Can someone tell me why there is nothing in my output file even though the target is in the file this script is searching, it even says saved in matches.txt Here is the script import re
# List of target public keys
target_public_keys = '0370e7c3d922008d9ccea410d560cd440a834a811a1ea74c2967637ca015a788a3'
# Regular expression pattern for matching public keys
pattern = r"\b(" + "|".join(target_public_keys) + r")\b"
# Output file to log matches
output_file = "matches.txt"
# List of files to search through
files_to_search =[
"Finish-him-Jack-wins.txt"]
# Open the output file for writing
with open(output_file, "w") as output:
for filename in files_to_search:
with open(filename, "r") as file:
content = file.read()
matches = re.findall(pattern, content)
if matches:
output.write(f"Matches found in {filename}:\n")
for match in matches:
output.write(match + "\n")
output.write("\n")
print("Search completed. Matching public keys saved in", output_file) Appreciate any help. |
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import bitcoin
import secp256k1 as ice
print("creating div....")
# target pubkey Compressed
target ="02ed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88"
N = 3
A_r = bitcoin.divide(target, N)# target / N
t_0 = ice.pub2upub(target)
T_1 = ice.point_sequential_increment(1, t_0).hex()
T_2 = ice.to_cpub(T_1)
B_r = bitcoin.divide(T_2, N)#(target + 1) / N
B_0 = ice.pub2upub(A_r)
B_1 = ice.pub2upub(B_r)
B_2 = ice.point_addition(B_0, B_1)
B_3 = B_2.hex()
C_0 = ice.point_subtraction(t_0 , B_2) #target - (A + B)
C_1 = C_0.hex()
C_r = ice.to_cpub(C_1)
A= str(A_r)
B= str(B_r)
C= str(C_r)
print("target:", target+"\n")
print("A:",A)
print("B:",B)
print("C:",C+"\n") Alright, as I was in the middle of the night on the roof looking at orion and ursa major ( I think ), since I couldn't find any green grass to touch for no apparent reason, lol. I thought it's best to use my phone to run scripts, so I downloaded Py 3 installed secp256k1 lib and ran your script, but it says no attribute for pub2upub in secp256k1, but here is another thing, start exec(open(mainpyfile).read(), __main__.__dict__) File "<string>", line 17 ^ SyntaxError: invalid non-printable character U+00A0 [Program finished] I will find a fix, just wanted to thank you for the time you spend responding. |
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That link is blocked for my ip, but I guess the only way is to pay it all as fees to a miner? Though that's a solution for the rightful owner.😉
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Just to share something I found out, so if you for example divide a point by 35, and then add 1 G to your point and divide the +1 of your point by 35, what do you think will happen?
Well the results of both divisions will have a distance equal of n/35, so if you after dividing p/35, start adding n/35 to your result, for each addition you'd get the result of +1p/35, add n/35 twice, you'd get +2p/35.
Do not fix on EC points, elliptic curve is just "graphical" interpretation of a numbers, all maths behind it works at finite field and obey all laws of "real" numbers math. If you divide any "real" number by 35, then divide number+1 by 35, then difference between results will be 1/35. This is also works in finite fields and for EC points. of course if after adding 9 to p, your k ends with 5, then dividing by 5 adding 9/5 won't give you the right key, that's how you can determine what the last digit of your k is.
This is not true, for example - 26/5 = 5,2; 35/5 = 7; 9/5 = 1,8. 26/5+9/5 = 5,2+1,8 = 7 = 35/5 you can also check this with EC points like: p95 = curve.fromPrivate(9).div(5)
p265 = curve.fromPrivate(26).div(5)
p355 = curve.fromPrivate(35).div(5)
print(p265, p355)
print(p265.add(p95))
output: 03F18D7188F16C93BB251A68DE95A057E02356BA0602854041BA4E170676E54169 025CBDF0646E5DB4EAA398F365F2EA7A0E3D419B7E0330E39CE92BDDEDCAC4F9BC
025CBDF0646E5DB4EAA398F365F2EA7A0E3D419B7E0330E39CE92BDDEDCAC4F9BC
Ok, but did I say otherwise? If we add 9 to 26 we get 35, dividing 35/5 = 7 + 9/5 = 8.8, and 8.8 is the wrong answer. As I said if our k ends with 5, adding 9/5 to result returns wrong key. And of course that when we are working with secp256k1, we are dealing with a prime and mod n. Ps, I am really uneducated totally, I just share my experience, please nobody take my posts as if I am arrogant, I am 0.00000001 while everyone here is beyond 100.🙂 |
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So after some research, I have found a new and yet useless solution, is there a script which could divide a point by a set range?
Like dividing target public key by 1000, 999, 998, 997 etc, we select start and end range for division and it should divide the target as many times as we specify.
