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Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 181. (Read 244669 times)

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Hello everyone, and especially those who still remember me)
Has anyone used OpenSSL to generate keys?
Probably not..

main.cpp (*Int class require some code changes within the SECP256k1 library to support BIGNUM* directly.)
Code:
#include "SECP256k1.h"
#include "Int.h"
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

const int numThreads = 4; // You can adjust this number based on your CPU cores

// Function to generate a random private key using BIGNUM
BIGNUM* generateRandomPrivateKey(const BIGNUM* minKey, const BIGNUM* maxKey) {
    BIGNUM* randomPrivateKey = BN_new();
    BN_rand_range(randomPrivateKey, maxKey);

    // Ensure the generated key is within the desired range
    while (BN_cmp(randomPrivateKey, minKey) < 0) {
        BN_rand_range(randomPrivateKey, maxKey);
    }

    return randomPrivateKey;
}

// Function to convert a BIGNUM to Int
Int bignumToBigInt(const BIGNUM* bignum) {
    char* bignumStr = BN_bn2dec(bignum);
    Int bigInt;
    bigInt.SetBase10(bignumStr);
    OPENSSL_free(bignumStr);
    return bigInt;
}

// Function to generate keys and check for a specific address
void generateKeysAndCheckForAddress(BIGNUM* minKey, BIGNUM* maxKey, std::shared_ptr secp256k1, const std::string& targetAddress) {
    while (true) {
        BIGNUM* randomPrivateKey = generateRandomPrivateKey(minKey, maxKey);

        // Convert the BIGNUM private key to an Int
        Int privateKey = bignumToBigInt(randomPrivateKey);

        // Continue with the rest of the address generation and checking logic
        Point publicKey;
        std::string caddr;
        std::string wifc;

        publicKey = secp256k1->ComputePublicKey(&privateKey);
        caddr = secp256k1->GetAddress(0, true, publicKey);
        wifc = secp256k1->GetPrivAddress(true, privateKey);

        // Display the generated address
        std::string message = "\r\033[01;33m[+] " + caddr;
        std::cout << message << "\e[?25l";
        std::cout.flush();

        // Check if the generated address matches the target address
        if (caddr.find(targetAddress) != std::string::npos) {
            time_t currentTime = std::time(nullptr);

            // Format the current time into a human-readable string
            std::tm tmStruct = *std::localtime(¤tTime);
            std::stringstream timeStringStream;
            timeStringStream << std::put_time(&tmStruct, "%Y-%m-%d %H:%M:%S");
            std::string formattedTime = timeStringStream.str();

            std::cout << "\n\033[32m[+] PUZZLE SOLVED: " << formattedTime << "\033[0m" << std::endl;
            std::cout << "\033[32m[+] WIF: " << wifc << "\033[0m" << std::endl;

            // Append the private key information to a file if it matches
            std::ofstream file("KEYFOUNDKEYFOUND.txt", std::ios::app);
            if (file.is_open()) {
                file << "\nPUZZLE SOLVED " << formattedTime;
                file << "\nPublic Address Compressed: " << caddr;
                file << "\nPrivatekey (dec): " << privateKey.GetBase10();
                file << "\nPrivatekey Compressed (wif): " << wifc;
                file << "\n----------------------------------------------------------------------------------------------------------------------------------";
                file.close();
            }

            // Free the BIGNUM and break the loop
            BN_free(randomPrivateKey);
            break;
        }

        // Free the BIGNUM
        BN_free(randomPrivateKey);

        // Convert the max key to an Int
        Int maxInt;
        maxInt.SetBase10(BN_bn2dec(maxKey));

        if (privateKey.IsGreater(&maxInt)) {
            break;
        }
    }
}


int main() {
    // Clear the console
    std::system("clear");

    time_t currentTime = std::time(nullptr);
    std::cout << "\033[01;33m[+] " << std::ctime(¤tTime) << "\r";
    std::cout.flush();

    BIGNUM* minKeyBN = BN_new(); // Initialize minKeyBN
    BIGNUM* maxKeyBN = BN_new(); // Initialize maxKeyBN

    // Configuration for the Puzzle
    // Set minKeyBN and maxKeyBN using the provided base 10 values
    BN_dec2bn(&minKeyBN, "62079069358943824031");
    BN_dec2bn(&maxKeyBN, "67079069358943924031");
    std::string targetAddress = "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so";

    // Initialize SECP256k1
    std::shared_ptr secp256k1 = std::make_shared();
    secp256k1->Init();

    // Create threads for key generation and checking
    std::vector threads;

    for (int i = 0; i < numThreads; ++i) {
        threads.emplace_back(generateKeysAndCheckForAddress, minKeyBN, maxKeyBN, secp256k1, targetAddress);
    }

    // Wait for all threads to finish
    for (std::thread& thread : threads) {
        thread.join();
    }

    // Cleanup BIGNUM variables
    BN_free(minKeyBN);
    BN_free(maxKeyBN);

    return 0;
}

Code:
SRC = Base58.cpp IntGroup.cpp main.cpp Random.cpp Timer.cpp \
      Int.cpp IntMod.cpp Point.cpp SECP256K1.cpp \
      hash/ripemd160.cpp hash/sha256.cpp hash/sha512.cpp \
      hash/ripemd160_sse.cpp hash/sha256_sse.cpp Bech32.cpp

OBJDIR = obj

OBJET = $(addprefix $(OBJDIR)/, \
        Base58.o IntGroup.o main.o Random.o Int.o Timer.o \
        IntMod.o Point.o SECP256K1.o \
        hash/ripemd160.o hash/sha256.o hash/sha512.o \
        hash/ripemd160_sse.o hash/sha256_sse.o Bech32.o)

CXX = g++
CXXFLAGS = -m64 -mssse3 -Wno-write-strings -O2 -I.

