Official comment

http://csrc.nist.gov/encryption/aes/aes_home.htm#comments OFFICIAL Comments - Anyone may provide NIST with OFFICIAL public comments on the AES candidate algorithms. NOTE THAT ALL COMMENTS RECEIVED SHALL BE MADE PART OF THE PUBLIC RECORD. A comment may be submitted by sending an e-mail message to AESFirstRound@nist.gov. OFFICIAL Comment http://www.aci.net/kalliste/bw1.htm to appear at http://zolatimes.com/ Title: Black and White Test of Cryptographic Algorithms Black and White Test of Cryptographic Algorithms by William H. Payne The purpose of this article is to explain the underlying principles of cryptography by examples, and to show some criteria that should be met by cryptographic algorithms before they are seriously considered for adoption. Cryptography is the art or science of scrambling plaintext into ciphertext with a key so that it cannot be read by anyone who does not possess the key. Digital information is stored in the form of 1s and 0s, called BINARY. Binary Numbers Let’s count from DECIMAL 0 to 15 in BINARY by adding 1 to the previous number. decimal: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 binary: 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 Notice that first we start with a single number position, which can be 0 or 1 in BINARY. This number position is a bit. Call this first bit b0. Notice that b0 is either 0 or 1. That is, b0 = 0 or b0 = 1. To get to DECIMAL 2, we have to introduce a second BINARY bit--call it b1. We have b1b0 = 10. Next, for DECIMAL 3, we have BINARY b1b0 = 11. binary: 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 numbered bits: b0 b0 b1b0 b1b0 b2b1b0 b2b1b0 b2b1b0 b2b1b0 b3b2b1b0 b3b2b1b0 b3b2b1b0 b3b2b1b0 b3b2b1b0 b3b2b1b0 b3b2b1b0 b3b2b1b0 Notice that the bit subscript represents a power of 2. That is, b0 really means b0*2^0, where * is multiplication, and ^ exponentiation (for example, 2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8). The subscript on b is the same as the power on 2. If we had b26, we would know its meaning was b26*2^26. If b26 = 0, then this value is 0. If b26 = 1, then this value is 2^26. Now look at "1111" (which in BINARY is equal to DECIMAL 15). In this case, b3b2b1b0 = 1111. The right-most BIT (b0) is the least-significant bit, because it corresponds to the lowest power of 2. Converting Binary Numbers to Decimal Numbers To convert a BINARY number ...b3b2b1b0 to a DECIMAL number Y, we simply write Y = b0 + b1 * 2 + b2 * 2^2 + b3 * 2^3 + ... The bits b0, b1, b2, b3 are limited to the values 0 and 1 ONLY. Performing the exponentiation of powers of 2 and reversing the bits gives Y = . . . + b3 * 8 + b2 * 4 + b1 * 2 + b0 . Most of us were brought-up to understand that the most significant digits are to the LEFT of the previous digit. For sake of economy of writing and easy conversion, binary numbers are frequently represented in base 16, or HEXADECIMAL, abbreviated HEX. Hexadecimal Numbers BinaryweightsHEXDECIMAL 8 4 2 1 0 0 0 1 1 1 1 0 2 2 1 1 3 3 1 0 0 4 4 1 0 1 5 5 1 1 0 6 6 1 1 1 7 7 1 0 0 0 8 8 1 0 0 1 9 9 1 0 1 0 A 10 1 0 1 1 B 11 1 1 0 0 C 12 1 1 0 1 D 13 1 1 1 0 E 14 1 1 1 1 F 15 Conversion from binary to hexadecimal or hexadecimal to binary is easy if you remember 1010 is A 1100 is C 1110 is E B is one greater than A, 1011. D is one greater than C, 1101. And F is one greater than E, 1111. Computer Memory Computer memory is frequently organized as BYTEs which are eight bits. Since one hexadecimal digit represents 4 bits, it takes two hexadecimal digits to represent one byte. There are 2^8 = 256 different binary values that can be represented in a byte. These 256 values (written in HEX for brevity) are: 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 20 21 22 23 24 25 26 27 28 29 2A 2B 2C 2D 2E 2F 30 31 32 33 34 35 36 37 38 39 3A 3B 3C 3D 3E 3F 40 41 42 43 44 45 46 47 48 49 4A 4B 4C 4D 4E 4F 50 51 52 53 54 55 56 57 58 59 5A 5B 5C 5D 5E 5F 60 61 62 63 64 65 66 67 68 69 6A 6B 6C 6D 6E 6F 70 71 72 73 74 75 76 77 78 79 7A 7B 7C 7D 7E 7F 80 81 82 83 84 85 86 87 88 89 8A 8B 8C 8D 8E 8F 99 91 92 93 94 95 96 97 98 99 9A 9B 9C 9D 9E 9F A0 A1 A2 A3 A4 A5 A6 A7 A8 A9 AA AB AC AD AE