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Lecture 12: Non-secret Key Cryptosystems (How Euclid, Fermat and Euler Created E-Commerce). David Evans http://www.cs.virginia.edu/evans. Real mathematics has no effects on war. No one has yet discovered any warlike purpose to be served by the theory of numbers.

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david evans http www cs virginia edu evans

Lecture 12:

Non-secret Key Cryptosystems

(How Euclid, Fermat

and Euler Created

E-Commerce)

David Evans

http://www.cs.virginia.edu/evans

Real mathematics has no effects on war. No one has yet discovered any warlike purpose to be served by the theory of numbers.

G. H. Hardy, The Mathematician’s Apology, 1940.

CS588: Security and Privacy

University of Virginia

Computer Science

applications of rsa
Applications of RSA
  • Privacy:
    • Bob encrypts message to Alice using EA
    • Only Alice knows DA
  • Signatures:
    • Alice encrypts a message to Alice using DA
    • Bob decrypts using EA
    • Knows it was from Alice, since only Alice knows DA
  • Things you use every day: ssh, SSL, DNS, ...

CS588 Spring 2005

public key applications privacy
Public-Key Applications: Privacy

Bob

Alice

  • Alice encrypts message to Bob using Bob’s Private Key
  • Only Bob knows Bob’s Private Key only Bob can decrypt message

Decrypt

Ciphertext

Encrypt

Plaintext

Plaintext

Bob’s Public Key

Bob’s Private Key

CS588 Spring 2005

signatures
Signatures

Bob

Alice

Signed

Message

Decrypt

Encrypt

  • Bob knows it was from Alice, since only Alice knows Alice’s Private Key
  • Non-repudiation: Alice can’t deny signing message (except by claiming her key was stolen!)
  • Integrity: Bob can’t change message (doesn’t know Alice’s Private Key)

Plaintext

Plaintext

Alice’s Public Key

Alice’s Private Key

CS588 Spring 2005

public key cryptography
Public-Key Cryptography
  • Private procedure: E
  • Public procedure: D
  • Identity: E (D(m)) = D (E(m)) = m
  • Secure: cannot determine E from D
  • But didn’t know how to find suitable E and D

CS588 Spring 2005

properties of e and d
Properties of E and D

Trap-door one way function:

  • D (E (M)) = M
  • E and D are easy to compute.
  • Revealing E doesn’t reveal an easy way to compute D

Trap-door one way permutation: also

  • E (D (M)) = M

CS588 Spring 2005

slide7
RSA

E(M) = Me mod n

D(C) = Cd mod n

n = pqp, qare prime

dis relatively prime to(p – 1)(q – 1)

ed 1 (mod (p – 1)(q – 1))

(red things are secret)

CS588 Spring 2005

properties of e and d1
Properties of E and D

Trap-door one way function:

  • D (E (M)) = M
  • E and D are easy to compute.
  • Revealing E doesn’t reveal an easy way to compute D

Trap-door one way permutation: also

  • E (D (M)) = M

CS588 Spring 2005

property 1 d e m m
Property 1: D (E (M)) = M

E(M) = Me mod n

D(E(M)) = (Me mod n)d mod n

= Med mod n (as in D-H proof)

Can we choosee, dandnwith this property: M  Med mod n

equivalently:1  Med-1mod n

CS588 Spring 2005

finding e d and n
Finding e, d and n
  • We are looking for e, d and n such that:Med-11mod n
  • Euler’s Theorem: for a and n relatively prime: a(n)1mod n
  • Next:
    • What is (n)
    • Proof of Euler’s Theorem
    • How it works for arbitrary M
    • Given (n) how do we find e and d

CS588 Spring 2005

euler s totient function
Euler’s Totient Function

 (n) = number of positive integers less than n that are relatively

prime to n

  • If n is prime, (n) = n – 1
    • Proof by contradiction
  • What if n = pq where p and q are prime?

