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Build Status License: GPL v3

cryptography-rsa-workshop

disclaimer

  • only for workshop purposes
    • for example: given implementation of RSA does not have padding

preface

  • goals of this workshop
    • introduction to asymmetric cryptography
    • mathematical basis for asymmetric cryptography
    • understanding purpose of trapdoor functions
    • introduction to RSA
      • common vulnerabilities
      • some basic attacks
    • basics of padding
    • basic knowledge of block ciphers
  • structure
    • package cryptography
      • decryption/encryption
      • signing/signature verification
    • package key - to generate RSA key pair: private and public
    • package prime - generating primes
    • exploits are shown in RsaExploitsTest
  • all used/needed math is described here: https://github.com/mtumilowicz/cryptography-math-basics

asymmetric cryptography

  • solve the problem of secure communications over an insecure network
  • symmetric cryptography context
    • to exchange messages => first mutually agree on a secret key k
      • what if every communication is monitored?
        • is it possible to exchange a secret key?
          • first reaction: not possible
            • reason: every piece of information is public
          • solution: public key (or asymmetric) cryptography
    • asymmetric ciphers - slower than symmetric ciphers
      • first use an asymmetric cipher to send the key to a symmetric cipher
      • then use symmetric key to transmit the actual file
  • analogy
    • Alice: buys a safe with a narrow slot in the top and puts it in a public location
    • Bob: writes his message and slips it through the slot
    • Alice: only a person with the key to the safe can retrieve Bob’s message
    • summary
      • public key: the safe
      • encryption algorithm: putting the message in the slot
      • decryption algorithm: opening the safe with the key
  • mathematical formulation
    • three sets
      • keys K
        • k = (kpriv, kpub) // private key and the public key
      • plaintexts M
      • ciphertexts C
    • for each kpub => exists encryption function e_kpub: M -> C
    • for each kpriv => exists decryption function d_kpriv: C -> M
    • (kpriv, kpub) e K => (d_kpriv) o (e_kpub) is identity on M
      • private key is sometimes called trapdoor information
        • it provides a trapdoor (shortcut) for computing the inverse function of e_kpub
          • it must be difficult to compute inverse function of e_kpub without a trapdoor information
    • encryption is a permutation of b-bit strings
      • {0, 1}^b -> {0, 1}^b
      • each key "chooses" some permutation

trapdoor function

  • is a function that
    • is easy to compute in one direction
    • believed to be difficult to find its inverse (without special information, called the "trapdoor")
  • analogy: padlock and its key
    • it is trivial to change the padlock from open to closed without using the key
    • opening the padlock easily requires the key to be used
      • key is the trapdoor
  • example
    • set some n, and define function as y = x^e mod n
      • bad trapdoor function
        • n = p (prime)
        • y = x^e mod p
        • we have to find inverse of e mod (p - 1)
          • Fermat's little theorem => we can perform calculations mod (p − 1) in the exponent
          • ed = 1 mod (p - 1)
            • it is solvable (for example using extended Euclidean algorithm) if gdc(e, p-1) = 1
        • y^d = (x^e)^d = x^ed = x mod p
        • solution is unique
          • suppose that we have two solution c1, c2
          • c1 ≡ c1^de ≡ (c1^e)^d ≡ y^d ≡ (c2^e)^d ≡ c2^de ≡ c2 mod p
        • summary: very easy to reverse
      • good trapdoor function (rsa)
        • n = pq, p,q - prime
        • y = x^e mod n
        • we have to find inverse of e mod φ(pq)
          • Euler's theorem => we can perform calculations mod φ(pq) in the exponent
          • ed = 1 mod φ(pq)
            • it is solvable (for example using extended Euclidean algorithm) if gdc(e, φ(pq)) = 1
            • however, calculating φ(pq) is as hard as factoring pq
          • d - decryption exponent
          • e - encryption exponent
        • solution is unique
          • c1 ≡ c1^de ≡ (c1^e)^d ≡ y^d ≡ (c2^e)^d ≡ c2^de ≡ c2 mod n
        • if we know the actual factors, we can use Euler’s theorem and write x as
          • x = y^d mod n
          • ed = 1 mod (p-1)(q-1)

