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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 02/08/2025, 08:34:10 UTC

It's not a theory, it's just basic logic, if a whole range violates a strict filter, it can be skipped, no need to generate or check those keys at all.
People who reject this are just ignoring the fundamentals. Filtering entire subranges is standard practice, not some controversial idea.

So why does everyone take sides on this rule and consider it useless at the very beginning of the discussion in the forum, causing the discussion to be fruitless and not continue its process?

That's just how it is; it doesn't make sense. You're free to skip any keys you want, but there aren't any keys that are more reasonable to skip than others
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 01/08/2025, 16:56:15 UTC
Quote
Initializing CUDA constants...
Total GPU memory: 4294639616 bytes
Free GPU memory: 3333160960 bytes
Available for tables: 2666528768 bytes
Max half_N: 10416128
Generating 20832256 points (10416128 positive, 10416128 negative)...
Generation time: 1409.00 ms
Speed: 14785.14 points/ms

First positive point:
  x = 3130e6f4bcf13065e63b05c3c26da883962d17e1d1a7b83deddfb82798c37fa8
  y = ff3191deb7f22286ffc93dbc46f0bb78489fb82fdf9c745f6d07a0f2cdafecaf

First negative point:
  x = 50132620dbf5bb47c30c603a60897e8e74873aac7e42f30c9d1ac42f22399410
  y = 0644bbbbe2809bcb11f620e540bdbd1366e8fb80d33b1488e85a0d41d6da2240

Generated 20832256 points successfully!

is good result for 1650ti?

Not much, on a 1650Ti you should be able to perform about 2 million additions per millisecond.

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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 23/07/2025, 22:38:12 UTC
with new vanity I'm getting this speed with 5090
VanitySearch v1.19 Linux with BitCrack integration
Difficulty: 1461501637330902918203684832716283019655932542976
Search: 1PWo3JeB9jrGwfHDNpdGK54CRas7fsVzXU [Compressed]
Current task START time: Wed Jul 23 17:22:14 2025
Number of CPU thread: 0
GPU: GPU #0 NVIDIA GeForce RTX 5090 (170x0 cores) Grid(1792x256)
[14.09 Gkeys/s][Total 2^34.71][00:00:02 RUN || END 00:20:41][Found 0]

Most likely, this version uses endos and -P, so 5 keys out of every 6 you’re checking are useless because they’re outside the range

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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 16/07/2025, 10:52:19 UTC
The Golden Ratio stuff might sound academic, but I'm just trying to find any edge that isn't pure brute force. These puzzles are getting impossible to crack without some mathematical insight.

What got you started analyzing Bitcoin puzzles? And more importantly - coffee or energy drinks for those late-night number crunching sessions? ☕

Actually, it's not very academic. I’d say it’s more delusional, not much different from people studying constellations in astrology.

I don’t understand what’s so disheartening about accepting randomness. In a world where so many things are biased and unfair, this puzzle is one of the few places where we get something close to true fairnes, and i like It.

I’ve simply focused on optimizing my code to compute hashes, which is the only way to increase the chances of finding the key (given fixed time and hardware), and I let the machines crunch the numbers.

My advice, if you want to participate in this lottery, is: stop wasting time and energy chasing absurd, imaginative solutions and dressing them up as holistic mathematics
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 07/07/2025, 16:25:52 UTC

Hmm but im against this. From what i discover now. Some long adress prefix actually have some significant distance between them.  

I can skip trilions of key before continue brute force .  And yes there a chance of missing it if the skip to far away. now im gathering bout quadrilions of keyspace and look the lowest distance between those prefix. And only skip half of the lowest distance that i discover  , so it will be consider safe skip

What you're discovering is simply how elements are distributed in a uniform distribution. You could search for prefixes across the entire space by jumping randomly, then calculate the distances between the prefixes using the jump index (instead of using the private key), you would get the same results (assuming the hash functions aren't rigged).

What you're doing is like trying to find the ace of hearts in a deck of 52 cards, and skipping 5 cards if you find an ace of another suit, because on average the distance between adiacent aces in a shuffled deck is around 10.6
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 04/07/2025, 18:03:50 UTC

N balls in an urn. one is blue, the rest are red.
Extract one ball, if it's blue = success. If it's red: failure.
Put the ball back in the urn.


