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CENG 520Lecture Note V
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Prime Numbers
prime numbers only have divisors of 1 and self
they cannot be written as a product of other numbers
note: 1 is prime, but is generally not of interest
eg. 2,3,5,7 are prime, 4,6,8,9,10 are not
prime numbers are central to number theory
list of prime number less than 200 is:
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 5961 67 71 73 79 83 89 97 101 103 107 109 113 127
131 137 139 149 151 157 163 167 173 179 181 191
193 197 199
Lecture slides by Lawrie Brown
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Prime Factorisation
to factor a number n is to write it as a product
of other numbers: n=a x b x c
note that factoring a number is relatively hard
compared to multiplying the factors together
to generate the number
the prime factorisation of a number n is when
its written as a product of primes
eg. 91=7x13 ; 3600=24x32x52
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Relatively Prime Numbers & GCD
two numbers a, b are relatively prime if have nocommon divisors apart from 1
eg. 8 & 15 are relatively prime since factors of 8 are 1,2,4,8and of 15 are 1,3,5,15 and 1 is the only common factor
conversely can determine the greatest commondivisor by comparing their prime factorizations andusing least powers
eg. 300=21x31x52 18=21x32 henceGCD(18,300)=21x31x50=6
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Fermat's Theorem
ap-1 = 1 (mod p)
where p is prime and gcd(a,p)=1
also known as Fermats Little Theorem
also have: ap = a (mod p)
useful in public key and primality testing
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Euler Totient Function (n)
when doing arithmetic modulo n
complete set of residues is: 0..n-1
reduced set of residues is those numbers (residues)
which are relatively prime to n
eg for n=10,
complete set of residues is {0,1,2,3,4,5,6,7,8,9}
reduced set of residues is {1,3,7,9}
number of elements in reduced set of residues is
called the Euler Totient Function (n)
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Euler Totient Function (n)
to compute (n) need to count number ofresidues to be excluded
in general need prime factorization, but
for p (p prime) (p)=p-1
for p.q (p,q prime) (p.q)=(p-1)x(q-1)
eg.
(37) = 36(21) = (31)x(71) = 2x6 = 12
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Euler's Theorem
a generalisation of Fermat's Theorem
a(n) = 1 (mod n)
for any a,n where gcd(a,n)=1
eg.a=3;n=10; (10)=4;
hence 34 = 81 = 1 mod 10
a=2;n=11; (11)=10;
hence 210 = 1024 = 1 mod 11 also have: a(n)+1 = a (mod n)
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Primality Testing
often need to find large prime numbers
traditionally sieve using trial division ie. divide by all numbers (primes) in turn less than the
square root of the number
only works for small numbers alternatively can use statistical primality tests based
on properties of primes for which all primes numbers satisfy property
but some composite numbers, called pseudo-primes, alsosatisfy the property
can use a slower deterministic primality test
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Chinese Remainder Theorem
used to speed up modulo computations
if working modulo a product of numbers
eg. mod M = m1
m2
..mk
Chinese Remainder theorem lets us work in
each moduli mi separately
since computational cost is proportional tosize, this is faster than working in the full
modulus M
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Chinese Remainder Theorem
can implement CRT in several ways
to compute A(mod M)
first compute all ai = A mod mi separately
determine constants ci below, where Mi = M/mi
then combine results to get answer using:
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Example
43216*27modulosolutionuniqueahavewouldproblemexampleOur
16mod727mod13
x
x
3*27)5(161
12*511
51*1116
111*1627
?27mod16mod 11
ii mM
473)16*5(*13)27*3(*7 x
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Primitive Roots
from Eulers theorem have a(n)mod n=1
consider am=1 (mod n), GCD(a,n)=1
must exist for m = (n) but may be smaller
once powers reach m, cycle will repeat
if smallest is m = (n) then a is called a primitive
root
ifp is prime, then successive powers ofa "generate"the group mod p
these are useful but relatively hard to find
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Powers mod 19
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Discrete Logarithms
the inverse problem to exponentiation is to find thediscrete logarithm of a number modulo p
that is to find i such that b = ai (mod p)
this is written as i = dloga b (mod p) ifa is a primitive root then it always exists, otherwise
it may not, eg.
