Public Key Cryptography

All the secret key algorithms  & the hash algorithms do the same thing

but public key algorithms look very different from each other.

What is common among them is each participant has two  key: public & private,

and most of them are based on modular arithmetic.

Modular Arithmetic

x mod n is the remainder of x when divided by n.

e.g., 8 mod 10 = 8,        18 mod 10 = 8,         24 mod 10 = 4
       8 mod 7 = 1,          18 mod 7 = 4,           24 mod 7 = 3

ü Addition:

Example: addition mod 10

8 + 8 = 6,

1 + 9 = 0,

7 + 6 = 3

See Fig. 6-1 for addition  mod 10 Table:

Encryption: Addition mod 10 can be used for encryption of digits.

Add k, a secret key between 1-9 to each digit. 

Example: if k = 7, then 1987 is encrypted to 8654.

Decryption: Add -k, the additive inverse of k, to each digit.

An additive inverse of x is the number you'd have to add to x to get 0.

Example:  if k = 7, then -k is 3 since 7+3 = 0

Thus 8654 will is decrypted to 1987.

In the above table (Fig. 6-1), each "0" is the intersection of k and -k,

e.g., 0 is the intersection of 3 and 7.

ü Multiplication:

Example: multiplication mod 10 :

 8 x 8 = 4, 1 x 9 = 9 , 7 x 6 = 2

See Fig. 6-2 for multiplication  mod 10 Table:

Encryption: multiplication by 1, 3, 7, 9 works as a cipher since it performs 1-1 mapping.

Example: if k = 7, then 1987 is encrypted to 7369

Decryption: is done by multiplying each digit by k^-1 , the multiplicative inverse of k.

It is the number to multiply by k to get 1.

Example:  if k = 7, then k^-1 is 3 since 7x3 = 1

In the above table (Fig. 6-2), each "1" is the intersection of k and k^-1.

Note that only {1,3,7,9} have multiplicative inverse mod 10.

à What is so special about the set {1,3,7,9}?

These numbers are relatively prime to 10,

i.e., they do not share with 10 any common factors other than 1.

Note that 9 is not a prime number but it is relatively prime to 10.

à Only numbers that are relatively prime to n have multiplicative inverses.

ü How many numbers less than n are relatively prime to n?

This quantity is referred to as:

Ø(n) and is called the totient function.

ü If n is prime: then {1,2, ..., n-1} are all relatively prime and thus Ø(n) = n-1.

ü If n = p.q where p and q are two distinct primes, then Ø(n) = (p-1)(q-1).

Why is that?

There are p multiples of q, q multiples of p and 0 that are not relatively prime to n=pq.

Thus Ø(n) = pq –(p+q-1) = (p-1)(q-1)

Example:  for n = 10 = 2.5, Ø(10) =(2-1).(5-1)=1.4=4,
which is the set {1,3,7,9}.

ê Exponentation:

Example: exponentiation mod 10

      4 ² = 6, 8 ⁸ = 6, 1⁹  = 1 , 7⁶  = 9

See Fig. 6-3  for exponentiation  mod 10 Table:

Amazing fact about Ø(n):

x ^m mod n = x ^{m mod Ø(n)} mod n

Since  Ø(10)=4, in Fig. 6-3:

x ^m mod 10 = x ^{m mod 4} mod 10

Thus the i^th column is identical to the i+4^th column,

e.g., 1^st =5^th =9^th and 3^rd = 7^th =11^th.

Special case:

if m = 1 mod Ø(n),  then  for any number x:

x ^m mod n = x mod n.

Example: For n =10, Ø(10)=4

Since 9  = 1 mod 4:

3⁹ mod 10 = 3 mod 10 = 3 &   6⁹ mod 10 = 6 mod 10 = 6

An exponentiative  inverse of e is the number  d  such that: e.d = 1 mod Ø(n)

Example: For n= 10, Ø(10)=4:

e=3 and d=7 are exponentiative inverses since: 3.7 = 21 = 1 mod 4

Encrypt/Decrypt:

· To encrypt m:  compute  c = m^e mod n

· To decrypt c:   compute  m = c^d mod n

Example:

encrypt m = 8: c = 8³ = 2

decrypt c = 2:  m = 2⁷ = 8

Sign/Verify:

· To sign m: compute s = m^d mod n

· To verify s: compute m = s^e mod n

Example:

sign m = 8: s = 8⁷ = 2

verify s = 2:  m = 2³ = 8

In public key cryptography:

<e, n> is public key

&

<d, n> is private key

Note that:

If n = p.q where p and q are two distinct primes, then Ø(n) = (p-1)(q-1).

The relation between e and d: e.d = 1 mod Ø(n)

To compute d by knowing e and n you must know Ø(n) which is

very hard to factor n to p & q

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RSA: Rivest, Shamir & Adleman

Key length:

Variable (long for security, short for efficiency).

