499 828AC9B453E0E653".
This is the secret code that can be transmitted or stored. Decrypting the ciphertext restores the original message "Your lips are smoother than vaseline". (Think how much better off Bill Clinton would be today, if Monica Lewinsky had used encryption on her Pentagon computer!)
DES is a block cipher--meaning it operates on plaintext blocks of a given size (64-bits) and returns ciphertext blocks of the same size. Thus DES results in a permutation among the 2^64 (read this as: "2 to the 64th power") possible arrangements of 64 bits, each of which may be either 0 or 1. Each block of 64 bits is divided into two blocks of 32 bits each, a left half block L and a right half R. (This division is only used in certain operations.)
Example: Let M be the plain text message M = 0123456789ABCDEF, where M is in hexadecimal (base 16) format. Rewriting M in binary format, we get the 64-bit block of text:
M = 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111
L = 0000 0001 0010 0011 0100 0101 0110 0111
R = 1000 1001 1010 1011 1100 1101 1110 1111
The first bit of M is "0". The last bit is "1". We read from left to right.
DES operates on the 64-bit blocks using key sizes of 56- bits. The keys are actually stored as being 64 bits long, but every 8th bit in the key is not used (i.e. bits numbered 8, 16, 24, 32, 40, 48, 56, and 64). However, we will nevertheless number the bits from 1 to 64, going left to right, in the following calculations. But, as you will see, the eight bits just mentioned get eliminated when we create subkeys.
Example: Let K be the hexadecimal key K = 133457799BBCDFF1. This gives us as the binary key (setting 1 = 0001, 3 = 0011, etc., and grouping together every eight bits, of which the last one in each group will be unused):
K = 00010011 00110100 01010111 01111001 10011011 10111100 11011111 11110001
The DES algorithm uses the following steps:
The 64-bit key is permuted according to the following table, PC-1. Since the first entry in the table is "57", this means that the 57th bit of the original key K becomes the first bit of the permuted key K+. The 49th bit of the original key becomes the second bit of the permuted key. The 4th bit of the original key is the last bit of the permuted key. Note only 56 bits of the original key appear in the permuted key.
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 |
Example: From the original 64-bit key
K = 00010011 00110100 01010111 01111001 10011011 10111100 11011111 11110001
we get the 56-bit permutation
K+ = 1111000 0110011 0010101 0101111 0101010 1011001 1001111 0001111
Next, split this key into left and right halves, C_{0} and D_{0} , where each half has 28 bits.
Example: From the permuted key K+, we get
C_{0}= 1111000 0110011 0010101 0101111
D_{0} = 0101010 1011001 1001111 0001111
With C_{0} and D_{0} defined, we now create sixteen blocks C_{n} and D_{n}, 1<=n<=16. Each pair of blocks C_{n} and D_{n} is formed from the previous pair C_{n-1} and D_{n-1}, respectively, for n = 1, 2, ..., 16, using the following schedule of "left shifts" of the previous block. To do a left shift, move each bit one place to the left, except for the first bit, which is cycled to the end of the block.
Iteration Number |
Number of Left Shifts |
1 |
1 |
2 |
1 |
3 |
2 |
4 |
2 |
5 |
2 |
6 |
2 |
7 |
2 |
8 |
2 |
9 |
1 |
10 |
2 |
11 |
2 |
12 |
2 |
13 |
2 |
14 |
2 |
15 |
2 |
16 |
1 |
This means, for example, C_{3} and D_{3} are obtained from C_{2} and D_{2}, respectively, by two left shifts, and C_{16} and D_{16} are obtained from C_{15} and C_{15}, respectively, by one left shift. In all cases, by a single left shift is meant a rotation of the bits one place to the left, so that after one left shift the bits in the 28 positions are the bits that were previously in positions 2, 3,..., 28, 1.
