Bitcoin puzzle transaction ~32 BTC prize to who solves it

[nc50lc]

-snip-
"3PdQoXyQwWmerpt3SbF7Hbh3aukXXXXXXX" P2SH ( sorry if i do "X" end address, it might be easier to cashout when you can do something with this main/test net addresses.

WHO KNOWS THE PUBLIC KEY 66? I CAN EVEN FIND THE P2SH ADDRESS, HAHA!  

Addresses are just encoding of the HASH160 of the public key, it's not something special that can be a solution to the puzzle.

Since puzzle #66's output contains its hash160 which is 20d45a6a762535700ce9e0b216e31994335db8a5, you can just derive the other address types from it.
You can even decode the address and derive other address types from it.

From HASH160, it's as simple as this:

• HASH160 of its pubKey: 20d45a6a762535700ce9e0b216e31994335db8a5
• Prepend OP_0 and size '0x14': 001420d45a6a762535700ce9e0b216e31994335db8a5
• HASH160 of the above: f0a433b3411b7c1812937977bebd25602f55b68f
• Prepend the version byte '0x05': 05f0a433b3411b7c1812937977bebd25602f55b68f
• Append the checksum: 05f0a433b3411b7c1812937977bebd25602f55b68f31c18fc0
• Encode to Base58: 3PdQoXyQwWmerpt3SbF7Hbh3aukC5w28GP

[kalos15btc]

Hi, so  i found 2 address similar 11 digits of hash160 in extraction of bit 130 that means if you scan on this range and have the private key of those one of two addresses we will split the 13 BTC ,

1st

1AoNf67iZUmz1Ck9eefUfTzWQJMN5pgcPG
02113ba90a97c020ade0f3d8d0369981a723fe2bc4352815df22dd3eafae13c5a5
6b7e582a29a549cc60b591279a963f02eff02f99

pk:00000000000000000000000000000000000de4cfcadfc034c963dd053d719e88

address to search in :116 bit range
1AoNf67iZrwMSYPTDbk3Sh1yXJCARQbD7a
039d0a0241abe2411f64b4f6d29f2e1b6c837b26b6bdded577c3fc93574d3d735c
6b7e582a29a7601b79761f9f153c300c3d988231
range: 804cfcadfc034c963dd053d719e88:fffffcadfc034c963dd053d719e88


2nd
1PtStkm2bWKryHVduVjciPUxVx9UeDcCXG
02dc52ba09b16bc5cbd25aca7c82dd924f81cd31ecf29ecb264fa2cc45393728b9
fb0d9859584e68c24c1698eea4d05d2822fe4b70
pk: ad0f6ba584b355089cf6ce9cc9774

address to search in 116 bit range
1PtStkm2bLM7EK7g1rnTLBxu6aLouVuULV
03ef06cec3b3e35f68ba78618e5a5cf8663cc1a3b685dcfd197c1c0030530b1293
fb0d9859584d782df3fe652d2da5a21c30f137f9
range: 804cfcadfc034c963dd053d719e88:fffffcadfc034c963dd053d719e88




if the pk found of one of those address we will split the prize.
they have the same 11 digits of hash160 that means maybe 90 pourcent is in that range

up, anyone interessing its same 11 digits hash160 so who can scan this range 116 and split the 13 btc ?

[citb0in]

Hi, so  i found 2 address similar 11 digits of hash160 in extraction of bit 130 that means if you scan on this range and have the private key of those one of two addresses we will split the 13 BTC
---SNIP---

what makes you think that a partial string of one hash has a reference to another?

[digaran]

Hi, so  i found 2 address similar 11 digits of hash160 in extraction of bit 130 that means if you scan on this range and have the private key of those one of two addresses we will split the 13 BTC
---SNIP---

what makes you think that a partial string of one hash has a reference to another?

This phase of learning, I call it kindergarten phase, I have also passed this era, I'm in preschool now. Lol.

They need to learn their lessons through experience if verbal education is  not effective.
😉

[kalos15btc]

Hi, so  i found 2 address similar 11 digits of hash160 in extraction of bit 130 that means if you scan on this range and have the private key of those one of two addresses we will split the 13 BTC
---SNIP---

what makes you think that a partial string of one hash has a reference to another?

This phase of learning, I call it kindergarten phase, I have also passed this era, I'm in preschool now. Lol.

They need to learn their lessons through experience if verbal education is  not effective.
😉

no one told you to give your opinion about how im searching the puzzle, old school new school and lessons, nobody told you your advise or what are you doing or you did pass this really ?? oh thats right so you are on the new school today, ok ,, you helped me thank you hhhhhhhhhhhhhhhhhhhhhhhhhh, omg, still people like you think like that nowadays so i stop posting new things im helping the community to solve the puzzle at least im doing something and sharing something and im not like you always ask for help and how to do that and never did give us something that help solving this puzzle. so relax bro and dont even talk this or about your experiences in old school, we are here to learn everyday a new thing we always learn

[citb0in]

@kalos15btc: would you be so kind and reply to my previous question, please? Certainly you have overlooked it. Looking forward to hear your thoughts.

