## A great article about des algorithm

Posted by in Security# http://orlingrabbe.com/des.htm

## How DES Works in Detail

DES is a

–meaning it operates on plaintext blocks of a given size (64-bits) and returns ciphertext blocks of the same size. Thus DES results in ablock cipheramong 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 blockpermutationLand a right halfR. (This division is only used in certain operations.)

Example:LetMbe the plain text messageM= 0123456789ABCDEF, whereMis in hexadecimal (base 16) format. RewritingMin 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 1111The first bit of

Mis “0″. The last bit is “1″. We read from left to right.DES operates on the 64-bit blocks using

keysizes 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:LetKbe the hexadecimal keyK= 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 11110001The DES algorithm uses the following steps:

## Step 1: Create 16 subkeys, each of which is 48-bits long.

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 keyKbecomes the first bit of the permuted keyK+. 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.PC-157 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 11110001we get the 56-bit permutation

K+ = 1111000 0110011 0010101 0101111 0101010 1011001 1001111 0001111Next, split this key into left and right halves,

andC_{0}, where each half has 28 bits.D_{0}

Example:From the permuted keyK+, we get

= 1111000 0110011 0010101 0101111C_{0}

= 0101010 1011001 1001111 0001111D_{0}With

andC_{0}defined, we now create sixteen blocksD_{0}andC_{n}, 1<=D_{n}<=16. Each pair of blocksnandC_{n}is formed from the previous pairD_{n}andC_{n-1}, respectively, forD_{n-1}= 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.nIteration Number of Number 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 1This means, for example,

andC_{3}are obtained fromD_{3}andC_{2}, respectively, by two left shifts, andD_{2}andC_{16}are obtained fromD_{16}andC_{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.D_{15}

Example:From original pair pairandC_{0}we obtain:D_{0}

= 1111000011001100101010101111C_{0}

= 0101010101100110011110001111D_{0}

= 1110000110011001010101011111C_{1}

= 1010101011001100111100011110D_{1}

= 1100001100110010101010111111C_{2}

= 0101010110011001111000111101D_{2}

= 0000110011001010101011111111C_{3}

= 0101011001100111100011110101D_{3}

= 0011001100101010101111111100C_{4}

= 0101100110011110001111010101D_{4}

= 1100110010101010111111110000C_{5}

= 0110011001111000111101010101D_{5}

= 0011001010101011111111000011C_{6}

= 1001100111100011110101010101D_{6}

= 1100101010101111111100001100C_{7}

= 0110011110001111010101010110D_{7}

= 0010101010111111110000110011C_{8}

= 1001111000111101010101011001D_{8}

= 0101010101111111100001100110C_{9}

= 0011110001111010101010110011D_{9}

= 0101010111111110000110011001C_{10}

= 1111000111101010101011001100D_{10}

= 0101011111111000011001100101C_{11}

= 1100011110101010101100110011D_{11}

= 0101111111100001100110010101C_{12}

= 0001111010101010110011001111D_{12}

= 0111111110000110011001010101C_{13}

= 0111101010101011001100111100D_{13}

= 1111111000011001100101010101C_{14}

= 1110101010101100110011110001D_{14}

= 1111100001100110010101010111C_{15}

= 1010101010110011001111000111D_{15}

= 1111000011001100101010101111C_{16}

= 0101010101100110011110001111D_{16}We now form the keys

, for 1<=K_{n}<=16, by applying the following permutation table to each of the concatenated pairsn. Each pair has 56 bits, butC_{n}D_{n}PC-2only uses 48 of these.PC-214 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 32Therefore, the first bit of

is the 14th bit ofK_{n}, the second bit the 17th, and so on, ending with the 48th bit ofC_{n}D_{n}being the 32th bit ofK_{n}.C_{n}D_{n}

Example:For the first key we have= 1110000 1100110 0101010 1011111 1010101 0110011 0011110 0011110C_{1}D_{1}which, after we apply the permutation

PC-2, becomes

= 000110 110000 001011 101111 111111 000111 000001 110010K_{1}For the other keys we have

= 011110 011010 111011 011001 110110 111100 100111 100101K_{2}

= 010101 011111 110010 001010 010000 101100 111110 011001K_{3}

= 011100 101010 110111 010110 110110 110011 010100 011101K_{4}

= 011111 001110 110000 000111 111010 110101 001110 101000K_{5}

= 011000 111010 010100 111110 010100 000111 101100 101111K_{6}

= 111011 001000 010010 110111 111101 100001 100010 111100K_{7}

= 111101 111000 101000 111010 110000 010011 101111 111011K_{8}

= 111000 001101 101111 101011 111011 011110 011110 000001K_{9}

= 101100 011111 001101 000111 101110 100100 011001 001111K_{10}

= 001000 010101 111111 010011 110111 101101 001110 000110K_{11}

= 011101 010111 000111 110101 100101 000110 011111 101001K_{12}

= 100101 111100 010111 010001 111110 101011 101001 000001K_{13}

= 010111 110100 001110 110111 111100 101110 011100 111010K_{14}

= 101111 111001 000110 001101 001111 010011 111100 001010K_{15}

= 110010 110011 110110 001011 000011 100001 011111 110101K_{16}So much for the subkeys. Now we look at the message itself.