LFLAGS = -lpthread -lssl -lcrypto

$(OBJDIR)/%.o : %.cpp
$(CXX) $(CXXFLAGS) -o $@ -c $<



VanitySearch: $(OBJET)
@echo Making Lottery...
$(CXX) $(OBJET) $(LFLAGS) -o LOTTO.bin && chmod +x LOTTO.bin

$(OBJET): | $(OBJDIR) $(OBJDIR)/hash

$(OBJDIR):
mkdir -p $(OBJDIR)

$(OBJDIR)/hash: $(OBJDIR)
cd $(OBJDIR) && mkdir -p hash

clean:
@echo Cleaning...
@rm -f obj/*.o
@rm -f obj/hash/*.o

Yes... Grin

But it is still slow. Even 5M keys/s per core muscles is not enough for such a large range. Undecided
copper member
Activity: 1330
Merit: 899
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It is difficult to specify which result is integer or not because they are all within the same curve, and can be represented by several pk.

1/2= 57896044618658097711785492504343953926418782139537452191302581570759080747169
3/2= 57896044618658097711785492504343953926418782139537452191302581570759080747170
1/2= 0.5
Secp256k1 curve, 1/2=
57896044618658097711785492504343953926418782139537452191302581570759080747169

3/2= 1.5
Secp256k1 curve, 3/2=
57896044618658097711785492504343953926418782139537452191302581570759080747170

When you operate mod n, 1.5 turns into 0.5+1, or half of n +1.  This is true for 1 up to n-1. Like 11/2 is just n/2+5.
So what about 51/2? It's  n/2+25, how about 701/2? It's n/2+350.  How about 1001/2? It's n/2+500.

Now moving forward, 10001/85= 117.65882
1/85= 0.011764706
Subtracting  0.011764706 - 117.65882 = 117.64706, not integer, now we want to know how to find 0.65882 of n, because 1/85 didn't give us 0.65882, it gives us 0.011764706, but  subtracting them gave us some clues, the answer is n.64706th+117.  We don't want our result to be a fraction, so we need to find the remainder of division mod n. 
Now going bigger, 1000001/85= 11764.718, 1000002/85=
 11764.729,   1000003/85= 11764.741,  1000004/85=
 11764.753.

See what happened?
0.011764706 1/85
0.011764718 1million and one/85, I added 0.0 - .
0.011764729 1million and two/85, added 0.0 - .
0.011764741 1m and three/85
0.011764753 1m and four/85
If you remove 0.0 from above fractions, you get the correct answer.
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New ideas will be criticized and then admired.

144M private keys are a drop in the ocean of keys. What's the point of generating public keys from private keys (it takes a long time). You need to add and multiply and divide public keys (XY coordinates). The process can of course be speeded up if you use C++. Not a single Python script has yet found complex solutions to a puzzle.
You have some unique creative designs from Jean Luc Pons, this is the highest level of programming skill and experience. Everything else can only be modified, new algorithms and new functions can be added to the C++ code. Look in the wrong direction... It's better to find the correct divisor for the unknown private key. Correct - this means the remainder of the division is zero!!! This is a good task, then there will be a result.
144 million keys using python script? Yeah it will take a few hours, at least for me.
I'm not sure about the highest programming skills since kangaroo is outdated, and in practice useless for large keys/ranges.
Also there is a script in python which divides a point by a start/end range, it's really useful to find the divisor, I have both scalar version and point version posted on project development board, there is also a version which operates with 2 targets and divides them by the set range, and on top of that there is a  subtraction function to sub the results of division, I suggest you play around with scalar version and change the range, also change the last digit of your targets  to at least have 10 keys ending with 0 through 9 and then repeat divide ranges with all of them.

Start by subtracting 2^129 from a known k in puzzle 130 range and then use fake #130 as first, and the result of subtraction as second target, start your range from 2, 32 and keep increasing it the next time, 32, 256, then 256, 2048 and so on, try to check the index number, if you see your scalar divided by 48 results in an integer, then that's your  divisor. Set multiples of 48 as start, end range and solve the key.

I'm working on a method to determine with 100% accuracy whether a point divided by a number results in an integer or not. Stay tuned.😉


It is difficult to specify which result is integer or not because they are all within the same curve, and can be represented by several pk.

1/2= 57896044618658097711785492504343953926418782139537452191302581570759080747169
3/2= 57896044618658097711785492504343953926418782139537452191302581570759080747170
copper member
Activity: 1330
Merit: 899
🖤😏

144M private keys are a drop in the ocean of keys. What's the point of generating public keys from private keys (it takes a long time). You need to add and multiply and divide public keys (XY coordinates). The process can of course be speeded up if you use C++. Not a single Python script has yet found complex solutions to a puzzle.
You have some unique creative designs from Jean Luc Pons, this is the highest level of programming skill and experience. Everything else can only be modified, new algorithms and new functions can be added to the C++ code. Look in the wrong direction... It's better to find the correct divisor for the unknown private key. Correct - this means the remainder of the division is zero!!! This is a good task, then there will be a result.
144 million keys using python script? Yeah it will take a few hours, at least for me.
I'm not sure about the highest programming skills since kangaroo is outdated, and in practice useless for large keys/ranges.
Also there is a script in python which divides a point by a start/end range, it's really useful to find the divisor, I have both scalar version and point version posted on project development board, there is also a version which operates with 2 targets and divides them by the set range, and on top of that there is a  subtraction function to sub the results of division, I suggest you play around with scalar version and change the range, also change the last digit of your targets  to at least have 10 keys ending with 0 through 9 and then repeat divide ranges with all of them.

Start by subtracting 2^129 from a known k in puzzle 130 range and then use fake #130 as first, and the result of subtraction as second target, start your range from 2, 32 and keep increasing it the next time, 32, 256, then 256, 2048 and so on, try to check the index number, if you see your scalar divided by 48 results in an integer, then that's your  divisor. Set multiples of 48 as start, end range and solve the key.

