Bitcoin puzzle transaction ~32 BTC prize to who solves it

[damiankopacz87]

Puzzle 71 is 2.36 sextillion possibilities.

Lets assume euromillion odds, that 1 in 140 million.

A 4090 should be capable of 6 billion keys/second.

That means, if you run it constantly, then approx every 47 minutes you have the equivalent chances of a single euro millions ticket.

1/140.000.000=7e^-9
2^70/6.000.000.000 /28s =7e^-9

47 minutes? Am I doing something wrong? It shows 28 seconds...

Anyway, it has to be caled gambling

BR
Damian

[GinnyBanzz]

Puzzle 71 is 2.36 sextillion possibilities.

Lets assume euromillion odds, that 1 in 140 million.

A 4090 should be capable of 6 billion keys/second.

That means, if you run it constantly, then approx every 47 minutes you have the equivalent chances of a single euro millions ticket.

1/140.000.000=7e^-9
2^70/6.000.000.000 /28s =7e^-9

47 minutes? Am I doing something wrong? It shows 28 seconds...

Anyway, it has to be caled gambling

BR
Damian

Ahhh, maths is not my strong point, think I miscalculated somewhere lol.

Well, 28seconds is far better odds, still gambling, but some semblance of a chance.

However, you'd need to calculate power costs to determine if its more financially feasible to try to crack the puzzle or just buy a bunch of euro millions tickets every week.

[damiankopacz87]

However, you'd need to calculate power costs to determine if its more financially feasible to try to crack the puzzle or just buy a bunch of euro millions tickets every week.

Ah, You nasty boot, You did it right, I was just checking.

Anyway, none of this is reasonable, logical nor clever I call it TAX from stupidity.

BR
Damian

[GinnyBanzz]

However, you'd need to calculate power costs to determine if its more financially feasible to try to crack the puzzle or just buy a bunch of euro millions tickets every week.

Ah, You nasty boot, You did it right, I was just checking.

Anyway, none of this is reasonable, logical nor clever I call it TAX from stupidity.

BR
Damian

Ahh haha, I just assume I'm wrong when it comes to maths usually.

It kind of is stupid when you think about it long enough, but it can be fun, it can help people learn new things (the power of cryptography, coding, etc), and, gives you a tiny bit of hope checking on your cracking instance to see if the message "KEY FOUND!" has appeared!

[Redni]

I made a Rust library that includes all b1000 puzzle data: https://github.com/oritwoen/boha

Code:

$ boha list b1000 --unsolved
$ boha show b1000/71
$ boha balance b1000/71

All 256 puzzles with addresses, solved keys (hex + WIF), transaction history, solver attribution where known. Data compiled into binary - no external API needed.
Might be useful for anyone building tools around these puzzles.

[hoanghuy2912]

How many graphics processing units (GPUs) are needed to solve puzzle 135 using the RC Kangaroo in the shortest time?

[gygy]

How many graphics processing units (GPUs) are needed to solve puzzle 135 using the RC Kangaroo in the shortest time?

The shortest possible time is the Planck time, but it’s completely impossible for you to solve it in that amount of time, no matter how many GPUs you have.

[filo1992]

I tested pool 135 t.me/puzzle135, and after winning the 115-bit reward, I was not paid the reward.

If someone can explain to me how to attach files I'll attach the whole conversation with the administrator.

PLEASE BE CAREFUL

[ee1234ee]

I tested pool 135, and after winning the 115-bit reward, I was not paid the reward.

If someone can explain to me how to attach files I'll attach the whole conversation with the administrator.

PLEASE BE CAREFUL

What kind of pool is it?What scanning software are they using for the pool you mentioned? How many GPUs did you use to find puzzle 115?
Are you sure it's a puzzle with a known public key? Nowadays, online public pools are all address based puzzle pools.



You can use   imgbb.com

[kTimesG]

I tested pool 135 t.me/puzzle135, and after winning the 115-bit reward, I was not paid the reward.

If someone can explain to me how to attach files I'll attach the whole conversation with the administrator.

PLEASE BE CAREFUL

What did you expect? Did you miss my PSA about not doing this, and about how shady the guy was (not even understanding basic things about algorithms, hiding Python code in executables, etc)?

