Topic: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED== - page 19. (Read 54326 times)

2^125 is big number.

I have a better chance then you do with more computations. Even if its centuries^millenia  vs millenia^millenia   There is no reason to call people p*******. Besides it being my favorite thing in the world. Unkind.

If you have any computations I would be willing to run a few here and there for you. They can be my Random or your random search. Odds will be the same IMO.

No need to be nasty. Have a pleasant day.

Thanks, please do tell what you don't understand, I also need to go over stuff a few times to actually realize my mistakes and find inspiration, sometimes when I talk about something with someone, I'd suddenly realize something new.

I will do my best if I can to explain anything which is unclear for you.

I like your ideas from this thread.

Don't understand them well however.

But I support your thinking here.

What do you mean the other half?  Of course spending money will do the trick of solving some of the puzzles, but as I said before, using high end hardware+ a lot of money to solve any of the puzzles is for dumb pussies, they wouldn't know any other way than spending cash to solve any problem they face.  Solving a puzzle by using the power of human mind is the ultimate honour one can achieve concerning this challenge.

And I don't know why so many people think that everything is in binaries, don't you know binary is the dictionary/vocabulary/language of computers? And math is the language of the universe and order, we humans do better with the things that were here before we existed, but trying to work with the things we created? Not a good idea, not only we can't learn anything from them, we'd become more dumb as many studies prove that human race is becoming dumber due to the excessive use of technology, lol. 
Imagine 50 years from now asking a 15 years old  what is 25 + 4? You would definitely see him reaching for his phone to use a calculator!

Good luck with -R flag, I assume that's  for random, only if people really knew how big a number like 2^125 is.😉

I actually thought I could see a pattern in binary for a few days. Turns out my eyes were playing tricks with math. They appear random.

Best bet here is with the -R flag until I think of something else. That and increase RAM, reason to spend $ the better half cant know about.

You guys need to stop subtracting and adding.

You are guys doing nothing but making the actual puzzle more complicated. You will most probably end on a valid public key that doesn't even exists in the whole range of #125. So it will not help.

This is 2^125-1= 1fffffffffffffffffffffffffffffff you just need to add 1 to it.
I didn't have  this
cfbcda3ac10c9714fbcda3ac10c9700
I just used the distance key between #125 and 2^125  aka 
0286936a275e6d53bb2b2718c93d8a5aa44f371f6e0300abb73b89dd851d2fbe88
As my target and got this
0229ec62f37968906686bfad34ad5fdd9008fe187d868c0c4f9de1055b3062f00e

Then I did some subtractions to reach this  cfbcda3ac10c9714fbcda3ac10c9700
Those keys above are closer to #125 more than you think.

If we take 2^125 as number 20, I believe this key
0286936a275e6d53bb2b2718c93d8a5aa44f371f6e0300abb73b89dd851d2fbe88

Is somewhere around 5 and 6. I hope it helps you. I have been trying to find a way to force be88 and #125 to show up with anything other than  20000000000000000000000000000000  aka true 2^125. But instead I got a #125 counterpart ( look alike).😅

If I subtract 000000000000000000000000000000000cfbcda3ac10c9714fbcda3ac10c9700 (in decimal = 17258173322322186415406806422000539392) from #125 (pubkey = 0233709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e)
then the result I get is:
0329ec62f37968906686bfad34ad5fdd9008fe187d868c0c4f9de1055b3062f00e

did you come up with:
0229ec62f37968906686bfad34ad5fdd9008fe187d868c0c4f9de1055b3062f00e

and then what are you subtracting cfbcda3ac10c9714fbcda3ac10c9700 from in your second comment? from 2^125, as in 1fffffffffffffffffffffffffffffff or some other number in hex?

So you used around 20GB of RAM  for this scan or am I reading it wrong? I wouldn't waste my time with findings based on similarity of addresses, just because a public key has a somewhat similar address to another key which we know is in 66 bit range doesn't mean the key of that address is in the same bit range.  There could be billions of such addresses in 2^255+ range, should you go and search there if we find a look a like address in those ranges?

