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Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 207. (Read 229766 times)

full member
Activity: 1162
Merit: 237
Shooters Shoot...
Quote
I rather give the whole world if I had the whole world to avoid experiencing the pain in my leg and my back again, nothing is worth more than our health.
No doubt. I had a massive blow out/bulging disc, and an annular tear in another disc. Left leg was numb for months. Couldn't get comfortable in any form or fashion or position. I finally went outside and rubbed a cup of dirt on my back...Now, I am back! lol.

Indeed, back pain is the worse. It still gives me issues from time to time.

Speedy recovery digaran!
copper member
Activity: 1330
Merit: 899
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Where did you get 8,192 chances from? Don't tell me, actually don't tell anyone. You know if I had the means I would have joined you right? When you think about me just visualize a laptop with 30 mk/s search capability!

If I haven't solved the key yet is due to my back, I can no longer sit behind the system to work, I have been at it for a month and I had half of one of my discs removed, so I'm feeling a bit numb in my leg as I type this.


I have got to get back to my trampoline, though I wouldn't call it that, it is more like triangulation method, for now I call it GPS for a lack of a better word.

You see, what I have is actually 2 public keys, when I subtract the 125 public key from each one of them I get the same key, one is inverse though, so one of them is -N. The part it gets interesting, when I add those 2 keys together it gives me the 125 key * 2.  And to my surprise I also have the distances between both of them with the *2 of 125, and obviously the distance between both of them with 125 key as well. Since I can't work on it right now, it remains unsolved.  Moreover I can't seem to figure the equations of adding and subtracting which to which and which from which yet because one or both are  probably -N.

So if you manage to help me out on this, I will give you 1BTC whole bitcoin.


I rather give the whole world if I had the whole world to avoid experiencing the pain in my leg and my back again, nothing is worth more than our health.😉
full member
Activity: 1162
Merit: 237
Shooters Shoot...
Quote
By the way fellow hunters, I have figured out that puzzle 125 is most likely between 65% and 85% of 125 bit range, according to my awful calculations, I am 90% certain it resides in 75% of 125 bit range area.  When it's solved we will know for sure, till then hang tight and keep looking hopelessly, not sure if we could even see it's private key though, unless the solver is a real gentle man unlike some people cough 120 solver cough.

Let's be honest, there are only a handful of people/groups looking for 125 using Kangaroo or random BSGS (hoping to get lucky), and no, I do not include you because I doubt you are putting in the work; besides playing with numbers lol. Which is fine, to learn and try applying your trampoline math to the challenge/curve. I'm not knocking you.

To be more honest, I could give people a 1/8,192 chance of solving #125, and earning 5 BTC, and they would not take it. People are strange creatures. We have a me me me and only me kinda vibe. People think they will get lucky and solve without helping or doing a lot of work, but even with a 1/8,192 chance of solving, they would not take it.

1/8,192 chance....
copper member
Activity: 1330
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when I next have an extended period of quiet and calm, to construct the new transaction carefully.
The sentence above is a puzzle of it's own, it keeps me up at nights ever since it was posted. lol.
What I would like to know, what quiet and calm? My man, you are a billionaire and one of the greatest inventors of all time, you should only have quiet and calm plus young virgins crawling around your palace! lol^2.

Ok now back to business, anybody here knows what happens to G when we multiply it by 1? I mean why is G multiplied by 1 is not G itself?

By the way fellow hunters, I have figured out that puzzle 125 is most likely between 65% and 85% of 125 bit range, according to my awful calculations, I am 90% certain it resides in 75% of 125 bit range area.  When it's solved we will know for sure, till then hang tight and keep looking hopelessly, not sure if we could even see it's private key though, unless the solver is a real gentle man unlike some people cough 120 solver cough.
newbie
Activity: 19
Merit: 2
This puzzle is very strange. If it's for measuring the world's brute forcing capacity, 161-256 are just a waste (RIPEMD160 entropy is filled by 160, and by all of P2PKH Bitcoin). The puzzle creator could improve the puzzle's utility without bringing in any extra funds from outside - just spend 161-256 across to the unsolved portion 51-160, and roughly treble the puzzle's content density.

If on the other hand there's a pattern to find... well... that's awfully open-ended... can we have a hint or two? Cheesy

I am the creator.

You are quite right, 161-256 are silly.  I honestly just did not think of this.  What is especially embarrassing, is this did not occur to me once, in two years.  By way of excuse, I was not really thinking much about the puzzle at all.