Did you know that if you're colorblind, you might actually be able to find the blue ball more easily than someone who can see colors? Let me teach you a trick: ask someone to assign a number to each ball, then apply a hash function to each number, and stick a label with the hash on the ball (the hash function must not be rigged).

Then, ask that person to give you the hash of the red ball, the winning one! Of course, since you're colorblind, you can't see colors, you can only read the hashes, but now you know which one is the winner.

At that point, to find the blue ball, you divide the N balls into blocks. When you find a hash in a block that starts with the same first X bits as the winning one's hash, you discard that whole block, because that prefix was already used in the block, and it's very unlikely that another ball in the same block would share it, unless the hash function is rigged.

With this method, you'll find the blue ball faster, thanks to uniform distribution, 1st grade math, and probabilistic search. But it works only if you’re not checking all the balls, because if you do that, then you don't gain any statistical advantage. And it only works if you're colorblind and can use hash functions and prefixes, if you can see colors, then you're stuck with the classic, old, and outdated probabilities of a uniform distribution.
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 04/07/2025, 08:30:51 UTC

 But it is true that if they tell you to roll a die 6 times, and the second time you got a 6, it's normal for you to bet that there won't be another 6 in the next 4 events.


No, sorry, it's not normal to me, not at all. But that's fine, maybe my small brain just can't grasp this thing about independent events (that apparently are not so independent), and probably that's why I don't understand the greatness of prefix search.

As for the script, there are pages in this thread where we try to explain to you that the result you're seeing isn't what you think it is, but again, that's fine.
We’re speaking two different languages, your math/probability knowledge and mine just don’t align. It doesn't seem like we're able to have a technical and constructive dialogue. I had already promised myself not to write about it anymore
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 04/07/2025, 00:34:11 UTC

The math is simple: for example, in a 4096-bit search space, an h160 prefix is ​​found on average once. If this were frequent (2 out of 3 prefixes in 4096 bits), the hash function would be broken. That's why this is a good guide for probabilistic searches. Anyone who says otherwise is wrong.


The math is simple: for example, in a 6 dice rolls, a "six" comes up on average once. If this were frequent (2 out of 3 times in 6 rolls), the die would be broken. That's why this is a good guide for probabilistic searches. Anyone who says otherwise is wrong.

So, if I'm rolling one of my 6 dice one at a time and a "six" shows up, say, on the fifth roll, then the last roll has a lower chance of being a "six", right? Otherwise the die is rigged. I guess we need to rewrite the definition of uniform distribution in math textbooks.
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 27/05/2025, 23:53:19 UTC
Is there a quick math trick to determine if two points share the same higher level endomorphic ring? Just by their X and Y?

By higher-level, I mean 149 or 631. 2 and 3 are obvious. I know that all of the points that share the same subgroup sum up to O. It's also possible to build up the subgroup entirely, but is this really required, in order to make such a check, that the other point is in the same subgroup?

If I understood your question right, maybe one possible way to simplify the check is to first verify whether both points are n-torsion points (nP = O and nQ = O). If not, you can already rule out that they belong to the same n-order subgroup. If they both are, i think you have to build up the entire subgroup as you said, i don't think there is a direct way to check. But at least the n-torsion check can help "quickly" exclude cases that definitely don’t match, or at least use less computation. Could this make sense?

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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 22/05/2025, 07:13:57 UTC

Look VS-bitcrack on my GitHub, It should works also with prefix
I will
Thanks you so much Smiley
D:\Zdown\VanitySearch-master\VanitySearch-master>VanitySearch.exe -t 0 -gpu -gpuId 0 --keyspace 100000000:1ffffffff 187swFMjz                       Unexpected --keyspace argument

I think keyspace mode (range) not included.


You have to use -start and -range, look at the readme, is there on purpose
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 22/05/2025, 00:10:29 UTC
Hi guys. Is it possible for someone to have an executable for Windows and GPU with range and prefix search?

Venitysearch doesn't have range. Vanity Bitcrack doesn't work well with prefix, Keyhunt Cuda doesn't have prefix search. Keyhunt have all and work amazing but only for CPU.

Is there one for Windows + GPU + Range + Prefix? Smiley

thanks to all

Look VS-bitcrack on my GitHub, It should works also with prefix
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 19/05/2025, 11:01:20 UTC
Akito did you modified the FixedPaul's code to specifically search in 71bit range and another thing i want to ask is, like nodemath code he is searching for prefix f6f543, if someone is able to find multiple of these prefixes in the range, will it help in any way to reach the target faster?

