Author

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 146. (Read 215539 times)

newbie
Activity: 25
Merit: 2
Anybody here familiar with Kangaroo? I have 1 stupid question. If let's say I put 1 million public address to search for private key and there's only 1 valid public address that fit the range. Will it take much more longer time to find the valid public key to get the private key? I tried just now with 100,000 public address, but the average time to solve shown unchanged.

I tried to put 1001 key with 1000 false public key and 1 puzzle 35 key. Why kangaroo can't solve it? Does that mean we can only put 1 public key at a time?

Looks like those who is searching for #66 are at 354d range, even myself also search at the same range. If my research correct, #66 range should be in between 354df - 358ae. Even this range will take ages to scan. LoL.
newbie
Activity: 8
Merit: 0
Hi, so  i found 2 address similar 11 digits of hash160 in extraction of bit 130 that means if you scan on this range and have the private key of those one of two addresses we will split the 13 BTC Smiley ,

1st

1AoNf67iZUmz1Ck9eefUfTzWQJMN5pgcPG
02113ba90a97c020ade0f3d8d0369981a723fe2bc4352815df22dd3eafae13c5a5
6b7e582a29a549cc60b591279a963f02eff02f99

pk:00000000000000000000000000000000000de4cfcadfc034c963dd053d719e88

address to search in :116 bit range
1AoNf67iZrwMSYPTDbk3Sh1yXJCARQbD7a
039d0a0241abe2411f64b4f6d29f2e1b6c837b26b6bdded577c3fc93574d3d735c
6b7e582a29a7601b79761f9f153c300c3d988231
range: 804cfcadfc034c963dd053d719e88:fffffcadfc034c963dd053d719e88


2nd
1PtStkm2bWKryHVduVjciPUxVx9UeDcCXG
02dc52ba09b16bc5cbd25aca7c82dd924f81cd31ecf29ecb264fa2cc45393728b9
fb0d9859584e68c24c1698eea4d05d2822fe4b70
pk: ad0f6ba584b355089cf6ce9cc9774

address to search in 116 bit range
1PtStkm2bLM7EK7g1rnTLBxu6aLouVuULV
03ef06cec3b3e35f68ba78618e5a5cf8663cc1a3b685dcfd197c1c0030530b1293
fb0d9859584d782df3fe652d2da5a21c30f137f9
range: 804cfcadfc034c963dd053d719e88:fffffcadfc034c963dd053d719e88




if the pk found of one of those address we will split the prize.
they have the same 11 digits of hash160 that means maybe 90 pourcent is in that range



up, anyone interessing Huh its same 11 digits hash160 so who can scan this range 116 and split the 13 btc ?

Hello! Can you find out how you get addresses with the same initial prefixes in which you say are in the same range ?!, what you do, subtract or add, can you explain! I saw your last post where you also found the same starting address prefixes which are both in the #66 range.
Even if they don't matter, just show how you calculate similar addresses and get their private key as well, show here!



1 - Convert the private key from hex to bytes
00000000000000000000000000000000000000000000000354d62e5f7a0d2eb2
b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x03T\xd6._z\r.\xb2'

2 - Create a signing key from the private key bytes using the SECP256k1 elliptic curve


3 - Get the corresponding public key
02b21a6b1590b145841a0dabbe71ea01e29ed60f0e468cff36445a9c92eb3a6375
VerifyingKey.from_string(b'\x02\xb2\x1ak\x15\x90\xb1E\x84\x1a\r\xab\xbeq\xea\x01\xe2\x9e\xd6\x0f\x0eF\x8c\xff6DZ\x9c\x92\xeb:cu', SECP256k1, sha1)

4 - Serialize the public key in compressed format (33 bytes)
b'\x02\xb2\x1ak\x15\x90\xb1E\x84\x1a\r\xab\xbeq\xea\x01\xe2\x9e\xd6\x0f\x0eF\x8c\xff6DZ\x9c\x92\xeb:cu'
02b21a6b1590b145841a0dabbe71ea01e29ed60f0e468cff36445a9c92eb3a6375

5 - Calculate the SHA-256 hash of the public key
b'\t\xb4\x87?D\'I\xef>\x86\xc7\x1d\x92\x86\xb1"\xa9\xdd\xf9v%\xa0\x03X\x88\xfb\x96%F\x0e\'\x16'
09b4873f442749ef3e86c71d9286b122a9ddf97625a0035888fb9625460e2716

6 - Calculate the RIPEMD-160 hash of the SHA-256 hash

b' \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb'
20d45a6a7625334252c8318a87ed303533c1c7bb

