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Group 5: Daryl, Etkin, Supartha, Rajendra and Aarthi 1

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Two Information Coding Schemes RSA Algorithm Privacy Authenticity Reed-Solomon Codes (Bursty) Noise Tolerance 2

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5 Trinity? Yes, it’s me.

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Encrypt messages with a symmetric-key cryptosystem (e.g. DES, AES, etc…) Requires prior agreement on a shared key over a secure channel What if Neo and Trinity have yet to meet? 6

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Mathematically-related public/private key pairs are generated Messages encrypted with public key Can only be decrypted with private key Infeasible to compute private key from public key alone No need to agree on a shared key! 7

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Rivest, Shamir and Adleman (1977) Based on difficulty of computing prime factors of large integers 8

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1. Pick two distinct primes p and q 2. Compute n = pq and ɸ (n) = (p – 1)(q – 1) 3. Pick e where 1 < e < ɸ (n) and gcd(e, ɸ (n)) = 1 4. Compute d where de ≡ 1 (mod ɸ (n)) 5. Public key is (n, e), private key is (n, d) 6. Encrypt with C ≡ M e (mod n) 7. Decrypt with M ≡ C d (mod n) 9 Setup Usage

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1. Pick two distinct primes p and q 2. Compute n = pq and ɸ (n) = (p – 1)(q – 1) 3. Pick e where 1 < e < ɸ (n) and gcd(e, ɸ (n)) = 1 4. Compute d where de ≡ 1 (mod ɸ (n)) 5. Public key is (n, e), private key is (n, d) 6. Encrypt with C ≡ M e (mod n) 7. Decrypt with M ≡ C d (mod n) 10

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Let e, d, n be integers with n ≠ 0 Fact: If gcd(e, n) = 1 (i.e. e and n are coprime) then there exists d such that de ≡ 1 (mod n) In other words, the multiplicative inverse of e (mod n) exists when gcd(e, n) = 1 11

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1. Pick two distinct primes p and q 2. Compute n = pq and ɸ (n) = (p – 1)(q – 1) 3. Pick e where 1 < e < ɸ (n) and gcd(e, ɸ (n)) = 1 4. Compute d where de ≡ 1 (mod ɸ (n)) 5. Public key is (n, e), private key is (n, d) 6. Encrypt with C ≡ M e (mod n) 7. Decrypt with M ≡ C d (mod n) 12

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1. Pick two distinct primes p and q 2. Compute n = pq and ɸ (n) = (p – 1)(q – 1) 3. Pick e where 1 < e < ɸ (n) and gcd(e, ɸ (n)) = 1 4. Compute d where de ≡ 1 (mod ɸ (n)) 5. Public key is (n, e), private key is (n, d) 6. Encrypt with C ≡ M e (mod n) 7. Decrypt with M ≡ C d (mod n) 13

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Definition: no. of integers 1 ≤ a ≤ n with gcd(a, n) = 1 Formula: For n = pq where p and q are primes 14

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Let x, y, m, n be integers with n ≥ 0 Fact: If x ≡ y (mod ɸ (n)), then m x ≡ m y (mod n) In other words, working in mod n requires that we work mod ɸ (n) in the exponent 15

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1. Pick two distinct primes p and q 2. Compute n = pq and ɸ (n) = (p – 1)(q – 1) 3. Pick e where 1 < e < ɸ (n) and gcd(e, ɸ (n)) = 1 4. Compute d where de ≡ 1 (mod ɸ (n)) 5. Public key is (n, e), private key is (n, d) 6. Encrypt with C ≡ M e (mod n) 7. Decrypt with M ≡ C d (mod n) 16

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1. Pick two distinct primes p and q 2. Compute n = pq and ɸ (n) = (p – 1)(q – 1) 3. Pick e where 1 < e < ɸ (n) and gcd(e, ɸ (n)) = 1 4. Compute d where de ≡ 1 (mod ɸ (n)) 5. Public key is (n, e), private key is (n, d) 6. Encrypt with C ≡ M e (mod n) 7. Decrypt with M ≡ C d (mod n) 17 (M e ) d ≡ M (mod n) and (M d ) e ≡ M (mod n)

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1. Pick p = 37 and q = Compute n = 1591 and ɸ (n) = Pick e = 71 gcd(e, ɸ (n)) = gcd(71, 1512) = 1 4. Compute d = 575 (Extended Euclidean Algorithm) de = ≡ 1 (mod 1512) 5. Public key is (n, e), private key is (n, d) 6. Encrypt: C ≡ M e ≡ ≡ 908 (mod 1591) 7. Decrypt: M ≡ C d ≡ ≡ 1234 (mod 1591) 18

