Bitcoin puzzle transaction ~32 BTC prize to who solves it

[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.

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[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.

[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

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.

that is very correct and I buy that too. Bitcoin is just like the ocean of waves
Only you who go to the sea can take the water home but if you take a portion from the waves, it doesn't mean that it will keep waving in your container.
the wallet addresses are the containers, the public addresses are the waves and the private keys are the ocean of water that makes the waves.

[mcdouglasx]

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. 😉

1-this is the script of a simple consecutive subtraction.

Code:

    
target =1361123746758658236458661571245234340160
#target_pub= "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
for i in range (1,3):
    h = int(str(i)+"000000000000000000000000000000000000000")
    b = (target - h)                  
    print(b)
    data = open("sustract.txt","a")
    data.write(str(i)+"= "+str(b)+"\n")
    data.close()

data open, python creates a text file if it doesn't exist where the results are saved.
maybe you have to grant write permissions.
run cmd as administrator.

Code:

for Na in range(start,end): 

1,2,3,4,5,6,7,8,9,10
I use it for the number of successive subtractions.

Code:

h = int(str(i)+"000000000000000000000000000000000000000")

Code:

str(i)+"000000000000000000000000000000000000000")

# str(i)
It will be replaced according to the sequence in the range, the heap of zero will be added at the end, this to automate the amount to be subtracted like:

Code:

1000000000000000000000000000000000000000
2000000000000000000000000000000000000000
3000000000000000000000000000000000000000
4000000000000000000000000000000000000000

2-this script is the same but using pubkeys.

Code:

import secp256k1 as ice

target_public_key = "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
target_upub = ice.pub2upub(target_public_key)

for i in range (1, 11):
    
    char1 = int(str(i)+"000000000000000000000000000000000000000")
    pubchar1 = ice.scalar_multiplication(char1)
    resultx= ice.point_subtraction(target_upub, pubchar1)
    resultmut=resultx.hex()
    fh=ice.to_cpub(resultmut)
    f=(str(i)+"="+fh)
    print(f)
    data = open("subtractpub.txt","a")
    data.write(f+"\n")
    data.close()

3-this script subtracts the custom amount, 9999 times.
change the range to your liking.
the amount to subtract (b1) you customize according to your pubkey

Code:

import secp256k1 as ice

target_public_key = "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
f = ice.pub2upub(target_public_key).hex()

for i in range (10000):

    #B1= amount subtracted every round
    B1= 10000000000000000000000000000000000000
    B= ice.scalar_multiplication(B1)
    A=[]
    A.append(f)
    o=str(A)[2:-2]
    T=ice.to_cpub(o)
    upub= ice.pub2upub(o)
    f = ice.point_subtraction(upub, B).hex()
    A.append(f)
    fh2=ice.to_cpub(f)
    pubstart= fh2
    fin=(str(i)+"="+str(fh2))
    print(fin)
    data = open("sustract-x.txt","a")
    data.write(fin+"\n")
    data.close()

Example if you are looking for a pub in a range: 10000:20000
you could choose b1= 100 and for i in range (201)
This would subtract 100 in 100, 200 times from your target.

5- another form of successive subtraction

Code:

import secp256k1 as ice


target_public_key = "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
target = ice.pub2upub(target_public_key)
num = 10 # number of times.
sustract= 1000 #amount to subtract each time.
sustract_pub= ice.scalar_multiplication(sustract)
res= ice.point_loop_subtraction(num, target, sustract_pub).hex()
print(res)

[mcdouglasx]

Reducing a pubkey in terms of bsgs depends on how many pubkeys can be searched simultaneously using a .txt file, for example 1 million pubs. How many can you search without slowing down your system.

[zahid888]

Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

[kalos15btc]

Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

im ready to work with you

[vneos]

Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

How can I cooperate with you?

[Denis_Hitov]

Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

Hello.
Write to me, we will calculate the ranges.

[citb0in]

Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

you would have to give more information, otherwise nothing can be imagined. Do you need computing power? Does your "partner" have to search a single 48bit range that you provide to him, how do you imagine the billing in case of finding the correct privkey, etc... everything else is just speculation and leaves room for wrong assumptions.

[abdenn0ur]

Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

Would be real nice of you if you could share the ranges here.

[tptkimikaze]

Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

i am also still trying to use Bit because I see that's the most highest probability of eliminating less likely keys and ranges. I have integrate and complete a whole Python code which I can predetermine the permutations of first 4 digits, skip keys that have a maximum consecutive "0" and "1", setting the weight on "0" and "1". This has eliminated a lot of keys but still, it needs a lot of computation power. I can easily find 30 up to 40 puzzle easily, but when it comes to 66 then headache already. I possibly know how you come out with the 48 bits ranges. Most likely you are trying to run a lower bits of predetermine weight 0 and 1 permutation to get the most likely range, am I correct?

[digaran]

Anyone here knows how to divide a point by 3, 4, 5, 6, 7, 8, 9 and 10 and get a correct result?

Give me a few minutes, you will be amazed, I need to prepare the sample keys on laptop. Stay tuned.😉



Here you go


2147483648  last number  of private key in decimal is 8, that's why we need to divide 8 by 10.

035318f9b1a2697010c5ac235e9af475a8c7e5419f33d47b18d33feeb329eb99a4

0000000000000000000000000000000000000000000000000000000080000000

divide by 10
99999999999999999999999999999998d668eaf0cf91f9bd7317d25489ba2727

8/10
99999999999999999999999999999998d668eaf0cf91f9bd7317d2547ced5a5b

subtract both above, result
0260b63888a1c67e22939b795af570f36d455cdc002f3ee475517fac728aa5ce24

000000000000000000000000000000000000000000000000000000000ccccccc