Example: From original pair pairC0 and D0 we obtain:
C_{0} = 1111000011001100101010101111
D_{0} = 0101010101100110011110001111
C_{1} = 1110000110011001010101011111
D_{1} = 1010101011001100111100011110
C_{2} = 1100001100110010101010111111
C_{2} = 0101010110011001111000111101
C_{3} = 0000110011001010101011111111
D_{3} = 0101011001100111100011110101
C_{4} = 0011001100101010101111111100
D_{4} = 0101100110011110001111010101
C_{5} = 1100110010101010111111110000
D_{5} = 0110011001111000111101010101
C_{6} = 0011001010101011111111000011
D_{6} = 1001100111100011110101010101
C_{7} = 1100101010101111111100001100
D_{7} = 0110011110001111010101010110
C_{8} = 0010101010111111110000110011
D_{8} = 1001111000111101010101011001
C_{9} = 0101010101111111100001100110
D_{9} = 0011110001111010101010110011
C_{10} = 0101010111111110000110011001
D_{10} = 1111000111101010101011001100
C_{11} = 0101011111111000011001100101
D_{11} = 1100011110101010101100110011
C_{12} = 0101111111100001100110010101
D_{12} = 0001111010101010110011001111
C_{13} = 0111111110000110011001010101
D_{13} = 0111101010101011001100111100
C_{14} = 1111111000011001100101010101
D_{14} = 1110101010101100110011110001
C_{15} = 1111100001100110010101010111
D_{15} = 1010101010110011001111000111
C_{16} = 1111000011001100101010101111
D_{16} = 0101010101100110011110001111
We now form the keys K_{n}, for 1<=n<=16, by applying the following permutation table to each of the concatenated pairs C_{n}D_{n}. Each pair has 56 bits, but PC-2 only uses 48 of these.
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 |
Therefore, the first bit of K_{n} is the 14th bit of C_{n}D_{n}, the second bit the 17th, and so on, ending with the 48th bit of K_{n} being the 32th bit of C_{n}D_{n}.
Example: For the first key we have C_{1}D_{1} = 1110000 1100110 0101010 1011111 1010101 0110011 0011110 0011110
which, after we apply the permutation PC-2, becomes
K_{1} = 000110 110000 001011 101111 111111 000111 000001 110010
For the other keys we have
K_{2} = 011110 011010 111011 011001 110110 111100 100111 100101
K_{3} = 010101 011111 110010 001010 010000 101100 111110 011001
K_{4} = 011100 101010 110111 010110 110110 110011 010100 011101
K_{5} = 011111 001110 110000 000111 111010 110101 001110 101000
K_{6} = 011000 111010 010100 111110 010100 000111 101100 101111
K_{7} = 111011 001000 010010 110111 111101 100001 100010 111100
K_{8} = 111101 111000 101000 111010 110000 010011 101111 111011
K_{9} = 111000 001101 101111 101011 111011 011110 011110 000001
K_{10} = 101100 011111 001101 000111 101110 100100 011001 001111
K_{11} = 001000 010101 111111 010011 110111 101101 001110 000110
K_{12} = 011101 010111 000111 110101 100101 000110 011111 101001
K_{13} = 100101 111100 010111 010001 111110 101011 101001 000001
K_{14} = 010111 110100 001110 110111 111100 101110 011100 111010
K_{15} = 101111 111001 000110 001101 001111 010011 111100 001010
K_{16} = 110010 110011 110110 001011 000011 100001 011111 110101
So much for the subkeys. Now we look at the message itself.
There is an initial permutationIP of the 64 bits of the message data M. This rearranges the bits according to the following table, where the entries in the table show the new arrangement of the bits from their initial order. The 58th bit of M becomes the first bit of IP. The 50th bit of M becomes the second bit of IP. The 7th bit of M is the last bit of IP.
IP
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 |
Example: Applying the initial permutation to the block of text M, given previously, we get
M = 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111
IP = 1100 1100 0000 0000 1100 1100 1111 1111 1111 0000 1010 1010 1111 0000 1010 1010
Here the 58th bit of M is "1", which becomes the first bit of IP. The 50th bit of M is "1", which becomes the second bit of IP. The 7th bit of M is "0", which becomes the last bit of IP.
Next divide the permuted block IP into a left half L_{0} of 32 bits, and a right half R_{0} of 32 bits.
Example: From IP, we get L_{0} and R_{0}
L_{0} = 1100 1100 0000 0000 1100 1100 1111 1111
R_{0} = 1111 0000 1010 1010 1111 0000 1010 1010
We now proceed through 16 iterations, for 1<=n<=16, using a function f which operates on two blocks--a data block of 32 bits and a key K_{n} of 48 bits--to produce a block of 32 bits. Let + denote XOR addition, (bit-by-bit addition modulo 2). Then for n going from 1 to 16 we calculate
L_{n} = R_{n-1}
R_{n} = L_{n-1} + f(R_{n-1},K_{n})
This results in a final block, for n = 16, of L_{16}R_{16}. That is, in each iteration, we take the right 32 bits of the previous result and make them the left 32 bits of the current step. For the right 32 bits in the current step, we XOR the left 32 bits of the previous step with the calculation f .
Example: For n = 1, we have
K_{1} = 000110 110000 001011 101111 111111 000111 000001 110010
L_{1} = R_{0} = 1111 0000 1010 1010 1111 0000 1010 1010
R_{1} = L_{0} + f(R_{0},K_{1})
It remains to explain how the function f works. To calculate f, we first expand each block R_{n-1} from 32 bits to 48 bits. This is done by using a selection table that repeats some of the bits in R_{n-1} . We'll call the use of this selection table the function E. Thus E(R_{n-1}) has a 32 bit input block, and a 48 bit output block.
