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RSA COSC 201 ST. MARY’S COLLEGE OF MARYLAND FALL 2012 RSA

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Public Key Encryption Bob wants to send Alice a super secret message. How? Alice generates 2 keys – a public encryption key and a private decryption key. Bob takes the encryption key, encodes the message, then sends it to Alice. Alice then can decrypt the message. No one else can decrypt the message.

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RSA How hard to crack is this kind of encryption – let’s take an example: 94021

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Four Problems At the core of RSA are four problems: Modular Exponentiation – compute x n (mod P) GCD Multiplicative inverse: solve AX ≡ 1 (mod P) for X Primality testing

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The Algorithm The receiver chooses two large primes p and q. p = 127, q = 211 Compute N = pq and N’ = (p-1)(q-1) N = 26797, N’ = 26460 Choose some e > 1 s.t. gcd(e, N’) = 1 e = 13379 Compute d, the multiplicative inverse of e, mod N’ d = 11099 The receiver then destroys p, q, and N’. Transmit e and N, keep d a secret. Encrypting – sender computes M e (mod N) and sends. M is the message. Decrypting – compute R d (mod N)

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Computing Modular Exponentiation Solution 0 – do what it says. Do the exponentiation, then mod it by P. Solution 1 – start with a result, multiply by X, then mod by P – keep going until we’ve done this N times. Solution 2 – observe that if N is even – x n = (x*x) n/2, if N is odd – x n = x * (x*x) n/2 public static long modpower(long x, long n, long p){ if (n == 0) return 1; long tmp = modpower((x*x) % p, n/2, p); if (n%2 != 0) tmp = (tmp*x) % p; return tmp; }

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Computing GCD Solution 1 – Euclid’s Algorithm – efficient, recursive, and 2300 years old. Subtract B from A continuously until A becomes less than B, then switch. Continue until B becomes 0. A is the gcd. Solution 2 – leverage modulus: public long gcd (long a, long b){ if (b == 0) return a; return gcd(b, a%b); }

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Basically, given A and N, solve for X where AX % N == 1 % N How to solve – leverage GCD from before ! private long x, y; public void fullGCD(long a, long b){ long x1, y1; if (b == 0){ x = 1; y = 0; }else{ fullGCD(b, a % b); x1 = x; y1 = y; x = y1; y = x1 – (a/b) * y1; } } public long inverse(long a, long n){ fullGCD(a, n); return x>0 ? x : x + n; } Compute Multiplicative Inverse

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Finally, Primality Testing Solution? Randomized Algorithm: Pick a random integer i from 2 to n-1 Compute gcd(i, n). If this is not 1, primality fails. Otherwise, repeat up to k times.

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