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Little explanation.
what we know: if we have two outputs:) if one , it doesn't matter.
1. We know:
a. partial WIF of privatekey
b. r1,s1,z1 and r2,s2,z2 ( two transactions)
2. What we can do?
a. first we must re-arrange the transactions outputs as : nonce is privatekey, and privatekeys is a nonce.! -> yes we can
b. we know partial WIF of privatekey as nonce - we can use Monte Carlo together with LLL
c. recentering for partial integers plus BKZ or LLL.
becouse: IF OP has partial WIF example knowns MSB , not known middle, and Known LSB - it is easy even for 23 missing characters.
what we know: if we have two outputs:) if one , it doesn't matter.
1. We know:
a. partial WIF of privatekey
b. r1,s1,z1 and r2,s2,z2 ( two transactions)
2. What we can do?
a. first we must re-arrange the transactions outputs as : nonce is privatekey, and privatekeys is a nonce.! -> yes we can
b. we know partial WIF of privatekey as nonce - we can use Monte Carlo together with LLL
c. recentering for partial integers plus BKZ or LLL.
becouse: IF OP has partial WIF example knowns MSB , not known middle, and Known LSB - it is easy even for 23 missing characters.
Becouse I can't : I do not know the partial WIF , if you know please share, and seconds there is only one output.
This is why I can't solve the famous puzzle.
You mean the unsolved puzzle output's "Public Key" right? not partial WIF.This is why I can't solve the famous puzzle.
Because otherwise this wont make any sense.
And BTW, specific puzzle output's partial WIF should be similar to the closest solved outputs' WIF since they're all encoded starting from bunch of zeroes.
Like #66's partial WIF, which should be: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3.......
Based from:
- #65: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZM21gaY8WN2CdwnTG57
- #64: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZ6FxoaD5r1kYegmtbaT
- #63: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYv5Z9J7hv7VYYN3XL3Y
- #62: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYpCemuaUp7NigjvtJug
- #61: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYmLDHsih379uP9zbHSD
I understand that you have difficulty with reading and understanding written text. I suggest either returning to school or, if it's a medical issue, seeking urgent medical attention. I have similar students in my class. They read quickly, don't understand what they read, and then express themselves saying nonsense. I always look forward to exam time, and then I proudly fail them and they repeat the year
@j2002ba2 has right. the QC wil not be used for "cracking" curve.
Any one who has minimal knowledge about "typing programing" of QC , knows that QC cannot be implement for decrypt anything.
Any one who has minimal knowledge about "typing programing" of QC , knows that QC cannot be implement for decrypt anything.
Becouse I can't : I do not know the partial WIF , if you know please share, and seconds there is only one output.
This is why I can't solve the famous puzzle. if you know math you should known that.
The principle is the same, there is a private key in a certain key range that we only know the corresponding address or public key. Whether it is a WIF missing characters or a puzzle with a key intentionally created in a certain range. The methods used for finding that key is the same too.This is why I can't solve the famous puzzle. if you know math you should known that.
No it is not the same. as I see you have no knowledge about this. topic close.
[/quote]
Becouse I can't : I do not know the partial WIF , if you know please share, and seconds there is only one output.
This is why I can't solve the famous puzzle. if you know math you should known that.
NotAtether:
you wrote: we all know that quantum computers will be able to defeat ECDSA, NO! , it is impossible. maybe in next thousend year.
[/quote]
Yeah, so you have tested it, with how many missing characters exactly? Was it 21 or 19 missing? THIS IS TRUE, you can't solve such a key "very quickly". I'd love to see how you can help! show us the magic.
[/quote]
digaran it is not magic , real normal math with implement some kinds additional operation on curve. it is not magic, there is no magic stuff. I will explain little more with recentering and partial integer signatures with LLL
nothing more. There is no magic there are only relations.
to OP: please show the public key and the wif
⭐ Merited by ecdsa123 (15)
Are you talking to me or advising others against me? Lol, even if OP shared both public key and the remaining of the WIF characters, nobody could solve the key in a few thousand years, a 256 bit private key is not something we could simply brute force.
THIS IS NOT TRUE! if on btc address are spent transactions , and you have remaining of the WIF characters, you can solve very quickly.