I will make up for two years of stupidity.  I will spend from 161-256 to the unsolved parts, as you suggest.  In addition, I intend to add further funds.  My aim is to boost the density by a factor of 10, from 0.001*length(key) to 0.01*length(key).  Probably in the next few weeks.  At any rate, when I next have an extended period of quiet and calm, to construct the new transaction carefully.

A few words about the puzzle.  There is no pattern.  It is just consecutive keys from a deterministic wallet (masked with leading 000...0001 to set difficulty).  It is simply a crude measuring instrument, of the cracking strength of the community.

Finally, I wish to express appreciation of the efforts of all developers of new cracking tools and technology.  The "large bitcoin collider" is especially innovative and interesting!

How to do the masked with leading 000...0001 to set difficulty in python code?
newbie
Activity: 19
Merit: 1
0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
02bcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb
02c994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb
0379be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
03bcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb
03c994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb


When you fuck around enough, you will find out! Lol, so I was playing with some tools and since I have no clue what I'm doing, I managed to find some twins for our beloved G. I call them alternative G spots. Funny they all have the same y coordinates.

From  
addr: c.   1MRxjnjFDhZfjtjgpxBNczsMGVEtYqfFyS    u.  1Jy6aHcRTWjiPN9DpjZztk2F8xY9iNsaj2
020000000000000000000000000000000000000000000000000000000000000001
040000000000000000000000000000000000000000000000000000000000000001483ada7726a3c 4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
```````
```````all hex range of x
```````
```````
to
addr: c.  1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw    u.   1F18os1CipgxyUj9wBMY4yK1HEVUaub2Nr
02ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
04ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff483ada7726a3c 4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8

there are addresses with public keys where the coordinate y is the same.

More interesting is the fact that, the public key 02ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff is the correspondence for the same addresses  1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw as 0200000000000000000000000000000000000000000000000000000001000003d0 and continuing like this, if we change the value of the y-coordinate of the public key, for each even/odd value it seems the compressed address is the same:
1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw
0400000000000000000000000000000000000000000000000000000001000003d00000000000000 000000000000000000000000000000000000000000000000002

1KmKgQHBMbequmyi9uP1yfa1vsNjdsjyEz
0400000000000000000000000000000000000000000000000000000001000003d00000000000000 000000000000000000000000000000000000000000000000003

1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw
0400000000000000000000000000000000000000000000000000000001000003d00000000000000 000000000000000000000000000000000000000000000000004

1KmKgQHBMbequmyi9uP1yfa1vsNjdsjyEz
0400000000000000000000000000000000000000000000000000000001000003d00000000000000 000000000000000000000000000000000000000000000000005
`````
`````an so on
1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw
0400000000000000000000000000000000000000000000000000000001000003d0fffffffffffff fffffffffffffffffffffffffffffffffffffffffffffffffff

I could conclude from here that any compressed public address exists as many times as there are for each uncompressed address and vice versa, and the mathematical space of the corresponding keys that generate them is much larger and can be treated as a three-dimensional space.




If you modify the SECP256K1 curve, for Bitcoin you obtain:
1. Non -valid addresses.
2. Valid addresses but with invalid private keys.
copper member
Activity: 1330
Merit: 899
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play a little with the application below and convince yourself

https://iancoleman.io/bitcoin-key-compression/

all keys are 100% valid

(for uncompressed keys remove the spaces after copy/paste from here)

Sorry to give you the bad news this way son! But you just posted the same keys with different encodings, they don't have the exact same hash.

You see, this key
Code:
0200000000000000000000000000000000000000000000000000000001000003d0
Has only 1 uncompressed key which is :
Code:
0400000000000000000000000000000000000000000000000000000001000003d03005f900bfcb00068a7051ed83f822487ce803fab076680c519f7f52ccb6b9f6
There are no valid key such as
Code:
0400000000000000000000000000000000000000000000000000000001000003d00000000000000000000000000000000000000000000000000000000000000004

Or
Code:
02fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

Whatever they are feeding you over at that site you linked, is not proper secp256k1, elliptic curve cryptography.

Though if you are fond of public keys with a lot of zeros, I can give you one.