I'm working on a method to determine with 100% accuracy whether a point divided by a number results in an integer or not. Stay tuned.😉
member
Activity: 93
Merit: 16
this python script generates about 144M private keys with the X and Y keys, which takes a long time to generate, so I'd like to know if it would be possible to speed up the process by keeping the calculation method used in the script, even if it's just a matter of generating the private keys while keeping the calculation method?
144M private keys are a drop in the ocean of keys. What's the point of generating public keys from private keys (it takes a long time). You need to add and multiply and divide public keys (XY coordinates). The process can of course be speeded up if you use C++. Not a single Python script has yet found complex solutions to a puzzle.
You have some unique creative designs from Jean Luc Pons, this is the highest level of programming skill and experience. Everything else can only be modified, new algorithms and new functions can be added to the C++ code. Look in the wrong direction... It's better to find the correct divisor for the unknown private key. Correct - this means the remainder of the division is zero!!! This is a good task, then there will be a result.
member
Activity: 206
Merit: 16
this python script generates about 144M private keys with the X and Y keys, which takes a long time to generate, so I'd like to know if it would be possible to speed up the process by keeping the calculation method used in the script, even if it's just a matter of generating the private keys while keeping the calculation method?

addressgeneration.py
Code:
from ecdsa import SigningKey, SECP256k1
import sha3
from binascii import unhexlify
import hashlib
from base58 import b58encode
import bech32
from cashaddress import convert
import datetime

def PrivateKeyComputation(p1: int, p2: int, p3: int, base: int, n: int):
    order = (2 ** 6) * 3 * 149 * 631 * p1 * p2 * p3
    prod = p1 * p2 * p3
    g = pow(base, prod, order + 1)
    privateSet = [None] * n

    for i in range(n):
        if i == 0:
            privateSet[0] = hex(g)
        else:
            k = (g * int(privateSet[i - 1], 16)) % (order + 1)
            privateSet[i] = hex(k)

    return privateSet

def CosetPrivateKeyComputation(p1: int, p2: int, p3: int, base: int, n: int):
    print("Coset PrivateKey Computation started at", datetime.datetime.now())
    prod = p1 * p2 * p3
    h = (2 ** 6) * 3 * 149 * 631
    order = h * prod
    g = pow(base, prod, order + 1)
    privateSet = [None] * n * 8

    for i in range(n):
        if i == 0:
            privateSet[0] = hex(g)
        else:
            k = (g * int(privateSet[i - 1], 16)) % (order + 1)
            privateSet[i] = hex(k)

    pows = [h, h*p1, h*p2, h*p3, h*p1*p2, h*p1*p3, h*p2*p3]

    for j in range(len(pows)):
        g = pow(base, pows[j], order + 1)
        for i in range(n):
            value = (g * int(privateSet[i], 16)) % (order + 1)
            privateSet[(j+1)*h+i] = hex(value)
    print("Coset PrivateKey Computation finished at", datetime.datetime.now())
    return privateSet

def CosetKeysFile(n: int, privateSet):
    print("Writing on txt file started at", datetime.datetime.now())
    f = open("secp256k1_keys.txt", "r+")
    f.seek(0)
    f.write('\t\tPrivateKey  \t\t\t\t\t\t\t\t\t\t  PublicKey-x \t\t\t\t\t\t\t\t\t\t   PublicKey-y  \n')

    for i in range(8*n):
        k = int(privateSet[i], 16).to_bytes(32, "big")
        k = SigningKey.from_string(k, curve=SECP256k1)
        K = k.get_verifying_key().to_string()

        f.write(str(i) + ')\t' + privateSet[i] + '\t' + K.hex()[0:64] + '\t' + K.hex()[64:128] + '\n')

    f.truncate()
    f.close()
    print("Writing on txt file finished at", datetime.datetime.now())

def UncompressedPublicKeyComputation(x, y):
    publicKey = '04' + str(x) + str(y)
    return publicKey

def CompressedPublicKeyComputation(x, y):
    if int(y, 16) % 2 == 0:
        publicKey = '02' + str(x)
    else:
        publicKey = '03' + str(x)

    return publicKey

def checksum_computation(string: str) -> hex:
    cs = hashlib.sha256(hashlib.sha256(unhexlify(string)).digest()).hexdigest()
    checksum = cs[:8]
    return checksum

def BitcoinClassicAddressComputation(publicKey):
    public_key_bytes = unhexlify(publicKey)

    sha256 = hashlib.sha256()
    sha256.update(public_key_bytes)
    hash_temp = sha256.digest()

    ripemd160 = hashlib.new('Ripemd160')
    ripemd160.update(hash_temp)
    hash2_temp = ripemd160.hexdigest()

    hash3_temp = '00' + hash2_temp

    checksum = checksum_computation(hash3_temp)

    hash_final = hash3_temp + str(checksum)
    hash_final_bytes = unhexlify(hash_final)
    address = b58encode(hash_final_bytes).decode("utf-8")
    return address

KeysFileGeneration.py
Code:
import AddressGeneration

p1 = 107361793816595537
p2 = 174723607534414371449
p3 = 341948486974166000522343609283189
base = 7
h = (2 ** 6) * 3 * 149 * 631


privateSet = AddressGeneration.CosetPrivateKeyComputation(p1, p2, p3, base, h)
PublicSet = AddressGeneration.CosetKeysFile(h, privateSet)
member
Activity: 239
Merit: 53
New ideas will be criticized and then admired.
is this puzzle still unsolved to date?
It's not one puzzle, there are 160 in total, some of them are already solved, #66 is not yet solved. #130 also not solved but these 2 are next in line.



Guys anyone here knows python? Of course you do, I'm stuck for days to make this happening, at first I wanted to have such function inside another script but I failed, so I thought of something else, the following script opens 2 text files, reads the keys in both and subtracts them from each other, that's the logic but whatever I did I couldn't get it to take public keys from each line of the files and do the subtraction with them all, it just reads the last line and returns only 1 result obviously, if you could provide a fix, it'd be great.

