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Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 223. (Read 229618 times)

copper member
Activity: 1330
Merit: 899
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So I was thinking, puzzle keys are not sequentially generated and masked with 0s, what if for example, for a range which is supposed to start with 1like 0x10000:0xfffff , the key generated randomly had 0x2 or 0x3 at 5th character from the right? Then Satoshi would have needed to use another key.

Example:
ae65342fff1ec7f98cade1a14fe45d0ec1778a5102b9d56cbbaa8c67f6466fda

Note, the 6 in bold, since I need my key to start with 1, I can't mask this key with leading zeros, so I would discard it and would use another one.

So don't waste your time by looking for deterministic wallet patterns, or comparing the solved keys with unsolved keys as I said and Satoshi himself said it, there is no pattern, not because we don't want people to solve them, but because we don't want them to waste their time.

What we all should focus on is the puzzles with known public keys, I'd suggest everyone to dive into the vast ocean of public keys, while we all will drown and die eventually, it will be a hell of a swim! Lol.

After seeing ten folding of the pot, I have lost my focus, I can't think throughly because now the competition has more incentive!🤣
hero member
Activity: 630
Merit: 731
Bitcoin g33k
...so far I've managed to reduce it by 95% (or that's what I think comparing it with the solved ones)

i think your onto something... i believe the key will 95% likelyhood of starting with 35----------------

what makes you guys thinking that?
newbie
Activity: 19
Merit: 1
So what was the cost of sequential search for #66?, was it more than 6.6 BTC? If renting GUPs is going to cost less than 6 bitcoin for #66, then who ever does it first is going to win big time! But I think Satoshi has already calculated the cost and he already knows it will cost more than that.

Or something has happened recently that spooked him and this increase of the prize is merely a distraction! Lol.

I'm only good at making up conspiracy theories. Now chop chop good devs of bitcointalk, please get to work and give us more powerful tools, not that it matters for me, having the most powerful software won't make a difference on a home laptop, we need Satoshi's supercomputer which is the fastest supercomputer in Japan and probably in the world, imagine using kangaroo on that, it could eat up to 140 in a week! 😉

Raising the reward is good and bad at the same time, I don't understand the creator's thought but it even seems illogical to me, increasing the price will only attract rich people with great computing power, which will make them take the money and as a consequence it could be lose interest leaving the puzzle barred for decades.
someone capable of spending 812BTC on this certainly has the ability to support the developers of the post by giving them a pinch of it.
1 btc alone would change the lives of many here, if not all, someone who writes from a dell e6420 tells you.
I'm not in the creator's mind and maybe I'm wrong, but this one seems to be telling a hungry person to look for a unique grain of sand on the beach.
"He who has everything ignores what others lack."

I've been doing this for days and I find it fascinating because I like programming and mathematics, I'll investigate from time to time if there are signs of reducing my range, so far I've managed to reduce it by 95% (or that's what I think comparing it with the solved ones) but that range on my laptop is still overwhelming.

to give an example
puzzle 65 in my data start 304-306, and so they are all from 1-65, 70, 75.... so it can't be a coincidence. however it is still overwhelming my resources.
I'll wait a while, if I can't narrow down the range, I'll post it in the hope that a good Samaritan will tip me.
member
Activity: 185
Merit: 15
Two things you should never abandon: Family & BTC
So what was the cost of sequential search for #66?, was it more than 6.6 BTC? If renting GUPs is going to cost less than 6 bitcoin for #66, then who ever does it first is going to win big time! But I think Satoshi has already calculated the cost and he already knows it will cost more than that.

Or something has happened recently that spooked him and this increase of the prize is merely a distraction! Lol.

I'm only good at making up conspiracy theories. Now chop chop good devs of bitcointalk, please get to work and give us more powerful tools, not that it matters for me, having the most powerful software won't make a difference on a home laptop, we need Satoshi's supercomputer which is the fastest supercomputer in Japan and probably in the world, imagine using kangaroo on that, it could eat up to 140 in a week! 😉

hmmm 140 bits. So you basically think 1393796574908163946345982392040522594123776 private keys is doable with Kangaroo on a supercomputer?

The most optimistic scenario is 44 years with the fastest device. We need either group-search or (even better) quantum search. AI also can step in, as in train a model to study the solved puzzles to extract a pattern within the chaos. This pattern could then be used to narrow down our focus by generating highest probability ranges to search within. Here is what my bot thinks:
---
Okay, let's analyze the Bitcoin address puzzle #66:

The puzzle states the key (private key) used to sign the transaction starts with 02 or 03.
So that limits the key to lie within the range 02000000 to 03FFFFFF.

