Cryptography and Network security # Lecture 5

Cryptography and Network security # Lecture 5

• 1.
  Lec-5: Cryptography &Network Security Mr. Islahuddin Jalal MS (Cyber Security) – UKM Malaysia Research Title – 3C-CSIRT Model for Afghanistan BAKHTAR UNIVERSITY ‫باخترپوهنتون‬ ‫د‬ Bakhtar University 1
• 2.
  Asymmetric Cryptography Bakhtar University2
• 3.
  Problems in symmetriccryptography • Key distribution • Two communicants already share a key by the use of ( KDC) • Neglected the essence of cryptography (Diffie and Martin Hellman) • Verifying a message comes intact from the claimed sender Bakhtar University 3
• 4.
  Public-Key Characteristics 1. Computationallyinfeasible to find decryption key • Knowing only algorithm and encryption key 2. 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 other used for decryption Bakhtar University 4
• 5.
  In Symmetric Cryptography •If Alice and Bob are physically apart and communicate, they have to agree on a key Meet personally, or Use trusted couriers • Alice needs one secret key for Bob, one for Carol, one for Dave and so on. This means that Storage of so many secret keys is not feasible. Bakhtar University 5
• 6.
  In Asymmetric KeyCryptography • 2 people who never met can communicate securely. Alice can securely communicate with all her friends by storing just a single private key. 2 keys are used • Public: known to everyone (for encryption or signature verification) • Private: known to receiver only (for decryption or signature generation) Bakhtar University 6
• 7.
  Public-Key Cryptography • Y= E(PUb, X ) • X = D(PRb, Y ) • Adversary can access PUb and Y, attempt to recover X or PRb; Impossible to alter the message without access to A’s private key. • Authenticate the source. Ensure data integrity. Authentication and Confidentiality. • Z = E(PUb, E(PRa, X)) • X = D(PUa, E(PRb, Z)) • Overhead: public key algorithm executed four times Bakhtar University 7
• 8.
  Public-Key Applications • Classifyuses into 3 categories • Encryption/Decryption (provide secrecy) • Digital Signatures (provide authentication) • Key Exchange (of session keys) Bakhtar University 8
• 9.
  Example: Party Awants to send a message to party B • When only confidentiality is needed Bakhtar University 9
• 10.
  Example: Party Awants to send a message to party B • When only authentication is needed Bakhtar University 10
• 11.
  Example: Party Awants to send a message to party B • When confidentiality and authentication are needed Bakhtar University 11
• 12.
  Type of Public-keycryptography •RSA Bakhtar University 12
• 13.
  RSA • by Rivest,Shamir & Adleman of MIT in 1977 • best known & widely used public-key scheme • Block cipher scheme: plaintext and ciphertext are integer b/w 0 to n-1 for some n. • uses large integers (eg. 1024 bits) • security due to cost of factoring large numbers 12/1/2017 Bakhtar University 13
• 14.
  RSA Key Setup •each user generates a public/private key pair by: • selecting two large primes at random - p, q • computing their system modulus N=p.q • note ø(N)=(p-1)(q-1) • selecting at random the encryption key e • where 1<e<ø(N), gcd(e,ø(N))=1 • solve following equation to find decryption key d • e.d=1 mod ø(N) and 0≤d≤N • publish their public encryption key: KU={e,N} • keep secret private decryption key: KR={d,p,q} 12/1/2017 Bakhtar University 14
• 15.
  RSA Use • toencrypt 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) 12/1/2017 Bakhtar University 15
• 16.
  RSA Example 1. Selectprimes: p=17 & q=11 2. Compute n = pq =17×11=187 3. Compute ø(n)=(p–1)(q-1)=16×10=160 4. Select e : gcd(e,160)=1; choose e=7 5. Determine d: de=1 mod 160 and d < 160 Value is d=23 since 23×7=161= (1×160)+1 6. Publish public key KU={7,187} 7. Keep secret private key KR={23,17,11} 12/1/2017 Bakhtar University 16
• 17.
  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 = 88 12/1/2017 Bakhtar University 17
• 18.
  RSA Key Generation •Select p, q • Calculate n=p x q • Calculate ǿ(n)= (p-1)(q-1) • Select integer e gcd(ǿ(n), e) =1; 1< e < ǿ(n) • Calculate d d e-1 mod(ǿ(n)) • Public Key PU= {e, n} • Private Key PR={d,n} 12/1/2017 Bakhtar University 18
• 19.
  Encryption by Bobwith Alice’s Public Key • Plaintext M<N • Ciphertext: C=Me mod n 12/1/2017 Bakhtar University 19
• 20.
  Decryption by Alicewith Alice’s private key • Ciphertext: C • Plaintext: M=Cd mod n 12/1/2017 Bakhtar University 20
• 21.
  RSA General Approach BakhtarUniversity 21
• 22.
  RSA Example Bakhtar University22
• 23.
  Thank You For YourPatience Bakhtar University 23