Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 183.

MisterZz

newbie

Activity: 11

Merit: 1

Quote from: digaran on September 15, 2023, 08:00:44 AM

Quote from: Kostelooscoin on September 15, 2023, 06:17:57 AM

Hello, I would like to know if it is possible to divide 2 public keys between them.
ex: 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 / 02c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 with "secp256k1 as ice" or even something else?

I can give you the result for your example, here it is :
Pub =
0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
Private =
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1

But if we don't know the private key of at least one of the points, we can never directly reach any meaningful result.
If we don't know k for one of the points, how can an algorithm/library know it?

There are ways to indirectly divide 2 points by using a k greater than 1 and smaller than the other. It would be nonsense/off topic if I try to explain it, so I just save everybody the headache and cut the nonsense.😉

Quote from: MisterZz on September 14, 2023, 02:55:38 PM

He is learning maths and programming, so I think is not a waste of time..

Hello to everyone

Agreed, we should never stop learning, for example learning how to quote someone properly, lol I mean OMG, look the mess you posted.
Jokes aside, welcome to the jungle.🤝

LOL I havn’t posted on forums in a long time, since AIO’s time I would say. Give me sometime 😌
Something I had made 😁
https://i.ibb.co/8m9sHPN/puzzle.png

mcdouglasx

member

Activity: 239

Merit: 53

New ideas will be criticized and then admired.

Quote from: albert0bsd on September 15, 2023, 08:37:05 AM

Quote from: mcdouglasx on September 15, 2023, 07:52:57 AM

but it is excellent, even better than C for testing. of new ideas,

I agree with this, personally i think that testing and proof of concept are the only thing that python is good for. But once that the proof of concept is proven to be useful or good then is time to move to another language. If speed is determining for the algorithm

Quote from: mcdouglasx on September 15, 2023, 07:52:57 AM

the fact that you think that there are no new things is simply the result of your own mental fatigue

What F.. mental Fatigue? I already shared here a lot of programs, ideas, users share with me their ideas on telegram and there is nothing new...

What i am tryting to said is if you have a vague idea without math background or without enough testing, then OPEN a new thread discuss their idea there and if the idea was proven to be good, then link that new topic here sharing a brief sumary of it and how to use.

What I want to say is that information should not be repressed, if information was repressed you could not have made your software, your software is made based on the work of others, like everything in life, one idea leads to another and so on until it is put to practical use. If the basic scripts bother you, ignore them (if you live in the city and the pedestrians bother you, you cannot eliminate the pedestrians, the solution is to move out of the city) and come back later, but we cannot expect others to have our own knowledge and denigrate them for not Know, we all ignore something in life.

mcdouglasx

member

Activity: 239

Merit: 53

New ideas will be criticized and then admired.

Quote from: digaran on September 15, 2023, 08:00:44 AM

Quote from: Kostelooscoin on September 15, 2023, 06:17:57 AM

Hello, I would like to know if it is possible to divide 2 public keys between them.
ex: 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 / 02c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 with "secp256k1 as ice" or even something else?

I can give you the result for your example, here it is :
Pub =
0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
Private =
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1

But if we don't know the private key of at least one of the points, we can never directly reach any meaningful result.
If we don't know k for one of the points, how can an algorithm/library know it?

The division does not depend directly on the private keys, each public key is itself the representation of a number.
There are ten (10) numbers represented by these symbols.
0123456789
in hex it would be 16
0123456789abcdef

in public key are
57896044618658097711785492504343953926418782139537452191302581570759080747168 (including decimals, although these are not valid for bitcoin, they exist and this is what makes division difficult)
represented differently (public keys), and this is the complicated part of the matter.

albert0bsd

hero member

Activity: 862

Merit: 662

Quote from: mcdouglasx on September 15, 2023, 07:52:57 AM

but it is excellent, even better than C for testing. of new ideas,

I agree with this, personally i think that testing and proof of concept are the only thing that python is good for. But once that the proof of concept is proven to be useful or good then is time to move to another language. If speed is determining for the algorithm

Quote from: mcdouglasx on September 15, 2023, 07:52:57 AM

the fact that you think that there are no new things is simply the result of your own mental fatigue

What F.. mental Fatigue? I already shared here a lot of programs, ideas, users share with me their ideas on telegram and there is nothing new...

What i am tryting to said is if you have a vague idea without math background or without enough testing, then OPEN a new thread discuss their idea there and if the idea was proven to be good, then link that new topic here sharing a brief sumary of it and how to use.

