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Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 254662 times)
digaran
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September 01, 2023, 01:42:38 AM
 #3341

Here is the solution for partnering with muscles ( those with hardware to scan ranges ) , before scanning, the one asking for the scan should give all his information to an escrow and then the one doing the scan should provide the key he has found to the escrow so that the escrow does the final key extraction and divide the prize according to contract.

Person who asks for the scan provides the key which was subtracted from puzzle key and also the offset which the muscle is going to search for, escrow can verify the authenticity of the keys by doing the subtraction himself. 😉

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lordfrs
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September 01, 2023, 09:00:43 AM
 #3342

@digaran indeed,, im about to breake puzzle 130 Smiley

----------------------------------------------------------------------------------------------------

Need someone to scan 89bit to 90

03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 # +

range
10005df8cf4d0f152b26859:49fd5df8cf4d0f152b26859
its easy range anyone have gpu can scan this in 15 min, we try this why not maybe im not wrong Smiley of course we split 13 btc Smiley


if im correct and we found that pk of this adress,, yes @digaran you can call me master of crypto  Grin




89-90 bits 20000000000000000000000:3ffffffffffffffffffffff

If you want to buy me a coffee

Btc = 3246y1G9YjnQQNRUrVMnaeCFrymZRgJAP7

Doge = DGNd8UTi8jVTVZ2twhKydyqicynbsERMjs
citb0in
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September 01, 2023, 09:17:59 AM
 #3343

@digaran indeed,, im about to breake puzzle 130 Smiley

----------------------------------------------------------------------------------------------------

Need someone to scan 89bit to 90

03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 # +

range
10005df8cf4d0f152b26859:49fd5df8cf4d0f152b26859
its easy range anyone have gpu can scan this in 15 min, we try this why not maybe im not wrong Smiley of course we split 13 btc Smiley


if im correct and we found that pk of this adress,, yes @digaran you can call me master of crypto  Grin




89-90 bits 20000000000000000000000:3ffffffffffffffffffffff

ok, I'm through, range scan done. Nothing found. Whats next?  Roll Eyes

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kalos15btc
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September 01, 2023, 10:58:15 AM
Last edit: September 01, 2023, 11:57:20 AM by kalos15btc
 #3344

@digaran indeed,, im about to breake puzzle 130 Smiley

----------------------------------------------------------------------------------------------------

Need someone to scan 89bit to 90

03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 # +

range
10005df8cf4d0f152b26859:49fd5df8cf4d0f152b26859
its easy range anyone have gpu can scan this in 15 min, we try this why not maybe im not wrong Smiley of course we split 13 btc Smiley


if im correct and we found that pk of this adress,, yes @digaran you can call me master of crypto  Grin






89-90 bits 20000000000000000000000:3ffffffffffffffffffffff

ok, I'm through, range scan done. Nothing found. Whats next?  Roll Eyes



where is the proof of the work §.?
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September 01, 2023, 03:48:28 PM
 #3345

I have no idea why so many are so obsessed about this adding and subtracting public key thingy. Maybe I know nothing much about it and makes me think that way. This adding/subtracting method in my opinion are totally different from Pollard Kangaroo idea or BSGS. From my understanding, by adding/subtracting public key, you can generate a bunch of valid public key, nothing wrong about it. But if 1 private key matching only 1 public key(of course it does), then where is the difference between adding/subtracting and private key bruteforcing? IMHO, it doesn't eliminate any possible private key combination at all. Please correct me if I am wrong.

because if you look for pk=1361129467683753853853498429727072845823
subtracting...

