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Author Topic: Simple question about Elliptic Curve Cryptography  (Read 510 times)
Delek (OP)
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April 15, 2015, 02:42:24 PM
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Hi! Smiley

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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There are several different types of Bitcoin clients. The most secure are full nodes like Bitcoin Core, but full nodes are more resource-heavy, and they must do a lengthy initial syncing process. As a result, lightweight clients with somewhat less security are commonly used.
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LiteCoinGuy
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April 15, 2015, 04:12:47 PM
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https://www.youtube.com/watch?v=iB3HcPgm_FI

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April 15, 2015, 04:41:18 PM
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Layman's Guide to Elliptic Curve Digital Signatures

“God does not play dice"
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April 15, 2015, 04:59:47 PM
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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.
Delek (OP)
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April 15, 2015, 05:13:13 PM
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OK, but consider that graph, what will happen if a multiplication reach that point?

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April 15, 2015, 05:26:28 PM
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more about ECC

http://www.johannes-bauer.com/compsci/ecc/?menuid=4


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