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

Quote from: Kostelooscoin on September 15, 2023, 06:17:57 AM

Hello, I would like to know if it is possible to divide 2 public keys between them.
ex: 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 / 02c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 with "secp256k1 as ice" or even something else?

I can give you the result for your example, here it is :
Pub =
0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
Private =
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1

But if we don't know the private key of at least one of the points, we can never directly reach any meaningful result.
If we don't know k for one of the points, how can an algorithm/library know it?

There are ways to indirectly divide 2 points by using a k greater than 1 and smaller than the other. It would be nonsense/off topic if I try to explain it, so I just save everybody the headache and cut the nonsense.😉

Quote from: MisterZz on September 14, 2023, 02:55:38 PM

He is learning maths and programming, so I think is not a waste of time..

Hello to everyone

Agreed, we should never stop learning, for example learning how to quote someone properly, lol I mean OMG, look the mess you posted.
Jokes aside, welcome to the jungle.🤝

mcdouglasx

member

Activity: 239

Merit: 53

New ideas will be criticized and then admired.

Quote from: albert0bsd on September 13, 2023, 06:24:16 PM

Personally i believe that this post just lost its purpose there is a lot of non-sense post or just some offtopic (Some moderator should close it)

A lot of discutions if some slowly python code do this or that, if you have something substantial to contribute to that discussion, it would be best to start a new thread.

In the field of the search of the solucion of this challenge there is no new algorithms or programs. We can resume the progress of the last year in the next.

- Puzzle/challenge 64 solve by Unknown a year ago ‎2022-09-09 privatekey f7051f27b09112d4
- Puzzle/challenge 120 solve by unknown on ‎2023-02-27 and the private key remains unknown.
- Puzzle/challenge 125 solve by unknown on ‎2023-07-09 and the private key remains unknown.

And there is not clue of who solve those and what hardware they use, the wallet 3Emiwzxme7Mrj4d89uqohXNncnRM15YESs is still untouch

Next challenges to be solve are 66 bits with just brute force/pool and puzzle 130 with kangaroo/pool Huh

Best idea to solve puzzle 66 proposed on this thread is giving less priotiry to some repeated patterns. (But the expected time that the user give was ridiculus)

And thats all.

I really don't share anything you say, if this post is deleted then there would only be more information scattered everywhere, according to what you say about python, you are right, it is slower than C, but it is excellent, even better than C for testing. of new ideas, the fact that you think that there are no new things is simply the result of your own mental fatigue, innovative ideas will always continue to exist, and yes, there is spam, I can't deny it, but on what website? there is no spam?

Kostelooscoin

member

Activity: 206

Merit: 16

Hello, I would like to know if it is possible to divide 2 public keys between them.
ex: 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 / 02c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 with "secp256k1 as ice" or even something else?

nomachine

member

Activity: 503

Merit: 38

Quote from: albert0bsd on September 13, 2023, 06:24:16 PM

Personally i believe that this post just lost its purpose there is a lot of non-sense post or just some offtopic (Some moderator should close it)

A lot of discutions if some slowly python code do this or that, if you have something substantial to contribute to that discussion, it would be best to start a new thread.

I think that you're right. That the further story is pointless. Regarding the speed of python programs. All are fast enough if someone knows the approximate range and public key. Wink

MisterZz

newbie

Activity: 11

Merit: 1

Quote from: kalos15btc on September 13, 2023, 04:53:17 PM

Quote from: Dexed on September 12, 2023, 03:21:56 PM

Quote from: BurtW on June 08, 2019, 10:11:23 AM

Quote from: mrxtraf on June 08, 2019, 06:49:04 AM

By the way, and you know why their puzzle addresses were removed from the 161st?
My thought is that this is the limit of public addresses. That is the further search in the theory goes on a circle.

Because all the funds from 161 through 255 were redundant (Bitcoin addresses only use 160 bits) the author of the puzzle moved all the funds from the 160-255 ranged down into the 1-160 range. It is explained up thread.