7 - Add the version byte (0x00 for mainnet) to the RIPEMD-160 hash
b'\x00'

8 - Extended RIPEMD-160 Hash
b'\x00 \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb'
0020d45a6a7625334252c8318a87ed303533c1c7bb

9 - Calculate the double SHA-256 checksum
b'\x01\x02l\xf90\xf6N\x8f\xeb\xca\xc8\xc2\x15\xd9Q\xb8i))\xb0\xce:\xb1\xba\x9e\xa4\xa1\x07_\x05\xe2\xa2'
b'\x01\x02l\xf9'

10 - Checksum: 01026cf9

11 - Append the checksum to the extended RIPEMD-160 hash
b'\x00 \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb\x01\x02l\xf9'
0020d45a6a7625334252c8318a87ed303533c1c7bb01026cf9

12 - Address (with checksum)
0020d45a6a7625334252c8318a87ed303533c1c7bb01026cf9

13 - Convert the bytes to a base58-encoded Bitcoin address
13zb1hQbWVnN3ag9GNS2vCraT8PQJDjVdr

provide an alternative, more straightforward method, if available instead of this ?

bro, how did you get the private key with the address 13ZB1HQBWNN3AG9GNS2VCRAT8PQJDJVDR, or did you happen to have it?If you compare it with the address #66 of the puzzle, then they have the same prefixes at the beginning of the address, also in the hash160. If you compare by range that this one and that one, both are in the #66 range. Maybe kalos15btc is right about something?!.



1 - Convert the private key from hex to bytes
00000000000000000000000000000000000000000000000354d62e5f7a0d2eb2
b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x03T\xd6._z\r.\xb2'

2 - Create a signing key from the private key bytes using the SECP256k1 elliptic curve


3 - Get the corresponding public key
02b21a6b1590b145841a0dabbe71ea01e29ed60f0e468cff36445a9c92eb3a6375
VerifyingKey.from_string(b'\x02\xb2\x1ak\x15\x90\xb1E\x84\x1a\r\xab\xbeq\xea\x01\xe2\x9e\xd6\x0f\x0eF\x8c\xff6DZ\x9c\x92\xeb:cu', SECP256k1, sha1)

4 - Serialize the public key in compressed format (33 bytes)
b'\x02\xb2\x1ak\x15\x90\xb1E\x84\x1a\r\xab\xbeq\xea\x01\xe2\x9e\xd6\x0f\x0eF\x8c\xff6DZ\x9c\x92\xeb:cu'
02b21a6b1590b145841a0dabbe71ea01e29ed60f0e468cff36445a9c92eb3a6375

5 - Calculate the SHA-256 hash of the public key
b'\t\xb4\x87?D\'I\xef>\x86\xc7\x1d\x92\x86\xb1"\xa9\xdd\xf9v%\xa0\x03X\x88\xfb\x96%F\x0e\'\x16'
09b4873f442749ef3e86c71d9286b122a9ddf97625a0035888fb9625460e2716

6 - Calculate the RIPEMD-160 hash of the SHA-256 hash

b' \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb'
20d45a6a7625334252c8318a87ed303533c1c7bb

7 - Add the version byte (0x00 for mainnet) to the RIPEMD-160 hash
b'\x00'

8 - Extended RIPEMD-160 Hash
b'\x00 \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb'
0020d45a6a7625334252c8318a87ed303533c1c7bb

9 - Calculate the double SHA-256 checksum
b'\x01\x02l\xf90\xf6N\x8f\xeb\xca\xc8\xc2\x15\xd9Q\xb8i))\xb0\xce:\xb1\xba\x9e\xa4\xa1\x07_\x05\xe2\xa2'
b'\x01\x02l\xf9'

10 - Checksum: 01026cf9

11 - Append the checksum to the extended RIPEMD-160 hash
b'\x00 \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb\x01\x02l\xf9'
0020d45a6a7625334252c8318a87ed303533c1c7bb01026cf9

12 - Address (with checksum)
0020d45a6a7625334252c8318a87ed303533c1c7bb01026cf9

13 - Convert the bytes to a base58-encoded Bitcoin address
13zb1hQbWVnN3ag9GNS2vCraT8PQJDjVdr

provide an alternative, more straightforward method, if available instead of this ?

bro, how did you get the private key with the address 13ZB1HQBWNN3AG9GNS2VCRAT8PQJDJVDR, or did you happen to have it?If you compare it with the address #66 of the puzzle, then they have the same prefixes at the beginning of the address, also in the hash160. If you compare by range that this one and that one, both are in the #66 range. Maybe kalos15btc is right about something?!.