Hi, so  i found 2 address similar 11 digits of hash160 in extraction of bit 130 that means if you scan on this range and have the private key of those one of two addresses we will split the 13 BTC
---SNIP---

what makes you think that a partial string of one hash has a reference to another?

[digaran]

so i stop posting new things

What are those, linking rmd 160 hashes to points on secp256k1 thinking if they look similar, they might be in same subgroup?

im helping the community to solve the puzzle at least im doing something and sharing something

You call that helping the community or helping yourself? Why don't you share the other half of info/keys which you are holding back from the community? Looks to me the only thing you are after is to split the 13 btc by having someone else searching the ranges you provided.

and im not like you always ask for help and how to do that

If I don't ask for help and guidance from those who know better and more than I do, would that make me a better person, how else could I learn them? If they don't help by sharing their knowledge, God will provide a different source, but if anyone teach and help another, they gain more instead, this is how the world works.

and never did give us something that help solving this puzzle.

Now I have to solve it for you? Don't you think if I could provide something to help solving a puzzle, I would use it myself first?
Have you ever asked something from me which I failed to answer, or from anyone? Is it a bad thing if I'm telling you there is no mathematical relations between points on a curve and hashes/ addresses, so that you could focus on something which has an actual mathematical solution?

Even if you don't ask for my opinion, it's my duty to point out misinformation to save others time, therefore if you post a long list of "useless" information, I will point it out again and again, that's what a community is, we help each other to the best of our abilities.

[mcdouglasx]

Hi, so  i found 2 address similar 11 digits of hash160 in extraction of bit 130 that means if you scan on this range and have the private key of those one of two addresses we will split the 13 BTC ,

1st

1AoNf67iZUmz1Ck9eefUfTzWQJMN5pgcPG
02113ba90a97c020ade0f3d8d0369981a723fe2bc4352815df22dd3eafae13c5a5
6b7e582a29a549cc60b591279a963f02eff02f99

pk:00000000000000000000000000000000000de4cfcadfc034c963dd053d719e88

address to search in :116 bit range
1AoNf67iZrwMSYPTDbk3Sh1yXJCARQbD7a
039d0a0241abe2411f64b4f6d29f2e1b6c837b26b6bdded577c3fc93574d3d735c
6b7e582a29a7601b79761f9f153c300c3d988231
range: 804cfcadfc034c963dd053d719e88:fffffcadfc034c963dd053d719e88


2nd
1PtStkm2bWKryHVduVjciPUxVx9UeDcCXG
02dc52ba09b16bc5cbd25aca7c82dd924f81cd31ecf29ecb264fa2cc45393728b9
fb0d9859584e68c24c1698eea4d05d2822fe4b70
pk: ad0f6ba584b355089cf6ce9cc9774

address to search in 116 bit range
1PtStkm2bLM7EK7g1rnTLBxu6aLouVuULV
03ef06cec3b3e35f68ba78618e5a5cf8663cc1a3b685dcfd197c1c0030530b1293
fb0d9859584d782df3fe652d2da5a21c30f137f9
range: 804cfcadfc034c963dd053d719e88:fffffcadfc034c963dd053d719e88




if the pk found of one of those address we will split the prize.
they have the same 11 digits of hash160 that means maybe 90 pourcent is in that range

up, anyone interessing its same 11 digits hash160 so who can scan this range 116 and split the 13 btc ?

one hash can be almost identical to another and this does not represent that their origin is similar. you can search in all ranges and you will always find matches in some hexes, and this does not imply that you are close to your objective.

[zahid888]

588 ranges are available for everyone. The ranges are hidden within the script to prevent different individuals from repeatedly scanning the same range. Please ensure that whoever is performing the scanning keeps the proof of work secure.
https://github.com/futuremoneybtc/VanBitCrackenS1-Distributed-Puzzle-66-Address-Scanner

[digaran]

588 ranges are available for everyone. The ranges are hidden within the script to prevent different individuals from repeatedly scanning the same range. Please ensure that whoever is performing the scanning keeps the proof of work secure.
https://github.com/futuremoneybtc/VanBitCrackenS1-Distributed-Puzzle-66-Address-Scanner

How do we know we are scanning a correct range, I mean what if someone scans all 588 ranges and finds nothing? You see, searching for addresses is a blind search no guarantee to land on the target unless you search the entire  original range one by one.

[kalos15btc]

588 ranges are available for everyone. The ranges are hidden within the script to prevent different individuals from repeatedly scanning the same range. Please ensure that whoever is performing the scanning keeps the proof of work secure.
https://github.com/futuremoneybtc/VanBitCrackenS1-Distributed-Puzzle-66-Address-Scanner

scaning with hash160 is 2 time faster ...

i run 588 ranges and have this error

C:\Users\Wafa>588_ranges.exe

=========================== VanBitCrackenS1-Distributed-Puzzle-66-Address-Scanner =================================

Enter the last range you are scanning:
Resumming your scan...
Sorry.. This Range does not exist in the list. Press any key to exit.

