The DES Algorithm Illustrated (2024)

The DES (Data Encryption Standard) algorithm is themost widely used encryption algorithm in the world. Formany years, and among many people, "secret code making" andDES have been synonymous. And despite the recent coup bythe Electronic Frontier Foundation in creating a $220,000machine to crack DES-encrypted messages, DES will live on ingovernment and banking for years to come through a life-extending version called "triple-DES."

How does DES work? This article explains the varioussteps involved in DES-encryption, illustrating each step bymeans of a simple example. Since the creation of DES, manyother algorithms (recipes for changing data) have emergedwhich are based on design principles similar to DES. Onceyou understand the basic transformations that take place inDES, you will find it easy to follow the steps involved inthese more recent algorithms.

But first a bit of history of how DES came about isappropriate, as well as a look toward the future.

The National Bureau of Standards Coaxes the Genie from the Bottle

On May 15, 1973, during the reign of Richard Nixon, theNational Bureau of Standards (NBS) published a notice in theFederal Register soliciting proposals for cryptographicalgorithms to protect data during transmission and storage.The notice explained why encryption was an important issue.

Over the last decade, there has been an accelerating increase in the accumulations and communication of digital data by government, industry and by other organizations in the private sector. The contents of these communicated and stored data often have very significant value and/or sensitivity. It is now common to find data transmissions which constitute funds transfers of several million dollars, purchase or sale of securities, warrants for arrests or arrest and conviction records being communicated between law enforcement agencies, airline reservations and ticketing representing investment and value both to the airline and passengers, and health and patient care records transmitted among physicians and treatment centers.

The increasing volume, value and confidentiality of these records regularly transmitted and stored by commercial and government agencies has led to heightened recognition and concern over their exposures to unauthorized access and use. This misuse can be in the form of theft or defalcations of data records representing money, malicious modification of business inventories or the interception and misuse of confidential information about people. The need for protection is then apparent and urgent.

It is recognized that encryption (otherwise known as scrambling, enciphering or privacy transformation) represents the only means of protecting such data during transmission and a useful means of protecting the content of data stored on various media, providing encryption of adequate strength can be devised and validated and is inherently integrable into system architecture. The National Bureau of Standards solicits proposed techniques and algorithms for computer data encryption. The Bureau also solicits recommended techniques for implementing the cryptographic function: for generating, evaluating, and protecting cryptographic keys; for maintaining files encoded under expiring keys; for making partial updates to encrypted files; and mixed clear and encrypted data to permit labelling, polling, routing, etc. The Bureau in its role for establishing standards and aiding government and industry in assessing technology, will arrange for the evaluation of protection methods in order to prepare guidelines.

NBS waited for the responses to come in. It receivednone until August 6, 1974, three days before Nixon'sresignation, when IBM submitted a candidate that it haddeveloped internally under the name LUCIFER. Afterevaluating the algorithm with the help of the NationalSecurity Agency (NSA), the NBS adopted a modification of theLUCIFER algorithm as the new Data Encryption Standard (DES)on July 15, 1977.

DES was quickly adopted for non-digital media, such asvoice-grade public telephone lines. Within a couple ofyears, for example, International Flavors and Fragrances wasusing DES to protect its valuable formulas transmitted overthe phone ("With Data Encryption, Scents Are Safe at IFF,"Computerworld 14, No. 21, 95 (1980).)

Meanwhile, the banking industry, which is the largestuser of encryption outside government, adopted DES as awholesale banking standard. Standards for the wholesalebanking industry are set by the American National StandardsInstitute (ANSI). ANSI X3.92, adopted in 1980, specifiedthe use of the DES algorithm.

Some Preliminary Examples of DES

DES works on bits, or binary numbers--the 0s and 1scommon to digital computers. Each group of four bits makesup a hexadecimal, or base 16, number. Binary "0001" isequal to the hexadecimal number "1", binary "1000" is equalto the hexadecimal number "8", "1001" is equal to thehexadecimal number "9", "1010" is equal to the hexadecimalnumber "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 theencryption, DES uses "keys" where are also apparently 16hexadecimal 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 anycase, 64 bits (16 hexadecimal digits) is the round numberupon 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 thesame secret DES key "0E329232EA6D0D73", the result is theoriginal plaintext "8787878787878787".

