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Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 182. (Read 215417 times)

member
Activity: 272
Merit: 20
the right steps towerds the goal
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
copper member
Activity: 1330
Merit: 899
🖤😏
I have been working on big numbers and public keys so much, now when I look at them, I'm just blank, totally with no idea what I'm looking at, now I understand why mathematicians are a bit off, they go crazy after a while.🤣
Need to rest to gain focus, this challenge is what I strive for.
member
Activity: 72
Merit: 43
I wrote a small python script to try to predict the next number of the #66 address

useless - you cannot predict the privkey of any puzzle address
quite right as any of other methods... i'm trying a new approach now but it will for sure fail or for the fun of things maybe not.
i'm trying random vs random but on smaller scale (like 16^9), chance 1 in 200 for each run on my current setup.
in testing i have got better results starting with sha256 hashes of random "strings" instead of simple RNG or sequential but still didnt found anything.
Good luck everyone...it's all about luck... as for now SHA family is safe
hero member
Activity: 630
Merit: 731
Bitcoin g33k
I wrote a small python script to try to predict the next number of the #66 address

useless - you cannot predict the privkey of any puzzle address
newbie
Activity: 26
Merit: 2
WanderingPhilospher
0x3b0812107ccf0168c:0x3b081212053d7cf17
0x3b0e4766c8585fd90:0x3b0e476850c6db61b
0x3b0fdd3a714e99cec:0x3b0fdd3bf9bd15577
0x3b01f9676c1298284:0x3b01f968f48113b0f
0x2f64750bb309d2b80:0x2f64750d3b784e40b
0x2f76405b2bc314700:0x2f76405cb4318ff8b
I tried
member
Activity: 185
Merit: 15
Two things you should never abandon: Family & BTC
Quote
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.

Ahhhh, gotcha...well it's not that you wouldn't have solved it with more resources. Smiley

At least you randomized the list somewhat. Sequential is brutal and defeating lol.

Lol yeah choosing sequential is like if you're deciding to fail before you even start.
full member
Activity: 1162
Merit: 237
Shooters Shoot...
Quote
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.

Ahhhh, gotcha...well it's not that you wouldn't have solved it with more resources. Smiley

At least you randomized the list somewhat. Sequential is brutal and defeating lol.
member
Activity: 185
Merit: 15
Two things you should never abandon: Family & BTC
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

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 Smiley

What do you mean it didn’t work?! So you were going through sequentially versus random?

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.
full member
Activity: 1162
Merit: 237
Shooters Shoot...
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

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 Smiley

What do you mean it didn’t work?! So you were going through sequentially versus random?
member
Activity: 185
Merit: 15
Two things you should never abandon: Family & BTC
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

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 Smiley
full member
Activity: 1162
Merit: 237
Shooters Shoot...
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.
So you only keep 33 m keys in a file and then search for what exactly? If you are searching small ranges, then you should  do it sequential. What is your speed in terms of keys/s ?

No, it has ranges, not keys. It's a pool of sorts that keeps track of ranges ran. And no, sequential is not the way to go lol.
I did a quick calculus, I can generate 1 trillion every 12 hours, so 2 trillions per 24 hours, now if I run my laptop 24/7, I could find the key after 30,000 + years or so give or take! Do you think people will be using bitcoin by then? Because I don't wanna waste 30,000 years  for unsure future investment.😂
copper member
Activity: 1330
Merit: 899
🖤😏
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.
So you only keep 33 m keys in a file and then search for what exactly? If you are searching small ranges, then you should  do it sequential. What is your speed in terms of keys/s ?

I did a quick calculus, I can generate 1 trillion every 12 hours, so 2 trillions per 24 hours, now if I run my laptop 24/7, I could find the key after 30,000 + years or so give or take! Do you think people will be using bitcoin by then? Because I don't wanna waste 30,000 years  for unsure future investment.😂
full member
Activity: 1162
Merit: 237
Shooters Shoot...
Meaning? We have a pool http://f4lc0n.com:8080/
Since you are part of the pool, what is one range size equal to?

Range size seems like an odd one.

325M left
14M complete

Not really that straight forward; like 2^65 / 2 ^ 40 range = 2 ^ 25 ranges to search
newbie
Activity: 26
Merit: 2
newbie
Activity: 8
Merit: 2
Hello,

I just created a simple solo-pool to find puzzle 66, 67 and 68.

It works with proof of work. 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 ranges to search. All possible ranges 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. Each range contains 1,1 trillion keys.