x = log3 4 mod 13 has no answer
x = log2 3 mod 13 = 4 by trying successive powers whilst exponentiation is relatively easy, finding
discrete logarithms is generally a hard problem
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Discrete Logarithms mod 19
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Private-Key Cryptography
traditional private/secret/single keycryptography uses one key
shared by both sender and receiver
if this key is disclosed communications arecompromised
also is symmetric, parties are equal
hence does not protect sender from receiverforging a message & claiming is sent by sender
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Public-Key Cryptography
probably most significant advance in the 3000
year history of cryptography
uses two keys a public & a private key
asymmetric since parties are not equal
uses clever application of number theoretic
concepts to function
complements rather than replaces private key
crypto
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Why Public-Key Cryptography?
developed to address two key issues:
key distribution how to have securecommunications in general without having to trust
a KDC with your key digital signatures how to verify a message
comes intact from the claimed sender
public invention due to Whitfield Diffie &Martin Hellman at Stanford Uni in 1976
known earlier in classified community
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Public-Key Cryptography
public-key/two-key/asymmetric cryptography involvesthe use oftwo keys:
a public-key, which may be known by anybody, and can beused to encrypt messages, and verify signatures
a related private-key, known only to the recipient, used todecrypt messages, and sign (create) signatures
infeasible to determine private key from public
is asymmetric because those who encrypt messages or verify signatures cannot
decrypt messages or create signatures
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Public-Key Cryptography
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Symmetric vs Public-Key
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Public-Key Cryptosystems
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Public-Key Applications
can classify uses into 3 categories:
encryption/decryption (provide secrecy)
digital signatures (provide authentication)
key exchange (of session keys)
some algorithms are suitable for all uses,
others are specific to one
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Public-Key Requirements
Public-Key algorithms rely on two keys where:
it is computationally infeasible to find decryption key
knowing only algorithm & encryption key
it is computationally easy to en/decrypt messages whenthe relevant (en/decrypt) key is known
either of the two related keys can be used for encryption,
with the other used for decryption (for some algorithms)
these are formidable requirements which onlya few algorithms have satisfied
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Public-Key Requirements
need a trapdoor one-way function one-way function has
Y = f(X) easy
X = f1
(Y) infeasible
a trap-door one-way function has
Y = fk(X) easy, if k and X are known
X = fk1(Y) easy, if k and Y are known
X = fk1(Y) infeasible, if Y known but k not known
a practical public-key scheme depends on a
suitable trap-door one-way function
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Security of Public Key Schemes
like private key schemes brute force exhaustivesearch attack is always theoretically possible
but keys used are too large (>512bits)
security relies on a large enough difference in
difficulty between easy (en/decrypt) and hard(cryptanalyse) problems
more generally the hard problem is known, but ismade hard enough to be impractical to break
requires the use ofvery large numbers
hence is slow compared to private key schemes
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RSA
by Rivest, Shamir & Adleman of MIT in 1977
best known & widely used public-key scheme
based on exponentiation in a finite (Galois) field over
integers modulo a prime
nb. exponentiation takes O((log n)3) operations (easy)
uses large integers (eg. 1024 bits)
security due to cost of factoring large numbers nb. factorization takes O(e log n log log n) operations (hard)
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RSA En/decryption
to encrypt a message M the sender:
obtains public key of recipient PU={e,n}
computes: C = Me mod n, where 0M
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RSA Key Setup
each user generates a public/private key pair by:
selecting two large primes at random: p, q
computing their system modulus n=p.q
note (n)=(p-1)(q-1)
selecting at random the encryption key e
where 1
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Why RSA Works
because of Euler's Theorem: a(n)mod n = 1 where gcd(a,n)=1
in RSA have: n=p.q
(n)=(p-1)(q-1) carefully chose e & d to be inverses mod (n)
hence e.d=1+k.(n) for some k
hence :
Cd
= Me.d
= M1+k.(n)
= M1
.(M(n)
)k
= M1.(1)k = M1 = M mod n
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RSA Example - Key Setup
1. Select primes:p=17 & q=11
