Bitcoin puzzle transaction ~32 BTC prize to who solves it

[fecell]

Code:

C:\_CuBubNix>wsl
user@DESKTOP-4A3SFP6:/mnt/c/_CuBubNix$ ./solver_kav0m -f 75
total bits: 75
range bits: 48
total iterations: 140,737,488,355,328
range step:       245,317,739,369,851
steps count:      77,000,000
total checks:     10,836,786,603,360,256,000,000

[!] Initializing GPU secp256k1...
Initializing secp256k1 for GPU...
secp256k1 constants initialized on GPU
Initializing precomputed table for base point G...
Table size: 256 entries (0.02 MB)
Processing batch: 2 to 256 (size: 254)
Computed 256 entries (100.0%)
Actual entries computed: 256 out of 256
Precomputed table initialized successfully in global memory
secp256k1 GPU initialization complete!

[!] Allocating GPU memory...

[!] Generating 77000000 base points...
[+] base points saved: basepoints_P75_step77000000.dat

[!] Generating 154000000 check points...
[+] checkpoints saved: checkpoints_P75_step77000000.dat
[+] Check points done in 223.12 s
[!] Building sorted check map (154000000 entries)...
[+] check map saved: checkmap_P75_step77000000.dat
[+] Check map built: 154000000 entries

[!] Creating bloom filter...
[+] Creating Bloom filter: 154000000 items, 2952198016 bits, 13 hashes, 369024752 bytes
[+] Bloom filter: 369024752 bytes
[+] bloom saved: bloom_P75_step77000000.dat
[!] Starting fresh

[!] Starting search...
[+] [284,800] S: 4384147563670.5 it/s, TE: 00:00:05, E: 00:00:05, ETA: 78y 4m 18d 22:19:09
[+] [582,912] S: 4487179977492.4 it/s, TE: 00:00:10, E: 00:00:10, ETA: 76y 7m 2d 00:37:06
[+] [882,560] S: 4529273222593.1 it/s, TE: 00:00:15, E: 00:00:15, ETA: 75y 10m 17d 06:01:10
[+] [1,180,800] S: 4544816919025.3 it/s, TE: 00:00:20, E: 00:00:20, ETA: 75y 7m 12d 12:58:35
[+] [1,481,856] S: 4562944268013.4 it/s, TE: 00:00:25, E: 00:00:25, ETA: 75y 3m 22d 21:40:21
[+] [1,787,392] S: 4586559085856.6 it/s, TE: 00:00:30, E: 00:00:30, ETA: 74y 11m 6d 09:01:25
[+] [2,092,160] S: 4601821799443.2 it/s, TE: 00:00:35, E: 00:00:35, ETA: 74y 8m 5d 16:15:00
[+] [2,397,184] S: 4613541927680.0 it/s, TE: 00:00:40, E: 00:00:40, ETA: 74y 5m 26d 10:29:49
[+] [2,701,312] S: 4621228818237.6 it/s, TE: 00:00:45, E: 00:00:45, ETA: 74y 4m 11d 05:10:42
[+] [3,005,184] S: 4626953023530.4 it/s, TE: 00:00:50, E: 00:00:50, ETA: 74y 3m 7d 15:18:54
[+] [3,310,464] S: 4633610024738.0 it/s, TE: 00:00:55, E: 00:00:55, ETA: 74y 1m 28d 16:38:07
[+] [3,616,896] S: 4640613488240.3 it/s, TE: 00:01:00, E: 00:01:00, ETA: 74y 17d 20:12:19
[+] [3,924,608] S: 4648118837239.4 it/s, TE: 00:01:05, E: 00:01:05, ETA: 73y 11m 9d 04:47:42
[+] [4,231,936] S: 4654095772874.0 it/s, TE: 00:01:10, E: 00:01:10, ETA: 73y 10m 4d 13:05:55
[+] [4,538,112] S: 4658143282004.4 it/s, TE: 00:01:15, E: 00:01:15, ETA: 73y 9m 11d 03:05:44
[+] [4,844,032] S: 4661454706113.2 it/s, TE: 00:01:20, E: 00:01:20, ETA: 73y 8m 22d 00:01:28
[+] [5,152,768] S: 4666893327847.9 it/s, TE: 00:01:25, E: 00:01:25, ETA: 73y 7m 20d 15:28:10
[+] [5,460,992] S: 4671219832224.0 it/s, TE: 00:01:30, E: 00:01:30, ETA: 73y 6m 25d 18:03:06
[+] [5,767,808] S: 4673999019513.5 it/s, TE: 00:01:35, E: 00:01:35, ETA: 73y 6m 9d 18:52:32
[+] [6,074,752] S: 4676591312679.1 it/s, TE: 00:01:40, E: 00:01:40, ETA: 73y 5m 24d 21:52:39
[+] [6,382,336] S: 4679429787516.5 it/s, TE: 00:01:45, E: 00:01:45, ETA: 73y 5m 8d 15:25:49
[+] [6,688,768] S: 4681210115712.4 it/s, TE: 00:01:50, E: 00:01:50, ETA: 73y 4m 28d 10:46:41

How are 77 bits cracked in seconds?!