Traceback (most recent call last):
  File "588_ranges.py", line 1205, in <module>
NameError: name 'exit' is not defined
[11720] Failed to execute script '588_ranges' due to unhandled exception!

C:\Users\Wafa>

[Denis_Hitov]

588 ranges are available for everyone. The ranges are hidden within the script to prevent different individuals from repeatedly scanning the same range. Please ensure that whoever is performing the scanning keeps the proof of work secure.
https://github.com/futuremoneybtc/VanBitCrackenS1-Distributed-Puzzle-66-Address-Scanner

Error!

=========================== VanBitCrackenS1-Distributed-Puzzle-66-Address-Scanner =================================

Enter the last range you are scanning: 3BCF5000000000000:3BCF5ffffffffffff
Resumming your scan...
Sorry.. This Range does not exist in the list. Press any key to exit.


Why can't I set my range?

[mcdouglasx]

increase the search speed of the Pubkeys, dividing the search range.

First of all, let me make it clear that this script is slow and basic, I did it like this in order to explain the range division method.

If you want to apply it quickly, modify keyhunt or other c++ code before compiling to make it work with this method.

The method consists in changing the parameters of the curve by x,y of a higher point (for the script I used pk#100).
by default the curve uses pubkey #1
what works in sequence 1,2,3,4....
pubkey #100
which works in sequence 100,200,300,400....

if we want to find pk# 3922
in a range of 2000:6000
in the traditional way we would need to advance +1900 pubkeys
In this method we will only need 19 pubkeys (1900/100).

the database is done by subtracting 1 consecutively from the target pubkey (in my case I need to subtract( 1), 100 times because I chose pubkey pk# 100 as divisor.

The trick is in your ability to use fast databases, the more pubkey you store in the database, the higher the divisor will be, therefore the search range is smaller.

save as brute_div.py

Code:

import os
import hashlib
import random
import binascii

#target pk decimal 3922

#Target_Upub "042f3342152eff6aca5e7314db6d3301a28d6a90ddcfd189f96babadc2a053d3929d0e4bef850141613e6bb3d3e2c480b3634af4332f125f4edfc65b88816f541d"

#Start_Range in decimal
Start_Range = 2000

#End_Range in decimal
End_Range   = 6000

#divisor pk in decimal 
boost = 100

#boost_upub "X=ed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88, y=e57a6f571288ccffdcda5e8a7a1f87bf97bd17be084895d0fce17ad5e335286e"




class Point:
    def __init__(self,
                 #change x and y for boost_upub
        x=0xed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88,
        y=0xe57a6f571288ccffdcda5e8a7a1f87bf97bd17be084895d0fce17ad5e335286e,
        p=2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 - 1):
        self.x = x
        self.y = y
        self.p = p

    def __add__(self, other):
        return self.__radd__(other)

    def __mul__(self, other):
        return self.__rmul__(other)

    def __rmul__(self, other):
        n = self
        q = None

        for i in range(256):
            if other & (1 << i):
                q = q + n
            n = n + n

        return q

    def __radd__(self, other):
        if other is None:
            return self
        x1 = other.x
        y1 = other.y
        x2 = self.x
        y2 = self.y
        p = self.p

        if self == other:
            l = pow(2 * y2 % p, p-2, p) * (3 * x2 * x2) % p
        else:
            l = pow(x1 - x2, p-2, p) * (y1 - y2) % p

        newX = (l ** 2 - x2 - x1) % p
        newY = (l * x2 - l * newX - y2) % p

        return Point(newX, newY)

    def toBytes(self):
        x = self.x.to_bytes(32, "big")
        y = self.y.to_bytes(32, "big")
        return b"\x04" + x + y


def getPublicKey(privkey):
    SECP256k1 = Point()
    pk = int.from_bytes(privkey, "big")
    publickey =(((SECP256k1 * pk).toBytes()).hex())
    return publickey




if __name__ == "__main__":
                    #range Div
    for i in range((Start_Range//boost), (End_Range//boost)):
        print(i)
        b = hex(i)[2:]
        c = str.format(b)
        pk= c.zfill(64)
        Seq_Bytes = binascii.unhexlify(pk)
        PublicKey = getPublicKey(Seq_Bytes)
        #print(PublicKey)
        Privatekey = (Seq_Bytes).hex()
        #print(Privatekey)
        
        data_base = 'data-base.txt'
        
        with open(data_base, 'r') as file:
            content= file.read()
            #search publickey in data-base
            if PublicKey in content:
                print("Range Found")
                print("Decimal:", int(Privatekey, base=16)*boost)
                

save as data-base.txt

Code:

