Bitcoin puzzle transaction ~32 BTC prize to who solves it

[digaran]

Ok let me explain, imagine #125 as number 750, if we subtract 500 (2^124, half of 2^125 or in our example 2^125 = 1000) if we subtract 500 from 750 we'd have 250, now all we have to do is to figure out what number is closest to our 250 (which we don't know it's exact value) and try to subtract that guessed number from our 1/3 of #125 puzzle, and the rest would be easy.

Guessing the right number is very hard. And you can always guess it wrong, subtract 800 from 750, and you will get -50. Then, assuming that your N is 1000000, you will try to break 999950, instead of trying to break 750. Also, as far as I know, people already start from some offset, so they are not trying to scan keys outside of that 500 to 1000 range, instead they just pick for example 550 to 560 range, and try to scan only that, and that alone takes a lot of time.

Though if you could explain how to figure out which key is greater or smaller than the other, that would be awesome!

You cannot figure it out, based only on a single public key. It is always relative to the base point. For example, if you have (basePoint,yourPoint) pair, you could have "yourPointPrivKey" as some even number, for example 250. But it is only even in a relation between this particular base point! So, only "basePoint*250=yourPoint" gives you "250" as an even number. However, you could use (basePoint*2,yourPoint) pair instead, and then your private key would be odd, and equal to 125. See? Your private key is not universally assigned to your public key alone, it is always relative to your base point. And for that reason, if there would be any algorithm that could give you a private key, based on some public key, it would take at least a pair of public keys as arguments.

For the same reason, there is no such thing as "which key is greater or smaller than the other". ECDSA is working like a clock, you can give me some two public keys, and then I could calculate "(firstPubKey+secondPubKey)/2", and then say that the distance between my point, and any of your two points, is identical.

Can you tell me why is this happening? Ok, if I subtract this

Code:

02A1940074961CDF60CB2A0E7BC7157A7970B05469C58EEB5AD1C0462CE0C17811

from #125 puzzle, the result is :

Code:

02ed01ff219ed5c1afc12d991a82e3063ddcee1fd53b46f7cad52a0d87a7112aed

And if I subtract this

Code:

0290ad85b389d6b936463f9d0512678de208cc330b11307fffab7ac63e3fb04ed4

which is actually 2^124 public key, if I subtract it from #125 puzzle the result is :

Code:

03ed01ff219ed5c1afc12d991a82e3063ddcee1fd53b46f7cad52a0d87a7112aed

I know why this is happening though, because the first subtraction is actually inverse addition, the distance between the first public key and the second one is exactly the private key of #125.

This is why I'm stuck and can't solve this puzzle, and this is one of the easiest of puzzles since we know exactly it starts with a 1.
But I am close God willing, it should be solved soon.

[GR Sasa]

Lol. This dogecoin @digaran actually trying to solve 125 bits which is currently the most hardest puzzle ever with some weird incorrect math HAHA.

He doesn't know that he needs to bruteforce at least 63 bits for a CHANCE that he gets the correct key using algorithm. Idk how subtraction and Multiplication would help u in such large range address to solve it lol.

it make absolutely no sense. Even if the pub key is known.

[garlonicon]

This is why I'm stuck and can't solve this puzzle, and this is one of the easiest of puzzles since we know exactly it starts with a 1.

All puzzles starts with "1". You can go through all keys, and subtract 2^N for each N-th key.

[BTC_Backdoor]

Can you tell me why is this happening? Ok, if I subtract this

Code:

02A1940074961CDF60CB2A0E7BC7157A7970B05469C58EEB5AD1C0462CE0C17811

from #125 puzzle, the result is :

Code:

02ed01ff219ed5c1afc12d991a82e3063ddcee1fd53b46f7cad52a0d87a7112aed

And if I subtract this

Code:

0290ad85b389d6b936463f9d0512678de208cc330b11307fffab7ac63e3fb04ed4

which is actually 2^124 public key, if I subtract it from #125 puzzle the result is :

Code:

03ed01ff219ed5c1afc12d991a82e3063ddcee1fd53b46f7cad52a0d87a7112aed

I know why this is happening though, because the first subtraction is actually inverse addition, the distance between the first public key and the second one is exactly the private key of #125.

