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Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 5. (Read 189853 times)

jr. member
Activity: 37
Merit: 1
Geniuses who found the key for address #66, can you tell me if the key is between 37396512239913013824 and 51680708854858323072?


The hexadecimal approximately probably 16.12% to found within the 66 bit range.  Roll Eyes
 
2B85F409A58861B43 into 3ffffffffffffffff  Huh

169,549,060,031,553,553,727.  Shocked


guess that big number.  Grin
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Geniuses who found the key for address #66, can you tell me if the key is between 37396512239913013824 and 51680708854858323072?
newbie
Activity: 19
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That's what is give me hope... I know is hard but let's try
By doing this in my half broken casio make me fell like Kepler  Grin Grin Grin Let's kill Tycho Brahe  Cheesy Cheesy Cheesy Cheesy

As I said way before, my hope is not to find the *exactly* address, but reduce the range because my hardware and my computer skills are not the top-notch, and I can't compete with people with gpu tho


It's more likely to find a $1 million lotto ticket using Dowsing than to solve this Puzzle.  Grin

I know, but at least I'm filling these lottery tickets using python  Grin Grin Grin
jr. member
Activity: 40
Merit: 6
If I would be the creator I would laugh so hard about some of the things discussed here.

Guys, a logarithm is an abstract concept, not some math function.

You get a thing called a "base change". In this case we're dealing with a change of base of an element from some position in a finite field (private keys) to an element in the same position in a finite group (EC public keys) and the problem is to solve for the position without a way to go back from the latter to the first (which is assumed to be hard, but not yet proven). And this in the best case that we even have such an element, and not some fingerprint of it (an address), which makes the problem levels of more absurdly difficult. WTF is with the real numbers field log2 discussion, it makes no sense, we already know the ranges double in size at each step, of course any polynomial regression or whatever is a straight line. Dividing 1 by (2**64) is four levels of magnitude below a double-precision IEEE floating point, so what errors do you expect, they will always be after the 64-th zero decimal digit in reality. Nevermind the fact that there's an infinity of real numbers between any two real numbers, so an infinity of computations. Take 7 as a private key and try to solve back from [1/4, 1/8) interval, mission impossible.

This is not an analytical problem, it's a group theory problem.


log2 - It's just a way of representing numbers in a different way. I'll ask you again, do you know what you're talking about?
I hope you are joking. That is NOT what a logarithm is.

Number: 7
Representing 7 in different ways:
Base 10: 7
Base 2: 0111
Base 7: 10
Base 16: 0x07
Base 1:

Shall I go on?

Logarithm of 7 in the infinite field of real numbers:
Base 2: 2.8073549221... (infinite number of decimals) - it counts how many bits you need to represent value 7
Base e: 1.9459101490553132... (infinite number of decimals) - it counts how many natural numbers you need that, raised to this power, yields 7

Logarithm of 7 in some modular finite field with N = 3:
log(7) = Huh mod 3 (oh no, this doesn't work, hmm... is it 0? 1? 2? damn, something is missing, maybe we didn't even define the neutral element yet? or not even an binary addition operator?)

So, do YOU know what you are talking about?
member
Activity: 286
Merit: 15
That's what is give me hope... I know is hard but let's try
By doing this in my half broken casio make me fell like Kepler  Grin Grin Grin Let's kill Tycho Brahe  Cheesy Cheesy Cheesy Cheesy

As I said way before, my hope is not to find the *exactly* address, but reduce the range because my hardware and my computer skills are not the top-notch, and I can't compete with people with gpu tho


It's more likely to find a $1 million lotto ticket using Dowsing than to solve this Puzzle.  Grin
brand new
Activity: 0
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If I would be the creator I would laugh so hard about some of the things discussed here.

Guys, a logarithm is an abstract concept, not some math function.

You get a thing called a "base change". In this case we're dealing with a change of base of an element from some position in a finite field (private keys) to an element in the same position in a finite group (EC public keys) and the problem is to solve for the position without a way to go back from the latter to the first (which is assumed to be hard, but not yet proven). And this in the best case that we even have such an element, and not some fingerprint of it (an address), which makes the problem levels of more absurdly difficult. WTF is with the real numbers field log2 discussion, it makes no sense, we already know the ranges double in size at each step, of course any polynomial regression or whatever is a straight line. Dividing 1 by (2**64) is four levels of magnitude below a double-precision IEEE floating point, so what errors do you expect, they will always be after the 64-th zero decimal digit in reality. Nevermind the fact that there's an infinity of real numbers between any two real numbers, so an infinity of computations. Take 7 as a private key and try to solve back from [1/4, 1/8) interval, mission impossible.

This is not an analytical problem, it's a group theory problem.


log2 - It's just a way of representing numbers in a different way. I'll ask you again, do you know what you're talking about?
newbie
Activity: 19
Merit: 0
member
Activity: 286
Merit: 15
newbie
Activity: 19
Merit: 0
If I would be the creator I would laugh so hard about some of the things discussed here.

Guys, a logarithm is an abstract concept, not some math function.

You get a thing called a "base change". In this case we're dealing with a change of base of an element from some position in a finite field (private keys) to an element in the same position in a finite group (EC public keys) and the problem is to solve for the position without a way to go back from the latter to the first (which is assumed to be hard, but not yet proven). And this in the best case that we even have such an element, and not some fingerprint of it (an address), which makes the problem levels of more absurdly difficult. WTF is with the real numbers field log2 discussion, it makes no sense, we already know the ranges double in size at each step, of course any polynomial regression or whatever is a straight line. Dividing 1 by (2**64) is four levels of magnitude below a double-precision IEEE floating point, so what errors do you expect, they will always be after the 64-th zero decimal digit in reality. Nevermind the fact that there's an infinity of real numbers between any two real numbers, so an infinity of computations. Take 7 as a private key and try to solve back from [1/4, 1/8) interval, mission impossible.

