1. Public Key Algorithms
……..
RAIT
M. Chatterjee

2. Public Key Cryptography
Two keys
Private key known only to individual
Public key available to anyone
Idea
Confidentiality:
encipher using public key,
decipher using private key
Integrity/authentication:
encipher using private key,
decipher using public one
RAIT
M. Chatterjee

3. Requirements
Given the appropriate key, it must be computationally easy to encipher or decipher a message
It must be computationally infeasible to derive the private key from the public key
It must be computationally infeasible to determine the private key from a chosen plaintext attack
RAIT
M. Chatterjee

4. 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
RAIT
M. Chatterjee

5. Public-Key Cryptography
is asymmetric because
those who encrypt messages or verify signatures cannot decrypt messages or create signatures
RAIT
M. Chatterjee

6. Public-Key Cryptography
RAIT
M. Chatterjee

7. 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
public invention due to Whitfield Diffie & Martin Hellman at Stanford in 1976
known earlier in classified community
RAIT
M. Chatterjee

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

9. Security of Public Key Schemes
like private key schemes brute force exhaustive search attack is always theoretically possible
but keys used are too large (>512 bits)
not comparable to symmetric key sizes
security relies on a large enough difference in difficulty between easy (en/decrypt) and hard (to cryptanalyze) problems
requires the use of very large numbers
hence is slow compared to secret key schemes
RAIT
M. Chatterjee

10. Diffie-Hellman Compute a common, shared key
Called a symmetric key exchange protocol
Based on discrete logarithm problem
Given integers n and g and prime number p, compute k such that n = gk mod p
Solutions known for small p
Solutions computationally infeasible as p grows large – hence, choose large p
RAIT
M. Chatterjee

11. Diffie-Hellman Key Exchange
a public-key distribution scheme
cannot be used to exchange arbitrary messages
rather it can establish a common key
known only to the two participants
value of key depends on the participants (and their private and public key information)
based on exponentiation in a finite (Galois) field (modulo a prime or a polynomial) - easy
security relies on the difficulty of computing discrete logarithms (similar to factoring) – hard
RAIT
M. Chatterjee

12. Generating the Diffie-Hellman public key
Diffie-Hellman Setup
Generating the Diffie-Hellman public key
The Diffie-Hellman system allows Alice and Bob to agree on a key even when Eve is listening to everything they say to each other.
Alice and Bob need to agree on a prime number p, which they can do by simply sending it to each other.
Eve is allowed to learn this number p. In practice the number p is often simply advertised somewhere public.
RAIT
M. Chatterjee

13. Generators & Public Key
Given a prime number p, it is possible to come up with a number g (the so-called generator) with a very interesting property. Every number between 1 and p-1 can be written as a power of g when calculating modulo p. For example, using p = 5 the generator is 2, because
20 = 1
21 = 2
22 = 4
23 = 3 (because 8 = 3 mod 5)
Alice and Bob agree in the same way on a generator g for the numbers between 1 and p-1.
The numbers p and g serve as the public key.
RAIT
M. Chatterjee

14. Diffie-Hellman Key Exchange
Shared key, public communication
No authentication of partners
What’s involved?
P is a prime (about 512 bits), and g < p
P and g are publicly known
RAIT
M. Chatterjee

15. Diffie-Hellman Key Exchange
Procedure
Alice Bob
pick secret Sa randomly pick secret Sb randomly
compute TA=gSa mod p compute TB=gSb mod p
send TA to Bob send TB to Alice
compute TBSa mod p compute TASb mod p
RAIT
M. Chatterjee

16. Alice and Bob reached the same secret gSaSb mod p, which is then used as the shared key.
RAIT
M. Chatterjee

17. DH Security - Discrete Logarithm Is Hard
T = gs mod p
Conjecture: given T, g, p, it is extremely hard to compute the value of s (discrete logarithm)
RAIT
M. Chatterjee

18. Diffie-Hellman Scheme
Security factors
Discrete logarithm very difficult.
Shared key (the secret) itself never transmitted.
RAIT
M. Chatterjee

19. Disadvantages: Expensive exponential operation
DoS possible.
The scheme itself cannot be used to encrypt anything – it is for secret key establishment.
No authentication, so you can not sign anything …
RAIT
M. Chatterjee

20. Bucket Brigade Attack...Man In The Middle
Alice Trudy Bob
gSa=123 gSx =654 gSb =255
123 --> >
< <--255
654Sa=123Sx Sx=654Sb
Trudy plays Bob to Alice and Alice to Bob
RAIT
M. Chatterjee

21. Encryption With Diffie-Hellman
Everyone computes and publishes <p, g, T>
T=gS mod p
Alice communicates with Bob:
Alice
Picks a random secret Sa
Computes gbSa mod pb
Use Kab = TbSa mod pb to encrypt message
Send encrypted message along with gbSa mod pb
RAIT
M. Chatterjee

22. Diffie-Hellman
Bob
(gbSa)Sb mod pb = (gbSb)Sa mod pb = TbSa mod pb = Kab
Use Kab to decrypt
RAIT
M. Chatterjee

23. Example of an attack N=11 g=7 Alice chooses x=3 Bob chooses y=9
Trudy chooses x=8 and y=6
RAIT
M. Chatterjee

24. Alice sends 2 to Bob….Trudy intercepts and sends 9
Alice computes 73 mod 11 =2
Bob computes 79 mod 11 =8
Trudy computes 78 mod 11 =9
and 76 mod 11 = 4
Alice sends 2 to Bob….Trudy intercepts and sends 9
Bob sends 8 to Alice….Trudy intercepts and sends 4
RAIT
M. Chatterjee

25. Thus Trudy is man-in-the-middle……..
Alice computes K1= 43 mod 11 =9
Bob computes K2= 99 mod 11 =5
Trudy computes K1 = 88 mod 11 =5
And K2= 26 mod 11 = 9
Thus Trudy is man-in-the-middle……..
RAIT
M. Chatterjee