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21  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 26, 2022, 06:13:19 PM
little experiment...

this code creates a sha256 key from str(random.random()), which is fed to the next "ice.get_sha256().hex()".
This is repeated, in this case + 2 times.
The created keys are concatenated ( keys = key1+key2+key3 ).
In the next step, the entire key is divided into 16-digit (64-bit) pieces and read out.
a filter keeps the keys in the desired range...

Happy hunting! XD

Code:
import time
import random
import secp256k1 as ice
from ctypes import c_int
from multiprocessing import Value, Lock, Process

y=1000000
cores=4
counter = Value(c_int)
counter_lock = Lock()     
def process1(number,counter,):
    while True:
        t = str(random.random())
        key1 = ice.get_sha256(t).hex()
        key2 = ice.get_sha256(key1).hex()
        key3 = ice.get_sha256(key2).hex()
        keys = key1+key2+key3
        for i in range(0, len(keys), 16):
            line = keys[i:i+16]
            if line.startswith("8") or line.startswith("9") or line.startswith("a") or line.startswith("b") or line.startswith("c") or line.startswith("d") or line.startswith("e") or line.startswith("f"):
                with counter_lock:
                        counter.value += cores
                        cv = str(counter.value)
                addr = ice.privatekey_to_address(0, True, int(line, 16))

                if addr.startswith("16jY7q"):
                    print('\n\n Pattern:',('0x'+line),addr,'\n')
                   
                if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
                    print ('\n\n TARGET FOUND!!:',addr,'\n\n')
                    file=open(u"16jY.Info.txt","a")
                    file.write('\n ' + ('0x'+line) + ' | ' + addr)
                    file.close()
                    wait = input("Press Enter to Exit.")
                    exit()
                   
                if (counter.value) % y == 0:
                    print(' CNT:',cv.zfill(10),' Recent:',('0x'+line),addr,end='\r')

if __name__ == '__main__':
    t = time.ctime()
    print('',t)
    number = y
    workers = []
    print('\n K E Y   -   C H O P P E R   -   64 \n')
    print('\n ===============|        |===== KEYS =======|============ ADDR ================| \n\n')
    for r in range(0,cores):
        p = Process(target=process1, args=(number,counter,))
        workers.append(p)
        p.start()

    for worker in workers:
        worker.join()           


donate BTC:  1DonateZNR9BUaCqJTgXCoyyCpRSosFujR



the generation of the keys on a time basis (ms) is too slow. the result was double, triple and multiple repetition of the keys in a row.
For this reason I switched to random.random()...
the better way ^^

 Sat Mar 26 20:40:24 2022

 K E Y   -   C H O P P E R   -   64

 =============|            |===== KEYS =======|============ ADDR ================|

 CNT: 0032000000  Recent:   0xaaf4c88e4a40a299     175NYisjDfbxUyoonY8rEbghWvBbyMGPRj

 Pattern: 0x9a5da0765a81d90f 16jY7qNwsgykvkAzm8oiwVSnC71KsZsXqM

 CNT: 0304000000  Recent:   0xb338bf39e2ba9922     1EfbHH9BPXSDuibzn3gCao9aQB6Chr7C3G

[moderator's note: consecutive posts merged]
22  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 24, 2022, 06:14:46 PM
Jolly Jocker,

From your info I have this:
[16jY7qLJnxtt8FSYMze4DRVMWQF47auU8M,8ec5edb376aeb55a]
[16jY7qLJnxht5HL2NBywuoC8jK34rz8NJv     ,9357254130f9270a]
[16jY7qLJnx4enVJC2Ao3wRv3tRusseu5Yb     ,af6f7a0015695aa4]
[16jY7qLJnxcZQ1ff7Y6ascX2FwkVfMEqcn     ,b0d6bbb249f4d96f]
[16jY7qLJnxoBx9Zhsa1tN62rfZXX1bgKNp    ,c4230300dfe089af]
[16jY7qLJnxPwmtRgVS3HBg2RtAoHe1koZF  ,d461baace35db135]
[16jY7qLJnx4f4yqNvmp7LtnaatiUuwgUwW   ,ebe2eb35503b43e4]
[16jY7qLJnx2ixrxCnTLSraerkgyB3YYAiT        ,edf9cd60356dcc19]
[16jY7qLJnxHBp3dqwV2kzYq1LucfZzgxsH    ,eeb9ab334df2770d]

my bet based on your data is that #64 is between edf9cd60356dcc19:eeb9ab334df2770d
but that is probably not true.

Edit: b is after H in the Base58 Alphabet do with that what you will

This database is sorted by Public key!!
from: 0200a89f5508695dd9c0900621681ff9b961d67772c323642a5e1f042d4cff3529
to   : 03ff94c3c4030faf55a10bda211e1d4d0891e3c68089e1ace2e62ce2f2f2eeb405

it's all in the text!
23  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 20, 2022, 07:12:30 PM
Why generate a template: "16jY7qLJn"?
There are still a lot of such addresses, this will not help in any way to find the private key from 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN
"16jY7qLJn........................................",  3ee4133d991............................................
"16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN", 3ee4133d991f52fdf6a25c9834e0745ac74248a4

and now I look at the pubkeys...

you understand?

These are a large part of the patterns, who was found in the range 2**63 to 2**64... ( a total of 662 were found ).. that's really meaningful ... who can tell you that the address you're looking for is really in this range!? ... we're all looking for ways to find this address... why shouldn't patterns help? you have to exhaust all possibilities... maybe someone with all the pubkeys can calculate a narrowing down of the range... if you find the pubkey for the said address, you will also find the privkey..."many ways lead to Rome"
24  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 20, 2022, 06:01:50 PM
I took the time to program a small tool to create this list..     Pattern: "16jY7qLJn"

maybe someone can do something with it, because the assignment of the pubkeys to the respective addresses is meaningful.
The list is sorted by pubkey. ( somewhat shortened due to lack of space )

