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Based on mcdouglas' post-claim comments, I can feel his frustration when he encountered EC's punch in face! He's been tricked and overpowered by the Illusion-of-Mastery, which made him post such a big claim. I am not demotivating him, no no!! I am happy about him that he is learning good maths here. Trying to divide like 58.5 and figuring what half would be, good good, great!!!! Majority of guys here possibly have been through lots of similar curve punches here. Someone rightly figured out here that EC cannot be broken from inside, whatsoever!!!! The only thing attached to it from outside is its point at infinity, WHICH CAN'T BE TOUCHED DUE TO BEING AT INFINITY SOMEHOW I GUESS, OR MAY BE IDK!! Whatever secret mcdouglasx has, I am sure many of guys here have already tried & tested and accordingly slapped in ass by EC! SORRY FOR THAT!! The possibilities of EC being broken by any BTC puzzle enthusiast are very slim! Because there are giant research & security institutes (I know few of them personally visited as well) which are rigorously and systematically approaching towards the solutions to ECDLP. I bet Satoshi must have a good laugh today, Cheers, it is 1st of Feb 24!!!
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I know exactly what range puzzle number 130 is in. It will reduce the key to be scanned by 50%. I will sell it to anyone who wants to buy it. proof is available. If anyone wants to know the range of PUZZLE 130 UN, I can tell you for money.
No offense, but it has already been identified couple of times. One can reduce the any given point by 50% by subtracting the starting bit point. If we are in the range of 2^129 and 2^130, if we subtract the public key point of 2^129 from the puzzle pubkey, the result is 50% down from the original pubkey. One can then start working on cracking it, and after cracking it, he can easily reach the original pubkey by simply adding 2^129 to it.
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You can find out the range from the public key.
This is serious stuff, you probably didn't realize what you said... HOW ON EARTH YOU FIGURE THE RANGE FROM PUBLIC KEY?
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Great NEWS!!!!! I found my rouge BTC. A simple list of collisions made things quite manageable, which was exactly I was looking for because BTC was placed on a strategic location. I thank all of you guys for your suggestions as well as criticism. All I say here is that, things won't always be the same! Elliptic Curve encryption system will soon be redundant! Enjoy your puzzle solving!
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If someone has any doubt or suspicions about what I am saying, they can simply leave, no big deal. I am not here to somehow skim or exploit anyone's programming capabilities, don't judge me. I will explicitly appreciate anyone's genuine help & reward accordingly. Those who think I was dumb making 1 billion keys are also wrong. 1 billion is what is left in USB Drive, however originally they were 1trillion. The reason I am not exposing my scheme is that I am still working on this thing by putting another BTC and moving forward with my research. I don't have any GPU neither want one & I am also not interested in 125 BTC puzzle rest assured. I am simply a victim of untoward incident like a flood, a tornado, or lightening or house on fire or robbery etc we have little to no control over it. you can call it dumb or whatever you like.
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Thanks for helping, it worked. Here comes the interesting part, that this Kangaroo algorithm takes only 1 public key. which is like finding 1 drop in an ocean as someone quoted. What about finding 1 billion drops instead of 1? Thats my point, when I say that 1 billion public keys increase the probability by 1 billion times. Hence, I am convinced if this Kangaroo algorithm takes 1 billion public keys instead of 1, the chances of finding one would increase exponentially. What do you say?
yes, the program can be modified to accept 1 billion keys and the chances do increase by the number of keys but also the speed drops proportionally with the number of keys. it's a tradeoff speed or keys Obviously with 1 billion key input its going to get super slow, but with manageable input it should be super fast. Can you elaborate further how can kangaroo algorithm take multiple pubkeys
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Thanks for helping, it worked. Here comes the interesting part, that this Kangaroo algorithm takes only 1 public key. which is like finding 1 drop in an ocean as someone quoted. What about finding 1 billion drops instead of 1? Thats my point, when I say that 1 billion public keys increase the probability by 1 billion times. Hence, I am convinced if this Kangaroo algorithm takes 1 billion public keys instead of 1, the chances of finding one would increase exponentially. What do you say?
