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Introduction to Modern Cryptography, Lecture 10

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RSA Signature generation/decryption: time O(n3) = O(k6) for k bits of security ... if these operations are on public data then no, it would be easy to break the ... – PowerPoint PPT presentation

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Title: Introduction to Modern Cryptography, Lecture 10


1
Introduction to Modern Cryptography, Lecture 10
  • Performance Improvements
  • Fast Arithmetic, Montegomery representation,
    Batch RSA,
  • Elliptic Curves

2
RSA
  • RSA security 2k require n bit modulus k2 bits
  • RSA Signature generation/decryption time O(n3)
    O(k6) for k bits of security (Simple arithmetic
    operations)
  • RSA Signature verification/encryption time O(n2)
    O(k4) for an n bit modulus (Simple arithmetic
    operations)

3
Fiat-Shamir
  • For k bits of security, modulus must be O(k2)
    bits in length
  • m (or t) must be at least k,
  • Signature generation O(n2 k)O(k5)
  • Signature verification O(n2 k) O(k5)

4
Improving Fiat-Shamir
  • Let the public key be only small primes 2, 3, 5,
    (if quadratic residue)
  • This means that verification now takes time equal
    to a constant number of full length
    multiplications (O(n2) O(k4))

5
Using fast arithmetic
  • If multiplication/division of n bit integers only
    takes time n log n then we have
  • RSA signature (decryption) O(n2 log n) O(k4
    log k)
  • Fiat-Shamir signature O(n log(n) k) O(k3 log
    k)
  • RSA / Fiat-Shamir verification O(n log n) O(k2
    log k)

6
Can we do better?
  • Can we do signature generation in less than k
    operations, say O(log(k)) operations?
  • Argument (false) if these operations are on
    public data then no, it would be easy to break
    the scheme by guessing what these operations are.

7
The trick
  • Combine lots of private operations together
  • Amortize the cost per private operation

8
Batch RSA
  • An RSA variant
  • When I send a signed message to Alice, I use one
    of the roots 3, 5, 7, , 997 or 2161, 216?,
    216??, , 217??? (all primes in the range)
  • Thus, to sign m1 and m2 I could extract the 3rd
    root of hash(m1) and the 5th root of hash(m2)

9
Batch RSA
The only expensive operation
10
What is going on?Can this always be done?
  • Yes (on Blackboard)
  • Cost per private operation is now down to O(log
    n) O(log k) mutiplications/divisions, i.e.,
    O(k2 log k) operations per private operation
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