Code:
# secp256k1 curve parameters
p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
a = 0
b = 7
Gx = 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
Gy = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
n = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

# Point addition and subtraction functions
def point_add(P, Q):
    if P == Q:
        return point_double(P)
    if P is None:
        return Q
    if Q is None:
        return P
    lam = ((Q[1] - P[1]) * pow(Q[0] - P[0], p-2, p)) % p
    x = (lam * lam - P[0] - Q[0]) % p
    y = (lam * (P[0] - x) - P[1]) % p
    return (x, y)

def point_sub(P, Q):
    if Q is None:
        return P
    # instead of using p - Q[1], use -Q[1] % p to correctly compute y coordinate
    Q_neg = (Q[0], (-Q[1]) % p)
    return point_add(P, Q_neg)

# Point doubling function
def point_double(P):
    if P is None:
        return None
    lam = ((3 * P[0] * P[0] + a) * pow(2 * P[1], p-2, p)) % p
    x = (lam * lam - 2 * P[0]) % p
    y = (lam * (P[0] - x) - P[1]) % p
    return (x, y)

def is_valid_point(point):
    # Check that the point is not the point at infinity
    if point is None:
        return False
    
    x, y = point
    # Check that the coordinates are within the allowed range
    if x < 0 or x >= p or y < 0 or y >= p:
        return False
    
    # Check that the point lies on the curve
    return (y*y - x*x*x - a*x - b) % p == 0

def decompress_point(compressed_key):
    if compressed_key.startswith(b'\x02') or compressed_key.startswith(b'\x03'):
        x = int.from_bytes(compressed_key[1:], byteorder='big')
        y_sq = (x * x * x + a*x + b) % p
        y = pow(y_sq, (p+1)//4, p)
        if (y*y) % p == y_sq:
            return (x, y)
        else:
            return None
    else:
        return None

def compress_point(point):
    x, y = point
    prefix = b'\x02' if y % 2 == 0 else b'\x03'
    return prefix + x.to_bytes(32, byteorder='big')

def point_subtraction(compressed_keys):
    P = None
    for compressed_key in compressed_keys:
        Q = decompress_point(compressed_key)
        
        # Check that the point is valid on the curve
        if not is_valid_point(Q):
            return None
        
        if P is None:
            P = Q
        else:
            P = point_sub(P, Q)
        
        # Check that the resulting point is not the point at infinity
        if P is None:
            return None
    
    if is_valid_point(P):
        return compress_point(P)
    else:
        return None

compressed_keys = []
file1 = open('file1.txt', 'r')
for line in file1:
    compressed_keys.append(bytes.fromhex(line.strip()))
file1.close()

file2 = open('file2.txt', 'r')
for line in file2:
    compressed_keys.append(bytes.fromhex(line.strip()))
file2.close()

result = point_subtraction(compressed_keys)

if result:
    file3 = open('result.txt', 'a')
    file3.write(result.hex() + "\n")
    file3.close()
else:
    print("The subtraction result is not a valid point on the curve.")

I know the problem is in line.strip section, just don't know how, btw I asked AI, it started by implementing sha256 for no reason, ended up with giving me some test cases from bitcoin wiki. Go figure!


Note, it's for academic purposes only ( whatever that means anyway ).😉

This code reads line by line from a file.
Code:
file= open("data-base_line_by_line.txt", "r")
lines = file.readlines()
for index, line in enumerate(lines):
    
    target = str(line.strip())
    print(target)

edit:
if you just want to subtract (lines in file1) - (lines in file2) use this:

Code:
import itertools
import secp256k1 as ice



with open('file1.txt', 'r') as f1, open('file2.txt', 'r') as f2:
   

    for line1, line2 in zip(f1, f2):
       
        x=  ice.pub2upub(str(line1.strip()))

        x2= ice.pub2upub(str(line2.strip()))

        res = ice.point_subtraction(x, x2).hex()
       
        result= ice.to_cpub(res)

        file3 = open('result.txt', 'a')
        file3.write(result + "\n")
        file3.close()
   

member
Activity: 93
Merit: 16
Coming soon.
Once upon a time,  there existed a WanderingPhilospher who also was a brute force developer, he used to be my mentor, he taught a lot of useful stuff to me. Will you be my new mentor? Lol

Do you have anything useful and fast that could do public key calculations to solve for private key? I have some ideas which I need to test extensively to make sure it could be used to solve DLP, God willing after a successful test run, I might bother you to develop something new!
Not funny at all! I already have a version with random divide of the public key.
I talked to WanderingPhilospher a couple of years ago, I remember well when I remade the starting keys for Kangaroo.
And probably everyone forgot that I was the first to post the GPU code for Kangaroo and this was before the release date of Kangaroo from Jean-Luc Pons.
here is the link: https://bitcointalksearch.org/topic/bitcoin-challenge-transaction-1000-btc-total-bounty-to-solvers-updated-5218972
At that time, I pushed this topic very hard, and soon a version from Jean-Luc Pons appeared. Thank him very much.
As promised, I published the fork code on Github https://github.com/alek76-2/VanitySearch
Enjoy it, and if you find solution 66 with this version, don’t forget about me  Smiley
copper member
Activity: 1330
Merit: 899
🖤😏
Coming soon.
Once upon a time,  there existed a WanderingPhilospher who also was a brute force developer, he used to be my mentor, he taught a lot of useful stuff to me. Will you be my new mentor? Lol

Do you have anything useful and fast that could do public key calculations to solve for private key? I have some ideas which I need to test extensively to make sure it could be used to solve DLP, God willing after a successful test run, I might bother you to develop something new!
member
Activity: 93
Merit: 16
I made a new version, while I’m checking how it works... It will be very difficult to find bit 66, even with the best GPUs.
OpenSSL is screwed into the program, with the ability to switch the OpenSSL functions used. Can also check different dlls by dropping them into the directory with the program. I disabled everything unnecessary - SSE, endomorphism, symmetry. Changed the GPU code - removed endomorphism and removed double (quadruples) checks of Ripemd160 (Match GPU - the Y coordinate is calculated in the curve). I'm checking it out, maybe I'll post it soon...
Sounds interesting as long as the source code is available so one can compile the program. Is it available for download ?
Today I checked the GPU code again, since I removed the unnecessary calculations of Ripemd160 (left 1 out of 6) and it works correctly. I needed to make sure again that everything was correct. In this code I also used functions from the Bitcoin client - to add a seed, the Rand_add() function. I'll post it on Github in the near future, because... With my computing resources it will not be possible to find a solution to bit 66. If they find it using my fork, then send me a tips Smiley. Coming soon.
I managed to get this speed on Tesla T4:
Code:
[ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ]
[                                                           ]
[===========================================================]
[                Changes by Alek76 modify 0.03              ]
[===========================================================]
[          Tips: 1NULY7DhzuNvSDtPkFzNo6oRTZQWBqXNE9         ]
[===========================================================]
[    Options added argv [-start] [-bits] and disable SSE    ]
[===========================================================]
[           Used OpenSSL Random number generator            ]
[===========================================================]
OpenSSL 3.0.2 15 Mar 2022 (Library: OpenSSL 3.0.2 15 Mar 2022)
[===========================================================]
[                OpenSSL add all algorithms                 ]
[===========================================================]
[                OpenSSL Used functions level 4             ]
[                                                           ]
[ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ]
                                                            

