Bitcoin puzzle transaction ~32 BTC prize to who solves it

[sssergy2705]

Just info.

Think in your own way and your own point of view about this WIF range

Do we have GPU crackers for WIF Private Keys? or do we have someone that can build 1? these have more potentials for brute forcing puzzle 66 from the Random Point of view for probably the last 16 - 18 characters ranges scanned sequentially, compared to the hexadecimal private keys because of the checksum involved in the WIF Private keys... this might be it, every hexadecimal key is valid but not all WIFs are which brings us to the point where scanning might be fun to play with

What is the point of this if what you see in every row is just representation of padded zeros?

That is the point right there "padded zeros" as explained earlier, the hexadecimal is just from 0 to F base16 whereas the WIF is a base58 equivalent of that base 16 hexadecimal representation... so what point does this make? Well, I can have 3.625 of the hexadecimal character represented as just 1 character of the WIF... this in otherwords simply means if you are able to hit the correct first 3 WIF characters, then you probably don't have too much range to scan compared to hitting the correct first 3 characters of the hexadecimal representation of the same key... you've got a long way to go bro... So i See some potentials in this WIF bruteforcing for the puzzle 66

Key search range 66
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q ZVfMsBQggk69993Lj3p
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q aCtgAeZDbST4e5pMroG

We can't do anything with this information.
In fact, it will be the same brute force of keys. There is no difference in what format to sort through the keys, even in binary, even in hex, even in base58.

How you can explain this ?

Puzzle 118 address is 1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6

this is the 0000000000000000000000000000000000af0f4d11574f5deee49961d9609ac6 address 1PEuXxTXfc3qJM4H7EKRbSb8TSPFtds9A5

Public key 02bc5f15678ceb70dad97a6b695b9e0df7c405142586931801c2df664563042fe7

I just random search WIF range on that, and i have lock the range area by Hex padded.

yeah it's suck it's like 0.0000001% chance hit the 4 same HEX value / WIF.


[EDIT 10:41 Indonesia Time]

Target

1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6 # 118

this, i found 3 word same address range, from range padded hex.

031e841aa39d5d92ae850400a2d8be24a245d53e9ba29dc86b86696e65290c32d7 = 1PJgNwNbpr1KRxdmPxPXMCWqdBT1bwgDEe

The public address has no binding either to the public key or to the private key.
You need to carefully study the process of obtaining bitcoin addresses.

[Tepan]

Just info.

Think in your own way and your own point of view about this WIF range

Do we have GPU crackers for WIF Private Keys? or do we have someone that can build 1? these have more potentials for brute forcing puzzle 66 from the Random Point of view for probably the last 16 - 18 characters ranges scanned sequentially, compared to the hexadecimal private keys because of the checksum involved in the WIF Private keys... this might be it, every hexadecimal key is valid but not all WIFs are which brings us to the point where scanning might be fun to play with

What is the point of this if what you see in every row is just representation of padded zeros?

That is the point right there "padded zeros" as explained earlier, the hexadecimal is just from 0 to F base16 whereas the WIF is a base58 equivalent of that base 16 hexadecimal representation... so what point does this make? Well, I can have 3.625 of the hexadecimal character represented as just 1 character of the WIF... this in otherwords simply means if you are able to hit the correct first 3 WIF characters, then you probably don't have too much range to scan compared to hitting the correct first 3 characters of the hexadecimal representation of the same key... you've got a long way to go bro... So i See some potentials in this WIF bruteforcing for the puzzle 66

Key search range 66
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q ZVfMsBQggk69993Lj3p
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q aCtgAeZDbST4e5pMroG

We can't do anything with this information.
In fact, it will be the same brute force of keys. There is no difference in what format to sort through the keys, even in binary, even in hex, even in base58.

How you can explain this ?

