Bitcoin puzzle transaction ~32 BTC prize to who solves it

[albert0bsd]

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

[1714263498]

1714263498

[1714263498]

1714263498

[Lolo54]

in 3 months maximum it will be discovered

[WanderingPhilospher]

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.

[Ovixx]

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.

[digaran]

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.🤣

[Ovixx]

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)

[digaran]

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.

[algorithm32]

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.

[Trancefan]

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?

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?

[digaran]

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.

[WanderingPhilospher]

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....

[digaran]

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.😉

[WanderingPhilospher]

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!

[digaran]

Speedy recovery digaran!

Thanks, it was 2/5 years ago though sitting long periods of time does hurt even now, a few days workout will build more muscle to support the spine.


Anyways, it seems there are no interesting parties wanting 1 bitcoin doing some simple (yeah right) math calculations, easy peasy. 😉

Anyhow, so I was thinking about a brute force/key solving method, it could work and not work but mentioning it doesn't hurt.

First we pick a small k, something like 3, then we start by multiplying a range of random numbers by our original number, when we reach the end of our search range without any hit, we'd pick another base key and select another set of random numbers until we again reach the end of our search range, and since multiplying only takes a few seconds to perform, we could scan the entire range in a few minutes and start again with new settings, this would increase the chance of a successful hit by many orders of magnitude.
Just today I realized that I have been performing what kangaroo and other tools are doing, by hand! Lol  when you have limited resources, your mind is the best asset to utilize.😉

[elvis13]

digaran

I want to help you if you tell me what to do. Do you have telegram? Where can you chat?

[digaran]

digaran

I want to help you if you tell me what to do. Do you have telegram? Where can you chat?

I had to lol a lot before typing this.
If I knew what to do, I would have done it myself already, though I have realized that what I'm doing is a simplified method of which the tools such as kangaroo are doing. I just add a number to my target p, and then subtract the result from my p, which seems to return my selected number.

Ok let me go technical on this, if we add 5 to 10, we will have 15, then subtracting 10 from 15 gives us 5, but in bitcoin's used curve, if  we simply subtract 15 from 10, we won't see -5, instead we will see 20, that's how this complicated elliptic curve works, what I have done:
I  have added 5 to 10 and 3 to 10,  sum of them is 28 now, then I'd subtract 10 from 28 to obtain 18, now imagine I don't know 10, 13, 15, 18 and 28, I just have 5 and 3. But I know the distance between 10 and 13, 10 and 15, 10 and 18.
Now can you tell me how I can reach 10, 13, 15, 18 and or 28 by only adding or multiplying 3 and 5 together?

Well, when we simplify it like that, we could multiply 3 by 5 to get to 15, but in our case if I knew what is half of 10, I wouldn't be here at all, I'm just randomly guessing some numbers and then I'd add and subtract them.

Today I found out if we add -20 to 10, we get 20, and adding -10 to 20, we get 10, it works other way as well.
If you are more confused than before, you should be, this is how cryptography works, confusion is the key!

[ilkerc]

Hello,

I just created a simple solo-pool to find puzzle 66, 67 and 68.

It works with proof of work. If you want to scan all private keys in puzzle 66; you need to do 36 quintillion scans in total. In case you do a random scan; previously generated private keys will be regenerated (random problem). This extends the scan time by x10.

Puzzle 66 HEX ranges as follows. It starts with 2 or 3. Any private key in this range is 17 characters long.

20000000000000000 to 3ffffffffffffffff

We take the first 7 characters and delete the rest for now. The result will be as follows.

2000000 to 3ffffff

We now have about 33 million possible ranges to search. All possible ranges are stored in the database. A random value will come up each time a scan job is called and will be marked as scanned when the scan is complete. Each range contains 1,1 trillion keys.

I can scan each key in about 10 minutes on NVIDIA 3090. This actually means about 1,1 trillion private keys. When the private key is scanned, it is marked as scanned. So it won't show up anymore.

https://btcpuzzle.info
https://github.com/ilkerccom/bitcrackrandomiser

[elvis13]

Meaning? We have a pool http://f4lc0n.com:8080/

[WanderingPhilospher]

Meaning? We have a pool http://f4lc0n.com:8080/

Since you are part of the pool, what is one range size equal to?

Range size seems like an odd one.

325M left
14M complete

Not really that straight forward; like 2^65 / 2 ^ 40 range = 2 ^ 25 ranges to search

[digaran]

Hello,

I just created a simple solo-pool to find puzzle 66.

It only works with BTC Puzzle 66. If you want to scan all private keys in puzzle 66; you need to do 36 quintillion scans in total. In case you do a random scan; previously generated private keys will be regenerated (random problem). This extends the scan time by x10.

Puzzle 66 HEX ranges as follows. It starts with 2 or 3. Any private key in this range is 17 characters long.

20000000000000000 to 3ffffffffffffffff

We take the first 7 characters and delete the rest for now. The result will be as follows.

2000000 to 3ffffff

We now have about 33 million possible private keys to search. All possible private keys are stored in the database. A random value will come up each time a scan job is called and will be marked as scanned when the scan is complete.

I can scan each key in about 10 minutes on NVIDIA 3090. This actually means about 1,1 trillion private keys. When the private key is scanned, it is marked as scanned. So it won't show up anymore.

So you only keep 33 m keys in a file and then search for what exactly? If you are searching small ranges, then you should  do it sequential. What is your speed in terms of keys/s ?
I did a quick calculus, I can generate 1 trillion every 12 hours, so 2 trillions per 24 hours, now if I run my laptop 24/7, I could find the key after 30,000 + years or so give or take! Do you think people will be using bitcoin by then? Because I don't wanna waste 30,000 years  for unsure future investment.😂