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Applied Cryptography

Applied Cryptography. Main goal Give some practical experience on cryptographic technics used today. Show how to use existing cryptographic software. Examination: Practical assignments and written exam Homepage containg latest course information: http://www.nada.kth.se/~marten/AC /2003

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Applied Cryptography

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  1. Applied Cryptography • Main goal • Give some practical experience on cryptographic technics used today. • Show how to use existing cryptographic software. • Examination: Practical assignments and written exam • Homepage containg latest course information: http://www.nada.kth.se/~marten/AC/2003 • Check course program (from homepage) for detailed information • Homepage mirror athttp://students.mii.lu.lv/user/AC/2003/ Mårten Trolin

  2. Requirements • Attend lectures (if you want to) • Collect at least 30 points • Two practical assignments give up to 20 points each • Written exam gives up to 20 points. Mårten Trolin

  3. Outline of course program • N.B. Course program is subject to change. Check the home page for the latest information • Lectures 1-5: Basic cryptographic functions and principles • Lectures 6-8: Smartcards in financial transactions • Lectures 9-11: SSL and PGP • Lectures 12: Passwords • Exam on April 29th. Mårten Trolin

  4. Why cryptography • Reason for using cryptography • Protect from eaves-dropping (confidentiality) • Ensure data is not modified (integrity) • Certify identity of sender (authenticity) • Requirements (application dependent) • Simple key management • Low hardware requirements (smart card applications, mobile phones) • Cost of bandwidth Mårten Trolin

  5. Simple example – substitution cipher • The key is a permutation of the letters of the alphabet, i.e. a bijection • Encryption is performed by substituting each letter for its corresponding letter • Decryption is the same as encryption with the difference that the inverse is used Mårten Trolin

  6. Substitution cipher – example • Example: Encrypt MY DOG ATE YOUR CAT using the key Mårten Trolin U

  7. Breaking the substitution cipher • Substitution ciphers are easily broken using frequency analysis • We use the fact that different letters (or combination of letters) occur with different probability • Example – breakTK IL KQ JKT TK IL TBST CR TBL OULRTCKJ • Frequency of letters in English: ETAOINSHRDLU • Most common two letter words: OF TO IN IS IT BE BY HE AS ON AT OR AN SO IF NO Mårten Trolin

  8. Symmetric vs. asymmetric cryptography • Symmetric ciphers – sender and recipient use the same key • Dkey(Ekey(m)) = m • Substitution cipher is an example of a symmetric cipher • Impractical for big systems – number of keys is quadratic in the number of users • The solution – asymmtric algorithms. Think of a locked mailbox! Different keys for encryption and decryption • Dprivate key(Epublic key(m)) = m Mårten Trolin

  9. Asymmetric cryptography • Each user has a public and a private key • The public key is published in a “phone book” • The private key is kept secret • Messages encrypted with the public key can be decrypted with the private key • To send a message to Mårten, look up Mårten’s public key in the “phone book”. • Mårten can then decrypt the message with his private key • Number of keys is linear in the number of users Mårten Trolin

  10. RSA • Asymmetric cryptographic algorithm published in 1978 • The most popular asymmetric algorithm used today • Now free to use – patent expired in 2000 • Relies on the hardness of factoring a number consisting of two primes Mårten Trolin

  11. The RSA algorithm – key generation • Generate two primes p, q and set n = pq • Choose e such that gcd(e,(p–1)(q–1)) = 1 • Compute d such that ed = 1 mod ((p–1)(q–1)) • The public key is the pair (e, n) • The private key is the pair (d, n) Mårten Trolin

  12. RSA – encryption and decryption • Message m – number 0 < m < n • Encryption E(m) = me mod n • Decryption D(m) = md mod n • Number theoretical exercise – check that D(E(m)) = m. Mårten Trolin

  13. Breaking RSA • If we can factor n we can break RSA • Suppose we know p, q such that pq = n • We can compute (p – 1)(q – 1) • It is now trivial to compute d = e-1 mod ((p – 1)(q – 1)) • The largest number that is (publicly) known to have been factored today is 512 bits • Other attacks exist for certain uses of RSA Mårten Trolin

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