Andzhig
Jr. Member
Offline
Activity: 183
Merit: 3
|
|
September 14, 2018, 09:39:38 PM Last edit: September 15, 2018, 06:47:53 PM by Andzhig |
|
So pampering in the Python 3 at the primary level. On a python, you can use gpu https://developer.nvidia.com/how-to-cuda-python but apparently have to rewrite (or write again) the bitcoin library. the fastest https://github.com/ofek/bit . In other things we have a lot of time to learn programming languages)) It would be interesting to look at the program, searching for a puzzle, move your mouse over the screen and look for the coveted prize)). Try it if you want such a script for ndv https://bitcointalk.org/index.php?topic=3102823.0 by copying the ndv program files into the c:\folder1 c:\folder2 etc, and create an empty c:\cmd1.cmd c:\cmd2.cmd For 2 cards. import secrets from bitcoin import * import subprocess import time
def fff(): nnn = str(secrets.choice("0123456789")) return nnn while True: a = 800 while a <= 1000: bbb = str(a) kkk = int(bbb+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()) ran = kkk myhex = "%064x" % ran myhex = myhex[:64] priv = myhex pub = privtopub(priv) pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed") addr = pubtoaddr(pubkey1) oy = """cd "C:\folder1" """ ey = "\nstart /min oclvanitygen.exe -D 0:0 -C -f addresses.txt -o found.txt " f=open("C:/cmd1.cmd","w") f.write (oy) f.write (ey) f.write (priv) f.close() subprocess.Popen([r"C:/cmd1.cmd"]) print(kkk,addr,priv)
bbb2 = str(a) kkk = int(bbb2+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()) ran = kkk myhex = "%064x" % ran myhex = myhex[:64] priv = myhex pub = privtopub(priv) pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed") addr = pubtoaddr(pubkey1) oy = """cd "C:\folder2" """ ey = "\nstart /min oclvanitygen.exe -D 1:0 -C -f addresses.txt -o found.txt " f=open("C:/cmd2.cmd","w") f.write (oy) f.write (ey) f.write (priv) f.close() subprocess.Popen([r"C:/cmd2.cmd"]) print(kkk,addr,priv) time.sleep(10.0) #delay between steps subprocess.call("taskkill /IM oclvanitygen.exe") a = a +1 pass
Stepping with a 10 second interval between steps. 80575324376405043 80547044120475565 80650225349114907 80618406564220162 80760456417194235 80798509910061778 80853179598579019 80880643941104019 80932464878676982 80985690149518090 81055712015396080 81096866333042872 81183973601624592 81155404309801184 81260793490764205 81295972493608040 81350689096804698 81318504516644022 *** For a random, another option (for collisions more suited). 1 10000000 17 10010000 33 10100000 49 10110000 65 11000000 81 11010000 97 11100000 113 11110000 100000000 00000000 00000000 00000000 00000000 00000000 00000000 15-40 57 2 10000001 18 10010001 34 10100001 50 10110001 66 11000001 82 11010001 98 11100001 114 11110001 10011101 00011000 10110110 00111010 11000100 11111111 11011111 22-34 56 44218742292676575 < 1 3 10000010 19 10010010 35 10100010 51 10110010 67 11000010 83 11010010 99 11100010 115 11110010 11010101 01111100 00111111 00110110 11001111 11000010 0010100 24-31 < 1 4 10000011 20 10010011 36 10100011 52 10110011 68 11000011 84 11010011 100 11100011 116 11110011 10001101 10111110 11011011 01010110 10110100 01111101 000100 23-31 < 1 5 10000100 21 10010100 37 10100100 53 10110100 69 11000100 85 11010100 101 11100100 117 11110100 11000000 00111100 01000111 00100011 11110001 10010011 01100 29-24 > 0 6 10000101 22 10010101 38 10100101 54 10110101 70 11000101 86 11010101 102 11100101 118 11110101 11101111 10101110 00010110 01001100 10111001 11100011 1100 22-30 < 1 7 10000110 23 10010110 39 10100110 55 10110110 71 11000110 87 11010110 103 11100110 119 11110110 11101010 00001110 00010100 00110100 00000001 00111010 000 33-18 > 0 8 10000111 24 10010111 40 10100111 56 10110111 72 11000111 88 11010111 104 11100111 120 11110111 10001010 11110101 00001111 00001011 10100100 11010101 00 26-24 > 0 9 10001000 25 10011000 41 10101000 57 10111000 73 11001000 89 11011000 105 11101000 121 11111000 10111010 00001011 10110101 10000000 10101111 10100110 1 24-25 < 1 10 10001001 26 10011001 42 10101001 58 10111001 74 11001001 90 11011001 106 11101001 122 11111001 10101101 11100110 11010111 11001110 00111011 10011011 17-31 < 1 11 10001010 27 10011010 43 10101010 59 10111010 75 11001010 91 11011010 107 11101010 123 11111010 11011001 10101100 00100001 01101010 01111001 0111010 23-24 < 1 12 10001011 28 10011011 44 10101011 60 10111011 76 11001011 92 11011011 108 11101011 124 11111011 10111011 00000110 00001110 00100011 01010101 000100 2 7-19 > 0 13 10001100 29 10011100 45 10101100 61 10111100 77 11001100 93 11011100 109 11101100 125 11111100 10010001 01111110 01010000 10100001 11100000 00101 26 -19 > 0 14 10001101 30 10011101 46 10101101 62 10111101 78 11001101 94 11011101 110 11101101 126 11111101 11100000 00101011 00110101 10100011 01011000 1111 22- 22 = 15 10001110 31 10011110 47 10101110 63 10111110 79 11001110 95 11011110 111 11101110 127 11111110 11010111 10100111 01100100 11111000 10110010 001 19-2 4 < 1 16 10001111 32 10011111 48 10101111 64 10111111 80 11001111 96 11011111112 11101111 128 11111111 10101000 10001000 01110001 01100011 01100011 11 23-19 > 0 10101001 11000011 01001101 01100110 00101101 1 20-21 < 1 1 00000000 17 00010000 33 00100000 49 00110000 65 01000000 81 01010000 97 01100000 113 01110000 11101001 10101110 01001001 00110011 11010110 18-22 40 1003651412950 < 1 2 00000001 18 00010001 34 00100001 50 00110001 66 01000001 82 01010001 98 01100001 114 01110001 10010110 10111111 00000110 00000111 1101001 19-20 < 1 3 00000010 19 00010010 35 00100010 51 00110010 67 01000010 83 01010010 99 01100010 115 01110010 10001000 11100000 10111110 10110011 010000 2 1-17 > 0 4 00000011 20 00010011 36 00100011 52 00110011 68 01000011 84 01010011 100 01100011 116 01110011 10111010 10111011 10101011 01010100 10011 14 -22 < 1 5 00000100 21 00010100 37 00100100 53 00110100 69 01000100 85 01010100 101 01100100 117 01110100 10011101 11101000 00100000 10100111 1100 19- 17 > 0 6 00000101 22 00010101 38 00100101 54 00110101 70 01000101 86 01010101 102 01100101 118 01110101 10010101 11011010 01000010 00101110 000 20-1 5 > 0 7 00000110 23 00010110 39 00100110 55 00110110 71 01000110 87 01010110 103 01100110 119 01110110 11010010 10011001 01100100 