3 pkc+rsa

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:
– 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

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:
– 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

7
Public-Key Characteristics
• Public-Key algorithms rely on two keys where:
– it is computationally infeasible to find decryption key
knowing only algorithm & encryption key
– it is computationally easy to en/decrypt messages when
the relevant (en/decrypt) key is known
– 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:
– 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

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 ø(n)=(p-1)(q-1)
• selecting at random the encryption key e
• where 1<e<ø(n), gcd(e,ø(n))=1
• solve following equation to find decryption key d
– e.d=1 mod ø(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:
– obtains public key of recipient PU={e,n}
– computes: C = Me mod n, where 0≤M<n
• to decrypt the ciphertext C the owner:
– uses their private key PR={d,n}
– 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 ø(n)=(p–1)(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:
– determine two primes at random - p, q
– select either e or d and compute the other
• primes p,q must not be easily derived from
modulus n=p.q
– means must be sufficiently large
– 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:
– brute force key search (infeasible given size of
numbers)
– mathematical attacks (based on difficulty of
computing ø(n), by factoring modulus n)
– chosen ciphertext attacks (given properties of
RSA)

21
Factoring Problem
• mathematical approach takes 3 forms:
– factor n=p.q, hence compute ø(n) and then d
– determine ø(n) directly and compute d
– 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
– use constant exponentiation time.
– 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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