Is it? I thought it uses almost 256 bits (the last possible private key is FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140)

OH, HIII, Good morning

Quote from: digaran on September 12, 2023, 05:20:10 PM

Quote from: lordfrs on September 03, 2023, 04:10:50 AM

Quote from: digaran on September 02, 2023, 05:45:02 AM

Quote from: mcdouglasx on August 20, 2023, 09:30:52 AM

Code:

from sympy import mod_inverse
import secp256k1 as ice
pub=ice.pub2upub('Here Compressed Public Key')
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))

def ters(Qx,Scalar):
   ScalarBin = str(bin(Scalar))[2:]
   le=len(ScalarBin)
   for i in range (1,le+1):
   if ScalarBin[le-i] == "0":
   Qx=ice.point_multiplication(k,Qx)
   else:
   Qx=ice.point_addition(Qx,neg1)
   Qx=ice.point_multiplication(k,Qx)
   return ice.point_to_cpub(Qx)

for x in range(1,65536):
   f= (ters(pub,x))
   data= open(“halfpub.txt”,”a”)
   data.write(f+”\n”)
   data.close()

Note this is where you decide how many bits should be reduced
for x in range(1,65536):
For example reducing 26 bits requires 67108864 to be generated, 1 of them is the correct 26 bit reduced key.

Will you release for public to use?

Yes this is my code, I can write bit reduction code in another way if you want. If you want, you can reduce bits by subtraction.

I'm developing a new algorithm for downgrading from a known bit range to the bit range I want. When I complete it by giving only 1 correct pubkey, I will share it here.

@lordfrs, I was thinking, why do we need to keep all the 65556 keys if we are reducing 16 bits? One other thing, if we multiply one of the keys from the end of the output file by 65556 we should see our target, correct? I haven't thought about on how to calculate, since we are subtracting one dividing by 2, one result if multiplied by 2^16 should reach the target or close to target, so why not discarding 90% of the generated keys and start multiplying the remaining 10% by 2^16? Could you add such operation to your script?

I guess if we could operate on private keys instead of public keys using your script, we could reach some results, ( as a test of course )

To test with scalar only, I use this script :

Code:

from sympy import mod_inverse
N = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

def ters(scalar):
k = mod_inverse(2, N)
scalar_bin = bin(scalar)[2:

bro please stop replying with those comments copy past from chatgpt you are doing nothing except replying and talk too much, you are doing nothing for real men, stop it bro, :) thank you
[/quote]

He is learning maths and programming, so I think is not a waste of time..

Hello to everyone

_Counselor

member

Activity: 110

Merit: 61

Quote from: Dexed on September 14, 2023, 07:33:50 AM

Quote from: albert0bsd on September 13, 2023, 06:24:16 PM

- Puzzle/challenge 64 solve by Unknown a year ago ‎2022-09-09 privatekey f7051f27b09112d4

If the solver is unknown, where is this private key came from?

When the solver move reward from 64 puzzle, it public key has been revealed; after that, finding a key in such a small range is trivial.

Dexed

newbie

Activity: 16

Merit: 0

Quote from: albert0bsd on September 13, 2023, 06:24:16 PM

- Puzzle/challenge 64 solve by Unknown a year ago ‎2022-09-09 privatekey f7051f27b09112d4

If the solver is unknown, where is this private key came from?

Kamoheapohea

jr. member

Activity: 47

Merit: 12

gmaxwell creator of 1000 BTC puzzl + Pinapple fund

Quote from: GR Sasa on September 14, 2023, 03:37:21 AM

Just for more info:

The owner of "3Emiwzxme7Mrj4d89uqohXNncnRM15YESs" will cashout his coins when he finishes targeting other puzzles.

In the meantime, he's aiming for #130 right now, and probably after he solves it in the next months, he will cashout all his coins at once. (Unless he deceides again to go on for #135).

EDIT: Hello Satoshi, maybe you could motivate us by increasing the prize by 10x times again? We would be greatful. We know that you have the ability to do it, but we need your Motivation so programmers can wake up and continute their work.

Thank you, your friend GR Sasa

There is already more than enough motivation. I think he should withdraw all the remaining funds. He knows the practical limits now. There will be no suprises.

GR Sasa

member

Activity: 194

Merit: 14

Just for more info:

The owner of "3Emiwzxme7Mrj4d89uqohXNncnRM15YESs" will cashout his coins when he finishes targeting other puzzles.

In the meantime, he's aiming for #130 right now, and probably after he solves it in the next months, he will cashout all his coins at once. (Unless he deceides again to go on for #135).

EDIT: Hello Satoshi, maybe you could motivate us by increasing the prize by 10x times again? We would be greatful. We know that you have the ability to do it, but we need your Motivation so programmers can wake up and continute their work.