[brainless]

No new news, is good news ?
All seems on vocation...

[bankeroft]

No new news, is good news ?
All seems on vocation...

or silence before the storm 

[zahid888]

What do you think about the puzzle creator's statement regarding the puzzle pattern—that it consists of sequential keys generated from a deterministic wallet, masked with leading zeros to lower the difficulty? Is there a flaw in the logic used to generate the private keys here?

Code:

import secp256k1 as ice
from mnemonic import Mnemonic
from bitcoin import *
from pathlib import Path
from bloomfilter import BloomFilter, ScalableBloomFilter, SizeGrowthRate
import time
import os


mys = Path(__file__).resolve()
fil = mys.parents[1] / 'found/found.txt'

myself = Path(__file__).resolve()
res = myself.parents[1] / 'datafiles/wordlist/english.txt'

myselff = Path(__file__).resolve()
ress = myselff.parents[1] / 'datafiles/BF/Addresses.bf'


with open(res) as f:
    wordlist = f.read().split('\n')

with open(ress, "rb") as fp:
    bloom_filter = BloomFilter.load(fp)
strength = 256
found = 0
phrase = 0
count = 0
while True:
    
    rbytes = os.urandom(strength // 8)
    h = hashlib.sha256(rbytes).hexdigest()

    b = ( bin(int.from_bytes(rbytes, byteorder="big"))[2:].zfill(len(rbytes) * 8) \
         + bin(int(h, 16))[2:].zfill(256)[: len(rbytes) * 8 // 32] )

    result = []
    for i in range(len(b) // 11):
        idx = int(b[i * 11 : (i + 1) * 11], 2)
        result.append(wordlist[idx])

    mnemonic = " ".join(result)
    phrase+=1
    backup_phrase = mnemonic
    mnemo = Mnemonic("english")
    seed = mnemo.to_seed(backup_phrase)
    master_key = bip32_master_key(seed)
    path = "m/44h/0h/0h/0/"   
    last_index = 8
    while last_index <= 160:
        index_path = path + str(last_index)
        indices = [int(i[:-1]) | 0x80000000 if i[-1] == "h" else int(i) for i in index_path.split("/") if i != "m"]
        derived_key = master_key
        for index in indices:
            derived_key = bip32_ckd(derived_key, index)
        private_key = bip32_extract_key(derived_key)
        compressed_private_key = encode_privkey(private_key, 'hex_compressed')
        hexa = compressed_private_key[:64]
        mask = 1 << last_index
        first_key = mask + int(hexa, 16) % mask
        first_key_hex = format(first_key, 'x').zfill(64)
        mask_hex = format(mask, 'x').zfill(64)
        hex = first_key_hex
        pvk = bytes.fromhex(str(hex))
        pvk2 = bytes.fromhex(str(hexa))
        addr1 = ice.privatekey_to_address(0, True, (int.from_bytes(pvk, "big")))
        addr2 = ice.privatekey_to_address(0, True, (int.from_bytes(pvk2, "big")))
        last_index += 1
        count += 1
        if addr2 in bloom_filter:
            print("BTC p2pkh comp:         \n" +addr2)
            print(hexa)
            print(backup_phrase + "\n"+ "\n")
            g=open(fil,"a")
            g.write("BTC p2pkh comp:         \n" +addr2)
            g.write("\n"+hexa)
            g.write("\n"+str(backup_phrase) + "\n"+ "\n")
            g.close()
            found += 1
        if addr1 in bloom_filter:
            print("BTC p2pkh comp:         \n" +addr1)
            print(hex)
            print(backup_phrase + "\n"+ "\n")
            g=open(fil,"a")
            g.write("BTC p2pkh comp:         \n" +addr1)
            g.write("\n"+hex)
            g.write("\n"+str(backup_phrase) + "\n"+ "\n")
            g.close()
            found += 1
        else: print('\n\n ================================ Deterministic wallet masked with leading Zeros ==================================\n\n'"Bip44 : "+mnemonic+"\nBipPH : "+index_path+"\nPhras : "+str(phrase)+"\ncount : "+str(count * 2)+"\nFound : "+str(found)+"\nPrivM : "+hexa+"\nMaskM : "+mask_hex+"\nMaskp : "+hex+"\nAddrp : "+addr1+"\nAddrM : "+addr2, end='\r')
        

[mumsidea]

Any new findings from Puzzle #71 research?