AF B0 B1 B2 B3 B4 B5 B6 B7 B8 B9 BA BB BC BD BE BF C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 CA CB CC CD CE CF D0 D1 D2 D3 D4 D5 D6 D7 D8 D9 DA DB DC DD DE DF E0 E1 E2 E3 E4 E5 E6 E7 E8 E9 EA EB EC ED EE EF FF F1 F2 F3 F4 F5 F6 F7 F8 F9 FA FB FC FD FE FF Representing Language on a Computer The problem of how to represent characters in a computer has been solved several ways. One way is the American Standard Code of Information Interchange (ASCII). ASCII represents characters as 7 bits. Here is table modified from a web site ( http://members.tripod.com/~plangford/index.html). Hex Char Description Hex Char Hex Char Hex Char ---------------------- --------- ---------- ---------- 00 NUL (null) 20 40 @ 60 ` 01 SOH (start of heading) 21 ! 41 A 61 a 02 STX (start of text) 22 " 42 B 62 b 03 ETX (end of text) 23 # 43 C 63 c 04 EOT (end of transmission) 24 $ 44 D 64 d 05 ENQ (enquiry) 25 % 45 E 65 e 06 ACK (acknowledge) 26 & 46 F 66 f 07 BEL (bell) 27 ' 47 G 67 g 08 BS (backspace) 28 ( 48 H 68 h 09 TAB (horizontal tab) 29 ) 49 I 69 i 0A LF (line feed, new line) 2A * 4A J 6A j 0B VT (vertical tab) 2B + 4B K 6B k 0C FF (form feed, new page) 2C , 4C L 6C l 0D CR (carriage return) 2D - 4D M 6D m 0E SO (shift out) 2E . 4E N 6E n 0F SI (shift in) 2F / 4F O 6F o 10 DLE (data link escape) 30 0 50 P 70 p 11 DC1 (device control 1) 31 1 51 Q 71 q 12 DC2 (device control 2) 32 2 52 R 72 r 13 DC3 (device control 3) 33 3 53 S 73 s 14 DC4 (device control 4) 34 4 54 T 74 t 15 NAK (negative acknowledge) 35 5 55 U 75 u 16 SYN (synchronous idle) 36 6 56 V 76 v 17 ETB (end of trans. block) 37 7 57 W 77 w 18 CAN (cancel) 38 8 58 X 78 x 19 EM (end of medium) 39 9 59 Y 79 y 1A SUB (substitute) 3A : 5A Z 7A z 1B ESC (escape) 3B ; 5B [ 7B { 1C FS (file separator) 3C < 5C \ 7C | 1D GS (group separator) 3D = 5D ] 7D } 1E RS (record separator) 3E > 5E ^ 7E ~ 1F US (unit separator) 3F ? 5F _ 7F DEL Now let us take two different one-word messages we might wish to cipher: "black" and "white". We can use the preceding table to find the ASCII codes for the characters in "black" and "white". messagecharacterASCII (Hex)Binary Message 1black62 6C 61 63 6B0110 0010 0110 1100 0110 0001 0110 0011 0110 1011 Message 2white77 68 69 74 650111 0111 0110 1000 0110 1001 0111 0100 0110 0101 But before doing this, we must understand the general "cipher problem." The Cipher Problem We have three elements in the encryption process: 1. The plaintext message 2. The key 3. The ciphertext Let’s start REAL SIMPLE. Let's consider a situation where the plaintext message, the key, and the ciphertext are all the same length. To make it even simpler, let's make each one only one bit long. So the key can be one of two possibilities (0 or 1), and so can the plaintext and the ciphertext. So, in all, there are 2*2*2 = 8 total possible encipherment circumstances. Let’s enumerate ALL 8 POSSIBILITIES. Possibilities Table PossibilityKeyPlaintextCiphertext 1000 2001 3010 4011 5100 6101 7110 8111 That’s it! There are no more possibilities than these 8. What does this mean for the encryption process--the "algorithm"? An ALGORITHM is a deterministic processes that accepts inputs and transforms them into outputs. "Deterministic" is important in that the same inputs ALWAYS produce the same output. Consider ANY algorithm which takes as its inputs the key values of 0 or 1 and the plaintext message values of 0 or 1. ANY algorithm can only produce one of the ciphertext outputs seen above. Image the following hypothetical but REAL SIMPLE algorithm: Hypothetical Algorithm: Find the value of the Key in column 2 of the Possibilities Table and the Plaintext message in column 3; then chose as the Ciphertext output whatever number is found in column 4 of that row. But there, of course, is a catch with a valid CRYPTOGRAPHIC ALGORITHM. Given the Key and the Ciphertext, one must be able to get back the Plaintext! A cryptographic algorithmic should have an INVERSE. So a cryptographic algorithm could not produce ALL of the eight combinations seen above for the reason that it is impossible to invert some of the possibilities. For example, some of the mappings are incompatible from