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exercise in the previous class

exercise in the previous class. give proof for the discussion in p.19. see http ://apal.naist.jp/~kaji/lecture /. chapter 4: cryptography. what we do, and what we do not in this class. cryptography is discusses in many contexts management politics history philosophy

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exercise in the previous class

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  1. exercise in the previous class • give proof for the discussion in p.19 see http://apal.naist.jp/~kaji/lecture/

  2. chapter 4:cryptography

  3. what we do, and what we do not in this class cryptography is discusses in many contexts • management • politics • history • philosophy • In this class, we focus on the technical aspects of cryptography.

  4. terminology encryption (暗号化) E p E(p) D D(c) c decryption (復号) plaintexts(平文,ひらぶん); make sense by themselves ciphertexts(暗号文); make no sense by themselves • cryptography (暗号) = pair of E and D such that D(E(p)) = p • many variations and confusions on the words: crypto  cipher, text  data, cryptography  encryption

  5. three types of cryptography • key-less cryptography • E(p) (resp. D(c)) is solely determined by p (resp. c). • no key ... the algorithms must be kept secret • security relies on the “gap of wisdom” of the recipients • “O, draconian devil”  “Leonardo da Vinci” • common-key cryptography • E and D must use the same key • public-key cryptography • E and D use different keys which are in special relation

  6. class plan • today: common-key cryptography • widely known algorithms • key agreement protocol • next: public-key cryptography • RSA • related algorithms June 4 (MON): exercise June 5 (TUE): test

  7. common-key cryptography symmetric-key―, classic ―, ... • E (resp. D) takes two inputs: key and plaintext (resp. ciphertext) • E(k, p): the ciphertext of p encrypted with the key k • D(k, c): the plaintext of c decrypted with the key k • D(k, E(k, p)) = p, but D(k’, E(k, p)) p if k’k k1 k2 p, if k1 = k2 p E c D ?, if k1k2

  8. substitution cipher substitution cipher (換字暗号): • encrypt: replace characters in plaintexts to different characters • decrypt: do the inverse replacement of encoding • key: the table of the character replacement ... plaintext A B C Y Z      ciphertext ... E K A Z G • the number of possible keys = 26! for English alphabet ... too many even for today’s computers • the statistics of the plaintexts can be observed in cipherexts

  9. frequency attack in a naive substitution cipher... • a character is always replaced to the identical character • in many data, there is bias on the frequencies of characters in English... • characters such as “e”, “t”, “a”, and “s” occur frequently • characters which occur frequently in a ciphertext = replacements of the above four frequent characters A.C. Doyle, 1903, The Adventure of the Dancing Men

  10. h x a c a b c d 8.4% 1.5% 2.7% 3.8% 8.6% 1.4% 2.8% 3.8% plaintext theory in modern english is a concept which originally derives from classical greek sketch of the frequency attack typical English texts information as a concept has many meanings the concept of information is ciphertext of unknowntext zpunim gt oncuit utqvgwp gw h antaubz spgap nigqgthvvm cuigluw eino → a → b → c → d

  11. many improvements The vulnerability (脆弱性) of the substitution cipher was well-known to cryptographers from early days... many improvements were considered... • one-to-many substitution • substitution of N-grams or words • use of multiple substitution tables • dynamically change the substitution table  Enigma

  12. Enigma • used by German military in the World War II • the substitution is determined by “rotor wheels” • the rotor wheels rotate as one character is processed A B D Enigma showed that machine power >> human power C

  13. DES (Data Encryption Standard) DES (Data Encryption Standard) • developed in the US in 70’s to secure classified data • not the “first-class” cryptography • “good security with reasonable cost” • insecure nowadays, but played important role in cryptology 1973 NBS solicited (公募する) encryption algorithms 1974 IBM submitted a candidate 1977 published as federal standard 1997 NIST (formerly NBS) solicited newer AES

  14. encryption of DES 56...# of bits 56 56 RK1 RK2 RK16 key round keys 48 48 48 32 R1 R2 R15 R16 R0 plaintext f f f IP IP IP-1 ciphertext 64 64 initial permutation L1 L2 L15 L16 L0 32 round 1 round 2 round 16

  15. Feistel structure • each round of DES has the Fesitel structure Li Ri RKi+1 f Li+1 Ri+1 • the Fesitel structure is easy to invert if RKi+1 is provided correctly • the inversion can be done with the same Feistel mechanism (with left and right exchanged) Ri+1 Li+1 RKi+1 f Ri Li

  16. decryption of DES RK16 RK15 RK1 key R1 R2 R15 R16 R0 ciphertext plaintext f f f IP IP IP-1 L1 L2 L15 L16 L0 inside this box is the same as the encryption  one circuit is used for both of encryption and decryption

  17. security of DES • theoretical attacks • differential analysis by Biham & Shamir (1990) • investigated at the design phase of DES... • linear analysis by Matsui (1993) • succeeded to break DES first time • exhaustive attacks • 22hours, 100K computers connected by network (1999) • 9days, FPGA-based parallel machine (2006) DES is not secure anymore!