Have recent searches provided any better insights or useful data?
Also, how are the results from prefix analysis looking? Are there any updated prefixes or density observations worth considering?

[OzBtcOz]

What do you think about the puzzle creator's statement regarding the puzzle pattern—that it consists of sequential keys generated from a deterministic wallet, masked with leading zeros to lower the difficulty? Is there a flaw in the logic used to generate the private keys here?

Code:

import secp256k1 as ice
from mnemonic import Mnemonic
from bitcoin import *
from pathlib import Path
from bloomfilter import BloomFilter, ScalableBloomFilter, SizeGrowthRate
import time
import os


mys = Path(__file__).resolve()
fil = mys.parents[1] / 'found/found.txt'

myself = Path(__file__).resolve()
res = myself.parents[1] / 'datafiles/wordlist/english.txt'

myselff = Path(__file__).resolve()
ress = myselff.parents[1] / 'datafiles/BF/Addresses.bf'


with open(res) as f:
    wordlist = f.read().split('\n')

with open(ress, "rb") as fp:
    bloom_filter = BloomFilter.load(fp)
strength = 256
found = 0
phrase = 0
count = 0
while True:
    
    rbytes = os.urandom(strength // 8)
    h = hashlib.sha256(rbytes).hexdigest()

    b = ( bin(int.from_bytes(rbytes, byteorder="big"))[2:].zfill(len(rbytes) * 8) \
         + bin(int(h, 16))[2:].zfill(256)[: len(rbytes) * 8 // 32] )

    result = []
    for i in range(len(b) // 11):
        idx = int(b[i * 11 : (i + 1) * 11], 2)
        result.append(wordlist[idx])

    mnemonic = " ".join(result)
    phrase+=1
    backup_phrase = mnemonic
    mnemo = Mnemonic("english")
    seed = mnemo.to_seed(backup_phrase)
    master_key = bip32_master_key(seed)
    path = "m/44h/0h/0h/0/"   
    last_index = 8
    while last_index <= 160:
        index_path = path + str(last_index)
        indices = [int(i[:-1]) | 0x80000000 if i[-1] == "h" else int(i) for i in index_path.split("/") if i != "m"]
        derived_key = master_key
        for index in indices:
            derived_key = bip32_ckd(derived_key, index)
        private_key = bip32_extract_key(derived_key)
        compressed_private_key = encode_privkey(private_key, 'hex_compressed')
        hexa = compressed_private_key[:64]
        mask = 1 << last_index
        first_key = mask + int(hexa, 16) % mask
        first_key_hex = format(first_key, 'x').zfill(64)
        mask_hex = format(mask, 'x').zfill(64)
        hex = first_key_hex
        pvk = bytes.fromhex(str(hex))
        pvk2 = bytes.fromhex(str(hexa))
        addr1 = ice.privatekey_to_address(0, True, (int.from_bytes(pvk, "big")))
        addr2 = ice.privatekey_to_address(0, True, (int.from_bytes(pvk2, "big")))
        last_index += 1
        count += 1
        if addr2 in bloom_filter:
            print("BTC p2pkh comp:         \n" +addr2)
            print(hexa)
            print(backup_phrase + "\n"+ "\n")
            g=open(fil,"a")
            g.write("BTC p2pkh comp:         \n" +addr2)
            g.write("\n"+hexa)
            g.write("\n"+str(backup_phrase) + "\n"+ "\n")
            g.close()
            found += 1
        if addr1 in bloom_filter:
            print("BTC p2pkh comp:         \n" +addr1)
            print(hex)
            print(backup_phrase + "\n"+ "\n")
            g=open(fil,"a")
            g.write("BTC p2pkh comp:         \n" +addr1)
            g.write("\n"+hex)
            g.write("\n"+str(backup_phrase) + "\n"+ "\n")
            g.close()
            found += 1
        else: print('\n\n ================================ Deterministic wallet masked with leading Zeros ==================================\n\n'"Bip44 : "+mnemonic+"\nBipPH : "+index_path+"\nPhras : "+str(phrase)+"\ncount : "+str(count * 2)+"\nFound : "+str(found)+"\nPrivM : "+hexa+"\nMaskM : "+mask_hex+"\nMaskp : "+hex+"\nAddrp : "+addr1+"\nAddrM : "+addr2, end='\r')
        

It's partially correct, but not all!
Please use below one! I reached up to 38 bit with it. It is Human Intellengence, not Artificial!

You only need 160.txt which includes all puzzle comp. add. and BIP39 words as kelimeler.txt.

