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RSA Encryption: Public-Key Cryptography Basics

This guide explains RSA encryption, a public-key cryptography method developed by Rivest, Shamir, and Adleman. Learn how RSA works, key generation steps, encryption, and decryption process with examples.

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RSA Encryption: Public-Key Cryptography Basics

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  1. 17.5.3 Encryption • A DBMS can use encryption to protect information in certain situations where the normal security mechanisms of the DBMS are not adequate. • For example, an intruder may steal tapes containing some data or tap a communication line. • By storing and transmitting data in an encrypted form, the DBMS ensures that such stolen data is not intelligible to the intruder.

  2. Cont. • The basic idea behind encryption is to apply an encryption algorithm, which may be accessible to the intruder, to the original data and a user-specified or DBA-specified encryption key, which is kept secret. • The output of the algorithm is the encrypted version of the data. • There is also a decryption algorithm, which takes the encrypted data and the encryption key as input and then returns the original data.

  3. Cont. • Without the correct encryption key, the decryption algorithm produces gibberish. • The main weakness of this approach is that authorized users must be told the encryption key, and the mechanism for communicating this information is vulnerable to clever intruders. • Another approach to encryption, called public-key encryption, has become increasingly popular in recent years.

  4. Cont. • The encryption scheme proposed by Rivest, Shamir, and Adleman, called RSA, is a well-known example of public-key encryption. • Each authorized user has a public encryption key, known to everyone, and a private decryption key (used by the decryption algorithm), chosen by the user and known only to him or her.

  5. RSA Encryption • In cryptography, RSA (which stands for Rivest, Shamir and Adleman who first publicly described it) is an algorithm for public-key cryptography. • It is the first algorithm known to be suitable for signing as well as encryption, and was one of the first great advances in public key cryptography. • In this method, one party (a bank customer, for example) uses a public key. Kp. • The other party uses a secret (private) key, Ks. • Both uses a number, N.

  6. Cont. • The RSA algorithm involves three steps: • key generation. • Encryption. • Decryption. • Key generation: • RSA involves a public key and a private key. • The public key can be known to everyone and is used for encrypting messages. • Messages encrypted with the public key can only be decrypted using the private key.

  7. Cont. • The keys for the RSA algorithm are generated in the following way: • Choose two distinct prime numbers p and q. • For security purposes, the integers p and q should be chosen uniformly at random and should be of similar bit-length. • Compute n = pq. • n is used as the modulus for both the public and private keys. • Compute m= (p − 1)(q − 1). • Choose an integer e such that 1 < e < m, and e and m share no divisors other than 1. (i.e. e and m are coprime).

  8. Cont. • In mathematics, two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1. (example 6, 35). • Find d, such that de % m = 1 • Publish e and n as the public key. • Keep d and n as the secret key. • Encryption : C = Pe % n • Decryption: P = Cd % n

  9. Example • Generate two large prime numbers, p and q. • To make the example easy to follow ,small numbers will be used, but this is not secure. • To find random primes, we start at a random number and go up ascending odd numbers until we find a prime. Lets have: p = 7, q = 19 • Let n = pq. • n = 7 * 19  = 133 • Let m = (p - 1)(q - 1) • m = (7 - 1)(19 - 1) = 6 * 18 = 108

  10. Cont. • Choose a small number, e coprime to m • e coprime to m, means that the largest number that can exactly divide both e and m (their greatest common divisor, or GCD) is 1. Euclid's algorithm is used to find the GCD of two numbers. • e = 2 => GCD(e, 108) = 2 (no)e = 3 => GCD(e, 108) = 3 (no)e = 4 => GCD(e, 108) = 4 (no)e = 5 => GCD(e, 108) = 1 (yes!)

  11. Cont. • Find d, such that de % m = 1 • This is equivalent to finding d which satisfies de = 1 + nm where n is any integer. • We can rewrite this as d = (1 + nm) / e. • Now we work through values of n until an integer solution for e is found: • n = 0 => d = 1 / 5 (no)n = 1 => d = 109 / 5 (no)n = 2 => d = 217 / 5 (no)n = 3 => d = 325 / 5            = 65 (yes!)

  12. Cont. • Public Key • n = 133 • e = 5 • Secret Key • n = 133 • d = 65

  13. Cont. • Encryption: • For this example, lets use the message "6". • C = Pe % n  = 65 % 133  = 7776 % 133  = 62

  14. Cont. • Decryption: • P = Cd % n  = 6265 % 133  = 62 * 6264 % 133  = 62 * (622)32 % 133  = 62 * 384432 % 133  = 62 * (3844 % 133)32 % 133  = 62 * 12032 % 133

  15. Cont. • We now repeat the sequence of operations that reduced 6265 to 12032 to reduce the exponent down to 1. •   = 62 * 3616 % 133  = 62 * 998 % 133  = 62 * 924 % 133  = 62 * 852 % 133  = 62 * 43 % 133  = 2666 % 133  = 6 • And that matches the plaintext we put in at the beginning.

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