Public key cryptography

Public key cryptography

RSA AlgorithmDiffie-Hellman key exchange

Public-Key Cryptography• traditional private/secret/single key

cryptography uses one key • Key is shared by both sender and

receiver • if the key is disclosed communications

are compromised • also known as symmetric, both parties

are equal – hence does not protect sender from receiver

forging a message & claiming is sent by sender

Disadvantages of Classical Cryptography

• Requires secure transmission of key value

• Requires a separate key for each group of people that wishes to exchange encrypted messages (readable by any group member)• For example, to have a separate key for

each pair of people, 100 people would need 4950 different keys.

• Cumbersome Handshaking involved in KDC

• probably most significant advance in the 3000 year history of cryptography

• uses two keys – a public key and a private key

• asymmetric since parties are not equal • uses clever application of number theory

concepts to function• complements rather than replaces

private key cryptography

Public-Key Cryptography

Why Public-Key Cryptography?

• developed to address two key issues:– key distribution – how to have secure

communications 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 U. in 1976

• Key Distribution requires that– Two communicants already share a

key,which somehow has been distributed to them

– Use of a key distribution center

• Electronic messages and documents would need the equivalent of signatures used in paper documents.

• public-key/two-key/asymmetric cryptography involves the use of two keys: – a public-key, which may be known by

anybody, and can be used to encrypt messages, and verify signatures

– a private-key, known only to the recipient, used to decrypt messages, and sign (create) signatures

• is asymmetric because– those who encrypt messages or verify

signatures cannot decrypt messages or create signatures

Public-Key Cryptography

Public-Key CryptographyEncryption

• Alice generates a key value (usually a number or pair of related numbers) which she makes public.

• Alice uses her public key (and some additional information) to determine a second key (her private key).

• Alice keeps her private key (and the additional information she used to construct it) secret.

• Bob (or Carol, or anyone else) can use Alice’s public key to encrypt a message for Alice.

• Alice can use her private key to decrypt this message.

• No-one without access to Alice’s private key (or the information used to construct it) can easily decrypt the message

An Example: Internet Commerce

• Bob wants to use his credit card to buy some books from Alice over the Internet.

• Alice sends her public key to Bob.• Bob uses this key to encrypt his

credit-card number and sends the encrypted number to Alice.

• Alice uses her private key to decrypt this message (and get Bob’s credit-card number).

Authentication

Public-Key Cryptosystems

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

Algorithm

Encryption / Decryption

Digital Signature

Key Exchange

RSA Y Y Y

Elliptic curve

Y Y Y

Diffie-Hellman

N N Y

DSS N Y N

THE RSA ALGORITHM

• Public key algorithm invented in 1977 by Ron Rivest, Adi Shamir & Leonard Adleman of MIT

• Best known & widely used public-key scheme

• Supports Encryption & Digital signatures

• Gets its security from integer factorization problem

Key Generation1. Generate two large random primes, p and q, of

approximately equal size such that their product n = pq is of the required bit length, e.g. 1024 bits.

2. Compute n = pq and phi φ(n) = (p-1)(q-1).3. Choose an integer e, 1 < e < φ(n), such that

gcd(e, φ) = 1. 4. Compute the secret exponent d, 1 < d < φ(n),

such that ed ≡ 1 mod φ(n). 5. The public key is (n, e) and the private key is (n,

d). Keep all the values d, p, q and φ(n) secret.• n is known as the modulus.• e is known as the public exponent or encryption exponent

or just the exponent.• d is known as the secret exponent or decryption

exponent.

Key Generation -detail

1. Use a random process to select two large prime numbers p and q. Compute the product n = p*q. This number is called the modulus, and is made publicly available.– RSA currently recommends a modulus

that’s at least 768 bits long.

2. Also compute the Euler totient φ(n) = φ(pq)= (p-1)*(q-1). Keep this number (as well as p and q) secret.

GCD & Relatively prime• Two numbers a and b which have no

common factors other than one are said to be coprime or relatively prime. For example, 4 and 9 are coprime but 15 and 25 are not.

• The greatest common divisor of two integers a and b is the largest integer that divides both numbers and is denoted by ‘gcd(a, b)’. For example, gcd(25, 15) = 5 and gcd(4, 9) = 1.

• a and b are coprime if and only if gcd(a, b) = 1.