Code:
-----BEGIN BITCOIN SIGNED MESSAGE-----
This is the public key "0300000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63" and this is the private key, known as half of N, the middle range 50% key "7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0"
-----BEGIN SIGNATURE-----
1LVAsnUyEtJgZ9HzLfbtiJZuZMzHLX1n6k
H/bLIdkWelLBFjDtbfmFSgEzJLMy+WyuD+8h0SbAlJEO4Ho8/KswngnulCU2E6jC4GafzN38+zKlOGhGRVkiXlE=
-----END BITCOIN SIGNED MESSAGE-----

Next time when you try to spread information, make it legit and verifiable. Posting some random invalid keys and insisting they are valid is not a way to go around these woods, you need to change your supplier I mean tool, what you are using is broken.
newbie
Activity: 30
Merit: 0
play a little with the application below and convince yourself

https://iancoleman.io/bitcoin-key-compression/

all keys are 100% valid

(for uncompressed keys remove the spaces after copy/paste from here)
copper member
Activity: 1330
Merit: 899
🖤😏

From  
addr: c.   1MRxjnjFDhZfjtjgpxBNczsMGVEtYqfFyS    u.  1Jy6aHcRTWjiPN9DpjZztk2F8xY9iNsaj2
020000000000000000000000000000000000000000000000000000000000000001
040000000000000000000000000000000000000000000000000000000000000001483ada7726a3c 4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
```````
```````all hex range of x
```````
```````
to
addr: c.  1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw    u.   1F18os1CipgxyUj9wBMY4yK1HEVUaub2Nr
02ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
04ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff483ada7726a3c 4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8

there are addresses with public keys where the coordinate y is the same.

More interesting is the fact that, the public key 02ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff is the correspondence for the same addresses  1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw as 0200000000000000000000000000000000000000000000000000000001000003d0 and continuing like this, if we change the value of the y-coordinate of the public key, for each even/odd value it seems the compressed address is the same:
1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw
0400000000000000000000000000000000000000000000000000000001000003d00000000000000 000000000000000000000000000000000000000000000000002

1KmKgQHBMbequmyi9uP1yfa1vsNjdsjyEz
0400000000000000000000000000000000000000000000000000000001000003d00000000000000 000000000000000000000000000000000000000000000000003

1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw
0400000000000000000000000000000000000000000000000000000001000003d00000000000000 000000000000000000000000000000000000000000000000004

1KmKgQHBMbequmyi9uP1yfa1vsNjdsjyEz
0400000000000000000000000000000000000000000000000000000001000003d00000000000000 000000000000000000000000000000000000000000000000005
`````
`````an so on
1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw
0400000000000000000000000000000000000000000000000000000001000003d0fffffffffffff fffffffffffffffffffffffffffffffffffffffffffffffffff

I could conclude from here that any compressed public address exists as many times as there are for each uncompressed address and vice versa, and the mathematical space of the corresponding keys that generate them is much larger and can be treated as a three-dimensional space.

Did you change the keys somehow? It seems most of them are invalid, I don't know where you got them, but you should put each public key inside a code bracket to better identify them.
To be exact, the number of compressed and uncompressed keys are exactly 50-50.
You don't happen to have any of their private keys do you? For a second I thought you just found hash collisions.🤣
newbie
Activity: 30
Merit: 0
0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
02bcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb
02c994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb
0379be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
03bcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb
03c994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb


When you fuck around enough, you will find out! Lol, so I was playing with some tools and since I have no clue what I'm doing, I managed to find some twins for our beloved G. I call them alternative G spots. Funny they all have the same y coordinates.

From  
addr: c.   1MRxjnjFDhZfjtjgpxBNczsMGVEtYqfFyS    u.  1Jy6aHcRTWjiPN9DpjZztk2F8xY9iNsaj2
020000000000000000000000000000000000000000000000000000000000000001
040000000000000000000000000000000000000000000000000000000000000001483ada7726a3c 4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
```````
```````all hex range of x
```````
```````
to
addr: c.  1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw    u.   1F18os1CipgxyUj9wBMY4yK1HEVUaub2Nr
02ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
04ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff483ada7726a3c 4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8

there are addresses with public keys where the coordinate y is the same.