For educational purposes only.😉
I did it like this by only changing the divisor because a sequential division doesn't make sense because it would end in a long float. import bitcoin
import secp256k1 as ice
print("Generating Div-Range...")
target= "03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852"
print("Target:",target)
Start_Range= 1000
end_Range= 2000
for i in range(Start_Range, end_Range+1):
Div = bitcoin.divide(target, i)
data = open("Div-Range.txt","a")
data.write(str(Div) +"\n")
data.close() I don't know what a long float is, but if I do the division the way I described, I might find new tricks. Anyways thanks for the code, I hope it works as intended, if it doesn't, there is an idiot AI which can help. 😉 Just to share something I found out, so if you for example divide a point by 35, and then add 1 G to your point and divide the +1 of your point by 35, what do you think will happen? Well the results of both divisions will have a distance equal of n/35, so if you after dividing p/35, start adding n/35 to your result, for each addition you'd get the result of +1p/35, add n/35 twice, you'd get +2p/35. Any point on the curve, no matter what the private key is acts like "1" in real numbers, for instance, if you add 9 to your p and then divide by 5, to get the correct result you'd need to add 9/5 to your result to have the correct key. ( of course if after adding 9 to p, your k ends with 5, then dividing by 5 adding 9/5 won't give you the right key, that's how you can determine what the last digit of your k is. Ps, about the division method, dividing a key by 1000 and then dividing it by 999, what do you think will happen? In EC, both results of both divisions will have a distance equal to "1". Just to show what I mean : 500/260 = 1.9>>>2<<<30769230769230769230769230769230769230769230769230769230769230769230769231 500/259 = 1.9>>>3<<<05019305019305019305019305019305019305019305019305019305019305019305019305 Note 2 and 3 pointed out with >>><<<, in EC that 2 turning to 3 will only add 1 to your result. |
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There's nothing to admit it is my own stupidity also admitting I'm wrong. I now know it's pure stupidity of subtracting a bunch of public key and by the end it just doesn't work that way. How great are you than? It's still the same couldn't even solve 1 puzzle, aren't you? So why are you mocking others silly mistake while you achieve absolutely nothing.
Who are you talking to, me or nomachine? I rarely use kangaroo, so I'm clueless as you were, but I guess you are doing it wrong, you said using keys from 64 bit while setting the range on 65 bit, you must at least have 1 key from 65 bit to find a collision in 65 bit range set. I'm so clueless that a few times I changed -d and saw "expected RAM" at a few thousand TBs. Lol😉 |
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So after some research, I have found a new and yet useless solution, is there a script which could divide a point by a set range?
Like dividing target public key by 1000, 999, 998, 997 etc, we select start and end range for division and it should divide the target as many times as we specify.
For educational purposes only.😉
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For keys with no available public key, the only possible solution is an especial hardware like an ASIC miner to grind sha256 and rmd160 hashes at a rate of trillions/s for the cheapest ASIC and thousands of T/s for advanced ASICs. Then you could wait 10 years and see if development can increase the hash per second for your ASIC or not.
So, other than engineers, a factory with billions in equipment and a few million to spend on prototypes, what else we don't have?
Oh and that's just for keys from 66 up to 80.
However if you know the public key, there is at least a chance that you could work out things mathematically.
Regarding unexposed keys, of course there is a mathematical solution which is finding hash 160 and then hash 256 collisions, but I haven't seen any tool doing that, and it's not discussed here at all.
Ps, I have some ideas about finding collisions, it is related to sha256 checksum operation, if I could change the checksum requirement from 8 first characters of the second hash, by making it 32 character, I could work on finding collisions, in fact if we could replace sha256 with rmd160 and code something to extract rmd160 double hash checksum by using the first 20 characters of either first or second hash, again we could start working on collisions.
But that's all just an idea, I don't know if it works or not.
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Anybody here familiar with Kangaroo? I have 1 stupid question. If let's say I put 1 million public address to search for private key and there's only 1 valid public address that fit the range. Will it take much more longer time to find the valid public key to get the private key? I tried just now with 100,000 public address, but the average time to solve shown unchanged.
I tried to put 1001 key with 1000 false public key and 1 puzzle 35 key. Why kangaroo can't solve it? Does that mean we can only put 1 public key at a time?
You can't find low range keys with kangaroo, 35 bit total range is less than 35 billion keys, I have tried with low ranges, my kangaroos start dying very fast, I can't even say goodbye. 🤣 I think more public keys you place in target file more you lose speed, but the speed reduction is insignificant even with few thousands less or more keys. |
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This paper talks about something about using prime number spirals to break private key cryptography.