I understand how it works, you don't have to answer.
member
Activity: 462
Merit: 24
[...]
You can use import secp256k1 as ice
https://github.com/iceland2k14/secp256k1
(You must have this function in the same folder where the command is executed for it to work.)
here's a fix I wrote months ago and requested a pull for that allows you to run icelands' library from any folder.
Thanks for the update.  Wink
copper member
Activity: 1330
Merit: 899
🖤😏
Nothing going on around these wood

We play with  numbers with at least 20 decimal places here.
If you could make one Giga (10^9) guesses per second, it would take:

10^20 / 10^9 = 10^11 seconds

This is equivalent to roughly 3.2 million years. So, even with an incredibly fast computer making a Giga guesses per second, it would take millions of years to guess a 20-decimal-place number.

We'd better start practicing crystal ball gazing to guess what the correct range of Puzzle 66 is Grin
According to my crystal ball, we'd only need 2339 years to completely scan the entire 66 bit range. Note that my crystal ball is uneducated and generates insensible results.😉
hero member
Activity: 630
Merit: 731
Bitcoin g33k
[...]
You can use import secp256k1 as ice
https://github.com/iceland2k14/secp256k1
(You must have this function in the same folder where the command is executed for it to work.)
here's a fix I wrote months ago and requested a pull for that allows you to run icelands' library from any folder.
newbie
Activity: 10
Merit: 0
Nothing going on around these wood
We'd better start practicing crystal ball gazing to guess what the correct range of Puzzle 66 is Grin
first sensible thought in last 10+ pages
member
Activity: 462
Merit: 24
Nothing going on around these wood

We play with  numbers with at least 20 decimal places here.
If you could make one Giga (10^9) guesses per second, it would take:

10^20 / 10^9 = 10^11 seconds

This is equivalent to roughly 3.2 million years. So, even with an incredibly fast computer making a Giga guesses per second, it would take millions of years to guess a 20-decimal-place number.

We'd better start practicing crystal ball gazing to guess what the correct range of Puzzle 66 is Grin
copper member
Activity: 1330
Merit: 899
🖤😏
provide an alternative, more straightforward method, if available instead of this ?

Skip steps 7 up to 13.



Nothing going on around these woods, lets make up new methods and try them to see if one of them works, here is something to work on.
End range :
400000000000000000000000000000000
Fake #130 key :

2c54a14a9556f03272bac1cd222396b57
Subtract 130 from end range : #2

13ab5eb56aa90fcd8d453e32dddc694a9
Subtract above #2 from 130 = #3 :

18a942952aade064e575839a44472d6ae
Subtract #3 from #2 = #4 :

4fde3dfc004d09758304567666ac4205
Multiply #4 by *4 = #5 :

13f78f7f0013425d60c1159d99ab10814
Subtract #5 from #2 = #6 :

4c30c9956a328fd37bd76abbcea736b

Figure out the rest and good luck. 😉
member
Activity: 462
Merit: 24
1 - Convert the private key from hex to bytes
00000000000000000000000000000000000000000000000354d62e5f7a0d2eb2
b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x03T\xd6._z\r.\xb2'

2 - Create a signing key from the private key bytes using the SECP256k1 elliptic curve


3 - Get the corresponding public key
02b21a6b1590b145841a0dabbe71ea01e29ed60f0e468cff36445a9c92eb3a6375
VerifyingKey.from_string(b'\x02\xb2\x1ak\x15\x90\xb1E\x84\x1a\r\xab\xbeq\xea\x01\xe2\x9e\xd6\x0f\x0eF\x8c\xff6DZ\x9c\x92\xeb:cu', SECP256k1, sha1)

4 - Serialize the public key in compressed format (33 bytes)
b'\x02\xb2\x1ak\x15\x90\xb1E\x84\x1a\r\xab\xbeq\xea\x01\xe2\x9e\xd6\x0f\x0eF\x8c\xff6DZ\x9c\x92\xeb:cu'
02b21a6b1590b145841a0dabbe71ea01e29ed60f0e468cff36445a9c92eb3a6375

5 - Calculate the SHA-256 hash of the public key
b'\t\xb4\x87?D\'I\xef>\x86\xc7\x1d\x92\x86\xb1"\xa9\xdd\xf9v%\xa0\x03X\x88\xfb\x96%F\x0e\'\x16'
09b4873f442749ef3e86c71d9286b122a9ddf97625a0035888fb9625460e2716