This example is neat and orderly because our plaintextwas exactly 64 bits long. The same would be true if theplaintext happened to be a multiple of 64 bits. But mostmessages will not fall into this category. They will not bean exact multiple of 64 bits (that is, an exact multiple of16 hexadecimal numbers).

For example, take the message "Your lips are smootherthan vaseline". This plaintext message is 38 bytes (76hexadecimal digits) long. So this message must be paddedwith some extra bytes at the tail end for the encryption.Once the encrypted message has been decrypted, these extrabytes are thrown away. There are, of course, differentpadding schemes--different ways to add extra bytes. Here wewill just add 0s at the end, so that the total message is amultiple of 8 bytes (or 16 hexadecimal digits, or 64 bits).

The plaintext message "Your lips are smoother thanvaseline" is, in hexadecimal,

"596F7572206C6970 732061726520736D 6F6F746865722074 68616E2076617365 6C696E650D0A".

(Note here that the first 72 hexadecimal digits representthe English message, while "0D" is hexadecimal for CarriageReturn, and "0A" is hexadecimal for Line Feed, showing thatthe message file has terminated.) We then pad this messagewith some 0s on the end, to get a total of 80 hexadecimaldigits:

"596F7572206C6970 732061726520736D 6F6F746865722074 68616E2076617365 6C696E650D0A0000".

If we then encrypt this plaintext message 64 bits (16hexadecimal 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 muchbetter off Bill Clinton would be today, if Monica Lewinskyhad used encryption on her Pentagon computer!)

How DES Works in Detail

DES is a block cipher--meaning it operates on plaintextblocks of a given size (64-bits) and returns ciphertextblocks of the same size. Thus DES results in a permutationamong the 2^64 (read this as: "2 to the 64th power") possible arrangements of 64 bits, each ofwhich may be either 0 or 1. Each block of 64 bits is dividedinto two blocks of 32 bits each, a left half block L and aright half R. (This division is only used in certainoperations.)

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-bitblock 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 readfrom 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 numbered8, 16, 24, 32, 40, 48, 56, and 64). However, we willnevertheless number the bits from 1 to 64, going left toright, in the following calculations. But, as you will see,the eight bits just mentioned get eliminated when we createsubkeys.

Example: Let K be the hexadecimal key K =133457799BBCDFF1. This gives us as the binary key (setting1 = 0001, 3 = 0011, etc., and grouping together every eightbits, 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:

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

The 64-bit key is permuted according to the followingtable, PC-1. Since the first entry in the table is "57",this means that the 57th bit of the original key K becomesthe first bit of the permuted key K+. The 49th bit of theoriginal key becomes the second bit of the permuted key.The 4th bit of the original key is the last bit of thepermuted key. Note only 56 bits of the original key appearin the permuted key.

 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

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, C₀ andD₀, where each half has 28 bits.

Example: From the permuted key K+, we get

C₀ = 1111000 0110011 0010101 0101111
D₀ = 0101010 1011001 1001111 0001111

With C₀ and D₀ defined, we now create sixteen blocks C_nand D_n, 1<=n<=16. Each pair of blocks C_n and D_n is formedfrom 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 bitone place to the left, except for the first bit, which iscycled to the end of the block.

 Iteration 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 1

This means, for example, C₃ and D₃ are obtained from C₂ andD₂, respectively, by two left shifts, and C₁₆ and D₁₆ areobtained from C₁₅ and D₁₅, respectively, by one left shift.In all cases, by a single left shift is meant a rotation ofthe bits one place to the left, so that after one left shiftthe bits in the 28 positions are the bits that werepreviously in positions 2, 3,..., 28, 1.