[i] RAND_add() Seed with CPU performance counter: 1695456415355946
Difficulty: 1461501637330902918203684832716283019655932542976
Search: 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so [Compressed]
Start Sat Sep 23 08:06:55 2023
Base Key: Randomly changed every 100000 Mkeys
Number of CPU thread: 0
GPU: GPU #0 Tesla T4 (40x64 cores) Grid(320x128)

[i] RAND_add() Seed with CPU performance counter: 1695456415480157
Bit 66 GPU Base Key 0: 2DAFFE6ACD76ABC1D
Bit 66 GPU Base Key 1: 27AD93FD5D885B767
Bit 66 GPU Base Key 2: 2D67AE2D1731671C5
Bit 66 GPU Base Key 3: 3C77CC7B5C7587A86
Bit 66 GPU Base Key 4: 277448DA9C423D147
Bit 66 GPU Base Key 5: 2F7EDC7B5C2B09F0F
Bit 66 GPU Base Key 6: 3DABCD2B602D43EA8
Bit 66 GPU Base Key 7: 238936C644C88AB7E
Bit 66 GPU Base Key 8: 3A97B3F0171947B59
Bit 66 GPU Base Key 9: 31B6CDD325D6B3FB7
Bit 66 GPU Base Key 40951: 3211CC54F75B5CD00
Bit 66 GPU Base Key 40952: 2327D47C30738E157
Bit 66 GPU Base Key 40953: 2A66575CC34FA37C6
Bit 66 GPU Base Key 40954: 2981DD1BA18306814
Bit 66 GPU Base Key 40955: 3775C5C27150FEA12
Bit 66 GPU Base Key 40956: 37A37E2863197B3DB
Bit 66 GPU Base Key 40957: 3D78C9919D6B6A072
Bit 66 GPU Base Key 40958: 22100282EE8370965
Bit 66 GPU Base Key 40959: 37E28F54F8EC9B7D2

[i] RAND_add() Seed with CPU performance counter: 1695456416086342
Bit 66 GPU Base Key 0: 33D16FF3A68C0BA73
Bit 66 GPU Base Key 1: 359F2BA0E89868D5E
Bit 66 GPU Base Key 2: 3DDDCE27056C63AD9
Bit 66 GPU Base Key 3: 2EFF53109E165A31F
Bit 66 GPU Base Key 4: 236DC50A204B42078
Bit 66 GPU Base Key 5: 2F128D7562A4DDC32
Bit 66 GPU Base Key 6: 2743F1524E422DF51
Bit 66 GPU Base Key 7: 284ECC8C4DC852018
Bit 66 GPU Base Key 8: 3932B5EC487EACE6A
Bit 66 GPU Base Key 9: 398E4AEE55E284239
Bit 66 GPU Base Key 40951: 3225C10410AF00667
Bit 66 GPU Base Key 40952: 229E0588B78FCDF88
Bit 66 GPU Base Key 40953: 25DAC65EBE4EF6B6B
Bit 66 GPU Base Key 40954: 2C5621F3B0D87CFB8
Bit 66 GPU Base Key 40955: 229EC31EBF4F800D2
Bit 66 GPU Base Key 40956: 31F29B5822AA4E982
Bit 66 GPU Base Key 40957: 27C891A611B7D7A20
Bit 66 GPU Base Key 40958: 2F1351CB9E5A7812F
Bit 66 GPU Base Key 40959: 22D025D76B453D748
[615.93 Mkey/s][GPU 615.93 Mkey/s][Total 2^36.06][Prob 0.0%][50% in 5.21538e+31y][Found 0]
Coming soon... Already available https://github.com/alek76-2/VanitySearch
hero member
Activity: 630
Merit: 731
Bitcoin g33k
I made a new version, while I’m checking how it works... It will be very difficult to find bit 66, even with the best GPUs.
OpenSSL is screwed into the program, with the ability to switch the OpenSSL functions used. Can also check different dlls by dropping them into the directory with the program. I disabled everything unnecessary - SSE, endomorphism, symmetry. Changed the GPU code - removed endomorphism and removed double (quadruples) checks of Ripemd160 (Match GPU - the Y coordinate is calculated in the curve). I'm checking it out, maybe I'll post it soon...
Sounds interesting as long as the source code is available so one can compile the program. Is it available for download ?
copper member
Activity: 1330
Merit: 899
🖤😏
Didn't work, still outputs a single result which is the result of last public keys in my files subtracted, this new method I'm working on is not yet tested, so I can't share it before testing it myself.

I managed to make up a scalar mod n version of my script, but as a py noob, I can't learn so many things in a short period of time. Thanks for the effort though.
Maybe I'm  unsuccessful because I'm using mobile instead of laptop? Lol I should find some place else for my coding guidance.

My new script can solve any key in scalar mode of course.
jr. member
Activity: 61
Merit: 6
You could just use it as

Code:
file1 = open('file1.txt', 'r')
for line in file1.read().splitlines():
    compressed_keys.append(bytes.fromhex(line))
file1.close()

file2 = open('file2.txt', 'r')
for line in file2.read().splitlines():
    compressed_keys.append(bytes.fromhex(line))
file2.close()

Good Luck with your studies  Wink
copper member
Activity: 1330
Merit: 899
🖤😏
is this puzzle still unsolved to date?
It's not one puzzle, there are 160 in total, some of them are already solved, #66 is not yet solved. #130 also not solved but these 2 are next in line.



