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Showing 8 of 8 results by viceversas
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Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
viceversas
on 16/05/2025, 18:41:14 UTC
I love the way the thread turned into occult poetry. I was hoping AI was smart enough yet, to build up smth really fast to solve at least 71, 72, 73 by now, since mockers serious people seem to had enough hate. After all, what's faster than running some non-working app? Simply ask AI for the solution, everyone's in a hurry after all.

I don't understand your obsession with treating others disrespectfully, you are becoming more and more like Digaran. If what is shared here bothers you so much, why not just leave people alone? If you want to correct others, there's no need to insult them or treat them as inferior.

Stop the trolling and keep your word… oh, right, you can't, because your need to troll is stronger. I recommend that anyone who sees this person trolling, insulting, or harassing someone report them or file a complaint in the reputation section.

or maybe just ignore him, stop making s* people famous.
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Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
viceversas
on 16/05/2025, 11:44:51 UTC
Currently, when I want to solve puzzle 71 using cuda, which is the best SOFTWARE to use?

Use Cyclone from Nomachine1 it do all !  Cheesy Cheesy Cheesy Cheesy Cheesy

what a D*
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Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
viceversas
on 29/04/2025, 19:24:00 UTC
please write your opinion, i m very interested.


You can't imagine what I've tried in the last six years.
I even went back in time, generating numbers from 2015—every second of that year.
From my archive of useless scripts:

Code:
import random
from datetime import datetime, timedelta

# List of target Puzzle, each corresponding to a range
target_numbers = [
    (1, 1), (2, 3), (3, 7), (4, 8), (5, 21), (6, 49), (7, 76), (8, 224), (9, 467), (10, 514),
    (11, 1155), (12, 2683), (13, 5216), (14, 10544), (15, 26867), (16, 51510),
    (17, 95823), (18, 198669), (19, 357535), (20, 863317), (21, 1811764),
    (22, 3007503), (23, 5598802), (24, 14428676), (25, 33185509),
    (26, 54538862), (27, 111949941), (28, 227634408), (29, 400708894),
    (30, 1033162084), (31, 2102388551), (32, 3093472814), (33, 7137437912),
    (34, 14133072157), (35, 20112871792), (36, 42387769980), (37, 100251560595),
    (38, 146971536592), (39, 323724968937), (40, 1003651412950),
    (41, 1458252205147), (42, 2895374552463), (43, 7409811047825),
    (44, 15404761757071), (45, 19996463086597), (46, 51408670348612),
    (47, 119666659114170), (48, 191206974700443), (49, 409118905032525),
    (50, 611140496167764), (51, 2058769515153876), (52, 4216495639600700),
    (53, 6763683971478124), (54, 9974455244496707), (55, 30045390491869460),
    (56, 44218742292676575), (57, 138245758910846492), (58, 199976667976342049),
    (59, 525070384258266191), (60, 1135041350219496382), (61, 1425787542618654982),
    (62, 3908372542507822062), (63, 8993229949524469768),
    (64, 17799667357578236628), (65, 30568377312064202855)
]

# Sort the target_numbers list by the first element of each tuple (the range start)
target_numbers.sort(key=lambda x: x[0])

# Specify the start and end date and times for the search
start_datetime_pre = datetime(2015, 1, 1, 0, 0, 0)
end_datetime_pre = datetime(2015, 1, 15, 19, 7, 14)
current_datetime = start_datetime_pre
time_step = timedelta(seconds=1)

# Initialize a set to keep track of found target numbers
found_targets = set()

# Function to find the seed for a single target number
def find_seed_for_target(target_num, current_time):
    num, target_number = target_num
    min_number = 2 ** (num - 1)
    max_number = (2 ** num) - 1

    low_seed = int(current_time.timestamp())
    high_seed = int(end_datetime_pre.timestamp())

    found_seed = None

    while low_seed <= high_seed:
        mid_seed = (low_seed + high_seed) // 2

        random.seed(mid_seed)
        generated_number = random.randint(min_number, max_number)

        if generated_number == target_number:
            found_seed = mid_seed
            break
        elif generated_number < target_number:
            low_seed = mid_seed + 1
        else:
            high_seed = mid_seed - 1

    return found_seed

# Iterate through the time range
while current_datetime <= end_datetime_pre:
    # Find seeds for all target numbers
    found_seeds = [find_seed_for_target(target, current_datetime) for target in target_numbers]

    # Print the results for each target number if found and not already printed
    for i, (num, target_number) in enumerate(target_numbers, start=1):
        if found_seeds[i - 1] is not None and target_number not in found_targets:
            linuxtime = found_seeds[i - 1]
            timestamp = datetime.fromtimestamp(linuxtime)
            formatted_time = timestamp.strftime('%Y-%m-%d %H:%M:%S')
            print(f"Puzzle {i} : Private Key : {target_number} | Timestamp: {formatted_time}")
            found_targets.add(target_number)