Puzzle 118 address is 1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6

this is the 0000000000000000000000000000000000af0f4d11574f5deee49961d9609ac6 address 1PEuXxTXfc3qJM4H7EKRbSb8TSPFtds9A5

Public key 02bc5f15678ceb70dad97a6b695b9e0df7c405142586931801c2df664563042fe7

I just random search WIF range on that, and i have lock the range area by Hex padded.

yeah it's suck it's like 0.0000001% chance hit the 4 same HEX value / WIF.


[EDIT 10:41 Indonesia Time]

Target

1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6 # 118

this, i found 3 word same address range, from range padded hex.

031e841aa39d5d92ae850400a2d8be24a245d53e9ba29dc86b86696e65290c32d7 = 1PJgNwNbpr1KRxdmPxPXMCWqdBT1bwgDEe

The public address has no binding either to the public key or to the private key.
You need to carefully study the process of obtaining bitcoin addresses.

Might be you need double the reading of my progress, it's not binding to pub or priv, just calculated the random and padded HEX on the same range, it's like HEX range -> check 1 by 1 hex to WIF -> Segwit Base 58 (If the address have the same targeted addresses, keep the range, if not skip the hex range *not saved for the next batch*).

This is the demo for 130 Puzzle

Puzzle 130 addresses 1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua

   2baf7cf4d51574f5deee49961d9609ac6 # 1FHMVBF6Z5ebm5LJKRpkZs8RjNJXUMyUVV # KwDiBf89QgGbjEhKnhXJuH85q VZKQivFjnRMPNnqDyFoSqrSaKfd
   2baf78fec51be4f59eaf6966bd9709ac6 # 1FXUvZnrcqRnJV1Cvre2ribDz6yKqQugjR # KwDiBf89QgGbjEhKnhXJuH85q VMiSmxd4F8urBwimS72wrJD2XYS
   2baf78fac5b3e4f6feaf6962bd9709ac6 # 1FqZVS2u9iqcycye9BmtVEcCGM6wCYXHM2 # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvjexRvRbBdvgeEZd
   2baf78fac5b3e4f6feaf6942bd9709ac6 # 1Fe7X3dKymfuae7xLDZ9Hk5dn9W5pf4SE8 # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvjAC4AFCc3v8Pooi
   2baf78fac5b3e4f6feaf69c2bf1709ac6 # 1FnUH4z8cgLuWBLFFWvXBDdQ1VfFc4SM1k # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvm9FrvdFhuoEb9k1
   2baf78fac5b3e4f6feaf69f2bf9a8a46c # 1FfUFwHRS4cEESP3JDBMQMaZxLCZTN8tY  # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvmtQXdpe3Zfy6UtD
   2baf78fac5b3e4d018100000000000000 # 1F2nhqcHQRKwjjokgkKkcSpLH4uucuQZSV # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPuxaPiBGv4PL2aFAnuFR8
   3baf70fac037e51011500040000000000 # 1FpLqZyycLiaLWQ6Lbe5zcXuypbX6F7q4C # KwDiBf89QgGbjEhKnhXJuH8Ma WVpmffaLXxzgafdFBXKhAAvtpYn
   3baf7efac999e3c95c4ea7660b9946c94 # 1Fcmac1xrLNMMr286BsFnWN35pFNgZeWWb # KwDiBf89QgGbjEhKnhXJuH8Ma XCqFfLS8nKhgd7RpHR2XZB5AKhx
   3baf7efac999e3cddc6ba764fb9b66cb4 # 1FqfgJmp8AKvDkAKbUcncgMA7kmDn9dJ2v # KwDiBf89QgGbjEhKnhXJuH8Ma XCqFfLS9Eo6LP4p7NDXN5sx82Js
   3bbff6cba4bd242d6ceb2777fb9866ca4 # 1FfPWje1WcZ47fAKuGBijKsqhvF8bHzf3f # KwDiBf89QgGbjEhKnhXJuH8Me C7XhwLTtDcX5XKNNEHm7o8DUj98
   3bbff6cb95bca56cdee22756fb970bcaa # 1F2UYUqkShEp8aa9CCE6mKN3yLHbKpEPud # KwDiBf89QgGbjEhKnhXJuH8Me C7XfgWcWBrqsVyALfRFgYf7PCe9

it's BSGS and manually count by my magic math.
 

