I've noticed a strange phenomenon in mod calculations. I wonder whether this particular question can be answered by someone fluent with maths:

4   8   16   3   7   11   15   2   6   10   14   1   5         
5   10   15   3   8   13   1   6                        
6   12   1   7                                    
7   14   4   11   1   8                              
8   16   7   15   6   14   5   13   4   12   3   11   2   10   1   9


Above numbers are running on mod 17. If we double a number, say 4, we double it 11 times until it results in 1.
Similarly if we pick number 5 and start doubling it, we need to double it 6 times to reach 1. and two times double for 6, and 4 time doubling with 7 and so on.

I wonder, if we somehow know the starting number, which we are doubling, can we determine or calculate number of doubles it requires to reach value 1?

This is surely related to EC calculations and I believe it certainly will help minimize size of private key bay many folds, depending on doubles!!!

I've been watching this forum for around two year, I did saw that advice that you were talking about, mentioning this same platform before  . Though I ignored but I have had that feeling of loosing and wasting time once in a while. I don't see similar life-balancing comments often here 
There are bashers here who might be as wealthy as filthy ones, who neither share their source of wealth neither they want other's to have such. One can easily divide all comments in two categories, Ones adding values in other's lives and others the bashers.

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 coz 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 then to develop a program and test it!

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!!!!!

Few pages earlier OP himself asked if someone needed help to upgrade system and stuff like that, I saw people bashing other people, sort of disappointing! I did asked him if I could have access to a strong CPU with more RAM and few TB HARD, I could have cracked puzzle 125, considering the time I ran those script on my dual core with upgraded 8 RAM.

To all who are using my script and have doubts!

First of all, the script is not some kind of magical idea or trick no nothing! It is simplest subtraction of equal amount in parts! The script won't generate any points itself into checkpoints.txt file. You have to provide it both checkpoints.txt and xy.txt files. It will only generate j_value.txt file in which it will be saving number of iterations (you'll know what they are in a minute). The script is a simplest possible kangaroo sort of thing and nothing else. Let me explain again:
Suppose our puzzle pub key has the private key 1593 and its bit range is between 1024 to 2048.
You have its public key x and y coordinates in decimal format in xy.txt file in same directory.
You have checkpoints.txt file in which you have public key x coordinate only in decimal (1 per line).
These are the public keys of starting private keys. In our example suppose they are 2 to 11. Now you have 10 x coordinates in checkpoints.txt file. (The important point why I gave it a shot is that I liked the idea of doubling the checkpoints while keeping it single because starting 10 private keys and last 10 private keys have same x coordinates. So if you subtract public key of 15 (15 is private key) from the public key of 10 (10 is private key) you'll get the public key of n-4 whose x coordinate must be same as of 5.
back to our example So you place initial 11 public keys x coordinates in checkpoints.txt file.
Script loads checkpoints.txt file and subtracts from it equal parts in one iteration and write 1 in j_value.txt file. I writes 2 after second iteration and so on. So if power goes out or anything happens and system is off, your script will start from where it left. Here is our example working:
pubkeys 2-11 are loaded.
You can change your settings, i'll just show what I did back there:
As we know the range it is between 1024 and 2048 (For puzzle 125 I divided the starting range into 100,000 parts, so scripts single iteration has 100,000 parts, (these are sort of subtraction jumps) in our example suppose we pick 16 equal jumps and we divide range into 10 parts so one subtraction is subtracting 102 (there is always be a whole number in EC don't worry about perfect numbers). Let's run it,

1593-102=1491 (script matches this point within checkpoints.txt file, NO MATCH)
1491-102=1389 (NO MATCH)
1389-102=1287=(NO MATCH)
.
.
.
165-102=63 (NO MATH)

1st iteration completed. The script writes 1 in j_value.txt file. NOW script subtracts 11 from your puzzle pub key and then starts iteration:

1593-11= 1582 (Now this replaces the starting point, now please don't asks how you would get back to original pub key, be mindful)

1582-102=1480 (NO MATCH)
1480-102=1378 (NO MATCH)
.
.
.
154-102=52 (NO MATCH)

It will certainly be touching the checkpoint BUT WHEN NO BODY CAN TELL FOR CERTAIN!

I saw someone commented on my script saying how script generated checkpoints and stuff,, please study pub key generation before using this script

No it won't work! First of all you need to understand how python thing works, any single extra space or comma would messup and script would show error! You can clearly see files nomenclatures inside the script itself. like xy.txt file not x,y.txt file,, checkpoints.txt etc... Now about your question, in xy.txt file x and y coordinates should be in decimal format with a single space between them, no comma or () just simply x coordinate single space and y coordinate both in one line!
Further in checkpoints.txt file, in every line there is only single x coordinate per line in decimal format!

Here is another idea, lets see if it can be done or has any merits to it.

If we know the exact start and end range of a public key, we could divide the end range by a number to reduce it down as much as possible, maybe making it a 6 digits number, now we only have to change our 6 digit number and multiply it by the number which we first divided our end range with, we need to change the numbers in a way that it never goes beyond our end range when multiplied.

Now, before doing that, we need to add and subtract 1 G to our target and divide the results by 2, e.g. if our target is 256, we add 1, 2, 3, 4 etc and divide them all by 2 also we will add n/2 to all the results just to be safe and keep all of the keys to compare with our multiplication result.  Eventually if we change our nonce, ( 6 digit number ) enough times and multiply it with powers of 2, 3 etc somewhere we should see a known key.

A simple example :

Start range 1, end range 512, our target is 95, now we divide 512 by 32 to get 16.

But first lets generate some offsets for our target.