01000111 01 18-16 > 0 8 00000111 24 00010111 40 00100111 56 00110111 72 01000111 88 01010111 104 01100111 120 01110111 11010100 10110110 01010100 01101100 0 17-16 > 0 9 00001000 25 00011000 41 00101000 57 00111000 73 01001000 89 01011000 105 01101000 121 01111000 10111000 01100010 10100110 00101110 17-15 > 0 10 00001001 26 00011001 42 00101001 58 00111001 74 01001001 90 01011001 106 01101001 122 01111001 11111010 10011111 11001110 1000111 10-21 < 1 11 00001010 27 00011010 43 00101010 59 00111010 75 01001010 91 01011010 107 01101010 123 01111010 11110110 01010011 00110101 100100 1 4-16 < 1 12 00001011 28 00011011 44 00101011 60 00111011 76 01001011 92 01011011 108 01101011 124 01111011 10111111 00010010 10101000 11110 13 -16 < 1 13 00001100 29 00011100 45 00101100 61 00111100 77 01001100 93 01011100 109 01101100 125 01111100 11011001 00010110 11001110 1000 14- 14 = 14 00001101 30 00011101 46 00101101 62 00111101 78 01001101 94 01011101 110 01101101 126 01111101 11010101 10000111 00001110 101 13-1 4 < 1 15 00001110 31 00011110 47 00101110 63 00111110 79 01001110 95 01011110 111 01101110 127 01111110 11010000 00001100 10011011 10 15-11 > 0 16 00001111 32 00011111 48 00101111 64 00111111 80 01001111 96 01011111 112 01101111 128 01111111 11111101 00101111 01110010 1 8-17 < 1 11011100 00101010 00000100 15-9 > 0 10101010 11011100 1010010 11-12 < 1 10110111 10010000 001111 1 0-12 < 1 11011101 00101001 10100 10 -11 < 1 11010010 11000101 0101 10- 10 = 10101110 10010011 111 7-12 < 1 11000010 00000011 01 12-6 > 0 10111011 00100111 1 6-11 < 1 11001001 00110110 8-8 = 11010001 1110011 6-9 < 1 10100100 110000 9 -5 > 0 10100011 00000 9- 4 > 0 10100111 1011 4-8 < 1 10010000 011 7-4 > 0 10000000 10 8-2 > 0 11101001 1 3-6 < 1 11100000 5-3 > 0 1001100 4-3 > 0 110001 3 -3 = 10101 2- 3 < 1 1000 3-1 > 0 111 3 11 2 1 1
gen these numbers import random from bit import * from PyRandLib import * rand = FastRand63() random.seed(rand())
c1 = str (random.choice("1")) b28 = "00000000000000000000000000000001111111111111111111111111" b29 = "00000000000000000000000000000000111111111111111111111111" b30 = "00000000000000000000000000000000011111111111111111111111" b31 = "00000000000000000000000000000000001111111111111111111111" b32 = "00000000000000000000000000000000000111111111111111111111" b33 = "00000000000000000000000000000000000011111111111111111111" b34 = "00000000000000000000000000000000000001111111111111111111" b35 = "00000000000000000000000000000000000000111111111111111111" b36 = "00000000000000000000000000000000000000011111111111111111" b37 = "00000000000000000000000000000000000000001111111111111111" b38 = "00000000000000000000000000000000000000000111111111111111" b39 = "00000000000000000000000000000000000000000011111111111111" b40 = "00000000000000000000000000000000000000000001111111111111" b41 = "00000000000000000000000000000000000000000000111111111111" b42 = "00000000000000000000000000000000000000000000011111111111" #c2 = "00000000000000000000000000000000000000000000000000000001" spisok =[b30]
while True: aa = 1 while aa <= 1: for element in (spisok): s = element d = ''.join(random.sample(s,len(s))) bina = (c1+d) b = int(c1+d,2) key = Key.from_int(b) addr = key.address if addr == "15c9mPGLku1HuW9LRtBf4jcHVpBUt8txKz": print ("found!!!",b,addr) s1 = str(b) s2 = addr f=open(u"C:/a.txt","a") f.write(s1) f.write(s2) f.close() pass else: print (s,bina,b,addr) aa = aa +1 pass
PyRandLib https://github.com/schmouk/CRandLib without it del from PyRandLib import * rand = FastRand63() random.seed(rand()) or #from PyRandLib import * #rand = FastRand63() #random.seed(rand()) *** angle "rate". 00000000000000000000001111111111111111111111111111111111 0000000000000000000000001111111111111111111111111111111 000000000000000000000001111111111111111111111111111111 00000000000000000000000000000111111111111111111111111 0000000000000000000000111111111111111111111111111111 000000000000000000000000000000000111111111111111111 00000000000000000000000000111111111111111111111111 0000000000000000000000001111111111111111111111111 000000000000000001111111111111111111111111111111 00000000000000000000000111111111111111111111111 0000000000000000000000000001111111111111111111 000000000000000000000000001111111111111111111 00000000000000000000001111111111111111111111 0000000000000000000111111111111111111111111 000000000000000000000001111111111111111111 00000000000000000000111111111111111111111 0000000000000000001111111111111111111111 000000000000000000011111111111111111111 00000000000000000000011111111111111111 0000000000000011111111111111111111111 000000000000000000011111111111111111 00000000000000000000111111111111111 0000000000000000001111111111111111 000000000000000001111111111111111 00000000000000000111111111111111 0000000000111111111111111111111 000000000000001111111111111111 00000000000001111111111111111 0000000000000011111111111111 000000000000011111111111111 00000000000000011111111111 0000000011111111111111111 000000000000000111111111 00000000000111111111111 0000000000111111111111 000000000011111111111 00000000001111111111 0000000111111111111 000000000000111111 00000011111111111 0000000011111111 000000111111111 00000000011111 0000000001111 000011111111 00000001111 0000000011 000111111 00000111 0000111 000111 00111 0001 111 11 1
|
|
|
|
|
|
Even in the event that an attacker gains more than 50% of the network's
computational power, only transactions sent by the attacker could be
reversed or double-spent. The network would not be destroyed.
|
|
|
Advertised sites are not endorsed by the Bitcoin Forum. They may be unsafe, untrustworthy, or illegal in your jurisdiction.
|
|
|
holy_ship
Jr. Member
Offline
Activity: 109
Merit: 1
|
|
September 15, 2018, 04:18:31 PM |
|
and you cant find the key unless you know exactly where it lies in the hexadecimal range
Do you really think that guy who found p55 and p56 "knew exactly" magic numbers? Why a fsck he waited for 3 months between p55 and p56? Why he will spend p57 not tomorrow, but after several months? The correct answer: because he brutforces whole zillion of addresses. And your speed 227MK is fscking awesome - it is almost the whole crapy-bla-bla-fastest-on-the-planet LBC. So this proggie is several times faster then VG from NDV.