Thank you, your friend GR Sasa

albert0bsd

hero member

Activity: 862

Merit: 662

Personally i believe that this post just lost its purpose there is a lot of non-sense post or just some offtopic (Some moderator should close it)

A lot of discutions if some slowly python code do this or that, if you have something substantial to contribute to that discussion, it would be best to start a new thread.

In the field of the search of the solucion of this challenge there is no new algorithms or programs. We can resume the progress of the last year in the next.

- Puzzle/challenge 64 solve by Unknown a year ago ‎2022-09-09 privatekey f7051f27b09112d4
- Puzzle/challenge 120 solve by unknown on ‎2023-02-27 and the private key remains unknown.
- Puzzle/challenge 125 solve by unknown on ‎2023-07-09 and the private key remains unknown.

And there is not clue of who solve those and what hardware they use, the wallet 3Emiwzxme7Mrj4d89uqohXNncnRM15YESs is still untouch

Next challenges to be solve are 66 bits with just brute force/pool and puzzle 130 with kangaroo/pool Huh

Best idea to solve puzzle 66 proposed on this thread is giving less priotiry to some repeated patterns. (But the expected time that the user give was ridiculus)

And thats all.

kalos15btc

jr. member

Activity: 50

Merit: 1

Quote from: Dexed on September 12, 2023, 03:21:56 PM

Quote from: BurtW on June 08, 2019, 10:11:23 AM

Quote from: mrxtraf on June 08, 2019, 06:49:04 AM

By the way, and you know why their puzzle addresses were removed from the 161st?
My thought is that this is the limit of public addresses. That is the further search in the theory goes on a circle.

Because all the funds from 161 through 255 were redundant (Bitcoin addresses only use 160 bits) the author of the puzzle moved all the funds from the 160-255 ranged down into the 1-160 range. It is explained up thread.

Is it? I thought it uses almost 256 bits (the last possible private key is FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140)

OH, HIII, Good morning

Quote from: digaran on September 12, 2023, 05:20:10 PM

Quote from: lordfrs on September 03, 2023, 04:10:50 AM

Quote from: digaran on September 02, 2023, 05:45:02 AM

Quote from: mcdouglasx on August 20, 2023, 09:30:52 AM

Code:

from sympy import mod_inverse
import secp256k1 as ice
pub=ice.pub2upub('Here Compressed Public Key')
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))

def ters(Qx,Scalar):
   ScalarBin = str(bin(Scalar))[2:]
   le=len(ScalarBin)
   for i in range (1,le+1):
   if ScalarBin[le-i] == "0":
   Qx=ice.point_multiplication(k,Qx)
   else:
   Qx=ice.point_addition(Qx,neg1)
   Qx=ice.point_multiplication(k,Qx)
   return ice.point_to_cpub(Qx)

for x in range(1,65536):
   f= (ters(pub,x))
   data= open(“halfpub.txt”,”a”)
   data.write(f+”\n”)
   data.close()

Note this is where you decide how many bits should be reduced
for x in range(1,65536):
For example reducing 26 bits requires 67108864 to be generated, 1 of them is the correct 26 bit reduced key.

Will you release for public to use?

Yes this is my code, I can write bit reduction code in another way if you want. If you want, you can reduce bits by subtraction.

I'm developing a new algorithm for downgrading from a known bit range to the bit range I want. When I complete it by giving only 1 correct pubkey, I will share it here.

@lordfrs, I was thinking, why do we need to keep all the 65556 keys if we are reducing 16 bits? One other thing, if we multiply one of the keys from the end of the output file by 65556 we should see our target, correct? I haven't thought about on how to calculate, since we are subtracting one dividing by 2, one result if multiplied by 2^16 should reach the target or close to target, so why not discarding 90% of the generated keys and start multiplying the remaining 10% by 2^16? Could you add such operation to your script?