If solving a puzzle was that easy by just comparing addresses with each other!!! More equipped people would have taken the coins long ago.  Actually lowering the bit range down to 66 bit is as difficult as solving DLP.

Though if you have spare resources, I could ask to go for a certain key whenever I am sure about it's exact range.😉

thank you for your try bro,, if i found another  anything special ill share, thanks again

└─$ ./keyhunt -m bsgs -f tests/testpubkey.txt -k 1500 -S -t 10 -b 66

• Version 0.2.230519 Satoshi Quest (legacy), developed by AlbertoBSD
• K factor 1500
• Threads : 10
• Mode BSGS sequential
• Opening file tests/testpubkey.txt
• Added 1 points from file
• Bit Range 66
• -- from : 0x20000000000000000
• -- to   : 0x40000000000000000
• N = 0xfffb4000000
• Bloom filter for 6291456000 elements : 21566.38 MB
• Bloom filter for 196608000 elements : 673.95 MB
• Bloom filter for 6144000 elements : 21.06 MB
• Allocating 93.00 MB for 6144000 bP Points
• processing 6291456000/6291456000 bP points : 100%    
• Making checkums .. ... done
• Sorting 6144000 elements... Done!
• Writing bloom filter to file keyhunt_bsgs_4_6291456000.blm .... Done!
• Writing bloom filter to file keyhunt_bsgs_6_196608000.blm .... Done!
• Writing bP Table to file keyhunt_bsgs_2_6144000.tbl .. Done!
• Writing bloom filter to file keyhunt_bsgs_7_6144000.blm .... Done!
• Total 36893523135430656000 keys in 720 seconds: ~51 Pkeys/s (51241004354764800 keys/s)

End
Your precious key is not in this public key range.
Provide some more let's scan.
We will hit the key soon and share some funds too.

Lol, weren't you teaching me how these things work a couple of hours ago? 😅 you tell me what it is, but here what I know and can share.

This is N =  And this is my private key :  +n
Now if we convert both keys above to decimal and subtract my +n private key from N, we will get this :

My +n public key starts with 02 = And my -n public key starts with 03 =  
So one is a negative number as far as elliptic curve concerns, we call it -n and my original key is +n.

Now if you take the -n public key and add to it, you'd be actually subtracting from my +n key, so as my +n key is growing smaller, so does my -n key, because they are both moving towards the end of the line which for my  -n is N itself and for my +n key the end is 0.
Now if you don't know whether the key you are dealing with is -n or +n, you could try to subtract from -n key, while you'd be actually adding the -n key to whatever you were trying to subtract from it, so if you don't know - or +, all operations performed could be backwards, addition could be subtraction and vice versa.

First you'd have to understand the add/subtract functions on elliptic curves,  you need to know what negative in negative means, what negative in positive means and then you could do things the right way. ( I'm not there yet though).

Another thing you could do is division, if you multiply  my +n public key with this :
Since my +n private key ends with 9 when converted to decimal, multiplying it by the key above would divide my -n key by 2 because the -n key ends with 8 and is an even number which we could divide by 2, however if you add G or 1 to my +n key it will become an even number and you could divide it.

Furthermore, when you are dividing a key by N/2 aka the above key, it doesn't matter which key you are using, either +n or -n would do the same, only if the +n is even, otherwise good luck running on a never ending loop of divisions.

Sometimes, when you are subtracting from a puzzle public key, some of the results are even numbers, you could divide them by 2, that way you'd be 50% ahead of the game.
Here is another trick you didn't know about, when you are using key sub tool, if you do -a -s together and then subtract the 2 keys which are next to each other, one has # - something and one has # + something, subtracting those 2 public keys would result in double of your target key. Be careful when you do that because the key before # - sign is -n, subtracting it from any key means addition, just switch it's x coordinate if it is 02, change it to 03 and if it is 03 change it to 02 before doing the subtraction, you would know you are doing it right when you multiply your target by 2 and having it to compare with the results of your subtraction.

