more about ECC
http://www.johannes-bauer.com/compsci/ecc/?menuid=4
Topic: Simple question about Elliptic Curve Cryptography (Read 553 times)
OK, but consider that graph, what will happen if a multiplication reach that point?
That will never happen for the group used in Bitcoin. That specific point, is not a point on the elliptic curve used by Bitcoin. Such a point, call it Q, would have order 2, i.e. 2Q = O, the point at infinity. However, the points on the elliptic curve Y^2 = X^3 + 7 over Fp form a cyclic group of order a huge prime (almost as large as p). Hence not divisible by 2 and therefore no elements of order 2.
Hi! 
I was reading about ECC in the mastering bitcoin book and this image is very clear about how the multiplication is done:
However, what will happen if a resulting NG exactly hits the most left point in the curve (intersection between the function and X axis)? The tangent line will not find any other point and the multiplication will fail?
Thanks a lot for your time.
I was reading about ECC in the mastering bitcoin book and this image is very clear about how the multiplication is done:
However, what will happen if a resulting NG exactly hits the most left point in the curve (intersection between the function and X axis)? The tangent line will not find any other point and the multiplication will fail?
Thanks a lot for your time.
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