20-751 ECOMMERCE TECHNOLOGY FALL 2003 COPYRIGHT © 2003 MICHAEL I. SHAMOS Cryptography.

20-751 ECOMMERCE TECHNOLOGY

FALL 2003

COPYRIGHT © 2003 MICHAEL I. SHAMOS

Cryptography


FALL 2003


Outline

• Information security• Encryption and keys• Symmetric encryption

– DES• Public-key cryptosystems

– RSA• Digital signatures• Digital certificates


FALL 2003


eCommerce Problems

1. Secrecy: keep data secret from unauthorized parties

2. Authentication: Verify identity of parties

3. Integrity: Verify that messages have not been altered

4. Nonrepudiation: Prove that a party engaged in a transaction

All these problems can be solved through cryptography


FALL 2003


Encryption

MATERIALWE WANT TOKEEP SECRET

UNREADABLEVERSION OFPLAINTEXT

DATA TO THEENCRYPTIONALGORITHM

MATHEMATICALSCRAMBLINGPROCEDURE

KEY TYPES:1. MESSAGES FROM THIS PERSON; OR2. MESSAGES FROM THIS SESSION; OR3. THIS MESSAGE

MIGHT BE:TEXTDATAGRAPHICSAUDIOVIDEOSPREADSHEET. . .

SOURCE: STEIN, WEB SECURITY

OBJECT: HIDE A MESSAGE (PLAINTEXT, CLEARTEXT) BY MAKING IT UNREADABLE (CIPHERTEXT)REQUIREMENT: MUST BE REVERSIBLE

INTERCHANGE KEY

SESSION KEY

ONE-TIME KEY


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Role of the Key in Cryptography

• The key is a parameter to an encryption procedure• Procedure stays the same, but produces different

results based on a given key

NOTE: THIS METHOD IS NOT USED IN ANY REAL CRYPTOGRAPHY SYSTEM.IT IS AN EXAMPLE INTENDED ONLY TO ILLUSTRATE THE USE OF KEYS.

S P E C I A L T Y B D F G H J K M N O Q R U V W X ZA B C D E F G H I J K L M N O P Q R S T U V W X Y Z

C O N S U L T I N G

D S R A V G H E R MEXAMPLE:


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Symmetric Encryption

SYMMETRIC =SAME KEY USED FORBOTH ENCRYPTIONAND DECRYPTION

SENDER AND RECIPIENT MUSTBOTH KNOW THE KEY.THIS IS A WEAKNESS

CALLED THE

KEY EXCHANGE PROBLEM



FALL 2003


Diffie-Hellman Key Exchange (1976)

• How can Alice and Bob exchange a secret key remotely without a secure communication channel?

• Idea: Alice and Bob each pick secret numbers x and y• The don’t exchange x and y; they exchange functions of

x and y that are difficult for Eve to invert• Alice and Bob use these function values to compute the

same shared secret key• Alice knows x, f (x), f (y)• Bob knows y, f (x), f (y)• Eve only knows f (x), f (y)• Find a function h where h(x, f (x), f (y)) = h(y, f (x), f (y))

ALICE CANCOMPUTE THIS

BOB CANCOMPUTE THIS

EVE CAN’T


FALL 2003


MULTIPLICATIONMOD 7

The Multiplicative Group mod n

• If n is prime, then the set of numbers 0, 1, 2, …, n-1 form a group under multiplication mod n

• If x and y are in the set, so is x y• Every non-zero element has a unique inverse. For every x,

there is exactly one y such that x y = 1

6 • 2 = 12WHEN DIVIDED BY 7GIVES REMAINDER 5

0 1 2 3 4 5 6

0 0 0 0 0 0 0 0

1 0 1 2 3 4 5 6

2 0 2 4 6 1 3 5

3 0 3 6 2 5 1 4

4 0 4 1 5 2 6 3

5 0 5 3 1 6 4 2

6 0 6 5 4 3 2 1

EACH ROW EXCEPTTHE ZERO ROWHAS EXACTLY ONE 1

EACH ELEMENT HASA UNIQUE INVERSE


FALL 2003


The Discrete Logarithm Problem

• An element g is a “generator” of the multiplicative group mod n if successive powers of g (mod n) produce all values from 1 to n-1

• Example: in the multiplicative group mod 7, 3 is a generator: 3, 32 = 2, 33 = 6, 34 = 4, 35 = 5, 36 = 1

• Given g and x, it is easy to compute gx

• BUT, given g and gx it is VERY DIFFICULT to compute x

• x is the DISCRETE LOGARITHM of gx in base g


FALL 2003


Diffie-Hellman• Alice creates two numbers: a large prime number n

and a generator g of the multiplicative group mod n

• (Easy. If n is prime, all 2 g < n are generators.)