Code:
('0200a89f5508695dd9c0900621681ff9b961d67772c323642a5e1f042d4cff3529', 'CCC4140031FEDA91', '16jY7qLJnMcUAdSQqtQzqcBMcrSRedgSk2')
('0200d8fb32f510c9c0cf659de7ef3ec5df5357d52ff8b356fca3aaa56e3b1e34d0', 'C845851F3595B29D', '16jY7qLJnrEmEdeY1ESqhcV2eejTYYihUz')
('02038120202a3157ab9e1b75bd185aee2d3fc3c09c4df356b0609562011a9d26e7', 'CCECAF00D74EFDAA', '16jY7qLJnJsNNyArhGgdcZBEKhR4DAXDvE')
('020404a74780b2c8f7a9a5f28e8b40a9eebe284097e05251f2037b4fa8fc97ab5d', 'C7445D5B38507F83', '16jY7qLJnepq968DgsxCqcfdeoqC3oX4yZ')
('020637952f64dfdd3c907857ed1f9b0f2b8b696136271384be96b458b2f5be38f0', 'CCEE88000BF3C9D3', '16jY7qLJnXHDFpJFrbEKPGVgS8h6XDjuLE')
('020799d410b4ca32b06dcb175268f437c0722885ad37e4fdbbc24145d1713417d1', 'A8C1DEBFC21902CC', '16jY7qLJna6grEq9tQz9FketLK6zDFLYNa')
('02091ed71dcd26e3b004409aa9211d8886663604eff9bac5ee9c14e9e6b0a89d89', '89F84ED1040B8782', '16jY7qLJnbwfCSSRXivZnCFG6phBAYvT1b')
('02094ddabdee0217b83b5abf285c60b7339ca71113762e227b4e6724983b9d8f89', '986EBB933EEE0962', '16jY7qLJnFpswGCQLBKoStQCoevvQb53G1')
('020953dd312efece156fade157f92ce57bccab87e85f272d5c0739b6cf6548520d', 'E2E1399B7F7EAE88', '16jY7qLJnMNGbuKo8w4ZunxcXtXjJHsf5j')
('020afa3ba6cc87ed60c56d2f4913a3804eda634bcf676bfe3e81786fed3c2b8470', 'B2885F30E6063B77', '16jY7qLJnQVihCA2Fcf6dDF9sYpA3TraHM')
('020b837269fbd708d433a694c32654bedd66175c2e397f273140b61a5ee916b021', 'E4B45971ECA7F5D2', '16jY7qLJnqQhrBCNFF5RJBsEjf6h7s2Pwq')
('020c183a237ba57a19c56b6ba9ed21d837307fad0ce369223ec02dc76df7e4f717', 'E687819A73B0D830', '16jY7qLJnjzpYWN9cqNXPsrvXKJhyNRbfd')
('020cbd5f57fbf0034809b3f112f1d33a2b5a8f580d110608532caeb95d30d5c98f', 'E570CE674AB8433D', '16jY7qLJn68WC24oCCbiApv1mmQPzsSWUv')
('020d33c9b1acb4e0a931b080c0f25d751f7bc9caa899d1907e8a79825ea40efbf6', 'BC6BBAAB1BEE0BE7', '16jY7qLJnZ44prYqrey4e99w61VZGZg65z')
('020d6089beec67cab0865f45690034b0ebeb2d7c5c40a4a2ed0aa822b7196ed895', 'C4DC9C0049D95DAB', '16jY7qLJnFLhrdYWZPdXmcg1ZE9pM7FisM')
('020ddbf6d854f92c21f6caf38ddac4ac0d7b547e38b1ff4d0e25e2a813f02924ca', 'CA6DF3000D542EEA', '16jY7qLJnZ4dcqeD8unXdpUczZTyza1y6F')
('0210988a7c2be0a7142eefe71b1dcdd358e61d15e47b293d58c24a8eb7c892d52d', 'A81975005CC66F98', '16jY7qLJnqiSjVcdAXdLpQSRVEVuDSpJM8')
('02116b732961730fa1bec1e1745c6de77787275400f73a3ed49032740afb6e323d', 'AFB0B40000B4CA6E', '16jY7qLJn1jqi1X7zf6f8iVBaJcpDoXWzk')
('0212608d7d304ffdb516ae21e3ce9b1a4b9aa1fd87bf722c0614acd3154352b6fb', 'A37E37006AFDBCA1', '16jY7qLJnMd3T5pEsjaMm9ErDZBCndfVBe')
('02139edcd79ec4ffdf7f4fe59d5d49ebcf30c66284a99ecfc88d21a9e6f3e487dc', 'EEB9AB334DF2770D', '16jY7qLJnxHBp3dqwV2kzYq1LucfZzgxsH')
('0216e507ed22039c2ec79f442bec4d3d696e0ea59d7826036ac50824582ec2a6b7', 'A9896478F913D4BA', '16jY7qLJnwbnmdMMYMyEfeXVC46UKNwGR2')
('0219b71b25ba5a74de65e76e79e9aa0ac53cb44381ffd9010e9f676a10b79c3d69', 'CC6C0C00A521FD62', '16jY7qLJnrqR3sy8SoMPrjyX53rJ6NC7Bw')
('021a336b45c301ef15486f4b00f6341ab619f2b632a697962da84d9f89de07e9be', 'E560CCCD46FD0290', '16jY7qLJnDjCcPohFResu1FTxwp7stW4PE')
('021b8742dc73ba324e92b463a828b685adcd04181271bf9384a27a4353f938cd68', 'EB78E99A80E1D123', '16jY7qLJnVvUHqZ96upirrHCHS7ETULczt')
('021c7bea5d119cef6415b6c5c810578fe97c4248972a0475e13c0ba006bda2370f', 'EB92B8000B510199', '16jY7qLJnh7VPS9dm85nnZvneCfMpX1DmZ')
('021cf1d2524d50a0fc6169b7f9a624894ae77121a2736ee4b082ad5f2b5a732830', 'EE27599A949748B2', '16jY7qLJnNYK62zykaE5TcEUoyRTfLoT76')
('021d37d5fa7a4c4c3de6a020538246d8e38e3cccb14b0688ad8782aeca7ae4738a', 'CFC8B60091586C5F', '16jY7qLJniTsjfigJpKXrG96cPW5ytfBTD')
('02237947d28b33047f1867bde2d3f6e3d0766d5bd7a05903d9ce847ab828215d99', 'E9D676668B88437C', '16jY7qLJnLm3z4BmPxrYpAdsiNFMP15JM2')
('02247f45a6466c92f701ca30b105a23d20589c5343c25c5364a9faa8823e547308', 'AA36FB000F56CFC6', '16jY7qLJnwZBquSc26xYSrkGnNp3UEHYWT')
('022502e3605f2ca83e0dacae1081ede84d4926da35b48810bcef4324e461a68950', 'C57FC2E3392D09BE', '16jY7qLJngSEfpYoYVwEFC7aja8HNkqYb9')
('02261e857fd8446ae01e9413753f806a66a1d22c072404749c052d9e8df8db5d37', 'D98CE6E78B188EC3', '16jY7qLJnAoKqcsm2mrGmWSJ7HiWrbSka9')
('0226fdc810f8dd4c150a23640a1470bfdea3fb05652af04da04ea5ab7026d4d7b9', '928D2067D6AE185D', '16jY7qLJnSwAQsS3pPsKEkDBcss8NQ3Pnz')
('02288e46da2ba67a3e1c08ed6d97a06e8b2c1d5d08bf3b4f81b5db33a21ecb2a15', 'CB45090018BEBCE8', '16jY7qLJn2wFyoYkSsr5vYG4itsygeQ1Kw')
('022f847a964fb5a740dec2071c4c984842483a96cca572ff8d7eaaa316c7e29a84', 'B0D6BBB249F4D96F', '16jY7qLJnxcZQ1ff7Y6ascX2FwkVfMEqcn')
('023243a9d2de809a6c8f181fd6bb4185daca6a427eef790af6100a8f7ffcf63d6d', 'E3AC980071C0B33B', '16jY7qLJnqGcDdJPQEiLmmYPJzc77sLmtA')
('0232dc8afe2e3503015a76387cdfa2633a1698bb40fecfffbc4d49638712e4e0a3', 'FD62B95C5BA1F037', '16jY7qLJnudSwk6jCSLLKzkZWgdydniDAe')
('023541e582fcf8a295141b92b9839113846881c3c013827a471f3f30b6de11dc17', 'F6D958F5C6675FAE', '16jY7qLJnEh6sMrStgBorbVECNSEsKLhEy')