6 - Calculate the RIPEMD-160 hash of the SHA-256 hash

b' \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb'
20d45a6a7625334252c8318a87ed303533c1c7bb

7 - Add the version byte (0x00 for mainnet) to the RIPEMD-160 hash
b'\x00'

8 - Extended RIPEMD-160 Hash
b'\x00 \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb'
0020d45a6a7625334252c8318a87ed303533c1c7bb

9 - Calculate the double SHA-256 checksum
b'\x01\x02l\xf90\xf6N\x8f\xeb\xca\xc8\xc2\x15\xd9Q\xb8i))\xb0\xce:\xb1\xba\x9e\xa4\xa1\x07_\x05\xe2\xa2'
b'\x01\x02l\xf9'

10 - Checksum: 01026cf9

11 - Append the checksum to the extended RIPEMD-160 hash
b'\x00 \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb\x01\x02l\xf9'
0020d45a6a7625334252c8318a87ed303533c1c7bb01026cf9

12 - Address (with checksum)
0020d45a6a7625334252c8318a87ed303533c1c7bb01026cf9

13 - Convert the bytes to a base58-encoded Bitcoin address
13zb1hQbWVnN3ag9GNS2vCraT8PQJDjVdr

provide an alternative, more straightforward method, if available instead of this ?
Translated into Python3 :
Code:
import hashlib
import ecdsa
import base58

# Private key in hexadecimal format
private_key_hex = "00000000000000000000000000000000000000000000000354d62e5f7a0d2eb2"

# Convert private key from hex to bytes
private_key_bytes = bytes.fromhex(private_key_hex)

# Create a signing key from the private key bytes using SECP256k1 curve
signing_key = ecdsa.SigningKey.from_string(private_key_bytes, curve=ecdsa.SECP256k1)

# Get the corresponding public key in compressed format
compressed_public_key = signing_key.get_verifying_key().to_string("compressed")

# Calculate the SHA-256 hash of the compressed public key
sha256_hash = hashlib.sha256(compressed_public_key).digest()

# Calculate the RIPEMD-160 hash of the SHA-256 hash
ripemd160_hash = hashlib.new('ripemd160', sha256_hash).digest()

# Add the version byte (0x00 for mainnet) to the RIPEMD-160 hash
extended_ripe160_hash = b'\x00' + ripemd160_hash

# Calculate the double SHA-256 checksum
checksum = hashlib.sha256(hashlib.sha256(extended_ripe160_hash).digest()).digest()[:4]

# Append the checksum to the extended RIPEMD-160 hash
address_bytes = extended_ripe160_hash + checksum

# Convert the bytes to a base58-encoded Bitcoin address
bitcoin_address = base58.b58encode(address_bytes).decode()

print("Bitcoin Address:", bitcoin_address)

The process of deriving a Bitcoin address from a private key involves several cryptographic operations and encoding steps, so there isn't a significantly shorter method to achieve this without skipping any essential steps.

While it may be possible to write more concise code or create a custom function to encapsulate the process, the core steps themselves are fundamental to Bitcoin address generation.

So there is no way to speed up more through the code. Unfortunately. Grin


p.s.
You can use import secp256k1 as ice
https://github.com/iceland2k14/secp256k1
(You must have this function in the same folder where the command is executed for it to work.)
This is the a custom function which encapsulate the process above :
Code:
import secp256k1 as ice

private_key_hex = "00000000000000000000000000000000000000000000000354d62e5f7a0d2eb2"
dec = int(private_key_hex, 16)
bitcoin_address = ice.privatekey_to_address(0, True, dec)
print("Bitcoin Address:", bitcoin_address)
or:
Code:
python3 -c "import secp256k1 as ice; private_key_hex = '00000000000000000000000000000000000000000000000354d62e5f7a0d2eb2'; dec = int(private_key_hex, 16); bitcoin_address = ice.privatekey_to_address(0, True, dec); print('Bitcoin Address:', bitcoin_address)"

Bitcoin Address: 13zb1hQbWVnN3ag9GNS2vCraT8PQJDjVdr

But this method is not much faster for me. It all depends on how it is used.
legendary
Activity: 3304
Merit: 3094
1 - Convert the private key from hex to bytes
00000000000000000000000000000000000000000000000354d62e5f7a0d2eb2
b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x03T\xd6._z\r.\xb2'

...

8 - Extended RIPEMD-160 Hash
b'\x00 \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb'
0020d45a6a7625334252c8318a87ed303533c1c7bb

...

