05_public - Public Key Encryption Karl-Johan Grinnemo...

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1 Public Key Encryption Karl-Johan Grinnemo grinnemo@kth.se 2004 Created by Dr. Johan Montelius 2005 – 2007 Revised by Dr. Johan Montelius and Dr. Jon-Olov Vatn 2009 Revised by Dr. Karl-Johan Grinnemo
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2 Symmetric Encryption • Also known as: – Shared key encryption – Secret key encryption • Same key for encryption and decryption • Sender and receiver need to agree on a key ----- ----- ----- Plaintext input Encryption Decryption ----- ----- ----- Plaintext output Ciphertext Shared, secret key
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3 Public Key Encryption (PKE) plaintext message, m ciphertext encryption algorithm decryption algorithm Bob’s public key plaintext message K (m) B + K B + Bob’s private key K B - m = K ( K (m) ) B + B -
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4 Requirements on Public-Key Algorithms • K - (K + (Plaintext)) = Plaintext • It is exceedingly difficult to deduce K - from K + • K + cannot be broken by a chosen plaintext attack
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5 The Modulo Operator • a = b (mod n) a = b + ξn, ξ is an integer – For example: • 1 = 13 (mod 12) (modular arithmetic ”clock” arithmetic) – 1 = 13 + (– 1) × 12 • 14 = 38 (mod 12) – 14 = 38 + (- 2) × 12 • -3 = 2 (mod 5) – -3 = 2 + (-1) × 5 • a = b (mod n) same remainder when divided by n
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6 Modular Addition • Definition: – a + b (mod n) = a + b if a + b < n – a + b (mod n) = (a + b) – floor((a + b)/n) × n otherwise • Example: a + b (mod 4) • For example: 12 + 13 (mod 4) = 26 – 6 × 4 = 2 0 1 2 3
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7 Modular Addition for PKE • Encryption: add recipients public key to plaintext – K + (m) = m + ξ (mod n), ξ is an Integer • Decryption: add recipients private key to ciphertext – m = K - (K + (m)) = m + ξ + ξ -1 (mod n)
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8 Requirements on Modular Addition from PKE • Needs to be one-to-one (injective function) – If a + b = c (mod n) and d + b = c (mod n) then a = d (a < n, b < n) • Needs to have an additive inverse – a + a -1 = 0 1 + 2 2 + 3 3 + 5 3 5 8 12 1 + 2 2 + 3 3 + 5 3 5 8 12 Injective function Non-injective function
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9 Example PKE using Modular Addition • Let K + (m) = m + 3 (mod 26), then K - (m) = K + (m) + 23 (mod 26) – 3 + 23 (mod 26) = 0 • This is a monoalphabetic substitution cipher (Caesar cipher) • Example: – plaintext = ”secret” 19,5,3,18,3,20 (position in the alphabet) – Encrypted plaintext: 22,8,6,21,6,23 (e.g., 19 + 3 (mod 26) = 22) – Decrypted plaintext: 19,5,3,18,3,20 (e.g., 22+23 (mod 26) = 19)
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10 Modular Multiplication • Definition: – a × b (mod n) = a × b if a × b < n – a × b (mod n) = a × b – floor((a × b)/n) × n • Example: a × b (mod 4) 1 2 3 0 3 2 0 2 0 2 3 2 1 0 1 0 0 0 0 0 3 2 1 0 × a b
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11 Requirements on Modular Multiplication from PKE • Needs to be one-to-one • Needs to have a multiplicative inverse – a × a -1 = 1 • Example: a × b (mod 4) 1 2 3 0 3 2 0 2 0 2 3 2 1 0 1 0 0 0 0 0 3 2 1 0 × No multiplicative inverse Not one-to-one b a
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12 Example PKE using Modular Multiplication • Let K + (m) = m × 3 (mod 4), then K - (m) = K + (m) × 3 (mod 4) – 3 × 3 (mod 4) = 1 • Example:
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This note was uploaded on 02/16/2011 for the course ICT 2 taught by Professor 2 during the Spring '11 term at Kungliga Tekniska högskolan.

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05_public - Public Key Encryption Karl-Johan Grinnemo...

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