Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 188.

digaran

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Quote from: james5000 on September 01, 2023, 08:23:35 PM

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.😅

lordfrs

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Quote from: james5000 on September 01, 2023, 08:23:35 PM

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.

james5000

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Activity: 69

Merit: 2

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.

Quote from: digaran on September 01, 2023, 06:47:09 PM

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.

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

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.😴

mcdouglasx

member

Activity: 239

Merit: 53

New ideas will be criticized and then admired.

Quote from: kalos15btc on September 01, 2023, 02:48:25 PM

Quote from: albert0bsd on September 01, 2023, 02:38:48 PM

Quote from: kalos15btc on September 01, 2023, 01:26:14 PM

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

Quote from: digaran on September 01, 2023, 02:17:59 PM

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 Huh

??

ECC and rmd160 are different, you won't find any similarity there.

the matches you find are just that, random matches.

citb0in

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Activity: 630

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Bitcoin g33k

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.

kalos15btc

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Quote from: albert0bsd on September 01, 2023, 02:38:48 PM

Quote from: kalos15btc on September 01, 2023, 01:26:14 PM

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

Quote from: digaran on September 01, 2023, 02:17:59 PM

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 Huh

??

albert0bsd

hero member

Activity: 862

Merit: 662

Quote from: kalos15btc on September 01, 2023, 01:26:14 PM

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

Quote from: digaran on September 01, 2023, 02:17:59 PM

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

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

Quote from: kalos15btc on September 01, 2023, 01:26:14 PM

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.

kalos15btc

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Activity: 50

Merit: 1

Quote from: digaran on September 01, 2023, 01:06:44 PM

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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Merit: 899

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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.😉

mcdouglasx

member

Activity: 239

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New ideas will be criticized and then admired.

Quote from: tptkimikaze on August 31, 2023, 03:43:35 PM

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

kalos15btc

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Activity: 50

Merit: 1

Quote from: citb0in on September 01, 2023, 04:17:59 AM

Quote from: lordfrs on September 01, 2023, 04:00:43 AM

Quote from: kalos15btc on August 31, 2023, 05:31:57 PM

@digaran indeed,, im about to breake puzzle 130

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

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

of course we split 13 btc

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 §.?

citb0in

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Activity: 630

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Bitcoin g33k

Quote from: lordfrs on September 01, 2023, 04:00:43 AM

Quote from: kalos15btc on August 31, 2023, 05:31:57 PM

@digaran indeed,, im about to breake puzzle 130

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

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

of course we split 13 btc

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

lordfrs

jr. member

Activity: 57

Merit: 1

Quote from: kalos15btc on August 31, 2023, 05:31:57 PM

@digaran indeed,, im about to breake puzzle 130

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

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

of course we split 13 btc

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

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

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. 😉

kalos15btc

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Activity: 50

Merit: 1

Quote from: digaran on August 31, 2023, 06:10:35 PM

Quote from: kalos15btc on August 31, 2023, 05:31:57 PM

@digaran indeed,, im about to breake puzzle 130

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

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

of course we split 13 btc

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

So you want others to search for you, if the key was found you split the prize that easy? Son, what happens if they say they have found the key and you go rent some GPUs to get it yourself and keep the prize? You see why nobody would search that range for you, unless they are willingly trying to help you.

Also there is no explanation about why you think that key is in that specific range. Should we be waiting for 587 other such ranges from you asking people to scan them for you? Lol.

Here is my 50satoshis, before attempting to search for a key, make sure it's in range.

you have concluded this ? and you judge me ?
no im not sure but at least im trying to solve it instead of doing nothing like you; you just reply like you know everything and you dont do nothing;; im helping here if i find the 130 i will leave tips for everyone even you Wink

so please dont judge like this

ps/ im the first one here in this thread who shared the gpu rent website; if im searching and want for my self i didint post the link;; i posted then 120 andd 125 cracked;; i have no probleme and im happy who will crack this puzzle;

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

Quote from: kalos15btc on August 31, 2023, 05:31:57 PM

@digaran indeed,, im about to breake puzzle 130

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

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

of course we split 13 btc

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

So you want others to search for you, if the key was found you split the prize that easy? Son, what happens if they say they have found the key and you go rent some GPUs to get it yourself and keep the prize? You see why nobody would search that range for you, unless they are willingly trying to help you.

Also there is no explanation about why you think that key is in that specific range. Should we be waiting for 587 other such ranges from you asking people to scan them for you? Lol.

Here is my 50satoshis, before attempting to search for a key, make sure it's in range.

kalos15btc

jr. member

Activity: 50

Merit: 1

@digaran indeed,, im about to breake puzzle 130

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

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

of course we split 13 btc

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

tptkimikaze

newbie

Activity: 25

Merit: 2

Quote from: albert0bsd on August 31, 2023, 04:31:49 PM

Quote from: tptkimikaze on August 31, 2023, 04:07:15 PM

Maybe I know too little information about it. I think I have some idea now, not sure if I am correct about it. By subtracting known 130 public key to lower bits to find the lower bits private key, once gotten that private key you do an inverse of lower bit private key back to 130 bits. Am I right?

Yes that is correct if you have some case like this:

Code:

P130 - PA = X
P130 - PB = Y
P130 - PC = Z
...
P130 - PM = N

if you found the privatekey of any X, Y, Z or N you only need to apply the inverse process to get the Private key of P130, you only need to be aware of what are the values of PA,PB PC and PM

the same appy for other operations like addition, multiplication and division

Regards!

Thanks. Maybe will try it later. More focusing on 66 now. If my idea and approach works and able to solve 66, then 67 and 68 should be also solvable in short period of time.

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 188. (Read 244595 times)