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cse443-lecture-6-cryptography

cse443-lecture-6-cryptography - Lecture 6 Cryptography...

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CMPSC 443 Introduction to Computer and Network Security - Spring 2012 - Professor Jaeger Lecture 6 - Cryptography CMPSC 443 - Spring 2012 Introduction Computer and Network Security Professor Jaeger www.cse.psu.edu/~tjaeger/cse443-s12
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CMPSC 443 Introduction to Computer and Network Security - Spring 2012 - Professor Jaeger Page Question Setup : Assume you and I don ʼ t know anything about each other, but we want to communicate securely. We want to establish a key that we can encrypt communication with each other. Q : Is this possible? 2 ?
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CMPSC 443 Introduction to Computer and Network Security - Spring 2012 - Professor Jaeger Page Diffie-Hellman Key Agreement The DH paper really started the modern age of cryptography, and indirectly the security community – Negotiate a secret over an insecure media – E.g., “in the clear” (seems impossible) – Idea: participants exchange intractable puzzles that can be solved easily with additional information. Mathematics are very deep – Working in multiplicative group G – Use the hardness of computing discrete logarithms in finite field to make secure – Things like RSA are variants that exploit similar properties 3
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CMPSC 443 Introduction to Computer and Network Security - Spring 2012 - Professor Jaeger Page Diffie-Hellman Protocol For two participants p 1 and p 2 Setup: We pick a prime number p and a base g (< p ) – This information is public – E.g., p=13 , g=4 Step 1: Each principal picks a private value x (< p-1 ) Step 2: Each principal generates and communicates a new value y = g x mod p Step 3: Each principal generates the secret shared key z z = y x mod p Where y is the value received from the other party. 4
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CMPSC 443 Introduction to Computer and Network Security - Spring 2012 - Professor Jaeger Page A protocol run ... p=17, g=6 Step 1) Alice picks x=4 Bob picks x=5 Step 2) Alice's y = 6^4 mod 17 = 1296 mod 17 = 4 Bob's y = 6^5 mod 17 = 7776 mod 17 = 7 Step 3) Alice's z = 7^4 mod 17 = 2401 mod 17 = 4 Bob's z = 4^5 mod 17 = 1024 mod 17 = 4 5
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CMPSC 443 Introduction to Computer and Network Security - Spring 2012 - Professor Jaeger Page Attacks on Diffie-Hellman This is key exchange, not authentication.
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