Code:

#!/usr/bin/env python3
"""
Electrum Çoklu Kelime Uyumlu & Bit Shifting Tabanlı Bitcoin Puzzle Dedektörü
------------------------------------------------------------------------
Desteklenen Kelime Sayıları: 12, 13, 15, 18, 21, 24
Koşul: 160.txt dosyasında Satır 25'ten büyük (Satır > 25) bir eşleşme bulursa kaydeder.
Canlı İzleme: Anlık eşleşen adres-hex matrisini ve toplam eşleşme sayısını ekranda gösterir.
Kurulum için: pip install coincurve base58
"""

import hashlib
import hmac
import base58
import struct
import os
import sys
import time
import secrets
import multiprocessing
from hashlib import pbkdf2_hmac
from coincurve import PublicKey

N = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

# --- KRİPTOGRAFİ VE ADRES TÜRETME FONKSİYONLARI ---

def private_key_to_public_key_compressed(priv_key_bytes):
    try:
        return PublicKey.from_secret(priv_key_bytes).format(compressed=True)
    except Exception:
        return None

def public_key_to_address(pub_key_bytes):
    if pub_key_bytes is None: return ""
    sha = hashlib.sha256(pub_key_bytes).digest()
    rip = hashlib.new('ripemd160', sha).digest()
    payload = b'\x00' + rip
    checksum = hashlib.sha256(hashlib.sha256(payload).digest()).digest()[:4]
    return base58.b58encode(payload + checksum).decode()

def private_key_to_wif_compressed(priv_key_bytes):
    extended_key = b'\x80' + priv_key_bytes
    extended_key += b'\x01'
    first_sha = hashlib.sha256(extended_key).digest()
    second_sha = hashlib.sha256(first_sha).digest()
    checksum = second_sha[:4]
    return base58.b58encode(extended_key + checksum).decode()

def electrum_mnemonic_to_seed(mnemonic, passphrase=""):
    return pbkdf2_hmac("sha512", mnemonic.encode("utf-8"), f"electrum{passphrase}".encode("utf-8"), 2048, 64)

def bip32_master_key_from_seed(seed_bytes):
    I = hmac.new(b"Bitcoin seed", seed_bytes, hashlib.sha512).digest()
    return I[:32], I[32:]

def derive_bip42_step(current_priv, current_chain, index):
    if index >= 0x80000000:
        data = b'\x00' + current_priv + struct.pack('>I', index)
    else:
        pub_comp = private_key_to_public_key_compressed(current_priv)
        if pub_comp is None: return current_priv, current_chain
        data = pub_comp + struct.pack('>I', index)
        
    I = hmac.new(current_chain, data, hashlib.sha512).digest()
    IL, IR = I[:32], I[32:]
    
    il_int = int.from_bytes(IL, 'big')
    parent_int = int.from_bytes(current_priv, 'big')
    
    if il_int >= N: 
        return current_priv, current_chain
    child_priv_int = (il_int + parent_int) % N
    if child_priv_int == 0: 
        return current_priv, current_chain
        
    return child_priv_int.to_bytes(32, 'big'), IR

def load_target_addresses(filename="160.txt"):
    if not os.path.exists(filename):
        print(f"❌ Hata: {filename} dosyası bulunamadı!")
        sys.exit(1)
    
    addresses_with_index = {}
    with open(filename, "r", encoding="utf-8") as f:
        for idx, line in enumerate(f, start=1):
            addr = line.strip()
            if addr:
                addresses_with_index[addr] = idx
    return addresses_with_index

# --- İŞÇİ ÇEKİRDEK (WORKER PROCESS) PROTOKOLÜ ---

def worker_process(worker_id, target_addresses, WORDS, word_choice, shared_counter, shared_preview, lock):
    seed_version_bytes = b"Seed version"
    bip44_fixed_indices = [0x8000002C, 0x80000000, 0x80000000, 0]
    
    while True:
        # Seçilen kelime sınırına göre dinamik kombinasyon üretimi
        word_list_sample = [secrets.choice(WORDS) for _ in range(word_choice)]
        seed_phrase = " ".join(word_list_sample)
        
        h = hmac.new(seed_version_bytes, seed_phrase.encode('utf-8'), hashlib.sha512).digest()
        # Electrum Standart/Legacy Cüzdan ön ek doğrulama kuralı
        if h[0] != 1:
            continue
            
        with lock:
            shared_counter.value += 1
            global_valid_count = shared_counter.value

        seed_bytes = electrum_mnemonic_to_seed(seed_phrase, "")
        master_priv, master_chain = bip32_master_key_from_seed(seed_bytes)
        
        current_priv = master_priv
        current_chain = master_chain
        for idx in bip44_fixed_indices:
            current_priv, current_chain = derive_bip42_step(current_priv, current_chain, idx)
            
        sample_hex_71 = ""
        current_seed_matches = []
        
