Public Key Algorithms …….. RAIT M. Chatterjee

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

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

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

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

Public-Key Cryptography RAIT M. Chatterjee

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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