Show Posts
|
Pages: [1]
|
free 8$ btc
addrss compress P2PSH : 34gbwEu34QBz7bozafSyStRqw9f6Jw7cTt
privkey hex : 2832ed74f2b5e35ee privkey wif : KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZfFoWMiwBt943V7CQeX
|
|
|
keyhunt not good
you should learn to configure it, defaul thread subrange is 32 bits, for small ranges with multiple threads you should lower the N value with "-n number" if the range is less than 1 Million keys you should use -n 0x10000 where 0x10000 its a 16 bits subrange per thread Look  Found i less than a second are you key hunt creator? nice too meet you It was my mistake, I apologize, dear friend ok , thank you I meant https://github.com/WanderingPhilosopher/KeyHuntCudaClienti use of first version keyhunt cuda Zero clues of how or what you ran, but with a single CPU core, using keyhunt-cuda, it is found pretty much as the program starts: KeyHunt-Cuda v1.08
COMP MODE : COMPRESSED COIN TYPE : BITCOIN SEARCH MODE : Single Address DEVICE : CPU CPU THREAD : 1 SSE : YES RKEY : 0 Mkeys MAX FOUND : 65536 BTC ADDRESS : 1E5V4LbVrTbFrfj7VN876DamzkaNiGAvFo OUTPUT FILE : Found.txt
Start Time : Sat May 25 10:11:49 2024 Global start : 20000000000000000 (66 bit) Global end : 21000000000000000 (66 bit) Global range : 1000000000000000 (61 bit)
[00:00:02] [CPU+GPU: 5.28 Mk/s] [GPU: 0.00 Mk/s] [C: 0.000000 %] [R: 0] [T: 10,706,944 (24 bit)] [F: 1]
BYE
SO maybe you entered wrong things/flags when trying to run the program. this is 1 milion key after start 200000000000f4240 and this is p2ph compressed public key : 1E5V4LbVrTbFrfj7VN876DamzkaNiGAvFo KeyHunt-Cuda.exe -t 0 -g --gpui 0 --gpux 24,256 -m address --coin BTC --range 20000000000000000:40000000000000000 1E5V4LbVrTbFrfj7VN876DamzkaNiGAvFo my keyhunt cuda speed is 60 Mk/s . In fact, Cuda should find it in less than 1 second, but it takes about 1:40 minute KeyHunt-Cuda.exe -t 0 -g --gpui 0 --gpux 24,256 -m address --coin BTC --range 20000000000000000:40000000000000000 1E5V4LbVrTbFrfj7VN876DamzkaNiGAvFo KeyHunt-Cuda v1.07 COMP MODE : COMPRESSED COIN TYPE : BITCOIN SEARCH MODE : Single Address DEVICE : GPU CPU THREAD : 0 GPU IDS : 0 GPU GRIDSIZE : 24x256 SSE : YES RKEY : 0 Mkeys MAX FOUND : 65536 BTC ADDRESS : 1E5V4LbVrTbFrfj7VN876DamzkaNiGAvFo OUTPUT FILE : Found.txt Start Time : Sun May 26 14:04:50 2024 Global start : 20000000000000000 (66 bit) Global end : 40000000000000000 (67 bit) Global range : 20000000000000000 (66 bit) GPU : GPU #0 Quadro P1000 (4x128 cores) Grid(24x256) [00:01:36] [CPU+GPU: 62.21 Mk/s] [GPU: 62.21 Mk/s] [C: 0.000000 %] [R: 0] [T: 5,976,883,200 (33 bit)] [F: 0] The problem is that Cuda performs the same task in all graphics cores in parallel and repetitively, and instead of the cores each helping the program, each of them repeats the same task as an island, for example, each core in parallel from the same start. He does and moves forward. And maybe in the best case, it divides the entire collection into the number of cores and each core starts working in that interval, in this case, once you exit the program, all your efforts will end in an unknown place, which will be used the next time you run the program. You don't know where to start Of course, it seems that you are using a higher version of this software v1.08. But anyway, I have to tell the truth. You have written a very good software. I really enjoyed your code ideas.