|
|
|
|
BurtW
Legendary
Offline
Activity: 2646
Merit: 1136
All paid signature campaigns should be banned.
|
|
September 15, 2018, 05:15:35 PM |
|
There is no magic formula. As the author of the puzzle stated the addresses were selected randomly. Random = no formula, got it? Continue to waste your time looking but you will never find it.
|
Our family was terrorized by Homeland Security. Read all about it here: http://www.jmwagner.com/ and http://www.burtw.com/ Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!
|
|
|
seoincorporation
Legendary
Offline
Activity: 3150
Merit: 2931
Top Crypto Casino
|
|
September 16, 2018, 06:04:17 PM |
|
I posted this puzzle transaction in January'15 https://bitcointalk.org/index.php?topic=932434.0(This tx was not sent by me) Let me translate from Russian into English the main point in my prevoius post: First output: take random number from 20 upto 21-1, use it as private key Second output: take random number from 21 upto 22-1, use it as private key Third output: take random number from 22 upto 23-1, use it as private key And so on... The Russians always get it first -snip- "1 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU73sVHnoWn 1BgGZ9tcN4rm9KBzDn7KprQz87SZ26SAMH (1)" "2 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S 1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb (3)" "3 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU76rnZwVdz 19ZewH8Kk1PDbSNdJ97FP4EiCjTRaZMZQA (7)" "4 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU77MfhviY5 1EhqbyUMvvs7BfL8goY6qcPbD6YKfPqb7e (8)" "5 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Dq8Au4Pv 1E6NuFjCi27W5zoXg8TRdcSRq84zJeBW3k (21)" "6 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Tmu6qHxS 1PitScNLyp2HCygzadCh7FveTnfmpPbfp8 (49)" "7 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7hDgvu64y 1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC (76)" "8 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU8xvGK1zpm 1M92tSqNmQLYw33fuBvjmeadirh1ysMBxK (224)" "9 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUB3vfDKcxZ 1CQFwcjw1dwhtkVWBttNLDtqL7ivBonGPV (467)" "10 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUBTL67V6dE 1LeBZP5QCwwgXRtmVUvTVrraqPUokyLHqe (514)" "11 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUGxXgtm63M 1PgQVLmst3Z314JrQn5TNiys8Hc38TcXJu (1155)" "12 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUW5RtS2JN1 1DBaumZxUkM4qMQRt2LVWyFJq5kDtSZQot (2683)" "13 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUspniiQZds 1Pie8JkxBT6MGPz9Nvi3fsPkr2D8q3GBc1 (5216)" "14 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFVfZyiN5iEG 1ErZWg5cFCe4Vw5BzgfzB74VNLaXEiEkhk (10544)" "15 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFY5iMZbuRxj 1QCbW9HWnwQWiQqVo5exhAnmfqKRrCRsvW (26867)" "16 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFbjHrFMWzJp 1BDyrQ6WoF8VN3g9SAS1iKZcPzFfnDVieY (51510)" "17 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFiHkRsp99uC 1HduPEXZRdG26SUT5Yk83mLkPyjnZuJ7Bm (95823)" "18 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFyWkjT5fywW 1GnNTmTVLZiqQfLbAdp9DVdicEnB5GoERE (198669)" "19 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rGP2jMrxCfX3 1NWmZRpHH4XSPwsW6dsS3nrNWfL1yrJj4w (357535)" "20 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rHfuE2Tg4nJW 1HsMJxNiV7TLxmoF6uJNkydxPFDog4NQum (863317)" "21 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rL6JJvw6XUry 14oFNXucftsHiUMY8uctg6N487riuyXs4h (1811764)" "22 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rP9Ja2dhtxoh 1CfZWK1QTQE3eS9qn61dQjV89KDjZzfNcv (3007503)" "23 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rVkthFNsQ6i7 1L2GM8eE7mJWLdo3HZS6su1832NX2txaac (5598802)" "24 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rtHyNcFoApRd 1rSnXMr63jdCuegJFuidJqWxUPV7AtUf7 (14428676)"
You got all the work done right there until the #24, nicely done I give it a shot with crunch, and to generate all the combinations is almost impossible: [... ~]$ crunch 52 52 abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789 + + + -t KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7r@@@@@@@@@@@ >> db_pk.txt Crunch will now generate the following amount of data: 9372849260642785280 bytes 8938645611422 MB 8729146104 GB 8524556 TB 8324 PB Crunch will now generate the following number of lines: 15143072536417990656
So, i don't have 8324PB to store that list, the way to do it would be to live verify those addys while generate them, if they have balance then store it, other way let it go. I will work on that code, sound fun.
|
|
|
|
Amnuazka
Newbie
Offline
Activity: 15
Merit: 0
|
|
September 16, 2018, 09:07:23 PM |
|
Congratulation! Next : 57 2^56 ~ 2^57
|
|
|
|
Andzhig
Jr. Member
Offline
Activity: 183
Merit: 3
|
|
September 17, 2018, 06:45:48 PM |
|
to generate all the combinations is almost impossible: in theory everything is not necessary. If you determine the number of zeros in 57 "100000000000000000000000000000000000000000000000000000000" (1+ 7 bytes) number of options to decrease. Eg pz16 in bin 1100100100110110 have 2 bytes 11001001 00110110 take the table 1 10000000 17 10010000 33 10100000 49 10110000 65 11000000 81 11010000 97 11100000 113 11110000 2 10000001 18 10010001 34 10100001 50 10110001 66 11000001 82 11010001 98 11100001 114 11110001 3 10000010 19 10010010 35 10100010 51 10110010 67 11000010 83 11010010 99 11100010 115 11110010 4 10000011 20 10010011 36 10100011 52 10110011 68 11000011 84 11010011 100 11100011 116 11110011 5 10000100 21 10010100 37 10100100 53 10110100 69 11000100 85 11010100 101 11100100 117 11110100 6 10000101 22 10010101 38 10100101 54 10110101 70 11000101 86 11010101 102 11100101 118 11110101 7 10000110 23 10010110 39 10100110 55 10110110 71 11000110 87 11010110 103 11100110 119 11110110 8 10000111 24 10010111 40 10100111 56 10110111 72 11000111 88 