I'm not sure. I know that Google search often brings me back to this forum. I also know that you have code you don't want to release publicly, or you release code as bait. Nothing specific which is the fastest. 

What would you do with the fastest Kangaroo program I have? To be fair, you still need 1,200 GPUs, even if kTimesG or RetiredCoder publicly release the fastest version.


Exactly. 

And what is to be learned? Definitely nothing about the 3-kang algorithm itself, all of that is public already. Like nomachine said, one would still need many, many GPUs to find high-bit solutions, no matter how well an implementation is optimized.

The only curiosity one might have would be "why is it faster" or "how did RetiredCoder manage to pull off 12.8 Gk/s on a RTX 4090", which might end up just as a terrible headache reading low-level code, or an "oh", not a revolutionary breakthrough.

Yes you are absolutely right, after one year of dreaming with babies, gigants and kangaroos, I understand that the jump table has to be the same for all herds  .

For me is very interesting the way you see the kangaroo methods conceptually. There is one part that is the method itself were we try to make as little as possible point additions, the other part is "optimizations" were we try to make those point additions as fast as possible, and use parallel computing to increase the point output.

I understand why is so important the concept of "K", I have an intuition about this that i would like to ask if it is correct. The kangaroo method by itself as in only 2 kangaroos (Tame and Wild) would solve the discrete logarithm in only sqrt(range) steps (k = 1) if both kangaroos started at the same point. This is sqrt(range)/2 for each kangaroo. This is related with the birthday paradox. Wen you use the first method in your program, and the kangaroos are separated, there is an overhead, because the path before both kangaroos meet is wasted. This accounts for an additional K points, so this method will solve in 2K jumps on average. What I have not been able to understand is how the 3 kangaroo method works. You start with one tamed an 2 wilds, and place them in a different position. This makes the kangaroos meet faster on average.

There is also the concept that if you start the path in the origin, and jump to the right, is the same as jumping to the left, so you are actually making, 2 paths at once.
 
For the 3 kangaroo methods this are my questions:

1) Where the 3 kangaroos should be placed at the beginning?
2) The wild kangaroos travel 1/4 of the jumps each and the tamed 1/2 of the jumps?

Thanks

Thanks for your answer. Yes, I am trying to understand SOTA+, but I believe I have to learn all the methods in order to understand SOTA+.  This means in order to understand SOTA+, I should understand 2 kangaroos, then 3-4 kangaroos, then Mirror, and then SOTA, and SOTA+. Is this correct?


This means the 2-3-4 kangaroo algorithm makes 2-3-4 long paths, while Gaudry-Schost variant makes many smaller paths until a distinguished point is found. I see your point, a normal kangaroo method can't go sideways, as the "maximum excursion" would not allow for the kangaroo to travel very far before going in to a loop?

But in general terms:

Mirror is faster than 3 kangaroos because it calculates 2 "cheap" points in every jump instead of 1, and SOTA is faster than mirror because it computes 4 "cheap" points instead of 1?

The trick in SOTA+ is how to detect and then get out of the cycles in a way that the overhead is not to big?

Thanks

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.

Yeah but if I am storing X points, I don't really care about having 2 in the database "will have 2 distinguished points instead of 1, the same X with n trailing zeroes and different Y, one with +Y and -Y"; I only need one of them stored to solve.

Yes I believe, you only need to store the X coordinate (or its hash) and the corresponding Kangaroo ID

I think it was misunderstood. Where did I mention that both DPs need to be stored?

I said that when a DP is found, then you basically have two DPs. In a single database entry, since they both have the same X, hence the same hash (if Y is not used to affect the hash of the DP entry).

It shouldn't even be possible to store both - DP database keys must be unique for fast search. So an insertion will fail when a collision is detected, or when the DP hash collides, or when some kangaroo type collides with itself (and dies if the landed point is the stored DP point, not its opposite).

Do you understand that the way you asked your question is plain stupid? And because you didn't put much effort into your question, I will answer deliberately on the same level: you need just one RTX 4090 to solve puzzle 135 eventually, in few hundreds of years. You could also use just zero RTX 4090 and go cpu-only, we're then in a time frame of many thousands of years.

I'm sure it's not precisely the answer you were looking for. Ask your questions with a bit more brain involved.