Download presentation

Presentation is loading. Please wait.

Published byMaverick Huddleston Modified over 3 years ago

1
RSA COSC 201 ST. MARY’S COLLEGE OF MARYLAND FALL 2012 RSA

2
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.

3
RSA How hard to crack is this kind of encryption – let’s take an example: 94021

4
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

5
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)

6
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; }

7
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); }

8
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

9
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.

Similar presentations

OK

Foundations of Network and Computer Security J J ohn Black CSCI 6268/TLEN 5550, Spring 2013.

Foundations of Network and Computer Security J J ohn Black CSCI 6268/TLEN 5550, Spring 2013.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google

Ppt on regional transport office delhi Ppt on ocean food chain Ppt on tata motors history Ppt on dc motor drives Seminar ppt on green cloud computing Ppt on non biodegradable waste list Ppt on computer malware software Marketing mix ppt on sony picture Ppt on generalized anxiety disorder Ppt on intelligent manufacturing systems