11010111 104 11100111 120 11110111 9 10001000 25 10011000 41 10101000 57 10111000 73 11001000 89 11011000 105 11101000 121 11111000 10 10001001 26 10011001 42 10101001 58 10111001 74 11001001 90 11011001 106 11101001 122 11111001 11 10001010 27 10011010 43 10101010 59 10111010 75 11001010 91 11011010 107 11101010 123 11111010 12 10001011 28 10011011 44 10101011 60 10111011 76 11001011 92 11011011 108 11101011 124 11111011 13 10001100 29 10011100 45 10101100 61 10111100 77 11001100 93 11011100 109 11101100 125 11111100 14 10001101 30 10011101 46 10101101 62 10111101 78 11001101 94 11011101 110 11101101 126 11111101 15 10001110 31 10011110 47 10101110 63 10111110 79 11001110 95 11011110 111 11101110 127 11111110 16 10001111 32 10011111 48 10101111 64 10111111 80 11001111 96 11011111 112 11101111 128 11111111 1 00000000 17 00010000 33 00100000 49 00110000 65 01000000 81 01010000 97 01100000 113 01110000 2 00000001 18 00010001 34 00100001 50 00110001 66 01000001 82 01010001 98 01100001 114 01110001 3 00000010 19 00010010 35 00100010 51 00110010 67 01000010 83 01010010 99 01100010 115 01110010 4 00000011 20 00010011 36 00100011 52 00110011 68 01000011 84 01010011 100 01100011 116 01110011 5 00000100 21 00010100 37 00100100 53 00110100 69 01000100 85 01010100 101 01100100 117 01110100 6 00000101 22 00010101 38 00100101 54 00110101 70 01000101 86 01010101 102 01100101 118 01110101 7 00000110 23 00010110 39 00100110 55 00110110 71 01000110 87 01010110 103 01100110 119 01110110 8 00000111 24 00010111 40 00100111 56 00110111 72 01000111 88 01010111 104 01100111 120 01110111 9 00001000 25 00011000 41 00101000 57 00111000 73 01001000 89 01011000 105 01101000 121 01111000 10 00001001 26 00011001 42 00101001 58 00111001 74 01001001 90 01011001 106 01101001 122 01111001 11 00001010 27 00011010 43 00101010 59 00111010 75 01001010 91 01011010 107 01101010 123 01111010 12 00001011 28 00011011 44 00101011 60 00111011 76 01001011 92 01011011 108 01101011 124 01111011 13 00001100 29 00011100 45 00101100 61 00111100 77 01001100 93 01011100 109 01101100 125 01111100 14 00001101 30 00011101 46 00101101 62 00111101 78 01001101 94 01011101 110 01101101 126 01111101 15 00001110 31 00011110 47 00101110 63 00111110 79 01001110 95 01011110 111 01101110 127 01111110 16 00001111 32 00011111 48 00101111 64 00111111 80 01001111 96 01011111 112 01101111 128 01111111
all bust will take 32678 options (for each of the first part of the table, each of the two parts), if we take variants with 8 zeros would need 6435 options, less than five times. it remains to determine how many zeros will be in 57.. somehow he's there for a cunning less than 2^x from a higher 2^x takes away pieces.. - 2^x remainder 11111111111111111111110000000000000000000000000000000000 72057576858058752 72057594037927936 17179869184 1111111111111111111111110000000000000000000000000000000 36028794871480320 36028797018963968 2147483648 111111111111111111111110000000000000000000000000000000 18014396361998336 18014398509481984 2147483648 11111111111111111111111111111000000000000000000000000 9007199237963776 9007199254740992 16777216 1111111111111111111111000000000000000000000000000000 4503598553628672 4503599627370496 1073741824 111111111111111111111111111111111000000000000000000 2251799813423104 2251799813685248 262144 11111111111111111111111111000000000000000000000000 1125899890065408 1125899906842624 16777216 1111111111111111111111110000000000000000000000000 562949919866880 562949953421312 33554432 111111111111111110000000000000000000000000000000 281472829227008 281474976710656 2147483648 11111111111111111111111000000000000000000000000 140737471578112 140737488355328 16777216 1111111111111111111111111110000000000000000000 70368743653376 70368744177664 524288 111111111111111111111111110000000000000000000 35184371564544 35184372088832 524288 11111111111111111111110000000000000000000000 17592181850112 17592186044416 4194304 1111111111111111111000000000000000000000000 8796076244992 8796093022208 16777216 111111111111111111111110000000000000000000 4398045986816 4398046511104 524288 11111111111111111111000000000000000000000 2199021158400 2199023255552 2097152 1111111111111111110000000000000000000000 1099507433472 1099511627776 4194304 111111111111111111100000000000000000000 549754765312 549755813888 1048576 11111111111111111111100000000000000000 274877775872 274877906944 131072 1111111111111100000000000000000000000 137430564864 137438953472 8388608 111111111111111111100000000000000000 68719345664 68719476736 131072 11111111111111111111000000000000000 34359705600 34359738368 32768 1111111111111111110000000000000000 17179803648 17179869184 65536 111111111111111110000000000000000 8589869056 8589934592 65536 11111111111111111000000000000000 4294934528 4294967296 32768 1111111111000000000000000000000 2145386496 2147483648 2097152 111111111111110000000000000000 1073676288 1073741824 65536 11111111111110000000000000000 536805376 536870912 65536 1111111111111100000000000000 268419072 268435456 16384 111111111111100000000000000 134201344 134217728 16384 11111111111111100000000000 67106816 67108864 2048 1111111100000000000000000 33423360 33554432 131072 111111111111111000000000 16776704 16777216 512 11111111111000000000000 8384512 8388608 4096 1111111111000000000000 4190208 4194304 4096 111111111100000000000 2095104 2097152 2048 11111111110000000000 1047552 1048576 1024 1111111000000000000 520192 524288 4096 111111111111000000 262080 262144 64 11111100000000000 129024 131072 2048 1111111100000000 65280 65536 256 111111000000000 32256 32768 512 11111111100000 16352 16384 32 1111111110000 8176 8192 16 111100000000 3840 4096 256 11111110000 2032 2048 16 1111111100 1020 1024 4 111000000 448 512 64 11111000 248 256 8 1111000 120 128 8 111000 56 64 8 11000 24 32 8 1110 14 16 2 000 00 0 although even in this case the number will be impressive...