Code:
import secp256k1 as ice


target_public_key = "03e067911ebf6bacf87a8088ab9344c95843aed80b070eed09f9d947c98dfc0249"
target = ice.pub2upub(target_public_key)
num = 136114 # number of times.
sustract= 10000000000000000000000000000000000 #amount to subtract each time.
sustract_pub= ice.scalar_multiplication(sustract)
res= ice.point_loop_subtraction(num, target, sustract_pub)
for t in range (num+1):
    h= (res[t*65:t*65+65]).hex()
    hc= ice.to_cpub(h)
    data = open("data-base.txt","a")
    data.write(str(hc)+"\n")
    data.close()
one of the pub in the result will be:
pk=9467683753853853498429727072845823

and looking for this         1361129467683753853853498429727072845823
is not the same as this               9467683753853853498429727072845823

digaran
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September 01, 2023, 06:06:44 PM
 #3346

So any update on the progress of finding this key?
 First offset = 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710
Half of above =
0291001b0dc6e5a2628cb4698eb00a6fb7dbd276dc2b214795f2fe52e61243aa9b
Half of 130?
0337374e00a32eaf009e9946035c0e69085627b60a844637d2b958dd83bcfa4383
The following is the subtracted key from #130
Second offset =
03d99bb89e8db75d20b882f13f8086fb39221858fa211de0346c926a93ae259b3a
Half of above?
03a3dc00bf5f7e7eec691569c7f67a15d3cdbb3a9994c9a5ec1430cffdb622cf9f

Now subtract half of first offset from half of #130 to get half of second offset.

Second offset is known, we need to work on first offset's half, use -1 divide by 2 script to reduce 18 bits from it, you'll have millions of new offsets and one of them is the target, now divide the #130 range by 2, subtract 18 bits from it and use the new range as your search range, input those millions offset keys and search the range.

Don't just try blind searching.😉

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kalos15btc
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September 01, 2023, 06:26:14 PM
 #3347

So any update on the progress of finding this key?
 First offset = 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710
Half of above =
0291001b0dc6e5a2628cb4698eb00a6fb7dbd276dc2b214795f2fe52e61243aa9b
Half of 130?
0337374e00a32eaf009e9946035c0e69085627b60a844637d2b958dd83bcfa4383
The following is the subtracted key from #130
Second offset =
03d99bb89e8db75d20b882f13f8086fb39221858fa211de0346c926a93ae259b3a
Half of above?
03a3dc00bf5f7e7eec691569c7f67a15d3cdbb3a9994c9a5ec1430cffdb622cf9f

Now subtract half of first offset from half of #130 to get half of second offset.

Second offset is known, we need to work on first offset's half, use -1 divide by 2 script to reduce 18 bits from it, you'll have millions of new offsets and one of them is the target, now divide the #130 range by 2, subtract 18 bits from it and use the new range as your search range, input those millions offset keys and search the range.

Don't just try blind searching.😉



its not blind searching

03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710
hash160
3aaccf438388c4aeb0b433c7b778f25cb6ab244c


hash160
3aaccf438388c31f14410f489939d3d5eac88f19
pk: 48ea48b7a25627365cff38d

13 same hash160 digits, and its two substraction of 130 its not one substraction, 13 digits means its 80 pourcent in that range like puzzle 66

03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 + 2786b52d106d22524ed9cf8a87d2
031ed6283a43d439eace1ee2815118cb6f16f475be60fa5ccc8598372d8c5f1995 # target

031ed6283a43d439eace1ee2815118cb6f16f475be60fa5ccc8598372d8c5f1995 # + 3.........................................
03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852 # target



digaran
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September 01, 2023, 07:17:59 PM
 #3348

its not blind searching

03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710
hash160
3aaccf438388c4aeb0b433c7b778f25cb6ab244c


hash160
3aaccf438388c31f14410f489939d3d5eac88f19
pk: 48ea48b7a25627365cff38d

13 same hash160 digits, and its two substraction of 130 its not one substraction, 13 digits means its 80 pourcent in that range like puzzle 66

03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 + 2786b52d106d22524ed9cf8a87d2
031ed6283a43d439eace1ee2815118cb6f16f475be60fa5ccc8598372d8c5f1995 # target

031ed6283a43d439eace1ee2815118cb6f16f475be60fa5ccc8598372d8c5f1995 # + 3.........................................
03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852 # target

Can you show one example where you could find a key based on rmd160 similarities? I'd like to learn this non existing equation where we can estimate EC points range by finding rmd160 similar starting characters.


