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Quote from: BitcoinADAB on July 26, 2021, 08:28:29 PM

Pollard's kangaroo / lambda / rho accelerator

It will lead to inner loops, but all solvable.
Profit: with one point addition, one will cover 6 points.

When will you be done with that project?

here is pubkey
02991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8
could you tell me first example if its x1 ? x2 ? x3 ?
if its x1 then whats x2 and x3 print pubkeys , it will help to vistors for understand about x1 x2 x3
thankx

Example: pubkey = 02991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8
This point becomes Point (x1, y1), but we don't know if it is Point 1, 2, 3, 4, 5 or 6.

from our offline server:

Code:

Point 1 (x1, y1)
x1 = 0x991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8
y1 = 0xeb3c392e5ac716a0cb40fa08e2616f47459e6a1cc0f2922836896a1ce5f631cc

Point 2 (x2, y2)
x2 = 0xa673e97568057fb5f41c35d6ed6c88ef97510d71222b3686ef892f4ccc2af536
y2 = 0xeb3c392e5ac716a0cb40fa08e2616f47459e6a1cc0f2922836896a1ce5f631cc

Point 3 (x3, y3)
x3 = 0xc06d5d9f69b4cb8d6f720d8f106b442956061673b01e9da1cb0886fe59dd2860
y3 = 0xeb3c392e5ac716a0cb40fa08e2616f47459e6a1cc0f2922836896a1ce5f631cc

Point 4 (x4, y4)
x4 = 0x991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8
y4 = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63

Point 5 (x5, y5)
x5 = 0xa673e97568057fb5f41c35d6ed6c88ef97510d71222b3686ef892f4ccc2af536
y5 = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63

Point 6 (x6, y6)
x6 = 0xc06d5d9f69b4cb8d6f720d8f106b442956061673b01e9da1cb0886fe59dd2860
y6 = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63

(Now we can say that the example point was Point 1, but that is not important.)

Remember:
x1 = x4 and x2 = x5 and x3 = x6
y1 = y2 = y3 and y4 = y5 = y6

Lowest x = x1 or x = x4
x = 0x991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8

Lowest y = y4 or y = y5 or y = y6
y = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63

That Point (x, y) would be the reference point to go on with. From that point you jump to another Point (x1, y1) according to your kangaroo / rho.
It doesn't matter if you jumped to Point 1 or 2 or 3 or 4 or 5 or 6, your reference point would be that Point (x, y) in all cases.

That makes kangaroo / rho faster. For example: A 'tame' that jumps to Point 2 will go on with Point 4. A 'wild' that jumps to Point 5 will also go on with Point 4 and we would have a solution.

But this only works if you have the full Bitcoin range (1 ... n) like in our project https://bitcointalk.org/index.php?topic=5347791.0 and not in a range like the puzzle #120 (2^119 ... 2^120 - 1).

So it works in the full range of 2^256 so what would be the expected operations?