Guys anyone here knows python? Of course you do, I'm stuck for days to make this happening, at first I wanted to have such function inside another script but I failed, so I thought of something else, the following script opens 2 text files, reads the keys in both and subtracts them from each other, that's the logic but whatever I did I couldn't get it to take public keys from each line of the files and do the subtraction with them all, it just reads the last line and returns only 1 result obviously, if you could provide a fix, it'd be great.

Code:
# secp256k1 curve parameters
p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
a = 0
b = 7
Gx = 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
Gy = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
n = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

# Point addition and subtraction functions
def point_add(P, Q):
    if P == Q:
        return point_double(P)
    if P is None:
        return Q
    if Q is None:
        return P
    lam = ((Q[1] - P[1]) * pow(Q[0] - P[0], p-2, p)) % p
    x = (lam * lam - P[0] - Q[0]) % p
    y = (lam * (P[0] - x) - P[1]) % p
    return (x, y)

def point_sub(P, Q):
    if Q is None:
        return P
    # instead of using p - Q[1], use -Q[1] % p to correctly compute y coordinate
    Q_neg = (Q[0], (-Q[1]) % p)
    return point_add(P, Q_neg)

# Point doubling function
def point_double(P):
    if P is None:
        return None
    lam = ((3 * P[0] * P[0] + a) * pow(2 * P[1], p-2, p)) % p
    x = (lam * lam - 2 * P[0]) % p
    y = (lam * (P[0] - x) - P[1]) % p
    return (x, y)

def is_valid_point(point):
    # Check that the point is not the point at infinity
    if point is None:
        return False
   
    x, y = point
    # Check that the coordinates are within the allowed range
    if x < 0 or x >= p or y < 0 or y >= p:
        return False
   
    # Check that the point lies on the curve
    return (y*y - x*x*x - a*x - b) % p == 0

def decompress_point(compressed_key):
    if compressed_key.startswith(b'\x02') or compressed_key.startswith(b'\x03'):
        x = int.from_bytes(compressed_key[1:], byteorder='big')
        y_sq = (x * x * x + a*x + b) % p
        y = pow(y_sq, (p+1)//4, p)
        if (y*y) % p == y_sq:
            return (x, y)
        else:
            return None
    else:
        return None

def compress_point(point):
    x, y = point
    prefix = b'\x02' if y % 2 == 0 else b'\x03'
    return prefix + x.to_bytes(32, byteorder='big')

def point_subtraction(compressed_keys):
    P = None
    for compressed_key in compressed_keys:
        Q = decompress_point(compressed_key)
       
        # Check that the point is valid on the curve
        if not is_valid_point(Q):
            return None
       
        if P is None:
            P = Q
        else:
            P = point_sub(P, Q)
       
        # Check that the resulting point is not the point at infinity
        if P is None:
            return None
   
    if is_valid_point(P):
        return compress_point(P)
    else:
        return None

compressed_keys = []
file1 = open('file1.txt', 'r')
for line in file1:
    compressed_keys.append(bytes.fromhex(line.strip()))
file1.close()

file2 = open('file2.txt', 'r')
for line in file2:
    compressed_keys.append(bytes.fromhex(line.strip()))
file2.close()

result = point_subtraction(compressed_keys)

if result:
    file3 = open('result.txt', 'a')
    file3.write(result.hex() + "\n")
    file3.close()
else:
    print("The subtraction result is not a valid point on the curve.")

I know the problem is in line.strip section, just don't know how, btw I asked AI, it started by implementing sha256 for no reason, ended up with giving me some test cases from bitcoin wiki. Go figure!


Note, it's for academic purposes only ( whatever that means anyway ).😉
member
Activity: 182
Merit: 33
is this puzzle still unsolved to date?
member
Activity: 93
Merit: 16
Please don't belive that you are the only one who think in that...  Roll Eyes

Yes, I checked some SSL libs on linux. Most of them works ok, The useful thing is when you tested some buggy version of it. but you need to check the source code to see where the bug is and use it to generate "weak" keys.

Yes, that's absolutely true. Here I spent a week digging through old versions of OpenSSL from their official source. Weak keys are possible if assembled incorrectly. For example, like in older versions of Bitcoin core bitcoin-0.2.0, where RandAddSeedPerfmon() for Linux does not work, not used function RAND_add().
Downloaded DLLs can be checked by saving them keys to file and checking for duplicates. But this is unlikely all, a very large range. In general, a good Random will not hurt to solve bit 66, since the versions and key generation programs for this puzzle are not known. Perhaps the purpose of its creation will help, but perhaps its purpose is also to test the strength of OpenSSL libraries. Although there are so many external events that it is not possible to re-generate duplicate keys, unless there are broken versions - removing some functions to generate a weak key.
Perhaps someone decided to check old versions and created this puzzle.
hero member
Activity: 862
Merit: 662
Has anyone used OpenSSL to generate keys?
Probably not..

Please don't belive that you are the only one who think in that...  Roll Eyes

Yes, I checked some SSL libs on linux. Most of them works ok, The useful thing is when you tested some buggy version of it. but you need to check the source code to see where the bug is and use it to generate "weak" keys.
member
Activity: 93
Merit: 16
Hello everyone, and especially those who still remember me)
I made a new version, while I’m checking how it works... It will be very difficult to find bit 66, even with the best GPUs.
OpenSSL is screwed into the program, with the ability to switch the OpenSSL functions used. Can also check different dlls by dropping them into the directory with the program. I disabled everything unnecessary - SSE, endomorphism, symmetry. Changed the GPU code - removed endomorphism and removed double (quadruples) checks of Ripemd160 (Match GPU - the Y coordinate is calculated in the curve). I'm checking it out, maybe I'll post it soon...
Code:
C:\VSearch-1.19 New Build check v0.03\win7 x64>VanitySearch.exe -stop -t 1 -bits 66 -r 5 -level 1 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so

[ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ]
[                                                           ]
[===========================================================]
[                Changes by Alek76 modify 0.03              ]
[===========================================================]
[          Tips: 1NULY7DhzuNvSDtPkFzNo6oRTZQWBqXNE9         ]
[===========================================================]
[    Options added argv [-start] [-bits] and disable SSE    ]
[===========================================================]
[           Used OpenSSL Random number generator            ]
[===========================================================]
[                OpenSSL add all algorithms                 ]
[===========================================================]
[                OpenSSL Used functions level 1             ]
[                                                           ]
[ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ]

Difficulty: 1461501637330902918203684832716283019655932542976
Search: 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so [Compressed]
Start Fri Sep 22 23:00:18 2023
Base Key: Randomly changed every 5 Mkeys
Number of CPU thread: 1

Bit 60 Base Key thId 0: 9F79EEEC674E61F < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 66 CPU Base Key thId 0: 29D1A41009D12513A
[0.89 Mkey/s][GPU 0.00 Mkey/s][Total 2^22.35][Prob 0.0%][50% in 3.60851e+34y][Found 0]
Bit 64 Base Key thId 0: E3643540371FB847 < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 64 Base Key thId 0: 80A2CD8B9A3C23BC < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 66 CPU Base Key thId 0: 3A53B397C7C3998F9
[0.85 Mkey/s][GPU 0.00 Mkey/s][Total 2^23.51][Prob 0.0%][50% in 3.768e+34y][Found 0]
Bit 66 CPU Base Key thId 0: 33A47F5D6B844E5CE
[0.82 Mkey/s][GPU 0.00 Mkey/s][Total 2^24.14][Prob 0.0%][50% in 3.90297e+34y][Found 0]
Bit 66 CPU Base Key thId 0: 23B25CA07BAC32C6F
[0.82 Mkey/s][GPU 0.00 Mkey/s][Total 2^24.58][Prob 0.0%][50% in 3.9043e+34y][Found 0]
Bit 65 Base Key thId 0: 12FF6314297C60ECB < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 66 CPU Base Key thId 0: 25B540E692C67500B
[0.82 Mkey/s][GPU 0.00 Mkey/s][Total 2^24.92][Prob 0.0%][50% in 3.90599e+34y][Found 0]
Bit 65 Base Key thId 0: 14F6F111D9ADFE665 < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 66 CPU Base Key thId 0: 28F5EB0F8DBAF6E80
[0.82 Mkey/s][GPU 0.00 Mkey/s][Total 2^25.19][Prob 0.0%][50% in 3.90512e+34y][Found 0]
Bit 66 CPU Base Key thId 0: 33755278B8367C2C3
[0.82 Mkey/s][GPU 0.00 Mkey/s][Total 2^25.42][Prob 0.0%][50% in 3.89525e+34y][Found 0]
Bit 64 Base Key thId 0: C63FFCFE4B85011C < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 66 CPU Base Key thId 0: 36CA7D64880F6EC8E
[0.82 Mkey/s][GPU 0.00 Mkey/s][Total 2^25.59][Prob 0.0%][50% in 3.89824e+34y][Found 0]
Bit 65 Base Key thId 0: 1602E2E612E2D1D5D < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 64 Base Key thId 0: 9CE5D3E01377163B < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 66 CPU Base Key thId 0: 22729A06418E0AA07
[0.82 Mkey/s][GPU 0.00 Mkey/s][Total 2^25.77][Prob 0.0%][50% in 3.90262e+34y][Found 0]
Bit 65 Base Key thId 0: 1E4BA74256A46FFF2 < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 65 Base Key thId 0: 19A395587DDFF6849 < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 66 CPU Base Key thId 0: 201909FF60F39CB91
[0.82 Mkey/s][GPU 0.00 Mkey/s][Total 2^25.89][Prob 0.0%][50% in 3.90116e+34y][Found 0]

Code:
C:\VSearch-1.19 New Build check v0.03\win7 x64>VanitySearch.exe -stop -t 2 -bits 66 -r 5 -level 4 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so

[ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ]
[                                                           ]
[===========================================================]
[                Changes by Alek76 modify 0.03              ]
[===========================================================]
[          Tips: 1NULY7DhzuNvSDtPkFzNo6oRTZQWBqXNE9         ]
[===========================================================]
[    Options added argv [-start] [-bits] and disable SSE    ]
[===========================================================]
[           Used OpenSSL Random number generator            ]
[===========================================================]
[                OpenSSL add all algorithms                 ]
[===========================================================]
[                OpenSSL RAND_screen() OK                   ]
[===========================================================]
[                OpenSSL Used functions level 4             ]
[                                                           ]
[ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ]


[i] RAND_add() Seed with CPU performance counter: 413322837980
Difficulty: 1461501637330902918203684832716283019655932542976
Search: 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so [Compressed]
Start Fri Sep 22 23:25:51 2023
Base Key: Randomly changed every 5 Mkeys
Number of CPU thread: 2

[i] RAND_add() Seed with CPU performance counter: 413323029166

Bit 64 Base Key thId 1: C3B95A81B85BFE25 < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 66 CPU Base Key thId 1: 20D9401B0A4FFDD06

[i] RAND_add() Seed with CPU performance counter: 413323029845

Bit 64 Base Key thId 0: A5BB18323A7CB6AB < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 66 CPU Base Key thId 0: 295244E678E41E12E
[1.82 Mkey/s][GPU 0.00 Mkey/s][Total 2^22.85][Prob 0.0%][50% in 1.76408e+34y][Found 0]
[i] RAND_add() Seed with CPU performance counter: 413333913173

Bit 66 CPU Base Key thId 0: 29EF09292123E8168

[i] RAND_add() Seed with CPU performance counter: 413333913233

Bit 65 Base Key thId 1: 1FC1E88D8E95BC854 < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 61 Base Key thId 1: 11B4BEA1EE95FF48 < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 66 CPU Base Key thId 1: 27D8A8F52B15BE829
[1.72 Mkey/s][GPU 0.00 Mkey/s][Total 2^23.76][Prob 0.0%][50% in 1.86995e+34y][Found 0]
[i] RAND_add() Seed with CPU performance counter: 413343522509