('023580db2376aed5d26e84ed62622f3b1c330ec38d89e461d24b0353e16a08c964', 'EC4A84CD52671107', '16jY7qLJngBWcvGgkZT11f8fWnACWzWm5Y')
('0236c4642837680fe51bd65ceca6573e398c277adc766c0ea6acbd664730d33d50', 'A36763009E5B89AA', '16jY7qLJnBwcNxttTgsV3rbgg6RDr9Pa8v')
('0236d820b5c428a1a2ffe3f26554442a11d50d7409fc98f45d637b4c2835529428', 'F7452826D3DD6883', '16jY7qLJndMYuvxfyMKtfjN3D3WBEmnFPG')
('0236e96a1ef34a2fe3fdaa82b36007e8a85280d04eaf95dd32939da14e5237124b', 'EDBF8801256286B6', '16jY7qLJntG5ggiueLP5abTnM4cbEsiUne')
('023a61e51bc1adb836460157410fc6cb7d883645f529e41d9a3d8852c14afb23a6', 'CFB4AF0019F17271', '16jY7qLJng5jx7EPC6ZDZtDGVoJ7msuuiN')
('023a91c27016d1d5d6297c31eaf27556e068d9cdd9388ba64e50d98ccb353b4d82', 'D451D25B05CC6B07', '16jY7qLJnq1zjdvv8H6t5mninvWmyX4KVg')
('023a9d81e99507892397bb64472356cbdc62cdd1ca39e47e8d10a601b08dd22513', 'E969E99B3C467DD0', '16jY7qLJn5JDg1XVasv8iExsHBic9nsB1j')
('023b3461ebe61c5d64063f550b570b890b2b6099bc03062d229fa1e1c7c9a6aba1', 'C00CD300FBF3F758', '16jY7qLJnrSUEX1TKieFZjtSmGxKQrsCmv')
('023b9d14a34d75d8408891ea2c9e9a48ee3777db5dd2717ec8a03194c2a67096f3', '90FDF4846B8DCD6F', '16jY7qLJnF5zbMPjJE4kQCey7cxSeL45ex')
('023c8d910f4ad782772832daab00efaf68584989b49496121808ee18f67783999e', 'A8F40F0071C70AAD', '16jY7qLJnLQnwrSLZ2pmvGWK27HvT1NEw8')
('023dd018c4deda3ea02c9457bb87ea713c98bb23a6e54192b458f216431957cab4', '939C913AE5F7BB7E', '16jY7qLJnk5SVUrPNbzEjh4FMxVwbTrCqo')
('023ded85782cd06746210d1502259b2cdb46f3192010607fb3f97e68bd1d832689', '8B64317D38AE452B', '16jY7qLJntfVwjTEEM9PniarTmYhvNRFNZ')
('024316e9f3a7d9a333fc3e000969e614d171d438ecb07192607ff30738a1f2cb46', 'B84910007E4DB91E', '16jY7qLJnppeixn5P2gpAhmXnwFfBmqRuy')
('02442f17c86bb87207954d14ba49b5a107371da5804170535e266f63d74877699c', 'CF2C010007881571', '16jY7qLJn4zVsaJ5KPupGyH45zTrYTkoTX')
('0245c14dbee0b509a45e5453d5ebdee88b28dbaadc50bc209b0754ce7a0fe35a8e', 'E8F0CCCD2B604E3D', '16jY7qLJnS9Pxwggb3Dhbkqb81EGAVE2Xw')
('02472d322767c26b07cd7533cb43f82cab870b9234823fdc765196b0e062882416', 'EE7573351CB3F997', '16jY7qLJn37f3UdwMBTzxNdfktSV21S7Jf')
('0248839a698590aa7d5467845703a49dccfb067b081014d1b9efa9bd41017c1138', 'E49EE99AA73144B4', '16jY7qLJnwX1wGwMiYWkFMQVRuFDWhedj1')
('024a955334f7ae7f30bf062a2cb75b00b5eb7077c63585957b10d1f20f053967ed', 'B98032B5127D5EB7', '16jY7qLJnvuj3VpCwYdwBJhZhHNKDvLUMR')
('024c4637811a5a086e7012dc45ec5d5697faf1425f8634ed5e80e0b488dcca6aff', 'A66C0400605976D3', '16jY7qLJn2hMVEanZjvheawPmAgEqtY4LY')
('024ceac8037faebc3f0a9edebf7491852120f874f42bbe88333516e979e384bb02', 'BEE9F000B1B49986', '16jY7qLJn42wpGQdknHH4NDqJayZriSHZP')
('025063ffb143ba18cf13a7c46e2cb145dc230670d6d7e567f553f680fa13fc66d0', 'E1EAF99A49E8ED26', '16jY7qLJnjvKWMtjgCfMU76RUJ8zuMrAQR')
('025069383100a5369963b7598f28cc6ef77961216f9dc3674cd27074728864e5c4', 'E92D94CE5B787C66', '16jY7qLJn211ERvdgQj1SLDX7mEag33fhm')
('0250bbc172ccf742cb072fa0a585fb108b36d8d37448dbae196fc493e9d820db60', '947DB4CCD369ED97', '16jY7qLJnCX3vYb1uECnKMC56BDEnLcuBp')
('02514528281519fd325c914b375d4038d847c20dda499448226796d932114cf83c', 'C858D70047F37DB5', '16jY7qLJndB4A4suAuJy6B8u7p3fix1iYP')
('025219e2f92f3d6df973ce3e338b2b160ff8de332ecfb431401f8893ff8304c611', 'C33EC500B6A76977', '16jY7qLJnwBdYKU6bCbCpgRAGPdMWFT6F4')
('0253cb189d8efe20f513e723423d7c396925c01c51009a4a43d26bcb7a4ac08d50', 'E3875C5568453B50', '16jY7qLJnZAuCGHQVDZJqa3ghqjSRkku9Q')
('02550d2c6770ad4f43e1a8c91f45f0fe0625f93c8ab7e0380294751ec7155cdee0', 'EDF9366773A5D6DC', '16jY7qLJnVtP5aD6MHfJ86b26KDznTmXw3')
('0258fcf8956e62dd86d0d731ce73a4f8e147b4ba6de7d21a90747fcdb3bd501196', 'AEE77A008E83B3C0', '16jY7qLJncTJ4BYHz2HEZsjnZ6p7xpRmEP')
('02593d6886e0bda3f5c1ec3abeb910f08162dfcdd985cfa8731b9a1e86911151a8', 'B283B80025F9880E', '16jY7qLJnqx6GJGoUJWErAw3PYMqyEXBrC')
('0259be98e63f4ee4d97732f8aaf9f2e6fa7d0f14c99b17b4177ddb989eeeffab2a', 'AF6F7A0015695AA4', '16jY7qLJnx4enVJC2Ao3wRv3tRusseu5Yb')
('025f402b0c6a1823140e2a89cebb47d72cf86674476f4fd151040777ee096b8e3c', 'B2C970001D41BAD2', '16jY7qLJnVrtPnva7NEWwWUn2U3E1T1Khv')
('025faa961d16a45930c002d11caedce4732435c1277b4fbd0f9e99e1cba7874423', 'E44DA217B4801D4C', '16jY7qLJnTHDryYf9YhcMNgZYD4s27i2ZE')
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('03d95ebc4c2aceebbb2785f08a0e3f5c815747db4b97579f03231ac398795a161d', 'FF696CFF78DD33C2', '16jY7qLJn6vYeDSgzKpdocFLhudWPsCLCV')
('03db196219c448ef4d0848fa29de5b16e08f5f69c8caf213d66a2add3a0679d047', 'A3AD1500AC0E306F', '16jY7qLJnstTgBAyGJEANvYUnCqS5SmyEK')
('03dc0100ddda788c36a68ebad00786638a2ccddcf45ea7f42af83df09a436bd44b', 'AF80F3004EE502E5', '16jY7qLJn2fXcGg2ntYSFShDm8tsyTorZC')
('03ddac12e9d7820b00062d76968f62568dff9a5a17fcd6c252c826fee1395abe1f', 'EBA374CCE415A459', '16jY7qLJnzdqiXpt7f484D39QQLmWufxiM')
('03de8d0611e7226fbab8f5efc202d65b2040c614232de7bddd2eeaddbad4da6552', 'E982E8021A7C3DFE', '16jY7qLJnAGNUChgoingyzkPi3GuBAGCNN')
('03dfaeac7f9269beaa6ee3c0bb9bf6d2c646fdbcc8acab52791f2e4432e25b00ad', 'EE045BA02E5672B9', '16jY7qLJnsZu4vZz2ceQcatusBAnXXRTBh')