You don’t need to modify the code, just use '-range 70'.

As for prefix, the answer is no, there’s no way to find the target in a "faster" way, but this has already been discussed and I don’t want to start a new argument. Someone here, however, will tell you that it is possible
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 13/05/2025, 09:20:31 UTC
How to edit VanitySearch by JLP to search in puzzle 71 space (not from 1)?

Anybody gone into c/c++ in this project?

Try JLP VanitySearch forked by @allinbit and @Ilker, starting from ranges what you want.
https://github.com/ilkerccom/VanitySearch

It hangs when given keyspace option.

There was need to modify the code. Works fine now.

If you want in my GitHub you can find VanitySearch-bitcrack repo, 50% faster than these
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 09/05/2025, 09:45:07 UTC

My theory is called Three Numbers. They are 4, 7, and 9. It is applicable in this puzzle.
 Those who don't believe can apply the two previous puzzles using these numbers, and everything will fall into place!
 For the script creators, I would also ask you to send me the working script in a private message if you create it.
 If you want, I can write examples for you and post them here.


This is a great method, I'm writing a script to find all the digits of Pi!
Finally, something very smart in this forum topic

Pi = 3.1415926535897932384 (first 20 digits)

3 = 7 - 4
1 = 4 + 4 - 7
4 = 4
1 = 4 + 4 - 7
5 = 9 - 4
9 = 9
2 = 9 - 7
6 = 9 - 7 + 4
5 = 9 - 4
3 = 7 - 4
5 = 9 - 4
8 = 4 + 4
9 = 9
7 = 7
9 = 9
3 = 7 - 4
2 = 9 - 7
3 = 7 - 4
8 = 4 + 4
4 = 4
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 05/05/2025, 09:11:02 UTC

You can see here also that Y number for puzzle 69 is low  Grin


You're computing a number that is simply sqrt(X) * 8192.
Where X is the normalized key position (from 0 to 1) in the puzzle space. What am I missing?   Cheesy
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 04/05/2025, 00:58:01 UTC

It's not working on windows 10 64bits, it says it's a 16 bit application

Try downloading the entire zipped folder, not the only .exe
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 03/05/2025, 23:05:04 UTC
Besides bitcrack which wasn't updated for the last 5+ years,  is there any similar tool using the cuda GPU but to generate randomly within a range?

You can check my GitHub, VanitySearch-bitcrack has a random mode

I don't see


https://github.com/FixedPaul/VanitySearch-Bitcrack

Thanks. Can you publish one release please? "Releases No releases published"

Yes i'll do, you can find compiled version in folder in meantime
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 03/05/2025, 22:53:59 UTC
Besides bitcrack which wasn't updated for the last 5+ years,  is there any similar tool using the cuda GPU but to generate randomly within a range?

You can check my GitHub, VanitySearch-bitcrack has a random mode

I don't see


https://github.com/FixedPaul/VanitySearch-Bitcrack
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 03/05/2025, 22:20:05 UTC
Besides bitcrack which wasn't updated for the last 5+ years,  is there any similar tool using the cuda GPU but to generate randomly within a range?

You can check my GitHub, VanitySearch-bitcrack has a random mode
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
fixedpaul
on 01/05/2025, 09:52:53 UTC


Attention!



Do you want to be the best mathematician?

Here's a new method to beat a sequential search that goes from 0 to N-1 !

It's called sequential-advanced, and it's based on solid mathematical and probabilistic theories. In fact, the probability that it does not find the key before the classic sequential method is 1/N — that's 1/2^68 if we look at puzzle 69, almost zero!!
This means that, compared to the classic sequential search, the advanced method will find the key first in 99.99999999% of cases! Incredible.

