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CSC 382: Computer Security

CSC 382: Computer Security

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CSC 382: Computer Security

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  1. CSC 382: Computer Security Classical Cryptography CSC 382: Computer Security

  2. Overview • Modular Arithmetic Review • What is Cryptography? • Transposition Ciphers • Substition Ciphers • Cæsar cipher • Vigènere cipher • Cryptanalysis: frequency analysis • Block Ciphers • DES CSC 382: Computer Security

  3. Modular Arithmetic Congruence • a = b (mod N) iff a = b + kn • Equivalently, a = b (mod N) iff N / (a – b) • ex: 37=27 mod 10 b is the residue of a, modulo N • Ints 0..N-1 are complete set of residues mod N CSC 382: Computer Security

  4. Laws of Modular Arithmetic • (a + b) mod N = (a mod N + b mod N) mod N • (a - b) mod N = (a mod N - b mod N) mod N • ab mod N = (a mod N)(b mod N) mod N • a(b+c) mod N = ((ab mod N)+(ac mod N)) mod N CSC 382: Computer Security

  5. What is Cryptography? Cryptography: The art and science of keeping messages secure. Cryptanalysis: the art and science of decrypting messages. Cryptology: cryptography + cryptanalysis CSC 382: Computer Security

  6. Plaintext Encryption Procedure Ciphertext Terminology • Plaintext: message to be encrypted. Also called cleartext. • Encryption: altering a message to keep its contents secret. • Ciphertext: encrypted message. CSC 382: Computer Security

  7. History of Cryptography Egyptian hieroglyphics ~ 2000 B.C.E. • Cryptic tomb enscriptions for regality. Spartan skytale cipher ~ 500 B.C.E. • Wrapped thin sheet of papyrus around staff. • Messages written down length of staff. • Decrypted by wrapped around = diameter staff. Cæsar cipher ~ 50 B.C.E. • Simple alphabetic substitution cipher. al-Kindi ~ 850 C.E. • Cryptanalysis using letter frequencies. CSC 382: Computer Security

  8. History of Cryptography Alberti’s polyalphabetic cipher 1467 Decryption of Zimmerman telegram 1917 • Leads US into World War I Japanese Purple Machine cracked 1937 • US breaks rotor machine for highest secrets. German Enigma machine cracked 1933-45 • Initially broken by Polish mathematician Rejewski • Variants broken at Bletchley Park in UK • Colossus, world’s 1st electronic computer. CSC 382: Computer Security

  9. Cryptosystem Formal Definition 5-tuple (E, D, M, K, C) • M set of plaintexts • K set of keys • C set of ciphertexts • E set of encryption functions e: M KC • D set of decryption functions d: C KM CSC 382: Computer Security

  10. Example: Cæsar cipher Letter shifting cipher (A=>D, B=>E, C=>F, …) 5-tuple • M = { all sequences of letters } • K = { i | i is an integer and 0 ≤ i ≤ 25 } • E = { Ek | kK and for all letters m, Ek(m) = (m + k) mod 26 } • D = { Dk | kK and for all letters c, Dk(c) = (26 + c – k) mod 26 } • C = M History: Cæsar’s key was 3. CSC 382: Computer Security

  11. Example: Cæsar cipher • Plaintext is HELLO WORLD • Change each letter to the third letter following it (X goes to A, Y to B, Z to C) • Key is 3, usually written as letter ‘D’ • Ciphertext is KHOOR ZRUOG CSC 382: Computer Security

  12. A Transposition Cipher Rearrange letters in plaintext. Example: Rail-Fence Cipher • Plaintext is HELLO WORLD • Rearrange as H L O O L E L W R D • Ciphertext is HLOOL ELWRD CSC 382: Computer Security

  13. Cryptosystem Security Dependencies • Quality of shared encryption algorithm E • Secrecy of key K CSC 382: Computer Security

  14. Cryptanalysis Goals • Decrypt a given message. • Recover encryption key. Adversarial models vary based on • Type of information available to adversary • Interaction with cryptosystem. CSC 382: Computer Security

  15. Cryptanalysis Adversarial Models • ciphertext only: adversary has only ciphertext; goal is to find plaintext, possibly key. • known plaintext: adversary has ciphertext, corresponding plaintext; goal is to find key. • chosen plaintext: adversary may supply plaintexts and obtain corresponding ciphertext; goal is to find key. CSC 382: Computer Security

  16. Classical Cryptography Sender & receiver share common key • Keys may be the same, or trivial to derive from one another. • Sometimes called symmetric cryptography. CSC 382: Computer Security

  17. Substitution Ciphers Substitute plaintext chars for ciphered chars. • Simple: Always use same substitution function. • Polyalphabetic: Use different substitution functions based on position in message. CSC 382: Computer Security

  18. Cryptanalysis of Cæsar Cipher Exhaustive search • If the key space is small enough, try all possible keys until you find the right one. • Cæsar cipher has 26 possible keys. CSC 382: Computer Security