|
|
|
Of course it is, and you are the main programmer and developer nice to meet you I have also written various programs for this work, soon I will post a complete version on GitHub Do you know how to search randomly in KeyhuntCuda version 2 ? Version 1 with -r key and the value of key space was sequential with random Like KeyHunt-Cuda.exe -t 0 -g --gpui 0 --gpux 24,256 -m address --coin BTC --range 20000000000000000:40000000000000000 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so -r 2000
|
|
|
keyhunt not good
you should learn to configure it, defaul thread subrange is 32 bits, for small ranges with multiple threads you should lower the N value with "-n number" if the range is less than 1 Million keys you should use -n 0x10000 where 0x10000 its a 16 bits subrange per thread Look  Found i less than a second are you key hunt creator? nice too meet you It was my mistake, I apologize, dear friend ok , thank you I meant https://github.com/WanderingPhilosopher/KeyHuntCudaClienti use of first version keyhunt cuda
|
|
|
i have a simple alghoritm with cpu in python , you can test it import bitcoin import ecdsa
def private_key_to_public_key(private_key): sk = ecdsa.SigningKey.from_string(bytes.fromhex(private_key), curve=ecdsa.SECP256k1) vk = sk.get_verifying_key() compressed_public_key = vk.to_string("compressed").hex() return compressed_public_key
bitcoin_address = bitcoin.pubtoaddr(public_key)
Why use Bitcoin and ECDSA imports? They're so slow, it feels like a waste of time. Instead, utilize ICE (import secp256k1 as ice) for this function and the Bitcoin address line: def private_key_to_public_key(private_key): priv_int = int(private_key, 16) return ice.scalar_multiplication(priv_int) and bitcoin_address = ice.pubkey_to_address(0, True, public_key) It's approximately 10 times faster than ECDSA. But even that is miserable if you attack the dinosaur numbers. The more you delve into Python, the more apparent it becomes that searching for Puzzle 66 through it is pointless. Perhaps someone knowingly obscures things by selling Python scripts as the ultimate solution.  In fact, what I meant by this comment was the problem of Key Hunt and I wrote different versions of these programs. And compared to Key Hunt, I put this code so that you can test and see the speed of a simple program with a CPUs and a powerful program with GPU that does not work properly. In general, I am very happy that you paid attention to this code and took your time, and I thank you. And also ice's idea was very good 
|
|
|
keyhunt not good i get a test with key hunt you can test it. key hunt speed in my pc is 60 Mk/s i search for this addrrss 1E5V4LbVrTbFrfj7VN876DamzkaNiGAvFo this privatekey is 200000000000f4240 search start of 20000000000000000 this private key is 1,000,000 key of start in fact keyhunt must be find in 2 sec , but fin at 100 Sec In the Cuda programming, an algorithm is executed in parallel, and in fact, this search operation in each core performs a similar task in parallel, and the cores do not help each other's process to increase the search speed, but each of them works as an island exactly the same way. They do something else and that's why this program has problems i have a simple alghoritm with cpu in python , you can test it # -*- coding: utf-8 -*-
""" Created on Thu Mar 14 17:50:54 2024
1000 Bitcoin Puzzle Scanner for 2^66 ~ 2^67
@author: Amin Solhi , Contacts => email: amin.solhi@gmail.com , +9891111842779 """ import bitcoin import ecdsa import secrets from timeit import default_timer as timer import datetime
global target_address global output_file global rng global private_key global ks global start global random_mode
target_address = "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so" output_file = "data.txt" rng = 20 # int(input("Enter Random Space Number 1 ~ 100 :")) private_key="20000000000000000" ks=0 start = timer() random_mod = True # True or False
print ("\nBTCGEN Bitcoin Puzzle #66 Scanner \n") print ("BTC Address : ",target_address) print ("OutPut File : ",output_file) print ("Randome Mod : ",f"{str(random_mod)}") if (random_mod): print ("Random Key : ",f'per {rng}K key') print ("Device : CPU") print ("Global Start: ",private_key) print ("Global END : 40000000000000000")