Not cool to post a suspicious link to some unknown blog, you should highlight the important parts and post them instead. But something tells me, if who ever wrote that paper knew how to crack private keys, he wouldn't publish it.😉 |
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Nothing going on around these wood
We play with numbers with at least 20 decimal places here. If you could make one Giga (10^9) guesses per second, it would take: 10^20 / 10^9 = 10^11 seconds This is equivalent to roughly 3.2 million years. So, even with an incredibly fast computer making a Giga guesses per second, it would take millions of years to guess a 20-decimal-place number. We'd better start practicing crystal ball gazing to guess what the correct range of Puzzle 66 is  According to my crystal ball, we'd only need 2339 years to completely scan the entire 66 bit range. Note that my crystal ball is uneducated and generates insensible results.😉 |
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It is fundamentally and structurally impossible for bitcoin to become "one world currency" whatever that even means.
You can't run a nation with financial transactions being processed every 10 minutes, you'd need instant transfer and a central authority to manage everything while keeping your money safe, with LN/other layers, you can't trust anyone because they are not governed by the laws of the nation.
This is bitcoin, it will remain like this forever, unless of course they change all the rules of the code in the future, not in our lifetime though.
So, relax and chill, don't overthink it.😉
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IMO, the only reason for all the hacks were to obtain information about satoshi and a few others directly in contact with him.
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And we want all hash 160 strings ever used for what?
Ah, sorry. So-called Brainwallets were popular in the mid-2010s. The tool I linked in the OP, Brainflayer (whose author is lurking here  ), was popular for snatching poorly protected addresses generated by a simple SHA256(passphrase). People still seem to spend a lot of time and effort making a complete HASH160 list - which is what Brainflayer needs as input - but it doesn't tell you how exactly, and the attack is then very similar to what you do with hashcat and leaked, and hashed passwords. Some were very successful with this vector a decade ago, and judging by how often the HASH160 question pops up, I assume some people aren't ready to give up just yet. So you'd want to make it easier for thieves to steal from people? Not that thieves don't know how to obtain all rmd160 hashes of all bitcoin addresses ever used, but you mentioning brainflayer specifically is not morally cool! While teaching something new. Anyways I hope people use standard wallets such as Core, electrum etc, to generate their private key/ addresses. Because using manual human input as a seed/passphrase to generate keys is just simply stupid. |
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provide an alternative, more straightforward method, if available instead of this ?
Skip steps 7 up to 13.
Nothing going on around these woods, lets make up new methods and try them to see if one of them works, here is something to work on. End range : 400000000000000000000000000000000 Fake #130 key : 2c54a14a9556f03272bac1cd222396b57 Subtract 130 from end range : #2 13ab5eb56aa90fcd8d453e32dddc694a9 Subtract above #2 from 130 = #3 : 18a942952aade064e575839a44472d6ae Subtract #3 from #2 = #4 : 4fde3dfc004d09758304567666ac4205 Multiply #4 by *4 = #5 : 13f78f7f0013425d60c1159d99ab10814 Subtract #5 from #2 = #6 : 4c30c9956a328fd37bd76abbcea736b Figure out the rest and good luck. 😉 |
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And we want all hash 160 strings ever used for what?
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I see a not very smart person who does not understand either mathematics or modular mathematics, but with all this with a crown on his head. your every post highlights your low level of intelligence.
What took you so long to realize this? I have been posting for more than 8, 9 months, and you just figured I'm uneducated with no knowledge of math and EC. 🤣 Off topic, but still. Guys, where diagran is trying to take you does not make any sense.
Not really off topic, if someone like me is misguided and is dragging others on to the wrong path, it's good for all to warn people. I just wonder who are these Guys you mentioned? Are they mentally retarded to a level where they can't figure out on their own that I'm useless and my posts are all garbage?
there is no difference to look for 1 key in range 2^130 or 4 in the range 2^128. in terms of the time spent by the currently known tools, you will spend the same amount of time.
The post you quoted has nothing to do with what you said, I don't know where you got 2^130 or 2^128, 1 key, 4 key, what are those? You see, usually when I communicate with smart people, they immediately realize I know nothing, and since they are smart with high levels of intelligence, they'd try to correct my mistakes and then teach me the right stuff, because they enjoy teaching other people, it also increases their knowledge after sharing what they already know. But a butthurt who thinks he knows better than everyone else hides behind a newbie account, insulting others, this is their joy, it also increases their butthurt pain and burn. ( this is how the universe works, action=reaction, share knowledge, gain knowledge, share/help with money, gain more money, spread hatred, absorb hatred ),
Where was I wrong exactly, implying my lack of understanding about math/ECC? Show me, then lets fix my mistakes together. ~digaran |
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Wallets might not support uncompressed public keys, but you can always send and spend by using public keys alone, that's the core function of bitcoin and can never change. Upub will require more fees though.
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Rmd-160 hash security is 80 bits AFAIK. IDK though, I might be wrong. Isn't keyhunt the same as any other brute force application when it comes to addresses/rmd160 hashes? I just can't understand the speed rates posted on this thread, peta keys/s , exa keys /s? I mean with exa keys/s per 1 PC, one could use 500 PCs with some extra EC math tricks to grind several puzzles in a month. Either nobody is able to properly use these tricks, or those capable are not aware of such tricks.
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