|
|
|
|
|
natedawg469
Newbie
Offline
Activity: 15
Merit: 0
|
|
September 18, 2018, 06:54:19 AM |
|
So pampering in the Python 3 at the primary level. On a python, you can use gpu https://developer.nvidia.com/how-to-cuda-python but apparently have to rewrite (or write again) the bitcoin library. the fastest https://github.com/ofek/bit . In other things we have a lot of time to learn programming languages)) It would be interesting to look at the program, searching for a puzzle, move your mouse over the screen and look for the coveted prize)). Try it if you want such a script for ndv https://bitcointalk.org/index.php?topic=3102823.0 by copying the ndv program files into the c:\folder1 c:\folder2 etc, and create an empty c:\cmd1.cmd c:\cmd2.cmd For 2 cards. import secrets from bitcoin import * import subprocess import time
def fff(): nnn = str(secrets.choice("0123456789")) return nnn while True: a = 800 while a <= 1000: bbb = str(a) kkk = int(bbb+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff())
ran = kkk myhex = "%064x" % ran myhex = myhex[:64] priv = myhex pub = privtopub(priv) pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed") addr = pubtoaddr(pubkey1) oy = """cd "C:\folder1" """ ey = "\nstart /min oclvanitygen.exe -D 0:0 -C -f addresses.txt -o found.txt " f=open("C:/cmd1.cmd","w") f.write (oy) f.write (ey) f.write (priv) f.close() subprocess.Popen([r"C:/cmd1.cmd"]) print(kkk,addr,priv)
bbb2 = str(a) kkk = int(bbb2+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()+fff()) ran = kkk myhex = "%064x" % ran myhex = myhex[:64] priv = myhex pub = privtopub(priv) pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed") addr = pubtoaddr(pubkey1) oy = """cd "C:\folder2" """ ey = "\nstart /min oclvanitygen.exe -D 1:0 -C -f addresses.txt -o found.txt " f=open("C:/cmd2.cmd","w") f.write (oy) f.write (ey) f.write (priv) f.close() subprocess.Popen([r"C:/cmd2.cmd"]) print(kkk,addr,priv) time.sleep(10.0) #delay between steps subprocess.call("taskkill /IM oclvanitygen.exe") a = a +1 pass
Stepping with a 10 second interval between steps. 80575324376405043 80547044120475565 80650225349114907 80618406564220162 80760456417194235 80798509910061778 80853179598579019 80880643941104019 80932464878676982 80985690149518090 81055712015396080 81096866333042872 81183973601624592 81155404309801184 81260793490764205 81295972493608040 81350689096804698 81318504516644022 *** For a random, another option (for collisions more suited). 1 10000000 17 10010000 33 10100000 49 10110000 65 11000000 81 11010000 97 11100000 113 11110000 100000000 00000000 00000000 00000000 00000000 00000000 00000000 15-40 57 2 10000001 18 10010001 34 10100001 50 10110001 66 11000001 82 11010001 98 11100001 114 11110001 10011101 00011000 10110110 00111010 11000100 11111111 11011111 22-34 56 44218742292676575 < 1 3 10000010 19 10010010 35 10100010 51 10110010 67 11000010 83 11010010 99 11100010 115 11110010 11010101 01111100 00111111 00110110 11001111 11000010 0010100 24-31 < 1 4 10000011 20 10010011 36 10100011 52 10110011 68 11000011 84 11010011 100 11100011 116 11110011 10001101 10111110 11011011 01010110 10110100 01111101 000100 23-31 < 1 5 10000100 21 10010100 37 10100100 53 10110100 69 11000100 85 11010100 101 11100100 117 11110100 11000000 00111100 01000111 00100011 11110001 10010011 01100 29-24 > 0 6 10000101 22 10010101 38 10100101 54 10110101 70 11000101 86 11010101 102 11100101 118 11110101 11101111 10101110 00010110 01001100 10111001 11100011 1100 22-30 < 1 7 10000110 23 10010110 39 10100110 55 10110110 71 11000110 87 11010110 103 11100110 119 11110110 11101010 00001110 00010100 00110100 00000001 00111010 000 33-18 > 0 8 10000111 24 10010111 40 10100111 56 10110111 72 11000111 88 11010111 104 11100111 120 11110111 10001010 11110101 00001111 00001011 10100100 11010101 00 26-24 > 0 9 10001000 25 10011000 41 10101000 57 10111000 73 11001000 89 11011000 105 11101000 121 11111000 10111010 00001011 10110101 10000000 10101111 10100110 1 24-25 < 1 10 10001001 26 10011001 42 10101001 58 10111001 74 11001001 90 11011001 106 11101001 122 11111001 10101101 11100110 11010111 11001110 00111011 10011011 17-31 < 1 11 10001010 27 10011010 43 10101010 59 10111010 75 11001010 91 11011010 107 11101010 123 11111010 11011001 10101100 00100001 01101010 01111001 0111010 23-24 < 1 12 10001011 28 10011011 44 10101011 60 10111011 76 11001011 92 11011011 108 11101011 124 11111011 10111011 00000110 00001110 00100011 01010101 000100 2 7-19 > 0 13 10001100 29 10011100 45 10101100 61 10111100 77 11001100 93 11011100 109 11101100 125 11111100 10010001 01111110 01010000 10100001 11100000 00101 26 -19 > 0 14 10001101 30 10011101 46 10101101 62 10111101 78 11001101 94 11011101 110 11101101 126 11111101 11100000 00101011 00110101 10100011 01011000 1111 22- 22 = 15 10001110 31 10011110 47 10101110 63 10111110 79 11001110 95 11011110 111 11101110 127 11111110 11010111 10100111 01100100 11111000 10110010 001 19-2 4 < 1 16 10001111 32 10011111 48 10101111 64 10111111 80 11001111 96 11011111112 11101111 128 11111111 10101000 10001000 01110001 01100011 01100011 11 23-19 > 0 10101001 11000011 01001101 01100110 00101101 1 20-21 < 1 1 00000000 17 00010000 33 00100000 49 00110000 65 01000000 81 01010000 97 01100000 113 01110000 11101001 10101110 01001001 00110011 11010110 18-22 40 1003651412950 < 1 2 00000001 18 00010001 34 00100001 50 00110001 66 01000001 82 01010001 98 01100001 114 01110001 10010110 10111111 00000110 00000111 1101001 19-20 < 1 3 00000010 19 00010010 35 00100010 51 00110010 67 01000010 83 01010010 99 01100010 115 01110010 10001000 11100000 10111110 10110011 010000 2 1-17 > 0 4 00000011 20 00010011 36 00100011 52 00110011 68 01000011 84 01010011 100 01100011 116 01110011 10111010 10111011 10101011 01010100 10011 14 -22 < 1 5 00000100 21 00010100 37 00100100 53 00110100 69 01000100 85 01010100 101 01100100 117 01110100 10011101 11101000 00100000 10100111 1100 19- 17 > 0 6 00000101 22 00010101 38 00100101 54 00110101 70 01000101 86 01010101 102 01100101 118 01110101 10010101 11011010 01000010 00101110 000 20-1 5 > 0 7 00000110 23 00010110 39 00100110 55 00110110 71 01000110 87 01010110 103 01100110 119 01110110 11010010 10011001 01100100 01000111 01 18-16 > 0 8 00000111 24 00010111 40 00100111 56 00110111 72 01000111 88 01010111 104 01100111 120 01110111 11010100 10110110 01010100 01101100 0 17-16 > 0 9 00001000 25 00011000 41 00101000 57 00111000 73 01001000 89 01011000 105 01101000 121 01111000 10111000 01100010 10100110 00101110 17-15 > 0 10 00001001 26 00011001 42 00101001 58 00111001 74 01001001 90 01011001 106 01101001 122 01111001 11111010 10011111 11001110 1000111 10-21 < 1 11 00001010 27 00011010 43 00101010 59 00111010 75 01001010 91 01011010 107 01101010 123 01111010 11110110 01010011 00110101 100100 1 4-16 < 1 12 00001011 28 00011011 44 00101011 60 00111011 76 01001011 92 01011011 108 01101011 124 01111011 10111111 