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2
Private-Key Cryptography
? 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 it is sent by
sender

3
Public-Key Cryptography
? public-key/two-key/asymmetric cryptography
involves the use of two keys:
�C a public-key, which may be known by anybody, and can be
used to encrypt messages, and verify signatures
�C a private-key, known only to the recipient, used to decrypt
messages, and sign (create) signatures
? is asymmetric because
�C those who encrypt messages or verify signatures cannot
decrypt messages or create signatures

Asymmetric key cryptography uses two separate keys: one
private and one public.
10.4
Asymmetric key cryptography
Locking and unlocking in asymmetric-key cryptosystem

10.5
10.1.2 General Idea
Figure 10.2 General idea of asymmetric-key cryptosystem

6
Why Public-Key Cryptography?
? developed to address two key issues:
�C key distribution �C how to have secure
communications in general without having to trust
a KDC with your key
�C digital signatures �C how to verify a message
comes intact from the claimed sender

7
Public-Key Characteristics
? Public-Key algorithms rely on two keys where:
�C it is computationally infeasible to find decryption key
knowing only algorithm & encryption key
�C it is computationally easy to en/decrypt messages when
the relevant (en/decrypt) key is known
�C either of the two related keys can be used for encryption,
with the other used for decryption (for some algorithms)

8
Public-Key Applications
? can classify uses into 3 categories:
�C encryption/decryption (provide secrecy)
�C digital signatures (provide authentication)
�C key exchange (of session keys)
? some algorithms are suitable for all uses,
others are specific to one

Symmetric-key cryptography is based on sharing secrecy;
asymmetric-key cryptography is based on personal secrecy.
10.9
Note

Plaintext/Ciphertext
Unlike in symmetric-key cryptography(symbols permuted
or substituted), plaintext and cipher text are treated as
integers in asymmetric-key cryptography.
Encryption/Decryption
10.10
Continued
C = f (Kpublic , P) P = g(Kprivate , C)

The main idea behind asymmetric-key cryptography is the
concept of the trapdoor one-way function.
10.11
10.1.4 Trapdoor One-Way Function
Functions
Figure 10.3 A function as rule mapping a domain to a range

Trapdoor One-Way Function (TOWF)
10.12
10.1.4 Continued
One-Way Function (OWF)
1. f is easy to compute.
2. f ?1 is difficult to compute.
3. Given y and a trapdoor, x can be
computed easily.

10.13
10.1.4 Continued
Example 10. 1
When n is large, n = p �� q is a one-way function. Given p and q , it is
always easy to calculate n ; given n, it is very difficult to compute p
and q. This is the factorization problem.

14
RSA
? By Rivest, Shamir & Adleman of MIT in 1977
? best known & widely used public-key scheme
? based on exponentiation in a finite field over
integers modulo a prime.
? uses large integers (e.g., 1024 bits)
? security due to cost of factoring large numbers

15
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
-define ?(n)=(p-1)(q-1)
? selecting at random the encryption key e
? where 1<e<?(n), gcd(e,?(n))=1
? solve following equation to find decryption key d
�C e.d=1 mod ?(n) and 0��d��n
? publish their public encryption key: PU={e,n}
? keep secret private decryption key: PR={d,n}

16
RSA Use
? to encrypt a message M the sender:
�C obtains public key of recipient PU={e,n}
�C computes: C = Me mod n, where 0��M<n
? to decrypt the ciphertext C the owner:
�C uses their private key PR={d,n}
�C computes: M = Cd mod n
? note that the message M must be smaller
than the modulus n (block if needed)

17
RSA Example - Key Setup
1. Select primes: p=17 & q=11
2. Compute n = pq =17 x 11=187
3. Compute ?(n)=(p�C1)(q-1)=16 x 10=160
4. Select e: gcd(e,160)=1; choose e=7
5. 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}

18
RSA Example - En/Decryption
? sample RSA encryption/decryption is:
? given message M = 88
? encryption:
C = 887 mod 187 = 11
? decryption:
M = 1123 mod 187 = 88

19
RSA Key Generation
? users of RSA must:
�C determine two primes at random - p, q
�C select either e or d and compute the other
? primes p,q must not be easily derived from
modulus n=p.q
�C means must be sufficiently large
�C typically guess and use probabilistic test
? exponents e, d are inverses, so use Inverse
algorithm to compute the other

20
RSA Security
? possible approaches to attacking RSA are:
�C brute force key search (infeasible given size of
numbers)
�C mathematical attacks (based on difficulty of
computing ?(n), by factoring modulus n)
�C chosen ciphertext attacks (given properties of
RSA)

21
Factoring Problem
? mathematical approach takes 3 forms:
�C factor n=p.q, hence compute ?(n) and then d
�C determine ?(n) directly and compute d
�C find d directly
? currently assume 1024-2048 bit RSA is secure

? Broadcast attack: If an entity sends the same
message with same encryption coe.(e) to
different recipients (moduli being n1,n2,n3)
? Let e=3 then,
? C1=P3mod n1
? C2=P3mod n2
? C3=P3mod n3
? Apply the Chinese Remainder Theorem to the 3
eqns. C��=P3mod (n1n2n3)
? P3< n1n2n3.
? C��=P3.
Hence get P

23
Timing Attacks
? developed by Paul Kocher in mid-1990��s
? cipher text only attack.
? Based on fast exponential algorithm.(guessing d from
Cd mod p)
? exploit timing variations in operations
? infer operand size based on the time taken by the
decrypting algorithm.
? Counter measures
�C use constant exponentiation time.
�C add random delays.

Chosen Cipher text Attacks
? based on the multiplicative property of RSA.
? attackers chooses cipher texts & gets
decrypted plaintext back.
? assume that intruder intercepts C=Pe mod n.
? Intruder chooses a random integer X in Zn*.
Calculate Y=C x Xe mod n
? He sends Y to Bob for decryption and get
Z=Ydmod n.
? With this intruder can find P easily.
24

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