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Key Exchange MethodsDiffie-Hellman and RSA
CPE 701 Research Case Study
Derek Eiler | April 2012
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OverviewToday’s discussion
•Background: “key” cryptography concepts
•Diffie-Hellman key exchange
•Public key infrastructure (PKI)
•RSA key pair generation
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BackgroundA few “key” concepts
•Encryption: plaintext -> ciphertext
•Decryption: ciphertext -> plaintext
•Cryptographic function: mathematicalfunction or algorithm used to encrypt/decrypt
•Key: parameter for a cryptographic function
•Symmetric vs. asymmetric keys
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So four people walk into a bar…Alice, Bob, Eve, and Mallory
•Alice and Bob want to speak privately over apublic channel
•Eve is always eavesdropping on Alice and Bob
•Mallory has malicious plans to interfere withAlice and Bob’s private conversation
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Diffie-Hellman key exchangeThe concept
Alice and Bob derive a shared secret key over a public channel (no prior arrangements) Publicly agree on two public values, 𝑝 and 𝑔 Each choose a private value, 𝑎 and 𝑏 Use clever math to compute a shared secret, 𝑘 Eve and Mallory never overhear enough information to derive the shared secret
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Diffie-Hellman key exchangeThe math: discrete logarithm problem
Let 𝑝 be a large prime number Let 𝑔 be an integer < 𝑝 For every number 𝑛 from 1… 𝑝−1 , inclusive, 𝑔 must have a power 𝑘 such that: 𝑛= 𝑔 𝑘 mod 𝑝 Solving the 𝑘 𝑡ℎ root mod 𝑝 is considered (but not proven) hard to do in polynomial time
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Diffie-Hellman key exchangeThe math: discrete logarithm in action
Solve for 𝑘, given values 𝑝, 𝑔, 𝑎′, 𝑏′ and knowing: 𝑘= 𝑔 𝑎𝑏 = 𝑔 𝑏𝑎 𝑎 ′ = 𝑔 𝑎 mod 𝑝 𝑏 ′ = 𝑔 𝑏 mod 𝑝 𝑘= (𝑏′) 𝑎 mod 𝑝 𝑘= (𝑎′) 𝑏 mod 𝑝 Finding 𝑘 is easy if 𝑎 or 𝑏 are known Quickly solved by brute force if 𝑝=3 and 𝑔=2 What if 𝑝=6,461,333,867 and 𝑔=3,691?
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Diffie-Hellman key exchangeExample using small numbers
Alice starts the exchange and tells Bob 𝑝=7, 𝑔=1 Privately, Alice chooses 𝑎=3 and Bob chooses 𝑏=5 Alice computes 𝑎 ′ = 𝑔 𝑎 mod 𝑝 and tells Bob the result 𝑎 ′ = 2 3 mod 7=1 Bob computes 𝑏 ′ = 𝑔 𝑏 mod 𝑝 and tells Alice the result 𝑏 ′ = 2 5 mod 7=4 Since k= 𝑔 𝑏𝑎 , Alice can compute 𝑘= (𝑏′) 𝑎 mod 𝑝 𝑘= (4) 3 mod 7=1 Since k= 𝑔 𝑎𝑏 , Bob can compute 𝑘= (𝑎′) 𝑏 mod 𝑝 𝑘= (1) 5 mod 7=1 Meanwhile, Eve doesn’t know 𝑎 or 𝑏 and can’t easily derive 𝑘
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RSA key generationThe concept
•Alice generates a pair of keys, publishing oneand keeping the other private
•Anyone may use the published key to encryptmessages intended for Alice
•Only Alice can decrypt messages encryptedwith the public key (unless the private key wascompromised somehow)
•Alice may also use the key pair to prove heridentity
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RSA key generationThe math: factoring problem
•Computing the product of two prime numbers iseasy (23*17 = 391)
•Factoring the product of two large primenumbers is “hard”
•Try factoring123,018,668,453,011,775,513,049,495,838,496,272,077,285,356,959,533,479,219,732,245,215,172,640,050,726,365,751,874,520,219,978,646,938,995,647,494,277,406,384,592,519,255,732,630,345,373,154,826,850,791,702,612,214,291,346,167,042,921,431,160,222,124,047,927,473,779,408,066,535,141,959,745,986,902,143,413
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RSA key generationThe math: public and private key pair
Calculate the product 𝑛=𝑝𝑞 where 𝑝 and 𝑞 are very large prime numbers (e.g. tens or even hundreds of digits long) Carefully choose exponents 𝑒 and 𝑑 such that we can publish the key 𝑘 𝑝𝑢𝑏 =(𝑛,𝑒) and retain the corresponding private key 𝑘 𝑝𝑟𝑖𝑣 =(𝑛,𝑑)
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RSA key generationThe math: “exponential” difficulty
Choose an 𝑒 such that 𝑒<𝑛 and 𝑒 is coprime to 𝑝−1 𝑞−1 How? For each 𝑒 in {2…𝑝𝑞}, test whether GCD 𝑒, 𝑝−1 𝑞−1 =1 until true.* Choose a 𝑑 such that 𝑒𝑑−1 is divisible by 𝑝−1 𝑞−1 Now publish 𝑘 𝑝𝑢𝑏 =(𝑛,𝑒) and retain the private key 𝑘 𝑝𝑟𝑖𝑣 =(𝑛,𝑑)
*Euclid’s or Stein’s algorithm are typically used to compute the GCD.
qrcode
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ReferencesSome light reading on the web