More interesting is the fact that, the public key 02ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff is the correspondence for the same addresses  1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw as 0200000000000000000000000000000000000000000000000000000001000003d0 and continuing like this, if we change the value of the y-coordinate of the public key, for each even/odd value it seems the compressed address is the same:
1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw
0400000000000000000000000000000000000000000000000000000001000003d00000000000000 000000000000000000000000000000000000000000000000002

1KmKgQHBMbequmyi9uP1yfa1vsNjdsjyEz
0400000000000000000000000000000000000000000000000000000001000003d00000000000000 000000000000000000000000000000000000000000000000003

1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw
0400000000000000000000000000000000000000000000000000000001000003d00000000000000 000000000000000000000000000000000000000000000000004

1KmKgQHBMbequmyi9uP1yfa1vsNjdsjyEz
0400000000000000000000000000000000000000000000000000000001000003d00000000000000 000000000000000000000000000000000000000000000000005
`````
`````an so on
1LCQXAGayRCWdAGrh7NKWEcyQNfbbxPdtw
0400000000000000000000000000000000000000000000000000000001000003d0fffffffffffff fffffffffffffffffffffffffffffffffffffffffffffffffff

I could conclude from here that any compressed public address exists as many times as there are for each uncompressed address and vice versa, and the mathematical space of the corresponding keys that generate them is much larger and can be treated as a three-dimensional space.

full member
Activity: 1162
Merit: 237
Shooters Shoot...
When playing with some iceland tools which is https://github.com/iceland2k14/kangaroo i got this speed on CPU
  
  • [24894.64 TeraKeys/s][Kang 28672][Count 2^29.29/2^29.10][Elapsed 06s][Dead 1][RAM 53.4MB/46.0MB]

Is it good bad fake idk but like i know iceland tools are good

iceland2k14 is proprietary and all his tools contain backdoor
do not use iceland2k14 libraries unless you want to loose your keys or findings
good luck

Proof what you said of GTFO.

Iceland tools are 100% trusted.

About the speed of kangaroo from that tool it is a calculated speed more o less about the expected speed, remember kangaroo it is a probabilistic algorithm and it can't give you exact speed only a near speed based on some math.

Regards

Albert0bsd, what do you mean you can't get exact speed with kangaroo? Do you mean in general or with Iceland's kangaroo? I do not know much about Iceland's version.
When I look at that info displayed, it doesn't seem right to me. Like a bad copy and paste job. If you look at the numbers:

Code:
[+] [24894.64 TeraKeys/s][Kang 28672][Count 2^29.29/2^29.10][Elapsed 06s][Dead 1][RAM 53.4MB/46.0MB]

First, there is no way it found 2^29.29 DPs in 6 seconds. Even if it was DP 1. And I am quite sure 2^29.29 would eat up more than 53MB of RAM.

But generally speaking, if you are using GPU Kangaroo by JLP or another GPU version, the speed shown, is accurate. My benchmarks are easy to follow. I have a display (so does JLPs) of total keys checked and number of DPs found. If you are running DP 25 and have completed 2^42 ops (GPU speed * number of cards = keys checked/operations) then you will normally have around 2^17 DPs. 2^42-2^25=2^17.
member
Activity: 117
Merit: 32
in 3 months maximum it will be discovered
hero member
Activity: 862
Merit: 662
When playing with some iceland tools which is https://github.com/iceland2k14/kangaroo i got this speed on CPU
  
  • [24894.64 TeraKeys/s][Kang 28672][Count 2^29.29/2^29.10][Elapsed 06s][Dead 1][RAM 53.4MB/46.0MB]

Is it good bad fake idk but like i know iceland tools are good

iceland2k14 is proprietary and all his tools contain backdoor
do not use iceland2k14 libraries unless you want to loose your keys or findings
good luck

Proof what you said of GTFO.

Iceland tools are 100% trusted.

About the speed of kangaroo from that tool it is a calculated speed more o less about the expected speed, remember kangaroo it is a probabilistic algorithm and it can't give you exact speed only a near speed based on some math.