13 - Convert the bytes to a base58-encoded Bitcoin address
13zb1hQbWVnN3ag9GNS2vCraT8PQJDjVdr

provide an alternative, more straightforward method, if available instead of this ?

Your process to generate the address from the hex private key is right, but when you are attacking private keys with brute force there are tons of ways to do it.

People do it the same way as you because they bruteforce the Hex Pk, ...000001, ....000002, etc. But if you want to know an alternate method let me explain one of my favorites:

You can start from the Address: 13zb1hQbWVnN3ag9GNS2vCraT8PQJDjVdr and get the RIPEMD-160 (20d45a6a7625334252c8318a87ed303533c1c7bb).

Sites like privatekeys.pw provide that Hash 160 as you can see in the next link: https://privatekeys.pw/address/bitcoin/13zb1hQbWVnN3ag9GNS2vCraT8PQJDjVdr

And if you want to get that RIPEMD-160 code by yourself you can use the next python script.

Code:
python3 -c "import binascii, hashlib, base58; hash160 = binascii.hexlify(base58.b58decode_check(b'13zb1hQbWVnN3ag9GNS2vCraT8PQJDjVdr')).decode()[2:]; print(hash160)"

Once you have the hash160 you can use tools like brainflayer to get the Private key from the hash160.

I want to be clear, is not the best way but is just a different way to do it.
copper member
Activity: 1330
Merit: 899
🖤😏

I see a not very smart person who does not understand either mathematics or modular mathematics, but with all this with a crown on his head. your every post highlights your low level of intelligence.
What took you so long to realize this? I have been posting for more than 8, 9 months, and you just figured I'm  uneducated with no knowledge of math and EC. 🤣

Off topic,  but still. Guys, where diagran is trying to take you does not make any sense.

Not really off topic, if someone like me is misguided and is dragging others on to the wrong path, it's good for all to warn people. I just wonder who are these Guys you mentioned? Are they mentally retarded to a level where they can't figure out on their own that I'm useless and my posts are all garbage? 


there is no difference to look for 1 key in range 2^130 or 4 in the range 2^128. in terms of the time spent by the currently known tools, you will spend the same amount of time.
The post you quoted has nothing to do with what you said, I don't know where you got 2^130 or 2^128, 1 key, 4 key, what are those?

You see, usually when I communicate with smart people, they immediately realize I know nothing, and since they are smart with high levels of intelligence, they'd try  to correct my mistakes and then teach me the right stuff, because they enjoy teaching other people, it also increases their knowledge after sharing what they already know. But a butthurt who thinks he knows better than everyone else hides behind a newbie account, insulting others, this is their joy, it also increases their butthurt pain and burn. ( this is how the universe works, action=reaction, share knowledge, gain knowledge, share/help with money, gain more money, spread hatred, absorb hatred ),



Where was I wrong exactly, implying my lack of understanding about math/ECC? Show me, then lets fix my mistakes together.

~digaran
hero member
Activity: 630
Merit: 731
Bitcoin g33k
offtopic
member
Activity: 272
Merit: 20
the right steps towerds the goal
1 - Convert the private key from hex to bytes
00000000000000000000000000000000000000000000000354d62e5f7a0d2eb2
b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x03T\xd6._z\r.\xb2'

2 - Create a signing key from the private key bytes using the SECP256k1 elliptic curve


3 - Get the corresponding public key
02b21a6b1590b145841a0dabbe71ea01e29ed60f0e468cff36445a9c92eb3a6375
VerifyingKey.from_string(b'\x02\xb2\x1ak\x15\x90\xb1E\x84\x1a\r\xab\xbeq\xea\x01\xe2\x9e\xd6\x0f\x0eF\x8c\xff6DZ\x9c\x92\xeb:cu', SECP256k1, sha1)

4 - Serialize the public key in compressed format (33 bytes)
b'\x02\xb2\x1ak\x15\x90\xb1E\x84\x1a\r\xab\xbeq\xea\x01\xe2\x9e\xd6\x0f\x0eF\x8c\xff6DZ\x9c\x92\xeb:cu'
02b21a6b1590b145841a0dabbe71ea01e29ed60f0e468cff36445a9c92eb3a6375

5 - Calculate the SHA-256 hash of the public key
b'\t\xb4\x87?D\'I\xef>\x86\xc7\x1d\x92\x86\xb1"\xa9\xdd\xf9v%\xa0\x03X\x88\xfb\x96%F\x0e\'\x16'
09b4873f442749ef3e86c71d9286b122a9ddf97625a0035888fb9625460e2716

