 # Seminar on (ENCODING AND DECODING TECHNIQUES)

Mar 26, 2015

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Slide 2 www.fakengineer.com Seminar on (ENCODING AND DECODING TECHNIQUES) Slide 3 www.fakengineer.com Introduction What is Cryptography Why we need Cryptography Slide 4 www.fakengineer.com Some terms associated with Cryptography Plaintext Ciphertext Encryption Decryption Key Slide 5 www.fakengineer.com Methods for Encryption/Decryption Conventional Methods Public key Methods Slide 6 www.fakengineer.com Conventional methods Character-level encryption Substitutional Transpositional Bit-level encryption Slide 7 www.fakengineer.com Data encryption standard (DES) Bit-level encryption Designed by IBM and adopted by the U.S government for nonmilitary use Encrypts a 64-bit plaintext using a 56-bit key The text is put through 19 different procedures to create a 64-bit ciphertext Slide 8 www.fakengineer.com Slide 9 Publickey Encryption Slide 10 www.fakengineer.com RSA Encryption Rivest, Shamir, Adleman encryption Public key encryption technique One party uses a public key, k p Other party uses a private key, k s Both use a number, N Slide 11 www.fakengineer.com Slide 12 Encryption algorithm (RSA) Encode the data to be encrypted as a number to create the plaintext P Calculate the ciphertext C=P k p modulo N Send C as ciphertext Slide 13 www.fakengineer.com Decryption algorithm (RSA) Receive C, the ciphertext Calculate plaintext P=C k s modulo N Decode P to the original data Slide 14 www.fakengineer.com Example of RSA Let K p=5, K s=77, and N=119 Let the character F is encoded as 6 (sixth char in alphabet) We calculate 6 Kp modulo 119 =41 At the receiver side, we calculate 41 Ks modulo 119=6 Decode 6 as F Slide 15 www.fakengineer.com Choosing K p, K s and N First choose 2 prime numbers p and q (we choose 7 and 17) Calculate N=p*q (N=7*17=119) Select K p such that it is not a factor of (p- 1)*(q-1) (we choose 5, not factor of 96) Select K s such that (K p* K s ) modulo (p- 1)*(q-1)=1 (4*96+1=385 and 385/5=77) Slide 16 www.fakengineer.com

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