Most common values are 512 & 1024 bits.

Block size:

plain text length is variable but less than key length &
cipher text length  equals key length.

Thus RSA is used for encrypting small amount of data, e.g., a secret key

and then use the secret key cryptography for encrypting/decrypting large amount of data.

RSA Algorithm:

Generate public & private keys pair:

1.  Choose two distinct large primes p and q.
     (Typically 256 bits each & keep them secret).

2.  Compute n = p.q & Ø(n) = (p-1)(q-1).
(Note that it is very hard to factor n into p & q ).

3.  Choose a number e that is relatively prime to Ø(n).

4.  Find a number d that is the exponentiative inverse of e
     i.e., e.d = 1 mod Ø(n).

5.  The public key <e,n>  &  the private key <d,n>.

Encrypt/Decrypt:

To encrypt a message m ( less than n): c = m^e mod n

To decrypt c: m = c^d mod n

This works since:

c^d mod n = (m^e)^d mod n
  = m^e.d mod n
  = m mod  n    // since e.d = 1 mod Ø(n)
= m                // since m < n

Sign/Verify:

To sign a message m ( less than n): s = m^d mod n

To verify s: m = s^e mod n

This also works since:

s^e mod n = m^e.d mod n = m mod n = m

Why is RSA Secure:

Every one knows the public key:  < e, n >.

To find the private key < d, n > you need to know Ø(n)

since e.d = 1 mod Ø(n).

To know Ø(n) you need  to  know p and q since

Ø(n) = (p-1).(q-1).

Thus to break RSA you should know:

how to factor n to find  p and q.

Factoring a big number like n is hard.

The best technique to factor 512 bit number takes 30,000 MIPS-years!

Efficiency of  RSA Operations:

Exponentiation

How to compute 123⁵⁴ mod 678?

123² = 123.123 = 15129 = 213 mod 678
123³ = 123.213 = 26199 = 435 mod 678
123⁴ = 123.435 = 53505 = 621 mod 678
......
123⁵⁴ =     ......                 = 87  mod 678

This requires 54  small number multiplications & 54 small number divisions.

How to compute 123³² mod 678?

123²   = 123.123 = 15129     =  213 mod 678
123⁴    = 213.213 = 45369     =  621 mod 678
123⁸    = 621.621 = 385641   =  537 mod 678
123¹⁶  = 537.537 = 288369   =  219 mod 678
123³²  = 219.219 = 47961     =  501 mod 678

This requires 5 multiplications & 5 divisions instead of 32.

To efficiently compute  123⁵⁴ : 54 is represented in binary as:

1   1              0          1      1         0

|               |           |           |           |

  ((((  (123²)123          ) ² )²123 )²123      )²

This requires 8 multiplications and 8 divisions instead of 54.

Each 1 requires 2 multiplications + 2 divisions &

each 0 requires 1 multiplication + 1 division.

Thus in the above we have 3 1s, 2 0s and that yields: 3.2 + 2.1 = 8.
Note that we ignore the leading 1.

Another example: y¹⁴ , 14 is represented in binary as:

1             1                 1                  0
           |              |               |
((             ( y²) y            )²y               )²

This requires 5 multiplications and 5 divisions instead of 32.

Generating RSA Keys

Finding e:

Two popular values for e are: 3 (2¹+ 1) and 65537 (2¹⁶ + 1).

Other common values: 5 (2²+ 1) and 17 (2⁴+ 1).

These make public key operations on message m faster
(encryption and signature verification is m^e ):

ü m³ requires 2 multiplications  & 2 divisions.

ü m⁶⁵⁵³⁷ requires 17 multiplications & 17 divisions

since the binary value of 65537  is 100..01 (15 zeros).

Finding n:

Randomly choose p and q and make sure that they are not 1 mod e.

n = p.q &

Ø(n) = (p-1)(q-1).

Finding Big Primes:

For a random 100 digit number (typical size for RSA), the chance is 1 in 230 being a prime.

How to test if a number p is prime? We use:

Fermat’s Theorem:

If p is prime and a <p, a^p-1 = 1 mod p

A primality test then is to pick a number a < p and compute a^p-1 mod p:

· If the answer is not 1, p is not prime.

· If it is 1 the probability that p is not a prime is ~ 1/10¹³

If the risk of 1 in 10¹³ is unacceptable, repeat the test using multiple values of a.

Finding d:

How to fine d such that e.d = 1 mod Ø(n) ?
Use Euclid algorithm (Section 7.4, page 187 of textbook).

The RSA keys:

Public key: <3 | 65537, n> Private key: <d , n>.

Precautions when using e=3:

· If a message m to be incrypted is small < then raising m to the power of 3 mod n will simply produce the value c = m³ < n. Thus any one can decrypt it by taking a cubec root of c to produce m

ü The problem can be avoided by padding the message with a random number before encyption so that m³ need to be reduced mod n.