  18. rumor of DES rumor, or urban legend: “NSA must settle a back-door in DES” NSA: National Security Agency • intelligence agency of the US • some activities not revealed • commitment to the Echelon system evidence? • the key length is shortened from the IBM proposal • some substitution tables in DES is replaced by NSA • NSA did know the differential analysis there is no way to verify what is true and what is not true...

  19. AES and others • DES is no more secure • there is no way to deny the bad rumor  the newer and stronger cryptography is needed 1997 NIST solicited Advanced Encryption Standard (AES) 15 candidate algorithms from 12 countries 1999 5 candidates passed the screening 2000 Rijndael, from Belgium, was selected as winner 2001 published as federal standard There are many other algorithms: Blowfish, IDEA, Camellia...

  20. key agreement Any common-key cryptography faces to one serious problem: How can we share a key with a person at remote place? • the sender and the receiver must have the same key • the key must not be known to anyone else solution... • use an expensive but secure communication channel • secret agent, registered mail, pigeon, etc... • utilize mathematical trick key agreement protocol

  21. ? key agreement protocol We consider a protocol between two users A and B: • the communication channel is not secure • an attacker C can wiretap (盗聴する) the communication, but does not modify data in the channel • after the protocol execution... • A and B know a certain information in common • C does not know the information

  22. Diffie-Hellman protocol Diffie-Hellman protocol; • is proposed by Diffie & Hellman in 1976 • makes use of the property that it is difficult to solve the discrete logarithm problem preliminary • Fq = {0, ..., q – 1} with q a big prime number • g, a generator of Fq (any nonzero aFq is written as a = gx mod q) • discrete logarithm problem (DLP): “given q, g and a, determine x with a = gx mod q”

  23. 6 5 4 3 2 1 0 1 2 3 4 5 6 example • F7 = {0, 1, 2, ..., 6} • g = 3 is a generator of F7 the answer of the DLP x 1 = 36 mod 7 2 = 32 mod 7 3 = 31 mod 7 4 = 34 mod 7 5 = 35 mod 7 6 = 33 mod 7 log3 1 = 6 log3 2 = 2 log3 3 = 1 log3 4 = 4 log3 5 = 5 log3 6 = 3 a no smart algorithm known today ... the only means to solve the problem is by exhaustive search ... nobody can solve the problem if q is large (> thousands bits)

  24. the protocol step 1: A and B agree the prime q and the generator g (in public) step 2a: A chooses random x, and sends mA= gx mod q to B step 2b: B chooses random y, and sends mB= gymod q to A step 3a: A computes(mB)xmod q =gxymod q step 3b: A computes(mA)y mod q = gxymod q determine q & g mA = gx mod q x mB = gy mod q y gxy mod q gxy mod q

  25. example How can we compute 3851 mod 197? • 3851 mod 197 = (3832 mod 197) (3816mod 197) (382mod 197) (381mod 197) mod 197 • 382nmod 197 = (38nmod 197)2mod 197 q = 197, g = 3 71 = 351mod 197 51 38 = 355mod 197 55 122 = 3851 mod 197 122 = 7155mod 197 381 382 384 388 3816 3832 mod 197

  26. security Is the protocol secure? determine q & g mA = gx mod q x • C finds q, g, mA and mB • C cannot know x and yunless he/she solves DLP • C cannot know the value of the shared gxy mod q mB = gy mod q y gxy mod q gxy mod q

  27. another security What happens if the attacker do more than wiretapping? • C communicates with A pretending B • C communicates with B pretending A A and B communicate with C, believing that he/she is communicating with a valid opponent.  man-in-the-middle attack(中間一致攻撃)

  28. summary • classification of cryptography • key-less, common-key and public-key • common-key cryptography • substitution cipher • DES • key-agreement protocol

  29. exercise Decrypt the following ciphertext. qiwaufmlyngcmwzyz c mcxaeyoqweocqyaocuwpwoqjwcqkeyogzkmmwe cod vyoqwezlaeqz, yoviyniqiakzcodzajcqiuwqwzlceqynylcqwyo c pceywqfajnamlwqyqyaoz. qiwaufmlyngcmwzicpwnamwqahwewgcedwdczqiwvaeud'zjaewmazqzlaeqznamlwqyqyaoviwewmaewqicoqvaikodewdocqyaozlceqynylcqw. qiwgcmwzcewnkeewoqufiwudwpwefqvafwcez, vyqizkmmwe cod vyoqweaufmlyngcmwzcuqweocqyog, cuqiakgiqiwfannkewpwefjakefwcezvyqiyoqiwyeewzlwnqypwzwczaocugcmwz.

  30. about test • June 4(Mon), 9:20AM, exercise • June 5 (Tue), 9:20AM, this room • you can bring books, notes and copies of slides • you can bring a calculator and/or PC • PC must be disconnectedfrom the network: download all needed material before the test starts • 本,ノート,資料,電卓,PC ...なんでも持ちこみ可 • PC 等の通信機能は使用不可 必要な資料類は事前にダウンロードしておくこと

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