• n and g are not secret

• Alice picks a random number x. x is secret. She sends gx to Bob. gx is not secret.

• Bob picks a random number y. y is secret. He sends

gy to Bob. gy is not secret.

• Alice computes K = (gy)x = gxy

• Bob computes K = (gx)y = gxy

• Alice and Bob now have a shared key K. Eve can’t compute K.


FALL 2003


Data Encryption Standard (DES)

• Symmetric, key-based encryption-decryption standard. No public keys

• Block cipher: operates on 64-bit blocks• Uses 56-bit key• 16 “rounds” -- key for each round is a 48-bit function

of the original 56-bit key. Each key bit participates in an average of 14 rounds

• Completely symmetric. Same algorithm decrypts.• Fast implementation in hardware: > 1 gigabit/second


FALL 2003


Data Encryption Standard (DES)

64 BITS OF MESSAGE INPUT PERMUTATION

INVERSE OF INPUTPERMUTATION

SUBKEYS:EACH IS A 48-BITFUNCTION OF A56-BIT KEY

OUTPUT: 64 BITS OFENCRYPTED TEXT

LEFT HALF OFBLOCK (32 BITS)

f IS A COMPLICATEDFUNCTION INVOLVINGVARIOUS PERMUTATIONS

SOURCE: SCHNEIER, APPLIED CRYPTOGRAPHY

IS EXCLUSIVE-OR


FALL 2003


Information Loss with Exclusive-OR

• x y = 1 if either x or y is 1 but not both:

• If x y = 1 we can’t tell which one is a 1• Can’t trace backwards to determine values

xy 0 1

0 0 1

1 1 0x

y


FALL 2003


Public-Key (Asymmetric) Encryption

1. USERS WANT TO SEND PLAINTEXT TO RECIPIENT WEBSITE

2. SENDERS USE SITE’S PUBLIC KEY FOR ENCRYPTION

3. SITE USES ITS PRIVATE KEY FOR DECRYPTION

4. ONLY WEBSITE CAN DECRYPT THE CIPHERTEXT. NO ONE ELSE KNOWS HOW


Symmetric encryption solves only the secrecy problemSomething else is needed for authentication, integrity and nonrepudiation


FALL 2003


Public-Key Encryption

• Alice wants to send Bob a secure message M.• Alice uses Bob’s public key to encrypt M.• Bob uses his private key to decrypt M.• Bob is the ONLY ONE who can do this,

so M is secure.• Problem: Anyone could have sent it. Was it really Alice?

ALICE’SCLEARTEXT

ALICE’SCODEDTEXT

ALICE’SCODEDTEXT

ALICE’SCLEARTEXT

TRANSM ISSION

BOB DECRYPTS WITHHIS PRIVATE KEY

ALICE ENCRYPTS WITHBOB’S PUBLIC KEY

BOB’SPUBLIC

KEY

BOB’SPRIVATE

KEY


FALL 2003


Digital Authentication• Alice wants to send Bob a message M so that Bob is sure Alice

is the sender.• Alice uses her own private key to encrypt M.• Bob uses Alice’s public key to decrypt M.• Alice is the ONLY ONE who could have sent it.• Problem 1: Anyone can read it! Problem 2: Replay attack!

ALICE’SCLEARTEXT

ALICE’SCODEDTEXT

ALICE’SCODEDTEXT

ALICE’SCLEARTEXT

TRANSM ISSION

BOB DECRYPTS WITHALICE’S PUBLIC KEY

ALICE ENCRYPTS WITHHER PRIVATE KEY

ALICE’SPRIVATE

KEY

ALICE’SPUBLIC

KEY


FALL 2003


Secure Authenticated Messages• Alice must send Bob a secret & authenticated message M so

Bob is sure it was sent by Alice. Use both encryption and signature.