 I have this rmd160 as my address
000000009f012a539f3df386f0bddbb874af55ec

I should get to work and find  this address  1111111111111111111114oLvT2

With rmd160 of
0000000000000000000000000000000000000000 ? Lol I'm done.
 

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albert0bsd
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September 01, 2023, 07:38:48 PM
 #3349

its not blind searching

No its not blind search it actually a dummy search

@kalos15btc There is no relationship if two publickey keys share a partial RMD hash even if you found an address with 159 bits exactly equals to another address, there is not going to be a relationshipt between the publickeys


I have this rmd160 as my address
000000009f012a539f3df386f0bddbb874af55ec

I should get to work and find  this address  1111111111111111111114oLvT2

With rmd160 of
0000000000000000000000000000000000000000 ? Lol I'm done.
 

This time I agree with you @digaran
kalos15btc
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September 01, 2023, 07:48:25 PM
 #3350

its not blind searching

No its not blind search it actually a dummy search

@kalos15btc There is no relationship if two publickey keys share a partial RMD hash even if you found an address with 159 bits exactly equals to another address, there is not going to be a relationshipt between the publickeys


I have this rmd160 as my address
000000009f012a539f3df386f0bddbb874af55ec

I should get to work and find  this address  1111111111111111111114oLvT2

With rmd160 of
0000000000000000000000000000000000000000 ? Lol I'm done.
 

This time I agree with you @digaran


you did not understand, im searching for substracted public keys, im searching for their range where exactly,, 110bits 112 118 109 102 bits,,  then like that public key 13 same hash160 digits, than means its in that range , so i scan the range if i found it there i can brake the 130,, thats how i think tell me if im wrong
exemple 10 substracted of 130
i found same 10 first lettre of uncompressed adress i have the pk, so i search in that range of pk the address similar in substracted if i found it i do + and find the target,,, am i wrong HuhHuhHuhHuh??
citb0in
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September 01, 2023, 07:56:51 PM
 #3351

can you try to be more specific, please? At best use a translator that will translate from your native language to english. I'm pretty sure this will be helpful for everyone. Oh, one more thing : try to give an -at least- one expressive example.

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mcdouglasx
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September 01, 2023, 09:59:16 PM
 #3352

its not blind searching

No its not blind search it actually a dummy search

@kalos15btc There is no relationship if two publickey keys share a partial RMD hash even if you found an address with 159 bits exactly equals to another address, there is not going to be a relationshipt between the publickeys


I have this rmd160 as my address
000000009f012a539f3df386f0bddbb874af55ec

I should get to work and find  this address  1111111111111111111114oLvT2

With rmd160 of
0000000000000000000000000000000000000000 ? Lol I'm done.
 

This time I agree with you @digaran


you did not understand, im searching for substracted public keys, im searching for their range where exactly,, 110bits 112 118 109 102 bits,,  then like that public key 13 same hash160 digits, than means its in that range , so i scan the range if i found it there i can brake the 130,, thats how i think tell me if im wrong
exemple 10 substracted of 130
i found same 10 first lettre of uncompressed adress i have the pk, so i search in that range of pk the address similar in substracted if i found it i do + and find the target,,, am i wrong HuhHuhHuhHuh??
ECC and rmd160 are different, you won't find any similarity there.

the matches you find are just that, random matches.


digaran
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September 01, 2023, 11:47:09 PM
 #3353

I'm bored, nothing interesting is happening around these woods no more.  Nobody has new and exciting ideas no more.

Something on mind for some time now, I wonder why 160 bit? Why not keep adding coins to 165, 170 etc?