Bit 66 CPU Base Key thId 0: 396121E3EFFA2A03C

[i] RAND_add() Seed with CPU performance counter: 413343523260

Bit 66 CPU Base Key thId 1: 2D60B4F910B675F54
[1.68 Mkey/s][GPU 0.00 Mkey/s][Total 2^24.31][Prob 0.0%][50% in 1.90682e+34y][Found 0]
[i] RAND_add() Seed with CPU performance counter: 413353124168

[i] RAND_add() Seed with CPU performance counter: 413353124942

Bit 64 Base Key thId 1: B461B1FC1240503B < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 66 CPU Base Key thId 1: 2C1BC11A5C0C595BA

Bit 66 CPU Base Key thId 0: 32DFF58084D0F0B48
[1.67 Mkey/s][GPU 0.00 Mkey/s][Total 2^24.72][Prob 0.0%][50% in 1.92559e+34y][Found 0]
[i] RAND_add() Seed with CPU performance counter: 413362823436

Bit 66 CPU Base Key thId 1: 2664445EC6A550E07

[i] RAND_add() Seed with CPU performance counter: 413362823661

Bit 66 CPU Base Key thId 0: 27E59959139581FB5
[1.62 Mkey/s][GPU 0.00 Mkey/s][Total 2^25.03][Prob 0.0%][50% in 1.98491e+34y][Found 0]
[i] RAND_add() Seed with CPU performance counter: 413372570037

[i] RAND_add() Seed with CPU performance counter: 413372575962

Bit 66 CPU Base Key thId 1: 2DA47EF2D2A90A017

Bit 66 CPU Base Key thId 0: 33A1C28B16D799A69
[1.62 Mkey/s][GPU 0.00 Mkey/s][Total 2^25.29][Prob 0.0%][50% in 1.98281e+34y][Found 0]
[i] RAND_add() Seed with CPU performance counter: 413382240693

[i] RAND_add() Seed with CPU performance counter: 413382241494

Bit 61 Base Key thId 1: 147BF7CC10BC669B < 20000000000000000 or > 3FFFFFFFFFFFFFFFF Rekey true

Bit 66 CPU Base Key thId 1: 2BC45552F10B6FAF2

Bit 66 CPU Base Key thId 0: 3BF647F72F5534012
[1.62 Mkey/s][GPU 0.00 Mkey/s][Total 2^25.40][Prob 0.0%][50% in 1.98292e+34y][Found 0]

Code:
C:\VSearch-1.19 New Build check v0.03\win7 x64>VanitySearch.exe -stop -t 2 -bits 28 -r 5 -level 1 12jbtzBb54r97TCwW3G1gCFoumpckRAPdY

[ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ]
[                                                           ]
[===========================================================]
[                Changes by Alek76 modify 0.03              ]
[===========================================================]
[          Tips: 1NULY7DhzuNvSDtPkFzNo6oRTZQWBqXNE9         ]
[===========================================================]
[    Options added argv [-start] [-bits] and disable SSE    ]
[===========================================================]
[           Used OpenSSL Random number generator            ]
[===========================================================]
[                OpenSSL add all algorithms                 ]
[===========================================================]
[                OpenSSL Used functions level 1             ]
[                                                           ]
[ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ]

Difficulty: 1461501637330902918203684832716283019655932542976
Search: 12jbtzBb54r97TCwW3G1gCFoumpckRAPdY [Compressed]
Start Fri Sep 22 23:32:33 2023
Base Key: Randomly changed every 5 Mkeys
Number of CPU thread: 2

Bit 28 CPU Base Key thId 0: F3DDC63

Bit 28 CPU Base Key thId 1: CFD7541
[1.82 Mkey/s][GPU 0.00 Mkey/s][Total 2^22.86][Prob 0.0%][50% in 1.76379e+34y][Found 0]
Bit 28 CPU Base Key thId 1: D299DE3

Bit 28 CPU Base Key thId 0: BAC1B81
[1.73 Mkey/s][GPU 0.00 Mkey/s][Total 2^23.78][Prob 0.0%][50% in 1.85735e+34y][Found 0]
Bit 28 CPU Base Key thId 1: CC24254

Bit 26 Base Key thId 0: 2165974 < 8000000 or > FFFFFFF Rekey true

Bit 28 CPU Base Key thId 0: CC18276
[1.69 Mkey/s][GPU 0.00 Mkey/s][Total 2^24.33][Prob 0.0%][50% in 1.8972e+34y][Found 0]
Bit 27 Base Key thId 1: 4B1D339 < 8000000 or > FFFFFFF Rekey true

Bit 28 CPU Base Key thId 1: D517B81

Bit 26 Base Key thId 0: 24181FC < 8000000 or > FFFFFFF Rekey true

Bit 26 Base Key thId 0: 2639DB4 < 8000000 or > FFFFFFF Rekey true

Bit 27 Base Key thId 0: 5992E85 < 8000000 or > FFFFFFF Rekey true

Bit 26 Base Key thId 0: 3AE84E6 < 8000000 or > FFFFFFF Rekey true

Bit 28 CPU Base Key thId 0: DF7A263
[1.68 Mkey/s][GPU 0.00 Mkey/s][Total 2^24.73][Prob 0.0%][50% in 1.91598e+34y][Found 0]
Bit 27 Base Key thId 1: 596F852 < 8000000 or > FFFFFFF Rekey true

Bit 27 Base Key thId 1: 5005795 < 8000000 or > FFFFFFF Rekey true

Bit 28 CPU Base Key thId 1: C626164

Bit 28 CPU Base Key thId 0: D72E1A1
[1.63 Mkey/s][GPU 0.00 Mkey/s][Total 2^24.90][Prob 0.0%][50% in 1.97098e+34y][Found 0]

  Addr :12jbtzBb54r97TCwW3G1gCFoumpckRAPdY
  Check:12jbtzBb54r97TCwW3G1gCFoumpckRAPdY

!!! Result.txt Found key: D916CE8
!!! Result.txt Found key: D916CE8
!!! Result.txt Found key: D916CE8
!!! Result.txt Found key: D916CE8
!!! Result.txt Found key: D916CE8
Has anyone used OpenSSL to generate keys?
Probably not..
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