  19. General Simple Substitution Cipher Key Space: All permutations of alphabet. Encryption: • Replace each plaintext letter x with K(x) Decryption: • Replace each ciphertext letter y with K-1(y) Example: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z K= F U B A R D H G J I L K N M P O S Q Z W X Y V T C E CRYPTO BQCOWP CSC 382: Computer Security

  20. General Substitution Cryptanalysis Exhaustive search impossible • Key space size is 26! =~ 4 x 1026 • Historically thought to be unbreakable. • Yet people solve them as newspaper puzzles every day… Solution: frequency analysis. Lesson: A large key space is necessary but not sufficient for security of a cryptosystem. CSC 382: Computer Security

  21. Cryptanalysis: Frequency Analysis Languages have different frequencies of • letters • digrams (groups of 2 letters) • trigrams (groups of 3 letters) • etc. Simple substitution ciphers preserve frequency distributions. CSC 382: Computer Security

  22. English Letter Frequencies CSC 382: Computer Security

  23. Additional Frequency Features • Digram frequencies • Common digraphs: EN, RE, ER, NT, TH • Trigram frequencies • Common trigrams: THE, ING, THA, ENT • Vowels other than E rarely followed by another vowel. • The letter Q is followed only by U. • Many others. CSC 382: Computer Security

  24. Countering Frequency Analysis Nulls • Insert additional symbols (numbers) which have no meaning in random places. Idiosyncratic spellings • Hacker speak: www.google.com/intl/xx-hacker Homophonic substitution • Each letter has multiple substitutions. These techniques increase difficulty of frequency analysis but don’t make it impossible. CSC 382: Computer Security

  25. Countering Frequency Analysis Primary weakness of simple substition: • Each ciphertext letter corresponds to only one letter of plaintext. Solution: polyalphabetic substitution • Use multiple cipher alphabets. • Switch between cipher alphabets from character to character in the plaintext. CSC 382: Computer Security

  26. Letter Frequency Distributions CSC 382: Computer Security

  27. Vigènere Cipher Use phrase instead of letter as key. Example: • Message THE BOY HAS THE BALL • Key VIG • Encipher using Cæsar cipher for each letter: key VIGVIGVIGVIGVIGV plain THEBOYHASTHEBALL cipher OPKWWECIYOPKWIRG Key space size is 26m. CSC 382: Computer Security

  28. G I V A G I V B H J W E L M Z H N P C L R T G O U W J S Y A N T Z B O Y E H T Tableau shown has relevant rows, columns only. Example encipherments: key V, letter T: follow V column down to T row (giving “O”) Key I, letter H: follow I column down to H row (giving “P”) Relevant Parts of Tableau CSC 382: Computer Security

  29. Useful Terms period: length of key • In earlier example, period is 3 tableau: table used to encipher and decipher • Vigènere cipher has key letters on top, plaintext letters on the left. CSC 382: Computer Security

  30. Simple Attacks • Chosen Plaintext • Choose plaintext of all a’s. • If long enough, it will be encrypted to the key. • Dictionary Attack • Guess key from dictionary and try decryption. • Brute Force • Try every possible key in turn. • Is there a ciphertext only attack that’s faster? CSC 382: Computer Security

  31. Vigènere Cryptanalysis • Find key length (period). • Break message into n parts, each part being enciphered using the same key letter. • Use frequency analysis to solve resulting simple substition ciphers. key VIGVIGVIGVIGVIGV plain THEBOYHASTHEBALL cipher OPKWWECIYOPKWIRG CSC 382: Computer Security

  32. Kaskski Test • Conjunction of key repetition with repeated portion of plaintext produces repeated ciphertext. • Example: key VIGVIGVIGVIGVIGV plain THEBOYHASTHEBALL cipher OPKWWECIYOPKWIRG Key and plaintext line up over the repetitions. • Distance between reptitions is 9 • Repeated phrase “OPK” at 1st and 10th positions. • Period is a multiple of 9 (1, 3 or 9.) CSC 382: Computer Security

  33. Example Vigènere Ciphertext ADQYS MIUSB OXKKT MIBHK IZOOO EQOOG IFBAG KAUMF VVTAA CIDTW MOCIO EQOOG BMBFV ZGGWP CIEKQ HSNEW VECNE DLAAV RWKXS VNSVP HCEUT QOIOF MEGJS WTPCH AJMOC HIUIX CSC 382: Computer Security

  34. Repetitions in Example CSC 382: Computer Security

  35. Estimate of Period • OEQOOG is probably not a coincidence • Two character repetitions may be chance. • Period may be 1, 2, 3, 5, 6, 10, 15, or 30 • Most others (7/10) have 2 in their factors • Almost as many (6/10) have 3 in their factors. • Begin with period of 2  3 = 6. CSC 382: Computer Security

  36. Letter Coincidence • Coincidence: Picking two letters at random from a message that are identical. • Probability of picking two a’s • Let there be n letters in the ciphertext. • Let there be na a’s in the ciphertext. • The probability of selecting two a’s at random CSC 382: Computer Security