00010010 10101000 11110 13 -16 < 1 13 00001100 29 00011100 45 00101100 61 00111100 77 01001100 93 01011100 109 01101100 125 01111100 11011001 00010110 11001110 1000 14- 14 = 14 00001101 30 00011101 46 00101101 62 00111101 78 01001101 94 01011101 110 01101101 126 01111101 11010101 10000111 00001110 101 13-1 4 < 1 15 00001110 31 00011110 47 00101110 63 00111110 79 01001110 95 01011110 111 01101110 127 01111110 11010000 00001100 10011011 10 15-11 > 0 16 00001111 32 00011111 48 00101111 64 00111111 80 01001111 96 01011111 112 01101111 128 01111111 11111101 00101111 01110010 1 8-17 < 1 11011100 00101010 00000100 15-9 > 0 10101010 11011100 1010010 11-12 < 1 10110111 10010000 001111 1 0-12 < 1 11011101 00101001 10100 10 -11 < 1 11010010 11000101 0101 10- 10 = 10101110 10010011 111 7-12 < 1 11000010 00000011 01 12-6 > 0 10111011 00100111 1 6-11 < 1 11001001 00110110 8-8 = 11010001 1110011 6-9 < 1 10100100 110000 9 -5 > 0 10100011 00000 9- 4 > 0 10100111 1011 4-8 < 1 10010000 011 7-4 > 0 10000000 10 8-2 > 0 11101001 1 3-6 < 1 11100000 5-3 > 0 1001100 4-3 > 0 110001 3 -3 = 10101 2- 3 < 1 1000 3-1 > 0 111 3 11 2 1 1
gen these numbers import random from bit import * from PyRandLib import * rand = FastRand63() random.seed(rand())
c1 = str (random.choice("1")) b28 = "00000000000000000000000000000001111111111111111111111111" b29 = "00000000000000000000000000000000111111111111111111111111" b30 = "00000000000000000000000000000000011111111111111111111111" b31 = "00000000000000000000000000000000001111111111111111111111" b32 = "00000000000000000000000000000000000111111111111111111111" b33 = "00000000000000000000000000000000000011111111111111111111" b34 = "00000000000000000000000000000000000001111111111111111111" b35 = "00000000000000000000000000000000000000111111111111111111" b36 = "00000000000000000000000000000000000000011111111111111111" b37 = "00000000000000000000000000000000000000001111111111111111" b38 = "00000000000000000000000000000000000000000111111111111111" b39 = "00000000000000000000000000000000000000000011111111111111" b40 = "00000000000000000000000000000000000000000001111111111111" b41 = "00000000000000000000000000000000000000000000111111111111" b42 = "00000000000000000000000000000000000000000000011111111111" #c2 = "00000000000000000000000000000000000000000000000000000001" spisok =[b30]
while True: aa = 1 while aa <= 1: for element in (spisok): s = element d = ''.join(random.sample(s,len(s))) bina = (c1+d) b = int(c1+d,2) key = Key.from_int(b) addr = key.address if addr == "15c9mPGLku1HuW9LRtBf4jcHVpBUt8txKz": print ("found!!!",b,addr) s1 = str(b) s2 = addr f=open(u"C:/a.txt","a") f.write(s1) f.write(s2) f.close() pass else: print (s,bina,b,addr) aa = aa +1 pass
PyRandLib https://github.com/schmouk/CRandLib without it del from PyRandLib import * rand = FastRand63() random.seed(rand()) or #from PyRandLib import * #rand = FastRand63() #random.seed(rand()) *** angle "rate". 00000000000000000000001111111111111111111111111111111111 0000000000000000000000001111111111111111111111111111111 000000000000000000000001111111111111111111111111111111 00000000000000000000000000000111111111111111111111111 0000000000000000000000111111111111111111111111111111 000000000000000000000000000000000111111111111111111 00000000000000000000000000111111111111111111111111 0000000000000000000000001111111111111111111111111 000000000000000001111111111111111111111111111111 00000000000000000000000111111111111111111111111 0000000000000000000000000001111111111111111111 000000000000000000000000001111111111111111111 00000000000000000000001111111111111111111111 0000000000000000000111111111111111111111111 000000000000000000000001111111111111111111 00000000000000000000111111111111111111111 0000000000000000001111111111111111111111 000000000000000000011111111111111111111 00000000000000000000011111111111111111 0000000000000011111111111111111111111 000000000000000000011111111111111111 00000000000000000000111111111111111 0000000000000000001111111111111111 000000000000000001111111111111111 00000000000000000111111111111111 0000000000111111111111111111111 000000000000001111111111111111 00000000000001111111111111111 0000000000000011111111111111 000000000000011111111111111 00000000000000011111111111 0000000011111111111111111 000000000000000111111111 00000000000111111111111 0000000000111111111111 000000000011111111111 00000000001111111111 0000000111111111111 000000000000111111 00000011111111111 0000000011111111 000000111111111 00000000011111 0000000001111 000011111111 00000001111 0000000011 000111111 00000111 0000111 000111 00111 0001 111 11 1
OK I have done the installation steps, how I actually run your program/code in Anaconda with numba?
|
|
|
|
Andzhig
Jr. Member
Offline
Activity: 183
Merit: 3
|
|
September 18, 2018, 08:42:20 AM Last edit: September 18, 2018, 11:42:15 AM by Andzhig |
|
OK I have done the installation steps, how I actually run your program/code in Anaconda with numba? Look in the (google) manual on numba jit. jit does not speed up every "own" random in python. acceleration example from google. from numba import jit import numpy import time #from bit import * #import secrets #from bitcoin import *
start_time = time.time()
@jit def inner_func(a_list, b_list): sum = 0 j = 0 for y in range(0, 16): for x in range(0, 16): p = a_list[j] - b_list[j] sum += p * p j += 1 return sum
@jit def outer_func(a_list, b_list): sum = 0 for g in range(0, 100000000): # 100 000 000 == 10^8 !!! sum += inner_func(a_list, b_list) return sum
def main(): maxint = numpy.iinfo(numpy.intc).max a_list = numpy.random.randint(maxint, size=256) b_list = numpy.random.randint(maxint, size=256) sum = outer_func(a_list, b_list) print(sum)
if __name__ == '__main__': main()
print("--- %s seconds ---" % (time.time() - start_time))
100 000 000 for a 30 second well cpu is not bad, but that's until we started converting the numbers into bitcoin addresses ((( example with the fastest Bitcoin library "bit" without it, the cycle 1000000 takes ~0.4 sec. with her (remove #) ~123,3! sec o_O import time from bit import * from numba import jit import random
@jit def ran(): a = random.randint(50028797018963968,66057594037927936) return a
start_time = time.time()
i = 1000000 while i >= 1: #ran() #k=Key() #k.address
i = i -1
print("--- %s seconds ---" % (time.time() - start_time)) example with a random access passage from "01111111111111111111111111111111111111111111111111111111" to "00000000000000000000000000000000000000000000000000000001" each shuffle 1000 times import random from bit import * from PyRandLib import * rand = FastRand63() random.seed(rand())