This is why I'm stuck and can't solve this puzzle, and this is one of the easiest of puzzles since we know exactly it starts with a 1.
But I am close God willing, it should be solved soon.

Ohh boy you are 100% correct. The real problem is one can only do first subtraction and after all, it doesn't reduces its size. Why? let me explain.
The private key which we are looking for is between 2^124 : 2^125.
Now you need to understand that amount of numbers or p.keys from 1 to 2^124 is equal to the amount between 2^124 : 2^125.
So when we subtract 2^124 from the pubkey which is between 2^124 : 2^125, then we are actually making it fall between the range of 1 to 2^124.
Let me take you to a step further. If the puzzle private key is above the half in range 2^124 : 2^125, then you can do one more subtraction. You can further subtract 2^123 from previous subtraction result. That's it.
So far, I figured that to actually work with this subtraction thing, one need a data base of public key from 2 to 2 Trillion for that matter or more, and perform various models of subtraction from puzzle pub key and check if resulting pubkey is falling within that 2 to 2 trillion result. That's the best shot I've reached so far.

[digaran]

Lol. This dogecoin @digaran actually trying to solve 125 bits which is currently the most hardest puzzle ever with some weird incorrect math HAHA.

He doesn't know that he needs to bruteforce at least 63 bits for a CHANCE that he gets the correct key using algorithm. Idk how subtraction and Multiplication would help u in such large range address to solve it lol.

it make absolutely no sense. Even if the pub key is known.

Dogecoin, really? To know that my math is incorrect you have to know the correct math, how else could you determine it is incorrect?
But yes, LOL indeed 2^125 times.

All puzzles starts with "1". You can go through all keys, and subtract 2^N for each N-th key.

Maybe you mean they all start with 1 in binary code? Otherwise puzzle key ranges differ from one another.

Now you need to understand that amount of numbers or p.keys from 1 to 2^124 is equal to the amount between 2^124 : 2^125.

I do realize that, one needs to understand certain things before starting to work on these puzzles!
I guess you are way off the map! 2 trillion keys? The actual number is 5 to 6 public keys in order to form 2 triangles and solve the key, more than that will confuse you, as it has confused me with a few million keys, I see hexadecimal characters wherever I look, after working on them for hours. Lol.

[BTC_Backdoor]

I guess you are way off the map! 2 trillion keys? The actual number is 5 to 6 public keys in order to form 2 triangles and solve the key, more than that will confuse you, as it has confused me with a few million keys, I see hexadecimal characters wherever I look, after working on them for hours. Lol.

LOL, these hexadecimals do often come into my dreams and dancing around as well.
I sure didn't got your point of TRIANGLES. What I was referring to trillion thing is this.. Suppose we are searching the private key which is within bit range of 16:32. I made a list of initial 6 public keys like I have a file in which 1-6 public keys are stored. Now I subtract 16 from the puzzle public key. Suppose the hidden private key is 21. So when I subtract 16 from it the resulting key at the same position but in lower bit size. Now 21-16=5 and now I check whether this public key is in my file in which I saved initial public keys. So this is simple thing I am referring as 2 trillions... Can you explain your TRIANGLE THING???

[digaran]

I guess you are way off the map! 2 trillion keys? The actual number is 5 to 6 public keys in order to form 2 triangles and solve the key, more than that will confuse you, as it has confused me with a few million keys, I see hexadecimal characters wherever I look, after working on them for hours. Lol.

LOL, these hexadecimals do often come into my dreams and dancing around as well.
I sure didn't got your point of TRIANGLES. What I was referring to trillion thing is this.. Suppose we are searching the private key which is within bit range of 16:32. I made a list of initial 6 public keys like I have a file in which 1-6 public keys are stored. Now I subtract 16 from the puzzle public key. Suppose the hidden private key is 21. So when I subtract 16 from it the resulting key at the same position but in lower bit size. Now 21-16=5 and now I check whether this public key is in my file in which I saved initial public keys. So this is simple thing I am referring as 2 trillions... Can you explain your TRIANGLE THING???

