M1 M2 M3 M4 Mn Plain Message (M) E E E E E C1 C2 C3 C4 Cn Cipher Message (C)
Symmetric cryptography It uses the same key for encryption and decryption. E.g. DES, 3DES, Blowfish Encrypt using key K Cipher text Sender Decrypt using key K Cipher text Plain text Receiver Send from Sender to Receiver Plain Text
Public key cryptography Public key is distributed whereas private key is kept secret. E.g. RSA, DSA Encrypt using B’s public key Cipher text A Decrypt using B’s private key Cipher text Plain text B Send from Sender to Receiver Plain Text
Hashing Function Hash Function Message digest Message Eg. a word “Linux conference” becomes EFDD2356. Typical Hash functions have an infinite domain, such as byte streams of arbitrary length and a finite range such as bit sequences of some fixed length.
Digital Signatures Hash Function Message digest Encrypt using Sender’s private key Digital Signature Message The message digest which is the hash value is Encrypted and anybody can check the signature using the public key.
The purpose of a Message authentication code is to authenticate a source of a message and and its integrity.
MAC Hash Function Message digest Encrypt using symmetric key MAC Message
NETWORKING PRODUCTS DIVISION, HCL Digital Certificates Country Name: State: Locality: Organizational Name: Common Name: E-mail address: Public key Certificate Sign using private key of Self or trusted Certification Authority(CA).
* Alice creates the cipher text c by exponentiation: c = m^e mod n, where e and n are Bob's public key. She sends c to Bob. To decrypt, Bob also exponentiates: m = c^d mod n; the relationship between e and d ensures that Bob correctly recovers m. Since only Bob knows d, only Bob can decrypt this message.
* Alice creates a digital signature s by exponentiation: s = m^d mod n , where d and n are Alice's private key. She sends m and s to Bob. To verify the signature, Bob exponentiates and checks that the message digest m is recovered: m = s^e mod n , where e and n are Alice's public key.
* The obvious way to do this attack is to factor the public modulus, n, into its two prime factors, p and q. From p, q, and e, the public exponent, the attacker can easily get d, the private exponent.
* The hard part is factoring n; the security of RSA depends on factoring being difficult. In fact, the task of recovering the private key is equivalent to the task of factoring the modulus: you can use d to factor n, as well as use the factorization of n to find d