6 - Calculate the RIPEMD-160 hash of the SHA-256 hash

b' \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb'
20d45a6a7625334252c8318a87ed303533c1c7bb

7 - Add the version byte (0x00 for mainnet) to the RIPEMD-160 hash
b'\x00'

8 - Extended RIPEMD-160 Hash
b'\x00 \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb'
0020d45a6a7625334252c8318a87ed303533c1c7bb

9 - Calculate the double SHA-256 checksum
b'\x01\x02l\xf90\xf6N\x8f\xeb\xca\xc8\xc2\x15\xd9Q\xb8i))\xb0\xce:\xb1\xba\x9e\xa4\xa1\x07_\x05\xe2\xa2'
b'\x01\x02l\xf9'

10 - Checksum: 01026cf9

11 - Append the checksum to the extended RIPEMD-160 hash
b'\x00 \xd4Zjv%3BR\xc81\x8a\x87\xed053\xc1\xc7\xbb\x01\x02l\xf9'
0020d45a6a7625334252c8318a87ed303533c1c7bb01026cf9

12 - Address (with checksum)
0020d45a6a7625334252c8318a87ed303533c1c7bb01026cf9

13 - Convert the bytes to a base58-encoded Bitcoin address
13zb1hQbWVnN3ag9GNS2vCraT8PQJDjVdr

provide an alternative, more straightforward method, if available instead of this ?
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Now ignore whatever you see before the dot "." Just look at whatever you see after the dot.😉 chop chop and good luck.
I see a not very smart person who does not understand either mathematics or modular mathematics, but with all this with a crown on his head. your every post highlights your low level of intelligence.
Ps
it was offtopic. but still. Guys, where diagran is trying to take you does not make any sense. for subtraction, division, multiplication, etc. even if it reduces the search range for a key, it proportionally increases the number of searched keys. it all makes no sense.
there is no difference to look for 1 key in range 2^130 or 4 in the range 2^128. in terms of the time spent by the currently known tools, you will spend the same amount of time.
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It would be great, also note that you should calculate the result private key mod n, your current script returns the correct public key but not private key.

Only thing it needs is to accept public key as target.
Amazing coding btw.👍, it's really joyful to see people are using their brains rather than giving up and quitting.
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New ideas will be criticized and then admired.
I'm not sure if it helps, but for 130 bits, we only need to divide the point by 2, 101 times. The issue is figuring out where to subtract 1 to avoid floating-point errors.
the problem of dividing by 2 is that you need 2**101 pubkeys (according to your approach).
update:
dividing by 3 you need 3**14 pubkeys, to reduce puzzle130 down to the equivalent of puzzle 105.

So reducing 25 bits, if dividing by 2 we need 2^25 public keys with 1 one of them to be the correct result, but dividing by 3 we need 3^14 keys, one of them would be correct, how did you calculate it?

I hope your scrip saves the results to a file, because I only see print, are we supposed to print thousands of keys on screen? 😅

He says he wants to divide 101 times, and since he doesn't know the pk he needs to do pk/2 and ( pk-1)/2 which results in 2x2x2x2.... 101 times (2**101)
by 3 is 3x3x3x3.... 14 times (3**14) equivalent to puzzle 105
Ok, can you tell us where to put our target public key in your script, where to put the number of times etc, please?

This script is just theory, not done to create massive continuous divisions, when I have time I'll do it.

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I'm not sure if it helps, but for 130 bits, we only need to divide the point by 2, 101 times. The issue is figuring out where to subtract 1 to avoid floating-point errors.
the problem of dividing by 2 is that you need 2**101 pubkeys (according to your approach).
update:
dividing by 3 you need 3**14 pubkeys, to reduce puzzle130 down to the equivalent of puzzle 105.

So reducing 25 bits, if dividing by 2 we need 2^25 public keys with 1 one of them to be the correct result, but dividing by 3 we need 3^14 keys, one of them would be correct, how did you calculate it?

I hope your scrip saves the results to a file, because I only see print, are we supposed to print thousands of keys on screen? 😅

He says he wants to divide 101 times, and since he doesn't know the pk he needs to do pk/2 and ( pk-1)/2 which results in 2x2x2x2.... 101 times (2**101)
by 3 is 3x3x3x3.... 14 times (3**14) equivalent to puzzle 105
Ok, can you tell us where to put our target public key in your script, where to put the number of times etc, please?
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