And is it really feasible brute forcing any key beyond 80 bit? With no pub exposed of course.

I guess from 81 to 159 will not be solved in the next decade, why bother keeping them in there when we can use them as extra incentive for higher ranges, like adding them to 165, 170, 175 etc?

Good luck to all.😴

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September 02, 2023, 01:23:35 AM
Last edit: September 02, 2023, 09:28:57 PM by Mr. Big
 #3354

Can someone please explain how to divide a point on the curve by 2?

I have been working on software for searching by public key, and I can divide by 2 by performing scalar multiplication by the inverse scalar, but this operation takes a long time. If anyone knows a faster way to divide, I can improve my code and release the tool for others to test.



I'm bored, nothing interesting is happening around these woods no more.  Nobody has new and exciting ideas no more.

Something on mind for some time now, I wonder why 160 bit? Why not keep adding coins to 165, 170 etc?

And is it really feasible brute forcing any key beyond 80 bit? With no pub exposed of course.

I guess from 81 to 159 will not be solved in the next decade, why bother keeping them in there when we can use them as extra incentive for higher ranges, like adding them to 165, 170, 175 etc?

Good luck to all.

Have you noticed that there are at most 2^160 private keys instead of 2^256? Any key above that will generate a repeated RIPEMD160 hash.
That's why there are no keys from 161 to 256 anymore; the creator of the challenge realized it was redundant.
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September 02, 2023, 08:26:55 AM
 #3355

Can someone please explain how to divide a point on the curve by 2?

I have been working on software for searching by public key, and I can divide by 2 by performing scalar multiplication by the inverse scalar, but this operation takes a long time. If anyone knows a faster way to divide, I can improve my code and release the tool for others to test.

If you're going to divide a point by 2, you multiply it by the inverse of 2 by N.

If you want to buy me a coffee

Btc = 3246y1G9YjnQQNRUrVMnaeCFrymZRgJAP7

Doge = DGNd8UTi8jVTVZ2twhKydyqicynbsERMjs
digaran
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September 02, 2023, 08:30:46 AM
 #3356

Can someone please explain how to divide a point on the curve by 2?

I have been working on software for searching by public key, and I can divide by 2 by performing scalar multiplication by the inverse scalar, but this operation takes a long time. If anyone knows a faster way to divide, I can improve my code and release the tool for others to test.
There is no such a thing as division in EC math,  there is only add, subtract and multiply, whatever you do, at the end you'll have to use * by inverse.


I wish there was a puzzle 161, so kalos could tackle it by finding rmd160 collisions, he seems to be interested in that.😅

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September 02, 2023, 08:54:16 AM
 #3357

I have been working on software for searching by public key, and I can divide by 2 by performing scalar multiplication by the inverse scalar, but this operation takes a long time. If anyone knows a faster way to divide, I can improve my code and release the tool for others to test.

If you're using division by a fixed number, first you can cache inverse of this number to avoid inversion on each step, second you can decompose inverse into fixed double-add chain, then optimize it with special formulas for point tripling and readditions. But it is will be still expensive opertaion.
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September 02, 2023, 09:35:56 AM
Last edit: September 02, 2023, 09:27:54 PM by Mr. Big
 #3358

Can someone please explain how to divide a point on the curve by 2?

I have been working on software for searching by public key, and I can divide by 2 by performing scalar multiplication by the inverse scalar, but this operation takes a long time. If anyone knows a faster way to divide, I can improve my code and release the tool for others to test.

If you're going to divide a point by 2, you multiply it by the inverse of 2 by N.

I'm currently doing it exactly like this, but it's proving to be very expensive this way.  Sad



My new software is for public keys, inspired by BSGS but a little different, using division and subtraction. Currently, I'm solving 35 bits in 3 seconds; this time is due to the delay in dividing a point. However, I believe that with the correct configuration, I will soon be able to break larger keys. I have some theories that have worked out very well in my tests, and I will be trying out more things to confirm.


