Post
Topic
Board Development & Technical Discussion
Re: Pollard's kangaroo ECDLP solver
by
MrFreeDragon
on 27/05/2020, 15:14:14 UTC
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1) for each private key there is only 1 point (n private keys, n points, n is the order of the curve)
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I know that EC collision is not proved as there is no example.
Yes, for each private key there is only 1 point, however not necessary that if we have n private key we will have n different points.

Simple example for the group of 10 elements:
Code:
Key   Point
1       7
2       6
3       5
4       7
5       3
6       2
7       7
8       0
9       9
0       7

You can see that for each key (0..9) there is only one Point, i.e. we do not receive 2 or 3 or more points for each key. We have only one point. However key 1,4,7 and 0 lead to Point 7 (collision). For group of 10 elements it is easy to check.

However how could you be sure that for group with almost 2^256 elements for every private key we have the unique Point?