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1. Pick two distinct primes p and q 2. Compute n = pq and ɸ (n) = (p – 1)(q – 1) 3. Pick e where 1 < e < ɸ (n) and gcd(e, ɸ (n)) = 1 4. Compute d where de ≡ 1 (mod ɸ (n)) 5. Public key is (n, e), private key is (n, d) 6. Sign M with S ≡ M d (mod n) ; Send (M, S) 7. Verify that M ≡ S e (mod n) 19 Reversed! Sign with private key Verify with public key

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1. Modular exponentiation ▪ Successive-Squaring 2. Computing d from e and ɸ (n) ▪ Extended Euclidean Algorithm 3. Finding large primes 20

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Successive-Squaring to Compute C ≡ M e (mod n) Let e = e k e k–1 … e 0 (binary representation of e) C := 1 For i := k, k – 1, …, 0 C := (C * C) mod n If e i = 1 Then C := (C * M) mod n End For Performance: O(log e) Memory: O(1) 21

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Extended Euclidean Algorithm: Since e and ɸ (n) are coprime, solving yields d = y satisfying 22 Find max. q i satisfyingand x i and y i satisfying When r k = 0, stop and output gcd(a, b) = r k-1 and x = x k-1 and y = y k-1 Similar to Euclidean Algorithm for gcd(a, b), but retain quotients q i at each step i to compute x i and y i

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Generate a large random integer Apply primal test repeatedly Primality Tests: Miller-Rabin Solovay-Strassen Fermat Primality Test Euler Witness, Euler Liar 23

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39 ???

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Noise is Natural Studied models in general Binary Symmetric Channel Binary Erasure Channel Noisy Typewriter Channel Continuous Output Channel … 41

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43 MC

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46 MC No Structure?? Have to store the whole mapping in a codebook

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47 G = n x k, where n =7, k=4 for example, Hamming Code is Linear =

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Applications: CDs, Space Communication, … Robust against Burst errors 49 From left: Gustave Solomon & Irving S. Reed

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A set of elements with two operations “Addition” and “Multiplication” defined on these elements. Closed under these two operations Basically all arithmetic operations are allowed Examples: Set of Real numbers, Set of Rational numbers… 56

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A field with finite number of elements. Example: {0,1} with modulo operations In general {0,1,2….p-1} is a field with p elements with modulo operations. (p is prime) How to construct fields with 8 elements? In general how to construct p r elements?? 57

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A field with 2 m elements can be constructed by extending the field GF(2) which is {0,1}. Let α denotes an additional element in GF(2 m ). Now GF(2 m ) ={ 0,1, α,α 2,….. α 2 m -1, α 2 m,…} To make the number of elements 2 m, we restrict α 2 m -1 = 1 = α 0 58

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GF(2 m )={0, α 0, α,…… α 2 m -2 } Any non-zero element in GF(2 m ) can be written as a polynomial of degree at most m-1. Coefficients are from GF(2) Also they can be mapped to binary values. 59

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An irreducible polynomial f(x) of degree m is said to be primitive if the smallest positive integer n for which f(x) divides x n +1 is n=2 m -1. Example: 1 + x + x 4 because it divides x n +1 for n=15 and not for other values less than 15. Used for construction GF(2 m ) RS codes use GF(2 m ). 60

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Linear Code Cyclic d = n-k+1 (Maximum Distance Separable) Can correct up to n-k erasures Can correct up to (n-k)/2 symbol errors 64

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This form of encoding is not in Systematic form Systematic form : Parity symbols message symbols Message polynomial α + α 3 x + α 5 x 2 Code Polynomial α 0 + α 2 x + α 4 x 2 + α 6 x 3 + αx 4 + α 3 x 5 + α 5 x 6 A generator polynomial g(x) is defined as g(x) = (x-α) (x-α 2 ) …………… (x-α 2t ) 65

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Shift the message polynomial m(x) by 2t positions by multiplying m(x) by x 2t. Define p(x) = x 2t m(x) (mod g(x)) The final codeword polynomial u(x) is u(x) = p(x) + x 2t m(x) 66

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Message polynomial α + α 3 x + α 5 x 2 Yielding αx 4 + α 3 x 5 + α 5 x 6 after multiplication with x 2t i.e. x 4 Take g(x) = (x- α)(x- α 2 )(x- α 3 )(x- α 4 ) = x 4 – α 3 x 3 + α 0 x 2 – αx + α 3 = α 3 + αx + α 0 x 2 + α 3 x 3 +x 4 Next divide α x 4 + α 3 x 5 + α 5 x 6 by g(x) to find the remainder p(x) = α 0 + α 2 x 4 + α 4 x 2 + α 6 x 3. Now u(x)=α 0 + α 2 x + α 4 x 2 + α 6 x 3 + αx 4 + α 3 x 5 + α 5 x 6 67