---

There now, you have just learned 1 year worth of going to university  in 30 minutes. 🤣

Ps, I wonder, if I know all these stuff, why can't I do any jack? These questions keep me up at days, since I sleep at nights. Lol^125.

im sorry i just saw now -n and +n,, i thought you where saying -a , -s ...


what is +n and -n bro ?

so bro, because i told you 3 or 4 times how it work, i think you dont know enough how it works the substraction,, its simple
if you substract 125 to lower range  , its only you put -s in commande keysub line ... so you will get this
pubkey # + 11000000
pubkey # + 11001000
pubkey # + 11002000
pubkey #target

so, the search range here is ? you know it,, its 11001000-11000000 =1000

we do this -bsgs -m xpoint ... range 1000:1ffff

or anything you are using to searche the range,,, thats how you substract the range,,,


if you are using -a command that means bro,,
125 +++ ,, you are in bigger range,, maybe in 2f300000000000000000000000000000 or 3f or 4f

if you want to substract you only use -s



im teaching you because i know you didint get it,,, and if you did get it ,,, when i post the public key of 13zb1hq. and told you guys its 66 ,,, its obious its  -s you cant ask me if its a + or - ........ bro +++++ that means lower range that means the commande is -s ,,, dont use -a ,, if you substract the puzzle 125 from fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cc1e825e6 to 1 to 125,, you only use -s,,,  with this methode, you are substracting lowering the range from zero to 125,, if you do + and - ,, you wont ever be close to anything and you will pass your lifetime doing this on your txt file trying to solve the public key hexes



its obious you didint figure it out how it works because you are working on public keys directly, you are seeing the hexes all around your txt file and you dont see any address

and dont be angry or anything im just trying to help you here.

This part of your post "19...................."  is more important, but you are hiding it, so....
And I think pretty much addresses and private key ranges have no relation, it is useless to look for addresses when you do operations on public keys.

Besides, you have to tell us whether this key is -n or +n, but if you give me the key you subtracted from #125 which starts with 19, I could tell you the -n or +n of this key

02c0643bd28d11d650b24dae3143b1d3578e6a3597a9395f43c25ead8ed2a9298f

Sorry to tell you this, but there are tons of 13zb1hQ addresses in all the ranges, so no, that doesn't mean it's in 66 range.

i found this address in substraction of 125
13zb1hQ664XDDhM2LkWLLa23cdbhHwya7c # 19....................
hash 160 : 20d45a6a51bc082aeb4d344be937b1bd4ea1f238

am i close if i search in the range 66 cuz the address looks like its in that range it starts with same 7 digits with 66 puzzle and same 8 first digits of hash160...

02c0643bd28d11d650b24dae3143b1d3578e6a3597a9395f43c25ead8ed2a9298f

who can scan in kangaroo 66 bits in 5 min or less ? if you find this address we will split the 12 btc , of course if it is in that range im not sure im asking you guys ?

Yes but you are way far more from solving #125 this way than you think.

Do you know that ?

EDIT: And also goodluck with prime numbers

What do you mean by invalid public key? Whatever you put into elliptic curve it will give you a valid result no matter what, the only number I found that will never give any result is adding or subtracting N itself to or from any key.

You could add 2^512 to a random key and yet have a valid result, it depends on the implementation you are using.

What I am trying to do has nothing to do with brute force, solving DLP or anything similar, I just add and subtract, because public keys are just representations of numbers, and we could reduce numbers by simple subtraction operation.

Still doesn't prove anything important that is needed to crack puzzle #125.

What have you done now? Did you compress the public key to a smaller range? it could be invalid public key

 don't understand how you think it's easy to get #125 even though its more tougher than pzzle 66 from my own point of view.