('03e17a13f46bcda242bfcff9f1ac40630ebbc54c75e877316088ed9486f87c52e3', '92464E0001412E8E', '16jY7qLJnHGW8iFkv4eGPKBeDQkAf2wk6N')
('03e238a449441839afdd1671eb69ccaeed152eaee78ad315b093f5fb3b0a6e7e65', '8A500564D1A344A2', '16jY7qLJnikKMRWYNo4U5AMLCSZpb99jTJ')
('03e27209fe924799801f1b0fba8a019cd9bee576ff7c5de78a325803397f70c3b2', 'C212630092232A6E', '16jY7qLJn7n2krf1zG38CN41ezuS3cSPX9')
('03e2e78935a2ac1aac187e05e31c53a93e3b9b8b2fc44c4d073b974a5c3c9426d3', '94D053333B23EB36', '16jY7qLJnhhupS32EEJhwv4KLHc4eZaCsL')
('03e50b44f5a613b4fac4b9ec386834c7953289c081e97b13f012eade770f1efb56', 'E0164B341EB51BA3', '16jY7qLJniyBP4SWLBR7Ys8WJRim2KXre5')
('03e5b694f5f8feb050335714013672de830eeb93ea1de4dc248cc5096a4a23d03e', 'C6140E007C5AE796', '16jY7qLJnCqaer4xgyukzgpkHF2NJsVjMn')
('03e6e3306e85bcd08bfc6266437f8211763813879ad1ee08e2d88339d640cc4615', 'CF4F71009D684449', '16jY7qLJnFd6CHEe8GzghvvwuytZ5EnxAi')
('03e7d9e46c0c9b48f94519c355369abc7f58548887e33870a8b5717497ad047204', '83A5912DABC2C017', '16jY7qLJn1XopLkqGvNp8gRRpKpfHLgML6')
('03e8480b4cb784584444690aab69aff14ce4ac88a89b032a275e83c8e60b264560', '80001A1A0B33FFFA', '16jY7qLJng79ZYaMRTnDxziUU9Pg2x4HAF')
('03e8c11d2f37e7e8dc651f47559733511d75d302b362a2706ee742347803f5cf73', '8443474BC9D2D542', '16jY7qLJnKSFLsRqHEGLX3iYwSvU6ySxZu')
('03e9f9079a3ea9a9151a418ad32e32cf90f533676f01fbda310751cb0c24c5a5f7', 'A929C6007A2A0B52', '16jY7qLJn4JfUt1o1XXXReJY7ba6ygFeRJ')
('03eb06f1794c77b99920cd7493d4fe98f2ac1e26d566dbadb84c61c636ed10ab64', 'AF644700AE15543A', '16jY7qLJny98qX7QiXX687hRw5wqGxXeS5')
('03eb07da0adc887d23a28ee384ff6374b12f947d4e3d15adb823a607570cdf6fc0', '84EE474DFBF8DE96', '16jY7qLJnME5xgbEhkdJsSHnhokjbBbu88')
('03eb8cae4a22cfc4028876a73f0cc2344a834ba9b2105a8c01381abd77c529acd2', 'A671AD0077C97FBB', '16jY7qLJnmcCa9Kc1G7PyM7gPkDGBe2BdC')
('03ec98713705b1f7a402441d53b506be08aa6b7247f559fbd66dee2a21a5d1c148', 'D71500004B498429', '16jY7qLJnKj8W1fCUcproVfnSYgMkqEyfV')
('03ee7353538debb1f526c91e49e2e4dcbba60e93a069eb1df6f09ece7d4408c0ad', 'C2DD3300C7A64DA4', '16jY7qLJnWCmSR4EECMMhtJYsA1FtWs8Px')
('03eea22459763640f05442cfb7d9eb96d785ff6a1e8fb13842cb5426d8d9bd8f33', 'B3CD2CB402F842BD', '16jY7qLJnANLkXBhTPCe5LhJ1BStnqfRw1')
('03efc1aa9ad6328a608271e0369994aff786555fed8e32ff2cdc501baf897b542c', 'E2E2300230502DAE', '16jY7qLJnC5BmpZdqGCT9iemdoKz1bi4Fa')
('03efdb001334f5dbfb6f28f879f09cb0a36f59ecbb920909a759921105d9d51c33', 'FE2865C6DBFEC383', '16jY7qLJnX2cZXJ78wV1ef42e7cLAZJ1Vn')
('03f27b25ef64dbd8a8ceaa40f872dd79c9d01fbcf059a671c8df57c708659d8ffe', 'E74C9334679B0B5E', '16jY7qLJn4SGRHYKER5NsnA9gcJngv5nog')
('03f3c0cc6752df6121e16e994ff44749b18b43ca6382de39a9a93a0bc2e8522c99', 'D83FEF5406698788', '16jY7qLJnkwTUBkJrbHyBPzUMSrXzSiQqt')
('03f4011b0c34b36bcea652457842ab87c1c5568b1b698da2b806c5ea596cd97629', 'CA2208005484586C', '16jY7qLJnvH3EisXyTXRN3QkxkCRwyaENc')
('03f6c1bb33160f1896e331b2989503cf4f715cd526ca37d483b2641a66474210f3', 'A3689B001C2D3919', '16jY7qLJns29hFr2KbW928ugg21Cx7P6Xi')
('03f6f9396a2649088d891308fff7cf066e78f86af0e0e336ed8e77bbbfd5c22b04', 'B817117EA76EA2D2', '16jY7qLJnncxUo9uzrBVntYQHEperUMoNm')
('03f7f23749aa519e182292b3f6ab0ffbe1a64a2f62c3e75e3f5fe72fc5f8e86f65', 'E927533456EB2F68', '16jY7qLJnBsaZBMfzSd5ZscK5PsNkonGMV')
('03f90682ae07302539e32a9b5b130690cd1ddc032c7dd3f24eddbc803bc57b8855', 'D45921D9B206053C', '16jY7qLJnT2uyE4umB8yLrv4ECoV54tfMN')
('03fb23c63144cd091f85c09b4c3b72905f6ea0a021e19f28553c378947532985ab', 'A89340004F2840CB', '16jY7qLJn2sqNyUckPHAxJx7wfHELLzLe7')
('03fb287e8c1afb1bad7e1841ac151a2cf3d15070010c354fba3eb12b6259b764fe', 'B51BDD55F149F10E', '16jY7qLJnEWbiTdYWuKHW6XSpduxEMEjes')
('03fc3acee9bc1a9b2e974e33e54926e7ad51001acf0751d6f08f0bbe76ad71e0d7', 'EA609001BF75F25F', '16jY7qLJnkjtRwseFG71YEMDyFxefE6uYR')
('03fcc448d808160a18acac6910fdb0863b9d07056788d164e0cc03d4c4e035f0ab', '8BB203974D48B13C', '16jY7qLJnHPYTL5daotAZJudvupQ3Wk2kw')
('03fce08b5215b860815fe5d9bf1a146d1d20a20e141926311f510890699d56134e', 'CA6DC9004F3E9E2F', '16jY7qLJn3nvJzFhdPwJP3GmckRkix3UvT')
('03fdb31a34ade12f59ac78d45faa1d683540b5b744e8009e6b79220e42fd6dff76', 'A10F2600708A8835', '16jY7qLJn6PmBE25TN1Fbe1yT1deVNBq7k')
('03ff4d34549c0b0feb2f350dc5c6d43fc5ea63ecb29a74afa6e9b1c9704e3c4ae4', 'E9421801FD3D5661', '16jY7qLJnjXrniqYWxF1oKSYRDTbda2nAU')
('03ff52c764bd785e1ff60bf6e5145765b5f633f81c27de0aa0dc87d89619d00060', '8DD4ECD2E4EB2CDD', '16jY7qLJnXfwnsfDQsDs72vJJu5XTmPDqJ')
('03ff94c3c4030faf55a10bda211e1d4d0891e3c68089e1ace2e62ce2f2f2eeb405', '8A6153AA9FBFE29C', '16jY7qLJnsEheSTVBw7q7KBfhmF8aT96yu')
donate BTC: 1DonateZNR9BUaCqJTgXCoyyCpRSosFujR
25  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 19, 2022, 10:27:39 PM
Traceback (most recent call last):
  File "main.py", line 3, in <module>
    import secp256k1 as ice
ModuleNotFoundError: No module named 'secp256k1'