[Denis_Hitov]

Hey guys, instead of wasting your eye sight on long and useless base58 WIFs which literally represent 0s in hexadecimal, let me share a little secret regarding public keys.

Here is how you can find half of your public key, it's not straight forward method but I bet many of you didn't know about it.

First we need to extract 1 and half of our public key then we can subtract our p which is 1 from it's 1.5 to get it's 0.5 half. Though we could just divide it by 2 without all this trouble, this is a hint to make you dive deeper in to this vast ocean of numbers and equations.

Target  pub:

Code:

03219b4f9cef6c60007659c79c45b0533b3cc9d916ce29dbff133b40caa2e96db8

Target priv:

Code:

0x800000000000

Our multiplier inverse or not doesn't matter, +n will give you +n result and vice versa.

Scalar : aka n/2+1 half n +1 or 1 and half of n.

Code:

7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2

If you multiply our target pub with scalar above, you will get this :
Pub2 :

Code:

02161c6cbee1483deaf6f9b395c817eb019228cda5afac5857295ba10959dffc96

Priv :

Code:

0xc00000000000

Now if we subtract target from pub2, we will get :

Pub3, half of target :

Code:

0313d1ffc481509beee68f17d8ff41c2590f4c85f15268605087eda8bab4e218da

Priv :

Code:

0x400000000000

We didn't even use division, *chop chop and good luck diving.😅


* = hurry! Get to work.

Now, if it were possible to calculate an even target pub or an odd one, then 130 puzzle could be solved in a few hours.

[digaran]

To install the `sympy` module on Windows, type CMD in the search bar and open Command Prompt and type
 pip install mpmath, first and then pip install sympy.
put ice dll and secp256k1.py in the same folder where you have the script.

Thanks, I did what you said and learned a few things about python commands, after a few errors now it says module secp256k1 has no attribute 'pub2upub'  even though I correctly pasted the upub in it's place.  I went to secp256k1.py and there was no mention of pub2upub. Lol this thing burned more energy than a week of dealing with EC math.

Now, if it were possible to calculate an even target pub or an odd one, then 130 puzzle could be solved in a few hours.

You can't figure out if a key is odd or even, that would make it easy to find every other digits to see if they are also odd or even.

But there is a way to make sure that the correct results are in your files and your tool had all the incorrect and correct keys.

Here is how, you know some keys can be divided by 2 a few times in a row, so when you divide a key you need to divide it by 2 at least 4 times, subtract 1 from all and do the same with all the keys.  This should occupy  less than 1 GB of storage/ RAM, and it should solve the low bit range keys in minutes, but I don't know why nobody is trying it.🤔

[bestie1549]

Hey guys, instead of wasting your eye sight on long and useless base58 WIFs which literally represent 0s in hexadecimal, let me share a little secret regarding public keys.

Here is how you can find half of your public key, it's not straight forward method but I bet many of you didn't know about it.

First we need to extract 1 and half of our public key then we can subtract our p which is 1 from it's 1.5 to get it's 0.5 half. Though we could just divide it by 2 without all this trouble, this is a hint to make you dive deeper in to this vast ocean of numbers and equations.

Target  pub:

Code:

03219b4f9cef6c60007659c79c45b0533b3cc9d916ce29dbff133b40caa2e96db8

Target priv:

Code:

0x800000000000

Our multiplier inverse or not doesn't matter, +n will give you +n result and vice versa.

Scalar : aka n/2+1 half n +1 or 1 and half of n.