040119630133edf190d14fdb00640f74cc4317e1ddb9dff1c9707eac3353ef9c21f16ab4e5e3a3ee6d743851876da07034caa4f5196106a0c803136d215420494c
04572dd1b78d9e4d185117860c71a224a05daee2b7bba651478ddbabadc71831c1d3fd83384c92bcaa326b75b10ef7cb33baf8b4d474cf8e5eac0e598a8a13200f
0462cc463c16ee8e8f16ea85175508278d4d5577e3556d3580ec8d90e3bbb97072cd294843e299c500846a6c91c913395b56b48efa2f9ab3fcb81863fe8a6a891e
040211625f596c1d65bd9f79fdd1fb4803f3165be3205f06e712aea0f191c08afa6ac2fd437b1bedadac240964a602346ea881d497f9a097a628f448006756c169
042144649d508cf5daa7b422f822e9125fd5a10965a9d32c6cc299bad46616771d1153178c0e33c445f03ef96b35067ceb76ddeacad9ddba1a74acd229d01ca76e
04627bbb09622ac837cef0922098ac6736db9599c855d32bb1d7f5d06d46dfb063fc5ee8575d03e1db328e6271f5e2c82eb7995ce60e80936976343faa67d329a7
04739fb43ab1b0715539c0b0ae43901bf2bc051a698ee8d50384fcc1f0ec37312ac12237043c8822262d4abe331de6ab4dff90d6b84c0b59e7a193e7636c6b7e28
04742771eed11bcc2c849c1ae6316eec9242dae6c581792b893802c01c2c6fb9b6cc3cc9dd51c8bcc37b93fd5b1c0d4026cecbf6b22637f4160229ddb0903505d7
0426ad559046e19dbf0e60953ed40a3958a357f6bc236422790cbcc35574fdab68af6fc5ef906721c0553c6a7460bc6894e5bb77e74dbc25e2184fb3f0fa795508
04abcf6e30599e5b9144892e59bbed1400f64fe0617d74b106fe7b61b250a84d165074e565f5908c234e798279d9ffff6394f8d94a2b7f0ee61e286b09a86b987a
0455ef868b6fc2923ee624fc0965f61852b3aeb38ce2c6b3bc54b86b3a60f0ab81ec198aeaab16b4cc5354d0adc66051fb76cff6df22dc1a0e72ce6e791622efab
04344133db83e7dc4dcf27c3b46a2d79e04abc9b55aa45bd7b8dfb7e8a3eeac809f801eb1f729b76ad8acf09b16134c052453dde983a2adc3a85deb8ed3ad23e99
0404f94305f92799f0a64a3cb37df2bac456d661d6445682770bac0ed66a5872124c48691fbd8683d6b94ddc7cad8d1d7e6baf796d6d00f3440511fc1313c054ab
04dcf8306eb1ba7186cc26e38aeb43e532087cfbbd1649f8efd28ceccccdb8420ac3db6bb8b26e11df8bb5dad7bde9061609229a7951c3ef50233030f40c0b51e4
0471be09bb1832a799c4e21bb8a2bb5ad0324c639ce6e62b6e69fa60bfd41c51bac5227fc5719a5b8f28c40767b54eb249950588dcc15e1656e4ff238d54dda472
0463131b929673be9d5e17d1828fb222927ef2472636f106abbaf824a82dbccb6254a25d1abfa90b3ff16fdd0c00b97bdfc47e37b2272c6eabbc93d5a6b7fa7641
045ffc18994da42fa4549627195ebe332c86a2c364f43b14f759710c544f6950a022eac0a879815faa220e61419db053cd4542ad2f9ba9f63f04331ad5c2c9bb88
0440e1770f17f5c972d06bf9c9334bed5377de24991d273e9913bcd99066b8a61931faaff6e2989e758f404509bf80ed818d14e217a99d6e14bfdea369e2731eac
042dad49af8702f39bcb353956a1d5601ba24fffbf58ab527fe9ac6ff9ea2ac295b5af68b8e604076255c0704e2cace8c0ed40361140dc02fe2773a62527e4675b
0462d73b4e6a3da9c8f794d01d884625df24f3d89be785db6bcc28bdf241c07bb49b163e85202a092a2cefa79fb82c684587efc3c30bb821526616ab282eed4930
0493e9256e4ddb4bfb6f4a928906e2ad04793e56423a22230ae645529d0712d6ca83b242dbd9b3dfec3640ea62f0e3989bd874f24790a75cd69756a8ed6eafdcea
0458d4408982dcd1740e8fcddb65400526a1d9e593314164bd7b59bf337f7ddb01a332c95298d23755085caf056f38c1c872077e0b020e6edd1237a6bebf8ac834
048706a73eb59fc9d192907dfdcc6f75b6454f10454a5e9584bdac978e8e0042cd8b1b5f566f1f78d6f528405a2b6a87e84265e2dd1e34c1ce7109286d5bdc3a49
04c9342036613fd8db140863797f0ee1a31925b606c773073abc5ccbea1a9888da521c2867487afdd02177d86bec1087fae4bba20c5cff325b93211944e5647f17
04b5918719a1d504d2b44a3d0e117c27984f2e536a03571db8752685cfdf6d34a09b01e8cb585cb7c8fa29828aac89905016beacb25e62ea356d54e9a618184abe
04540f37a73add0efafa3713cfc198fb26c6134bcf12eac07e9e2c4501caee8a44e2d1e8d39c6f5e65db36976d6e09689e089934bcae1b954955b58209d97cde5a
04c78eaf601afdfc202123d22808496189b90816371fb2c4752773faf063cac063095325b1e0c22dc5df6e84c516d396f9bacb42cf6c1fe9c76df83b3f3eb42de7
04e1c060c136db028fd879d15103661e3a3d8fb0b87d271b36b7356a6b525c5815de2eb43caa91741f5c656aafd0507460d233c4e44db1f1f6203e8254b410a7b0
04ebc9d95a8a237b64212e9dfa607fe4a37a05fbfe7c5594c8ae472c4f6ffeaa5a39406dc0e311a47ccf15fb458f8b48d38fec2fd86b4afc05b1747957e7f1162a
04d7b22b292bf93e7ff00e1aa79f25f30b317c9edc910a1845ea6a06a9f25b7ca1d00428edebff7fdd0ae65db06730fed9edfeef354e28d77f6ebeee2ef0d92d30
041e577a749883a5fb855938097d0180a1dce21791e10b2c0a3fe6cea1cd03b532a147662f856f5e15dba530007278e8787b575612d5dcd79019417ec3ac41c003
044febed821ad1ea18493871e3db83c695a2f2b08f173b6e72d9bd10b649e4e50e60d253688a0d107ea384a6b6f1820e5604411be23cd3c5066b2af7b0e4f62ef0
04942ea85310dfdc3e1f7993e747bdac04ecc8692b94847bd3e044cdf22cd77e6ebbf2f7c8e688b6e9b4510bbe58c41b1df33e9afa5428ce51e02e624b9ea155e5
0476af379a7214118fa13884cf3c6ad6d87e3f6ebab3c3bbb44caddfc53de3347522d0565c4df8aec2ed307fa43d6d2bb99437a626896c50888310e9acbeb6d71c
043e040511233b0f5a53df4d5ab08541372db4ecef0170b566807c49236ce82075f9337271dd44384377a35ebf56753b221fda08e01b790e02b9603827600df3d3
04c193bedf5c0a676b3e0cca406968a9683865becedeee8c49d829f1c3e075be483277c4b1eb5009616fa1f24db03a6e0f95cf20f247a4f3773f1e76631b57f8ee
04404f0dd6941953131953b53c67fdfab7f67c24bc92fb4d69a9479a9b6c9175fb857c75e7f40a48d3ab1560851d42b0377e7d3e7986efd570147917bdf24d87b3
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I currently don't have what it takes to code in C++ using the quick search methods, my resources are limited to the basics.
I currently don't have an up-to-date pc to do the relevant tests (the disadvantages of the third world).