This module appears frequently on the last few pages, and there is also a download link... so again for everyone ^^

"[faster "bit" there is a library from "ice" https://github.com/iceland2k14/secp256k1 there it is necessary to throw its libraries into the folder with the script.]"
26  Local / Anfänger und Hilfe / Re: Suche nach dem Bitcoin Puzzle #64 für kostenlose und legale Bitcoins on: March 17, 2022, 09:00:20 PM

Gegeben:
1Lets1xxxx1use1xxxxxxxxxxxy2EaMkJ
1fuLL1xxxx1power1xxxxxxxxxxzatvCK
1of1xxxxx1anonymity1xxxxxxxz9JzFN
1See1xxxx1memo1xxxxxxxxxxxxxBuhPF
1dot1xxxxx1sv1xxxxxxxxxxxxxwYqEEt
1topic1xxx1hmwyda1xxxxxxxxxvo8wMn
1xxxxxxxxxxxxxxxxxxxxxxxxxy1kmdGr

Frage an Euch, mit welchem Tool könnten diese Adressen erzeugt worden sein?

Mit gar keinem Tool. Das sind ausgedachte Adressen. Ausschließlich die Checksum (hinten) wurde berechnet um gültige Adressen zu erhalten. Den privat-Key dazu hat niemand, auch der Ersteller nicht. Deshalb ist das Guthaben auch noch auf den Adressen erhalten und kann aktuell nicht ausgegeben werden - selbst wenn man es wollte.

Jupp... und oftmals ergeben die Pubkeys sogar 0 und sind sogenannte "Black Hole" Adressen, die nie ausgegeben werden können...
Selbst im ETH gibt es so etwas.. https://www.blockchain.com/eth/address/0x0000000000000000000000000000000000000000 ... aus dem Fenster geworfenes Vermögen... schade schade
27  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 17, 2022, 11:50:05 AM
you can select the number of cores via ' a = ' in the top of the scrypt ..
I have to deal with the upload first... I haven't done it yet..

i just see that " a " wasn't a good choice for the cores, since a is already used in scrypt.. change from "a = 4" to "c =4" and in the part "for r in range(a): " to " for r in range(c): " important!!

Although it works with "a" , it could lead to errors!


The corrected code !

Code:
import time
import random
import secp256k1 as ice
from time import sleep
from multiprocessing import Process, Value

c = 4
y = 2097152
counter = Value('L')
def process1(number,counter,):
    while True:
        a = random.randint(2**63, 2**64)
        for x in range(number):
            x = x
            bina = bin(a)[2:]
            zeros = bina.count("0")
            if zeros >= 16:
                if zeros <= 43:
                    addr = ice.privatekey_to_address(0, True, a)
                    with counter.get_lock():
                            counter.value += 1
                    
                    if addr.startswith('16jY7q'):
                        print('\n Pattern found: '+hex(a)+' | '+ addr)
                        sleep(1)
                        print("\n continue...\n")
                        
                    if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
                        file=open(u"16jY.Info.txt","a")
                        file.write('\n '+hex(a)+' | '+ addr)
                        file.close()
                        wait = input("Press Enter to Exit.")
                        sleep(1)
                        exit()
                        
                    if x == 0:
                        print ('', hex(a), bina, zeros, '', str(counter.value))
                
                
            a = a +1
        pass

if __name__ == '__main__':
    t = time.ctime()
    print('',t)
    number = y
    workers = []
    print('\n\n BINARY MULTI PY \n #64: 8000000000000000...ffffffffffffffff\n\n')
    print('                                                                                    zeros')
    print(' == PRIVKEY HEX == |____________________________ BINARY ____________________________|  | CNT == \n')
    for r in range(c):
        p = Process(target=process1, args=(number,counter,))
        workers.append(p)
        p.start()

    for worker in workers:
        worker.join()          

 


donate BTC: 1DonateZNR9BUaCqJTgXCoyyCpRSosFujR
28  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 16, 2022, 05:54:37 AM

Code:
import sys
import time
import random
#from tqdm import tqdm
from time import sleep
import secp256k1 as ice

print('\n      B I N A R Y   S E E D   S C A N   Q U A D R O   \n')
print(' 11_________________0____________________________________________\n')
x=16000
y=131071
list=[]
x0=0b100000000000000000000000000000000000000000000
for n in range(y):
    x1 = (x0+n)
    a1 = bin(random.randrange(131072,262143))#[2:].zfill(18)
    a1 = (a1)[2:]
    for i in range(x):
        a0 = str("1")           
        a2 = str("0")
        a3 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a3 = (a3)[3:]#.zfill(44)
        a4 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a4 = (a4)[3:]#.zfill(44)
        a5 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a5 = (a5)[3:]#.zfill(44)
        a6 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a6 = (a6)[3:]#.zfill(44)
        binb = "".join(a0+a1+a2+a3)
        binc = "".join(a0+a1+a2+a4)
        bind = "".join(a0+a1+a2+a5)
        bine = "".join(a0+a1+a2+a6)
        list.append(binb)
        list.append(binc)
        list.append(bind)
        list.append(bine)
        if len(list)==x:
            line_count=0
            for bina in (list):
                bina!='\n'
                addr = ice.privatekey_to_address(0, True, int(bina,2))
                line_count+=1
                                 
                if addr.startswith('16jY7'): # 16jY7
                    ran = int(bina,2)
                    print('',bina, '\n')
                    sys.stdout.write('\r ' + hex(ran))
                    sys.stdout.write(' ' + '| ' + addr + ' ' + str(n))
                    sleep(1)
                    sys.stdout.write('\n\n continue...\n\n')
                    sleep(1)

                if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
                    ran = int(bina,2)
                    print('\n\n Target found!!' + hex(ran) + ' \n ' + addr,'\n')
                    file=open(u"16j.Target.Info.txt","a")
                    file.write('\n ' + hex(ran) + ' | ' + addr)
                    file.close()
                    sleep(2)
                    wait = input("Press Enter to Exit.")
                    sleep(1)
                    exit()
                   
                if line_count==x:
                    list.clear()
                   
        if len(list)==0:
            print(' scan... ', str(n), end='\r')                       

Ca 42000 keys/s from c000000000000000 to ffffffffffffffff... The sector between  11_________________0 comprise a range of 131071 and will be processed.
The rest of this line are random numbers from 4 different generators..
The counter reflects the number of loops...

      B I N A R Y   S E E D   S C A N   Q U A D R O

 11_________________0____________________________________________

 1110010110110011011010011110001111111110100110111011100101000111

 0xe5b369e3fe9bb947 | 16jY7mM6wrpuAy53vKkh7QXm1JNdWK9Myu 19

 continue...

 1110010011000111001010001111000110100111001000001001110001000100

 0xe4c728f1a7209c44 | 16jY7Ezk5et98CVBWYEigGRwWovzhEtaQ3 121

 continue...

 scan...  192

Hi. I think your idea interesting. In this article another method how to recover privkey with add pubkeys what you think about this ?
https://toadstyle.org/cryptopals/66.txt

K is prrivkey, b is a picies of privkey:

For example, with n = 6, trace(58) gives me:

  # k = 111010
  # i = 2, b = 1
  add(1Q, 1Q)
  add(2Q, 1Q)
  # i = 3, b = 1
  add(3Q, 3Q)
  add(6Q, 1Q)
  # i = 4, b = 0
  add(7Q, 7Q)
  # i = 5, b = 1
  add(14Q, 14Q)
  add(28Q, 1Q)
  # i = 6, b = 0
  add(29Q, 29Q)

And here's trace(62):

  # k = 111110
  # i = 2, b = 1
  add(1Q, 1Q)
  add(2Q, 1Q)
  # i = 3, b = 1
  add(3Q, 3Q)
  add(6Q, 1Q)
  # i = 4, b = 1
  add(7Q, 7Q)
  add(14Q, 1Q)
  # i = 5, b = 1
  add(15Q, 15Q)
  add(30Q, 1Q)
  # i = 6, b = 0
  add(31Q, 31Q)





asic miner
github.com/achow1o1

update: Kiss
1J8wBriLsHvE52Jah75bKU1VSxj9btzLeq
1EhRDeGsoQVkAsiCtbDUbNnRW8a977bduS

Update2
2021.03.02
1CAsB6DdK6ezYzefzRe3SuQ1bZztsgaiQS
12JQGbJUHohtfoPbbZKbF8Z8hW5wYde1KN
1DKsKYzpqaTAg3Q4KrCBJ7AS4R1YyQrisE
1GymA7nkPZVBz3crj2NuFfGZsngTTX51Sb

How you crack them bro ? Thx



Enother interesting code - https://github.com/jairopaiva/BitcoinExprCracker

Search metadata for finding privkey... Needs modification for work


Thanks a lot Bro!

But, I have to fight my way through this first... and it's not so easy, especially since English is not my mother tongue... so it needs a while ^^

you know, g00gle translater .... lol XD



A little toy... 4 cores in working.. Happy Hunting!!