Code:

7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2

If you multiply our target pub with scalar above, you will get this :
Pub2 :

Code:

02161c6cbee1483deaf6f9b395c817eb019228cda5afac5857295ba10959dffc96

Priv :

Code:

0xc00000000000

Now if we subtract target from pub2, we will get :

Pub3, half of target :

Code:

0313d1ffc481509beee68f17d8ff41c2590f4c85f15268605087eda8bab4e218da

Priv :

Code:

0x400000000000

We didn't even use division, *chop chop and good luck diving.😅


* = hurry! Get to work.

Now, if it were possible to calculate an even target pub or an odd one, then 130 puzzle could be solved in a few hours.

The tools already developed achieve these operations better...
all we need is something to work better than these tools or programs...
if we can write all these ideas out in a code and run the scripts, we would be able to work faster on the CPU 100 times faster than the codes already built to run on GPUs.
BSGS works great, Kangaroo is awesomely perfect...
these programs are great but we can programmatically edit their functions to our desires or the codes will completely break down which eventually brings us  to the point where all we need is functions to calculate these operations programmatically to our own tastes and commands.
the 1s and 0s idea is a great idea that worked but still, we needed it to go faster...
the BitCrack, Kangaroo programs works super fast but what do we get when the bit ranges keeps increasing? these programs, though very fast now feels super slow because the way the keys starts to get even bigger and bigger and the ranges wider and wider which now brings us to the moment of truth...

There are only two possible explanations: either no one told us, or no one knows.
Because as we are undoubtedly gathering, the anomaly is not systemic, and it's not creating any fluctuations in even the most simplistic equations.

The problem is choice.

Our function is now to return to the source code, allowing a temporary dissemination of the code, reinserting the prime program.
Failure to comply with this process will result in a cataclysmic system crash in the brain because we won't be able to break the curve without getting to it's source.

Which brings us at last to the moment of truth, wherein the fundamental flaw is ultimately expressed, and the anomaly revealed as both beginning, and end... and the curve at one point is a reflection of another point... wherein we can call it a mirror.

There are two doors. One door is bruteforcing or as we all call it, BSGS, Kangaroo... whatever you like you can call it... and the other door is owning, buying, mining. this door only costs you some resources like money or vis-a-vis finance but the first door only gives us Hope, it is the quintessential human delusion, simultaneously the source of our greatest strength, and our greatest weakness.

[Tepan]

Hey guys, instead of wasting your eye sight on long and useless base58 WIFs which literally represent 0s in hexadecimal, let me share a little secret regarding public keys.

Here is how you can find half of your public key, it's not straight forward method but I bet many of you didn't know about it.

First we need to extract 1 and half of our public key then we can subtract our p which is 1 from it's 1.5 to get it's 0.5 half. Though we could just divide it by 2 without all this trouble, this is a hint to make you dive deeper in to this vast ocean of numbers and equations.

Target  pub:

Code:

03219b4f9cef6c60007659c79c45b0533b3cc9d916ce29dbff133b40caa2e96db8

Target priv:

Code:

0x800000000000

Our multiplier inverse or not doesn't matter, +n will give you +n result and vice versa.

Scalar : aka n/2+1 half n +1 or 1 and half of n.

Code:

7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2

If you multiply our target pub with scalar above, you will get this :
Pub2 :

Code:

02161c6cbee1483deaf6f9b395c817eb019228cda5afac5857295ba10959dffc96

Priv :

Code:

0xc00000000000

Now if we subtract target from pub2, we will get :

Pub3, half of target :

Code:

0313d1ffc481509beee68f17d8ff41c2590f4c85f15268605087eda8bab4e218da

Priv :

Code:

0x400000000000

We didn't even use division, *chop chop and good luck diving.😅


* = hurry! Get to work.

Now, if it were possible to calculate an even target pub or an odd one, then 130 puzzle could be solved in a few hours.

Yeah absolutely 1 hours 4 minutes, but he's doin nothing with random speculation ahaha, see.
just computing use 50% brain and computing power, not 100% brain enough.