You simply contributed to his DP database for P135, like all the other participants. Even if he ever pays you back for your obvious costs (losses at this point).

For anyone thinking "oh, but DPs for a 115 bits puzzle don't work for 135": use your brain.

[filo1992]

I tested pool 135, and after winning the 115-bit reward, I was not paid the reward.

If someone can explain to me how to attach files I'll attach the whole conversation with the administrator.

PLEASE BE CAREFUL

What kind of pool is it?What scanning software are they using for the pool you mentioned? How many GPUs did you use to find puzzle 115?
Are you sure it's a puzzle with a known public key? Nowadays, online public pools are all address based puzzle pools.



You can use   imgbb.com

Effective time: 30 days.

Except that after modifying the program to no avail, I rebooted my entire system with 17 GPUs and new settings, and in 4 days I found the solution.

I tested pool 135 t.me/puzzle135, and after winning the 115-bit reward, I was not paid the reward.

If someone can explain to me how to attach files I'll attach the whole conversation with the administrator.

PLEASE BE CAREFUL

What did you expect? Did you miss my PSA about not doing this, and about how shady the guy was (not even understanding basic things about algorithms, hiding Python code in executables, etc)?

You simply contributed to his DP database for P135, like all the other participants. Even if he ever pays you back for your obvious costs (losses at this point).

For anyone thinking "oh, but DPs for a 115 bits puzzle don't work for 135": use your brain.

I know it's a hard lesson to learn, but at least there's actual evidence of the pool system being unfair.

[sxiclub]

What kind of pool is it?What scanning software are they using for the pool you mentioned? How many GPUs did you use to find puzzle 115?
Are you sure it's a puzzle with a known public key? Nowadays, online public pools are all address based puzzle pools.

You can use   imgbb.com

Seems to be a pool of Russian origin, or at least the participants in the Telegram channel speak Russian. In this thread, user Torin Keepler promoted the $200 challenge for finding a 115-bit PVK.

[ee1234ee]

What kind of pool is it?What scanning software are they using for the pool you mentioned? How many GPUs did you use to find puzzle 115?
Are you sure it's a puzzle with a known public key? Nowadays, online public pools are all address based puzzle pools.

You can use   imgbb.com

Seems to be a pool of Russian origin, or at least the participants in the Telegram channel speak Russian. In this thread, user Torin Keepler promoted the $200 challenge for finding a 115-bit PVK.

A privately initiated challenge for only $200
I thought it was problem 115? Unfortunately, I am unable to manage my Telegram account as I am unable to access it.

[optioncmdPR]

The puzzles are solved first by application of logic on its respective bitspace boundaries,
yielding the first few bytes of the specifically designed key and substantially increasing
the viability of solving the remainder of the key using brute force.
Lets apply this claim to puzzle 64 as a real life example.
note: arbitrary calculation for full precision is necessary

        LOGIC
Divide upper boundary by lower boundary:
set decimal precision to twice the expected digit length of the private key , -1.
In this example, then, 38 digits for precision.

Upper boundary: 2^64 -1  18446744073709551615   
Lower boundary: 2^63      9223372036854775808

   18446744073709551615 ÷ 9223372036854775808
=  1.99999999999999999989157978275144955659

To quote the creator "the odds of guessing any private key with balance is like
winning powerball , 9 times in a row "

So we next take the first 20 least significant bits, reassign the decimal point,
and raise to the power of 9, divide the result by two.  The first 4 bytes( -1 at end byte ) are
the beginning of private key.   



take first 20 least significant bits    1.99999999999999999989157978275144955659
                                                                                      ^^^^^^^^^^^^^^^^
 
              reassign decimal point      .89157978275144955659
                                                    ^   

                 raise to 9th power         .89157978275144955659^9
 
                                                    .355993368271774728639363164404970728311
           
                        divide by 2            .1779966841358873643196815822024853641555
                                                     ^^^^^^
                 
                   first 4 bytes -1            17799667xxxxxxxxxxxx
                                                              ( trivial brute-force )
          brute force the remainder

         fun: should you increase the precision on calculations will notice familiar bytes for the # 65 key
                If you look at the key # 63 calculations will see bytes for # 64 etc..  In summary , with enough
                insight , the keys are absolutely 100% solve able by logic. There is more to be unearthed. Dont give up!     