Add 1 to 95 = 96/2 = 48
Add 3 to 95 = 98/2 = 49
Add 5 to 95 = 100/2 = 50
Sub 1 from 95 = 94/2 = 47
Sub 15 from 95 = 80/2 = 40

Now we have 48, 49, 50, 47 and 40, then we multiply 16 by 2, 3, 4, 5, 6, 7, etc until we see one of the keys above.

Imagine if we generate different offsets, like billions of them, then we could simply keep multiplying to find one of our targets.

---

This just came to my mind today, it's not cooked well , yet!😉

[/quote]

I already tried and test it and much more that I don't remember but someone refer it I already know how it will slap our ass  !
The reason this and every similar approach won't work is strictly due to the modular behaviour of calculations!
Which is why BSGS Kangaroo has so far been the best approach is because is increases the chances of luck! However, any other way around addition and subtraction would not work, BECAUSE there is no full stop, when your calculations go around the curve and passes your mark IT WON'T RING A BELL OR SHOW ANY ALERT!!!!!!!! It would still be showing you the point and you'll be happy assuming it as per your assumption but it would change itself from being ODD to EVEN, that where everyone go nuts when we assume elliptically!!

Just copy the code and paste in notepad and save it as myscript.py (it would run on python), save in your folder, open folder, type CMD in address bar above and press ENTER, command prompt terminal will open, type there
>python myscript.py
and hit enter the script will run, you'll see step or jumps prints, chill!

Here is the code doing exactly what digaran explained earlier... I used on puzzle 125, chant it when someone else found it. I am not optimistic of it! you can have it if you've huge ram... it works on more more of a simple phenomenon rather then complex... In the script checkpoint.txt file has only public key x coordinates in decimal format. Why only x because, if you have public  keys of private keys from 1 to 1 billion, they are actually two billion public keys, first billion from 1 to 1 billion and the last billion,, both billion have same x coordinate,, so kind of really big help there ... The script first loads the file checkpoints.txt. so in this file I kept 80 million public keys from 2 - 80000000. The file size was 5.88 GB, been running on 8 GB RAM system. Here is how script works with a simple example you can modify relevant figures in script as per your needs...
Suppose our public key has the private key of 87 (which we don't know obviously) we have checkpoints from 1-5 (which is really big if we compare 80 million with 125 puzzle key distance),, so what the script does it, is subtracts a set jump from pubkey and checks the resulting point within checkpoints, you can set your own jump.. suppose you want to set jump size as 20, here is how script works:

87-20=67 (not in checkpoints)
67-20=47  (not in checkpoints)
47-20=27(not in checkpoints)
27-20=7(not in checkpoints)
7-20=-13(not in checkpoints)

As you know the range you can set the step size to suite the range, in our case, it is suitable to have 5 steps if we choose its size as 20...

After finishing 5 stepts, the script subtracts 5 from 87 and saves the progress in j_value.txt file, after every iteration it changes the value in j_value.txt file. So if you shutdown or power goes off, next time script restarts from where it left...
After first iteration script subtract 10 from 87-10=77
(10 because you have first 5 and last 5)
next iteration is like
77-20=57 (not in checkpoints)
57-20=37  (not in checkpoints)
37-20=17(not in checkpoints)
17-20=-3(matched)
-3 is matching because its inverse point is 3 and both have same x coordinate...
If a match is found, the script save that point in results.txt file.
I forget to mention xy.txt file has the decimal x and y coordinate of puzzle pub key.
Here is whole script: Chill
7-20=-13(not in checkpoints)

entende de programação?
[/quote]

Sim, como programador Python, tenho um forte conhecimento dos conceitos de programação e da linguagem Python.

Sim, sou brasileiro! Como posso ajudá-lo? Aqui estão muitos caras de merda por aí, tome cuidado com eles.

Bro you are 100% correct about your vision as well as utility! Just tell me from which website one should buy bitmap ordinals??

You rightly and correctly said it. Everyone here is getting technical around bitmap ordinals and not being humans. Being a human what you do with bitmap ordinals. Well you wish to join the ride and have a piece of it.. I've been looking for unclaimed bitmap ordinals and found out that all metaverse has been claimed. I wish if could have one! Alas!

Would you plz be specific, like solve for what? What is expected result conditions? How result should look like, can explain how the successful result should be some how

Someone bought some serious amount of bitmap land before people like us who've missed mighty Bitcoin even touched it. On July 13, yes 3 days ago.

Why on earth you said that above puzzles are already solved, however they aren't? We've been trying to solve a single puzzle 124 for past 8 months are so and now you suddenly claimed that all puzzles with pubkeys are solved. Are you checking their balances on Ethereum or some other cheap drug Blockchain?

Thank you for your interest in the ongoing puzzle and your willingness to contribute to the efforts to crack the keys. You are correct in noting that the current conclusion is that brute force remains the only viable method for cracking the keys. However, this does not mean that progress has not been made or that there is no hope of success.

Advancements in hardware technology have made it possible to perform brute force attacks much more quickly and efficiently than in the past. By using specialized hardware and parallel processing techniques, it is now possible to greatly reduce the time required to crack a key.

Personally speaking, your contribution could help me fund the acquisition of such a hardware, thereby accelerating the pace of progress and increasing the likelihood of success. With your support, I could potentially make significant strides in cracking the keys and solving this puzzle.

Furthermore, the potential benefits of cracking the keys could be significant. It could pave the way for the development of new encryption methods and security protocols, which could have far-reaching implications for the security of sensitive data across industries.

I deeply admire and appreciate your consideration and hope that you will choose me to support my effort at 15Y5kMbf8Niu6ALFMUhgP9x9RVcVoeBknm to crack the keys and solve this intriguing puzzle.

Regards,
Unplugged Coin
15Y5kMbf8Niu6ALFMUhgP9x9RVcVoeBknm