To install the `sympy` module on Windows, type CMD in the search bar and open Command Prompt and type
 pip install mpmath, first and then pip install sympy.
put ice dll and secp256k1.py in the same folder where you have the script.

Thanks, I did what you said and learned a few things about python commands, after a few errors now it says module secp256k1 has no attribute 'pub2upub'  even though I correctly pasted the upub in it's place.  I went to secp256k1.py and there was no mention of pub2upub. Lol this thing burned more energy than a week of dealing with EC math.

Now, if it were possible to calculate an even target pub or an odd one, then 130 puzzle could be solved in a few hours.

You can't figure out if a key is odd or even, that would make it easy to find every other digits to see if they are also odd or even.

But there is a way to make sure that the correct results are in your files and your tool had all the incorrect and correct keys.

Here is how, you know some keys can be divided by 2 a few times in a row, so when you divide a key you need to divide it by 2 at least 4 times, subtract 1 from all and do the same with all the keys.  This should occupy  less than 1 GB of storage/ RAM, and it should solve the low bit range keys in minutes, but I don't know why nobody is trying it.🤔

just remember bro, what are you doing in right now it's already done with someone who was not in this forum, believe me or not, u just need develop it no being amateur with code flying inside.

just ensure the progress and don't be a jerk about someone idea.

[digaran]

just ensure the progress and don't be a jerk about someone idea.

Sorry but I'm not really being a jerk when I say what you are doing is waste of time, I wasted 2 month on WIFs, then I found out there is no shortcut unless we know the checksum which is 6 base58 characters even though the checksum is not part of the hex private key but knowing it could change everything.

Apology for calling you a spammer, you could post a one line example instead of 20 lines.

Btw, if I wanted to spend time on learning how to code, I couldn't learn the little I know about ECC.  Imagine I spent 1 hour to download and install packages and then copy them one by one to the directory of my script just to realize I am not built for such things, every person has his own sets of skills, learning how to code is not mine to master.

[bestie1549]

just ensure the progress and don't be a jerk about someone idea.

Sorry but I'm not really being a jerk when I say what you are doing is waste of time, I wasted 2 month on WIFs, then I found out there is no shortcut unless we know the checksum which is 6 base58 characters even though the checksum is not part of the hex private key but knowing it could change everything.

Apology for calling you a spammer, you could post a one line example instead of 20 lines.

Btw, if I wanted to spend time on learning how to code, I couldn't learn the little I know about ECC.  Imagine I spent 1 hour to download and install packages and then copy them one by one to the directory of my script just to realize I am not built for such things, every person has his own sets of skills, learning how to code is not mine to master.

codes make life easy bro, copying and pasting could be easily done with a few codes... the files your notepads and editors can't open... say 100GB of pubkeys in a single file, you can easily manipulate the data in the file even if you have 1GB RAM... the point is the things you are busy spending hours to calculate, you could just build a program that would run it 1 million times before you are done doing 3. that's why it's called coding or programming... you program it to your taste. you could almost leave your system running all on it's own if you have the programs needed without even opening anything as long as you've written the program to fulfil your choice of operation. double clicking is too slow and I bet if you knew coding, we wouldn't be stuck with puzzle 66 or puzzle 130 by now... You have the ideas needed but you're not flexible enough to write the programs needed to carry out the operations. That's why I said sometimes ideas might be half in the bearer's mind but it takes another mind thinking towards the same direction to make it a full idea...

you know all the divisions and G factors to multiply with and but if a code should handle those ideas you'd definitely be solving these puzzles like it's nothing... you don't necessarily need to learn the coding yourself if you don't want to, but just try to get closer to someone that knows how to translate these ideas to programs and you'll see for yourself how you won't be needing to spend hours dividing and multiplying keys manually. just always make sure there's an argparse or a command line args which is defined already in places where you need to constantly make some edits on the code so you don't have to always edit the code itself every time. we all are learning and until we move on to the other side of the world can we ever stop learning.