Example: From original pair pair C₀ and D₀ we obtain:

C₀ = 1111000011001100101010101111
D₀ = 0101010101100110011110001111

C₁ = 1110000110011001010101011111
D₁ = 1010101011001100111100011110

C₂ = 1100001100110010101010111111
D₂ = 0101010110011001111000111101

C₃ = 0000110011001010101011111111
D₃ = 0101011001100111100011110101

C₄ = 0011001100101010101111111100
D₄ = 0101100110011110001111010101

C₅ = 1100110010101010111111110000
D₅ = 0110011001111000111101010101

C₆ = 0011001010101011111111000011
D₆ = 1001100111100011110101010101

C₇ = 1100101010101111111100001100
D₇ = 0110011110001111010101010110

C₈ = 0010101010111111110000110011
D₈ = 1001111000111101010101011001

C₉ = 0101010101111111100001100110
D₉ = 0011110001111010101010110011

C₁₀ = 0101010111111110000110011001
D₁₀ = 1111000111101010101011001100

C₁₁ = 0101011111111000011001100101
D₁₁ = 1100011110101010101100110011

C₁₂ = 0101111111100001100110010101
D₁₂ = 0001111010101010110011001111

C₁₃ = 0111111110000110011001010101
D₁₃ = 0111101010101011001100111100

C₁₄ = 1111111000011001100101010101
D₁₄ = 1110101010101100110011110001

C₁₅ = 1111100001100110010101010111
D₁₅ = 1010101010110011001111000111

C₁₆ = 1111000011001100101010101111
D₁₆ = 0101010101100110011110001111

We now form the keys K_n, for 1<=n<=16, by applying thefollowing permutation table to each of the concatenatedpairs C_nD_n. Each pair has 56 bits, but PC-2 only uses 48 ofthese.

 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_nD_n, thesecond bit the 17th, and so on, ending with the 48th bit ofK_n being the 32th bit of C_nD_n.

Example: For the first key we haveC₁D₁ = 1110000 1100110 0101010 1011111 1010101 0110011 0011110 0011110

which, after we apply the permutation PC-2, becomes

K₁ = 000110 110000 001011 101111 111111 000111 000001 110010

For the other keys we have

K₂ = 011110 011010 111011 011001 110110 111100 100111 100101
K₃ = 010101 011111 110010 001010 010000 101100 111110 011001
K₄ = 011100 101010 110111 010110 110110 110011 010100 011101
K₅ = 011111 001110 110000 000111 111010 110101 001110 101000
K₆ = 011000 111010 010100 111110 010100 000111 101100 101111
K₇ = 111011 001000 010010 110111 111101 100001 100010 111100
K₈ = 111101 111000 101000 111010 110000 010011 101111 111011
K₉ = 111000 001101 101111 101011 111011 011110 011110 000001
K₁₀ = 101100 011111 001101 000111 101110 100100 011001 001111
K₁₁ = 001000 010101 111111 010011 110111 101101 001110 000110
K₁₂ = 011101 010111 000111 110101 100101 000110 011111 101001
K₁₃ = 100101 111100 010111 010001 111110 101011 101001 000001
K₁₄ = 010111 110100 001110 110111 111100 101110 011100 111010
K₁₅ = 101111 111001 000110 001101 001111 010011 111100 001010
K₁₆ = 110010 110011 110110 001011 000011 100001 011111 110101

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 permutation IP of the 64 bits ofthe message data M. This rearranges the bits according tothe following table, where the entries in the table show thenew arrangement of the bits from their initial order. The58th bit of M becomes the first bit of IP. The 50th bit ofM becomes the second bit of IP. The 7th bit of M is thelast bit of IP.

 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

Example: Applying the initial permutation to the blockof 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 bitof IP. The 50th bit of M is "1", which becomes the secondbit of IP. The 7th bit of M is "0", which becomes the lastbit of IP.

Next divide the permuted block IP into a left half L₀of 32 bits, and a right half R₀ of 32 bits.

Example: From IP, we get L₀ and R₀

L₀ = 1100 1100 0000 0000 1100 1100 1111 1111
R₀ = 1111 0000 1010 1010 1111 0000 1010 1010

We now proceed through 16 iterations, for 1<=n<=16, usinga function f which operates on two blocks--a data block of32 bits and a key K_n of 48 bits--to produce a block of 32bits. Let + denote XOR addition, (bit-by-bit additionmodulo 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 L₁₆R₁₆. Thatis, in each iteration, we take the right 32 bits of theprevious result and make them the left 32 bits of thecurrent step. For the right 32 bits in the current step, weXOR the left 32 bits of the previous step with thecalculation f .