2. Calculate n =pq=17 x 11=187
3. Calculate (n)=(p1)(q-1)=16x10=160
4. Select e: gcd(e,160)=1; choose e=75. Determine d: de=1 mod 160 and d< 160
Value is d=23 since 23x7=161= 10x160+1
6. Publish public key PU={7,187}
7. Keep secret private key PR={23,187}
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RSA Example - En/Decryption
sample RSA encryption/decryption is:
given message M = 88 (nb. 88
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Exponentiation
can use the Square and Multiply Algorithm
a fast, efficient algorithm for exponentiation
concept is based on repeatedly squaring base
and multiplying in the ones that are needed to
compute the result
look at binary representation of exponent
only takes O(log2 n) multiples for number n eg. 75 = 74.71 = 3.7 = 10 mod 11
eg. 3129 = 3128.31 = 5.3 = 4 mod 11
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Exponentiation
c = 0; f = 1
for i = k downto 0
do c = 2 x c
f = (f x f) mod n
if b i == 1 then
c = c + 1
f = (f x a) mod n
return f
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Efficient Encryption
encryption uses exponentiation to power e
hence if e small, this will be faster
often choose e=65537 (216-1)
also see choices of e=3 or e=17
but if e too small (eg e=3) can attack
using Chinese remainder theorem & 3 messages
with different modulii if e fixed must ensure gcd(e,(n))=1
ie reject any p or q not relatively prime to e
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Efficient Decryption
decryption uses exponentiation to power d
this is likely large, insecure if not
can use the Chinese Remainder Theorem
(CRT) to compute mod p & q separately. thencombine to get desired answer
approx 4 times faster than doing directly
only owner of private key who knows valuesof p & q can use this technique
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RSA Key Generation
users of RSA must:
determine two primes at random - p, q
select either e or d and compute the other
primes p,q must not be easily derived frommodulus n=p.q
means must be sufficiently large
typically guess and use probabilistic test exponents e, d are inverses, so use Inverse
algorithm to compute the other
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RSA Security
possible approaches to attacking RSA are:
brute force key search - infeasible given size of
numbers
mathematical attacks - based on difficulty ofcomputing (n), by factoring modulus n
timing attacks - on running of decryption
chosen ciphertext attacks - given properties of RSA
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Factoring Problem
mathematical approach takes 3 forms: factor n=p.q, hence compute (n) and then d
determine (n) directly and compute d
find d directly
currently believe all equivalent to factoring have seen slow improvements over the years
as of May-05 best is 200 decimal digits (663) bit with LS (LatticeSieve)
biggest improvement comes from improved algorithm cf QS(Quadratic Sieve) to GHFS(Generalized Number Field Sieve) to
LS
currently assume 1024-2048 bit RSA is secure ensure p, q of similar size and matching other constraints
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Progress in Factoring
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Progress inFactoring
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Diffie-Hellman Key Exchange
first public-key type scheme proposed
by Diffie & Hellman in 1976 along with the
exposition of public key concepts
note: now know that Williamson (UK CESG)
secretly proposed the concept in 1970
is a practical method for public exchange of a
secret key
used in a number of commercial products
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Diffie-Hellman Key Exchange
a public-key distribution scheme
cannot be used to exchange an arbitrary message
rather it can establish a common key
known only to the two participants
value of key depends on the participants (and theirprivate and public key information)
based on exponentiation in a finite (Galois) field
(modulo a prime or a polynomial) - easy security relies on the difficulty of computing discrete
logarithms (similar to factoring) hard
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Diffie-Hellman Setup
all users agree on global parameters:
large prime integer or polynomial q
a being a primitive root mod q
each user (eg. A) generates their key
chooses a secret key (number): xA < q
compute their public key: yA = axA mod q
each user makes public that key yA
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Diffie-Hellman Key Exchange
shared session key for users A & B is KAB:KAB = a
xA.xB mod q
= yAxB mod q (which B can compute)
= yBxA mod q (which A can compute)
KAB is used as session key in private-key encryptionscheme between Alice and Bob
if Alice and Bob subsequently communicate, they will
have the same key as before, unless they choosenew public-keys
attacker needs an x, must solve discrete log
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Diffie-Hellman Example
users Alice & Bob who wish to swap keys:
agree on prime q=353 and a=3
select random secret keys:
A chooses xA=97, B chooses xB=233 compute respective public keys:
