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I'm developing brute force software following a slightly more coherent logic, I'll post the code soon.  support the project:  1JamesJ2H2myei94NswaBATqEsBhATENSU Cool project, keeping an eye on this one! Btw what project? first show us what you got, then we'll talk about support, in case you haven't noticed, people are here to find money not give it away. unfortunately by posting the code people with more knowledge will gain an advantage, so first I'll break down the 66, 67 and 68 puzzle before explaining how it works. If you manage to solve puzzles 66, 67 and 68, what guarantees would we have that someone would see your code? If you already have the code, why do you want sponsorship? Personally, if I invest in a project, I do it on my own. If I have positive results, I will take my profit from there. So, I just wish you luck and that's it. I can't trust you and I think that most of the members on this forum would have no reason to trust you. My code currently runs on the CPU, but we all know that for the next puzzles it needs to be made to run on the GPU, if anyone with CUDA knowledge is interested in splitting 50/50 we can work together. I guarantee that with a rate of 300 Mk/s we break the 66, 67 and 68 in one week for CPU coded in which lang? How many keys/s do you get with your computer? |
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I wrote a small python script to try to predict the next number of the #66 address
from sklearn.linear_model import LinearRegression
# Training numbers
train_sequence = [1,3,7,8,21,49,76,224,467,514,1155,2683,5216,10544,26867,51510,95823,198669,357535,863317,1811764,3007503,5598802,14428676,33185509,54538862,111949941,227634408,400708894,1033162084,2102388551,3093472814,7137437912,14133072157,20112871792,42387769980,100251560595,146971536592,323724968937,1003651412950,1458252205147,2895374552463,7409811047825,15404761757071,19996463086597,51408670348612,119666659114170,191206974700443,409118905032525,611140496167764,2058769515153876,4216495639600700,6763683971478124,9974455244496707,30045390491869460,44218742292676575,138245758910846492,199976667976342049,525070384258266191,1135041350219496382,1425787542618654982,3908372542507822062,8993229949524469768,17799667357578236628,30568377312064202855]
# Forming the training data and target values
X_train = [[2**i] for i in range(len(train_sequence))] # Powers of 2
y_train = train_sequence
# Creating and training the model
model = LinearRegression()
model.fit(X_train, y_train)
# Predicting the next number
next_number = model.predict([[2**len(train_sequence)]])
next_number_decimal = int(next_number) # Converting the predicted number to decimal
next_number_hex = hex(next_number_decimal) # Converting the predicted number from decimal to hexadecimal
print("The next number in the sequence (hexadecimal) is:", next_number_hex)
print("The next number in the sequence (decimal) is:", next_number_decimal)
If helpfull 1N97k1LdzjwdG6riN9ksJBhgDQyajV14GU
The next number in the sequence (hexadecimal) is: 0x370ce49e46646a000 P66
The next number in the sequence (decimal) is: 63468747843896254464
The next number in the sequence (hexadecimal) is: 0x6e15318735d5b8000 P67
The next number in the sequence (decimal) is: 126916812407624663040
The next number in the sequence (hexadecimal) is: 0xdc25cb5914b880000 P68
The next number in the sequence (decimal) is: 253812941535081660416
The next number in the sequence (hexadecimal) is: 0x1b846fefcd27e20000 P69
The next number in the sequence (decimal) is: 507605199789995720704
The next number in the sequence (hexadecimal) is: 0x3708966444e0900000 P70 : 349b84b6431a6c4ef1
The next number in the sequence (decimal) is: 1015189716299823448064
The goal mate is not give you the exact number, but to get a number that is fairly close to searched number so you can small search in smaller range. As you can se in line #70 the number that predicted is not far away from the actual one. So… a range greater than 2^65 is not far? OMG! xD |
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So I was thinking about a new method, I call it GPS tracking, it simply uses triangulation tactic to pinpoint the location of the key. The question though, how do I derive the key given not 3 but 9 interconnected values? I have the mutual distances between several keys, I have tried at least 50 combinations and all lead to dead end, but I know there is a way to estimate the closest location coordinates of the key I'm looking for.
For example, if I know the distances between 3 objects, I should be able to determine the nearest point to at least one of them, right?
