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Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 409225 times)
pbies
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August 07, 2024, 11:10:36 AM
 #5641

I think we all need a new topic/reminder - where it will be written how many puzzles have been solved and how many are left - then Q & A for n00b`s about what you use as a puzzle solver(which scripts are useless and why), what is used as a wallet if the puzzle is solved and why, how does the bots work and what is a solution against bots(MARA's Marathon Slipstream)  Tongue

Maybe after that close this topic? ... Too much time is wasted on (the same) explanations in this topic.

All of us have it here:

https://privatekeys.pw/puzzles/bitcoin-puzzle-tx

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blueman1985
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August 09, 2024, 04:36:00 AM
 #5642

Hello friends, I have just registered on this site. I apologize if I am not asking my question in the right place. I have a question for you dears. I was on the site https://frikiscape.com looking for a puzzle solution, but suddenly after I pressed the start button, a few minutes later I was shown a private key address along with two compressed and uncompressed addresses with their private keys. Does this mean that I have achieved an address with balance? Because I checked both addresses and both had zero balance. And my next question is, if I have solved the puzzle, what are the next steps? That is, what should I do in order to be able to withdraw the number of bitcoins that exist?
AlanJohnson
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August 09, 2024, 05:00:06 AM
 #5643

Hello friends, I have just registered on this site. I apologize if I am not asking my question in the right place. I have a question for you dears. I was on the site https://frikiscape.com looking for a puzzle solution, but suddenly after I pressed the start button, a few minutes later I was shown a private key address along with two compressed and uncompressed addresses with their private keys. Does this mean that I have achieved an address with balance? Because I checked both addresses and both had zero balance. And my next question is, if I have solved the puzzle, what are the next steps? That is, what should I do in order to be able to withdraw the number of bitcoins that exist?

If you ask such questions  it's not for you and you have no idea what this whole "puzzle thing" is all about...
kTimesG
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August 09, 2024, 11:57:20 AM
 #5644

New version of Python Kangaroo!

Using gmpy2 to accelerate a little bit the field math yields around 400k ops/s so please do not use this to search for high-bit puzzles.

I added some different strategies for sampling tames and wilds from the search interval which affects the theoretical complexity. My goal was to validate the following strategies:
- interval [0, b) using van Oorschot parameters for parallel collision should average 2* sqrt(b) ops
- interval [-b/2, b/2] using equivalence classes and birthday paradox should average 1.36 * sqrt(b) ops
- the 3-kangaroo method using negation symmetry should have resulted in 1.72 ... 1.81 * sqrt(b) ops

So what's the catch? After doing around 1000 test runs using random private keys with misc. strategies the average solving complexity (the sqrt(b) factor excluding the DP overhead) doesn't really match expectations:

- when DP = 0 (all points are DP) and we use the negation symmetry we can converge to almost 1.36 * sqrt(b) ops on average, which I think it is close to the minimal possible complexity conjectured by Pollard, e.g. 1.25 sqrt(b)
- when DP > 0, if we only jump forward, symmetry fails for walks, and complexity seems to go back to 2.0 sqrt(b) or worse.

How to fix? Jump side-by-side in deterministic way, so use Y sign to determine if jump goes forward or backward. Now we can end up with cycles in the jump travel. Most of them are 2-cycles (next jump goes to the same element as previous jump) which can easily be detected, and if so, jump again with the previous distance instead of jumping back to the previous point...

THREE kangaroos!

The papers analyse how we can use a third walk and solve the key whenever ANY two of the three walks collide, when only traveling forward. The trick is that we also start with a wild kangaroo that starts from -P plus some offset. Then we have three types of collision:

T / W1: solve for k = tameDist - wildDist
T / W2: solve for k = tameDist + wildDist (e.g. P traveled "backward" since -P walked forward)
W1/W2: if offsets don't sum up to 0, then k = (w1 - w2) * 2**-1

Because we hash DP by X we can easily solve the 4 different cases of each type of collision and check if it's the solution.

Tames should start at around -b/2 + b*8/10 == b*3/10  -this makes average distance to P or -P less than 0.2*b and means we would have to travel shorter walks on average.