        for index in range(160):
            raw_priv, _ = derive_bip42_step(current_priv, current_chain, index)
            raw_priv_int = int.from_bytes(raw_priv, 'big')
            
            target_bit_length = index + 1
            shift_amount = 256 - target_bit_length
            
            shifted_int = raw_priv_int >> shift_amount
            legal_base = 1 << index  
            manipulated_int = legal_base | shifted_int
            
            manipulated_int = manipulated_int % N
            if manipulated_int == 0:
                manipulated_int = 1
                
            manipulated_priv = manipulated_int.to_bytes(32, 'big')
            
            if index == 70: 
                sample_hex_71 = manipulated_priv.hex()
            
            pub_comp = private_key_to_public_key_compressed(manipulated_priv)
            if pub_comp is None: continue
            
            addr_compressed = public_key_to_address(pub_comp)
            
            if addr_compressed in target_addresses:
                line_number = target_addresses[addr_compressed]
                priv_hex = manipulated_priv.hex()
                
                # Arayüz matris yapısı için kısaltma
                short_hex = f"{priv_hex[:8]}...{priv_hex[-8:]}"
                short_addr = f"{addr_compressed[:6]}...{addr_compressed[-6:]}"
                current_seed_matches.append(f"Hex: {short_hex} = {short_addr} (Satır {line_number})")
                
                if line_number > 25:
                    priv_wif = private_key_to_wif_compressed(manipulated_priv)
                    
                    with lock:
                        print(f"\n🎯 🎯 HEDEF BULUNDU! SATIR {line_number} ÜZERİNDE EŞLEŞME! 🎯 🎯")
                        print(f"🔑 Seed ({word_choice} Kelime): {seed_phrase}")
                        print(f"🏠 Adres: {addr_compressed}")
                        
                        with open("winnerseed.txt", "a", encoding="utf-8") as wf:
                            wf.write(f"=== 💎 PUZZLE MATCH ({word_choice} Words) | Zaman: {time.strftime('%Y-%m-%d %H:%M:%S')} ===\n")
                            wf.write(f"Geçerli Seed Phrase : {seed_phrase}\n")
                            wf.write(f"Eşleşen Adres       : {addr_compressed}\n")
                            wf.write(f"160.txt Satır No    : {line_number}\n")
                            wf.write(f"🔑 HEX PRIVATE KEY    : {priv_hex}\n")
                            wf.write(f"🔑 COMPRESSED WIF KEY : {priv_wif}\n")
                            wf.write("=" * 75 + "\n\n")
                        print("💾 Detaylar 'winnerseed.txt' dosyasına kaydedildi.\n")
                
        if global_valid_count % 10 == 0 or len(current_seed_matches) > 0:
            shared_preview['seed'] = seed_phrase
            shared_preview['p71_hex'] = sample_hex_71
            shared_preview['matches'] = current_seed_matches
            shared_preview['total_matches'] = len(current_seed_matches)

# --- ANA MOTOR KONTROL PANELİ ---

def main():
    print("🚀 Esnek Kelime Seçenekli Electrum Bit Shifting Puzzle Dedektörü")
    print("───────────────────────────────────────────────────────────────────────────────")
    
    valid_choices = [12, 13, 15, 18, 21, 24]
    print(f"💡 Kullanılabilir Electrum kelime varyasyonları: {valid_choices}")
    
    try:
        word_choice = int(input("👉 Kaç kelimelik seed aramak istiyorsunuz?: ").strip())
        if word_choice not in valid_choices:
            print("❌ Geçersiz seçim! Program kapatılıyor.")
            return
    except ValueError:
        print("❌ Lütfen sadece sayısal bir değer giriniz!")
        return

    target_addresses = load_target_addresses("160.txt")
    total_targets = len(target_addresses)
    print(f"🎯 {total_targets} adet hedef bulmaca adresi hafızaya yüklendi.")
    
    words_file = "kelimeler.txt"
    if not os.path.exists(words_file):
        print(f"❌ Hata: '{words_file}' bulunamadı!")
        return
    with open(words_file, "r", encoding="utf-8") as f:
        WORDS = [line.strip() for line in f if line.strip()]
    print(f"✅ {len(WORDS)} kaynak kelime havuzu yüklendi.")
    