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://btcpuzzle.info
https://github.com/ilkerccom/bitcrackrandomiser
copper member
Activity: 1330
Merit: 899
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digaran

I want to help you if you tell me what to do. Do you have telegram? Where can you chat?

I had to lol a lot before typing this.
If I knew what to do, I would have done it myself already, though I have realized that what I'm doing is a simplified method of which the tools such as kangaroo are doing. I just add a number to my target p, and then subtract the result from my p, which seems to return my selected number.

Ok let me go technical on this, if we add 5 to 10, we will have 15, then subtracting 10 from 15 gives us 5, but in bitcoin's used curve, if  we simply subtract 15 from 10, we won't see -5, instead we will see 20, that's how this complicated elliptic curve works, what I have done:
I  have added 5 to 10 and 3 to 10,  sum of them is 28 now, then I'd subtract 10 from 28 to obtain 18, now imagine I don't know 10, 13, 15, 18 and 28, I just have 5 and 3. But I know the distance between 10 and 13, 10 and 15, 10 and 18.
Now can you tell me how I can reach 10, 13, 15, 18 and or 28 by only adding or multiplying 3 and 5 together?

Well, when we simplify it like that, we could multiply 3 by 5 to get to 15, but in our case if I knew what is half of 10, I wouldn't be here at all, I'm just randomly guessing some numbers and then I'd add and subtract them.

Today I found out if we add -20 to 10, we get 20, and adding -10 to 20, we get 10, it works other way as well.

If you are more confused than before, you should be, this is how cryptography works, confusion is the key!
newbie
Activity: 26
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digaran

I want to help you if you tell me what to do. Do you have telegram? Where can you chat?
copper member
Activity: 1330
Merit: 899
🖤😏

Speedy recovery digaran!
Thanks, it was 2/5 years ago though sitting long periods of time does hurt even now, a few days workout will build more muscle to support the spine.



Anyways, it seems there are no interesting parties wanting 1 bitcoin doing some simple (yeah right) math calculations, easy peasy. 😉

Anyhow, so I was thinking about a brute force/key solving method, it could work and not work but mentioning it doesn't hurt.

First we pick a small k, something like 3, then we start by multiplying a range of random numbers by our original number, when we reach the end of our search range without any hit, we'd pick another base key and select another set of random numbers until we again reach the end of our search range, and since multiplying only takes a few seconds to perform, we could scan the entire range in a few minutes and start again with new settings, this would increase the chance of a successful hit by many orders of magnitude.

Just today I realized that I have been performing what kangaroo and other tools are doing, by hand! Lol  when you have limited resources, your mind is the best asset to utilize.😉


full member
Activity: 1162
Merit: 237
Shooters Shoot...
Quote
I rather give the whole world if I had the whole world to avoid experiencing the pain in my leg and my back again, nothing is worth more than our health.
No doubt. I had a massive blow out/bulging disc, and an annular tear in another disc. Left leg was numb for months. Couldn't get comfortable in any form or fashion or position. I finally went outside and rubbed a cup of dirt on my back...Now, I am back! lol.

Indeed, back pain is the worse. It still gives me issues from time to time.

Speedy recovery digaran!
copper member
Activity: 1330
Merit: 899
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Where did you get 8,192 chances from? Don't tell me, actually don't tell anyone. You know if I had the means I would have joined you right? When you think about me just visualize a laptop with 30 mk/s search capability!

If I haven't solved the key yet is due to my back, I can no longer sit behind the system to work, I have been at it for a month and I had half of one of my discs removed, so I'm feeling a bit numb in my leg as I type this.


I have got to get back to my trampoline, though I wouldn't call it that, it is more like triangulation method, for now I call it GPS for a lack of a better word.

You see, what I have is actually 2 public keys, when I subtract the 125 public key from each one of them I get the same key, one is inverse though, so one of them is -N. The part it gets interesting, when I add those 2 keys together it gives me the 125 key * 2.  And to my surprise I also have the distances between both of them with the *2 of 125, and obviously the distance between both of them with 125 key as well. Since I can't work on it right now, it remains unsolved.  Moreover I can't seem to figure the equations of adding and subtracting which to which and which from which yet because one or both are  probably -N.

So if you manage to help me out on this, I will give you 1BTC whole bitcoin.


I rather give the whole world if I had the whole world to avoid experiencing the pain in my leg and my back again, nothing is worth more than our health.😉
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