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763 ELECTRONIC PAYMENT SYSTEMS
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COPYRIGHT © 2002 MICHAEL I. SHAMOS
Electronic Payment Systems
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763
Lecture 6
Epayment Security II
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
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
SOURCE: STEIN,
WEB SECURITY
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763 ELECTRONIC PAYMENT SYSTEMS
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COPYRIGHT © 2002 MICHAEL I. SHAMOS
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’S
CLEAR
TEXT
ALICE’S
CODED
TEXT
ALICE’S
CODED
TEXT
ALICE’S
CLEAR
TEXT
TRANSM ISSION
BOB DECRYPTS WITH
HIS PRIVATE KEY
ALICE ENCRYPTS WITH
BOB’S PUBLIC KEY
BOB’S
PUBLIC
KEY
BOB’S
PRIVATE
KEY
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
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’S
CLEAR
TEXT
ALICE’S
CODED
TEXT
ALICE’S
CODED
TEXT
ALICE’S
CLEAR
TEXT
TRANSM ISSION
BOB DECRYPTS WITH
ALICE’S PUBLIC KEY
ALICE ENCRYPTS WITH
HER PRIVATE KEY
ALICE’S
PRIVATE
KEY
ALICE’S
PUBLIC
KEY
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
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’S
CODED
TEXT
ALICE’S
CODED
TEXT
(AUTHENTICATED)
ALICE’S
CLEAR
TEXT
BOB DECRYPTS WITH
ALICE’S PUBLIC KEY
ALICE ENCRYPTS WITH
HER PRIVATE KEY
ALICE ENCRYPTS WITH
BOB’S PUBLIC KEY
ALICE’S
CODED AND
SIGNED TEXT
ALICE’S
CODED AND
SIGNED TEXT
T
R
A
N
S
M
I
T
ALICE’S
CLEAR TEXT
(DECRYPTED AND
AUTHENTICATED)
BOB DECRYPTS WITH
HIS PRIVATE KEY
BOB’S PUBLIC
ALICE’S PUBLIC
BOB’S PRIVATE
ALICE’S PRIVATE
4 KEYS
NEEDED:
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
One

Way Trapdoor Function
•
A function that is easy to compute
•
Computationally difficult to invert
without knowing the
secret
(the “trapdoor”)
•
Example:
f
(x, y) = x•y
•
Given
f
(x, y), it is difficult to find either x or y
•
Given
f
(x, y) and x (the secret), it is easy to find y
•
Any one

way trapdoor function can be used in public

key cryptography.
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763 ELECTRONIC PAYMENT SYSTEMS
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COPYRIGHT © 2002 MICHAEL I. SHAMOS
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
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
RSA Encryption
•
Select two large prime numbers p, q (e.g. 1024 bits)
•
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
–
Then x
e
•d
= x (mod n) (by Fermat’s “Little” Theorem)
•
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
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763 ELECTRONIC PAYMENT SYSTEMS
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COPYRIGHT © 2002 MICHAEL I. SHAMOS
MULTIPLICATION
MOD 7
INVERSE OF 5 IS 3
Multiplicative Inverses
Over Finite Fields
•
The inverse e

1
of a number e satisfies e

1
• e = 1
•
The inverse of 5 is 1/5
•
If we only allow numbers from 0 to n

1 (mod n), then for special
values of n, each e has a unique inverse
6
•
2 = 12
WHEN DIVIDED BY 7
GIVES REMAINDER 5
EACH ROW EXCEPT
THE ZERO ROW
HAS EXACTLY ONE 1
EACH ELEMENT HAS
A UNIQUE INVERSE
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763 ELECTRONIC PAYMENT SYSTEMS
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COPYRIGHT © 2002 MICHAEL I. SHAMOS
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)
•
DEMO
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
RSA Example
p = 61; q = 53
n = pq = 3233 (
modulus
, can be given to others)
e = 17 (
public exponent
, can be given to others)
d = 2753 (
private exponent
, kept secret!)
PUBLIC KEY = (3233, 17)
PRIVATE KEY = (3233, 2753)
To encrypt 123, compute 123
17
(mod 3233) =
337587917446653715596592958817679803 mod 3233 = 855
To decrypt 855, compute 855
2753
(mod 3233) = 123
(intermediate value has 8072 digits)
SOURCE:
FRANCIS LITTERIO
37 digits
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Trapdoor Functions for Cryptogrpahy
•
Any one

way trapdoor function
f
(
x
) can be used for
public

key cryptography
•
Alice wants to send message m to Bob
•
Bob’s public key
e
is a parameter to the trapdoor
function
f
e
(
x
) (the inverse
f
e