· If a message m is send to 3 receipients with public keys: <3, n₁>, <3, n₂>, <3, n₃> and if a bad guy sees:

c₁ = m³ mod n₁ ,

c₂ = m³ mod n₂

c₃ = m³ mod n_3.

Then he can use the Chinnese Remainder Theorem (CRT) to compute:

c = m³ mod n₁n₂n₃ from c₁ , c₂ , c₃ values.

Since m is smaller than each of n_i , m³ will be smaller than n₁n₂n₃

Thus the bad guy gets m by computing the

ü The problem can be avoided by padding the message with the ID of each recipient.

Optimizing RSA Private Key Operations:

· In RSA, d and n=pq are on the order of 512-bit numbers

while p and q are on the order of 265-bits.

· RSA private key operations takes a message c in the order of 512-bits and computes:

m = c^d mod n.

· We can use the CRT to speed up the RSA operations using mod p and mod q instead of mod n.

i.e. using 256-bits instead of 512-bits operations.

The following is known as Garner’s Formula:

m = (((a-b)( q^-1 mod p)) mod p).q + b

where:

a = c_p ^dp mod p & b = c_q ^dq mod q

c_p = c mod p & c_q = c mod q

d_p = d mod (p-1) & d_q = d mod (q-1)

Use Euclid’s Algorithm to compute q^-1 mod p

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Diffie-Hellman

Alice and Bob agree on:  p (large prime) & g < p.

Alice Bob

Pick S_A (512-bit random number)                 Pick S_B (512-bit random number)
Compute T_A = ( gS_A) mod p                           Compute T_B = (gS_B) mod p

T_A >>> ……. <<< T_B

Compute X =   T_B ^SA mod p Compute Y = T_A ^SB mod p

X is the same as Y!

why?

X =  (T_B) ^SA  = (g^SB) ^SA
Y =  (T_A) ^SB  = (g^SA) ^SB

No one can compute g ^(S_A^S_B⁾ by knowing g ^SA   & g ^SB

The bucket Brigade / Man-in-theMiddle Attach

Alice Mr. X  Bob

Pick S_A           Pick S_X          Pick S_B

Compute: T_A = g^SA mod   T_X = g^SX mod p              T_B = g^SB mod p

T_{A           >>} T_{A  ..} T_{X         >>} T_X
                      T_{X            <<} T_{X  ..} T_{B        <<} T_B

Compute: K_AX  = T_X ^SA mod p        K_AX  =  T_A ^SX mod p K_BX  =  T_X ^SB mod p
K_BX  =  T_B ^SX mod p

Possible Defense

· Each person i picks S_i   and computes  T_i = g^Si mod p and

keeps S_i  private & makes T_i  public

· If Alice like to communicate with Bob, she finds T_B and computes:
                                   K_AB = T_B ^SA mod p

Then tells Bob she likes to communicate with him.

· Bob finds T_A   and then   computes:

         K_BA = T_A ^SB mod p

This requires PKI  (Public Key Infrastructure) to manage  T_i

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El-Gammal Signature

ü Each person has long-term public/private key pair:

Private:  S
Public: < g, p, T > where T  = gS mod p

ü For each message m to be signed:

generate a  per-message public/private key pair:

Private: S_m
Public:  T_m = g^Sm mod p

ü Compute the message digest: d_m  = MD ( m  | T_m )

ü Compute the message signature:  X = S_m + d_m ^S mod  ( p -1 )

ü Send:  <m, T_m , X >

Verify:

Received:   <m, T_m , X> & knows the sender public : < g, p, T >

ü Compute the message digest: d_m  = MD ( m  | T_m )

ü Compute: Y1 = g^X     and    Y2 = T_m T ^dm

If   Y1 = Y2   then the signature is   correct.

Why?

Since X = S_m + d_m ^S mod  ( p -1 ),

T_m = g^Sm mod p

T = gS mod p

Y1 = g ^X = g^Sm ^{+ d_mS} = g^Sm g^d_m^S = T_m g^{S dm}= T_m T ^dm  = Y2

Digital Signature Standard (DSS)

Proposed by NIST based on a modified version of El-Gammal algorithm.

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Zero Knowledge Proof Systems

Example:

Graph isomorphism problem:

Two graphs are isomorphic if we can rename the vertices of one to get a graph identical to the other.

This is a well-known NP-complete problem.

f > A > 1

g > B > 2

h > C > 3

i > D > 4

j > E > 5

Alice specifies a large graph G_A and renames the vertices to produce another isomorphic graph G_B.

Public Key: ( G_{A ,} G_B )
Private Key: G_A ßà G_B

To prove to Bob that she is Alice:

ü She renames the vertices to produce a set of isomorphic graphs:

G₁ G_{2  ....} G_k

and sends them to Bob.

ü Bob asks Alice to show him for each i the mapping between:

G_i and either G_A or G_B but not both (Otherwise Bob may know her private key!)