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5.1 INTRODUCTION
In the real world , key management is the hardest part of
cryptology .Cryptanalysts often attack cipher system through their key
management .One of the problems facing symmetric cipher systems is key
distribution Keys must be distributed in secret , since knowledge of the key
gives knowledge of the massage .From the other side, keys are shared by
pairs , and could work well in small networks , but raised tremendously as
network grows up , since every pair of users must exchange keys .
The total numbers of key exchanges required in n-person network is
n(n-1)/2 .In six person network , 15 key exchange are required ,
in 1000 – person network ,nearly 500,000 key exchanges are requied .
Public key system are invented to over come the problems of key
distribution faced by symmetric (one - key) cipher systems .We summarize
the mentioned problems as follows :
 First key must be distributed in secret .
 Second : If key is compromised , then stranger can share the system
as member .
 Third : number of key increases rapidly as users increased .
Public key cryptography based on one-way hash function (trap-door
function),that is depend on two keys public key used for encryption process
and it is available for every one in the network, the second key is the secret
key which is used for decryption processes , and every person in the
network has his own secret keys .In another meaning users have their own
secret keys and share others with public keys .Since public key is known
for every one , then A can communicate with B as follows :
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 A gets public key of B from database.
 A encrypt his message using B public key and send it .
 B decrypt the message using his own secret key .
5.2 ONE-WAY HAASAH FUNCTION:
The notion of one- way function is central to public key
cryptography .One way function are easy to compute , but it is significantly
hard (computationally) to reverse. That is given x , it is easy to compute
f(x) , but given f(x), it is hard to compute x .Breaking plate is a good
example of a one-way hash function , it is easy to smash a plate into
thousands of tiny pieces back together into a plate . One-way functions are
not useful in public key cryptography, in public key. We need a special
type of one –way function, that is trap-door one-way function, which has a
secret door (secret key) used to reverse the other direction .i.e giving f(x)
and some secret Value we can deduce.
5.3 PUBLIC KEY CRY TOGRAPHY:
The concept of public key or (exponential ciphers) was invented by
Whitfield Diffie and Martin Hellman and independently by Ralph Merkle
at 1976. Since 1976, numerous Public key cryptography algorithms have
been proposed, many of these are insecure, others are impractical, only a
few of them are secure and practical. Only three algorithms work well for
both encryption and digital signature, RSA, EIGamal, and Rabin. All of
them are much slower than symmetricp algorithms by 1000 limes.
Public key cryptography used two different keys, public key for
encrypting process, and secret key for decrypting process. Any one can use
public key and encrypt a message, but only those who have secret key are
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allowed to decrypt the message. Mathematically, the process is based on
the trapdoor one-way function.Encryption is the
easy direction, while
decryption is the hard direction unless he has the secret key.
Public key system encrypt a message block M c 
0, n by computing
1
the exponential
e
C=M mod n
Where e (public key) and are the keys of encryption transformation ,
M is restored by the same operation using different exponential d (secret
key):
d
M=C mod n
By symmetry, Encryption and decryption are commulative and malual
inverses, thus substituting 5.2 by 5.1:
M=(Md mod n)e mod n
= Mde mod n = M
Encryption and decryption can be implemented using fast exponentiation
algorithm.
Fastexp (a,z,n); rturn x = a mod n
Begin
Al := ai zi := zi x=1
While (zi<> 0) do
Begin
While (zI mod 2 =0) do
Begin
zI : = zI div 2
al:= (al*al) mod n;
end;
z1:=z1-1;
x : =(x*a1) mod n;
end;
fastexp: = x;
end;
Figure 5-1 Fast exponentiation
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Using above algorithm then:
C = fastexp (m ,e , n)
M = fastexp (c ,d , n)
5.4 EXPONENTIAL CIPHER:
Encryption and decryption transformations are based on modular
exponentiation. Modular arithmetic is easier to work with on computers,
because, it restricts She range of all intermediate values and the result.
Exponentiation in modular arithmetic is performed without huge
intermediate results. For example, to calculate a8 mod n, don't use the
naive approach and perfonn seven multiplications and one huge modular
reduction:
a8 mod n = (a . a. a. a . a . a. a. a) mod n
Instead, perform three smaller multiplications and three smaller modular
reductions:
a8 modn = ((a2 mod n)2 mod n)2 mod n
also
a16 modn = (((a2 mod n)2 mod n)2 mod n)2 mod n
a25 mod n= (a . a24) mod n
=(a. a8 . a16) mod n
=(((a2 . a2)2)2 . (((a2)2)2)2) mod n
= ((((a2- a)2)2)2. a) mod n
=(((((((a2 mod n) .a) mod n)2 mod n)2 mod n)2 mod n) . a) mod n
Inverses is a problem of finding an integer x such that:
ax mod n = 1
a-1 = x mod
n.
For example 3 and 7 are multiplicative inverses mod 10, because
21 mod 10 = 1. In general a-1 = x mod n has a unique solution if a and n
are relatively prime. If n is prime number then every number in the range
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(1, n-1) is relatively prime to n and has exactly one inverse modulo in that
range. Femat's theorem and Euler's generalization can solve such a
problem.