Well, you are guessing the number, you could as well be subtracting 2 from 21 and not knowing it, and then going to search below 10 in hopes of finding your "5".

About triangulation, it is similar to how GPS works, but instead of sending and receiving signals, we add and subtract, you basically need only to know the value of 2 keys out of 6 keys, you have to find mutually related keys and find your way by adding and subtracting your 2 known keys to and from 4 unknown mutually related keys to find one of them!

It is rather complex, if I had the whole idea pictured in front of me, I could have explained it better, this however is easy to do on #125 because we know the first character is 1, and if I manage to solve it, the method won't work for any other puzzle, unless another puzzle starts with 1. Lol.


I'm going to call this method, "PPP" short for PinPointing  
Position!

I have managed to figure out that #125 starts with obviously 1 and the second character is definitely greater than 4, but I know for sure it is greater than 14000000000000000000000000000000  I'm still confused about the way numbers work in EC, but if for example I use 2^124 aka 10000000000000000000000000000000  as one of my 2 base point keys, I just need to add and subtract other secret keys by the same numbers to see if I can find a few of their mutual friends, that is how I'd know which one is smaller than my target.

But now I could be subtracting a -n key from my target and actually be adding their difference to my target, e.g adding -70 to +50  will result in -20.

[BTC_Backdoor]

Subtracting -70 from +50 is resulting in +120.
I am still not getting trangular thing, can you explain with a simple example pl

[_Counselor]

Yeah, if you know that K lies between A and B, you can subtract A from K, and thereby "reduce" the key, but any other subtractions is just a blind guessing.
As for the any number manipulations, secp256k1 operations obey all the basic math laws, i.e. A - 2*A = -A; A - -A = A+A; A+B+C = A + (B+C) = (A+B) + C; A*(B+C) = A*B + A*C and so on.

[digaran]

Subtracting -70 from +50 is resulting in +120.
I am still not getting trangular thing, can you explain with a simple example pl

Lol, I corrected my mistake. As for showing examples, if I had a working example, I would have used it already to solve the puzzle!
But here how it works in my mind, we have 2 and 4 as our base known points, and two unknown keys, we pick 15 and 20  for simplicity, we start by subtracting 2 from 15 and 4 from 15.

Now we have two new mutual keys, 11 and 13, if we subtract 11 from 20 we will have 9 which by subtracting 2*4 = 8 from it, we will have half of our first base point "2" ( aka 1).

Second step, we subtract 13 from 20 to have 7, now if we subtract 2+4 = 6 from it, we still have the half of our first base point "2" ( aka 1).

Therefore we have a triangle 2, 4 and 1, our known keys, and then we have 7, 9, 11, 13, 15, 20! Wait that turned into a threesome of triangles, 🤣 anyways the quantity of triangles is insignificant  compared to the quality of our equation.

The important part, if we do 2+4 = 6 *4 = 24, we now have 20 + 4 = 24, because we know 4, by adding it to x= 20(unknown key) we can observe whether any of our results are mutually related or not.

More complicated equation, we do the following, 2*4 = 8*2 = 16, now adding 4 to it, we will observe our x= 20 again.

Trick is picking your base points carefully, otherwise you will be doing things blindly. Find mutual keys that's all.😉

[garlonicon]

Maybe you mean they all start with 1 in binary code? Otherwise puzzle key ranges differ from one another.

Of course it is in binary. Each key is two times harder to solve than the previous one. You have 2^125 in the next puzzle for a reason. It is always 2^N for each puzzle. And how 2^125 looks in binary? Of course it is just one with 125 zeroes.