However, later I will share faster methods, ECC tricks.

[tptkimikaze]

588 ranges are available for everyone. The ranges are hidden within the script to prevent different individuals from repeatedly scanning the same range. Please ensure that whoever is performing the scanning keeps the proof of work secure.
https://github.com/futuremoneybtc/VanBitCrackenS1-Distributed-Puzzle-66-Address-Scanner

How do we know we are scanning a correct range, I mean what if someone scans all 588 ranges and finds nothing? You see, searching for addresses is a blind search no guarantee to land on the target unless you search the entire  original range one by one.

There's a phrase in Chinese sounds like this "There are no cookies fall from the sky".

If someone can scan 48 bits in an hour, let's assume the key is in the 588 range, this means he just need 24.5 days to completely scan all 588 ranges. If he is so confident with those range, nobody will want to share such high probability ranges. It is a zero risk while he get 40+% if he share it because whoever find it need give him that. If the key wasn't there, he has nothing to lose. The final conclusion is he has no confident at all on those 588 range. If I have those highly confident ranges, I can easily rent a GPU to scan all those ranges. Why would I want to share the 60% jackpot with someone out there? Doesn't make sense at all.

[citb0in]

588 ranges are available for everyone. The ranges are hidden within the script to prevent different individuals from repeatedly scanning the same range. Please ensure that whoever is performing the scanning keeps the proof of work secure.
https://github.com/futuremoneybtc/VanBitCrackenS1-Distributed-Puzzle-66-Address-Scanner

How do we know we are scanning a correct range, I mean what if someone scans all 588 ranges and finds nothing? You see, searching for addresses is a blind search no guarantee to land on the target unless you search the entire  original range one by one.

There's a phrase in Chinese sounds like this "There are no cookies fall from the sky".

If someone can scan 48 bits in an hour, let's assume the key is in the 588 range, this means he just need 24.5 days to completely scan all 588 ranges. If he is so confident with those range, nobody will want to share such high probability ranges. It is a zero risk while he get 40+% if he share it because whoever find it need give him that. If the key wasn't there, he has nothing to lose. The final conclusion is he has no confident at all on those 588 range. If I have those highly confident ranges, I can easily rent a GPU to scan all those ranges. Why would I want to share the 60% jackpot with someone out there? Doesn't make sense at all.

that's the point, absolutely correct. That's why everyone realized that zahid888 trying to setup a big scam (not even mentioning he offers a binary executable without source code on his github repo)

[tptkimikaze]

588 ranges are available for everyone. The ranges are hidden within the script to prevent different individuals from repeatedly scanning the same range. Please ensure that whoever is performing the scanning keeps the proof of work secure.
https://github.com/futuremoneybtc/VanBitCrackenS1-Distributed-Puzzle-66-Address-Scanner

How do we know we are scanning a correct range, I mean what if someone scans all 588 ranges and finds nothing? You see, searching for addresses is a blind search no guarantee to land on the target unless you search the entire  original range one by one.

