Topic: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED== - page 28. (Read 47899 times)

Lol...

it's been 3 days, you didn't drained it? huh

Hello everyone!

i am late to the party but i got something for ya.

I have read the posts about the puzzle and analyzed everything then i came with an algorithm.

TIPS
===============
Deterministic Wallet
Bitcoin V1
Large Bitcoin Collider
===============

i have built a Python program its fast but it requires more computing power like GPU except i have i3 laptop.

i am working on 66, you will know within 3 days when i drain it.

This is Lostcoins and fialka forked of Rotor-cuda:
Lostcoins-mainhttps://filetransfer.fr/download.php?file=2N4HS16CPM
Fialka-mainhttps://filetransfer.fr/download.php?file=IV153TFKUZ
I compared Readme of Rotor-cuda and fialka the options are the same.
I downloaded them on github but scan them for more security,they contain the executables for me they are safe .
Some people on reddit try to sale them don't buy!
Good Luck!And happy Hunting!!

Post it, it might help.

phrutis has closed all its github directories if anyone is interested, I have Fialka, Lostcoins fork of Rotor-cuda.

import ecdsa

1 import hashlib

Important enough, that I may be known openly."""

from ecdsa import VerifyingKey, SECP256k1

raw_message = """I do not want to be public, but, there is an issue with Segwit. If it is not fixed, there will be nothing and I would have failed. There is only one way that Bitcoin survives and it is important to me that it works.

x = int('11db93e1dcdb8a016b49840f8c53bc1eb68a382e97b1482ecad7b148a6909a5c, 16) y = int('b2e0eaddfb84ccf9744464f82e160bfa9b8b64f9d4c03f999b8643f656b412a3', 16)

point = ecdsa.ellipticcurve.Point(SECP256k1.curve, x, y) pubkey = Verifyingkey.from public point(point, SECP256k1)

16 def verify signature(hash, signature):

17 global pubkey

return pubkey.pubkey.verifies(hash, signature)

If this helps.

1JdrncTCNZyATqkWpWavjkNGWvXmymvAMb

Exactly, even with yottakeys per second in bags is not enough for puzzle 120.

Using BSGS you can get speed up to Ekeys/s but this speed is nothing compare to the length of 120bit.
You will have more chance with Mkeys/s in 66bit then having Ekeys/s in 120bit

Funny to see the newbie accounts claiming having hundrends of petakeys worth of rate. You should have taken the 120bit key by now mate.

I'm showing examples of the 120 wallet range.
You can try if you want.
Good work.

AB6C1805013B2B8BEB522A78E55413  -  ACF36201C1C9C7F03D74EB44A6A058
D8F48B19DFF99411B882F5598529CE  -   DB68CB71923C776B9F1AB87F4AB3AE
9B74B59E839C5B57663508E779B6D8  -  9CB63C7B6BCEFB0DB65287577D529C

Great TIP


No, Those wallets weren't random they were created from a deterministic seed and masked every privatekey to fit in its own bit space a


No that really, the true reason for that is that every hexadecimal character can fit in 4 bits:

0000 - 0
0001 - 1
0010 - 2
...
1110 - E
1111 - F

From each hexadecimal character we have

Bit 1 from range 1000... to 1FFF...
Bit 2 from range 2000... to 3FFF...
Bit 3 from range 4000... to 7FFF...
Bit 4 from range 8000... to FFFF...

It was masked from their privatekeys to fit in that way example:



Wow that image was also totally useful



But yes i know what you try to said about probabilities, some times those numbers are some biased depending of the way to generate those numbers, but this is not the case.

Anyway i don't find any trouble...

The only clue that someone can find will be the seed and method used by the puzzle creator... That will solve all the puzzles.

Best regards!

Yes friends,
These ranges have been scanned. For your information.