Let E be such that the 48 bits of its output, written as 8 blocks of 6 bits each, are obtained by selecting the bits in its inputs in order according to the following table:
E BIT-SELECTION TABLE
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 |
Thus the first three bits of E(R_{n-1}) are the bits in positions 32, 1 and 2 of R_{n-1} while the last 2 bits of E(R_{n-1}) are the bits in positions 32 and 1.
Example: We calculate E(R_{0}) from R_{0} as follows:
R_{0} = 1111 0000 1010 1010 1111 0000 1010 1010
E(R_{0}) = 011110 100001 010101 010101 011110 100001 010101 010101
(Note that each block of 4 original bits has been expanded to a block of 6 output bits.)
Next in the f calculation, we XOR the output E(R_{n-1}) with the key K_{n}:
K_{n} + E(R_{n-1}).
Example: For K_{1} ,E(R_{0}), we have
K_{1} = 000110 110000 001011 101111 111111 000111 000001 110010
E(R_{0}) = 011110 100001 010101 010101 011110 100001 010101 010101
K_{1}+E(R_{0}) = 011000 010001 011110 111010 100001 100110 010100 100111.
We have not yet finished calculating the function f . To this point we have expanded R_{n-1} from 32 bits to 48 bits, using the selection table, and XORed the result with the key K_{n} . We now have 48 bits, or eight groups of six bits. We now do something strange with each group of six bits: we use them as addresses in tables called "S boxes". Each group of six bits will give us an address in a different S box. Located at that address will be a 4 bit number. This 4 bit number will replace the original 6 bits. The net result is that the eight groups of 6 bits are transformed into eight groups of 4 bits (the 4-bit outputs from the S boxes) for 32 bits total.
Write the previous result, which is 48 bits, in the form:
K_{n} + E(R_{n-1}) =B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}B_{7}B_{8},
where each B_{i} is a group of six bits. We now calculate
S_{1}(B_{1})S_{2}(B_{2})S_{3}(B_{3})S_{4}(B_{4})S_{5}(B_{5})S_{6}(B_{6})S_{7}(B_{7})S_{8}(B_{8})
whereS_{i}(B_{i})referres to the output of the i-thS box.
To repeat, each of the functions S1, S2,..., S8, takes a 6-bit block as input and yields a 4-bit block as output. The table to determine S_{1} is shown and explained below:
S1
Column Number
Row No. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
0 |
14 |
4 |
13 |
1 |
2 |
15 |
11 |
8 |
3 |
10 |
6 |
12 |
5 |
9 |
0 |
7 |
1 |
0 |
15 |
7 |
4 |
14 |
2 |
13 |
1 |
10 |
6 |
12 |
11 |
9 |
5 |
3 |
8 |
2 |
4 |
1 |
14 |
8 |
13 |
6 |
2 |
11 |
15 |
12 |
9 |
7 |
3 |
10 |
5 |
0 |
3 |
15 |
12 |
8 |
2 |
4 |
9 |
1 |
7 |
5 |
11 |
3 |
14 |
10 |
0 |
6 |
13 |
If S_{1} is the function defined in this table and B is a block of 6 bits, then S_{1}(B) is determined as follows: The first and last bits of B represent in base 2 a number in the decimal range 0 to 3 (or binary 00 to 11). Let that number be i. The middle 4 bits of B represent in base 2 a number in the decimal range 0 to 15 (binary 0000 to 1111). Let that number be j. Look up in the table the number in the i-th row and j-th column. It is a number in the range 0 to 15 and is uniquely represented by a 4 bit block. That block is the output S_{1}(B) of S_{1} for the input B. For example, for input block B = 011011 the first bit is "0" and the last bit "1" giving 01 as the row. This is row 1. The middle four bits are "1101". This is the binary equivalent of decimal 13, so the column is column number 13. In row 1, column 13 appears 5. This determines the output; 5 is binary 0101, so that the output is 0101. Hence S_{1}(011011) = 0101.
The tables defining the functions S_{1},...,S_{8} are the following:
S1
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 |
S2
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 |
S4
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 |
S5
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 |
S6
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 |
S7
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 |
S8
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 |
Example: For the first round, we obtain as the output of the eight S boxes:
K_{1} + E(R_{0}) = 011000 010001 011110 111010 100001 100110 010100 100111.