an inverse standpoint, because same values of the key and ciphertext can lead to two different values of the plaintext. Notice how the following pairs of possibilities conflict: 1 and 3; 2 and 4; 5 and 7; 6 and 8. PossibilityKeyPlaintextCiphertext 1000 3010 2001 4011 5100 7110 6101 8111 But there two different cryptographic algorithms that could be made from the Possibilities Table, both of which have inverses: Cryptographic Algorithm 1 PossibilityKeyPlaintextCiphertext 1000 4011 6101 7110 Cryptographic Algorithm 2 PossibilityKeyPlaintextCiphertext 2001 3010 5100 8111 Of course, the output of Algorithm 2 is merely the same as the output of Algorithm 1, with 0s and 1s switched. (This is called a logical NOT operation.) Logic and Its Electronic Representation Logic, sometimes called Boolean logic when it is dealing with 0s and 1s, has several elementary rules. In computers, TRUE is usually represented by a 1. FALSE is represented by a 0. Electrically a 1 is usually, but not always, represented by a HIGH VOLTAGE. A zero by a LOW VOLTAGE. The three basic operations in logic are NOT, AND, and OR: Logical OperationInput(s)Output NOT0 1 1 0 AND000 010 100 111 OR 000 011 101 111 A derivative operation called an EXCLUSIVE-OR, abbreviated XOR, is defined as follows: Logical OperationInput(s)Output XOR000 011 101 110 In XOR, if the two input bits have the the same value, they sum to 0. If they have different values, they sum to 1. Now look back at Cryptographic Algorithm 1. It is, in fact, the exclusive-or (XOR) of the key and plaintext. Algorithm 1: Ciphertext Output = Key XOR Plaintext. Cryptographic Algorithm 2, meanwhile, is just the NOT of Algorithm 1. Algorithm 2: Ciphertext Output = NOT (Key XOR Plaintext). The REALLY IMPORTANT property of the XOR is THAT IT HAS AN INVERSE. By contrast, logical AND does not have an inverse for the reason that if the Key and (Key AND Plaintext) are both 0, then the Plaintext itself is ambiguously either 0 or 1. Logical OperationKey InputPlaintext InputKey AND Plaintext AND000oops 010oops 100 111 Likewise, logical OR does not have an inverse for the reason that if the Key and (Key OR Plaintext) are both 1, then the Plaintext itself is ambiguously either 0 or 1. Logical OperationKey InputPlaintext InputKey OR Plaintext OR 000 011 101oops 111oops So logical AND and OR don’t work well for a crypto algorithm, but the XOR does because it has an inverse. How to Create Two Keys for Deniability The XOR works even better from a legal standpoint. Imagine the following conversation: Ciphercop: We have the ciphertext 0 and we CAUGHT you with the key with a bit value of 1, so you sent a plaintext 1. Citizen: No I did not! You PLANTED the key with bit value 1. The real key bit is a 0, and I sent 0 as the plaintext. Let’s generate a key for the REAL WORLD crypto messages "black" and "white", messagecharacterASCII (Hex)Binary Message 1black62 6C 61 63 6B0110 0010 0110 1100 0110 0001 0110 0011 0110 1011 Message 2white77 68 69 74 650111 0111 0110 1000 0110 1001 0111 0100 0110 0101 and see if we can produce a REAL EXAMPLE of a SECOND KEY. Here’s a key, which we will call key 1: key 1: 1010 0101 1100 0011 1110 0111 1111 0000 0110 1001 Key 1 doesn’t look too random. Each group of four bits is followed by its logical NOT (e.g. NOT(1010) = 0101, etc.). Which leads to another lesson. To claim that a sequence of 0s and 1s is random requires statistical testing. Otherwise, the state of the sequence is UNKNOWN. Here's another key, which we will call key 2: key 2: 1011 0000 1100 0100 1110 1111 1110 0111 0110 0111 These two keys produce the same ciphertext for the two different messages "black" and "white". black0110 0010 0110 1100 0110 0001 0110 0011 0110 1011 key 11010 0101 1100 0011 1110 0111 1111 0000 0110 1001 ciphertext (XOR)1100 0111 1010 1100 1000 0110 1001 0011 0000 0010 white0111 0111 0110 1000 0110 1001 0111 0100 0110 0101 key 21011 0000 1100 0100 1110 1111 1110 0111 0110 0111 ciphertext (XOR)1100 0111 1010 1100 1000 0110 1001 0011 0000 0010 So when the ciphercops FALSELY accuse you of encrypting "black", you SCREAM "Bull pucky!", and produce key 2 to show that you, IN FACT, encrypted "white". Then sue the government--pro se, of course. (See http://jya.com/whpfiles.htm.) The recipe for producing the second key in this example is simple. Take two plaintext messages of the same length. Encrypt one of them with an arbitrary key that yields a ciphertext of the same length. XOR the ciphertext with the second plaintext message. The result is the second key. Store this one for plausible deniability. So from the standpoint of plausible deniability it is BEST to have TWO KEYS for any given encryption: 1. The REAL KEY 2. The key you can CLAIM was the REAL KEY. Just in case you get caught. "Have we gone beyond the bounds of reasonable dishonesty?" --CIA memo (The CIA quote is from Weird History 101, by John Richard Stephens, page 55.) None of us want to get caught going beyond the bounds of reasonable dishonesty. Thus far two criteria of a worth candidate for cryptographic algorithm have been established. Criterion 1: The ciphertext is invertible with the help of a key back into the plaintext. Criterion 2: There is ALWAYS a second key. Just in case you get caught. Plaintext and Ciphertext Sizes The plaintext and ciphertext should be the same size. First, note that if the plaintext is longer than the ciphertext, then the ciphertext is not invertible. For example, let’s suppose that the plaintext is two bits long and the ciphertext is one bit long. PlaintextCiphertext 0 00 or 1 0 11 or 0 1 0oops 1 1oops After the first two ciphertext bits have been assigned to plaintext pairs, the next two plaintext pairs (10,11) must conflict with this assignment. The ciphertext thus correspondents to more than one plaintext possibility. We run into problems for the reason that we cannot establish a one-to-one correspondence between the plaintext and cipher text and, therefore, can’t possibly have an inverse. Second, if the plaintext is shorter than the ciphertext, then the ciphertext can't be trusted. It may include too much information. For example, let’s suppose that the plaintext is one bit long, the key is one bit long, and the cipher text is two bits long. Iran Cryptographic Algorithm KeyPlaintextCiphertext 0 0 00 0 1 10 1 0 11 1 1 01 Not only is the above algorithm invertible, but now the crypto key has been sent along with the ciphertext in the second bit position! That is, the first bit in the ciphertext is is the value of (key XOR plaintext). The second bit is the key itself. So if you XOR the two ciphertext bits with each other, you recover the plaintext bit. You might ask who would be audacious enough to pull such stunt. The Great Satan, of course. For the story of how the National Security Agency (NSA) bugged the encryption equipment that was sold by a Swiss company to 140 nations around the world, see the following links: http://www.aci.net/kalliste/speccoll.htm http://caq.com/cryptogate http://www.qainfo.se/~lb/crypto_ag.htm http://jya.com/whpfiles.htm And the Great Satan got caught. No plan B. Or in crypto parlance, no second key. So we have a third criterion for a cryptographic algorithm we might wish to adopt. Criterion 3: The length of the plaintext equals the length of the ciphertext. In simple terms, if more bits come out of a crypto algorithm than go in, WATCH OUT! Otis Mukinfuss and the Advanced Encryption Standard Bruce Hawkinson (BHAWKIN@sandia.gov) WAS Sandia National Laboratories Lab News editor some years ago. In one editorial, Hawkinson wrote that while we was traveling for Sandia, he spent his motel time looking up strange names in the phone book. One name I recall mentioned was Steely Gray who was a government office furniture salesman. Hawkinson concluded his article by writing his all-time favorite name was Otis Mukinfuss. Hawkinson was no longer editor of Sandia’s Lab News shortly thereafter. J. Orlin Grabbe has done an excellent job writing about cryptographic algorithms in Cryptography and Number Theory for Digital Cash. One inescapable conclusion from Grabbe’s internet article is that from a layman’s standpoint public key cryptography is an incomprehensible mess. A Muckinfuss. The National Institute of Standards and Technology (NIST) is holding a CONTEST to select an