    # Move to the next second
    current_datetime += time_step


Puzzle 1 : Private Key : 1 | Timestamp: 2015-01-08 09:33:37
Puzzle 2 : Private Key : 3 | Timestamp: 2015-01-12 02:20:26
Puzzle 3 : Private Key : 7 | Timestamp: 2015-01-12 02:20:26
Puzzle 4 : Private Key : 8 | Timestamp: 2015-01-01 05:32:54
Puzzle 5 : Private Key : 21 | Timestamp: 2015-01-10 05:57:01
Puzzle 10 : Private Key : 514 | Timestamp: 2015-01-01 02:46:31
Puzzle 6 : Private Key : 49 | Timestamp: 2015-01-04 16:08:31
Puzzle 7 : Private Key : 76 | Timestamp: 2015-01-01 00:26:36
Puzzle 12 : Private Key : 2683 | Timestamp: 2015-01-01 22:35:35
Puzzle 9 : Private Key : 467 | Timestamp: 2015-01-12 01:35:26
Puzzle 11 : Private Key : 1155 | Timestamp: 2015-01-05 14:58:56
Puzzle 8 : Private Key : 224 | Timestamp: 2015-01-13 00:32:15
Puzzle 13 : Private Key : 5216 | Timestamp: 2015-01-01 23:52:16
Puzzle 14 : Private Key : 10544 | Timestamp: 2015-01-05 03:59:22
Puzzle 16 : Private Key : 51510 | Timestamp: 2015-01-04 16:54:18
Puzzle 17 : Private Key : 95823 | Timestamp: 2015-01-02 16:43:39
Puzzle 15 : Private Key : 26867 | Timestamp: 2015-01-02 13:47:12
Puzzle 22 : Private Key : 3007503 | Timestamp: 2015-01-08 15:41:51


this looks like a script for randstorm, cmiiw
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Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
viceversas
on 28/04/2025, 09:18:30 UTC
I'm not going to prove anything to anyone, but here is the Millisecond Puzzle 30 solver

Code:
import random
import os
import time
import secp256k1 as ice

puzzle = 30
target = "d39c4704664e1deb76c9331e637564c257d68a08"
lower_range_limit = 2 ** (puzzle - 1)
upper_range_limit = (2 ** puzzle) - 1

start_time = time.time()

for x in range(10000000):
    #Random seed Config
    #constant_prefix = b''  #back to no constant
    constant_prefix = b'yx\xcb\x08\xb70l'
    prefix_length = len(constant_prefix)
    length = 8
    ending_length = length - prefix_length
    ending_bytes = os.urandom(ending_length)
    random_bytes = constant_prefix + ending_bytes
    random.seed(random_bytes)
    dec = random.randint(lower_range_limit, upper_range_limit)
    h160 = ice.privatekey_to_h160(0, True, dec).hex()
    if h160 == target:
        HEX = "%064x" % dec
        caddr = ice.privatekey_to_address(0, True, dec)
        wifc = ice.btc_pvk_to_wif(HEX)
        print("Bitcoin address Compressed: " + caddr)
        print("Private Key (decimal): " + str(dec))
        print("Private key (wif) Compressed : " + wifc)
        print(f"Random seed: {random_bytes}")
        break

end_time = time.time()
execution_time_ms = (end_time - start_time) * 1000

print("Execution Time (ms):", execution_time_ms)

Bitcoin address Compressed: 1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps
Private Key (decimal): 1033162084
Private key (wif) Compressed : KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M8diLSC5MyERoW
Random seed: b'yx\xcb\x08\xb70l\xf1'
Execution Time (ms): 2.977609634399414


There are no rules for how any puzzle should be solved — the rule is that there are no rules Grin



even less in mine lol

Code:
Bitcoin address Compressed: 1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps
Private Key (decimal): 1033162084
Private key (wif) Compressed : KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M8diLSC5MyERoW
Random seed: b'yx\xcb\x08\xb70l\xf1'
Execution Time (ms): 1.6162395477294922
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Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
viceversas
on 28/04/2025, 07:41:05 UTC
In theory, modular traversal is bijective and randomized. So, generally speaking, if the target keys are randomly distributed across the keyspace, modular traversal can be as efficient as linear traversal. But, for clustered targets like in the 2^69 puzzle, modular traversal is not efficient because the targets were clustered near the starting point (within a small range of the keyspace).

just my 2cents.
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Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
viceversas
on 24/04/2025, 18:35:25 UTC
Uso Bitcrack atualmente, qual vc acha melhor ? esta bem testado ?

english please.
Post
Topic
Board Development & Technical Discussion
Re: Probabilistic search of prefixes vs random+sequential
by
viceversas
on 23/04/2025, 17:48:43 UTC
great!

imo, in the context of prefix filtering for Bitcoin private key recovery, Hamming distance tolerance can be used to widen the match range without heavily sacrificing efficiency. Maybe combining not only prefix address or hash160, but both.  Smiley
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Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
viceversas
on 17/04/2025, 20:31:50 UTC
But what are you going to do with it?


I need a script that can recover the beginning of a WIF where some characters are missing — for example, WIF500, etc  Tongue

[/quote]

https://github.com/NoMachine1/WIFHunter

There's no manual for this. If you can't figure out how it works on your own, then you don't need the script.  Grin
[/quote]

https://imgur.com/a/EwUM3kj

this is great! Thank you.