Pollard's paper for 3 and 4 kangaroos suggest that the complexity should be 1.81 sqrt(b) or even 1.71 sqrt(b) for 4 kangaroos. I wasn't able to reach this complexity by using the suggested mean jump size (0.375*sqrt(b)). This strategy should in theory keep the same complexity when using DP > 0 and be parallelizable with same van Oorschot methodology. Both of these statements are in the paper, but I didn't manage to have success with them. So if anyone knows more about this 3 or 4 kangaroos, please explain:

- what is the maximum expected/allowed walk distance for a tame and for a wild_1 and wild_2?
- does the parallelization using van Oorschot (e.g. start = idx * v + z) apply or not? has anyone ever actually studied this?

Again - do not search for puzzles using this script. It is 100% just of theoretical interest.

CODE: https://pastebin.com/8Z69sRcU

Couldn't embed in post as it was too long...

nomachine
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August 09, 2024, 05:26:27 PM
Last edit: August 09, 2024, 06:29:34 PM by nomachine
 #5645


Again - do not search for puzzles using this script. It is 100% just of theoretical interest.

CODE: https://pastebin.com/8Z69sRcU

Couldn't embed in post as it was too long...

This can be converted to a .so file using Cython. I achieved an acceleration of over 150K hops per second, so it should reach around 550K hops per second.

Here’s an example of what your setup.py might look like:

copy/paste code to puzzle.pyx

setup.py
Code:
from setuptools import setup, Extension
from Cython.Build import cythonize

extensions = [
    Extension(
        "puzzle",
        ["puzzle.pyx"],
        extra_compile_args=["-O3", "-march=native", "-flto", "-funroll-loops"],
        extra_link_args=["-flto"],
    ),
]

setup(
    ext_modules=cythonize(
        extensions,
        compiler_directives={
            "boundscheck": False,
            "wraparound": False,
            "cdivision": True,
            "nonecheck": False,
        },
    ),
)

Build the .so File

Code:
python3 setup.py build_ext --inplace

This command compiles the .pyx file into a .so file. The --inplace flag tells the compiler to place the resulting .so file in the same directory as your source file.


puzzle.py
Code:
import puzzle

if __name__ == '__main__':
    puzzle.run_puzzle(48, '0291bee5cf4b14c291c650732faa166040e4c18a14731f9a930c1e87d3ec12debb', herd_size=1024, dp=6)



Private Key: 0xade6d7ce3b9b
Ops: 11631360 Stored: 11631276
Speed: 449407 ops/s

Optimization flags like -O3, -march=native, -flto will generally improve performance.But it can be worse or it can be buggy.

However, the problem is that the kangaroo algorithm needs to attack Puzzle 130 right away, which is beyond Python's capabilities.

These are such large numbers that you can only get depression from them.

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kTimesG
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August 09, 2024, 06:21:23 PM
 #5646


Again - do not search for puzzles using this script. It is 100% just of theoretical interest.

CODE: https://pastebin.com/8Z69sRcU


This can be converted to a .so file using Cython.

These are such large numbers that you can only get depression from them.
The goal was to find the best parameters to reduce the complexity, as in, the minimum average number of group operations, which does not depend on the problem size. In no way should anyone attempt to use the script as a production executable, even if built in Cython. It is still a naive approach, done in RAM.

So what do you think about the 3-walks kangaroos, what should the correct parameters be to reach 1.7 or 1.8 sqrt(b) ops? The script is broken for that strategy, after some fiddling with the intervals and the way tames and wilds are sampled sometimes it goes at about 1.9 sqrt(b) but it should in theory be a lot lower than that.

I think the wilds and anti-wilds colliding between them is the trick, but they don't do that as often as a tame/wild collision... so it must depend on where they start from.

For DP = 0 the script can solve on average in 1.36 or 1.46 sqrt(N) using the appropriate strategies, but obviously that only makes sense up to a point. We need something that is sub-2 sqrt(b) and also works with DP > 0. And not using 3 kangaroos, but many tens of thousands.

nomachine
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August 09, 2024, 06:34:23 PM
 #5647

I don't think so much about this topic anymore. I don't have enough number of GPU cards. Some 500-600 pieces are beyond my reach. Grin

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Akito S. M. Hosana
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August 09, 2024, 06:39:29 PM
Last edit: August 09, 2024, 07:02:21 PM by Akito S. M. Hosana
 #5648

500-600 pieces

How much can be mined with 500 RTX 4090 GPUs if you have your own power station? Undecided
nomachine
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August 09, 2024, 07:10:45 PM
Last edit: August 09, 2024, 07:31:10 PM by nomachine
 #5649

500-600 pieces

How much can be mined with 500 RTX 4090 GPUs if you have your own power station? Undecided

About 11,000 satoshis per hour.