Regards
sr. member
Activity: 345
Merit: 250
Thanks for the response guys, I was just more curious than anything.
full member
Activity: 1162
Merit: 237
Shooters Shoot...
yes go to github https://github.com/iceland2k14/quick download quick or just pubdiv.py instructions on page are very clear how to.



so to reduce 2 you need to generate 2 power 2, 4 keyspaces 2 power 4 etc

Power   Value
0   1
1   2
...
50   1,125,899,906,842,624


Going back to this... So you drop it 50 bit... Get your insane number of public keys. Theoretically if you then Kangaroo it and get the private key in the range of 75...What's the following process?
To add to it...
Yes, first you would have to generate 2^75 pubkeys, and then run the kangaroo program, 2^75 times, once for each pubkey generated.
jr. member
Activity: 76
Merit: 4
yes go to github https://github.com/iceland2k14/quick download quick or just pubdiv.py instructions on page are very clear how to.



so to reduce 2 you need to generate 2 power 2, 4 keyspaces 2 power 4 etc

Power   Value
0   1
1   2
...
50   1,125,899,906,842,624


Going back to this... So you drop it 50 bit... Get your insane number of public keys. Theoretically if you then Kangaroo it and get the private key in the range of 75...What's the following process?
Firstly it would take at least a year and a very large hard drive to generate that many keys. but if you did and somehow found the key on 75 then you  convert that key to decimal and multiply this by 1,125,899,906,842,624 and convert back to hex for the 125 key.
member
Activity: 185
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Two things you should never abandon: Family & BTC
The puzzle is not fun. There are use only power of GPU not power your Brain.

If you want fun, play video games. This is a serious thing. A security testing, increasingly hard, mind boggling, enthusiastically coding, mathematically challenging and most of all: very lucrative job. If you're a puss, you'll be run down by it.

Come down friend. There are not mathematically is just Random. I try many theory mathematics and found some rule of puzzle. There are never complete.
If you like lottery of bitcoin. Try your luck.

Did I understand correctly? We are engaged in breaking elliptical curves - it's really fascinating! As soon as we do this, all cryptocurrencies will be worth - zero! People are a strange animal!))

This will never happen .. even on 160 bit range .. will never happen .. i am more likely to become the president of the US without having its nationality than we breaking the elliptic curve .. you're dealing with numbers as much as almost the number of atoms of the known universe .. but for solving puzzles, it can happen, good luck everyone.
I understand that!))) but you see - not everyone here thinks so! They probably want to discover new mathematics!

Yeah, most people don't know that Beating Satoshi is impossible without quantum devices
sr. member
Activity: 345
Merit: 250
yes go to github https://github.com/iceland2k14/quick download quick or just pubdiv.py instructions on page are very clear how to.



so to reduce 2 you need to generate 2 power 2, 4 keyspaces 2 power 4 etc

Power   Value
0   1
1   2
...
50   1,125,899,906,842,624


Going back to this... So you drop it 50 bit... Get your insane number of public keys. Theoretically if you then Kangaroo it and get the private key in the range of 75...What's the following process?
newbie
Activity: 13
Merit: 1
The puzzle is not fun. There are use only power of GPU not power your Brain.

If you want fun, play video games. This is a serious thing. A security testing, increasingly hard, mind boggling, enthusiastically coding, mathematically challenging and most of all: very lucrative job. If you're a puss, you'll be run down by it.

Come down friend. There are not mathematically is just Random. I try many theory mathematics and found some rule of puzzle. There are never complete.
If you like lottery of bitcoin. Try your luck.

Did I understand correctly? We are engaged in breaking elliptical curves - it's really fascinating! As soon as we do this, all cryptocurrencies will be worth - zero! People are a strange animal!))

This will never happen .. even on 160 bit range .. will never happen .. i am more likely to become the president of the US without having its nationality than we breaking the elliptic curve .. you're dealing with numbers as much as almost the number of atoms of the known universe .. but for solving puzzles, it can happen, good luck everyone.
I understand that!))) but you see - not everyone here thinks so! They probably want to discover new mathematics!
member
Activity: 185
Merit: 15
Two things you should never abandon: Family & BTC
The puzzle is not fun. There are use only power of GPU not power your Brain.

If you want fun, play video games. This is a serious thing. A security testing, increasingly hard, mind boggling, enthusiastically coding, mathematically challenging and most of all: very lucrative job. If you're a puss, you'll be run down by it.

Come down friend. There are not mathematically is just Random. I try many theory mathematics and found some rule of puzzle. There are never complete.
If you like lottery of bitcoin. Try your luck.

Did I understand correctly? We are engaged in breaking elliptical curves - it's really fascinating! As soon as we do this, all cryptocurrencies will be worth - zero! People are a strange animal!))

This will never happen .. even on 160 bit range .. will never happen .. i am more likely to become the president of the US without having its nationality than we breaking the elliptic curve .. you're dealing with numbers as much as almost the number of atoms of the known universe .. but for solving puzzles, it can happen, good luck everyone.
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