Loading in 5 sec....

David Evans cs.virginia/evansPowerPoint Presentation

David Evans cs.virginia/evans

- 124 Views
- Uploaded on
- Presentation posted in: General

David Evans cs.virginia/evans

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

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

- 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

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

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

- 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

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

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

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

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

- We are looking for e, d and n such that:Med-11mod 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

(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

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

If n is prime and a is not divisible by nan-11mod n

CS588 Spring 2005

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

For a and n relatively prime:

a(n)1mod n

Partial Proof:

If n is prime, (n) = n – 1 and an - 11mod n

by Fermat’s Little Theorem

What if n is not prime?

CS588 Spring 2005

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

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

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

CS588 Spring 2005

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 nQED.

CS588 Spring 2005

What if M is not relatively prime to n?

- We are looking for e, d and n such that:
Med-11mod 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

- 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

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 (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) + 1M mod n

CS588 Spring 2005

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

de 1 mod (n)

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

Hence,

Med-1mod n = Mk * (n)mod n

CS588 Spring 2005

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

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

Adam Glaser and Portman Wills

CS588 Fall 2001 PS4

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.

CS588 Spring 2005

- 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

- “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

- “(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

- 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

- 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

- 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