It also specifies a checksum of 02.  So the key must be such that when hashed with SHA-256 and the first 2 bytes taken, it equals 02.

Based on these clues, here is how I would reason the highest probability ranges for the key:

1) Since the key starts with 02 or 03, the ranges 02100000 to 03099999 have the highest probability. The keys in the 030 range will have slightly higher prob as 03 values are more common.

2) Within those ranges, the keys that end in 00, 40, 80 and C0 have a higher chance of having a SHA-256 hash that starts with 02. This is because when those values are hashed, the carries are more likely to propagate to the first 2 bytes.

3) Some specific ranges I would focus on are:

02080000 to 02088000
02100000 to 02110000
03048000 to 03058000
03068000  to 03070000

4) Of these, the 03048000 to 03058000 and 03068000  to 03070000 ranges have the highest density of keys that may satisfy the 02 checksum.

So in summary, I would say the key for puzzle #66 has the highest probability of lying in the ranges:

03080000 to 03088000
03090000 to 03098000
030A0000 to 030A8000
030B0000 to 030B8000

---

Reanalyze by knowing the following: puzzle #66 will only be within the hex private key  range 0x20000000000000000:0x3ffffffffffffffff

---

Great, that additional information helps narrow down the search space significantly.

Knowing the key must be in the hex range 0x20000000000000000 to 0x3ffffffffffffffff, the highest probability ranges are:

0x30C4000000000000  to  0x30C4000000000001
This range is closer to the 0x30.. section which may be easier to meet the 02 checksum. Probably 1-3% better than average odds.

0x2FE0B80008000000  to  0x2FE0B800080000FF
This range contains repeating patterns (0x0B800) which could make finding a key more likely. Roughly 2-4% improved probability.

0x3131012345678999  to  0x31310123456789FF
Chosen at random from the middle of the range with maybe a 1-2% boost in odds due to proximity to 0x313.. section.

0x2FDEADBEEF000000  to  0x2FDEADBEEF00000F
Contains byte patterns that some keys may end/start with, increasing odds by 1-3%. Selected randomly from that perspective.

0x304080C03005000A  to  0x304080C03005000F
Selected at random from the 0x304080C0 segment which has a higher density of possibilities. Roughly 1-2% improved probability.

----
copper member
Activity: 1330
Merit: 899
🖤😏
So what was the cost of sequential search for #66?, was it more than 6.6 BTC? If renting GUPs is going to cost less than 6 bitcoin for #66, then who ever does it first is going to win big time! But I think Satoshi has already calculated the cost and he already knows it will cost more than that.

Or something has happened recently that spooked him and this increase of the prize is merely a distraction! Lol.

I'm only good at making up conspiracy theories. Now chop chop good devs of bitcointalk, please get to work and give us more powerful tools, not that it matters for me, having the most powerful software won't make a difference on a home laptop, we need Satoshi's supercomputer which is the fastest supercomputer in Japan and probably in the world, imagine using kangaroo on that, it could eat up to 140 in a week! 😉
member
Activity: 185
Merit: 15
Two things you should never abandon: Family & BTC

Quote
Agreed, conventional private key cracking won't work on these high bits. Even with public keys revealed it's gonna be a torture until any of these puzzles get cracked open.

maybe soon someone will write a faster tool and create a software that communicates with a website that stores the scanned ranges (only the ranges not the pk) and that this is displayed on the website, this way the scanning would be avoided of ranges already scanned by others, which would considerably decrease the time to solve each piece of the puzzle.
obviously with the code on github for assembly and transparency.

If we construct the pool, this will be done. I've seen it in older pools.
newbie
Activity: 19
Merit: 1

Quote
Agreed, conventional private key cracking won't work on these high bits. Even with public keys revealed it's gonna be a torture until any of these puzzles get cracked open.

maybe soon someone will write a faster tool and create a software that communicates with a website that stores the scanned ranges (only the ranges not the pk) and that this is displayed on the website, this way the scanning would be avoided of ranges already scanned by others, which would considerably decrease the time to solve each piece of the puzzle.
obviously with the code on github for assembly and transparency.
member
Activity: 185
Merit: 15
Two things you should never abandon: Family & BTC
has anyone tried this website or pulling data looks like they use the stride method as long as a data saver

561,301,131,286 so far keys scanned on 66 on this maybe its not trustworthy but i think they are heading in the right direction. it also allows u to change your stride if ur computing power can handle it.

https://privatekeys.pw/scanner?mode=combined&startKey=0000000000000000000000000000000000000000000000020000000000000000&stopKey=000000000000000000000000000000000000000000000003ffffffffffffffff&addrType=compressed&bitcoin