Code:
import time
import random
import secp256k1 as ice
from time import sleep
from multiprocessing import Process, Value

a = 4
y = 2097152
counter = Value('L')
def process1(number,counter,):
    while True:
        a = random.randint(2**63, 2**64)
        for x in range(number):
            x = x
            bina = bin(a)[2:]
            zeros = bina.count("0")
            if zeros >= 16:
                if zeros <= 43:
                    addr = ice.privatekey_to_address(0, True, a)
                    with counter.get_lock():
                            counter.value += 1
                   
                    if addr.startswith('16jY7q'):
                        print('\n Pattern found: '+hex(a)+' | '+ addr)
                        sleep(1)
                        print("\n continue...\n")
                       
                    if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
                        file=open(u"16jY.Info.txt","a")
                        file.write('\n '+hex(a)+' | '+ addr)
                        file.close()
                        wait = input("Press Enter to Exit.")
                        sleep(1)
                        exit()
                       
                    if x == 0:
                        print ('', hex(a), bina, zeros, '', str(counter.value))
               
               
            a = a +1
        pass

if __name__ == '__main__':
    t = time.ctime()
    print('',t)
    number = y
    workers = []
    print('\n\n BINARY MULTI PY \n #64: 8000000000000000...ffffffffffffffff\n\n')
    print('                                                                                    zeros')
    print(' == PRIVKEY HEX == |____________________________ BINARY ____________________________|  | CNT == \n')
    for r in range(a):
        p = Process(target=process1, args=(number,counter,))
        workers.append(p)
        p.start()

    for worker in workers:
        worker.join()           

Quote

  Wed Mar 16 22:50:29 2022


 BINARY MULTI PY
 #64: 8000000000000000...ffffffffffffffff


                                                                                                                                                zeros
 == PRIVKEY HEX == |____________________________ BINARY ____________________________|  | CNT ==

 0x8fb6f979c971deb7 1000111110110110111110010111100111001001011100011101111010110111 23  1
 0xd71cef3d4d5dbb35 1101011100011100111011110011110101001101010111011011101100110101 24  2
 0xbf232d596f8c506f 1011111100100011001011010101100101101111100011000101000001101111 29  79
 0xb303436dcb92d933 1011001100000011010000110110110111001011100100101101100100110011 32  293

 Pattern found: 0xd71cef3d4d64dbb4 | 16jY7h2kCHJmhW42oUFyCeyucfrsRSNM9t

 continue...


 Pattern found: 0xb303436dcba91968 | 16jY7qGbzuSoT18AEyfZ1tyEsv8uzN5Bmp

 continue...

 0xe85e879a14758f47 1110100001011110100001111001101000010100011101011000111101000111 31  8239375
 0xa9b982acb6cf7e30 1010100110111001100000101010110010110110110011110111111000110000 30  8330122
 0xed25182830ace349 1110110100100101000110000010100000110000101011001110001101001001 37  8381965
 0xc162d7c463b14442 1100000101100010110101111100010001100011101100010100010001000010 37  8607642



[moderator's note: consecutive posts merged]
29  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 12, 2022, 04:41:06 PM
Test Code for the 64 bit range... ca. 310000000 keys/hr...

Code:
import random
from threading import Thread
import secp256k1 as ice
from time import sleep
print('\n ========= ================= ==ADDRESS HUNTER== ================= ==========\n\n')
y=1000000
count=-2000000
def process1(number):
    line_count=0
    num = random.randrange(0x8000000000000000,0xffffffffffffffff)
    for i in range(number):
        num+=i
        addr = ice.privatekey_to_address(0, True, num)
        lineA = (addr, hex(num))
        line_count+=1
                    
        if '16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN' in lineA:
            file=open(u"BTH.Target.Info.txt","a")
            file.write('\n ' + str(lineA))
            file.close()
            wait = input("Press Enter to Exit.")
            exit()
                        
        if line_count == y:
            print(' current line: ', str(lineA))
            line_count=0
                    
while True:
    count+=(2*y)
    number = y
    threads = []
    print('\n Total Scanned Lines: ' + str(count) + '\n')
    for n in range(0,2):
        t = Thread(target=process1, args=(number,))
        threads.append(t)
        t.start()
        
    for t in threads:
        t.join()

donate BTC:  1DonateZNR9BUaCqJTgXCoyyCpRSosFujR
I tested your code, but the wrong ddd code doesn't collide even with 2 hex, which is a big problem



this is what your code should look like which works at 100% delete tqdm slowing code 10%
import random
from tqdm import tqdm
import secp256k1 as ice
for x in tqdm(range(100000000)):
 num = random.randrange(0xd2c00,0xd2cff)
 add = ice.privatekey_to_address(0,True,num)
 if "1HsMJxNiV7TLxmoF6uJNkydxPFDog4NQum" in (add):
  f=open("von.txt","a")
  f.write(str(num)+"-"+(add)+"\n")
  f.close()

I see you didn't understand the principle of my code...

This code (slightly modified) for the search for that Pubkey of 16jY7.... with 4 Threads make more as 380000000 Keys in 30 minutes...
I'll keep working on it... I'm sure there's more to come^^

Code:
import time
import random
import secp256k1 as ice
from time import sleep
from threading import Thread
from multiprocessing import Value
t = time.ctime()
print('',t)
print('\n |========== ========== ==========! PUBKEY HUNTER !========== ========== ==========||============\n\n')
print(' [*] create privkeys....', end='\r')

list=[]
a=2
y=1048576
counter = Value('L')
def process1(counter,number,n):
    ran = random.randrange(0x8000000000000001,0xffffffffffffffff)
    for i in range(number):
        ran+=1
        pub = ice.privatekey_to_h160(0, True, ran).hex()
        lineA = pub, hex(ran)
        list.append(lineA)
        if len(list)==y:
            for line in list:
                line!='\n'
                with counter.get_lock():
                    counter.value += 1
           
                if '3ee4133d991f52fdf6a25c9834e0745ac74248a4' in line:
                    threads.clear()
                    a=0
                    f=open("16jY7.txt","a")
                    f.write(str(line) +"\n")
                    f.close()
                    print('\n\n Target found!! \n' + str(line))
                    exit()
               
                if (counter.value)%y == 0:
                    print(' [ Recent-Line:', str(line),'] Total:', str(counter.value))#, end='\r')
                    list.clear()
                       
while len(list)<=y:
    number = y
    threads = []
    for n in range(a):
        t = Thread(target=process1, args=(counter,number,n))
        threads.append(t)
        t.start()

    for t in threads:
        t.join()

ok i take it back, I won't go into that anymore there are only 2 options 64 address has total epic fail hex or in this range is not at all happy hunting

The range doesn't matter in principle, we're just unlucky that we came too late and the lower areas have already been solved... every higher range reduces the chance of finding the desired address. Of course, luck plays a big part...
The "fun" factor and further learning how to use Python inspire the whole thing...

so don't fret Smiley Happy hunting!! despite everything ...
30  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 12, 2022, 01:38:29 PM
Test Code for the 64 bit range... ca. 310000000 keys/hr...

Code:
import random
from threading import Thread
import secp256k1 as ice
from time import sleep
print('\n ========= ================= ==ADDRESS HUNTER== ================= ==========\n\n')
y=1000000
count=-2000000
def process1(number):
    line_count=0
    num = random.randrange(0x8000000000000000,0xffffffffffffffff)
    for i in range(number):
        num+=i
        addr = ice.privatekey_to_address(0, True, num)
        lineA = (addr, hex(num))
        line_count+=1
                    
        if '16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN' in lineA:
            file=open(u"BTH.Target.Info.txt","a")
            file.write('\n ' + str(lineA))
            file.close()
            wait = input("Press Enter to Exit.")
            exit()
                        
        if line_count == y:
            print(' current line: ', str(lineA))
            line_count=0
                    
while True:
    count+=(2*y)
    number = y
    threads = []
    print('\n Total Scanned Lines: ' + str(count) + '\n')
    for n in range(0,2):
        t = Thread(target=process1, args=(number,))
        threads.append(t)
        t.start()
        
    for t in threads:
        t.join()

donate BTC:  1DonateZNR9BUaCqJTgXCoyyCpRSosFujR
I tested your code, but the wrong ddd code doesn't collide even with 2 hex, which is a big problem



this is what your code should look like which works at 100% delete tqdm slowing code 10%
import random
from tqdm import tqdm
import secp256k1 as ice
for x in tqdm(range(100000000)):
 num = random.randrange(0xd2c00,0xd2cff)
 add = ice.privatekey_to_address(0,True,num)
 if "1HsMJxNiV7TLxmoF6uJNkydxPFDog4NQum" in (add):
  f=open("von.txt","a")
  f.write(str(num)+"-"+(add)+"\n")
  f.close()

I see you didn't understand the principle of my code...