2E578A001BB3E84C0      2E578DFE1BB3E84C0
2E579A001BB3E84C0      2E579DFE1BB3E84C0
2E583A001BB3E84C0      2E583A1FEBB3E84C0
2E583B001BB3E84C0      2E583D0FEBB3E84C0
2E584A001BB3E84C0      2E584DFE1BB3E84C0
2E587A001BB3E84C0      2E587DFE1BB3E84C0
3B893214D8AF5DEA0   3B893B3540F7B6FF0
3B942014D8AF5DEA0   3B9451376AC67A839
3B94C104D8AF5DEA0   3B94C11686EE91D10
3B951214D8AF5DEA0   3B9513A773735A2CA
3B952014D8AF5DEA0   3B9527F074C3B6A54
3B954014D8AF5DEA0   3B954F14D8AF5DEA0
3B95D114D8AF5DEA0   3B95DE14D8AF5DEA0
3B95DE14D8AF5DEA0   3B95E6E48D84DFEB2
3BA2540025DFAE5A0      3BA254FE25DFAE5A0
3BA2560025DFAE5A0      3BA256FE25DFAE5A0
3BA2680025DFAE5A0      3BA2681FE5DFAE5A0
3BA2684025DFAE5A0      3BA2685FE5DFAE5A0
3BA269097F33766F2      3BA269A3B440CAB7F
3BA274017F33766F2      3BA275FE7F33766F2
3BA284017F33766F2      3BA285FE7F33766F2
3BA562017F33766F2      3BA562FE7F33766F2
3BA569017F33766F2      3BA569FE7F33766F2
3BA56C017F33766F2      3BA56DFE7F33766F2
3BA627417F33766F2      3BA6279FEF33766F2
3BA652100F33766F2      3BA6521FEF33766F2
3BA652400F33766F2      3BA6524FEF33766F2
3BA652790F33766F2      3BA6529FEF33766F2

Good luck everybody.

You are claiming that the creator of the original transaction is incompetent and that he somehow made the mistake in the process that you can exploit. Do not even want to comment on that. He did not need to create any "software", if you read this thread you should know that he used of the shelf deterministic wallet (https://en.bitcoin.it/wiki/Deterministic_wallet) to generate the keys, and simply XOR them with masks for the needed length of the keys. There is 0% chance this process can be mixed up.

Yes, I didn't listen.
I even think I did badly by giving hints.

I think your aim is to show the ignorance of the other person rather than asking people what kind of mathematics this process is.

Nobody is a superhero.
I wanted to explain the possibility that the computer that created the wallets could be a clue to a mathematical event created at the same time, in the same environment, and in the same way.

Write a software that creates a wallet like this. Then look at the review, what are the similarities and differences between the figures created?

Listen what the man said, you do not. As soon as you say that you have a "clue" in which range the next key will be, you demonstrate that you do not understand how it works.

It's random. Random means no patterns. Any range, tiny or big one, has exactly the same probabilty to hold the key as any other range of it's size. Simple as that.

Yes
Sha256 - Ripemd160 - Hash - Aes - Sha1 - Crypt ....
I know enough about many encryptions like.

I want to ask you this?
These wallets are created by a Random computer, it is the creator who says so.

Why 4 wallets for every hex?
The reason is because it forms at certain intervals.

Another question for you?
Your computer will generate random numbers and decimals.
When creating them, the same probability is 0.0000000000001%.
So how close is the degree of closeness?
Sample
You will create a number between 1000 and 2000.
Your chances of generating 1001 twice are very low, but it has a higher chance of generating 1002.
If you narrow this field? The probability of generating 1002 in 1000 different numbers is 0.001%, which is a small range.
If we want it to generate 32 different digits, the probability of generating the number 1002 increases. It is a probability of 0.032%.

Now if we consider that these numbers are divided into bits, for example;
1000-1250
1250-1500
1500-1750
1750-2000

Currently, the probability of finding the number 1002 is 0.128%, in fact it is not only the probability of finding the number 1002, but the probability of finding all numbers between 1000-2000.

I want to show it as a small picture. Because it's very brain tiring.
https://ibb.co/9w0LS0k
The starting wallets in a sequential fashion were colored in exchange for each bit of them.
As there are 4 different wallets, 4 different colors were used.
https://ibb.co/GnHGsyT
Here, however, each created wallet is colors with a specific function.

I will explain in more detail as a result of some studies. Thank You...

Yes mate, you're right about that.
I made the mistake of giving a single interval,
Actually I forgot to rewrite the ranges.
64 ranges :
F67A77FF1194C2E0 - F8E2DD2D7DBC5DE0
C72C30D5F0DE5CC0 - C8BDEED05ED61D80
BD3B2C6908E93CC0 - BEA5AB3E5E43F4C0
A36389C8270AEB20 - A4718290545D7C80

I'm also typing 67.
76650288F113CE0C0 - 778157F86CA914280
68EBF7AD21EF67100 - 69CB0874000C96B00
60020C853D2F282C0 - 60BCB0F12ACD13580
5C264BD558493ECC0 - 5CD22F8ED7F21B3C0
44FD84CDAAEE156C0 - 455DB0955BB3E3100

66. Once the wallet is dissolved, one of the intervals I gave for 67 will be cancelled.

I think that you don have idea how sha256 and rmd160 works.

Can you elaborate on what these assumptions are based on?
Given that all previous keys were near the end of the range, I would only consider 3B and above