c1 = str (random.choice("1")) b1 = "01111111111111111111111111111111111111111111111111111111" b2 = "00111111111111111111111111111111111111111111111111111111" b3 = "00011111111111111111111111111111111111111111111111111111" b4 = "00001111111111111111111111111111111111111111111111111111" b5 = "00000111111111111111111111111111111111111111111111111111" b6 = "00000011111111111111111111111111111111111111111111111111" b7 = "00000001111111111111111111111111111111111111111111111111" b8 = "00000000111111111111111111111111111111111111111111111111" b9 = "00000000011111111111111111111111111111111111111111111111" b10 = "00000000001111111111111111111111111111111111111111111111" b11 = "00000000000111111111111111111111111111111111111111111111" b12 = "00000000000011111111111111111111111111111111111111111111" b13 = "00000000000001111111111111111111111111111111111111111111" b14 = "00000000000000111111111111111111111111111111111111111111" b15 = "00000000000000011111111111111111111111111111111111111111" b16 = "00000000000000001111111111111111111111111111111111111111" b17 = "00000000000000000111111111111111111111111111111111111111" b18 = "00000000000000000011111111111111111111111111111111111111" b19 = "00000000000000000001111111111111111111111111111111111111" b20 = "00000000000000000000111111111111111111111111111111111111" b21 = "00000000000000000000011111111111111111111111111111111111" b22 = "00000000000000000000001111111111111111111111111111111111" b23 = "00000000000000000000000111111111111111111111111111111111" b24 = "00000000000000000000000011111111111111111111111111111111" b25 = "00000000000000000000000001111111111111111111111111111111" b26 = "00000000000000000000000000111111111111111111111111111111" b27 = "00000000000000000000000000011111111111111111111111111111" b28 = "00000000000000000000000000001111111111111111111111111111" b29 = "00000000000000000000000000000111111111111111111111111111" b30 = "00000000000000000000000000000011111111111111111111111111" b31 = "00000000000000000000000000000001111111111111111111111111" b32 = "00000000000000000000000000000000111111111111111111111111" b33 = "00000000000000000000000000000000011111111111111111111111" b34 = "00000000000000000000000000000000001111111111111111111111" b35 = "00000000000000000000000000000000000111111111111111111111" b36 = "00000000000000000000000000000000000011111111111111111111" b37 = "00000000000000000000000000000000000001111111111111111111" b38 = "00000000000000000000000000000000000000111111111111111111" b39 = "00000000000000000000000000000000000000011111111111111111" b40 = "00000000000000000000000000000000000000001111111111111111" b41 = "00000000000000000000000000000000000000000111111111111111" b42 = "00000000000000000000000000000000000000000011111111111111" b43 = "00000000000000000000000000000000000000000001111111111111" b44 = "00000000000000000000000000000000000000000000111111111111" b45 = "00000000000000000000000000000000000000000000011111111111" b46 = "00000000000000000000000000000000000000000000001111111111" b47 = "00000000000000000000000000000000000000000000000111111111" b48 = "00000000000000000000000000000000000000000000000011111111" b49 = "00000000000000000000000000000000000000000000000001111111" b50 = "00000000000000000000000000000000000000000000000000111111" b51 = "00000000000000000000000000000000000000000000000000011111" b52 = "00000000000000000000000000000000000000000000000000001111" b53 = "00000000000000000000000000000000000000000000000000000111" b54 = "00000000000000000000000000000000000000000000000000000011" b55 = "00000000000000000000000000000000000000000000000000000001" #xx = "00000000000000000000000000000000000000000000000000000001"
while True: spisok = [b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11,b12,b13,b14,b15,b16,b18,b19,b20,b21,b22,b23,b24,b25,b26,b27,b28,b30,b31,b32,b33,b34,b35,b36,b37,b38,b39,b40,b41,b42,b43,b44,b45,b46,b47,b48,b49,b50,b51,b52,b53,b54,b55] for element in (spisok): for spisok in range(1000): s = element d = ''.join(random.sample(s,len(s))) bina = (c1+d) b = int(c1+d,2) key = Key.from_int(b) addr = key.address if addr == "15c9mPGLku1HuW9LRtBf4jcHVpBUt8txKz": print ("found!!!",b,addr) s1 = str(b) s2 = addr f=open(u"C:/a.txt","a") f.write(s1) f.write(s2) f.close() pass else: print (s,bina,b,addr) pass "print" slows down speed so you can not use it (#bina = (c1+d),pass #print (s,bina,b,addr)) and shorten the "spisok" but use @jit here useless bottle neck "bitcoin lib".
|
|
|
|
adaseb
Legendary
Offline
Activity: 3752
Merit: 1710
|
|
September 29, 2018, 11:44:21 PM |
|
Looks like the LBC is behind and someone else is solving these puzzles.
I am pretty sure some large farm of >1000 GPUs or so switched from mining ETH to solving these private keys, already 2 of them were solved beating LBC pretty much.
Its too bad because with each solved puzzle, the work required basically doubles and so does the time. Probably not worth using the GPU to solve from here on out, unless that Vanitygen software gets improved and gets a faster speed. I think the most I could get out of that was like 75MH/s from an RX 470 highly overclocked.
|
|
|
|
holy_ship
Jr. Member
Offline
Activity: 109
Merit: 1
|
|
October 01, 2018, 02:29:00 AM |
|
Looks like the LBC is behind and someone else is solving these puzzles.
Looks like you've been sleeping for too long Stupid rico is still continuing on depleted 55-range. Of course not a single sane person will participate in LBC now. I am pretty sure some large farm of >1000 GPUs or so switched from mining ETH to solving these private keys, already 2 of them were solved beating LBC pretty much.
Maybe a Martian Colony of Sectoids awoke to snatch our precious 32BTC! p.s. If you do a simple math, you'll see that the guy who took puzzle 55 and puzzle 56 only have speed of 1000 megakeys. That's less than LBC had several months ago. If rico wasn't such a worthless project manager, this could be LBC. Now sane people use BitCrack, it makes 1000megakeys with only two gtx1080ti.
|
|
|
|
bayu7adi
|
|
October 01, 2018, 02:57:33 AM |
|
This puzle has 32 BTC prizes, then what is the motive behind the person who gave 32 BTC to use as a public donation for those who can decode? 32 BTC is a lot, it's like a treasure that can be solved by anyone, even though it's not easy.
|
Mr.Stork | | E X C H A N G E S E R V I C E | ████████▄▄▄▄▄ ██████▄█████████▀ █████▀██▀▀█████▄ ████████████████ █████▐█▄▀▄██▄▀▄██ █████▄█▄▄▄▄██▄▄▄ █████▀██████████ ███████▀▀██████▌ ███████████▀██▀ ███████▄████▄▄ █████▄█████████▄ ███▄█████████████▄ ░░▄███████████████▄ | |
| | | ██████ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██████ | ▄▄███▄▄ ▄█████████▄ ▄███████████▄ ███▀██▀██▀███ ▀███████████▀ ▀█████████▀ ▄█████████████▄ █████████████████ ███████████████████ ▐███████████████████▌ █████████████████████ █████████████████████ ▀▀███████████████▀▀ | | ██████ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██████ |
|
|
|
GoldTiger69
|
|
October 01, 2018, 03:13:32 AM |
|
This puzle has 32 BTC prizes, then what is the motive behind the person who gave 32 BTC to use as a public donation for those who can decode? 32 BTC is a lot, it's like a treasure that can be solved by anyone, even though it's not easy.