Nope, no pattern, so the "triangulation" method is useless, you can fin some collision with almost 10 chars, and guess what? in a near area of 2^40 or maybe more, you will fine nothing. |
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======================================
Matching Characters: 10
13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so
^^^^^^^^^ ^
13zb1hQbWxg82eQtkF3pF9HFYTF3KjUj14 3BCF540CE40D82701
^^^^^^^^^ ^
======================================
Is output of wich programm? |
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Don't waste anytime on this,I found this during Covid lockdown when I was looking into my old wallets and keys I had and needed some help,I finished with them and thought this would be a cool thing to try,it's a scam,the OP put in a clearly fake picture,all burned instead of water damaged LOL,the Address is real the key is fake for that address,no one is so stupid to post the the real key,sure we can't break it now but in 10 years who knows,so the way he had it set up was you would join his group and be given a "range" and when you got 90% through the range you would contact him and then he would finish it up,this guy was on crack or something thinking people would actually do it,the range he sent you with HIS program is probably real but it was not searching for the key in the picture,only the real key is known to him and a few others maybe,garbage don't waste your time,if the OP is out there and reads this come back on here and talk,prove me wrong please others think your an idiot too,he won't.
You mean the 500 WIF? of the user Phrutis? or (something like that), this is not another Key, I remember that 500wif is 12 chars missing. |
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Man really wish I'd have done at least 1 tx on this thing. I know it's a long shot, but still: Any chance you ever signed a message to prove ownership of that address? Thats a good idea, but ssadly this was a offline wallet from the start and never used, only ever deposited into. Thanks tho  S. Are you still trying?, cause I have my own programms for that, I can join to the search ASAP!  |
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Hello,
Due to the situation that happened in this challenge - I allowed myself to refresh the topic - I corrected the name of the thread, added a new date to "important dates", changed the image of the remaining amount of BTC up for grabs, and restored the functioning of my website zielar.pl from which you can now restart download the VanitySearch builds, because I saw somewhere earlier that someone mentioned that the links are out of date. It's a fact - they were outdated because I didn't extend the hosting :-) Nevertheless, everything is fine now. I'm also considering re-entering with some power... therefore, to shorten the necessary time to refresh the news about currently running programs - give me the details of any latest and greatest versions you are currently using (VanitySearch, Kangaroo) - and whether poolem 66bit also works pool fighting for 125?
Regards
<3 |
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There are over $30m in bitcoins waiting to be collected, though for someone like him, this is some change in his pocket, still it's a lot of money and nobody in the world would do this other than Satoshi.
Asking him directly for money will surely upset him as it would upset anyone! Imagine if he gave some money to somebody around here and people find out about it, there will be no puzzle solving discussions anymore but just people asking constantly for BTC, you have stated your case once, and if he is reading these topics as you suggested, if he wants to help you out, he should/would contact you privately and ask for an anonymous address as well as asking you not to tell anyone.
I doubt he'd do that though, I just said what I would do, but not for any random guy saying he just needs it, I would investigate your situation first and if I know you have contributed to the system then I would consider to lift some of your financial burdens.
Hope the best for you all.
Ps, chop chop people, think deeper, this is an ocean and we need to act like the whales, going as deep as possible, stop waiting for others to make useless tools so you could use them, start building your own algorithms, those need only our minds, and they require days of trial and error, I myself have discovered at least 10 ways to solve them, but the complexity is severe and needs to be worked on for months.
Remember, any mathematical problem is like a dot at the center of a circle, there are infinite solutions for them, so get to work and find your solutions.
The only one who can give you the knowledge is your maker, you just need to ask him properly.😉
Yes it is... for now I'm stuck in Point Multiplication with precomputed tables in C++.  |
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The amount of space that 2^67 lines of text would take up in a .txt file depends on the length of each line.
line: 20000000000000000(17 char) 1LgpDjsqkxF9cTkz3UYSbdTJuvbZ45PKvx (34 char) 17+34 = 51
total number of characters as follows:
2^67 lines * 51 characters per line = 3,689,348,814,741,248 characters
Each character takes up 1 byte of storage in a .txt file, so we can convert the total number of characters to bytes as follows:
3,689,348,814,741,248 bytes = 3,689,348,814,741 kilobytes (KB)
3,689,348,814,741 KB = 3,600,116,533 megabytes (MB)
3,600,116,533 MB = 3,515,203 terabytes (TB)
Therefore, if each line contains 51 characters, 2^67 lines of text would take up approximately 3,515,203 terabytes of space in a .txt file.
As of 2021, it is estimated that Google has over 2.5 million servers located in data centers around the world, with a total storage capacity of around 10 exabytes (10,000,000 terabytes) or more.
So, let us persuade Google to allocate 35% of its storage space to accommodate the data for our website.
The worst discussion I have seen in the entire thread till date is in these 2 to 3 pages.
