Transcript NET 311Information Security

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NET 311
INFORMATION SECURITY
Networks and
Communication
Department
Lecture 4: Asymmetric Ciphers
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lecture contents:
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17-Jul-15
Public-Key Cryptography
Why Public-Key Cryptography?
RSA Cryptography
Networks and Communication Department
Private-Key Cryptography
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traditional private/secret/single key cryptography
uses one key
 shared by both sender and receiver
 if this key is disclosed communications are
compromised
 also is symmetric, parties are equal
 hence does not protect sender from receiver forging a
message & claiming is sent by sender
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Networks and Communication Department
Public-Key Cryptography
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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
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Why Public-Key Cryptography?
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 Developed
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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
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Public-Key Cryptography
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 related 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
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Public-Key Cryptography
Asymmetric-Key Cryptography
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Asymmetric-key (public key cryptography) uses two keys:
one private and one public.
There are two algorithms:
 RSA
 ELGAMAL
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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Modular arithmetic
This example is
modulo 7
 The numbers allowed
are 0 to 6
 After 6, numbers
“wrap around”
7 (mod 7) = 0
3+3 (mod 7)= 6
4+4 (mod 7)= 1
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6
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5
2
4
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RSA
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The most common public key algorithm is RSA,
named for its inventors Rivest, Shamir, and Adelman.
It uses two numbers:
 public key
 d private
e
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The two keys, e and d, have a special relationship to
each other.
RSA
RSA – selecting keys
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Bob uses the following steps to select the private and
public keys:
 Bob
chooses two very large prime numbers p and q
 Bob multiplies p and q to find n  n=p x q
 Bob calculates another number ф = (p -1) X (q -1)
 Bob chooses a random encryption number e,
where 1< e< ф and gcd(e, ф)=1
 He then calculates d so that d X e mod ф = 1
 Bob announces e and n to the public; he keeps ф and d
secret.
RSA – Encryption and Decryption
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Encryption
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Decryption
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Restriction
P < n , if not, the plaintext needs to be divided into
blocks to make P less than n.
RSA Exercise
Encrypt the following message “NO” By Using RSA
algorithm.
Assume that p=17, q =11, and e= 7
Hint, each character will be encoded according to this
algorithm (A - Z)  (00 – 25)
 Find d , then encrypt each character individually .
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The solution
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Step1: calculate n, where n= p x q
= 17 x 11 = 187
Step2: calculate ɸ where ɸ = (p-1) x (q-1)
= 16 X 10 =160
Step3: calculate d, where e is given
e . d mod ɸ=1
7d mod 160 =1  d=23
 This
is because 23 X 7 = 161 = ( 1 X 160) + 1
The solution
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Step5: encode the plaintext.
P=“NO”= 13 14
Step6: C= pe mod n
C(N)= (13 ) 7 mod 187 = 106
C(O)=(14) 7 mod 187= 108
Step7: P= Cd mod n
P(106)= (106) 23 mod 187 =13
P(108)= (108) 23 mod 187 =14
Example 2
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Using e=13 ,d=37, n=77 in the RSA algorithm
encrypt the message "IF" using the value of 00 to
25 for letters A to Z, do the encryption character by
character.
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Networks and Communication Department
Example 3
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
Using e=7 ,d=23, n=187 in the RSA algorithm encrypt
the message M = 88.
C= 88 7 
7
4
2
 88 mod 187 = [(88 mod 187) × (88 mod 187)
× (881 mod 187)] mod 187
881 mod 187 = 88
882 mod 187 = 7744 mod 187 = 77
884 mod 187 = 59,969,536 mod 187 = 132
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 88 mod 187 = (88 × 77 × 132) mod 187 =
894,432 mod 187 = 11
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RSA- Application
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RSA is useful for short messages but not for long
messages.
RSA is used in digital signature , and authentication
algorithms.
Diffie-Hellman Procedure
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Step 1: Alice chooses a large random number a and calculates
R1 = ga mod p
Step 2: Bob chooses another large random number b and
calculates R2 = gb mod p
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Step 3: Alice sends R1 to Bob
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Step 4: Bob sends R2 to Alice
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Step 5: Alice calculates K = (R2) a mod p
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Step 6: Bob also calculates K = (R1) b mod p
Diffie-Hellman Example
By giving g and p the following values
g = 7 and p = 23, calculate the key.
The steps are as follows:
1.
Alice chooses a = 3 and calculates R1 = 73 mod 23 = 21.
2. Bob chooses b = 6 and calculates R2 = 76 mod 23 = 4.
3. Alice sends the number 21 to Bob.
4. Bob sends the number 4 to Alice.
5. Alice calculates the symmetric key K = 43 mod 23 = 18.
6. Bob calculates the symmetric key K = 216 mod 23 = 18.
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Diffie-Hellman Example
The value of K is the same for both Alice and Bob;
gab mod p = 718 mod 23 = 18.
Therefore, we can say that symmetric (shared) key
in the Diffie-Hellman protocol is K = gab mod p.
Diffie-Hellman characteristics
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In the Diffie-Hellman cryptosystem, two parties
create a symmetric session key to exchange data
without having to remember or store the key for
future use.
They do not have to meet to agree on the key; it can
be done through the Internet.
The two parties need to choose two numbers p and
g, both number announced to the public.
 P
large prime number
 g random number
Diffie-Hellman Cont.
Diffie-Hellman is a very sophisticated symmetric-key
creation algorithm.
 If a and b are very large numbers, it is extremely difficult
for Eve to find the key, knowing only p and g.
 An intruder needs to determine a and b if R1 and R2 are
intercepted.
 The key will be changed each time the two parities need to
communicate.
Diffie-Hellman