rsa

  • asymmetric encryption algorithm
  • named after its inventors: Ron Rivest, Adi Shamir, and Leonard Adleman
  • rsa = good trapdoor function explained above (product of two large prime numbers)
  • encryption is faster if e is small and decryption is faster if d is small
    • most common e is 65537
  • we are encrypting / decrypting numbers - not letters
    • example
      • encrypting: "hello"
      • observation: all of the information is already stored in binary
        • encoding standards like ASCII or Unicode are used for humans to understand
      • this means that "hello" already exist as number
  • RSA relies on the size of its key to be difficult to break
    • longer RSA key => more secure it is
  • prime generating problem
    • how to test a given number n for being prime?
      • maybe use Fermat’s little theorem?
        • take A: gcd(A,n) == 1
          • if A^(n−1) ≠ 1 mod n => then n is composite
            • otherwise, it is prime with some probability
            • repeat for many As to increase the likelihood of being prime
        • why it's wrong?
          • take n = 561
            • this is composite number and fulfills Fermat's theorem for any A
          • family of such numbers are called Carmichael numbers
      • Miller Rabin primarity test
        • let p be an odd prime
        • p−1 = 2^k q, gcd(a,p)=1 => one of the following two conditions is true
          • a^q is congruent to 1 modulo p
            • q = p-1 / 2^k
          • one of a^q, a^2q , a^4q ,..., a^2^(k−1)q is congruent to −1 modulo p
        • proof
          • n = 2^k * q + 1
          • a^(n-1) = 1 mod n
          • a^(n-1) - 1 = 0 mod n
          • (a^(n-1 / 2) - 1)(a^(n-1 / 2) + 1) = 0 mod n
          • (a^(n-1 / 4) - 1)(a^(n-1 / 4) + 1)(a^(n-1 / 2) + 1) = 0 mod n
          • (a^(n-1 / 2^k) - 1)(a^(n-1 / 2^k) + 1)...(a^(n-1 / 2) + 1) = 0 mod n
            • we can expand it until n-1 / 2^k is odd
          • if n divides at least one multiplier => probably prime
            • Euclid's lemma: if p prime <=> p|ab => p|a or p|b
            • so we check this one by one
            • each number in the list is the square of the previous number
              • n-1 / 2^k, n-1 / 2^(k-1), n-1 / 2^(k-2)
        • if n is composite then running k iterations of the Miller–Rabin test will declare n probably prime with a probability at most 4^(−k)
        • prime number density
          • φ(n) is the number of prime numbers ≤ n
          • prime number theorem states that n / ln(n) is a good approximation of φ(n)
          • it means the probability that a randomly chosen number is prime is 1 / ln(n)
            • there are n positive integers ≤ n
            • approximately n / ln(n) primes
            • n / ln(n) / n = 1 / ln(n)
          • probability to find a prime number of 1024 bits: (ln(2¹⁰²⁴)) = (1 / 710)
            • primes are odd (except 2), we can increase this probability by 2
            • to generate a 1024 bits prime number, we have to test 355 numbers randomly generated
  • elliptic curves cryptography pros over rsa
    • smaller key sizes and signatures
      • example
        • 256-bit ECC key ~ 3072-bit RSA key
        • 384-bit ECC key ~ 7680-bit RSA key
      • lower storage requirements for both private keys and public keys
    • less computational power for key generation, encryption, and decryption
      • valuable for low-power devices such as IoT devices and mobile phones

padding

  • structure of a message can give attackers clues about its content
  • padding: adding randomized data to hide the original formatting
    • using the word padding for RSA is by now rather incorrect
      • RSA without padding is also called Textbook RSA
    • old padding schemes for RSA did simply extend the message before converting a number
    • newer schemes actually alter the message itself as well
      • example: OAEP
        • entire message is randomly transformed before RSA modular exponentiation
        • the same message encrypted multiple times looks different each time
  • padding oracles

block ciphers

  • if block ciphers act on short blocks, how do we encrypt a long message?

    • electronic codebook mode (ECB)
      • encrypt each block separately
      • example
        • codebook
          • 00 -> 11
          • 01 -> 00
          • 10 -> 01
          • 11 -> 10
        • plaintext: 00|11|00|01|00
        • cipher: 11|10|11|00|11
      • cons
        • patterns
    • cipher block chaining (CBC)
      • each block of plaintext is XORed with the previous ciphertext block before being encrypted
      • cipher block chaining uses what is known as an initialization vector (IV) of a certain length
      • pros
        • decryption of a block of ciphertext to depend on all the preceding ciphertext blocks
  • summary

    alt text

vulnerabilities

  • TLS 1.3 no longer supports RSA
  • criticism: https://www.youtube.com/watch?v=lElHzac8DDI
  • using prime factorization, researchers managed to crack a 768 bit key RSA algorithm
    • recommendations: a minimum key length of 2048 bits now
      • many organizations have been using keys of length 4096 bits
  • p and q must be globally unique
    • if p or q ever gets reused in another RSA moduli => can be easily factored using the GCD algorithm
  • RSA primitive is based on modular exponentiation
    • this operation is homomorphic
    • c1=md1, c2=md2 => (m1m2)^d=c1c2=c
    • to fix it - it is the essential to break this "homomorphism"
      • padding
    • example: RsaExploitsTest "sign then verify - product attack"
  • small exponent
    • for example: 3 // most common exponent is 65537
    • suppose you’re using a 2048-bit modulus N and exchanging a 256-bit key
      • message m is simply the key without padding => m³ < N => take the cube root
    • example: RsaExploitsTest "encode / decode - e < n"

digital signature

  • solves a problem analogous to the purpose of a pen-and-ink signature on a physical document
  • assymetric cryptography vs digital signatures
    • consider an analogy: bank deposit vaults vs signet rings
      • in today’s world signet rings and wax images obviously would not provide much security
  • digital signatures are at least as important as public key cryptosystems
  • significant use-case
    • your computer receives program and system upgrades over the Internet
    • how can your computer tell that an upgrade comes from a legitimate source?
      • example: the company that wrote the program?
    • solution: digital signature
      • original program comes equipped with the company’s public verification key
      • company uses its private signing key to sign the upgrade
      • your computer can use the public key to verify the signature before installing it on your system
  • it is quite inefficient to sign a large digital document D
    • it takes a lot of time to sign each b bits of D
    • resulting digital signature ~ as large as the original document
    • solution: use a hash function
      • hash: (arbitrary size documents) -> {0,1}^k
        • it should be very difficult to find D and D' whose hash(D) and hash(D') are the same
      • rather than signing document D sign the hash hash(D)
      • for verification: compute and verify the signature on hash(D)
  • setup
    • the same as for RSA encryption
    • encryption
      • e = encryption exponent
      • d = decryption exponent
    • signing
      • d = signing exponent
        • sign document D by computing S ≡ D^d (mod N)
      • e = verification exponent
        • compute S^e mod N and verify that it is equal to D