FrançaisGet started

ElGamal

ElGamal encryption: public key and secret, encryption with a one-time random value, the pair (c1, c2), decryption through the inverse of the mask, and the danger of reusing the random value.

  • 22 cards
  • 6 topics
  • Free
Study this deck

The outline, topic by topic

Each topic is weighted the way the exam weighs it.

  1. Concepts
  2. Questions
  3. Sort and link
  4. The numbers
  5. Scenarios
  6. Understand

A few cards from the deck

  • Public value h: what is it?

    Show the answer

    The power g^x mod p computed with the recipient's secret x, published with p and g so that anyone can write to them.

    With p = 23, g = 5 and x = 6, h = 5⁶ mod 23 = 8. Recovering x from h takes a discrete logarithm.

  • What does this decryption program print?

    ```python
    p, x = 23, 6
    c1, c2 = 10, 14
    mask = pow(c1, x, p)
    m = c2 * pow(mask, -1, p) % p
    print(mask, m)
    `

    • 6 10
    • 6 4
    • 4 10
    • 8 10
    Show the answer

    6 10

    The mask is 10⁶ mod 23 = 6, its inverse is 4 (6 × 4 = 24 ≡ 1), and the message is 14 × 4 = 56 ≡ 10. 6 4 forgets the multiplication by c2, 4 10 prints the inverse instead of the mask.

  • With the secret x, the public value is h = … mod p; to encrypt, a random k is drawn and c1 = … mod p is computed.

    Show the answer

    With the secret x, the public value is h = g^x mod p; to encrypt, a random k is drawn and c1 = g^k mod p is computed.

    x stays with the recipient; k stays with the sender. Only h and c1 travel, and neither lets you easily go back to its exponent.

  • Ephemeral value k: what is it?

    Show the answer

    A number drawn at random by the sender for one message only, which must never be used twice.

    It builds the mask h^k. If it repeats, two messages are hidden by the same value and give each other away.

Study this deck