1
(
x
) is easy to compute
knowing Bob’s private key d but difficult without
d
)
•
Alice computes
f
e
(
m
), sends it to Bob
•
Bob computes
f
e

1
(
f
e
(
m
)) =
m
(easy if
d
is known)
•
Eavesdropper Eve can’t compute
m
=
f
e

1
(
f
e
(
m
))
without the trapdoor
d
to find the inverse
f
e

1
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Digital Signatures
•
A handwritten signature is a function of the signer
only, not the message
•
Handwritten signatures can be copied and forged
•
The digital equivalent of a handwritten signature
would be
useless
in eCommerce
•
Must be able to
–
Compare it with the “real” signature; AND
–
Must be sure it isn’t copied or forged
•
How can A prove his identity over the Internet?
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Digital Signatures
•
A digital signature is a function of
both
the
signer
and
the
message
•
A digital signature is a digest of the message
encrypted with the signer’s private key
MESSAGE M (LONG)
HASH
SIG
USE SECURE HASH ALGORITHM (SHA)
TO PRODUCE HASH (MESSAGE DIGEST)
ENCRYPT HASH USING SIGNER’S PRIVATE KEY
PRIVATE KEY
OF MR. A
THIS IS THE DIGITAL SIGNATURE
OF MR. A ON MESSAGE M
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Authentication by Digital Signature
MESSAGE (LONG)
HASH
HASH
RECIPIENT USES SHA
TO COMPUTE HASH
RECIPIENT DECRYPTS SIG
WITH 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
=?
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Digital Signature
•
Message digest encrypted with signer’s private key
MESSAGE (LONG)
SIG
APPEND SIGNATURE TO MESSAGE; SEND BOTH
MESSAGE (LONG)
HASH
SIG
USE SHA TO PRODUCE HASH (MESSAGE DIGEST)
ENCRYPT HASH WITH SIGNER’S PRIVATE KEY
Recipient decrypts SIG with signer’s public key.
Recipient computes the message digest.
If it matches the SIG, the SIG is genuine
AND the message has not been altered!
PRIVATE KEY
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Discrete Logarithms
•
If a
b
= c, we say that log
a
c = b
•
Example: 2
32
= 4294927296 so log
2
(4294927296) = 32
•
Computing a
b
and log
a
c are both easy for real numbers
•
In a finite field, it is easy to calculate c = a
b
mod p but
given c, a and p it is
very difficult
to find b
•
This is the “discrete logarithm” problem
•
Analogy: Given x it is easy to find two real numbers y, z
such that x = y
•
z
•
Given an integer n it is hard to find two
integers
p, q
such that n = p
•
q
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Diffie

Hellman Key Exchange
•
Object: allow Alice and Bob to exchange a secret key
•
Protocol has two public parameters: a prime p and a
number g < p such that given 0 < n < p there is some k
such that g
k
= n (g is called a
generator
)
•
Alice and Bob generate random private values a, b
between 1 and p

2
•
Alice’s public value is g
a
(mod p); Bob’s is g
b
(mod p)
•
Alice and Bob share their public values
•
Alice computes (g
b
)
a
(mod p) = g
ba
•
Bob computes (g
a
)
b
(mod p) = g
ab
=
g
ba
•
Let key = g
ab
. Now both Alice and Bob have it.
•
No one else can compute it

they don’t know a or b
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
El Gamal Encryption
•
Based on the discrete logarithm
•
Bob’s public key is (p, q, r)
•
Bob’s private key is s such that r = q
s
mod p
•
Alice sends Bob the message m by picking a random
secret number k and sending
(a, b) = (q
k
mod p, mr
k
mod p)
•
Bob computes
b (a
s
)

1
mod p = mr
k
(q
ks
)

1
= mq
ks
(q
ks
)