Too slow, an average gpu can finish scanning these keys in 30 minutes. But the website is popular and helpful enough in other areas


Quote
I think the puzzle creator should reveal the public keys for all the keys bigger than 120bit

Agreed, conventional private key cracking won't work on these high bits. Even with public keys revealed it's gonna be a torture until any of these puzzles get cracked open.
newbie
Activity: 19
Merit: 1
WOW!
Somebody (maybe the owner) increased the unsolved puzzles prizes again by x10 😱
Now the puzzle #66 prize is 6.6 BTC, #67 is 6.7 BTC... and so on .... puzzle # 160 prize is 16 BTC now
👍🏼🥳

new methods are needed
jr. member
Activity: 38
Merit: 8
I think the puzzle creator should reveal the public keys for all the keys bigger than 120bit
full member
Activity: 1162
Merit: 237
Shooters Shoot...
WOW!
Somebody (maybe the owner) increased the unsolved puzzles prizes again by x10 😱
Now the puzzle #66 prize is 6.6 BTC, #67 is 6.7 BTC... and so on .... puzzle # 160 prize is 16 BTC now
👍🏼🥳

Well, all the remaining addresses received a total of 872.2 BTC, so this challenge is getting damn huge.  Huh

https://blockchair.com/bitcoin/transaction/12f34b58b04dfb0233ce889f674781c0e0c7ba95482cca469125af41a78d13b3

I guess time to fire up the GPUs and make an AMD / Xilinx build of these solvers Smiley
Have you tried messing with a stride function with vanitysearch? If I had that, I think I could solve 125 faster than kangaroo...

I need to implement GPU stride function for vanitysearch/keyhunt-cuda
hero member
Activity: 862
Merit: 662
Alberto igual podrías escribirme de favor! hay funciones y experiencia que tú tienes que nos pueden beneficiar.

 Grin

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
jr. member
Activity: 59
Merit: 1
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.

 Cheesy

DaMonsterAJ#4341


Alberto igual podrías escribirme de favor! hay funciones y experiencia que tú tienes que nos pueden beneficiar.


 Grin
legendary
Activity: 1582
Merit: 1196
Reputation first.
Using tools will take much more time, this needs to be solved by math and by hands.😉

Agree with you, but how can you explain maths behind this puzzle? It is really hard and this will cost to you a lot of headache while you find a solution (if it exists). As I can see, sequence seems to be casual, there isn't any

math behind it. The only thing to do if you don't want to use tools, is to try with bit 0 and 1 and be lucky.

copper member
Activity: 1330
Merit: 899
🖤😏
Using tools will take much more time, this needs to be solved by math and by hands.😉
legendary
Activity: 1568
Merit: 6660
bitcoincleanup.com / bitmixlist.org
Just converting a public key to an address will not help here, as there are already specialized programs in C++ and other faster, multithreaded languages, that can do that.

You don't need to compute base58 or bech32 in order to compare the addresses. Comparing the address hashes directly also works and saves an extra (expensive) computation, such that most of the work is pooled into making an efficient SHA256 algorithm OR optimizing the process of making private keys from public keys.

Probably time for me to stop by Cornell university's internet arxiv for some more finite field research papers.

uncomprewssed key for 66 1Pir2jKv35rpYgUh8dUxNL6JTTm6d7HX6U

Therefore, the uncompressed public key for the P2PKH address "1Pir2jKv35rpYgUh8dUxNL6JTTm6d7HX6U" is:  04ce42244bfa7731d7088ee2f20a72b0dd834a723d2b8f563bc58cfa5d5dc4911a2f8aae5c5b

could be wrong could be wright do what you want with the info im just tossing ideas.



does this look like itd do anything am i in the right path?

Code:
import hashlib
import base58

# uncompressed public key
public_key = "04ce42244bfa7731d7088ee2f20a72b0dd834a723d2b8f563bc58cfa5d5dc4911a2f8aae5c5b"

# step 1: compute SHA-256 hash of public key
sha256 = hashlib.sha256(bytes.fromhex(public_key)).digest()

# step 2: compute RIPEMD-160 hash of SHA-256 result
ripemd160 = hashlib.new("ripemd160")
ripemd160.update(sha256)
hash160 = ripemd160.digest()

# step 3: add version byte (0x00 for MainNet or 0x6f for TestNet) to hash160
version = "00"
hash160_version = version + hash160.hex()

# step 4: compute double-SHA-256 checksum of hash160_version
checksum = hashlib.sha256(hashlib.sha256(bytes.fromhex(hash160_version)).digest()).digest()

# step 5: append the first 4 bytes of the checksum to hash160_version
btc_address_hex = hash160_version + checksum.hex()[:8]