But even if you really want to calculate everything in hexadecimal, then still, there is the same pattern, just it is harder to see. If you check the first digit in the private key, it is always in the same range:

Code:

0001 -> 1
001. -> 2 or 3
01.. -> from 4 to 7
1... -> from 8 to f

And then, it is simply repeated every four keys, just because if you have 16 hexadecimal digits, then they take 4 bits. It is that simple.

If you really want to see those leading ones in every single key, you can just multiply known keys, and see:

Code:

...0001 -> already starts with one
...0003 * 8 = ...0018 -> leading one
...0007 * 4 = ...001c -> leading one
...0008 * 2 = ...0010 -> leading one
...0015 -> already starts with one
...0031 * 8 = ...0188 -> leading one
...004c * 4 = ...0130 -> leading one
...00e0 * 2 = ...01c0 -> leading one
...01d3 -> already starts with one

Now, can you see that every key has a leading one in binary? If you use hexadecimal, then you can just multiply it by 8,4,2,1, and that pattern repeats for every key in the puzzle. So, if you can see some key that has some leading one in hexadecimal, it doesn't mean it is easier. It is exactly as hard as other keys, because for every N-th output, you can just subtract 2^N. It always works.

For 2^125, you can simply see the leading one in your range, just because 125/4=31+1/4. That means, you have 31 hexadecimal numbers, and the leading one. But still, if you want to see the leading one in 115-bit puzzle, you can multiply it by 4, and you will see it (because 115/4=28+3/4, but 117/4=29+1/4, so you can move it by two bits to the left by multiplying it by 4).

[digaran]

I give up,

Hey, I don't know who you are but don't give up yet!

Here I will give you some clues to keep going.

Imagine puzzle 125 is  number 25.

Now instead of subtracting a smaller key, we'd subtract it from a bigger key, we'd subtract it from 30, aka 2^125 = 0000000000000000000000000000000020000000000000000000000000000000
And the result =  0286936a275e6d53bb2b2718c93d8a5aa44f371f6e0300abb73b89dd851d2fbe88
Now the result is around 1/5 of our target, give or take. Now forget about puzzle #125 and focus on this new key.😉

[vjudeu]

Now instead of subtracting a smaller key, we'd subtract it from a bigger key, we'd subtract it from 30

It doesn't matter that much. If you have 123456 as your solution, and you know it is between 100000 and 200000, then it doesn't matter if you will try to solve 23456 (by subtracting lower range) or 76544 (by subtracting higher range). In the worst case, you have 100000 values to check. Also, people usually use subranges, and scan for example keys from 111000 to 112000. Sometimes they are lucky, scan subrange from 123000 to 124000, and reach the right answer, but still, in a typical scenario, your initial subtraction doesn't change that much.

[digaran]

Do people really think Satoshi is reading all the posts here every day?
I don't know if anyone here is a giver, unfortunately we are here to take the loots.

Welcome to the jungle, and happy hunting.😉

Wow guys! I just discovered something, everytime I double a key I can get double the distance with my original base key!

I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit?  A good thing about mod function in EC, you could keep halving for eternity without reaching destination, as we are moving towards bitcoin halving, we should start key halving to solve some puzzles before the halving event! 😉

[elvis13]

Digaran @
You can't offer anything useful. An endless stream of insanity.!

[vjudeu]

I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit?

You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation.

Code:

SHA-256("125-bit")=c383a1ae19ff4401f72fdfbb4ffeb6fc7a38c6692b07b188edcbdc31b0160ee7
           mask125=0000000000000000000000000000000020000000000000000000000000000000
(hash%mask125)=key=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee7
             key/2=7fffffffffffffffffffffffffffffff6a73d1a7ed2828e256cf1d5f40262814
             key-1=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee6
         (key-1)/2=000000000000000000000000000000000d1c63349583d8c476e5ee18d80b0773

See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits.

[batareyka]

digaran
If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers
I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve.
There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve.
There are special numbers on the curve that can be used to get exactly 1 digit less.
A third of the number you are looking for.
A quarter of the sought number.
You can get the mirror part of the number you are looking for, which can also be compared with the original.
And this is not even a 10th part of what is there.
You can get a lot.