1
= m
•
(Bob knows s; nobody else can do this)
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Elliptic Curve Cryptography (ECC)
•
An elliptic curve is the set of points (x, y) satisfying
y
2
+ axy + by = x
3
+ cx
2
+ dx + e
x
y
An elliptic curve has the property that a
line drawn between two points of the curve
intersects the curve at a single point.
(Warning: need to include the point at infinity.)
This allows us to define P + Q so that the sum
is always another point on the curve.
If the sum P + Q is always on the curve, so are
the points P, P + P, P + P + P, . . .
= P, 2P, 3P, 4P, . . .
ONLINE TUTORIAL
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763 ELECTRONIC PAYMENT SYSTEMS
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COPYRIGHT © 2002 MICHAEL I. SHAMOS
Elliptic Curve Operations
SOURCE:
INTEGRITY SCIENCES
The point at infinity
O
is an identity element
for addition
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Elliptic Curves Over Finite Fields
•
Select a large prime number p
•
Choose two non

negative integers a and b with
4a
2
+ 27b
2
†
0 (mod p)
•
The pairs (x, y) with x, y < p that satisfy
y
2
= x
3
+ ax + b (mod p)
are the
elliptic group mod p
–
addition is closed and associative (x + y) + z = x + (y + z)
–
there is an identity element
O
such that x +
O
= x
–
every element x has an inverse x

1
such that x + x

1
=
O
•
If y = k x (mod p), then given k and x it is
easy
to find y
but given x and y it is
computationally hard
to find k
•
So elliptic curves can be used for cryptography
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Elliptic Curves for El Gamal
•
Multiplication in the elliptic group corresponds to exponentiation
of real numbers
•
Solving y = k x (mod p) for k in the elliptic group is similar to
solving c = a
b (
mod p) for b in El Gamal (discrete logarithm)
•
Choose a special point g of the group (called a generator)
•
Bob’s private key is s; Bob’s public key is (g, s
g)
•
A plaintext message m is transformed to a point x in the group
•
Alice encrypts x by picking a random value k and sending
(k
g, x + k
s
g)
•
Bob decrypts by computing (x + k
s
g)

(k
g)
s = x
Alice sent him these
Bob knows s (his private key)
g and sg are public; Alice knows x and k
Can’t find s
from g and sg
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Security of ECC versus RSA
GRAPHIC:
RICHARD SOUTHERN
ECC Advantages
1. The elliptic curve logarithm
problem is harder than the
discrete logarithm problem.
2. Key size in ECC is much
smaller for a given security
level.
3. ECC is complicated; fewer
people understand it.
4. ECC is not patented.
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763 ELECTRONIC PAYMENT SYSTEMS
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COPYRIGHT © 2002 MICHAEL I. SHAMOS
Birthday Attacks
•
Dave’s birthday is Jan. 29. How many people must
be in a room for the probability to be > 1/2 that
someone else was born on Jan. 29?
•
Probability that 1 person was not born on Jan. 29 =
364/365.
•
Probability that
n
people were not born on Jan. 29 is
p(
n
) = (364/365)
n
. Now choose
n
so that p(
n
) < 0.5
•
log p(
n
) <
n
log (364/635)
•
n
> log(1/2)/log(364/365)
253
•
If
n
= 183 (half of 366), p(
n
) = 0.6053. Less then
40% chance that someone else has same birthday
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Birthday Probabilities
•
Suppose a year has
d
days. How many people must
be in a room for the probability to be > 1/2 that some
pair of people have the same birthday?
•
Label the people 1 …
n
•
Probability that person
i
has no birthday in common
with people 1 …
i

1 is (
d

i + 1
)/d, so
•
If
d
= 365 and
n
= 23, p(
n
)
0.4927
•
If
d
= 365 and
n
= 50, p(
n
)
0.0296
•
For large
d
, taking
n
1.17 gives
p(
n
)
> 1/2
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763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Attacking Hash Algorithms
•
If two strings M and M* can be found such that
H
(M) =
H
(M*) then a hash algorithm can be
compromised
•
Let M = PO for $100; M* = PO for $100,000
•
John digitally signs
H
(M), so it can’t be altered!
•
If
H
(M*) =
H
(M) then we can “prove” in court that John
signed the $100,000 PO
•
Birthday attack: If the hash length is
b
bits, then
d
= 2
b
;
= 2
b/2
•
Try about 2
b/2
small variations of the message. Prob.
~ 50% we will find one that hashes to the same value
•
If the digest is 64 bits, try 2
32
variations. Possible!
20

763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
Major Ideas
•
Digital signature = message digest encrypted with signer’s private
key
•
Dual signature: two people sign a document without being able to
read the other person’s content
•
Blind signature: one person signs a document without being able
to read it
•
Any trapdoor function can be used for public

key cryptography
•
Great care must be used with public

key systems to avoid
protocol failure (allowing cracking through mistakes)
•
Elliptic

curve cryptography (ECC) is replacing RSA
–
Shorter keys for the same level of security
20

763 ELECTRONIC PAYMENT SYSTEMS
FALL 2002
COPYRIGHT © 2002 MICHAEL I. SHAMOS
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