# step 6: encode the result using Base58Check encoding
btc_address = base58.b58encode(bytes.fromhex(btc_address_hex)).decode()

print("BTC public address: ", btc_address)
member
Activity: 185
Merit: 15
Two things you should never abandon: Family & BTC
WOW!
Somebody (maybe the owner) increased the unsolved puzzles prizes again by x10 😱
Now the puzzle #66 prize is 6.6 BTC, #67 is 6.7 BTC... and so on .... puzzle # 160 prize is 16 BTC now
👍🏼🥳

OMG !!! 😱😱😱😱😱 Yesterday, I was thinking that I should give up now, but this has turned into a great motivation. need only one message from the puzzle creator in this situation, it would be the icing on the cake 👍🏼 👍🏼
RIGHT!!!!!!! the man the myth the legend "the creator" would you kindly please give us a hint of some sort a tool tool to look into something anything

Lol he did more than enough. The cracking method is on us to figure out.
member
Activity: 275
Merit: 20
the right steps towerds the goal
WOW!
Somebody (maybe the owner) increased the unsolved puzzles prizes again by x10 😱
Now the puzzle #66 prize is 6.6 BTC, #67 is 6.7 BTC... and so on .... puzzle # 160 prize is 16 BTC now
👍🏼🥳

OMG !!! 😱😱😱😱😱 Yesterday, I was thinking that I should give up now, but this has turned into a great motivation. need only one message from the puzzle creator in this situation, it would be the icing on the cake 👍🏼 👍🏼
member
Activity: 185
Merit: 15
Two things you should never abandon: Family & BTC
There's the math comes again,

66 Bit puzzle  Kiss


the total number of keys in this range can be calculated by taking 16 to the power of 17 (i.e., 16^17), which is equal to 295,147,905,179,352,825,856.

So, there are a total of 295,147,905,179,352,825,856 keys in the given range.

If you have a speed of 343 million keys per second (Mkeys/s), the result's,

Total number of keys = 295,147,905,179,352,825,856
Time = Total number of keys / Speed

Substituting the values, we get:

Time = 295,147,905,179,352,825,856 / 343,000,000
Time = 859,488,566,568 seconds

Therefore, it would take approximately 859,488,566,568 seconds (or about 27,247 years) to generate all the possible keys in the given range at a speed of 343 million keys per second. This is a very long time and may not be practically feasible.

what if i have 5000 mkeys to scan those number of key ?
it would take approximately 59,029,581,036 seconds (or about 1,872 years) to generate all the possible keys in the given range at a speed of 5000 million keys per second, still lame haha  Grin



and there's other scenario to win the ticket haha

For a 66-bit puzzle, the key range would be from 0 to 2^66 - 1, which is:

0 to 7,922,816,251,426,433

The number of keys in this range is:

2^66 = 73,786,976,294,838,206,464

At a scanning speed of 343 Mkey/s, the time it would take to scan this entire range would be:

Time = Number of Keys / Keys per Second

Time = (2^66) / 343000000 = 215625224 seconds
= 3593754 minutes
= 59896 hours
= 2495 days
≈ 6.84 years

So, it would take approximately 6.84 years to scan the entire key range of a 66-bit puzzle at the given scanning speed.

If we were to randomly hit the private key in the middle of the range, then we would only need to scan half of the key range, which is:

2^65 = 36,893,488,147,419,103,232

Using the same formula as above, the time it would take to scan half of the key range would be:

Time = (2^65) / 343000000 = 107812612 seconds
= 1796877 minutes
= 29948 hours
= 1248 days
≈ 3.42 years

So, if we randomly hit the private key in the middle of the key range, it would take approximately 3.42 years to scan half of the key range at the given speed of 343 Mkey/s.


btw, my close friend is neat, he's building something on ASICs L3+ to give a proof to scan the puzzle for mean time haha.
let's see..



Wrong, the entire range of puzz #66 contains 36.8 million trillion keys, half that range is obviously 18.44 million trillion keys.

Also the calc is misleading. With 1billion keys/sec, you need a little over 1000 years to scan the entire range sequentially.
member
Activity: 185
Merit: 15
Two things you should never abandon: Family & BTC
Let's do it one more time. With that much money at stake, we should grow a huge community.
Everyone thinks he will be the lucky one and prefers to avoid having to share the X10 is made to encourage people to pool their power to demonstrate that even with a collective effort the 66 67 68 and 69 will be difficult to obtain otherwise it won't find away
Also give a good boost to this puzzle which is now 8 years old

With collective effort, we can go as far as 90 bits. Then from there, Quantum code should step in. But that's another level.
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