The last comment x2 |
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Alberto igual podrías escribirme de favor! hay funciones y experiencia que tú tienes que nos pueden beneficiar. Hi, there is no need to write in Spanish i can read English without need to translate it, so if you want to speak in Spanish you can do it in PM or telegram. Regards I can’t send DM to you, cause my rank. |
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Is there gonna be an updated topic saying 900 bitcoin puzzle or what?
I say that we should not make noise, only those of us who are already aware. |
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id love to help in any way i have 3 3070ti gpus i would constintly run 24/7 for someone if you can push me in the right direction if youd want to split the price if the chance came one of hit. in conclusion what im saying is im a script kiddie who cant code worth shit but this is too apealing not to learn how or help atleast. under the correct cicomstances id be willing to basically give 24\7 acess to my computing power "probably not much" to help in this if anyone wants to help let me know.P.s ive tried kangaroo its too confusing i cant even get it to work thats how "not hip " i am lol... best of luck to all though ive been reading this thread and multipule others about it for a few months now with the right teacher i feel like i could be somewhat of help.
Hello, I've been following this thread for a year. I stocked all of them, including many archives that were deleted from the internet, along with their source code.
I just came up with a new idea and I need a little help with it.
Do you have fast point extraction algorithm for python ?
Thanks..
Not: sorry for my bad english, i don't know english i use google translate
Guys! Add me to discord, we can work together. I'm working with python, C++ and Rust. Need help with some coding functions, and more power will really help. DaMonsterAJ#4341 Alberto igual podrías escribirme de favor! hay funciones y experiencia que tú tienes que nos pueden beneficiar. |
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When playing with some iceland tools which is https://github.com/iceland2k14/kangaroo i got this speed on CPU - [24894.64 TeraKeys/s][Kang 28672][Count 2^29.29/2^29.10][Elapsed 06s][Dead 1][RAM 53.4MB/46.0MB]
Is it good bad fake idk but like i know iceland tools are good iceland2k14 is proprietary and all his tools contain backdoor do not use iceland2k14 libraries unless you want to loose your keys or findings good luck Really? 
Hello, I've been following this thread for a year. I stocked all of them, including many archives that were deleted from the internet, along with their source code.
I just came up with a new idea and I need a little help with it.
Do you have fast point extraction algorithm for python ?
Thanks..
Not: sorry for my bad english, i don't know english i use google translate
you mean point multiplication x,y for g in the elliptic curve? you can DM me! |
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I can do 1 quintillion/sec 1,000,000,000,000,000,000 but the range of 110 alone is 649037107316853453566312041152512 or
six hundred forty nine nonillion ,
thirty seven octillion ,
one hundred seven septillion ,
three hundred sixteen sextillion ,
eight hundred fifty three quintillion ,
four hundred fifty three quadrillion ,
five hundred sixty six trillion ,
three hundred twelve billion ,
forty one million ,
one hundred fifty two thousand ,
five hundred twelve
so it would take me 649,037,107,316,853 seconds or 7,635,730,674 days 20,919,810 (a little over 20 million years, I better stop smoking) just to go all the way through 110 with my 1070 and a whole lot of luck. I will try with kangaroo with an old gt1030 which does about 370 Trillion/ sec and about as much electricity as an led bulb. have you scanned any of 110?
with wich programm?  |
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But it doesn't have multithread, and it's not as fast as others.
The problem with multithread is that the output can be in different order than the Input file, if that doesn't matter to your output then it is OK. Why you only need the output address or rmd hash ? Regards! Doesn't matter cause it's for: Experimental purposes with different languages and scripts, learn many more!!!, and have fun with the impossible luck. BTW, my DM inbox is open. Thanks for reply! |
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Anyone knows a C program that can convert hexadecimal integer like "0000000000000000000000000000000000000000000000000000f7051f27b09112d4" to hash 160 or btc address, but from a text file or better in parallel in bash? The only one I've found something similar is brainlayer https://github.com/ryancdotorg/brainflayeryour_generator | brainflayer -v -b example.blf But it doesn't have multithread, and it's not as fast as others. Thank you, happy holidays, and greetings. |
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Hello why is pairgen so fast and how does it work? with vanitygen : thor@thor-All-Series:~/van$ ./vanitygen 1MLEvxNe
Difficulty: 873388193410
987.22 Kkey/s] [total 4036352] [Prob 0.0%] [50% in 7.1d] [total 4036352] [Prob 0.0%] [50% in 7.1d
with pairgen : thor@thor-All-Series:~/pairgen$ ./pairgen 40
n = 40bits
difficulty = 1234604
distinguished = 11bits
threads = 4
init | ...|
find_Y |...Y|
refine_Y[7] |.................................Y|
refine_Y[0] |...S...