[mcdouglasx]

To install the `sympy` module on Windows, type CMD in the search bar and open Command Prompt and type
 pip install mpmath, first and then pip install sympy.
put ice dll and secp256k1.py in the same folder where you have the script.

Thanks, I did what you said and learned a few things about python commands, after a few errors now it says module secp256k1 has no attribute 'pub2upub'  even though I correctly pasted the upub in it's place.  I went to secp256k1.py and there was no mention of pub2upub. Lol this thing burned more energy than a week of dealing with EC math.

Now, if it were possible to calculate an even target pub or an odd one, then 130 puzzle could be solved in a few hours.

You can't figure out if a key is odd or even, that would make it easy to find every other digits to see if they are also odd or even.

But there is a way to make sure that the correct results are in your files and your tool had all the incorrect and correct keys.

Here is how, you know some keys can be divided by 2 a few times in a row, so when you divide a key you need to divide it by 2 at least 4 times, subtract 1 from all and do the same with all the keys.  This should occupy  less than 1 GB of storage/ RAM, and it should solve the low bit range keys in minutes, but I don't know why nobody is trying it.🤔

Code:

target_public_key = "03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852"
pub = ice.pub2upub(target_public_key)

replace the line with this, the pub should be compressed, don't forget to copy secp256k1.py and ice_secp256k1.dll in the same folder where your script is located.

[unpluggedcoin]

There are two doors. One door is bruteforcing or as we all call it, BSGS, Kangaroo... whatever you like you can call it... and the other door is owning, buying, mining. this door only costs you some resources like money or vis-a-vis finance but the first door only gives us Hope, it is the quintessential human delusion, simultaneously the source of our greatest strength, and our greatest weakness.

About the DOORs, I read on this forum at some other threads about Backdoors, and Oracle Black box,, that may be true or false God Knows, but as far I tried for some one and half year around Elliptic Curve maths, there is no way to somehow calculate or comeup with existing mathematical functions of the curve like Addition, Subtraction, Division and Multiplication that can replace any bruteforce method, NOT A CHANCE!! Few months earlier I started working on the point at infinity! The ultimate master source behind all Elliptic Curves. Though I am still lost in this jungle along with you and didn't reached anywhere but I suspect there might be a third Door behind this point at infinity!!! It is also related with Euclidean key to Universe as Stephen Hawkins said, i guess!!!!!

[GR Sasa]

Warning: Please stay On-Topic, otherwise you will be getting banned

[Tepan]

just ensure the progress and don't be a jerk about someone idea.

Sorry but I'm not really being a jerk when I say what you are doing is waste of time, I wasted 2 month on WIFs, then I found out there is no shortcut unless we know the checksum which is 6 base58 characters even though the checksum is not part of the hex private key but knowing it could change everything.

Apology for calling you a spammer, you could post a one line example instead of 20 lines.

Btw, if I wanted to spend time on learning how to code, I couldn't learn the little I know about ECC.  Imagine I spent 1 hour to download and install packages and then copy them one by one to the directory of my script just to realize I am not built for such things, every person has his own sets of skills, learning how to code is not mine to master.

that's the point of the puzzle, makes you work, it's not a newborn puzzle everyday like on X, it's found and launch from 2015, someone do BSGS like month and year, 2 month trying is nothing.

[Tepan]

Warning: Please stay On-Topic, otherwise you will be getting banned

I still on the topic, others use some math with computating ideas, i just use basic bruteforce idea and powered base by proof of wallet import format.

just clarification, the puzzle it's owned by native script and secured by public key and hidden private key.

we need just to know what the singularity the solved problem the puzzle upper from 60.

[digaran]

Anyone knows what is going on with my python windows?

Aside from adding ^^^^^^^ under ECC operations on screen, everytime I fix a missing lib/module, it returns error about some line 447 or 464 in secp256k1 module, error about pkg-config which I can't install because no such file, even if I download it and add it to my folder, it wont work. Is this related to the py version? I have python 3.10, and also 3.11.  Is there a place where I could paste my script to have access to everything instead of installing and moving needed files one by one? Lol

I have some old scripts which some moron AI helped me to code back then, if I can get python scripts running without any errors I could test them.