I guess if we could operate on private keys instead of public keys using your script, we could reach some results, ( as a test of course )

To test with scalar only, I use this script :

Code:

from sympy import mod_inverse
N = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

def ters(scalar):
   k = mod_inverse(2, N)
   scalar_bin = bin(scalar)[2:]
   result = 2
   for i in range(len(scalar_bin)):
   if scalar_bin[i] == '0':
   result = (result * k) % N
   else:
   result = ((result * k) % N + N - 1) % N
   return hex(result)[2:].zfill(64)

for x in range(1, 20):
   f = ters(x)
   print(f)

It was a lucky shot at AI generated code and it worked.😅
Btw, the first "result" in the script above, represents your target, I couldn't risk the AI to mess one thing she has done right. So result = 2, replace 2 with your own scalar in decimal.

bro please stop replying with those comments copy past from chatgpt you are doing nothing except replying and talk too much, you are doing nothing for real men, stop it bro,

thank you

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

Quote from: lordfrs on September 03, 2023, 04:10:50 AM

Quote from: digaran on September 02, 2023, 05:45:02 AM

Quote from: mcdouglasx on August 20, 2023, 09:30:52 AM

Code:

from sympy import mod_inverse
import secp256k1 as ice
pub=ice.pub2upub('Here Compressed Public Key')
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))

def ters(Qx,Scalar):
   ScalarBin = str(bin(Scalar))[2:]
   le=len(ScalarBin)
   for i in range (1,le+1):
   if ScalarBin[le-i] == "0":
   Qx=ice.point_multiplication(k,Qx)
   else:
   Qx=ice.point_addition(Qx,neg1)
   Qx=ice.point_multiplication(k,Qx)
   return ice.point_to_cpub(Qx)

for x in range(1,65536):
   f= (ters(pub,x))
   data= open(“halfpub.txt”,”a”)
   data.write(f+”\n”)
   data.close()

Note this is where you decide how many bits should be reduced
for x in range(1,65536):
For example reducing 26 bits requires 67108864 to be generated, 1 of them is the correct 26 bit reduced key.

Will you release for public to use?

Yes this is my code, I can write bit reduction code in another way if you want. If you want, you can reduce bits by subtraction.

I'm developing a new algorithm for downgrading from a known bit range to the bit range I want. When I complete it by giving only 1 correct pubkey, I will share it here.

@lordfrs, I was thinking, why do we need to keep all the 65556 keys if we are reducing 16 bits? One other thing, if we multiply one of the keys from the end of the output file by 65556 we should see our target, correct? I haven't thought about on how to calculate, since we are subtracting one dividing by 2, one result if multiplied by 2^16 should reach the target or close to target, so why not discarding 90% of the generated keys and start multiplying the remaining 10% by 2^16? Could you add such operation to your script?

I guess if we could operate on private keys instead of public keys using your script, we could reach some results, ( as a test of course )

To test with scalar only, I use this script :

Code:

from sympy import mod_inverse
N = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

def ters(scalar):
   k = mod_inverse(2, N)
   scalar_bin = bin(scalar)[2:]
   result = 2
   for i in range(len(scalar_bin)):
   if scalar_bin[i] == '0':
   result = (result * k) % N
   else:
   result = ((result * k) % N + N - 1) % N
   return hex(result)[2:].zfill(64)

for x in range(1, 20):
   f = ters(x)
   print(f)

It was a lucky shot at AI generated code and it worked.😅
Btw, the first "result" in the script above, represents your target, I couldn't risk the AI to mess one thing she has done right. So result = 2, replace 2 with your own scalar in decimal.

albert0bsd

hero member

Activity: 862

Merit: 662

Quote from: Dexed on September 12, 2023, 03:21:56 PM

Quote from: BurtW on June 08, 2019, 10:11:23 AM

Because all the funds from 161 through 255 were redundant (Bitcoin addresses only use 160 bits) the author of the puzzle moved all the funds from the 160-255 ranged down into the 1-160 range. It is explained up thread.

Is it? I thought it uses almost 256 bits (the last possible private key is FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140)

The maximum value of a Privatekey is the value that you pointed, what they actually want to said is that address is a reduction of the public key, it is a 160 bit hash, so if you are capable to fin a collision on a 160 bits hash, that mean that you are capable to find any collision, regarless that those private keys are up to 256 bits

Dexed

newbie

Activity: 16

Merit: 0

Quote from: BurtW on June 08, 2019, 10:11:23 AM

Quote from: mrxtraf on June 08, 2019, 06:49:04 AM

By the way, and you know why their puzzle addresses were removed from the 161st?
My thought is that this is the limit of public addresses. That is the further search in the theory goes on a circle.

Because all the funds from 161 through 255 were redundant (Bitcoin addresses only use 160 bits) the author of the puzzle moved all the funds from the 160-255 ranged down into the 1-160 range. It is explained up thread.

Is it? I thought it uses almost 256 bits (the last possible private key is FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140)

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

Kostelooscoin

member

Activity: 206

Merit: 16

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 183. (Read 244669 times)