ALICE’SCODEDTEXT

ALICE’SCODEDTEXT

(AUTHENTICATED)

ALICE’SCLEARTEXT

BOB DECRYPTS WITHALICE’S PUBLIC KEY

ALICE ENCRYPTS WITHHER PRIVATE KEY

ALICE ENCRYPTS WITHBOB’S PUBLIC KEY

ALICE’SCODED AND

SIGNED TEXT

ALICE’SCODED AND

SIGNED TEXT

T R A NSMI

T

ALICE’SCLEAR TEXT

(DECRYPTED ANDAUTHENTICATED)

BOB DECRYPTS WITHHIS PRIVATE KEY

BOB’S PUBLIC

ALICE’S PUBLIC

BOB’S PRIVATE

ALICE’S PRIVATE

4 KEYSNEEDED:


FALL 2003


Rivest-Shamir-Adelman (RSA)

• It is easy to multiply two numbers but apparently hard to factor a number into a product of two others.

• Given p, q, it is easy to compute n = p • q• Example: p = 5453089; q = 3918067• Easy to find n = 21365568058963• Given n, hard to find two numbers p, q with p • q = n• Now suppose n = 7859112349338149

What are p and q such that p • q = n ?• Multiplication is a one-way function• RSA exploits this fact in public-key encryption


FALL 2003


RSA Encryption

• Select two large prime numbers p, q (> 100 digits)• Let n = p • q• Choose a small odd integer e that does not divide

m = (p - 1)(q - 1). Then x(p-1)(q-1) = 1 (mod n)• Compute d = e-1(mod m)

– That is, d • e gives remainder 1 when divided by m

• Public key is the pair (e, n)• Private key is the pair (d, n)• Knowing (e, n) is of no help in finding d. Still need p

and q, which involves factoring n• DEMO


FALL 2003


RSA Encryption

• Message M is a number

• To encrypt message M using key (e, n):• Compute C(M) = M

e (mod n)

• To decrypt message C using key (d, n):• Compute P(C) = C

d (mod n)

• Note that P(C(M)) = C(P(M)) = (M e)d (mod n)

= M e•d (mod n) = M

because e • d = 1 and m = (p-1)(q-1)


FALL 2003


Message Digest (Hash)

• A message digest is a “fingerprint” of a message• Much shorter than the original message (e.g. 160 bits)• Easy to compute• Can’t recover the message from the digest• Changing the message changes the digest

MESSAGE (VERY LONG)

DIGEST

DIGEST CAN BE USED TO VERIFY THATTHE MESSAGE HAS NOT BEEN ALTERED


FALL 2003


Single Step of SHA-1

Operates on 16-word (512-bit) blocks

Expands 16 words to 80 words Wt

Performs 80 operations as shown for t = 0..79

a, b, c, d, e are special constants

Kt are special constants

SOURCE: SCHNEIER, APPLIED CRYPTOGRAPHY

INITIALLY CONSTANTS

80 WORDS INPUT HERE, 1 EACH STEP

MAGIC CONSTANTS

“<<< 5” means “cyclic left shift 5 bits”

+ + ++

REVISEDCONSTANTSFOR NEXTSTEP


FALL 2003


Digital Signature• A function of both the message AND the signer’s private key

(different for every message)

MESSAGE (LONG)

HASH

SIG

USE SECURE HASH ALGORITHM (SHA) TO PRODUCE HASH (MESSAGE DIGEST)

ENCRYPT HASH USING SIGNER’S PRIVATE KEYPRIVATE KEY

MESSAGE (LONG)SIG

APPEND SIGNATURE TO MESSAGE; SEND BOTHDIGITALLYSIGNEDMESSAGE


FALL 2003


Authentication by Digital Signature

MESSAGE (LONG)

HASHHASH

RECIPIENT USES SHATO COMPUTE HASH

RECIPIENT DECRYPTS SIGWITH SIGNER’S PUBLIC KEY

MESSAGE (LONG)SIG

IF HASHES ARE EQUAL, MESSAGE IS AUTHENTIC.

WHY? IF ANY BIT OF M OR SIG IS ALTERED, HASH CHANGES.

RECIPIENT RECEIVES SIG + MESSAGE

=?


FALL 2003


X.509 Version 2 Certificate

SOURCE: FORD & BAUM,SECURE ELECTRON IC COMMERCE

VERSION # OF X.509

UNIQUE # ASSIGNED BY CA

EXAMPLES: MD5RSA,sha1RSA

USUALLY A DOMAIN NAME

EXAMPLES: RSA


FALL 2003


Cryptography Applications

1. Secrecy: encryption

2. Authentication: digital certificates

3. Integrity: hash functions, message digests

4. Nonrepudiation: digital signatures


FALL 2003


QA&

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20-751 ECOMMERCE TECHNOLOGY FALL 2003 COPYRIGHT © 2003 MICHAEL I. SHAMOS Cryptography.

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