For example, for a 35-bit key, I reduce the challenge #35 to 3 bits using my code. You input the public key and the 3-bit key '111,' and it returns 0x4aed21170, which is the equivalent private key of the public key.

Code:
key: 111
found: 00000000000000000000000000000000000000000000000000000004aed21170
Total time: 0 h, 0 m, 2 s

"111101" would be the key for #38.
Code:
key: 111101
found: 00000000000000000000000000000000000000000000000000000022382facd0
Total time: 0 h, 0 m, 13 s
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September 02, 2023, 10:15:26 AM
 #3359

So any update on the progress of finding this key?
 First offset = 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710
Half of above =
0291001b0dc6e5a2628cb4698eb00a6fb7dbd276dc2b214795f2fe52e61243aa9b
Half of 130?
0337374e00a32eaf009e9946035c0e69085627b60a844637d2b958dd83bcfa4383
The following is the subtracted key from #130
Second offset =
03d99bb89e8db75d20b882f13f8086fb39221858fa211de0346c926a93ae259b3a
Half of above?
03a3dc00bf5f7e7eec691569c7f67a15d3cdbb3a9994c9a5ec1430cffdb622cf9f

Now subtract half of first offset from half of #130 to get half of second offset.

Second offset is known, we need to work on first offset's half, use -1 divide by 2 script to reduce 18 bits from it, you'll have millions of new offsets and one of them is the target, now divide the #130 range by 2, subtract 18 bits from it and use the new range as your search range, input those millions offset keys and search the range.

Don't just try blind searching.😉
Hi! Can you show me how you do it?, to remove on video. I can't understand how you make the -1 divide by 2 scenario, and how you get the millionth offsets.
digaran
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September 02, 2023, 10:45:02 AM
Last edit: September 02, 2023, 11:06:11 AM by digaran
 #3360

So any update on the progress of finding this key?
 First offset = 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710
Half of above =
0291001b0dc6e5a2628cb4698eb00a6fb7dbd276dc2b214795f2fe52e61243aa9b
Half of 130?
0337374e00a32eaf009e9946035c0e69085627b60a844637d2b958dd83bcfa4383
The following is the subtracted key from #130
Second offset =
03d99bb89e8db75d20b882f13f8086fb39221858fa211de0346c926a93ae259b3a
Half of above?
03a3dc00bf5f7e7eec691569c7f67a15d3cdbb3a9994c9a5ec1430cffdb622cf9f

Now subtract half of first offset from half of #130 to get half of second offset.

Second offset is known, we need to work on first offset's half, use -1 divide by 2 script to reduce 18 bits from it, you'll have millions of new offsets and one of them is the target, now divide the #130 range by 2, subtract 18 bits from it and use the new range as your search range, input those millions offset keys and search the range.

Don't just try blind searching.
Hi! Can you show me how you do it?, to remove on video. I can't understand how you make the -1 divide by 2 scenario, and how you get the millionth offsets.

Hi there, I can't make a video but here is the script you can use to reduce bits.

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.


My new software is for public keys, inspired by BSGS but a little different, using division and subtraction. Currently, I'm solving 35 bits in 3 seconds; this time is due to the delay in dividing a point. However, I believe that with the correct configuration, I will soon be able to break larger keys. I have some theories that have worked out very well in my tests, and I will be trying out more things to confirm.


For example, for a 35-bit key, I reduce the challenge #35 to 3 bits using my code. You input the public key and the 3-bit key '111,' and it returns 0x4aed21170, which is the equivalent private key of the public key.

Code:
key: 111
found: 00000000000000000000000000000000000000000000000000000004aed21170
Total time: 0 h, 0 m, 2 s

"111101" would be the key for #38.
Code:
key: 111101
found: 00000000000000000000000000000000000000000000000000000022382facd0
Total time: 0 h, 0 m, 13 s
Will you release for public to use?

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