[GinnyBanzz]

The puzzles are solved first by application of logic on its respective bitspace boundaries,
yielding the first few bytes of the specifically designed key and substantially increasing
the viability of solving the remainder of the key using brute force.
Lets apply this claim to puzzle 64 as a real life example.
note: arbitrary calculation for full precision is necessary

        LOGIC
Divide upper boundary by lower boundary:
set decimal precision to twice the expected digit length of the private key , -1.
In this example, then, 38 digits for precision.

Upper boundary: 2^64 -1  18446744073709551615   
Lower boundary: 2^63      9223372036854775808

   18446744073709551615 ÷ 9223372036854775808
=  1.99999999999999999989157978275144955659

To quote the creator "the odds of guessing any private key with balance is like
winning powerball , 9 times in a row "

So we next take the first 20 least significant bits, reassign the decimal point,
and raise to the power of 9, divide the result by two.  The first 4 bytes( -1 at end byte ) are
the beginning of private key.   



take first 20 least significant bits    1.99999999999999999989157978275144955659
                                                                                      ^^^^^^^^^^^^^^^^
 
              reassign decimal point      .89157978275144955659
                                                    ^   

                 raise to 9th power         .89157978275144955659^9
 
                                                    .355993368271774728639363164404970728311
           
                        divide by 2            .1779966841358873643196815822024853641555
                                                     ^^^^^^
                 
                   first 4 bytes -1            17799667xxxxxxxxxxxx
                                                              ( trivial brute-force )
          brute force the remainder

         fun: should you increase the precision on calculations will notice familiar bytes for the # 65 key
                If you look at the key # 63 calculations will see bytes for # 64 etc..  In summary , with enough
                insight , the keys are absolutely 100% solve able by logic. There is more to be unearthed. Dont give up!     

The keys are absolutely NOT 100% solvable by logic, that's a ridiculous thing to suggest.

[ArtificialLove]

Is there anything that would allow you to use a mask for the random hexadecimals? e.g. --keyspace rrrrBrrrr:rrrrFrrrr (where r is random characters)
There's a project on GitHub called "KeyScanner" and its older tags show such a feature but it appears abandoned since, I was also wondering about Python and OpenCL, is there anything? There's one for Sol inspired by VanityGen but nothing for BTC, which is weird.

Thanks!

[optioncmdPR]

The puzzles are solved first by application of logic on its respective bitspace boundaries,
yielding the first few bytes of the specifically designed key and substantially increasing
the viability of solving the remainder of the key using brute force.
Lets apply this claim to puzzle 64 as a real life example.
note: arbitrary calculation for full precision is necessary

        LOGIC
Divide upper boundary by lower boundary:
set decimal precision to twice the expected digit length of the private key , -1.
In this example, then, 38 digits for precision.

Upper boundary: 2^64 -1  18446744073709551615   
Lower boundary: 2^63      9223372036854775808

   18446744073709551615 ÷ 9223372036854775808
=  1.99999999999999999989157978275144955659

To quote the creator "the odds of guessing any private key with balance is like
winning powerball , 9 times in a row "

So we next take the first 20 least significant bits, reassign the decimal point,
and raise to the power of 9, divide the result by two.  The first 4 bytes( -1 at end byte ) are
the beginning of private key.   



take first 20 least significant bits    1.99999999999999999989157978275144955659
                                                                                      ^^^^^^^^^^^^^^^^
 
              reassign decimal point      .89157978275144955659
                                                    ^   

                 raise to 9th power         .89157978275144955659^9
 
                                                    .355993368271774728639363164404970728311
           
                        divide by 2            .1779966841358873643196815822024853641555
                                                     ^^^^^^
                 
                   first 4 bytes -1            17799667xxxxxxxxxxxx
                                                              ( trivial brute-force )
          brute force the remainder

         fun: should you increase the precision on calculations will notice familiar bytes for the # 65 key
                If you look at the key # 63 calculations will see bytes for # 64 etc..  In summary , with enough
                insight , the keys are absolutely 100% solve able by logic. There is more to be unearthed. Dont give up!     