print('\n')
def remove_zeros(input_string): result = "" zero = "" for char in input_string: if char != "0": zero = "finish" if zero =="finish" : result += char return result
t="" t +="0"*47
def h(a): #a = a[:1] + '0' + a[1:] if (len(a) < 64): #a = a[:1] + '0' + a[1:] a = '0' + a[:] if (len(a) < 64): a = h(a) return a
def generate_random_priv(): p=str(secrets.choice(range(2, 4))) return (p+secrets.token_hex(8))
def generate_private_key(num_hex): num_decimal = int(num_hex, 16) num_decimal += 1 num_hex = h(str(f'{num_decimal:x}')) return (num_hex)
def private_key_to_public_key(private_key): sk = ecdsa.SigningKey.from_string(bytes.fromhex(private_key), curve=ecdsa.SECP256k1) vk = sk.get_verifying_key() compressed_public_key = vk.to_string("compressed").hex() return compressed_public_key
def onmain(): #start = timer() global private_key global rng global ks global start global target_address global i for _ in range(rng*1000):#while True: ks+=1 private_key = generate_private_key(private_key)#secrets.randbelow(32)) # Generate a random private key public_key = private_key_to_public_key(private_key) # Convert private key to compressed public key #print (private_key) #print(i,"--- ",timer()-start,"seconds ---" ,) # Generate Bitcoin address from public key bitcoin_address = bitcoin.pubtoaddr(public_key) if (bitcoin_address == target_address): f = open(output_file, "a") f.write('\nprivate key int: ' + private_key +'\nBitcoin address: ' + bitcoin_address+'\n_________\n') f.close() print(f"\nFound matching Bitcoin address for private key: {private_key}") input("") print(f"\r[Total : {(ks/1000000)} Mk/{int(timer()-start)}s] [Private key hex: {(remove_zeros(private_key))}] ", end="")
def onmain_random(): #start = timer() global private_key global rng global ks global start global target_address global i global random_str for _ in range(rng*1000):#while True: ks+=1 private_key = generate_private_key(private_key)#secrets.randbelow(32)) # Generate a random private key public_key = private_key_to_public_key(private_key) # Convert private key to compressed public key #print (private_key) #print(i,"--- ",timer()-start,"seconds ---" ,) # Generate Bitcoin address from public key bitcoin_address = bitcoin.pubtoaddr(public_key) if (bitcoin_address == target_address): f = open(output_file, "a") f.write('\nprivate key int: ' + private_key +'\nBitcoin address: ' + bitcoin_address+'\n_________\n') f.close() print(f"\nFound matching Bitcoin address for private key: {private_key}") input("") print(f"\r[{str(datetime.timedelta(seconds=int(timer()-start)))}] [Total : {(ks/1000000)} Mk] [R: {i}] [Private key hex: {(remove_zeros(private_key))}] ", end="")
def main(): global private_key global i global random_str random_str ="" i=0 if (random_mod): while True: i+=1 private_key=generate_random_priv() onmain_random() else: while True: i+=1 onmain()
if __name__ == "__main__": main()
|
|
|
My question is: How can you tell from the public key that the public key is multi-bit? How can we determine the bit length of a public key
for example : 19zK7RvpE7UE7nDaRtLcTAw98tyyYr8Pz7
|
|
|
Hi All, I've been reading this and other related topics for a few weeks and I'm trying to understand the complexity of the different algorithms and approaches. For Pollard's Kangaroo algorithm, someone gave an estimate of 2^66.05 operations needed for solving #130 ( https://bitcointalk.org/index.php?topic=5244940.2740 not sure how they reached this conclusion, but from reading about the algorithm it has a similar complexity to BSGS). This equals ~7.6x1019 operations needed. For BSGS, from reading about it, is has a complexity of O(n 1/2). Taking the square root of the range (2 130 - 2 129 - 1) equals ~ 2.6x1019 operations needed. Is this correct? Is BSGS more efficient, or is it just a wrong way of calculating the efficiency? I think I'm missing something in estimating the total "operations needed" for these algorithms. Is there a better/correct way of doing this? Considering HW limitations, I