This is not an analytical problem, it's a group theory problem.

Funnier than that is the circles of apocalypse  Grin Grin Grin

I agree with you. It's so many problems, I will humongous lists if I start go in depth.
My idea is to see if the pseudo-randomicity of the numbers gave some clues, like I did as a work 10 years ago.
Its the only route I'm thinking tbh bc I have a old laptop and find a job is horrible rn.
But even than that, my jupyter is killing my patience and floating number problem is a REAL pain in the a**.

as a side note: I was also telling my date I was reviewing GF(n) to solve a puzzle and I confronted a quite skeptical reaction like "you are not a procrastinator person to do this as a hobby".... oh well  Grin Grin Grin
member
Activity: 286
Merit: 15

I can't find any way to add the rule backward (child structure to parent key)
I have zero idea how to do it  Huh.... even chatgpt and mistral said is impossible ( i have questions about it  Roll Eyes , take as a grain of salt)

And no, a potato pc can't do all this smiths... I'm saying because I have a simple potato laptop and god knows how bad is going

The result will always be the same, whether you use the range as a decimal number and generate random numbers there, or whether you use log(2) as a different notation. You could also chase the numbers through other algorithms. Some will work faster and that is the only advantage.
jr. member
Activity: 40
Merit: 6
If I would be the creator I would laugh so hard about some of the things discussed here.

Guys, a logarithm is an abstract concept, not some math function.

You get a thing called a "base change". In this case we're dealing with a change of base of an element from some position in a finite field (private keys) to an element in the same position in a finite group (EC public keys) and the problem is to solve for the position without a way to go back from the latter to the first (which is assumed to be hard, but not yet proven). And this in the best case that we even have such an element, and not some fingerprint of it (an address), which makes the problem levels of more absurdly difficult. WTF is with the real numbers field log2 discussion, it makes no sense, we already know the ranges double in size at each step, of course any polynomial regression or whatever is a straight line. Dividing 1 by (2**64) is four levels of magnitude below a double-precision IEEE floating point, so what errors do you expect, they will always be after the 64-th zero decimal digit in reality. Nevermind the fact that there's an infinity of real numbers between any two real numbers, so an infinity of computations. Take 7 as a private key and try to solve back from [1/4, 1/8) interval, mission impossible.

This is not an analytical problem, it's a group theory problem.
newbie
Activity: 19
Merit: 0
member
Activity: 286
Merit: 15
So.....here is a new toy.

puzzle 66 private key is from
65.00000000000000 to
65.99999999999999  Log(2)

Code:
import math
import sys
import secp256k1 as ice


# Function to convert Log(2) to decimal number
def log2_to_decimal(log2_value):
    return 2 ** log2_value

# Range of Log(2) values
start_log2 = 65.00000000000000
end_log2 = 65.99999999999999

# Number of decimal places for Log(2)
decimal_places = 15

target_caddr = "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so"

# Iterate through the range and calculate decimal numbers
for i in range(int(start_log2 * 10**decimal_places), int((end_log2 + 10**-decimal_places) * 10**decimal_places)):
    log2_value = i / 10**decimal_places
    decimal_number = log2_to_decimal(log2_value)
    HEX = "%064x" % int(decimal_number)
    dec = int(HEX, 16)
    caddr = ice.privatekey_to_address(0, True, dec)
    message = "\r[+] {}".format(dec)
    messages = []
    messages.append(message)
    output = ''.join(messages) + "\r"
    sys.stdout.write(output)
    sys.stdout.flush()
    if caddr == target_caddr:
       wifc = ice.btc_pvk_to_wif(HEX)
       print(wifc)
       break

This  experimental log(2) method skips decimal numbers on the fly as desired in "decimal_places" . . .
Maybe someone can guess on luck what the "decimal_places" number is (from 1 to 17) .  Grin
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member
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If you guys are saying that pubkey is enough to take over any puzzle tx,
then it can be done with any tx that is in mempool.

It is possible, but it requires knowledge of the private key range with a minimum precision width of Puzzle 30 scale, so that you could brute-force the private key on a potato PC. The key thing here is that we know the exact ranges of the keys for puzzles. And on too big like Puzzle 130, that factor doesn't help us either. Due to lack of hardware.
newbie
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full member
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If you guys are saying that pubkey is enough to take over any puzzle tx,
then it can be done with any tx that is in mempool.
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I formulated the question: is it possible to create a number system in which:

1=1
2=3
3=7
4=8
5=21
6=49
.....

Also I have a question, how did the creator access the addresses above puzl#160 repeatedly? I.e. he was transferring btc from these addresses to smaller addresses. Did he really save every key to the address and "manually" (using an algorithm) spell them out?
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wish me luck, current speed is 100K keys per sec.

100kk/s is very low. 
You're better off using keyhunt by alberto
You may get +1Mk/s even on a potato CPU



Perhaps his algorithm works exponentially? That is, the time to find the key decreases exponentially.
brand new
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hi all i read the post since one year sorry for my english i found a interesting think but i does takes me further. in every elliptic curve like y^2= x^3+7 there is something interesting like :
if P(1,y1) - k times--> Q(-29/3 ,y2)
   P(2,y3) --k times--> Q(-3,y2)
 so on there is a simple math here where k is always too know independent from whcih curve we work .
 i don't want to give more information this operation is 10 times faster then k*G= and find the x value 


I think to solve the equation we need to introduce a third Z axis
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