## Step 2: Encode each 64-bit block of data.

There is an

initial permutationIPof the 64 bits of the message dataM. 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 ofMbecomes the first bit ofIP. The 50th bit ofMbecomes the second bit ofIP. The 7th bit ofMis the last bit ofIP.IP58 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

Example:Applying the initial permutation to the block of textM, 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 1010Here the 58th bit of

Mis “1″, which becomes the first bit ofIP. The 50th bit ofMis “1″, which becomes the second bit ofIP. The 7th bit ofMis “0″, which becomes the last bit ofIP.Next divide the permuted block

IPinto a left halfof 32 bits, and a right halfL_{0}of 32 bits.R_{0}

Example:FromIP, we getandL_{0}R_{0}

= 1100 1100 0000 0000 1100 1100 1111 1111L_{0}

= 1111 0000 1010 1010 1111 0000 1010 1010R_{0}We now proceed through 16 iterations, for 1<=

<=16, using a functionnwhich operates on two blocks–a data block of 32 bits and a keyfof 48 bits–to produce a block of 32 bits.K_{n}Let + denote XOR addition, (bit-by-bit addition modulo 2). Then forngoing 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

= 16, ofn. 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 calculationL_{16}R_{16}.f

Example:For= 1, we haven

= 000110 110000 001011 101111 111111 000111 000001 110010K_{1}

=L_{1}= 1111 0000 1010 1010 1111 0000 1010 1010R_{0}

=R_{1}+L_{0}(f,R_{0})K_{1}It remains to explain how the function

works. To calculatef, we first expand each blockffrom 32 bits to 48 bits. This is done by using a selection table that repeats some of the bits inR_{n-1}. We’ll call the use of this selection table the functionR_{n-1}E. ThusE() has a 32 bit input block, and a 48 bit output block.R_{n-1}Let

Ebe 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 TABLE32 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 1Thus the first three bits of

E() are the bits in positions 32, 1 and 2 ofR_{n-1}while the last 2 bits ofR_{n-1}E() are the bits in positions 32 and 1.R_{n-1}

Example:We calculateE() fromR_{0}as follows:R_{0}

= 1111 0000 1010 1010 1111 0000 1010 1010R_{0}

E() = 011110 100001 010101 010101 011110 100001 010101 010101R_{0}(Note that each block of 4 original bits has been expanded to a block of 6 output bits.)

Next in the

calculation, we XOR the outputfE() with the keyR_{n-1}:K_{n}

+K_{n}E().R_{n-1}Example:For,K_{1}E(), we haveR_{0}

= 000110 110000 001011 101111 111111 000111 000001 110010K_{1}

E() = 011110 100001 010101 010101 011110 100001 010101 010101R_{0}

+K_{1}E() = 011000 010001 011110 111010 100001 100110 010100 100111.R_{0}We have not yet finished calculating the function

. To this point we have expandedffrom 32 bits to 48 bits, using the selection table, and XORed the result with the keyR_{n-1}. 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 “K_{n}S boxes“. Each group of six bits will give us an address in a differentSbox. 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 theSboxes) for 32 bits total.Write the previous result, which is 48 bits, in the form:

+K_{n}E() =R_{n-1},where eachB_{1}B_{2}B_{3}B_{4}B_{5}B_{6}B_{7}B_{8}is a group of six bits. We now calculateB_{i}

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})where

referres to the output of theS_{i}(B_{i})-thiSbox.To repeat, each of the functions

, takes a 6-bit block as input and yields a 4-bit block as output. The table to determineS1, S2,…, S8is shown and explained below:S_{1}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 13If

is the function defined in this table andS_{1}is a block of 6 bits, thenBis determined as follows: The first and last bits ofS_{1}(B)represent in base 2 a number in the decimal range 0 to 3 (or binary 00 to 11). Let that number beB. The middle 4 bits ofirepresent in base 2 a number in the decimal range 0 to 15 (binary 0000 to 1111). Let that number beB. Look up in the table the number in thej-th row andi-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 outputjofS_{1}(B)for the inputS_{1}. For example, for input blockB= 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. HenceB(011011) = 0101.S_{1}The tables defining the functions

are the following:S_{1},…,S_{8}S114 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 13S215 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 9S310 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 12S47 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 14S52 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 3S612 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 13S74 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 12S813 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 eightSboxes:

+K_{1}E() = 011000 010001 011110 111010 100001 100110 010100 100111.R_{0}

= 0101 1100 1000 0010 1011 0101 1001 0111S_{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})The final stage in the calculation of

is to do a permutationfPof theS-box output to obtain the final value of:f

=fP()The permutationS_{1}(B_{1})S_{2}(B_{2})…S_{8}(B_{8})Pis defined in the following table.Pyields a 32-bit output from a 32-bit input by permuting the bits of the input block.P16 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 eightSboxes:

= 0101 1100 1000 0010 1011 0101 1001 0111we getS_{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})

= 0010 0011 0100 1010 1010 1001 1011 1011f=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 0100In the next round, we will have

=L_{2}, which is the block we just calculated, and then we must calculateR_{1}=R_{2}, and so on for 16 rounds. At the end of the sixteenth round we have the blocksL_{1}+ f(R_{1}, K_{2})andL_{16}. We thenR_{16}the order of the two blocks into the 64-bit blockreverse

and apply a final permutationR_{16}L_{16}IPas defined by the following table:^{-1}IP40 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^{-1}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,

= 0100 0011 0100 0010 0011 0010 0011 0100L_{16}

= 0000 1010 0100 1100 1101 1001 1001 0101R_{16}We reverse the order of these two blocks and apply the final permutation to

= 00001010 01001100 11011001 10010101 01000011 01000010 00110010 00110100R_{16}L_{16}

= 10000101 11101000 00010011 01010100 00001111 00001010 10110100 00000101IP^{-1}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.

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