Y[0] = 63960b937332a82b859a35801456876de48650d0
Y[1] = bcf69bf233f413a4f6be6a26ab6c6d3260da5b3d
time = 3827ms
--RESULT-(SECRET)---------------------------------------------------------------
priv_key[1] = 2DCA02F4E17AF59D4F53FA47C88A3DA5ECB2DBD4BC55102A63A7D1B5F92BCDCC
priv_key[2] = C9B3646080483325B03337D29B64F78ED2CB3B72A8DD3AEC1D9435E84E5A01DE
WIF [1] = KxkieiAUmkvB29hNgnkCnGY7EZVrW3h2QJSgW1BnMNSNxC717C5h
WIF [2] = L3ynpSthAk4eHLe88bJ9srgFQWYeoitgHH7hVpMKX2cJeGg7q3nx
--RESULT-(SIGNATURES)-----------------------------------------------------------
message = "This is a real Bitcoin address."
sig[1] = H/eQdTr9ZNaQtpZBlFqgxXJwdFAxcaPqMLiZo10Fp8zNP6g54Y7/8K5ZucmRqIB3iI2ZtFBS6wdrfag2zwKtLsY=
sig[2] = H1mIBPAu+Q1D4k1leUaz6l0IF4jXIUG8yU8EXWieaeeQWnV/QutXdZO6r0Q3Dt7rbGJevamYDkYIwoB/j/1ykPk=
--RESULT-(PUBLIC)---------------------------------------------------------------
pub_key[1] = 0262128BE3997ADD392C7A27C5D751E64F7B06039C66F7DB79F91DA2F27EE328
pub_key[2] = 034D9D75D9B9330B77342C09242C028029DAF850C720E6182A9D5A74BDFCE9E7
bonus = 1bits
shared = 10 chars
hash160 [1] = df060c1e73c2d4fb8463696a6e4df867aa70d1bf
hash160 [2] = df060c1e73b63bb1035571ccdf3fdb1f191ed1fb
shared = 8 chars
addr[1] = 1MLEvxNeJHjv2Stbsm7Nfaa9nP9A3MFCAU
addr[2] = 1MLEvxNeFyxwMeZFyaddeD67ag1rEYVAgf
warning: verify the keys/addresses before use!
in less than 10 secondsHi, how you do for search using this 1MLEvxNe as prefix in pairgen? |
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match 8 characters on bit range 64 then the key is in that range
match first 12 characters on 66 then key is in that range
i matched 8 characters on 2?
i matched 9 characters on 3?
i matched 10 characters on 3 ?
i matched 12 characters on 3?
and thats where i am now
Last year I found more than 2K addresses with 8 coincidence in all bit 64 range. Guess what? No matters, no pattern. |
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so none of addresses in 12 bit space matches 4 characters of that address?
if no, the it means random know something we don't
Bigger bit more characters appears. |
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So today i made an experiment to see the probability of matching characters of address in the bitrange that is not the challenge address but i was surprised that the script solved the puzzle in seconds and did not find any other address except the target ii ran it with puzzle 12 and see i have set it to 1 million tries import random
import bitcoin
import sys
def hex2addr(hx,ds):
decoded_private_key = ds
private_key=hx
compressed_private_key = private_key + '01'
# Multiply the EC generator point G with the private key to get a public key point
public_key = bitcoin.fast_multiply(bitcoin.G, decoded_private_key)
# Encode as hex, prefix 04
hex_encoded_public_key = bitcoin.encode_pubkey(public_key,'hex')
# Compress public key, adjust prefix depending on whether y is even or odd
(public_key_x, public_key_y) = public_key
if (public_key_y % 2) == 0:
compressed_prefix = '02'
else:
compressed_prefix = '03'
hex_compressed_public_key = compressed_prefix + bitcoin.encode(public_key_x, 16)
ad =bitcoin.pubkey_to_address(public_key)
# Generate compressed bitcoin address from compressed public key
adr= bitcoin.pubkey_to_address(hex_compressed_public_key)
return adr
def check(ad):
if "1DBau" in ad:
print(ad)
st = "0"
n = 1000000
while n>0:
ps = random.getrandbits(12)
pt = str(ps).rstrip("L")
ps = int(pt)
hx = hex(ps).lstrip("0x").rstrip("L")
tp = 64-len(hx)
hxs = st*tp+hx
ads = hex2addr(hxs,ps)
check(ads)
n =n-1
print("done") output └──╼ $python dif2.py
1DBaumZxUkM4qMQRt2LVWyFJq5kDtSZQot
1DBaumZxUkM4qMQRt2LVWyFJq5kDtSZQot 2^12 = 4,096 that's why |
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