Sir plz, don't banned me just learn here.😅

[mcdouglasx]

Anyone knows what is going on with my python windows?

Aside from adding ^^^^^^^ under ECC operations on screen, everytime I fix a missing lib/module, it returns error about some line 447 or 464 in secp256k1 module, error about pkg-config which I can't install because no such file, even if I download it and add it to my folder, it wont work. Is this related to the py version? I have python 3.10, and also 3.11.  Is there a place where I could paste my script to have access to everything instead of installing and moving needed files one by one? Lol

I have some old scripts which some moron AI helped me to code back then, if I can get python scripts running without any errors I could test them.


Sir plz, don't banned me just learn here.😅

Try to install visual studio code to start programming, it shows errors in the code, you can easily download libraries, you can use it for python or another language, the problem is that it can be slow when executing the codes, but with it you can edit the codes , avoid errors save them and then run it in the command window.
As for the puzzle, notice that the code you try to execute to reduce the pubkey is wrong, it uses an upub key to convert it to an Upub key,  that code is used to convert compressed to uncompressed.
Original

Code:

from sympy import mod_inverse
import secp256k1 as ice
pub=ice.pub2upub('0433709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e2a1c304a39a77775d3579d077b6ee5e4d26fd3ec36f52ad674a9b47fdd999c48')
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))


def ters(Qx,Scalar):
     ScalarBin = str(bin(Scalar))[2:]
     le=len(ScalarBin)
     for i in range (1,le+1):
        if ScalarBin[le-i] == "0":
            Qx=ice.point_multiplication(k,Qx)
        else:
            Qx=ice.point_addition(Qx,neg1)
            Qx=ice.point_multiplication(k,Qx)
     return ice.point_to_cpub(Qx)


for x in range(1,65536):
         print(ters(pub,x))

Fixed

Code:

from sympy import mod_inverse
import secp256k1 as ice
pub=ice.pub2upub('Here Compressed Public Key')
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))


def ters(Qx,Scalar):
     ScalarBin = str(bin(Scalar))[2:]
     le=len(ScalarBin)
     for i in range (1,le+1):
        if ScalarBin[le-i] == "0":
            Qx=ice.point_multiplication(k,Qx)
        else:
            Qx=ice.point_addition(Qx,neg1)
            Qx=ice.point_multiplication(k,Qx)
     return ice.point_to_cpub(Qx)


for x in range(1,65536):
         f= (ters(pub,x))
         data= open(“halfpub.txt”,”a”)
         data.write(f+”\n”)
         data.close()

[unpluggedcoin]

codes make life easy bro, copying and pasting could be easily done with a few codes... the files your notepads and editors can't open... say 100GB of pubkeys in a single file, you can easily manipulate the data in the file even if you have 1GB RAM... the point is the things you are busy spending hours to calculate, you could just build a program that would run it 1 million times before you are done doing 3. that's why it's called coding or programming... you program it to your taste. you could almost leave your system running all on it's own if you have the programs needed without even opening anything as long as you've written the program to fulfil your choice of operation. double clicking is too slow and I bet if you knew coding, we wouldn't be stuck with puzzle 66 or puzzle 130 by now... You have the ideas needed but you're not flexible enough to write the programs needed to carry out the operations. That's why I said sometimes ideas might be half in the bearer's mind but it takes another mind thinking towards the same direction to make it a full idea...

you know all the divisions and G factors to multiply with and but if a code should handle those ideas you'd definitely be solving these puzzles like it's nothing... you don't necessarily need to learn the coding yourself if you don't want to, but just try to get closer to someone that knows how to translate these ideas to programs and you'll see for yourself how you won't be needing to spend hours dividing and multiplying keys manually. just always make sure there's an argparse or a command line args which is defined already in places where you need to constantly make some edits on the code so you don't have to always edit the code itself every time. we all are learning and until we move on to the other side of the world can we ever stop learning.