It is a test, and back then BTC wasn't that expensive.
|
|
|
|
adaseb
Legendary
Offline
Activity: 3752
Merit: 1710
|
|
October 01, 2018, 07:45:56 AM |
|
Looks like the LBC is behind and someone else is solving these puzzles.
Looks like you've been sleeping for too long Stupid rico is still continuing on depleted 55-range. Of course not a single sane person will participate in LBC now. I am pretty sure some large farm of >1000 GPUs or so switched from mining ETH to solving these private keys, already 2 of them were solved beating LBC pretty much.
Maybe a Martian Colony of Sectoids awoke to snatch our precious 32BTC! p.s. If you do a simple math, you'll see that the guy who took puzzle 55 and puzzle 56 only have speed of 1000 megakeys. That's less than LBC had several months ago. If rico wasn't such a worthless project manager, this could be LBC. Now sane people use BitCrack, it makes 1000megakeys with only two gtx1080ti. I might have to look into the BitCrack program, if a pair of 1080ti can reach 1GH/s then it might still be worth scanning for. The LBC is still scanning the 55th range probably because they can't skip over and jump ahead, they probably didn't realise that someone would be doing this solo. It would be best if there was some pool for searching for these keys, then it would be fair for everybody. But now it seems risky because you start scanning, find nothing after a month or so and then someone already found the key right ahead of you. No idea how many people now are scanning for these keys instead of mining ETH.
|
|
|
|
lrossy
Newbie
Offline
Activity: 2
Merit: 0
|
|
November 07, 2018, 05:57:22 PM |
|
Hi guys, I'm trying to understand the procedure to solving this puzzle, starting from the easy ones that were solved already. Im clearly a little over my head here, and I cannot figure out what I am doing wrong. I have this code so far, which gives me the same results as this website ( https://brainwalletx.github.io/#generator), but they never match what I see here. So clearly i'm doing something wrong let bitcoin = require('bitcoinjs-lib');
let pvk_seed = '000000000000000000000000000000000011'; let hash = bitcoin.crypto.sha256(pvk_seed); const keyPair = bitcoin.ECPair.fromPrivateKey(hash); const { address } = bitcoin.payments.p2pkh({ pubkey: keyPair.publicKey });
let wif = keyPair.toWIF();
console.log(`${pvk_seed} = Address ${address}, WIF : ${wif}`)
That code is using 3 to generate PVK, ive tried 3 in Decimal, Binary, Hex, nothing works. Thanks for any tips
|
|
|
|
arulbero
Legendary
Offline
Activity: 1915
Merit: 2074
|
|
November 07, 2018, 06:33:45 PM |
|
Hi guys, I'm trying to understand the procedure to solving this puzzle, starting from the easy ones that were solved already. Im clearly a little over my head here, and I cannot figure out what I am doing wrong. I have this code so far, which gives me the same results as this website ( https://brainwalletx.github.io/#generator), but they never match what I see here. So clearly i'm doing something wrong let bitcoin = require('bitcoinjs-lib');
let pvk_seed = '000000000000000000000000000000000011'; let hash = bitcoin.crypto.sha256(pvk_seed); const keyPair = bitcoin.ECPair.fromPrivateKey(hash); const { address } = bitcoin.payments.p2pkh({ pubkey: keyPair.publicKey });
let wif = keyPair.toWIF();
console.log(`${pvk_seed} = Address ${address}, WIF : ${wif}`)
That code is using 3 to generate PVK, ive tried 3 in Decimal, Binary, Hex, nothing works. Thanks for any tips If you put 0000000000000000000000000000000000000000000000000000000000000003 here -> https://www.bitaddress.org/bitaddress.org-v3.3.0-SHA256-dec17c07685e1870960903d8f58090475b25af946fe95a734f88408cef4aa194.html(wallet details) you'll get 2 addresses: Bitcoin Address (Uncompressed) 1NZUP3JAc9JkmbvmoTv7nVgZGtyJjirKV1 Bitcoin Address Compressed 1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb In this case the address in puzzle transaction is the last one. Look here : https://github.com/bitcoinjs/bitcoinjs-lib/issues/155
|
|
|
|
BurtW
Legendary
Offline
Activity: 2646
Merit: 1136
All paid signature campaigns should be banned.
|
|
November 07, 2018, 07:07:12 PM |
|
Hi guys, I'm trying to understand the procedure to solving this puzzle, starting from the easy ones that were solved already. Im clearly a little over my head here, and I cannot figure out what I am doing wrong. I have this code so far, which gives me the same results as this website ( https://brainwalletx.github.io/#generator), but they never match what I see here. So clearly i'm doing something wrong let bitcoin = require('bitcoinjs-lib');
let pvk_seed = '000000000000000000000000000000000011'; let hash = bitcoin.crypto.sha256(pvk_seed); const keyPair = bitcoin.ECPair.fromPrivateKey(hash); const { address } = bitcoin.payments.p2pkh({ pubkey: keyPair.publicKey });
let wif = keyPair.toWIF();
console.log(`${pvk_seed} = Address ${address}, WIF : ${wif}`)
That code is using 3 to generate PVK, ive tried 3 in Decimal, Binary, Hex, nothing works. Thanks for any tips If you put 0000000000000000000000000000000000000000000000000000000000000003 here -> https://www.bitaddress.org/bitaddress.org-v3.3.0-SHA256-dec17c07685e1870960903d8f58090475b25af946fe95a734f88408cef4aa194.html(wallet details) you'll get 2 addresses: Bitcoin Address (Uncompressed) 1NZUP3JAc9JkmbvmoTv7nVgZGtyJjirKV1 Bitcoin Address Compressed 1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb In this case the address in puzzle transaction is the last one. Look here : https://github.com/bitcoinjs/bitcoinjs-lib/issues/155I did a quick write up on compressed versus uncompressed Bitcoin addresses here: https://bitcointalk.org/index.php?topic=1573035.msg45500313#msg45500313in case you want more technical details.
|
Our family was terrorized by Homeland Security. Read all about it here: http://www.jmwagner.com/ and http://www.burtw.com/ Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!
|
|
|
lrossy
Newbie
Offline
Activity: 2
Merit: 0
|
|
November 07, 2018, 07:25:34 PM |
|
Thank you arulbero and BurtW!!
Now i can get to the good stuff. Really appreciate it!
|
|
|
|
|
BurtW
Legendary
Offline
Activity: 2646
Merit: 1136
All paid signature campaigns should be banned.
|
|
November 07, 2018, 08:19:15 PM |
|
My answer was a general answer to a general question about the two forms of Bitcoin address and was not meant to be a technical paper or exact description of how to create Bitcoin addresses (that is what the wiki is for).
So, I started out with "Leaving out some small details:".
Thanks for filling in a few of the technical details.
BTW, even with the additional details, the description is still incomplete because you left out the checksum in the hashing description.
Anyone who wants to know all the nitty gritty details can look it up.
|
Our family was terrorized by Homeland Security. Read all about it here: http://www.jmwagner.com/ and http://www.burtw.com/ Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!
|
|
|
|