    num_cores = 3
    print(f"🚀 Hibrit Dedektif Motoru {num_cores} Çekirdek ({word_choice} Kelime Modu) ile Başlatılıyor...\n")
    
    shared_counter = multiprocessing.Value('i', 0)
    manager = multiprocessing.Manager()
    shared_preview = manager.dict()
    shared_preview['seed'] = "Filtreleme Başlatılıyor..."
    shared_preview['p71_hex'] = "0000000000000000"
    shared_preview['matches'] = []
    shared_preview['total_matches'] = 0
    
    lock = multiprocessing.Lock()
    start_time = time.time()
    processes = []
    
    for i in range(num_cores):
        p = multiprocessing.Process(
            target=worker_process, 
            args=(i, target_addresses, WORDS, word_choice, shared_counter, shared_preview, lock)
        )
        processes.append(p)
        p.start()
        
    try:
        while True:
            time.sleep(0.4)
            elapsed = time.time() - start_time
            current_count = shared_counter.value
            speed = current_count / elapsed if elapsed > 0 else 0
            
            # Konsol yenileme
            sys.stdout.write("\033[H\033[J")
            sys.stdout.write(f"🎯 Hedef Bulmaca Adres Havuzu : {total_targets} | Aktif Mod: {word_choice} Kelime\n")
            sys.stdout.write(f"📈 [FİLTRE] İşlenen Geçerli Electrum Seed Seti: {current_count:,} | Süre: {elapsed:.1f} sn | Hız: {speed:.2f} set/sn\n")
            sys.stdout.write(f"───────────────────────────────────────────────────────────────────────────────\n")
            sys.stdout.write(f"🔑 [CANLI İZLENEN GEÇERLİ SEED ({word_choice} KELİME)]:\n   👉 {shared_preview.get('seed')}\n\n")
            
            sys.stdout.write(f"🔍 [ANLIK SEED MATRİSİNDEKİ EŞLEŞMELER]:\n")
            matches_list = shared_preview.get('matches', [])
            if matches_list:
                for match_line in matches_list[:6]:
                    sys.stdout.write(f"   • {match_line}\n")
                if len(matches_list) > 6:
                    sys.stdout.write(f"   • ... ve {len(matches_list) - 6} adres eşleşmesi daha var.\n")
            else:
                sys.stdout.write(f"   (Bu evrende henüz bir 160.txt adres çakışması saptanmadı.)\n")
                
            sys.stdout.write(f"\n📊 Toplam Eşleşme = {shared_preview.get('total_matches', 0)}\n")
            sys.stdout.write(f"───────────────────────────────────────────────────────────────────────────────\n")
            sys.stdout.write(f"🧮 [MATEMATİKSEL ARALIK DOĞRULAMA (Puzzle #71 Örneği)]:\n")
            sys.stdout.write(f"   • Üretilen Hex : {shared_preview.get('p71_hex')}\n")
            sys.stdout.write(f"───────────────────────────────────────────────────────────────────────────────")
            sys.stdout.flush()
            
    except KeyboardInterrupt:
        print("\n👋 Tarama durduruluyor, çekirdekler kapatılıyor...")
        for p in processes:
            p.terminate()
            p.join()

if __name__ == "__main__":
    multiprocessing.freeze_support()
    main()

[optioncmdPR]

 Im just, like, terribly frustrated by this riddle. This is mockery.

If starting from the top of a 45 degree descending staircase consisting of 255 steps,
why the hell is the 64th step down marked as "step 65, ladder position = 30.568377312064202855" Also at the top of the staircase a sign that says "position = 2^.001 ^9" and at the bottom step another one saying "position = 1.0000032733"  ?

[asweaq]

I'm certain this puzzle wasn't created using the Electrum Bitcoin client, nor any other desktop client. It's most likely generated using code generation or a cold wallet. You can check its transaction history to see this. If a client-side wallet were used, all addresses (balances) would be consolidated into a single transaction. However, after examining most of the creator's transaction history (including hidden transactions unrelated to the puzzle), I found that none of his transfers occurred within a single block. Furthermore, he prioritized outgoing transactions based on their entry dates.


Furthermore, the puzzle creator has a trading habit: all transaction fees are integers, both before and after the puzzle's creation. I remain convinced that this puzzle doesn't use BIP32 or BIP44. It must be hiding some kind of pattern or algorithm. The initial input to this puzzle is also strange. If it were a regular exchange, the address 173ujrhEVGqaZvPHXLqwXiSmPVMo225cqT would have a very simple input and a history of gambling.

https://mempool.space/tx/a898efd3a1bd4a3408d2d88584fd9c0e39f33f3f80871f59298ff98af3dafe33,

https://mempool.space/tx/6f47f4c009db22dad29a64ff9d2c54654b201453a318eb1866739c02dfb5ff84,

If it's an exchange, is the person managing the private key involved in gambling? Or did someone, for fun, enter a gambling address when purchasing from the exchange?