Well said! So far the ideas that Diagran has shared, I can easily resemble them with mine and I have had python scripts on all of em! the point is, with these methods, you are trying either to reduce the size of key, or break ECDSA in any form, like if you figure odd and even of a point without doubling it (we can know odd and even through doubling and adding G to doubled point makes it odd) but we can't figure it out with some other form or manipulation!! I tried lots and lots of python scripts on these and other various forms of calculus.... I am done with these as I realized why they say in modular you'll enter a loop from where you can't escape! So I decided to approach this circle of elliptic curve from outside instead of inside. The only think attached to elliptic curve that can be considered outside of this circle is POINT AT INFINITY! I working on point at infinity, someone earlier referred two doors, I think the real door is behind point at infinity! Though I am not anywhere yet but figured few observations and working on it!
From my experience of write and testing scripts on ideas like that of diagran, it will still be optimal to test the idea manually instead of spending hours to fine tuning a script and again matching, calculating, validating various observations untillyou stuck at dead end!
I don't recommend building programms at initial stage of ideas, Even with long x and y coordinates decimals and hex string you'd be better off calculating and testing your ideas manually then to develop a program and test it!

[digaran]

for x in range(1,65536):
         f= (ters(pub,x))
         data= open(“halfpub.txt”,”a”)
         data.write(f+”\n”)
         data.close()

It seems I was using ice secp256k1 files from 2022, after downloading the latest one from github I managed to get it running, but first public key I provided the script only generated 780 keys and returned error no 13, permission denied to open the txt file, my second attempt with another key generated 65535 keys and stopped, now what happened with the first run, why would it want to open the txt file?

Btw, how can I change the number of subtraction? For example which ones should I change if I wanted to subtract 50 instead of 1 and then divide?


k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))


def ters(Qx,Scalar):
     ScalarBin = str(bin(Scalar))[2:]
     le=len(ScalarBin)
     for i in range (1,le+1):
        if ScalarBin[le-i] == "0":
            Qx=ice.point_multiplication(k,Qx)
        else:
            Qx=ice.point_addition(Qx,neg1)
            Qx=ice.point_multiplication(k,Qx)
     return ice.point_to_cpub(Qx)


Should I replace all 1s with 50? Lol, I'm noob.

Your help is appreciated.👍

2nd edit:
A hint for interested parties, 2^255, find it's associated private key and double it mod n. Cheers. 😉

[Tepan]

Hey guys, I just think to learn the script how this address are calculated.

because bitcoin has its own native scripting language,
While funds are usually locked to a public key alone, they can also be secured with a script.

we just need to know how the address calculated, no reverse math in here.

soon i'll update what i learn about the pubkey and the address.

because this puzzle was working on single signature (P2PKH).

Peace.  

eg. i need to learn.

[bestie1549]

Hey guys, I just think to learn the script how this address are calculated.

because bitcoin has its own native scripting language,
While funds are usually locked to a public key alone, they can also be secured with a script.

we just need to know how the address calculated, no reverse math in here.

soon i'll update what i learn about the pubkey and the address.

because this puzzle was working on single signature (P2PKH).

Peace. 

Awesome bro... We'd be waiting for the updates from you. share let us all learn. thanks

[Tepan]

Hey guys, I just think to learn the script how this address are calculated.

because bitcoin has its own native scripting language,
While funds are usually locked to a public key alone, they can also be secured with a script.

we just need to know how the address calculated, no reverse math in here.

soon i'll update what i learn about the pubkey and the address.

because this puzzle was working on single signature (P2PKH).

Peace. 

Awesome bro... We'd be waiting for the updates from you. share let us all learn. thanks

Because i read more and figure the ways.
bitcoin is never moved, is just locked and unlocked.
The only change is who can spend that amount.