I am pleased it worked and you have recovered your BTC. Can I claim the bounty? My BTC address 1NcFJzayiHy4iPFz4QR9BqP8XHuSDCv5qx The chances may increase exponentially with more addresses but of course it slows down with a billion keys to check. The keyhunt program is much faster. Plus a billion addresses may seem like a lot but it is a tiny drop in the ocean. Well done again on recovering your btc. By "it worked" I meant that the script worked. It says I have to wait 1.7 million year before it actually finds the key.
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I was wondering, have you guys ever seen a wallet to generate a private key consisting of only decimal numbers without a single hex character? Well I haven't, and it is the case with only hex chars as well, so is there any way (possible way) to skip brute forcing all hex, all decimal keys and just search through the mixed keys? And what would happen to our search range if we could eliminate such keys, as in how many keys could we skip brute forcing? Isn't that an idea worth exploring?
In my case (keyhunt) and I think that it is a general case, hexadecimal is only used as output only for "friendly" representation. Internally the program works with decimal numbers a.k.a. "Big number" variable like mpz_t (from libgmp) or Int for (libsecp256k1 library). So it should be the same performance. Hi Alberto! I hope you are doing great! I posted my 1 BTC lost case at this forum here https://bitcointalk.org/index.php?topic=5448332.msg62070769#msg62070769I visited your keyhunt options for my issue but since I am not a fulltime programmer I am confused which one is best suited to my identical situation. I'll gladly share with you 0.1 BTC on recovery of my 1 BTC. I am in possession of 1 billion public keys. Those public keys have 1 billion private keys which are a sequence like starting private key for instance is 503, 504, 505, 506,..... upto 1 billion added. I know the start and end range like that in BTC puzzle but its' not perfect like zeros and fffs, I'll not be sharing specifics for obvious reasons. Plus I know the public key holding the funds. I am convinced that if Kangaroo algorithm is modified to take this 1 billion public keys in some way and find only one private key. Rest is a matter of adding or subtracting around it within 1 billion one can easily reach the private key holding funds.
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just download the code to a directory(folder) change to that directory don't alter any code. It is all run from the command line.
Open a terminal / powershell window and type in the command line
python3 kangaroo.py 00008:0000ff 02049370a4b5f43412ea25f514e8ecdad05266115e4a7ecb1387231808f8b45963 cut & paste the above to get the idea.
python 3 calls the program kangaroo.py
00008 is the start of the range in hexadecimal. Here you put in your start range of the first key you generated yours might be much bigger 1234567abcde or whatever.
: is a separator
000ff is the end of the range (1000 decimal is a different number to 1000 hex again put in your end range
then put in the public key of the first key you generated and wait.
If successful, a file containing the key will be generated as well as being shown on the screen
I have tested this program and it works. good luck.
Thanks for helping, it worked. Here comes the interesting part, that this Kangaroo algorithm takes only 1 public key. which is like finding 1 drop in an ocean as someone quoted. What about finding 1 billion drops instead of 1? Thats my point, when I say that 1 billion public keys increase the probability by 1 billion times. Hence, I am convinced if this Kangaroo algorithm takes 1 billion public keys instead of 1, the chances of finding one would increase exponentially. What do you say?
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still dont know what the author wants..
@Unplugged Taste tell us, do you know the public key of the address containing 1 BTC or not ??
from your posts i believe you dont know which public key is the correct one....
i posted on the first page a script that converts all the public keys you have into addresses and compares them to the address containing your money. it will return only the address and it's public key
IF you already know the public key of the address why would you need to check all the public keys at once??
Perhaps you misunderstood what I said. Let me clear you that I know the public key & address carrying funds. The reason of me stressing on 1 billion public keys around it and trying to find any one private key of any of them is that it is increasing chances by 1 billion times. Why? Because they are in a sequence like 1,2,3,4,... etc. Its like there are 1 billion private keys carrying that 1 BTC because any one could easily lead to the principle private key easily.