It's not worth it. You can buy 200 Antminers T21 for the same amount of GPUs.

BTC: bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8
holy_ship
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August 15, 2024, 06:26:30 PM
 #5650

Hey guys, can

1. Pay-to-Taproot (P2TR)
2. CoinJoin
3. Blind Signatures

help with p66 withdrawal?

Cricktor
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August 15, 2024, 08:05:05 PM
 #5651

~~~
How about
4. Sprinkle some fairy dust over unicorn poop?
Just kidding...

Simple answer, no, that won't solve the problem of having to reveal the public key of the puzzle's P2PKH-UTXOs while spending them. You have to move aka spend the UTXOs first and if you send the spending transaction publicly, not undercover as presumably e.g. slipstream.mara.com appears to offer, any bots lurking and observing public mempools will see the public key and can and likely will try to find the private key in the time until the next block is found.

If you're unlucky the next block can take minutes while your spending transaction sits in public mempools, dozens of minutes if you're really unlucky. Plenty of time to find the private key and replace your transaction with another one with higher fee. In the end the stealing bots will turn all 66-puzzle's coins into transaction fees and some mining pool will benefit from it.

albert0bsd
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August 15, 2024, 11:07:02 PM
 #5652

help with p66 withdrawal?

No, none of them help

₿ SIGNATURE SPACE AVAILABLE    HERO MEMBER    PM bc1qjyhcjacmpc9pg8wes82lktyrjcwk4sdvqtm7ky
brainless
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August 16, 2024, 04:02:31 PM
 #5653

help with p66 withdrawal?

No, none of them help
Only I can help to withdrawal, you need share PK with me, I will withdraw and send you 50% share,

13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
pbies
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August 16, 2024, 04:17:44 PM
 #5654

help with p66 withdrawal?

No, none of them help
Only I can help to withdrawal, you need share PK with me, I will withdraw and send you 50% share,

I am reporting to be the first one to use your service.

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albert0bsd
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August 16, 2024, 04:51:39 PM
 #5655

Guys learn to quote properly without unnecesary sub-quotes

Only I can help to withdrawal, you need share PK with me, I will withdraw and send you 50% share,

I can try it, i only wan the 10% y send you the 90% share

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madogss
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August 18, 2024, 02:14:17 PM
 #5656

"Hello everyone, how are you? First, I apologize if what I write is not very clear, as I am using translation and I am not very proficient in English. However, I found the key for puzzle number 66 and I downloaded the wallet from https://guarda.com/app. Through browsing this forum, I found that there are some who might steal the funds when withdrawing. First, what should I do since I don't know any programming languages and I found it after a long time of searching using keyhunt python GUI? Secondly, does this apply to puzzle 67, 68, and the other puzzles up to 160? https://imgur.com/a/Iq0GhqG

You have to change to qr code as well it's still showing puzzle 64's address.
albert0bsd
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August 18, 2024, 04:15:37 PM
 #5657

You have to change to qr code as well it's still showing puzzle 64's address.

Those fuckers believe that they can scam us

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vneos
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August 19, 2024, 11:08:18 AM
 #5658

"Hello everyone, how are you? First, I apologize if what I write is not very clear, as I am using translation and I am not very proficient in English. However, I found the key for puzzle number 66 and I downloaded the wallet from https://guarda.com/app. Through browsing this forum, I found that there are some who might steal the funds when withdrawing. First, what should I do since I don't know any programming languages and I found it after a long time of searching using keyhunt python GUI? Secondly, does this apply to puzzle 67, 68, and the other puzzles up to 160? https://imgur.com/a/Iq0GhqG

Cool, can you explain this? https://imgur.com/a/oZGoCHU
The pubkey seems belong to puzzle 64 lol
pbies
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August 19, 2024, 12:20:58 PM
 #5659



Waiting for someone to help me move the balance without losing it.

BTC: bc1qmrexlspd24kevspp42uvjg7sjwm8xcf9w86h5k
alexxino
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August 19, 2024, 01:27:08 PM
 #5660

Aaaaaand 66 points go to Poland!

If this is really true and not a nice fake image generation then....Congrats!!
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