Post
Topic
Board Development & Technical Discussion
Re: Solving ECDLP with Kangaroos: Part 1 + 2 + RCKangaroo
by
Veliquant
on 05/03/2025, 19:38:21 UTC
Quote
Mirror = takes advantage of group symmetry - you have both P and -P with the same X. If DP is on X and a DP is found, then you basically have two DPs found instead of one. Also this helps with more collisions.

This doesn't make sense to me. Why would you need the same point in the database. I don't see how this is useful.

In general terms, first you bring the wild kangaroo to the (-N/2, N/2) range, and you are not sure if it landed on the positive side or the negative side of the range, so you need a way to calculate the next jump, taking in to consideration both possibilities but having the same resulting landing point. This means you begin in one point, calculate the negative (-Y) at no cost and then you choose the one with smaller Y to calculate the jump.

In practice if the tamed kangaroo lands on the positive or negative point, it will be a collision, and you will catch it at the next distinguished point.

Also you have the same range N and you double the amount of calculated points but with only 1 point computation per 2 points result cost.

About the database now 1 distinguished point will "Include" twice as much points in the trailing paths. Remember the X coordinate is the same for both points so the probability of finding one distinguished point stays the same. Also as KtimesG says you will have 2 distinguished points instead of 1, the same X with n trailing zeroes and different Y, one with +Y and -Y.