CS588 Spring 2005

totient products
Totient Products

For primes,pandq:n = pq

(n) = numbers < nnot relatively prime to pq

= pq – 1 ; numbers less thanpq

– (q – 1) ; size of p, 2p, …, (q –1)p

– (p – 1) ; size of q, 2q, …, (p –1)q

= pq – 1 – (q – 1) – (p – 1)

= pq – (p + q) + 1

= (p – 1) (q – 1) = (p)(q)

CS588 Spring 2005

fermat s little theorem
Fermat’s Little Theorem

If n is prime and a is not divisible by nan-11mod n

CS588 Spring 2005

fermat s little theorem proof
Fermat’s Little Theorem Proof

If n is prime and a is not divisible by n:

{a mod n, 2a mod n, … , (n-1)a mod n} = {1, 2, …, (n – 1) }

Product of all elements in sets:

a 2a  …  (n – 1) a (n – 1)! mod n

(n – 1)!an-1 (n – 1)! mod n

an-1  1 mod nQED.

CS588 Spring 2005

euler s theorem
Euler’s Theorem

For a and n relatively prime:

a(n)1mod n

Partial Proof:

If n is prime, (n) = n – 1 and an - 11mod n

by Fermat’s Little Theorem

What if n is not prime?

CS588 Spring 2005

euler s theorem cont
Euler’s Theorem, cont.

For a and n relatively prime:

a(n)1mod n

(n) = number of numbers < nnot relatively prime ton

We can write those numbers as:

R = { x1, x2, … , x(n)}

CS588 Spring 2005

proving euler s theorem
Proving Euler’s Theorem

R = { x1, x2, … , x(n)} multiply bya mod n:

S = { ax1 mod n, ax2 mod n, …, ax (n) mod n}

S is a permutation of R:

    • a is relatively prime to n
    • a is relatively prime to all xi
    • axi is relatively prime to n
  • Hence all elements of S are in R.
  • There are no duplicates in S.

If aximod n = axjmod n then i = j. since a is relatively prime to n

CS588 Spring 2005

proving euler s theorem1
Proving Euler’s Theorem

x1 x2 … x(n)

= ax1 mod n ax2 mod n … ax(n) mod n

 (ax1 ax2 … ax(n)) mod n

a(n) x1 x2 … x (n) mod n

1 a (n)mod n QED.

CS588 Spring 2005

recap

What if M is not relatively prime to n?

Recap
  • We are looking for e, d and n such that:

Med-11mod n

  • Euler’s Theorem: 1 a (n)mod n

for a and n relatively prime

  • If n is prime,  (n) = n – 1.
  • For p and q prime, (pq) =  (p) (q)

n = pq

ed – 1 =  (n) = (p-1)(q-1)

CS588 Spring 2005

m and n
M and n
  • Suppose M and n not relatively prime:

gcd (M, n)  1

  • Since n = pq and p and q are prime:

gcd (M, p)  1 OR gcd (M, q)  1

Case 1: M = cp

gcd (M, q) = 1 (otherwise M is multiple of both p and q, but M < pq).

So, M(q) 1 mod q

(by Euler’s theorem, since M and q are relatively prime)

CS588 Spring 2005

m and n cont
M and n, cont

Case 1: M = cp

gcd (M, q) = 1 (otherwise M is multiple of both p and q, but M < pq).

So, M (q) 1 mod q

(by Euler’s theorem, since M and q are relatively prime)

M(q) 1 mod q

(M(q))(p) 1 mod q

M (q)(p) 1 mod q

M(n) 1 mod q

CS588 Spring 2005

m and n1
M and n

M (n) 1 mod q

M (n) = 1 + kq for some k

M = cp recall gcd (M, p)  1

M M (n) = (1 + kq)cp

M(n) + 1 =cp + kqcp = M + kcn

M(n) + 1M mod n

CS588 Spring 2005

where s ed
Where’s ED?

ed – 1 = (n) = (p-1)(q-1)

  • So, we need to choose e and d:

ed =  (n) + 1 = n – (p + q)

  • Pick random d, relatively prime to  (n)

gcd (d,  (n)) = 1

  • Since d is relatively prime to  (n)it has a multiplicative inverse e:

de  1 mod  (n)

CS588 Spring 2005

identity
Identity

de  1 mod  (n)

So,d * e = (k *  (n)) + 1 for some k.