But let's put it to the test with a script

Code:
import hashlib
import random
import time

TOTAL_SIZE = 10000
SIMULATIONS = 500

def generate_h160(data):
    return hashlib.new('ripemd160', str(data).encode()).hexdigest()

def sequential(dataset, target_hash):
    checks = 0
    for i in range(len(dataset)):
        checks += 1
        if generate_h160(dataset[i]) == target_hash:
            return {"checks": checks, "found": True}
    return {"checks": checks, "found": False}

def sequential_advanced(dataset, target_hash):
    checks = 0
    for i in range(1, len(dataset)):
        checks += 1
        if generate_h160(dataset[i]) == target_hash:
            return {"checks": checks, "found": True}
    checks += 1  # check 0 at the end
    if generate_h160(dataset[0]) == target_hash:
        return {"checks": checks, "found": True}
    return {"checks": checks, "found": False}

def compare_methods():
    results = {
        "sequential": {"wins": 0, "total_checks": 0, "total_time": 0},
        "sequential_advanced": {"wins": 0, "total_checks": 0, "total_time": 0},
        "ties": 0
    }

    for i in range(SIMULATIONS):
        dataset = list(range(TOTAL_SIZE))
        random.shuffle(dataset)
        target = random.choice(dataset)
        target_hash = generate_h160(target)

        start = time.perf_counter()
        res_seq = sequential(dataset, target_hash)
        time_seq = time.perf_counter() - start

        start = time.perf_counter()
        res_adv = sequential_advanced(dataset, target_hash)
        time_adv = time.perf_counter() - start

        results["sequential"]["total_checks"] += res_seq["checks"]
        results["sequential_advanced"]["total_checks"] += res_adv["checks"]
        results["sequential"]["total_time"] += time_seq
        results["sequential_advanced"]["total_time"] += time_adv

        if res_seq["checks"] < res_adv["checks"]:
            results["sequential"]["wins"] += 1
        elif res_seq["checks"] > res_adv["checks"]:
            results["sequential_advanced"]["wins"] += 1
        else:
            results["ties"] += 1

        print(f"Simulation {i+1}: Sequential = {res_seq['checks']} checks in {time_seq:.6f}s | Sequential-Advanced = {res_adv['checks']} checks in {time_adv:.6f}s")

    # Calcolo medie
    avg_time_seq = results["sequential"]["total_time"] / results["sequential"]["wins"] if results["sequential"]["wins"] > 0 else float('inf')
    avg_time_adv = results["sequential_advanced"]["total_time"] / results["sequential_advanced"]["wins"] if results["sequential_advanced"]["wins"] > 0 else float('inf')
    avg_checks_seq = results["sequential"]["total_checks"] / results["sequential"]["wins"] if results["sequential"]["wins"] > 0 else float('inf')
    avg_checks_adv = results["sequential_advanced"]["total_checks"] / results["sequential_advanced"]["wins"] if results["sequential_advanced"]["wins"] > 0 else float('inf')

    print(f"""
=== FINAL RESULTS ===
Wins:
Sequential: {results['sequential']['wins']}
Sequential-Advanced: {results['sequential_advanced']['wins']}
Ties: {results['ties']}

Total Checks:
Sequential: {results['sequential']['total_checks']}
Sequential-Advanced: {results['sequential_advanced']['total_checks']}

Total Time:
Sequential: {results['sequential']['total_time']:.6f} seconds
Sequential-Advanced: {results['sequential_advanced']['total_time']:.6f} seconds

Averages (Total Time / Wins):
Sequential : {avg_time_seq:.6f} seconds/victory
Sequential-Advanced : {avg_time_adv:.6f} seconds/victory

Checks per Win:
Sequential : {avg_checks_seq:.2f} checks/win
Sequential-Advanced : {avg_checks_adv:.2f} checks/win
""")

if __name__ == '__main__':
    compare_methods()


Result:

Code:
=== FINAL RESULTS ===
Wins:
Sequential: 0
Sequential-Advanced: 500
Ties: 0

Total Checks:
Sequential: 2397028
Sequential-Advanced: 2396528

Total Time:
Sequential: 15.394807 seconds
Sequential-Advanced: 12.848593 seconds

Averages (Total Time / Wins):
Sequential : inf seconds/victory
Sequential-Advanced : 0.025697 seconds/victory

Checks per Win:
Sequential : inf checks/win
Sequential-Advanced : 4793.06 checks/win

Look at that! 500 to 0 — unbelievable, right? Never use the normal sequential search again.

And don’t use the prefix-method either, because against the original sequential approach, you might lose sometimes.
With this new method, instead, you’re practically unbeatable!

But how does it work?

Simple: the advanced method searches sequentially starting from 1 instead of 0, and it checks 0 only at the end.

Checkmate to the sequential method from 0 to N-1!
I've found a method that retires it — one that lets me find the key faster in 99.999999% of cases.

Right?

Puzzle 69 solved as soon as I posted 100% winrate method. Will accept donation too