Euler Totient

φ(n) is the no. of positive integers less than n and relatively prime to n.

e.g. φ(12)=4 as the four integers {1,5,7,11} are coprime to 12.

φ(7)=6 as the 6 integers {1,2,3,4,5,6} are coprime to 7.

i.e. for any prime p, φ(p) = p-1

Key Generation -detail3. Randomly choose a public key e that

has no factors in common with φ(n)= (p-1)*(q-1). so that gcd(e, φ(n)) = 1, 1 < e < φ(n);[Euclid’s algorithm: gcd(a,b)=gcd(b,r) where r=a mod b e.g. gcd(18,12)=gcd(12,6)=gcd(6,0)=6 i.e. a = qb + r]

4. Solve following equation to find decryption key d ,e.d ≡ 1 mod φ(n) and d < n; where d is the multiplicative inverse of e in mod φ(n);d ≡ e-1mod φ(n)

• ‘mod’ as a congruence relation: The notation ‘a ≡ b (mod n)’ means a and b have the same remainder when divided by n, or, equivalently,

• (a) n|a − b, or• (b) a − b = nk for some integer k .• We say that a is congruent to b modulo n,

where n is the modulus of the congruence. The two ways of using ‘mod’ are related:

• a ≡ b (mod n) <==> a mod n = b mod n.

• Compute a private key d so that e*d leaves a remainder of 1 when divided by φ(n) . Find a value for d such that φ(n) divides (ed-1).Note that d is easy to compute only if one knows the value of φ(n). This is essentially the same as knowing the values of p and q.

Using RSA• publish their public encryption key:

KU={e,n} • keep secret private decryption key:

KR={d,p,q} • to encrypt a message M the sender:

– obtains public key of recipient KU={e,n}

– computes: C=Me mod n, where 0≤M<n• to decrypt the ciphertext C the

owner:– uses their private key KR={d,p,q} – computes: M=Cd mod n

RSA Example1. Select primes: p =17 & q =112. Compute n = pq =17×11=1873. Compute ø(n)=(p–1)(q-

1)=16×10=1604. Select e : gcd(e,160)=1; choose e =75. Determine d: de=1 mod 160 and d <

160 Value is d=23 since 23×7=161= 1×160+1

6. Publish public key KU={7,187}7. Keep secret private key KR={23,17,11}

• Use a property of modular arithmetic:(a x b) mod n = [ (a mod n) x (b mod n) ]

mod n

• modulo of a product = modulo of the product of its multipliers’ modulos

• Strategy - reduce xy into xaxbxc where y=abc using powers of 2 for a, b, c . Then apply above property

• given message M = 88 (nb. 88<187)

• Encryption:• C = 887 mod 187 = [(884 mod 187)( 882

mod 187)( 881 mod 187)] mod 187881 mod 187 =88882 mod 187 = 7744 mod 187 = 77884 mod 187 = 59,969,536 mod 187=132

= (88 x 77 x 132) mod 187 = 894,432 mod 187 = 11•

• Decryption:• M = 1123 mod(187) = [(111

mod 187)( 112 mod 187)( 114 mod 187)( 118 mod 187)( 118 mod 187)] mod (187)

= (11 x 121 x 55 x 33 x 33) mod(187) = 88

• Q. p=11,q=3 M=7

Comparison

DES RSA

Symmetric Key Encryption algorithm

Block size – 64 bits

Contains complex operations(S-boxes)

Bit wise operations used (Shift and

XOR)

There are some operations on key

16 rounds of same operations

Calculations are based on permutation

Public Key Encryption algorithm

Block size – 1024 bits

Only mathematical calculations used

No bit wise operations

No operations on key.

Consists a single round operation

Calculations are based on ‘primitive

logarithm’.

Diffie-Hellman Key Exchange

• Whittfield Diffie and Martin Hellman are called the inventors of Public Key Cryptography.

• Diffie-Hellman Key Exchange is the first Public Key Algorithm published in 1976.

Discrete Logarithms• What is a logarithm?

• log10100 = 2 because 102 = 100

• In general if logmb = a then ma = b

• Where m is called the base of the logarithm

• A discrete logarithm can be defined for integers only

• In fact we can define discrete logarithms mod p also where p is any prime number

Primitive Roots

• If x n = a then a is called the n-th root of x

• For any prime number p, if we have a number a such that powers of a mod p generate all the numbers between 1 to p-1 then a is called a Primitive Root of p.