[eggsylacer]

Furthermore, the puzzle creator has a trading habit: all transaction fees are integers, both before and after the puzzle's creation. I remain convinced that this puzzle doesn't use BIP32 or BIP44. It must be hiding some kind of pattern or algorithm. The initial input to this puzzle is also strange. If it were a regular exchange, the address 173ujrhEVGqaZvPHXLqwXiSmPVMo225cqT would have a very simple input and a history of gambling.

https://mempool.space/tx/a898efd3a1bd4a3408d2d88584fd9c0e39f33f3f80871f59298ff98af3dafe33,

https://mempool.space/tx/6f47f4c009db22dad29a64ff9d2c54654b201453a318eb1866739c02dfb5ff84,

If it's an exchange, is the person managing the private key involved in gambling? Or did someone, for fun, enter a gambling address when purchasing from the exchange?

there is no hidden pattern here, CLOSED.

[NUCLEAR7.1]

Furthermore, the puzzle creator has a trading habit: all transaction fees are integers, both before and after the puzzle's creation. I remain convinced that this puzzle doesn't use BIP32 or BIP44. It must be hiding some kind of pattern or algorithm. The initial input to this puzzle is also strange. If it were a regular exchange, the address 173ujrhEVGqaZvPHXLqwXiSmPVMo225cqT would have a very simple input and a history of gambling.

https://mempool.space/tx/a898efd3a1bd4a3408d2d88584fd9c0e39f33f3f80871f59298ff98af3dafe33,

https://mempool.space/tx/6f47f4c009db22dad29a64ff9d2c54654b201453a318eb1866739c02dfb5ff84,

If it's an exchange, is the person managing the private key involved in gambling? Or did someone, for fun, enter a gambling address when purchasing from the exchange?

there is no hidden pattern here, CLOSED.

You are wrong. It exists because I discovered a deterministic structure that the creator left unintentionally.

[eggsylacer]

Furthermore, the puzzle creator has a trading habit: all transaction fees are integers, both before and after the puzzle's creation. I remain convinced that this puzzle doesn't use BIP32 or BIP44. It must be hiding some kind of pattern or algorithm. The initial input to this puzzle is also strange. If it were a regular exchange, the address 173ujrhEVGqaZvPHXLqwXiSmPVMo225cqT would have a very simple input and a history of gambling.

https://mempool.space/tx/a898efd3a1bd4a3408d2d88584fd9c0e39f33f3f80871f59298ff98af3dafe33,

https://mempool.space/tx/6f47f4c009db22dad29a64ff9d2c54654b201453a318eb1866739c02dfb5ff84,

If it's an exchange, is the person managing the private key involved in gambling? Or did someone, for fun, enter a gambling address when purchasing from the exchange?

there is no hidden pattern here, CLOSED.

You are wrong. It exists because I discovered a deterministic structure that the creator left unintentionally.

Instead of looking for something that isn't here, better learn how to determine which next point on the curve follows this point: 135 puzzle

Code:

02145d2611c823a396ef6712ce0f712f09b9b4f3135e3e0aa3230fb9b6d08d1e16

->

03a8c204d9e0cd0e7f6da825d55b5c2b9d0093f96650bf37e67bc802189b3bc478

or

0281be9db6e23f09c91ba700bda1b36a5ca5e4abc035725bf2b916bf8666bb5886

[Kingofdarkweb]

After all this time spent researching, I have discovered how to solve all these puzzles; based on a calculation, I can deduce the next number in the sequence. I will not tell anyone how, nor do I wish to discuss it. To anyone who doubts this process, here is puzzle number 71—the first two numbers are 69.

[NUCLEAR7.1]

After all this time spent researching, I have discovered how to solve all these puzzles; based on a calculation, I can deduce the next number in the sequence. I will not tell anyone how, nor do I wish to discuss it. To anyone who doubts this process, here is puzzle number 71—the first two numbers are 69.

Unfortunately, I beat you to it, and your prediction is wrong; it's not 69. 

[Kingofdarkweb]

After all this time spent researching, I have discovered how to solve all these puzzles; based on a calculation, I can deduce the next number in the sequence. I will not tell anyone how, nor do I wish to discuss it. To anyone who doubts this process, here is puzzle number 71—the first two numbers are 69.

Unfortunately, I beat you to it, and your prediction is wrong; it's not 69. 

trust me you will see when someone find just wait and back to reply to my msg im sure 100% starting with 69 and i've another 5 numbers good luck to you and take care about bots will steal it.....

[ineed200billionnow]

Hi everyone, I just discovered this puzzle a few months ago, and I've been trying to crack it ever since. This may seem like a stupid question, but I'm curious. In every method I've seen online for finding the keys to the puzzle, people assume the key is compressed rather than uncompressed. Is there a reason for that, or did the creator state that all the keys are compressed?

[Cricktor]

...

Sure, your genius shines so bright, we're all blinded. Now, tell me, why didn't you claim puzzles #135, #140, #145, #150, #155 and #160 already? I mean, you don't need MARA Slipstream for those, so you can't excuse yourself that you still don't have a client code to submit transactions privatly at Slipstream service.