Advanced Encryption Standard to replace the current Data Encryption Standard (DES). Click through the candidates to view some additional examples of Mukinfusses. So another criterion has been established for a cryptographic algorithm to be considered for adoption. Criterion 4: The crypto algorithm must be simple and CONVINCE EVERYONE that it is ONLY PERFORMING ITS SIMPLE INTENDED FUNCTION. While we are at the NIST web site, the NIST Advanced Encryption Standard contest reminds me of a the plot of a recent movie, The Game, starring Michael Douglas and Sean Penn: The film is a thriller directed by David Fincher (Se7en). "The Game" is what begins when a high-powered businessman named Nicholas Van Orton (Douglas) receives the birthday gift of a lifetime from his brother he alienated years ago (Penn). What Nicholas gets is entry into a mysterious new form of entertainment provided by C.R.S. (Consumer Recreational Services) simply called "The Game." It proves to be an all-consuming contest with only one rule: there are no rules. By the time Van Orton realizes he is in, it is too late to get out. ... (See http://www.movietunes.com/soundtracks/1997/game/.) NIST does not appear to publish any criteria for winning the AES contest! Look at http://www.nist.gov/public_affairs/confpage/980820.htm and decide for yourself. Perfect Cryptography Here we have described a process of encrypting a plaintext by XORing it with a key of the same length. This encryption technique is called a "one-time pad", or Vernam cipher. Just as long as each key is only used once, the encryption technique is perfectly secure. The one-time pad described here satisfies all criteria mentioned so far: 1. The ciphertext is invertible with the help of a key back into the plaintext. 2. There is ALWAYS a second key. Just in case you get caught. 3. The length of the plaintext equals the length of the ciphertext. 4. The crypto algorithm must be simple and CONVINCE EVERYONE that it is ONLY PERFORMING ITS SIMPLE INTENDED FUNCTION. I add Criterion 5: The length of the key must equal the length of the plaintext. Extensive mathematics or complication fails Criterion 4. Public key cryptograpy that uses the RSA algorithm MAY fail Criterion 1 if the message divides the product of the two prime numbers, p and q, used in the modulus. Most crypto algorithms are designed so that the key cannot be recovered from a plaintext-ciphertext pair. Therefore, they fail Criterion 2. Criterion 3 is much more difficult to ensure against. Black and White Hats American western movie producers used to aid their audiences in identification of the heroes and villains. The heroes wore white hats. The villains, black hats. US government agencies adopted the term ‘black hatter’ to describe an employee whose job it is to break into THINGS. Or screw them up: http://www.jya.com/whp1.htm A ‘white hatter’ is one who analyzes THINGS to make sure they cannot be broken into. And they can’t be screwed up. But the empirical fact is that the ‘black hatters’ can figure-out methods to transmit the key on a covert channel, tell the ‘white hatters’ they did this. And the ‘white hatters’ can’t find out how they did it. Algorithmic Processes Suppose the key is five bits: 1 0 1 0 1 Suppose the plaintext is six bits: 1 1 1 1 1 1 And the ciphertext is also six bits: 1 0 1 1 0 1 Ask the cryptographer give you a key which changes ONLY the sixth bit of the ciphertext, as in the following: 1 0 1 1 0 0 You like the other 5 bits just fine. If the cryptographer can’t, then you might look for another algorithm to adopt. Conclusion We have five criteria to judge the outcome of the NIST Advanced Encryption Standard contest. If none of the algorithms pass the five tests, we will not be discouraged. We know that Gilbert S. Vernam and Joseph O. Mauborgne solved the crytptography problem in 1918, when they created the one-time pad. (See "What is a One-Time Pad?".) William H. Payne P.O. Box 14838 Albuquerque, New Mexico 87191 505-292-7037 Embedded Controller Forth. Generalized Feedback Shift Registers Here are some links to some of my students: Ted Lewis (Friction-Free Economy) John Sobolewski (HPCERC) -30-