Man in the Middle AttackThe man-in-the-middle attack is a protocol for an eavesdropper E to intercept a messageexchange between A and B. The attack is premised on a Diffie-Hellman key exchange, butthe principle applied to any public key cryptosystem for which the keys used for public keyexchange is not certified with a cerification authority.We assume that A and B have agreed on a prime p and a primitive element a of Z/pZ, andthat E is positioned between A and B. Having observed this Diffie-Hellman initializationE prepares for the man-in-the-middle attack.1. A chooses a secret key x, creates a public key a x , and sends it to B, which is interceptedby E.2. E chooses a private integer z at random, and creates the alternative public key a zwhich she sends to B, pretending to be A. At the same time she sends same key a z to A,now posing as B.3. Now E has established a common session key a xz with A and common session keya yz with B. Message exchanges between A and B pass through E and can be deciphered,read, modified, re-enciphered, and resent in transit.The breakdown of the key exchange protocol is due to lack of identity authentication ofthe communicating parties. If, for instance the public key (a, a x , p) of A could be confirmedwith an independent certification authority, then B would not have confused E with A.ExercisesThe RSA cryptosystem is based on the difficulty of factoring large integers into its compositeprimes.Based on Fermat’s little theorem, we know that a m ≡ 1 mod p exactly when p−1 dividesm. Therefore we recover the identity a u ≡ a mod p where u is of the form 1 + (p − 1)r.Now given any e such that e and p − 1 have no common divisors, there exists a d such thated ≡ 1 mod p − 1. In other words, u = ed is of the form 1 + (p − 1)r. This means that themapa ↦→ a e mod pfollowed bya e mod p ↦→ (a e mod p) d mod p ≡ a ed mod p = a mod pare inverse maps. This only works for a prime p.Exercise 8.1 Use SAGE to find a large prime p and to compute inverse exponentiationpairs e and d. The following functions are of use:Diffie–Hellman 75
andom prime, gcd, xgcd, and inverse mod.The RSA cryptosystem is based on the fact that for primes p and q and any integer e withno common factors with p − 1 and q − 1, it is possible to find an d 1 such thated 1 ≡ 1 mod (p − 1),ed 2 ≡ 1 mod (q − 1).Using the Chinese remainder theorem, it is possible to then find the unique d such thatd = d 1 mod (p − 1) and d = d 2 mod (q − 1)in the range 1 ≤ d < (p − 1)(q − 1). This d has the property thata ed ≡ a mod n.The send a message securely, the public key (e, n) is used. First we encoding the message asan integer a mod n, then form the ciphertext a e mod n. The recipient recovers the messageusing the secret exponent d.Exercise 8.2 Use your exponents e, d, verify the identities mod p:for various random values of a.(a e ) d ≡ a mod n, (a d ) e ≡ a mod n, and a ed ≡ a mod n,Note that after construction of d, the primes p and q are not needed, but that withoutknowing the original factorization of n, Fermat’s little theorem does not apply, and findingthe inverse exponent for e is considered a hard problem.Exercise 8.3 Use the above factorization to reproduce the private key L (generated but notprinted above) for this K.Exercise 8.4 Why is the choice for which key is the public key and which key is theprivate key arbitrary? Practice encoding, decoding, enciphering, and deciphering with theRSA cryptosystem. Why do the member functions enciphering and deciphering returnthe same values?An ElGamal cryptosystem is based on the difficulty of the Diffie–Hellman problem:Given a prime p, a primitive element a of (Z/pZ) ∗ = {c ∈ Z/pZ : c ≠ 0}, and elementsc 1 = a x and c 2 = a y , find the element a xy in (Z/pZ) ∗ .Exercise 8.5 Recall the discrete logarithm problem: Given a prime p, a primitive elementa of (Z/pZ) ∗ , and an element c of (Z/pZ) ∗ , find an integer x such that c = a x . Explainhow a general solution to the discrete logarithm problem for p and a implies a solution tothe Diffie–Hellman problem.76 Chapter 8. Public Key <strong>Cryptography</strong>
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Author (David R. Kohel) /Title (Cry
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CONTENTS1 Introduction to Cryptogra
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PrefaceWhen embarking on a project
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information. We introduce here some
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ut strings in A ∗ map injectively
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CHAPTERTWOClassical Cryptography2.1
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LV MJ CW XP QO IG EZ NB YH UA DS RK
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As a special case, consider 2-chara
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Note that if d k = 1, then we omit
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ExercisesSubstitution ciphersExerci
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Ciphertext-only AttackThe cryptanal
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of size n, suppose that p i is the
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- Page 42 and 43: CHAPTERFIVEBlock CiphersData Encryp
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- Page 46 and 47: The Advanced Encryption Standard al
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Solution. Now we can verify that e
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which has no common factors with p
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sage: p = 2^32+61sage: m = (p-1).qu
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sage: a5 := a^n5sage: c5 := c^n5sag
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The application of this function E
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5. (∗) How many elements a of G h
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1. The value f(0) of the polynomial