  37. Index of Coincidence Probability of chosing two identical letters Coincidence probabilities for two letters: • English plaintext: 0.0667 • Random English letters: 1/26 @ 0.0385 CSC 382: Computer Security

  38. English Letter Frequencies CSC 382: Computer Security

  39. Coincidence Counting Plaintext Plaintext/Ciphertext Simple Language: f(A)=0.75, f(B)=0.25 Simple Cipher: Swap A’s and B’s CSC 382: Computer Security

  40. Index of Coincidence Shorter Key Longer Key 0.0667 Friedman Test Expected IC • Random: 0.0385 • Plaintext: 0.0667 Expected IC by period • 2: 0.052 • 3: 0.047 • 4: 0.045 • 5: 0.044 • 10: 0.041 0.0385 CSC 382: Computer Security

  41. Compute I.C. for Example For our ciphertext, IC = 0.043 • Indicates a key of slightly more than 5. • A statistical measure, so it can be in error, but it agrees with the previous estimate (6). If the key has m characters, then every mth character is enciphered with the same shift. • The string of letters won’t be recognizable. • But its letter frequencies should be the same as English as it’s a monoalphabetic ciphertext. CSC 382: Computer Security

  42. Splitting Into Alphabets Alphabet IC AIKHOIATTOBGEEERNEOSAI 0.069 DUKKEFUAWEMGKWDWSUFWJU 0.078 QSTIQBMAMQBWQVLKVTMTMI 0.078 YBMZOAFCOOFPHEAXPQEPOX 0.056 SOIOOGVICOVCSVASHOGCC 0.124 MXBOGKVDIGZINNVVCIJHH 0.043 Divide cipher into 6 (period) alphabets. IC indicates single alphabet, except #4 and #6. CSC 382: Computer Security

  43. Frequency Examination ABCDEFGHIJKLMNOPQRSTUVWXYZ 1 31004011301001300112000000 2 10022210013010000010404000 3 12000000201140004013021000 4 21102201000010431000000211 5 10500021200000500030020000 • 01110022311012100000030101 HMMMHMMHHMMMMHHMLHHHMLLLLL Unshifted frequencies (H high, M medium, L low) CSC 382: Computer Security

  44. Begin Decryption • First matches characteristics of unshifted alphabet • Third matches if I shifted to A • Sixth matches if V shifted to A • Substitute into ciphertext (bold are substitutions) ADIYS RIUKB OCKKL MIGHK AZOTO EIOOL IFTAG PAUEF VATAS CIITW EOCNO EIOOL BMTFV EGGOP CNEKI HSSEW NECSE DDAAA RWCXS ANSNP HHEUL QONOF EEGOS WLPCM AJEOC MIUAX CSC 382: Computer Security

  45. Look For Clues AJE in last line suggests “are”, meaning second alphabet maps A into S: ALIYS RICKB OCKSL MIGHS AZOTO MIOOL INTAG PACEF VATIS CIITE EOCNO MIOOL BUTFV EGOOP CNESI HSSEE NECSE LDAAA RECXS ANANP HHECL QONON EEGOS ELPCM AREOC MICAX CSC 382: Computer Security

  46. Next Alphabet MICAX in last line suggests “mical” (a common ending for an adjective), meaning fourth alphabet maps O into A: ALIMS RICKP OCKSL AIGHS ANOTO MICOL INTOG PACET VATIS QIITE ECCNO MICOL BUTTV EGOOD CNESI VSSEE NSCSE LDOAA RECLS ANAND HHECL EONON ESGOS ELDCM ARECC MICAL CSC 382: Computer Security

  47. Got It! QI means that U maps into I, as Q is always followed by U: ALIME RICKP ACKSL AUGHS ANATO MICAL INTOS PACET HATIS QUITE ECONO MICAL BUTTH EGOOD ONESI VESEE NSOSE LDOMA RECLE ANAND THECL EANON ESSOS ELDOM ARECO MICAL CSC 382: Computer Security

  48. Countering Frequency Analaysis • Observation: If Vigènere key is very long, frequency analysis won’t work. • Problem: Long keys are hard to remember. • Solution: Use multiple encryptions. • Encrypting with a key m and key n is same as encryption by key whose length is least common multiple of m and n. • If m and n are relatively prime, then the least common multiple is mn. CSC 382: Computer Security

  49. Rotor Machines Use multiple rounds of Vigènere substitution. • Machine contains multiple cylinders. • Each cylinder has 26 states (ciphers). • Cylinders rotate to change states on different schedules. • m-cylinder machine has 26m substitution ciphers. CSC 382: Computer Security

  50. Enigma Machine • 3 rotors: 17576 substitutions. • 3 rotors can be used in any order: 6 combinations. • Plug board: 6 pairs of letters can be swapped. • Total keys ~ 1016 CSC 382: Computer Security