There's a phrase in Chinese sounds like this "There are no cookies fall from the sky".

If someone can scan 48 bits in an hour, let's assume the key is in the 588 range, this means he just need 24.5 days to completely scan all 588 ranges. If he is so confident with those range, nobody will want to share such high probability ranges. It is a zero risk while he get 40+% if he share it because whoever find it need give him that. If the key wasn't there, he has nothing to lose. The final conclusion is he has no confident at all on those 588 range. If I have those highly confident ranges, I can easily rent a GPU to scan all those ranges. Why would I want to share the 60% jackpot with someone out there? Doesn't make sense at all.

that's the point, absolutely correct. That's why everyone realized that zahid888 trying to setup a big scam (not even mentioning he offers a binary executable without source code on his github repo)

There's one evil point I even don't want to point out because I want to assume he is a good man. It's not only nothing to lose. If let's say someone fall into this and scan all the ranges and doesn't get the key, we even help him to eliminate all those 588 ranges for free which only he know it.

[digaran]

increase the search speed of the Pubkeys, dividing the search range.

First of all, let me make it clear that this script is slow and basic, I did it like this in order to explain the range division method.

If you want to apply it quickly, modify keyhunt or other c++ code before compiling to make it work with this method.

The method consists in changing the parameters of the curve by x,y of a higher point (for the script I used pk#100).
by default the curve uses pubkey #1
what works in sequence 1,2,3,4....
pubkey #100
which works in sequence 100,200,300,400....

if we want to find pk# 3922
in a range of 2000:6000
in the traditional way we would need to advance +1900 pubkeys
In this method we will only need 19 pubkeys (1900/100).

the database is done by subtracting 1 consecutively from the target pubkey (in my case I need to subtract( 1), 100 times because I chose pubkey pk# 100 as divisor.

The trick is in your ability to use fast databases, the more pubkey you store in the database, the higher the divisor will be, therefore the search range is smaller.

save as brute_div.py

Code:

import os
import hashlib
import random
import binascii

#target pk decimal 3922

#Target_Upub "042f3342152eff6aca5e7314db6d3301a28d6a90ddcfd189f96babadc2a053d3929d0e4bef850141613e6bb3d3e2c480b3634af4332f125f4edfc65b88816f541d"

#Start_Range in decimal
Start_Range = 2000

#End_Range in decimal
End_Range   = 6000

#divisor pk in decimal 
boost = 100

#boost_upub "X=ed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88, y=e57a6f571288ccffdcda5e8a7a1f87bf97bd17be084895d0fce17ad5e335286e"




class Point:
    def __init__(self,
                 #change x and y for boost_upub
        x=0xed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88,
        y=0xe57a6f571288ccffdcda5e8a7a1f87bf97bd17be084895d0fce17ad5e335286e,
        p=2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 - 1):
        self.x = x
        self.y = y
        self.p = p

    def __add__(self, other):
        return self.__radd__(other)

    def __mul__(self, other):
        return self.__rmul__(other)

    def __rmul__(self, other):
        n = self
        q = None

        for i in range(256):
            if other & (1 << i):
                q = q + n
            n = n + n

        return q

    def __radd__(self, other):
        if other is None:
            return self
        x1 = other.x
        y1 = other.y
        x2 = self.x
        y2 = self.y
        p = self.p

        if self == other:
            l = pow(2 * y2 % p, p-2, p) * (3 * x2 * x2) % p
        else:
            l = pow(x1 - x2, p-2, p) * (y1 - y2) % p

        newX = (l ** 2 - x2 - x1) % p
        newY = (l * x2 - l * newX - y2) % p

        return Point(newX, newY)

    def toBytes(self):
        x = self.x.to_bytes(32, "big")
        y = self.y.to_bytes(32, "big")
        return b"\x04" + x + y


def getPublicKey(privkey):
    SECP256k1 = Point()
    pk = int.from_bytes(privkey, "big")
    publickey =(((SECP256k1 * pk).toBytes()).hex())
    return publickey




if __name__ == "__main__":
                    #range Div
    for i in range((Start_Range//boost), (End_Range//boost)):
        print(i)
        b = hex(i)[2:]
        c = str.format(b)
        pk= c.zfill(64)
        Seq_Bytes = binascii.unhexlify(pk)
        PublicKey = getPublicKey(Seq_Bytes)
        #print(PublicKey)
        Privatekey = (Seq_Bytes).hex()
        #print(Privatekey)
        
        data_base = 'data-base.txt'
        
        with open(data_base, 'r') as file:
            content= file.read()
            #search publickey in data-base
            if PublicKey in content:
                print("Range Found")
                print("Decimal:", int(Privatekey, base=16)*boost)
                

save as data-base.txt

Code:

040119630133edf190d14fdb00640f74cc4317e1ddb9dff1c9707eac3353ef9c21f16ab4e5e3a3ee6d743851876da07034caa4f5196106a0c803136d215420494c
04572dd1b78d9e4d185117860c71a224a05daee2b7bba651478ddbabadc71831c1d3fd83384c92bcaa326b75b10ef7cb33baf8b4d474cf8e5eac0e598a8a13200f
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I currently don't have what it takes to code in C++ using the quick search methods, my resources are limited to the basics.
I currently don't have an up-to-date pc to do the relevant tests (the disadvantages of the third world).

However, later I will share faster methods, ECC tricks.

Can you explain what this script does?  I mean do we need to just provide it with a target upub, set a range and change G to whatever we want ( stride ) and then what about the database, where should we get our database keys from?

[mcdouglasx]

increase the search speed of the Pubkeys, dividing the search range.

First of all, let me make it clear that this script is slow and basic, I did it like this in order to explain the range division method.

If you want to apply it quickly, modify keyhunt or other c++ code before compiling to make it work with this method.

The method consists in changing the parameters of the curve by x,y of a higher point (for the script I used pk#100).
by default the curve uses pubkey #1
what works in sequence 1,2,3,4....
pubkey #100
which works in sequence 100,200,300,400....

if we want to find pk# 3922
in a range of 2000:6000
in the traditional way we would need to advance +1900 pubkeys
In this method we will only need 19 pubkeys (1900/100).

the database is done by subtracting 1 consecutively from the target pubkey (in my case I need to subtract( 1), 100 times because I chose pubkey pk# 100 as divisor.

The trick is in your ability to use fast databases, the more pubkey you store in the database, the higher the divisor will be, therefore the search range is smaller.

save as brute_div.py

Code:

import os
import hashlib
import random
import binascii

#target pk decimal 3922

#Target_Upub "042f3342152eff6aca5e7314db6d3301a28d6a90ddcfd189f96babadc2a053d3929d0e4bef850141613e6bb3d3e2c480b3634af4332f125f4edfc65b88816f541d"

#Start_Range in decimal
Start_Range = 2000

#End_Range in decimal
End_Range   = 6000

#divisor pk in decimal 
boost = 100

#boost_upub "X=ed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88, y=e57a6f571288ccffdcda5e8a7a1f87bf97bd17be084895d0fce17ad5e335286e"




class Point:
    def __init__(self,
                 #change x and y for boost_upub
        x=0xed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88,
        y=0xe57a6f571288ccffdcda5e8a7a1f87bf97bd17be084895d0fce17ad5e335286e,
        p=2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 - 1):
        self.x = x
        self.y = y
        self.p = p

    def __add__(self, other):
        return self.__radd__(other)

    def __mul__(self, other):
        return self.__rmul__(other)

    def __rmul__(self, other):
        n = self
        q = None

        for i in range(256):
            if other & (1 << i):
                q = q + n
            n = n + n

        return q

    def __radd__(self, other):
        if other is None:
            return self
        x1 = other.x
        y1 = other.y
        x2 = self.x
        y2 = self.y
        p = self.p

        if self == other:
            l = pow(2 * y2 % p, p-2, p) * (3 * x2 * x2) % p
        else:
            l = pow(x1 - x2, p-2, p) * (y1 - y2) % p

        newX = (l ** 2 - x2 - x1) % p
        newY = (l * x2 - l * newX - y2) % p

        return Point(newX, newY)

    def toBytes(self):
        x = self.x.to_bytes(32, "big")
        y = self.y.to_bytes(32, "big")
        return b"\x04" + x + y


def getPublicKey(privkey):
    SECP256k1 = Point()
    pk = int.from_bytes(privkey, "big")
    publickey =(((SECP256k1 * pk).toBytes()).hex())
    return publickey




if __name__ == "__main__":
                    #range Div
    for i in range((Start_Range//boost), (End_Range//boost)):
        print(i)
        b = hex(i)[2:]
        c = str.format(b)
        pk= c.zfill(64)
        Seq_Bytes = binascii.unhexlify(pk)
        PublicKey = getPublicKey(Seq_Bytes)
        #print(PublicKey)
        Privatekey = (Seq_Bytes).hex()
        #print(Privatekey)
        
        data_base = 'data-base.txt'
        
        with open(data_base, 'r') as file:
            content= file.read()
            #search publickey in data-base
            if PublicKey in content:
                print("Range Found")
                print("Decimal:", int(Privatekey, base=16)*boost)
                

save as data-base.txt

Code:

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I currently don't have what it takes to code in C++ using the quick search methods, my resources are limited to the basics.
I currently don't have an up-to-date pc to do the relevant tests (the disadvantages of the third world).