Example: For n = 1, we have

K₁ = 000110 110000 001011 101111 111111 000111 000001 110010
L₁ = R₀ = 1111 0000 1010 1010 1111 0000 1010 1010
R₁ = L₀ + f(R₀,K₁)

It remains to explain how the function f works. Tocalculate f, we first expand each block R_n-1 from 32 bits to48 bits. This is done by using a selection table thatrepeats some of the bits in R_n-1 . We'll call the use ofthis selection table the function E. Thus E(R_n-1) has a 32bit input block, and a 48 bit output block.

Let E be such that the 48 bits of its output, writtenas 8 blocks of 6 bits each, are obtained by selecting thebits in its inputs in order according to the followingtable:

 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 inpositions 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(R₀) from R₀ as follows:

R₀ = 1111 0000 1010 1010 1111 0000 1010 1010
E(R₀) = 011110 100001 010101 010101 011110 100001 010101 010101

(Note that each block of 4 original bits has beenexpanded to a block of 6 output bits.)

Next in the f calculation, we XOR the outputE(R_n-1) with the key K_n:

Example: For K₁ , E(R₀), we have

K₁ = 000110 110000 001011 101111 111111 000111 000001 110010
E(R₀) = 011110 100001 010101 010101 011110 100001 010101 010101
K₁+E(R₀) = 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 48bits, using the selection table, and XORed the result withthe key K_n . We now have 48 bits, or eight groups of sixbits. We now do something strange with each group of sixbits: we use them as addresses in tables called "S boxes".Each group of six bits will give us an address in adifferent S box. Located at that address will be a 4 bitnumber. This 4 bit number will replace the original 6 bits.The net result is that the eight groups of 6 bits aretransformed into eight groups of 4 bits (the 4-bit outputsfrom the S boxes) for 32 bits total.

Write the previous result, which is 48 bits, inthe form:

where each B_i is a group of six bits. We now calculate

where S_i(B_i) referres to the output of the i-th Sbox.

To repeat, each of the functions S1, S2,..., S8, takesa 6-bit block as input and yields a 4-bit block as output.The table to determine S₁ is shown and explained below:

 S1 Column NumberRowNo. 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 S₁ is the function defined in this table and B is a blockof 6 bits, then S₁(B) is determined as follows: The firstand last bits of B represent in base 2 a number in thedecimal range 0 to 3 (or binary 00 to 11). Let that numberbe i. The middle 4 bits of B represent in base 2 a numberin the decimal range 0 to 15 (binary 0000 to 1111). Letthat 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 15and is uniquely represented by a 4 bit block. That block isthe output S₁(B) of S₁ for the input B. For example, forinput block B = 011011 the first bit is "0" and the last bit"1" giving 01 as the row. This is row 1. The middle fourbits are "1101". This is the binary equivalent of decimal13, so the column is column number 13. In row 1, column 13appears 5. This determines the output; 5 is binary 0101, sothat the output is 0101. Hence S₁(011011) = 0101.

The tables defining the functions S₁,...,S₈ arethe 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 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 S3 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 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 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 theoutput of the eight S boxes:

K₁ + E(R₀) = 011000 010001 011110 111010 100001 100110 010100 100111.

S₁(B₁)S₂(B₂)S₃(B₃)S₄(B₄)S₅(B₅)S₆(B₆)S₇(B₇)S₈(B₈) = 0101 1100 1000 0010 1011 0101 1001 0111

The final stage in the calculation of f is to do apermutation P of the S-box output to obtain the final valueof f:

The permutation P is defined in the following table. Pyields a 32-bit output from a 32-bit input by permuting thebits 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:

we get

R₁ = L₀ + f(R₀ , K₁ )

= 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 L₂ = R₁, which is theblock we just calculated, and then we must calculate R₂ =L₁ + f(R₁, K₂), and so on for 16 rounds. At the end of thesixteenth round we have the blocks L₁₆ and R₁₆. We thenreverse the order of the two blocks into the 64-bit block

and apply a final permutation IP^-1 as defined bythe 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 thepreoutput block as its first bit, bit 8 as its second bit,and so on, until bit 25 of the preoutput block is the lastbit of the output.

Example: If we process all 16 blocks using the methoddefined previously, we get, on the 16th round,

L₁₆ = 0100 0011 0100 0010 0011 0010 0011 0100
R₁₆ = 0000 1010 0100 1100 1101 1001 1001 0101

We reverse the order of these two blocks and applythe final permutation to

R₁₆L₁₆ = 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 inwhich the subkeys are applied.