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The syndrome is the result of a parity check performed on the received polynomial r(x) to determine whether r(x) is a valid codeword. The syndromes are basically evaluations of the received polynomial r(x) at α, α 2, α 3,… α 2t. S i = r(α i ), i=1,2,….2t If r(x) is a valid codeword then we get all the S i evaluate to zero. Any non-zero S i indicates the presence of errors. 68

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The errors introduced by the channel during transmission can be modeled as a polynomial e(x) over the field GF(2 m ). Hence r(x) = u(x) + e(x). The problem finding e(x) from r(x) (or the syndromes) is decoding. 69

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Given any collection of k coefficients from the polynomial u(x), it is possible to construct the remaining coefficients of u(x). If at most ‘t’ of the coefficients of u(x) are incorrect, it is possible to reconstruct all the coefficients of u(x) correctly. Polynomial multiplication/division using FFT takes O(klogk) where k is the max of degrees of the polynomial. 70

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Error polynomial e(x) = e 0 + e 1 x +….e n x n Suppose the received polynomial r(x) has ν errors in it at the locations i 1,i 2,…i ν. The magnitude of error at these locations are e i j. Then syndromes can be written in the form Sj = e 1 j x l j + e 2 j x l j + ….+e ν j x l j j=1,2,…2t where X l = α il 71

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1. Calculate Syndromes 2. Find the error locator polynomial Peterson-Gorenstein-Zierler Decoder 3. Find error locations Chien Search 4. Find error values Forney’s Algorithm 73

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74 Non- Linear System!! Finding S j for j = 1 to 2t

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75 Intuition: The roots of this polynomial are inverses of the error locations Help to find the locations where an error has occurred Expanding Λ(x):

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76 For x = X l -1 and for any 1 ≤ l ≤ ν Multiplying throughout by Y l X l (j+ν)

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77 Sum over l = 1 to t

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78 Equation (1) – (4) now form a system of Linear Equations

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79 Solved for Λ i s by finding the largest value of ν for which M ν is non-singular starting from ν = t Overall this algorithms runs in polynomial time ν ≤ t; 2t = n – k = O(n) (could also be O(1) for large n & k) Use polynomial-time algorithms for matrix determinants and inversion

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81 For i ≥ ν, Y i X i = 0 The Middle matrix is singular M μ is singular

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1. Calculate Syndromes 2. Find the error locator polynomial Peterson-Gorenstein-Zierler Decoder 3. Find error locations Chien Search 4. Find error values Forney’s Algorithm 82

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Find roots of Error Locator Polynomial, Λ(x), by exhaustive search Evaluate Λ(α i ) for i = 1, 2, …, 2t Find all i where Λ(α i ) = 0 α i is a root of Λ(x) Error locations will also be of the form: α j Here, α j = α -1 and j = 2t – i. If number of errors found is ≥ t, abort process 83

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Number of iterations = O(n) Linear time algo Correctness of Chien Search: Viewing Λ(x) as a polynomial over a finite field 85

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1. Calculate Syndromes 2. Find the error locator polynomial Peterson-Gorenstein-Zierler Decoder 3. Find error locations Chien Search 4. Find error values Forney’s Algorithm 86

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87 Convert to matrix form Vandermonde Matrix

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88 Defining the Syndrome polynomial: Defining the Error Evaluator polynomial: where, b is the degree of the smallest root of the generating function of the code &

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Computing coefficients of Ω(x) takes ≈ ν 2 /2 multiplications Computing each Y i takes 2ν+1 multiplications and one reciprocal Total computation time needed ≈ 2.5ν 2 multiplications ν = O(n) Operations needed O(n 2 ) Polynomial Time algorithm 89

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1. Calculate Syndromes 2. Find the error locator polynomial Peterson-Gorenstein-Zierler Decoder 3. Find error locations Chien Search 4. Find error values Forney’s Algorithm 90

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r(x) = u(x) + e(x) Decoding techniques help determine e(x) completely Hence, u(x) = r(x) – e(x) = Message sent is recovered 91 We are done!!

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Other more efficient (implementation wise) algorithms for decoding: Berlekamp-Massey Decoder (LFSR and iterative correction) Euclidean Algorithm (Values and locations simultaneously determined using iterative GCD of polynomials) Decoders implemented as dedicated chips by manufacturers (Hardware and Software) 92

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RSA: 1. Evgeny Milanov, RSA algorithm, 2. Kenneth Rose, Elementary Number Theory and its applications, 5 th Ed., Pearson International 3. Trappe & Washington, Introduction to Cryptography with Coding Theory, 2 nd Ed., Pearson International Reed-Solomon Codes: 1. Bernard Sklar, Reed Solomon error correction, solomon/elementLinks/art_sklar7_reed-solomon.pdf solomon/elementLinks/art_sklar7_reed-solomon.pdf 2. V. Guruswami, Introduction to Coding Theory, CMU, 3. John Gill, EE 387 Note #7, Stanford University, 4. Wikipedia 93

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Thanks for your attention!! 94

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