However, later I will share faster methods, ECC tricks.

Can you explain what this script does?  I mean do we need to just provide it with a target upub, set a range and change G to whatever we want ( stride ) and then what about the database, where should we get our database keys from?

basically by changing Gx, Gy for example by pubkey of pk #1000
the sequence of the curve would be 1000,2000,3000,4000...
and you will wonder what happens with 1001,1002,1003...1999.
for that you subtract 1, 1000 times from the target and store it in the database.

If you search for 10924, and subtract 1000 times 1, the database would look like this: 10923,10922,10921...10000...9999,9998.
therefore you would find the pk 10000 in 10 steps, and you would know that your target is in the range 10000-10999

you can create the data-base.txt like this:

Code:

import secp256k1 as ice


target_public_key = "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
target = ice.pub2upub(target_public_key)
num = 100 # number of times.
sustract= 1 #amount to subtract each time.
sustract_pub= ice.scalar_multiplication(sustract)
res= ice.point_loop_subtraction(num, target, sustract_pub)
for t in range (num+1):
    h= res[t*65:t*65+65]
    data = open("data-base.txt","a")
    data.write(str(h.hex())+"\n")
    data.close()

[zahid888]

Alright, let me explain everything clearly. For the past few months, I've been collecting all the ranges for puzzle 66, a significant part of which belongs to me. I've scanned certain ranges into the diffrent pools and performed many offline experiments as well. I've also gathered numerous ranges from different individuals' posts, where I trusted that these ranges were authentic. I used to save them, but all of these ranges turned out to be insufficient.

Then, I came across a text file from somewhere. When I added it to my database, it introduced a deceptive alteration in the scanned percentage, which was a major portion of the scanned ranges. I spent several days attempting to determine whether these ranges were genuinely scanned or if they were fake. After being unable to ascertain the truth, I accepted it as genuine and included it in my database. Following that, I only needed to scan 588 more ranges to achieve 100% completion of scanning.

This is the one file that has perplexed me with the question of "yes or no..

As far as it goes, why don't I complete all these ranges myself? I've already invested a significant amount of my money and time here. I would like to say one more thing that might not enter your minds, where we are completely powerless, upper forces can assist us, but to benefit from that power, we have to share things with others, after seeing our kindness, that power also shows us mercy . Therefore, I thought it's better to share my hard work with people, and whoever is destined will receive it. (even I removed 60-40 % too) Well, Maybe these words won't make sense to you, but that's okay, keep demotivating us... Looking at some people, it also feels like why am I sharing my hard work with those who have such negative thoughts about me, well... well, all five fingers aren't the same. By the way, these days, old folks are hardly visible. Maybe because the current generation has become quite disrespectful.

[tptkimikaze]

Alright, let me explain everything clearly. For the past few months, I've been collecting all the ranges for puzzle 66, a significant part of which belongs to me. I've scanned certain ranges into the diffrent pools and performed many offline experiments as well. I've also gathered numerous ranges from different individuals' posts, where I trusted that these ranges were authentic. I used to save them, but all of these ranges turned out to be insufficient.

Then, I came across a text file from somewhere. When I added it to my database, it introduced a deceptive alteration in the scanned percentage, which was a major portion of the scanned ranges. I spent several days attempting to determine whether these ranges were genuinely scanned or if they were fake. After being unable to ascertain the truth, I accepted it as genuine and included it in my database. Following that, I only needed to scan 588 more ranges to achieve 100% completion of scanning.

This is the one file that has perplexed me with the question of "yes or no..

As far as it goes, why don't I complete all these ranges myself? I've already invested a significant amount of my money and time here. I would like to say one more thing that might not enter your minds, where we are completely powerless, upper forces can assist us, but to benefit from that power, we have to share things with others, after seeing our kindness, that power also shows us mercy . Therefore, I thought it's better to share my hard work with people, and whoever is destined will receive it. (even I removed 60-40 % too) Well, Maybe these words won't make sense to you, but that's okay, keep demotivating us... Looking at some people, it also feels like why am I sharing my hard work with those who have such negative thoughts about me, well... well, all five fingers aren't the same. By the way, these days, old folks are hardly visible. Maybe because the current generation has become quite disrespectful.

(588 more ranges to achieve 100% completion of scanning)

Do you think these words make sense for any individual who have a sensible mind? If you have invest already so much, why don't you invest another less than USD 2k and less than a month time for a 100% scan ? Another USD 2k with a return of USD 170k at least. This is not rocket science or difficult business math to do. Don't look at someone account as newbie and said we're the new generation. I know BTC when it is worth less than USD 100 per pieces and Mt Gox haven't go bankrupt yet. The day I joined bitcointalk is the time we're still using USB miner to mine BTC and I even own couple of faucet during those days. What is current generation or older generation? I don't like to stay at forum anymore because it's not anymore like those old days and full of scammers. I can easily find back one of my Bitcointalk account which can go back roughly 6-7 years ago and I am just lazy to do so. Not new generation are disrespectful, it is your action and plan that doesn't make sense at all for any sensible human being.