1
  1. Does the point ever change for different keys or other functions that use the curve?

  2. If the point never changes is the only curve input parameter the index (r) that is used in the scalar multiplication?

  3. I saw that the base point (G) is (x, -4/5) or hex 5866666666666666666666666666666666666666666666666666666666666666 source. But link there to here where the coordinates of the base point are stated as [15112221349535400772501151409588531511454012693041857206046113283949847762202, 46316835694926478169428394003475163141307993866256225615783033603165251855960] how is the connection from the hex value to the point?

1
  1. Yes. There is nothing special about the base point G that you mention. It's just an arbitrary point agreed upon by convention to be used to transform private keys (scalars) into public keys via scalar multiplication. Scalar multiplication with points other than G happens during the Diffie Hellman exchange which is part of the generation of stealth addresses. There is also another base point H which is used for creating Pedersen Commitments. H is specifically chosen such that the discrete logarithm with respect to G is unknown (i.e. H = nG, but H was chosen such that n will never be known).

  2. Because of (1.), you now know that scalar multiplication is done with arbitrary points and not just with G.

  3. EC points exist in a 2D finite field, so can be represented with two coordinates. However, because the Ed25519 curve is symmetrical across the y-axis, it's possible to represent a point simply with the y-coordinate and a sign bit that indicates which of the two possible x-coordinates is the correct one. This is called compressed representation, and is how Monero EC points are stored on the blockchain.

5866666666666666666666666666666666666666666666666666666666666666 is a little endian hex representation (i.e. the byte order is reversed) of the decimal number 46316835694926478169428394003475163141307993866256225615783033603165251855960, which is the y-coordinate stated in the link you've mentioned.

|improve this answer|||||

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.