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Would someone please see the script that I think is best suited for my needs. The script is generated by chatGPT not me. So if you suggest any changes in it please be specific like which should be replaced with which line of code. Thanks. import multiprocessing import hashlib import binascii from Crypto.Util.number import long_to_bytes, bytes_to_long
start_hex = '0000000000000000000000000000000000000000000000000000000000000000' end_hex = '0000000000000000000000000000000000000000000000000000000000000000' start = int(start_hex, 16) end = int(end_hex, 16) num_parts = 4
def kangaroo(start, end, pubkeys): G = 2 hash_size = 256 max_iters = 2**32
def H(x): if isinstance(x, bytes): return hashlib.sha256(x).digest() else: return hashlib.sha256(x.to_bytes((x.bit_length() + 7) // 8, byteorder='big')).digest()
def kangaroo_algorithm(x1, x2, step): x = x1 xs = set() for i in range(max_iters): x = (x + step) % end h = bytes_to_long(H(long_to_bytes(x))) if h in pubkeys: return x, pubkeys[h] xs.add(x) if len(xs) == hash_size: xs.clear() y = x2 for j in range(i): y = (y + step) % end if y in xs: return x, None return None, None
step = end // hash_size x1 = start x2 = start + step private_key, public_key = kangaroo_algorithm(x1, x2, step) if private_key is not None: private_hex = hex(private_key)[2:] private_hex = '0' * (64 - len(private_hex)) + private_hex return (private_key, private_hex, public_key) else: return None
if __name__ == '__main__': pubkeys_file = 'pubkeys.txt' foundkeys_file = 'foundkeys.txt'
with open(pubkeys_file, 'r') as f: pubkeys = {int(line.strip(), 16) for line in f}
pool = multiprocessing.Pool(processes=num_parts)
start_ranges = [start + i * ((end - start) // num_parts) for i in range(num_parts)] end_ranges = [start + (i + 1) * ((end - start) // num_parts) for i in range(num_parts)] args = [(start_ranges[i], end_ranges[i], pubkeys) for i in range(num_parts)]
results = pool.starmap(kangaroo, args) foundkeys = [result for result in results if result is not None]
with open(foundkeys_file, 'w') as f: for result in foundkeys: f.write(f"{result[0]}\n") print(f"Found private key: {result[1]} for public key: {result[2]}")
print("Done.")
I ran the script its just blinking since there are no print statements added along... I don't know whether it is working correctly or not.
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First of all I generated a random private key from within a range, like that of BITCOIN Puzzle range. I generated its public key and I than started adding 1 million in the private key and generated its public key. Again I added 1 million in the private key and generated its public key. I continued this for 1k times. So basically from first private key and last private there is 1 billion private keys. In case you didn't get my point: Suppose I randomly chosen private key as 980 I added 1 million in private key = 1,000,980 Again added 1 million = 2,000,980, ... ... ... Up till I added 1 billion in my first private key.
From start to end of private keys, I generated all public keys! YES, I still have 1 billion public keys of those 1 billion private keys. Ironically I lost all my private keys. Not a single public key is released since no transaction has ever happened.
What was the purpose to iterate 1k times to add 1m to the initial and then intermediate random key of this iteration? To me it makes no sense when you used also all intermediate private keys anyway (or I can't follow your procedure). If I understand you correctly you have a starting private key from some range and you stepped over a consecutive range of 1 billion (1k times 1m) private keys. All private keys are only one unit apart from each other. Then you generated from all those 1 billion private keys the respective public keys. It's a fair assumption that you only initially stored the starting private key and not all the intermediate ones. You did that by some script or program and kept only the output of the public keys. You had no or poor backup and somehow lost your program or script and with it your initial private key. Well, shit happens. Out of curiosity, when did you do that and how much was 1 BTC worth at that time? I placed 1 BTC in any one of those 1 billion private keys that I am looking for.