This code (slightly modified) for the search for that Pubkey of 16jY7.... with 4 Threads make more as 380000000 Keys in 30 minutes...
I'll keep working on it... I'm sure there's more to come^^

Code:
import time
import random
import secp256k1 as ice
from time import sleep
from threading import Thread
from multiprocessing import Value
t = time.ctime()
print('',t)
print('\n |========== ========== ==========! PUBKEY HUNTER !========== ========== ==========||============\n\n')
print(' [*] create privkeys....', end='\r')

list=[]
a=2
y=1048576
counter = Value('L')
def process1(counter,number,n):
    ran = random.randrange(0x8000000000000001,0xffffffffffffffff)
    for i in range(number):
        ran+=1
        pub = ice.privatekey_to_h160(0, True, ran).hex()
        lineA = pub, hex(ran)
        list.append(lineA)
        if len(list)==y:
            for line in list:
                line!='\n'
                with counter.get_lock():
                    counter.value += 1
           
                if '3ee4133d991f52fdf6a25c9834e0745ac74248a4' in line:
                    threads.clear()
                    a=0
                    f=open("16jY7.txt","a")
                    f.write(str(line) +"\n")
                    f.close()
                    print('\n\n Target found!! \n' + str(line))
                    exit()
               
                if (counter.value)%y == 0:
                    print(' [ Recent-Line:', str(line),'] Total:', str(counter.value))#, end='\r')
                    list.clear()
                       
while len(list)<=y:
    number = y
    threads = []
    for n in range(a):
        t = Thread(target=process1, args=(counter,number,n))
        threads.append(t)
        t.start()

    for t in threads:
        t.join()

31  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 11, 2022, 06:47:45 PM
start the scrypt with double click .. simply to use in Python...
You need:
faster "bit" there is a library from "ice" https://github.com/iceland2k14/secp256k1 there it is necessary to throw its libraries into the folder with the script.

donate BTC:  1DonateZNR9BUaCqJTgXCoyyCpRSosFujR

you dont have GPU ? any GPU ?

This code is only for CPU, working with two Threads...

Sure, I use GPU too, but I don't can coding for Cuda applications like, vanitygen, Kangaroo, etc...

i will sugggest same working your code at c and gpu, above link use git scripts, working 10005 better then you python code , ttoal same function,
even if your cpu have gpu inside, like i3-6100 process have gpu, its also work on this rotar, and auto use gpu at cpu based command,
https://github.com/phrutis/Rotor-Cuda, for fully random at gpu like bitcrack is BitCrack2, use that, you will forget python
dont forget merit me.... Smiley

I know Rotor-Cuda... and many more cuda programs... but I enjoy coding myself...

I do not have enough sendable merit. Sorry Dude!
32  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 11, 2022, 05:56:16 PM
start the scrypt with double click .. simply to use in Python...
You need:
faster "bit" there is a library from "ice" https://github.com/iceland2k14/secp256k1 there it is necessary to throw its libraries into the folder with the script.

donate BTC:  1DonateZNR9BUaCqJTgXCoyyCpRSosFujR

you dont have GPU ? any GPU ?

This code is only for CPU, working with two Threads...

Sure, I use GPU too, but I don't can coding for Cuda applications like, vanitygen, Kangaroo, etc...
33  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 11, 2022, 03:31:25 PM
start the scrypt with double click .. simply to use in Python...
You need:
faster "bit" there is a library from "ice" https://github.com/iceland2k14/secp256k1 there it is necessary to throw its libraries into the folder with the script.

donate BTC:  1DonateZNR9BUaCqJTgXCoyyCpRSosFujR
34  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 11, 2022, 03:08:58 PM
Test Code for the 64 bit range... ca. 310000000 keys/hr...

Code:
import random
from threading import Thread
import secp256k1 as ice
from time import sleep
print('\n ========= ================= ==ADDRESS HUNTER== ================= ==========\n\n')
y=1000000
count=-2000000
def process1(number):
    line_count=0
    num = random.randrange(0x8000000000000000,0xffffffffffffffff)
    for i in range(number):
        num+=i
        addr = ice.privatekey_to_address(0, True, num)
        lineA = (addr, hex(num))
        line_count+=1
                    
        if '16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN' in lineA:
            file=open(u"BTH.Target.Info.txt","a")
            file.write('\n ' + str(lineA))
            file.close()
            wait = input("Press Enter to Exit.")
            exit()
                        
        if line_count == y:
            print(' current line: ', str(lineA))
            line_count=0
                    
while True:
    count+=(2*y)
    number = y
    threads = []
    print('\n Total Scanned Lines: ' + str(count) + '\n')
    for n in range(0,2):
        t = Thread(target=process1, args=(number,))
        threads.append(t)
        t.start()
        
    for t in threads:
        t.join()

donate BTC:  1DonateZNR9BUaCqJTgXCoyyCpRSosFujR
35  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: March 02, 2022, 10:48:38 PM
The problem with Python is the "print" output, which costs time/processing power, without this, 12000 to 13500 keys/s are generated.. (12000/s = 1036800000 keys/24 hr)
With the print output only 1455.26it/s are generated, which is much too slow.. even 50000 or 100000 keys/s are too slow with a range of 9223372036854775807 possible keys..
Even GPU scanners with more than 150000000 keys/s need more than 1000 years ( 1949.80 ) for this range... (4730400000000000 keys/ year)


36  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: February 25, 2022, 09:26:14 PM
I have revised this code for 64 bit, this code changes the starting point every 1000000 counters..
the problem in this code, is the truncation of the "zeros" what causes the skipping...it has pros and cons...
the advantage is that you can use it as a filter, the average of the zeros for the address you are looking for is between 22 and 43...

!! zeros = bina.count("100") .. this function only counts the zeros in binary code!!! also zeros = bina.count("0") or you count the ones ... ones = bina.count("1") !!

It's important to understand the functions so that you know what the code does and you can make conscious changes that work...

Code:
from bit import *
import random
from time import sleep

count=0

while True:
    a = random.randint(2**63, 2**64)
 #  while a <= 2**64:
    for x in range(1000000):
        x = x
        bina = bin(a)[2:]
        zeros = bina.count("0")
     #  if x in range(1):
        if x == 0:
            print(" !!!NEW LOOP!!!")
            sleep(2)
        if zeros >= 1:     # "00000000000000000000000000000011111111111111111111111111"
            if zeros <= 63: # "00000000000000000000000000000011111111111111111111111111"
                key = Key.from_int(a)
                addr = key.address
                count+=1
                print (hex(a), bina, zeros, str(count))
                
                if addr.startswith('16jY7qLJnxb'):
                    print(' '+hex(a)+'|'+ addr)
                    file=open(u"16jY.Info.txt","a")
                    file.write('\n '+hex(a)+' | '+ addr)
                    file.close()
                    sleep(3)
                    wait = input("Press Enter to continue.")
                    print(" continue...")
                    sleep(1)
             #  else:
             #      print (' Scan Nr.: ', str(count), end='\r') #pass
        a = a +1
    pass

Thank you very much for your attention, my friend.
always a pleasure dude!! Smiley
37  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: February 25, 2022, 06:05:36 PM
I have revised this code for 64 bit, this code changes the starting point every 1000000 counters..
the problem in this code, is the truncation of the "zeros" what causes the skipping...it has pros and cons...
the advantage is that you can use it as a filter, the average of the zeros for the address you are looking for is between 22 and 43...