DES Modes of Operation

The DES algorithm turns a 64-bit message block M into a64-bit cipher block C. If each 64-bit block is encryptedindividually, then the mode of encryption is calledElectronic Code Book (ECB) mode. There are two other modesof DES encryption, namely Chain Block Coding (CBC) andCipher Feedback (CFB), which make each cipher blockdependent on all the previous messages blocks through aninitial XOR operation.

Cracking DES

Before DES was adopted as a national standard, duringthe period NBS was soliciting comments on the proposedalgorithm, the creators of public key cryptography, MartinHellman and Whitfield Diffie, registered some objections tothe use of DES as an encryption algorithm. Hellman wrote:"Whit Diffie and I have become concerned that the proposeddata encryption standard, while probably secure againstcommercial assault, may be extremely vulnerable to attack byan intelligence organization" (letter to NBS, October 22,1975).

Diffie and Hellman then outlined a "brute force" attackon DES. (By "brute force" is meant that you try as many ofthe 2^56 possible keys as you have to before decrypting theciphertext into a sensible plaintext message.) Theyproposed a special purpose "parallel computer using onemillion chips to try one million keys each" per second, andestimated the cost of such a machine at $20 million.

Fast forward to 1998. Under the direction of JohnGilmore of the EFF, a team spent $220,000 and built amachine that can go through the entire 56-bit DES key spacein an average of 4.5 days. On July 17, 1998, they announcedthey 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 theDepartment of Justice, denied it was possible for the FBI tocrack DES: "Let me put the technical problem in context:It took 14,000 Pentium computers working for four months todecrypt a single message . . . . We are not just talkingFBI and NSA [needing massive computing power], we aretalking about every police department."

Responded cryptograpy expert Bruce Schneier: " . . .the FBI is either incompetent or lying, or both." Schneierwent on to say: "The only solution here is to pick analgorithm with a longer key; there isn't enough silicon inthe galaxy or enough time before the sun burns out to brute-force triple-DES" (Crypto-Gram, Counterpane Systems, August15, 1998).

Triple-DES

Triple-DES is just DES with two 56-bit keys applied.Given a plaintext message, the first key is used to DES-encrypt the message. The second key is used to DES-decryptthe encrypted message. (Since the second key is not theright key, this decryption just scrambles the data further.)The twice-scrambled message is then encrypted again with thefirst key to yield the final ciphertext. This three-stepprocedure is called triple-DES.

Triple-DES is just DES done three times with two keysused in a particular order. (Triple-DES can also be donewith three separate keys instead of only two. In eithercase the resultant key space is about 2^112.)

"Cryptographic Algorithms for Protection of Computer DataDuring Transmission and Dormant Storage," Federal Register38, No. 93 (May 15, 1973).

Data Encryption Standard, Federal Information ProcessingStandard (FIPS) Publication 46, National Bureau ofStandards, U.S. Department of Commerce, Washington D.C.(January 1977).

Carl H. Meyer and Stephen M. Matyas, Cryptography: A NewDimension in Computer Data Security, John Wiley & Sons, NewYork, 1982.

Dorthy Elizabeth Robling Denning, Cryptography and DataSecurity, Addison-Wesley Publishing Company, Reading,Massachusetts, 1982.

D.W. Davies and W.L. Price, Security for Computer Networks:An Introduction to Data Security in Teleprocessing andElectronics Funds Transfer, Second Edition, John Wiley &Sons, New York, 1984, 1989.

Miles E. Smid and Dennis K. Branstad, "The Data EncryptionStandard: Past and Future," in Gustavus J. Simmons, ed.,Contemporary Cryptography: The Science of InformationIntegrity, IEEE Press, 1992.

Douglas R. Stinson, Cryptography: Theory and Practice, CRCPress, Boca Raton, 1995.

Bruce Schneier, Applied Cryptography, Second Edition, JohnWiley & Sons, New York, 1996.

Alfred J. Menezes, Paul C. van Oorschot, and Scott A.Vanstone, Handbook of Applied Cryptography, CRC Press, BocaRaton, 1997.

This article appeared in Laissez FaireCity Times, Vol 2, No. 28.

Homepage: http://orlingrabbe.com/
Laissez Faire City Times: http://zolatimes.com/