Fermit's theorem:
Let n is a prime number; then for every a such that gcd (a,n) =1:
a n-1 mod n =1
Totient Function:
For n = p q and p, q are prime
Ø(n)=(p-1)(q-1)
Where Ø(n) is Euler totient function, the number of'elements in tlie
reduced set of residues modulo n.
Example: let p = 3, q = 5, 11 = p
, q = 15
'.
Ø (15) = (3-1) (5-1) =8
There are eight elements in the reduced set of residues modulo 15 (1, 2, 4,
7, 8, 11, 13, 14) are relatively prime to 15
5.5 POHLIG-HELLMAN CIPHERS:
Pohlig-Helman scheme is not a symmetric algorithm because different keys
are used for encryption and decryption. It is not a public key scheme,
because the keys are easily derived from each other, both encryption and
decryption keys must kept secret .In the Pohlig-Helman scheme, the
modulus is chosen to be a large prime P.
because P is prime,
then Ø(P) =P-1 , which is trivially derived from P, thus the scheme can
only be used for conventional encryption where e and d are both kept
secret.The enciphering and deciphering functions are thus given by :
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C=Me mod p and
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M = Cd mod p
Where e d = l ( mod some complex number )
Example:
Let p = l l , whence Ø (p) = p – l = 10; chose d = 7 and compute
e = inv(7,10) = 3
,
e
M=5
C = Me mod p = 53 mod l l = 4
M = Cd mod p = 47 mod l l = 5
5.6 RSA CIPHER SYSTEM:
One of the most well known and popular public key systems is the
RSA system, named after the first letters of the surnames of its designers
(Rivest, Shamir and Adleman of the Massachusetts lnstitute of Technology
MIT). The RSA based on the fact that it is relatively easy to calculate the
product of tow prime numbers, but giving the product it far more
complicated. First two prime numbers are generated (p and q of
length 100 -200 digit ), and their product is calculated and denoted by :
n = p* q
Chose e (encryption key ) randomly, relatively prime to
p and q and
satisfy the following expression:
3 < e < (p – l) (q – l) and gcd [ e , (p – 1 ) * (p – 1 ) ] = 1
The value of e is used to determine another, d ( decryption key ) for which:
e d = 1 (mod (p -1) (q – 1))
d = e-1 mod ((p – 1) * (q – 1))
The public key consists of the pair (e. n). The encipherment of atext is
performed by taking the binary representation of a message divided into
blocks and denoted by M. the cipher block C is computed by raising the
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decimal value of M To the power of e and taking the remainder of a
division by n:
Encrypt:
C = Me mod n
ciphertext
Decrypt
M = Cd mod n
plaintext
Public key gets its efficiency from the difficulty of factorizing large prime
numbers. Furthermore, it is almost impossible to calculate the value of d if
only the public key (e, n) is known. In order to calculate d then p and q
must be known too.
Example
Let p = 3, q = 17 ;
n=p*q
, 3 x 17 = 51
(p – 1 ) (q – 1) = 2 x 16 = 32
Find a number e between 3 and 32 which has no factor in common with 32.
Let e = 7; d can be determined using equation 5.13, then d = 23
e d = 7 x 23 = 161 mod 32 = 1
Let M = 2 then using (5.14) C = 27 mod 51 = 26 and
2623 mod 51 = 261 . 262 . 264.2616 (mod 51 ) = 16 x 13 x 16 x 1 (mod 51) = 2
1. Cenerate two large prime mumbers p and q.
2. Calculate their product n. = p* q.
3. Determine encryption key e such that ;
3 < e < (p – 1) (q – 1) and Ø (n) = (p – 1) * (q – 1)
gcd ( e, (p – 1) ( q – 1 ) ) = 1
4. Calculate d = d = e-1 mod (( p – 1 ) * ( q – 1 ) = e-1 mod Ø (n)
5. Encrypt : C = Me mod n
6.Decrypt : M = Cd mod n
Figure 5-3 RSA algorithm
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Example 2:
Let p = 53 and q 61 , then pq 3233, and Ø (n)= (p – 1) (q -1) =52 X 61=
3120, the value of e must be chosen some where between 3 and 3233.
Assume e =71, we can calcuilate d with d =e–1 mod (p -1) (q – 1) = 71–1
mod 3120 = 157.
Assuume that the message is given by M = RENAISSANCE. And the
alphabet is represented by decimal values a = 00, b = 01, c = 02, ete. with
space = 26, And divided into 4 –digit blocks, then we can proceeds as
follows:
M = RE
1704
NA
IS
SA
1300 0818 1800
M1
M2
M3
NC
1302
M4
M5
E
0426
M6
Now encrypt:
C1 = M171 mod 3233, = 1704 71 mod 3233 = 3106
C2 =M271 mod 3233 , C3 = M371m 3233, etc…
C = 3106 0100 0931 2691
C1
C2
C3
C4
1984 2927
C5
C6
And decrypt:
M1 = C1791 mod 3233 =3106791 mod 3233
M1 = C1791 mod 3233, M2 = C2791 mod 3233, M3 = C3791mod 3233, etc