Private-Key CryptographyPublic-Key CryptographySlide 3Slide 4Why Public-Key Cryptography?Public-Key CharacteristicsSlide 7Slide 8Public-Key CryptosystemsPublic-Key ApplicationsSecurity of Public Key SchemesRSA (Rivest, Shamir, Adleman MIT 1977 )RSA UseWhy RSA WorksRSA ExampleRSA Example contRSA Key GenerationRSA SecurityFactoring ProblemLong-range factoring prediction (from Applied Cryptography by B. Schneier)RSA recommended Key sizes (from Applied Cryptography by B. Schneier)Private-Key Cryptography•traditional private/secret/single key cryptography uses one key •shared by both sender and receiver •if this key is disclosed communications are compromised •also is symmetric, parties are equal •hence does not protect sender from receiver forging a message & claiming is sent by senderPublic-Key Cryptography•the most significant advance in the 3000 year history of cryptography •uses two keys – a public & a private key•asymmetric since parties are not equal •uses clever application of number theoretic concepts to function•complements rather than replaces private key cryptoPublic-Key Cryptography•public-key/two-key/asymmetric cryptography involves the use of two keys: –a public-key, which may be known by anybody, and can be used to encrypt messages, and verify signatures –a private-key, known only to the recipient, used to decrypt messages, and sign (create) signatures•is asymmetric because–those who encrypt messages or verify signatures cannot decrypt messages or create signaturesPublic-Key CryptographyWhy Public-Key Cryptography?•developed to address two key issues:–key distribution – how to have secure communications in general without having to trust a KDC with your key–digital signatures – how to verify a message comes intact from the claimed sender•public invention due to Whitfield Diffie & Martin Hellman at Stanford Uni in 1976–known earlier in classified communityPublic-Key Characteristics•Public-Key algorithms rely on two keys with the characteristics that it is:–computationally infeasible to find decryption key knowing only algorithm & encryption key–computationally easy to en/decrypt messages when the relevant (en/decrypt) key is known–either of the two related keys can be used for encryption, with the other used for decryption (in some schemes)Encryption using aPublic-Key System7Authentication using aPublic-Key System8Public-Key CryptosystemsPublic-Key Applications•can classify uses into 3 categories:–encryption/decryption (provide secrecy)–digital signatures (provide authentication)–key exchange (of session keys)•some algorithms are suitable for all uses, others are specific to oneSecurity of Public Key Schemes•like private key schemes brute force exhaustive search attack is always theoretically possible •but keys used are too large (>512bits) •The public-key algorithms are based on a known hard problem. The its just made too hard to do in practise •RSA Problem: Given n=pq, with p and q primes. Find p and q.•requires the use of very large numbers•hence is slow compared to private key schemesRSA (Rivest, Shamir, Adleman MIT 1977 )Each user generates a public/private key pair by: 1. selecting two large primes at random : p, q 2. computing their system modulus N=pq note ø(N)=(p-1)(q-1) 3. selecting at random the encryption key e where 1<e<ø(N), gcd(e,ø(N))=1 4. solve following equation to find decryption key d ed=1 mod ø(N) and 0≤d≤N 5. publish their public encryption key: KU={e,N} 6. keep secret private decryption key: KR={d,p,q}RSA Use•to encrypt a message M the sender:–obtains public key of recipient KU={e,N} –computes: C=Me mod N, where 0≤M<N•to decrypt the ciphertext C the owner:– uses their private key KR={d,p,q} –computes: M=Cd mod N •note that the message M must be smaller than the modulus N (block if needed)Why RSA Works•because of Euler's Theorem:•aø(n)mod N = 1 –where gcd(a,N)=1•in RSA have:–N=p.q–ø(N)=(p-1)(q-1) –carefully chosen e & d to be inverses mod ø(N) –hence e.d=1+k.ø(N) for some k•hence :Cd = (Me)d = M1+k.ø(N) = M1.(Mø(N))q = M1.(1)q = M1 = M mod NRSA Example1. Select primes: p=17, q=112. Compute n = pq =17×11=1873. Compute ø(n)=(p–1)(q-1)=16×10=1604. Select e : gcd(e,160)=1; choose e=75. Determine d: de=1 mod 160 and d < 160 Value is d=23 since 23×7=161= 10×160+16. Publish public key KU={7,187}7. Keep secret private key KR={23,17,11}RSA Example cont•sample RSA encryption/decryption is: •given message M = 88 (nb. 88<187)•encryption:C = 887 mod 187 = 11 •decryption:M = 1123 mod 187 = 88RSA Key Generation•users of RSA must:–determine two primes at random p, q –select either e or d and compute the other•primes p,q must not be easily derived from modulus N=pq–means must be sufficiently large–typically guess and use probabilistic testRSA Security•three approaches to attacking RSA:–brute force key search (infeasible given size of numbers)–mathematical attacks (based on difficulty of computing ø(N), by factoring modulus N)–timing attacks (on running of decryption)Factoring Problem•mathematical approach takes 3 forms:–factor N=pq, hence find ø(N) and then d–determine ø(N) directly and find d–find d directly•currently believe all equivalent to factoringLong-range factoring prediction (from Applied Cryptography by B. Schneier)Year Key length in bits1995 10242005 20482015 40962025 81922035 16,3842045 32,768RSA recommended Key sizes(from Applied Cryptography by B. Schneier)Year Individual Corporation Government1995 768 1280 15362000 1024 1280 15362005 1280 1536 20482010 1280 1536 20482015 1536 2048