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Asymmetric encryption

Asymmetric encryption . Asymmetric encryption . Asymmetric encryption, often called "public key" encryption, allows Alice to send Bob an encrypted message without a shared secret key there  is  a secret key, but only Bob knows what it is, and he does not share it with anyone, including Alice.

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Asymmetric encryption

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  1. Asymmetric encryption

  2. Asymmetric encryption • Asymmetric encryption, often called "public key" encryption, allows Alice to send Bob an encrypted message without a shared secret key • there is a secret key, but only Bob knows what it is, and he does not share it with anyone, including Alice

  3. Asymmetric encryption

  4. Asymmetric encryption • Bob creates a pair of keys, one of which he keeps secret and one of which he sends to Alice. • Alice composes a confidential message and encrypts it using the key that Bob has sent to her. • Alice sends the encrypted data to Bob. • Bob uses his secret key to decrypt the data and reads the confidential message.

  5. Asymmetric encryption • Asymmetric algorithms include a "key generation" protocol that Bob uses to create his key pair, as shown by Figure 15-2. Following the protocol results in the creation of a pair of keys that have a mathematical relationshipthe exact detail of the protocol and the relationship between the keys is different for each algorithm.

  6. Asymmetric encryption

  7. Creating Asymmetric Keysusing RSA Algorithm • 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 Or d = e -1 (mod ( (n)) • publish their public encryption key: KU={e,N} • keep secret private decryption key: KR={d,p,q}

  8. Example of RSA Ciphertext C = Memod n Plaintext M = Cdmod n Example: • p = 17 , q = 11 • n = 178 • ø(N)= 160 • e = 7 • D = 23 • PU = { 7 , 187 } ; PR = { 23 , 187 } • M= 88; C =88 7mod 187 = 11 • C = 1123mod 187 = 88

  9. Encryption

  10. Decryption

  11. Solve this example • Start with p= 61 q=53

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