But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.

I wish you success.

[digaran]

Digaran @
You can't offer anything useful. An endless stream of insanity.!

Look who is talking! Why would I offer something which could help breaking ECC? I can only hint at them, but there are already resourceful people around here explaining everything in details.

I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit?

You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation.

Code:

SHA-256("125-bit")=c383a1ae19ff4401f72fdfbb4ffeb6fc7a38c6692b07b188edcbdc31b0160ee7
           mask125=0000000000000000000000000000000020000000000000000000000000000000
(hash%mask125)=key=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee7
             key/2=7fffffffffffffffffffffffffffffff6a73d1a7ed2828e256cf1d5f40262814
             key-1=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee6
         (key-1)/2=000000000000000000000000000000000d1c63349583d8c476e5ee18d80b0773

See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits.

I already know that, we don't want +n key to be odd, halving it would halve the entire -n key which is a number near 2^256, and when we reach the middle range the halving reverses. Complex stuff! 😉

digaran
If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers
I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve.
There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve.
There are special numbers on the curve that can be used to get exactly 1 digit less.
A third of the number you are looking for.
A quarter of the sought number.
You can get the mirror part of the number you are looking for, which can also be compared with the original.
And this is not even a 10th part of what is there.
You can get a lot.



But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.

I wish you success.

Mirror, quarter, third, one digit less, I'm not interested in those, I'm interested to find new ways and equations to solve a puzzle.

I live in this jungle, and I rather not pursuit shadows, they take you to dark parts of the woods.

Good luck to you too, may the help of God be with you all brothers! We can do this. 🤣

[lordfrs]

Digaran @
You can't offer anything useful. An endless stream of insanity.!

Look who is talking! Why would I offer something which could help breaking ECC? I can only hint at them, but there are already resourceful people around here explaining everything in details.

I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit?

You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation.

Code:

SHA-256("125-bit")=c383a1ae19ff4401f72fdfbb4ffeb6fc7a38c6692b07b188edcbdc31b0160ee7
           mask125=0000000000000000000000000000000020000000000000000000000000000000
(hash%mask125)=key=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee7
             key/2=7fffffffffffffffffffffffffffffff6a73d1a7ed2828e256cf1d5f40262814
             key-1=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee6
         (key-1)/2=000000000000000000000000000000000d1c63349583d8c476e5ee18d80b0773

See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits.

I already know that, we don't want +n key to be odd, halving it would halve the entire -n key which is a number near 2^256, and when we reach the middle range the halving reverses. Complex stuff! 😉

digaran
If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers
I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve.
There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve.
There are special numbers on the curve that can be used to get exactly 1 digit less.
A third of the number you are looking for.
A quarter of the sought number.
You can get the mirror part of the number you are looking for, which can also be compared with the original.
And this is not even a 10th part of what is there.
You can get a lot.



But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.

I wish you success.

Mirror, quarter, third, one digit less, I'm not interested in those, I'm interested to find new ways and equations to solve a puzzle.

I live in this jungle, and I rather not pursuit shadows, they take you to dark parts of the woods.

Good luck to you too, may the help of God be with you all brothers! We can do this. 🤣

My algorithm is working, it will be smoother with a few minor adjustments. I tried with Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds. It took 1 second for the collider application to find the pubkey. Now writing the code to reverse the Algorithm .

I don't like working, my brain is almost numb from the blood pressure pills
i can't think

I'm sorry my English is bad, I'm writing from translate, but I hope you understand what I mean.

[digaran]

My algorithm is working, it will be smoother with a few minor adjustments. I tried with Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds. It took 1 second for the collider application to find the pubkey. Now writing the code to reverse the Algorithm .

If your algorithm is working for one key, it should work for all other keys, unless it's not working, if you can solve a known key and not any other key, you are doing things in a wrong way. Manage to find a key which you don't know it's private key, then you can claim success.

Hope you feel better.