The keys are absolutely NOT 100% solvable by logic, that's a ridiculous thing to suggest.

Well I don't believe that the described " deterministic wallet" has anything to do with ecdsa or 32 byte ECC points, and that is why I am able to entertain such outrageous absurdities. ( good sir )

[0xastraeus]

Is there anything that would allow you to use a mask for the random hexadecimals? e.g. --keyspace rrrrBrrrr:rrrrFrrrr (where r is random characters)
There's a project on GitHub called "KeyScanner" and its older tags show such a feature but it appears abandoned since, I was also wondering about Python and OpenCL, is there anything? There's one for Sol inspired by VanityGen but nothing for BTC, which is weird.

Thanks!

Are you talking about something like this:

Code:

import random

BIT_GAP = 2 ** 26

first = random.getrandbits(70) | (1 << 70)
second = first + BIT_GAP
print(f"{first:x}:{second:x}")

outputs:

Code:

69a9101cbb05e768d8:69a9101cbb09e768d8

it's going to search from 69a9101cbb05e768d8:69a9101cbb09e768d8

[ArtificialLove]

Is there anything that would allow you to use a mask for the random hexadecimals? e.g. --keyspace rrrrBrrrr:rrrrFrrrr (where r is random characters)
There's a project on GitHub called "KeyScanner" and its older tags show such a feature but it appears abandoned since, I was also wondering about Python and OpenCL, is there anything? There's one for Sol inspired by VanityGen but nothing for BTC, which is weird.

Thanks!

Are you talking about something like this:

Code:

import random

BIT_GAP = 2 ** 26

first = random.getrandbits(70) | (1 << 70)
second = first + BIT_GAP
print(f"{first:x}:{second:x}")

outputs:

Code:

69a9101cbb05e768d8:69a9101cbb09e768d8

it's going to search from 69a9101cbb05e768d8:69a9101cbb09e768d8

Something rather like VanitySearch, KeyHunt, etc. or at least something like SolVanityGen (Python + OpenCL) where I can keep modifying the Python code or enter new masks with every search, I couldn't find anything, except for that old repository for KeyScanner which had exactly that but then the whole thing was abandoned it seems.

Thanks for your earlier answer by the way, and also FixedPaul, appreciated.

[0xastraeus]

Is there anything that would allow you to use a mask for the random hexadecimals? e.g. --keyspace rrrrBrrrr:rrrrFrrrr (where r is random characters)
There's a project on GitHub called "KeyScanner" and its older tags show such a feature but it appears abandoned since, I was also wondering about Python and OpenCL, is there anything? There's one for Sol inspired by VanityGen but nothing for BTC, which is weird.

Thanks!

Are you talking about something like this:

Code:

import random

BIT_GAP = 2 ** 26

first = random.getrandbits(70) | (1 << 70)
second = first + BIT_GAP
print(f"{first:x}:{second:x}")

outputs:

Code:

69a9101cbb05e768d8:69a9101cbb09e768d8

it's going to search from 69a9101cbb05e768d8:69a9101cbb09e768d8

Something rather like VanitySearch, KeyHunt, etc. or at least something like SolVanityGen (Python + OpenCL) where I can keep modifying the Python code or enter new masks with every search, I couldn't find anything, except for that old repository for KeyScanner which had exactly that but then the whole thing was abandoned it seems.

Thanks for your earlier answer by the way, and also FixedPaul, appreciated.

No problem,

the only actual program I can think of that does something similar would be ecloop https://github.com/vladkens/ecloop and using the -d offset:size flag. If you want a cuda program you may have to grab the source and do it yourself or stick with python to create the range and use subprocess to call whatever program you want.

but here's an example of ecloop -d:

Code:

./ecloop rnd -f data/btc-puzzles-hash -d 20:32 -r 400000000000000000:7fffffffffffffffff -o ./found_71.txt
threads: 12 ~ addr33: 1 ~ addr65: 0 ~ endo: 0 | filter: list (160)
----------------------------------------
[RANDOM MODE] offs: 20 ~ bits: 32

0000000000000000 0000000000000000 000000000000006c 730000000001e0a6
0000000000000000 0000000000000000 000000000000006c 730ffffffff1e0a6