Anyway, STFU and don't waste our time with your nonsense.

...

You could answer your question yourself with minimal effort because all solved puzzles so far have addresses that stem from P2PKH compressed public key derivation.

[Kingofdarkweb]

...

Sure, your genius shines so bright, we're all blinded. Now, tell me, why didn't you claim puzzles #135, #140, #145, #150, #155 and #160 already? I mean, you don't need MARA Slipstream for those, so you can't excuse yourself that you still don't have a client code to submit transactions privatly at Slipstream service.

Anyway, STFU and don't waste our time with your nonsense.

...

You could answer your question yourself with minimal effort because all solved puzzles so far have addresses that stem from P2PKH compressed public key derivation.

No one can simply withdraw the coins from the wallet; the person who sent them used a bot linked to the wallet. If you find the key and attempt to send the coins, the bot alters the transaction signature and redirects the funds to a different wallet. If you doubt my words, we will soon see whether I am right about Puzzle 71. I will not post another comment until we find out the truth. Once the solution to Puzzle 71 is found within the specific range I identified with 100% certainty, please come back and comment on this post again.

[NUCLEAR7.1]

...

Sure, your genius shines so bright, we're all blinded. Now, tell me, why didn't you claim puzzles #135, #140, #145, #150, #155 and #160 already? I mean, you don't need MARA Slipstream for those, so you can't excuse yourself that you still don't have a client code to submit transactions privatly at Slipstream service.

Anyway, STFU and don't waste our time with your nonsense.

...

You could answer your question yourself with minimal effort because all solved puzzles so far have addresses that stem from P2PKH compressed public key derivation.

No one can simply withdraw the coins from the wallet; the person who sent them used a bot linked to the wallet. If you find the key and attempt to send the coins, the bot alters the transaction signature and redirects the funds to a different wallet. If you doubt my words, we will soon see whether I am right about Puzzle 71. I will not post another comment until we find out the truth. Once the solution to Puzzle 71 is found within the specific range I identified with 100% certainty, please come back and comment on this post again.

Nonsense, because we would have discovered that, but nothing happened. As you can see, puzzles 67, 68, 115, 120, 125, and 130 were recently solved, and the creator didn't steal the reward. Successful withdrawals were documented. and puzzles 66 and 69, which were stolen because those who found them had little knowledge of what would happen if they broadcast the transaction publicly.  Why do you think the creator would waste their time, wait years, maintain the bot, etc., to steal 7 Bitcoins that are rightfully theirs when they could withdraw their Unsolved Reward 916.52 BTC ($59,095,553.88) in a single day?   

[bankeroft]

After all this time spent researching, I have discovered how to solve all these puzzles; based on a calculation, I can deduce the next number in the sequence. I will not tell anyone how, nor do I wish to discuss it. To anyone who doubts this process, here is puzzle number 71—the first two numbers are 69.

wrong, the first two are 6A

[Kingofdarkweb]

...

Sure, your genius shines so bright, we're all blinded. Now, tell me, why didn't you claim puzzles #135, #140, #145, #150, #155 and #160 already? I mean, you don't need MARA Slipstream for those, so you can't excuse yourself that you still don't have a client code to submit transactions privatly at Slipstream service.

Anyway, STFU and don't waste our time with your nonsense.

...

You could answer your question yourself with minimal effort because all solved puzzles so far have addresses that stem from P2PKH compressed public key derivation.

No one can simply withdraw the coins from the wallet; the person who sent them used a bot linked to the wallet. If you find the key and attempt to send the coins, the bot alters the transaction signature and redirects the funds to a different wallet. If you doubt my words, we will soon see whether I am right about Puzzle 71. I will not post another comment until we find out the truth. Once the solution to Puzzle 71 is found within the specific range I identified with 100% certainty, please come back and comment on this post again.

Nonsense, because we would have discovered that, but nothing happened. As you can see, puzzles 67, 68, 115, 120, 125, and 130 were recently solved, and the creator didn't steal the reward. Successful withdrawals were documented. and puzzles 66 and 69, which were stolen because those who found them had little knowledge of what would happen if they broadcast the transaction publicly.  Why do you think the creator would waste their time, wait years, maintain the bot, etc., to steal 7 Bitcoins that are rightfully theirs when they could withdraw their Unsolved Reward 916.52 BTC ($59,095,553.88) in a single day?   

Let me tell you: you don't know who created these puzzles or what the point of them is. The creator gained nothing from them other than leveraging his cloud mining sites to profit by renting out capacity to people. This is irrefutable proof; you will soon see the truth of my words. Just wait and see whether I am right or not.