S_{1}(B_{1})S_{2}(B_{2})S_{3}(B_{3})S_{4}(B_{4})S_{5}(B_{5})S_{6}(B_{6})S_{7}(B_{7})S_{8}(B_{8}) = 0101 1100 1000 0010 1011 0101 1001 0111
The final stage in the calculation of f is to do a permutation P of the S-box output to obtain the final value of f:
f = P(S_{1}(B_{1})S_{2}(B_{2})...S_{8}(B_{8}))
The permutation P is defined in the following table. P yields a 32-bit output from a 32-bit input by permuting the bits of the input block.
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 |
Example: From the output of the eight S boxes:
S_{1}(B_{1})S_{2}(B_{2})S_{3}(B_{3})S_{4}(B_{4})S_{5}(B_{5})S_{6}(B_{6})S_{7}(B_{7})S_{8}(B_{8}) = 0101 1100 1000 0010 1011 0101 1001 0111
we get
f = 0010 0011 0100 1010 1010 1001 1011 1011
R_{1} = L_{0} + f(R_{0} , K_{1} )
= 1100 1100 0000 0000 1100 1100 1111 1111
+ 0010 0011 0100 1010 1010 1001 1011 1011
= 1110 1111 0100 1010 0110 0101 0100 0100
In the next round, we will have L_{2} = R_{1}, which is the block we just calculated, and then we must calculate R_{2} =L_{1} + f(R_{1}, K_{2}), and so on for 16 rounds. At the end of the sixteenth round we have the blocks L_{16} and R_{16}. We then reverse the order of the two blocks into the 64-bit block
R_{16}L_{16}
and apply a final permutation IP^{-1} as defined by the following table:
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 |
That is, the output of the algorithm has bit 40 of the preoutput block as its first bit, bit 8 as its second bit, and so on, until bit 25 of the preoutput block is the last bit of the output.
Example: If we process all 16 blocks using the method defined previously, we get, on the 16th round,
L_{16} = 0100 0011 0100 0010 0011 0010 0011 0100
R_{16} = 0000 1010 0100 1100 1101 1001 1001 0101
We reverse the order of these two blocks and apply the final permutation to
R_{16}L_{16} = 00001010 01001100 11011001 10010101 01000011 01000010 00110010 00110100
IP^{-1} = 10000101 11101000 00010011 01010100 00001111 00001010 10110100 00000101
which in hexadecimal format is
85E813540F0AB405.
This is the encrypted form of M = 0123456789ABCDEF: namely, C = 85E813540F0AB405.
Decryption is simply the inverse of encryption, follwing the same steps as above, but reversing the order in which the subkeys are applied.
The DES algorithm turns a 64-bit message blockM into a 64-bit cipher block C. If each 64-bit block is encrypted individually, then the mode of encryption is called Electronic Code Book (ECB) mode. There are two other modes of DES encryption, namely Chain Block Coding (CBC) and Cipher Feedback (CFB), which make each cipher block dependent on all the previous messages blocks through an initial XOR operation.
Before DES was adopted as a national standard, during the period NBS was soliciting comments on the proposed algorithm, the creators of public key cryptography, Martin Hellman and Whitfield Diffie, registered some objections to the use of DES as an encryption algorithm. Hellman wrote: "Whit Diffie and I have become concerned that the proposed data encryption standard, while probably secure against commercial assault, may be extremely vulnerable to attack by an intelligence organization" (letter to NBS, October 22, 1975).
Diffie and Hellman then outlined a "brute force" attack on DES. (By "brute force" is meant that you try as many of the 2^56 possible keys as you have to before decrypting the ciphertext into a sensible plaintext message.) They proposed a special purpose "parallel computer using one million chips to try one million keys each" per second, and estimated the cost of such a machine at $20 million.
Fast forward to 1998. Under the direction of John Gilmore of the EFF, a team spent $220,000 and built a machine that can go through the entire 56-bit DES key space in an average of 4.5 days. On July 17, 1998, they announced they had cracked a 56-bit key in 56 hours. The computer, called Deep Crack, uses 27 boards each containing 64 chips, and is capable of testing 90 billion keys a second.
Despite this, as recently as June 8, 1998, Robert Litt, principal associate deputy attorney general at the Department of Justice, denied it was possible for the FBI to crack DES: "Let me put the technical problem in context: It took 14,000 Pentium computers working for four months to decrypt a single message . . . . We are not just talking FBI and NSA [needing massive computing power], we are talking about every police department."
Responded cryptograpy expert Bruce Schneier: " . . . the FBI is either incompetent or lying, or both." Schneier went on to say: "The only solution here is to pick an algorithm with a longer key; there isn't enough silicon in the galaxy or enough time before the sun burns out to brute- force triple-DES" (Crypto-Gram, Counterpane Systems, August 15, 1998).
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