Just to be clear: there's 1 BTC controlled by a single private key from your range of 1 billion consecutive private keys located somewhere in the private key space of Bitcoin? Still, your 1k steps don't make any sense to me in the context of your other details. Feel free to correct me where I'm wrong. So far you are the only one who has perfectly understood what I am saying! You are 100% correct, there's 1 BTC controlled by a single private key within the range. It was around 2 months ago.
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1. Adding numbers to private keys ? Is that even possible/safe ? How would that generate a public key ? Hmm RSA, eclipctive curve ? hmmm..
2. 1 BTC each in one of these billion addresses ? Euh are you aware bitcoin only has 21 million coins ?! LOL.
This smells and sounds fishy to me ! =D
Perhaps you misunderstood what I said. There is only 1 BTC in only 1 address. That 1 address that I already know & its public key has 1 billion public keys around it. you can turn any number to become your private key or any sequence of number some people are asking whether it is possible or not!
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In simplest form... Within given range start and end bit,,,, There are 1 billion private keys in sequence. and We have their 1 billion public keys. So the reason I am stressing on it is that because it like finding 1 billion private keys in given range instead of finding 1, don't you think it got to be easy? The range is similar to that of bitcoin puzzle but it is not straight forward zeros and fffs, it is more than 2^122 i guess!
2^122 ranges are extremely hard to crack, even if you have the public keys and can use Kangaroo, because it took an insanely long time (and who knows how much hardware) to break puzzle #120 which has a similar range. What is the algorithm you used to create each subsequent private key from the first? In puzzle#120 you are looking for 1 private key and in mine it is 1 billion.
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In simplest form... Within given range start and end bit,,,, There are 1 billion private keys in sequence. and We have their 1 billion public keys. So the reason I am stressing on it is that because it like finding 1 billion private keys in given range instead of finding 1, don't you think it got to be easy? The range is similar to that of bitcoin puzzle but it is not straight forward zeros and fffs, it is more than 2^122 i guess!
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I highly appreciate all of you answers and I'll try to answer your questions. With regard to checking the balance in the address of public key, I already have it. The reason I stressed on 1 billion public keys is the fact that they are in sequence of private keys like private key 503,504,505,506,507,..... I was only saving private key after 1 million. 1 billion Public keys that I have they are not 1 million apart they are the sequence of 1 billion private keys. I know the position of the one in which the BTC have but I am still keeping all 1 billion hoping that there might be some algorithm which can take this sequence of public keys and try to find 1 single private key within given range.
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Hi everyone! I hope you are all doing great.
I am not a programmer. Just started running ready made python scripts given by Chat GPT. But now I realized robots cannot replace humans whatsoever Tech advancements may happen. Here is my situation. I am private keys lost case. AND I just don't want to handover matters into other's hands for obvious reasons. My case is super identical. I created a bitcoin puzzle myself with my own BTCs and in the process I lost my own private keys, very funny hah! BUT its true. First of all I generated a random private key from within a range, like that of BITCOIN Puzzle range. I generated its public key and I than started adding 1 million in the private key and generated its public key. Again I added 1 million in the private key and generated its public key. I continued this for 1k times. So basically from first private key and last private there is 1 billion private keys. In case you didn't get my point: Suppose I randomly chosen private key as 980 I added 1 million in private key = 1,000,980 Again added 1 million = 2,000,980, ... ... ... Up till I added 1 billion in my first private key.
From start to end of private keys, I generated all public keys! YES, I still have 1 billion public keys of those 1 billion private keys. Ironically I lost all my private keys. Not a single public key is released since no transaction has ever happened. I placed 1 BTC in any one of those 1 billion private keys that I am looking for. I will be glad to share 0.1 BTC for genuine help specifically for python scripts would be great. Currently I am considering modifying Kangaroo algorithm to suit my needs, no luck so far. Looking forward
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