Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 206. (Read 229769 times)

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
https://media.tenor.com/UiUFIuFzuH0AAAAd/memes-wrong-number.gif

I wrote a small python script to try to predict the next number of the #66 address

Here is what i exactly did:

1. Generate the entire 64 range divided as 17,666,000 sub-ranges sorted sequentially at first.
2. Randomize the sub-ranges completely
3. Search each sub-range line by line, so that each sub-range completed gets me 1 in 17.6 million chance of winning.

After someone solved 64 and the pvt key was revealed, i found out that right sub-range after randomization was in line 1 million.

i.e: i was never meant to solve it if i kept going forward with this plan.

I did something like that with puzzle 64 using keyhunt-cuda.. it was a file of 17,666,000 ranges .. after 64 got solved, i was curious as to where the correct range lied in my file, you know where it was?

Line number 1+ million !!!!!

Doesn't work. But you could be more lucky and find the correct range in line number 10 or something in your file

Hello,

I just created a simple solo-pool to find puzzle 66.

It only works with BTC Puzzle 66. If you want to scan all private keys in puzzle 66; you need to do 36 quintillion scans in total. In case you do a random scan; previously generated private keys will be regenerated (random problem). This extends the scan time by x10.

Puzzle 66 HEX ranges as follows. It starts with 2 or 3. Any private key in this range is 17 characters long.

20000000000000000 to 3ffffffffffffffff

We take the first 7 characters and delete the rest for now. The result will be as follows.

2000000 to 3ffffff

We now have about 33 million possible private keys to search. All possible private keys are stored in the database. A random value will come up each time a scan job is called and will be marked as scanned when the scan is complete.

I can scan each key in about 10 minutes on NVIDIA 3090. This actually means about 1,1 trillion private keys. When the private key is scanned, it is marked as scanned. So it won't show up anymore.

https://github.com/ilkerccom/bitcrackrandomiser

Hello,

I just created a simple solo-pool to find puzzle 66.

It only works with BTC Puzzle 66. If you want to scan all private keys in puzzle 66; you need to do 36 quintillion scans in total. In case you do a random scan; previously generated private keys will be regenerated (random problem). This extends the scan time by x10.

Puzzle 66 HEX ranges as follows. It starts with 2 or 3. Any private key in this range is 17 characters long.

20000000000000000 to 3ffffffffffffffff

We take the first 7 characters and delete the rest for now. The result will be as follows.

2000000 to 3ffffff

We now have about 33 million possible private keys to search. All possible private keys are stored in the database. A random value will come up each time a scan job is called and will be marked as scanned when the scan is complete.

I can scan each key in about 10 minutes on NVIDIA 3090. This actually means about 1,1 trillion private keys. When the private key is scanned, it is marked as scanned. So it won't show up anymore.