yA=397
mod 353 = 40 (Alice)
yB=3233
mod 353 = 248 (Bob)
compute shared session key as: KAB= yB
xA mod 353 = 24897
= 160 (Alice)
KAB= yAxB mod 353 = 40
233= 160 (Bob)
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Key Exchange Protocols
users could create random private/public D-Hkeys each time they communicate
users could create a known private/public D-H
key and publish in a directory, then consultedand used to securely communicate with them
both of these are vulnerable to a meet-in-the-Middle Attack
authentication of the keys is needed
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Man-in-the-Middle Attack
1. Darth prepares by creating two private / public keys2. Alice transmits her public key to Bob
3. Darth intercepts this and transmits his first public key to Bob.Darth also calculates a shared key with Alice
4. Bob receives the public key and calculates the shared key (withDarth instead of Alice)
5. Bob transmits his public key to Alice
6. Darth intercepts this and transmits his second public key toAlice. Darth calculates a shared key with Bob
7. Alice receives the key and calculates the shared key (with Darthinstead of Bob)
Darth can then intercept, decrypt, re-encrypt, forward allmessages between Alice & Bob
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ElGamal Cryptography
public-key cryptosystem related to D-H
so uses exponentiation in a finite (Galois)
with security based difficulty of computing
discrete logarithms, as in D-H
each user (eg. A) generates their key
chooses a secret key (number): 1 < xA < q-1
compute their public key: yA = axA mod q
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ElGamal Message Exchange
Bob encrypt a message to send to A computing
represent message M in range 0
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ElGamal Example
use field GF(19) q=19 and a=10 Alice computes her key:
A chooses xA=5 & computes yA=105mod 19 = 3
Bob send message m=17 as (11,5) by
chosing random k=6
computing K = yAkmod q = 3
6mod 19 = 7
computing C1 = akmod q = 10
6mod 19 = 11;
C2 = KM mod q = 7.17 mod 19 = 5
Alice recovers original message by computing: recover K = C1
xA mod q = 115mod 19 = 7
compute inverse K-1 = 7-1 = 11
recover M = C2 K-1 mod q = 5.11 mod 19 = 17
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Elliptic Curve Cryptography
majority of public-key crypto (RSA, D-H) use
either integer or polynomial arithmetic with
very large numbers/polynomials
imposes a significant load in storing and
processing keys and messages
an alternative is to use elliptic curves
offers same security with smaller bit sizes
newer, but not as well analysed
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Real Elliptic Curves
an elliptic curve is defined by an equation intwo variables x & y, with coefficients
consider a cubic elliptic curve of form
y2 =x3 + ax+ b
where x,y,a,b are all real numbers
also define zero point O
consider set of points E(a,b) that satisfy
have addition operation for elliptic curve
geometrically sum of P+Q is reflection of theintersection R
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Real Elliptic Curve Example
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Finite Elliptic Curves
Elliptic curve cryptography uses curves whose
variables & coefficients are finite
have two families commonly used:
prime curves Ep(a,b) defined over Zp use integers modulo a prime
best in software
binary curves E2m(a,b) defined over GF(2n) use polynomials with binary coefficients
best in hardware
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Elliptic Curve Cryptography
ECC addition is analog of modulo multiply
ECC repeated addition is analog of moduloexponentiation
need hard problem equiv to discrete log Q=kP, where Q,P belong to a prime curve
is easy to compute Q given k,P
but hard to find k given Q,P known as the elliptic curve logarithm problem
Certicom example: E23(9,17)
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ECC Diffie-Hellman
can do key exchange analogous to D-H
users select a suitable curve Eq(a,b)
select base point G=(x1,y1)
with large order n s.t. nG=O
A & B select private keys nA
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ECC Encryption/Decryption
several alternatives, will consider simplest
must first encode any message M as a point on the
elliptic curve Pm
select suitable curve & point G as in D-H
each user chooses private key nA
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ECC Security
relies on elliptic curve logarithm problem
fastest method is Pollard rho method
compared to factoring, can use much smaller
key sizes than with RSA etc
for equivalent key lengths computations are
roughly equivalent
hence for similar security ECC offers significant
computational advantages
Comparable Key Sizes for Equivalent
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Comparable Key Sizes for Equivalent
Security
Symmetric
scheme
(key size in bits)
ECC-based
scheme
(size ofn in bits)
RSA/DSA
(modulus size in
bits)
56 112 512
80 160 1024
112 224 2048
128 256 3072
192 384 7680
256 512 15360
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