• Then for any integer b and a primitive root a of prime number p we can find a unique exponent i such that

b = a i mod p• The exponent i is referred to as the

discrete logarithm or index, of b for the base a.

• 2 as primitive root

m 1 2 3 4 5 6 7 8 9 10

2m mod 11 2 4 8 5 10 9 7 3 6 1

m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

2m mod 19 2 4 8 16 13 7 14 3 6 17 15 11 3 6 12 5 10 1

3m mod 19 3 9 8 5 15 7 2 6 18 16 10 11 14 4 12 17 13 1

Diffie-Hellman Algorithm

• Five Parts1. Global Public Elements2. User A Key Generation3. User B Key Generation4. Generation of Secret Key by User A5. Generation of Secret Key by User B

Global Public Elements

• p Prime number• g g < p and g is a

primitive root of p• The global public elements are also

sometimes called the domain parameters

User A Key Generation

• Select private XA XA < p

• Calculate public YA YA = g XA mod p

User B Key Generation

• Select private XB XB < p

• Calculate public YB YB = g XB mod p

Generation of Secret Key by User A

• K = (YB)XA mod p

Generation of Secret Key by User B

• K = (YA)XB mod p

Diffie-Hellman Key Exchange

• Alice and Bob agree upon and make public two• numbers g and p, where p is a prime and g is a

primitive root mod p. Note: Anyone has access to these numbers.

• The Exchange:1. Alice chooses a random number a (secret)and

computes u =g a (mod p), and sends u to Bob.2. Bob chooses a random number b (secret) and

computes v= g b (mod p), and sends v to Alice.3. Bob computes the key kb using secret b= u b mod

p4. Alice computes the key ka using secret a=v a mod p

Now, both Alice and Bob have the same key,namely ka = kb = gab (mod p).

• kb = u b mod p

=(g a mod p)b mod p =(g a)b mod p by modular

arithmetic

= g ab mod p =(g b)a mod p =(g b mod p)a mod p = v a mod p = ka

Example• Suppose Alice and Bob agree to• use p = 47 and g = 5.• Alice chooses a number between

0 and 46,• say a = 18.• Bob chooses a number between 0

and 46,• say b = 22.

Example• Alice publishes ga (mod p), i.e.• u = 518 (mod 47) = 2.• Bob publishes gb (mod p), i.e.• v = 522 (mod 47) = 28.• If Alice wants to know the secret

key k, she takes Bob’s public number,v =28 and raises it to her private number, a = 18 (taking the result mod 47).

• This gives her: 2818 (mod 47) = 24.

Example

• If Bob wants to know the secret key, he takes Alice’s public number, u = 2, and raises it to his private number, b = 22 (taking the result mod 47).

• This gives him: 222 (mod 47) = 24.• Thus, Alice and Bob have agreed

upon a secret key, k = 24.

Security of DH exchange• Opponent Eve has the following

ingredients p,g,u,v to work with• If Eve wants to compute k, then she

would need either a or b.• Otherwise, Eve would need to solve a Discrete Logarithm Problem.

– There is no known algorithm to accomplish this in a reasonable amount of time.

– E.g. for attacking user B’s secret key,the opponent must compute b=indg,p(v)

Man in the middle attack• Susceptibility: If Eve can intercept u

and v, it is possible for her to substitute her own u’ and v’.

• If she can intercept all communication• between Alice and Bob, then she can

substitute her own messages.• In 1992, the exchange was modified

to prevent the man-in-the-middle attack described above.

Example for m-i-t-m-a

• Alice wants to talk to Bob, sends her public key to Bob.

• But Eve intercepts it, and replaces Alice’s public key with hers. She sends this to Bob.

• Bob thinks Alice wants to talk to him. He sends his public key to her.

• But Eve intercepts and replaces!• Then Eve sets up shared keys with

both!

What to do?

• If the public keys are certified (e.g., VeriSign) then Alice and Bob verifies that they got the right public keys!

• User Authentication: Alice encrypts the message, m, with her private key a, call it ma. ma = Ea (m)

• Alice encrypts ma with Bob’s public key, v, and sends the message to Bob. ma v = Ev (ma)

• Bob recovers ma using his private key b and recovers m by using Alice’s public key u. m = Du(Db(mav))

• Thus, Bob is sure that only Alice could have sent the message.