Hence,

Med-1mod n = Mk * (n)mod n

CS588 Spring 2005

d e m m
D (E (M)) = M

Med-1mod n = Mk * (n)mod n

Euler says 1 M(n)mod n.

So 1 Mk * (n)mod n

1  Med-1mod n

M Med mod n

QED.

CS588 Spring 2005

properties of e and d2
Properties of E and D

Trap-door one way function:

  • D (E (M)) = M
  • E and D are easy to compute.
  • Revealing E doesn’t reveal an easy way to compute D

Trap-door one way permutation: also

  • E (D (M)) = M

CS588 Spring 2005

movie break

Movie Break

Adam Glaser and Portman Wills

CS588 Fall 2001 PS4

questionable statements in rsa paper finalists
1) "The reader is urged to  find a way to "break" the system. Once the method has withstood all attacks for a sufficient length of time it may be used with a reasonable amount of confidence." The authors appear to advocating the same method of validation that they called "fruitless" earlier in the paper (referring to the NBS certification). 2)  RSA seems to gloss over the whole PKI issue. They suggest either a single authority to hold all the keys or publishing a book to all the users. I\'d also like to point out that RSA like to "excessively" to quotation marks for no apparent "purpose."

1. The problem is mentioned on page 4 and 6 of the paper: The trusted distribution of the public portion of the key. If one were to modify the public keys in transport or to attack a central repository, then it would be impossible to be sure of the authenticity of the keys. (The suggestion of having a telephone book is not even possibly applicable due to the need to securely deliver it to all users from trusted central source). This can be seen as present problem with the loss of keys by Microsoft (resulting in forced revocation) and the limited trust one can put in Verisign due to limited checks done. 2. The assumption in conclusion that a protocol is secure due to lack of success in attacks for some period of time. The recent attacks upon SHA/MD5 show that even 10 years could be insufficient time to prove security.

Questionable Statements in RSA Paper: Finalists

CS588 Spring 2005

only two submissions
Only Two Submissions
  • This is pathetic!
  • There will be a Short Quiz in class Tuesday
    • Closed book, closed notes
    • Covers material in RSA paper and new paper handed out today
    • Andrew and Aleks are exempt

CS588 Spring 2005

two questionable statements in rsa paper
Two “Questionable” Statements in RSA Paper
  • “The need for a courier between every pair of users has thus been replaced by the requirement for a single secure meeting between each user and the public file manager when the user joins the system.”

(p. 6)

CS588 Spring 2005

two questionable statements in rsa paper1
Two “Questionable” Statements in RSA Paper
  • “(The NBS scheme (DES) is probably somewhat faster if special-purposed hardware encryption devices are used; our scheme may be faster on a general-purpose computer since multiprecision arithmetic operations are simpler to implement than complicated bit manipulations.)”

(p. 4)

CS588 Spring 2005

who really invented rsa
Who really invented RSA?
  • General Communications Headquarters, Cheltenham (formed from Bletchley Park after WWII)
  • 1969 – James Ellis asked to work on key distribution problem
  • Secure telephone conversations by adding “noise” to line
  • Late 1969 – idea for PK, but function

CS588 Spring 2005

rsa diffie hellman
RSA & Diffie-Hellman
  • Asks Clifford Cocks, Cambridge mathematics graduate, for help
  • He discovers RSA (four years early)
  • Then (with Malcolm Williamson) discovered Diffie-Hellman
  • Kept secret until 1997!
  • NSA claims they had it even earlier

CS588 Spring 2005

charge
Charge
  • Reread the parts of RSA paper you didn’t understand the first time
  • Work on your project!
  • Short Quiz on RSA material and Encrypted Searches paper in class Tuesday
    • Closed-book, closed-notes, open-T-shirt
  • Next time: RSA Properties 2, 3 and 4

CS588 Spring 2005

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