!! zeros = bina.count("100") .. this function only counts the zeros in binary code!!! also zeros = bina.count("0") or you count the ones ... ones = bina.count("1") !!

It's important to understand the functions so that you know what the code does and you can make conscious changes that work...

Code:
from bit import *
import random
from time import sleep

count=0

while True:
    a = random.randint(2**63, 2**64)
 #  while a <= 2**64:
    for x in range(1000000):
        x = x
        bina = bin(a)[2:]
        zeros = bina.count("0")
     #  if x in range(1):
        if x == 0:
            print(" !!!NEW LOOP!!!")
            sleep(2)
        if zeros >= 1:     # "00000000000000000000000000000011111111111111111111111111"
            if zeros <= 63: # "00000000000000000000000000000011111111111111111111111111"
                key = Key.from_int(a)
                addr = key.address
                count+=1
                print (hex(a), bina, zeros, str(count))
                
                if addr.startswith('16jY7qLJnxb'):
                    print(' '+hex(a)+'|'+ addr)
                    file=open(u"16jY.Info.txt","a")
                    file.write('\n '+hex(a)+' | '+ addr)
                    file.close()
                    sleep(3)
                    wait = input("Press Enter to continue.")
                    print(" continue...")
                    sleep(1)
             #  else:
             #      print (' Scan Nr.: ', str(count), end='\r') #pass
        a = a +1
    pass
38  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: February 25, 2022, 04:43:25 PM

Code:
import sys
import time
import random
#from tqdm import tqdm
from time import sleep
import secp256k1 as ice

print('\n      B I N A R Y   S E E D   S C A N   Q U A D R O   \n')
print(' 11_________________0____________________________________________\n')
x=16000
y=131071
list=[]
x0=0b100000000000000000000000000000000000000000000
for n in range(y):
    x1 = (x0+n)
    a1 = bin(random.randrange(131072,262143))#[2:].zfill(18)
    a1 = (a1)[2:]
    for i in range(x):
        a0 = str("1")            
        a2 = str("0")
        a3 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a3 = (a3)[3:]#.zfill(44)
        a4 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a4 = (a4)[3:]#.zfill(44)
        a5 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a5 = (a5)[3:]#.zfill(44)
        a6 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a6 = (a6)[3:]#.zfill(44)
        binb = "".join(a0+a1+a2+a3)
        binc = "".join(a0+a1+a2+a4)
        bind = "".join(a0+a1+a2+a5)
        bine = "".join(a0+a1+a2+a6)
        list.append(binb)
        list.append(binc)
        list.append(bind)
        list.append(bine)
        if len(list)==x:
            line_count=0
            for bina in (list):
                bina!='\n'
                addr = ice.privatekey_to_address(0, True, int(bina,2))
                line_count+=1
                                
                if addr.startswith('16jY7'): # 16jY7
                    ran = int(bina,2)
                    print('',bina, '\n')
                    sys.stdout.write('\r ' + hex(ran))
                    sys.stdout.write(' ' + '| ' + addr + ' ' + str(n))
                    sleep(1)
                    sys.stdout.write('\n\n continue...\n\n')
                    sleep(1)

                if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
                    ran = int(bina,2)
                    print('\n\n Target found!!' + hex(ran) + ' \n ' + addr,'\n')
                    file=open(u"16j.Target.Info.txt","a")
                    file.write('\n ' + hex(ran) + ' | ' + addr)
                    file.close()
                    sleep(2)
                    wait = input("Press Enter to Exit.")
                    sleep(1)
                    exit()
                    
                if line_count==x:
                    list.clear()
                    
        if len(list)==0:
            print(' scan... ', str(n), end='\r')                        

Ca 42000 keys/s from c000000000000000 to ffffffffffffffff... The sector between  11_________________0 comprise a range of 131071 and will be processed.
The rest of this line are random numbers from 4 different generators..
The counter reflects the number of loops...

      B I N A R Y   S E E D   S C A N   Q U A D R O

 11_________________0____________________________________________

 1110010110110011011010011110001111111110100110111011100101000111

 0xe5b369e3fe9bb947 | 16jY7mM6wrpuAy53vKkh7QXm1JNdWK9Myu 19

 continue...

 1110010011000111001010001111000110100111001000001001110001000100

 0xe4c728f1a7209c44 | 16jY7Ezk5et98CVBWYEigGRwWovzhEtaQ3 121

 continue...

 scan...  192



B I N A R Y   S E E D   S C A N   Q U A D R O   

 11_________________0____________________________________________

Traceback (most recent call last):
  File "11.py", line 41, in <module>
    addr = ice.privatekey_to_address(0, True, int(bina,2))
AttributeError: module 'secp256k1' has no attribute 'privatekey_to_address'

You must use this https://github.com/iceland2k14/secp256k1 for my code.....
39  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: February 25, 2022, 03:06:33 PM

Code:
import sys
import time
import random
#from tqdm import tqdm
from time import sleep
import secp256k1 as ice

print('\n      B I N A R Y   S E E D   S C A N   Q U A D R O   \n')
print(' 11_________________0____________________________________________\n')
x=16000
y=131071
list=[]
x0=0b100000000000000000000000000000000000000000000
for n in range(y):
    x1 = (x0+n)
    a1 = bin(random.randrange(131072,262143))#[2:].zfill(18)
    a1 = (a1)[2:]
    for i in range(x):
        a0 = str("1")            
        a2 = str("0")
        a3 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a3 = (a3)[3:]#.zfill(44)
        a4 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a4 = (a4)[3:]#.zfill(44)
        a5 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a5 = (a5)[3:]#.zfill(44)
        a6 = bin(random.randrange(x1,0b111111111111111111111111111111111111111111111))
        a6 = (a6)[3:]#.zfill(44)
        binb = "".join(a0+a1+a2+a3)
        binc = "".join(a0+a1+a2+a4)
        bind = "".join(a0+a1+a2+a5)
        bine = "".join(a0+a1+a2+a6)
        list.append(binb)
        list.append(binc)
        list.append(bind)
        list.append(bine)
        if len(list)==x:
            line_count=0
            for bina in (list):
                bina!='\n'
                addr = ice.privatekey_to_address(0, True, int(bina,2))
                line_count+=1
                                
                if addr.startswith('16jY7'): # 16jY7
                    ran = int(bina,2)
                    print('',bina, '\n')
                    sys.stdout.write('\r ' + hex(ran))
                    sys.stdout.write(' ' + '| ' + addr + ' ' + str(n))
                    sleep(1)
                    sys.stdout.write('\n\n continue...\n\n')
                    sleep(1)

                if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
                    ran = int(bina,2)
                    print('\n\n Target found!!' + hex(ran) + ' \n ' + addr,'\n')
                    file=open(u"16j.Target.Info.txt","a")
                    file.write('\n ' + hex(ran) + ' | ' + addr)
                    file.close()
                    sleep(2)
                    wait = input("Press Enter to Exit.")
                    sleep(1)
                    exit()
                    
                if line_count==x:
                    list.clear()
                    
        if len(list)==0:
            print(' scan... ', str(n), end='\r')                        

Ca 42000 keys/s from c000000000000000 to ffffffffffffffff... The sector between  11_________________0 comprise a range of 131071 and will be processed.
The rest of this line are random numbers from 4 different generators..
The counter reflects the number of loops...

      B I N A R Y   S E E D   S C A N   Q U A D R O

 11_________________0____________________________________________

 1110010110110011011010011110001111111110100110111011100101000111

 0xe5b369e3fe9bb947 | 16jY7mM6wrpuAy53vKkh7QXm1JNdWK9Myu 19

 continue...

 1110010011000111001010001111000110100111001000001001110001000100

 0xe4c728f1a7209c44 | 16jY7Ezk5et98CVBWYEigGRwWovzhEtaQ3 121

 continue...

 scan...  192
40  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: February 22, 2022, 07:47:00 PM
jeniferangel79,

take a look at when the puzzle is from and what the BTC was worth at that time... surely there is a reason, security plays a big role in the crypto universe... and see what and how far the programmers get solving the puzzle is certainly an interesting aspect for future crypto projects... ^^
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