am still not familiar with how different Kangaroo implementations work, I see that the JeanLucPons' based on VanitySearch requires GPU power (I assume more cores means better, but not sure what is the requirement for VRAM, CPU, RAM). For BSGS, seems like keyhunt is the best, and works better with more RAM and more CPU cores. Am I missing anything? Are there any other/better tools available for searching the ranges with known public keys? In fact, for block 66, a private key must be generated with the 81129638414569788207641586040831 number, and this process is slow with a CPU, but using a graphics processor speeds it up. I have written a simple Python program below that randomly generates private key within the range of block 66, and compares its public key at every moment with the public key of block number 66. If found, it displays a message and saves it in the data.txt file in the root of the program. import bitcoin import ecdsa import secrets from timeit import default_timer as timer start = timer() t="" for _ in range(47):# generate 47 character ( 0 ) t = t + "0" def generate_private_key(): p=str(secrets.choice(range(2, 4)))# generate random( 2 or 3) for fist number return (t+p+secrets.token_hex(8))#return { 47 character ( 0 ) + random (2 or 3 ) + generate random (16 character in hexadecimal) }
def private_key_to_public_key(private_key): sk = ecdsa.SigningKey.from_string(bytes.fromhex(private_key), curve=ecdsa.SECP256k1) vk = sk.get_verifying_key() compressed_public_key = vk.to_string("compressed").hex() return compressed_public_key
def main(): target_address = "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so" # this is Target wallet address , just p2pkh compressed while True:# #for _ in range(100000):#while True: private_key = generate_private_key()#secrets.randbelow(32)) # Generate a random private key public_key = private_key_to_public_key(private_key) # Convert private key to compressed public key # Generate Bitcoin address from public key bitcoin_address = bitcoin.pubtoaddr(public_key) print(private_key,' ',bitcoin_address) if (bitcoin_address == target_address): f = open("data.txt", "a") f.write('\nprivate key int: ' + private_key +'\nBitcoin address: ' + bitcoin_address+'\n_________\n') f.close() print(f"Found matching Bitcoin address for private key: {private_key}") break print("--- %s seconds ---" ,timer()-start)
if __name__ == "__main__": main()
|
|
|
I still don't understand, by whom it will be broken, I also wrote software and I am searching randomly.
What should we do to transfer, add the private key in online wallets and go to the transfer form, use phone apps, which wallet software should we use to withdraw the balance, or write a program for the Bitcoin transfer form or Should we add its in Bitcoin core???
If you initiate a transaction you will reveal the public key. Since it belongs to a 66-bit public key, you can quickly find the private address with kangaroo or bsgs because you know the range and it is "narrow". Then you will start your transaction and generate a double expense. To win you will "bid" a higher transaction fee to get more confirmations faster. If there is any mistake in this, someone correct me. I am grateful to the members of this group and I thank them very much for providing good information and guidance. I ask many questions more for a deep understanding of the issue rather than seeking a solution myself; of course, my intention is not self-praise, I just can't be a bystander to the problems. As far as I understand, you mean that powerful individuals and teams like mining pools or similar ones have very high processing power and can accomplish our efforts to search for block 66, which may take several years, in less than a day? And while we have requested the transfer of Bitcoin, they may also register it with a higher fee, and more people repeat this process, creating something like an auction? Are they still unaware of the existence of the Bitcoin puzzle so they can think about it sooner?
|
|
|
I still don't understand, by whom it will be broken, I also wrote software and I am searching randomly.
What should we do to transfer, add the private key in online wallets and go to the transfer form, use phone apps, which wallet software should we use to withdraw the balance, or write a program for the Bitcoin transfer form or Should we add its in Bitcoin core???
|
|
|
quote] Hundreds of years. Once you crack it you'll want to spend it, so you'll have to disclose the public key over the network. Once everyone sees the public key and knows it has a 66-bit private key, they will crack the private key in a few seconds at most and double spend, making your efforts completely futile. So I guess the reward for ruining a lots of hardware and paying for the power goes straight to zero, since everyone will fight to double spend, making the fee higher and higher and the net profit going to zero. [/quote]
So all the bitcoins that are from 66 bits to 160 bits have no financial value?
|
|
|
I have already built a system that generates a wif and public key by entering a hexadecimal value. Of course, it has many features and does many things
I don't know what you know about crypto math and calculations, but I'll give you an example: the two public addresses below are both the same and correspond to a common hexadecimal private key, they just have a different generation process.
P2PKH Address : 1HT7xU2Ngenf7D4yocz2SAcnNLW7rK8d4E
P2PSH Address : 3J98t1WpEZ73CNmQviecrnyiWrnqRhWNLy
Probably, by giving the hexadecimal address of a special wallet, you will not be able to open that wallet. Because we have talked about this process on this page and you did not know it
|
|
|
look you forget a 0 before 2
correct value is =>> Public key hex : 0400000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000
and this is =>> Public key ( wallet address p2pkh) : 16QaFeudRUt8NYy2yzjm3BMvG4xBbAsBFM
so , max private key hex : fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffe in clearly : 2^256 -1 or 16^64 -1
if calculate fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffe public key hex , result is : 04e0a65740c1c5148111a8305938c56322fbf345a18cb90cf48a788961c6b3ff918aaf0580fd9e8 be571e97bef4a067fa7911da0ff0e1a5f307fea63e3ecd903af and publick key : 1JHzdjEL4xQ7yifAaQQKAfqPyAN92vfstP
and another
public key hex : ( 0 )
public key ( wallet address p2pkh ) : 1FYMZEHnszCHKTBdFZ2DLrUuk3dGwYKQxh
SHA256 6e340b9cffb37a989ca544e6bb780a2c78901d3fb33738768511a30617afa01d RipeMD-160 9f7fd096d37ed2c0e3f7f0cfc924beef4ffceb68
Add Version Byte 009f7fd096d37ed2c0e3f7f0cfc924beef4ffceb68 Double
SHA256 7524061a0ca9cca8e0c2881fd447c20c821751e357dc7d960cd03cfc3b4cece0
Add 4 Byte Checksum 009f7fd096d37ed2c0e3f7f0cfc924beef4ffceb687524061a Address 1FYMZEHnszCHKTBdFZ2DLrUuk3dGwYKQxh
|
|
|
orginaly this value: var P = new bigInt('fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f', 16); //2²⁵⁶ - 2³² - 2⁹ - 2⁸ - 2⁷ - 2⁶ - 2⁴ - 1 var N = new bigInt('ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff', 16); var ZERO = new bigInt(0); var SEVEN = new bigInt(7); var x = new bigInt('79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798', 16); var y = new bigInt('483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8', 16); var ONE = new bigInt(1); var TWO = new bigInt(2); var THREE = new bigInt(3); var Pp1d4 = new bigInt('3fffffffffffffffffffffffffffffffffffffffffffffffffffffffbfffff0c', 16); var Nd2p1 = new bigInt('7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1', 16);
but i changed to this var P = new bigInt('ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff', 16); //2²⁵⁶ - 2³² - 2⁹ - 2⁸ - 2⁷ - 2⁶ - 2⁴ - 1 var N = new bigInt('ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff', 16); var ZERO = new bigInt(0); var SEVEN = new bigInt(7); var x = new bigInt('ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff', 16); var y = new bigInt('ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff', 16); var ONE = new bigInt(1); var TWO = new bigInt(2); var THREE = new bigInt(3); var Pp1d4 = new bigInt('3fffffffffffffffffffffffffffffffffffffffffffffffffffffffbfffff0c', 16); var Nd2p1 = new bigInt('ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff', 16);
and now , result : i generate by any (private key hex) this wallet address :
Public Key Hex : 4000000000000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000
address: 16QaFeudRUt8NYy2yzjm3BMvG4xBbAsBFM
balance : 0.01075050 BTC
|
|
|
|