Здравствуйте. Отвечая на вопросы. Это оборудование от компании, объявившей о банкротстве, и она реализует товары со складов. Что касается цены, я недавно подтвердил, что в зависимости от количества штук еще можно договориться о цене. Больше информации, пожалуйста, в личных сообщениях.
example of DLEQ:
Code:
from cryptography.hazmat.primitives.asymmetric import ec
def dleq(g1, h1, n1, x1, y1, g2, h2, n2, x2, y2):
# Check if g1, h1, g2, and h2 are elements of their respective multiplicative groups
curve1 = ec.SECP256R1() # You can change this to a different elliptic curve if needed
curve2 = ec.SECP384R1() # You can change this to another elliptic curve if needed
if not (g1 in curve1 and h1 in curve1 and g2 in curve2 and h2 in curve2):
raise ValueError("g1, h1, g2, and h2 must be elements of their respective multiplicative groups")
# Calculate the left-hand side and right-hand side of the DLEQ equations for both curves
lhs1 = pow(g1, x1, n1)
rhs1 = pow(h1, y1, n1)
lhs2 = pow(g2, x2, n2)
rhs2 = pow(h2, y2, n2)
# Compare the left-hand side and right-hand side for both curves
return lhs1 == rhs1 and lhs2 == rhs2
# Example usage
g1 = ec.SECP256R1().generator() # Generator of the SECP256R1 elliptic curve
h1 = 123456789 # Example element of the multiplicative group for curve 1
n1 = ec.SECP256R1().order() # Order of the multiplicative group for curve 1
x1 = 987654321 # Secret integers for curve 1
y1 = 543210987
g2 = ec.SECP384R1().generator() # Generator of the SECP384R1 elliptic curve
h2 = 987654321 # Example element of the multiplicative group for curve 2
n2 = ec.SECP384R1().order() # Order of the multiplicative group for curve 2
x2 = 246813579 # Secret integers for curve 2
y2 = 135792468
result = dleq(g1, h1, n1, x1, y1, g2, h2, n2, x2, y2)
print("DLEQ result:", result)
def dleq(g1, h1, n1, x1, y1, g2, h2, n2, x2, y2):
# Check if g1, h1, g2, and h2 are elements of their respective multiplicative groups
curve1 = ec.SECP256R1() # You can change this to a different elliptic curve if needed
curve2 = ec.SECP384R1() # You can change this to another elliptic curve if needed
if not (g1 in curve1 and h1 in curve1 and g2 in curve2 and h2 in curve2):
raise ValueError("g1, h1, g2, and h2 must be elements of their respective multiplicative groups")
# Calculate the left-hand side and right-hand side of the DLEQ equations for both curves
lhs1 = pow(g1, x1, n1)
rhs1 = pow(h1, y1, n1)
lhs2 = pow(g2, x2, n2)
rhs2 = pow(h2, y2, n2)
# Compare the left-hand side and right-hand side for both curves
return lhs1 == rhs1 and lhs2 == rhs2
# Example usage
g1 = ec.SECP256R1().generator() # Generator of the SECP256R1 elliptic curve
h1 = 123456789 # Example element of the multiplicative group for curve 1
n1 = ec.SECP256R1().order() # Order of the multiplicative group for curve 1
x1 = 987654321 # Secret integers for curve 1
y1 = 543210987
g2 = ec.SECP384R1().generator() # Generator of the SECP384R1 elliptic curve
h2 = 987654321 # Example element of the multiplicative group for curve 2
n2 = ec.SECP384R1().order() # Order of the multiplicative group for curve 2
x2 = 246813579 # Secret integers for curve 2
y2 = 135792468
result = dleq(g1, h1, n1, x1, y1, g2, h2, n2, x2, y2)
print("DLEQ result:", result)
Привет всем.
Это гостевой аукцион.
На сегодняшний день у нас есть для продажи со складов торговой компании, находящейся в банкротстве:
20 штук двигателей Limbach L550E - 50 л.с.
без транспорта 0,5 BTC / штука
с транспортом в европейские страны, Турцию, Китай: дополнительно 0,5 BTC / штука
мы продаем не менее 3 штук за раз.
Это гостевой аукцион.
На сегодняшний день у нас есть для продажи со складов торговой компании, находящейся в банкротстве:
20 штук двигателей Limbach L550E - 50 л.с.
без транспорта 0,5 BTC / штука
с транспортом в европейские страны, Турцию, Китай: дополнительно 0,5 BTC / штука
мы продаем не менее 3 штук за раз.
Привет всем.
Это гостевой аукцион.
На сегодняшний день у нас есть для продажи со складов торговой компании, находящейся в банкротстве:
20 штук двигателей Limbach L550E - 50 л.с.
без транспорта 0,5 BTC / штука
с транспортом в европейские страны, Турцию, Китай: дополнительно 0,5 BTC / штука
мы продаем не менее 3 штук за раз.
Это гостевой аукцион.
На сегодняшний день у нас есть для продажи со складов торговой компании, находящейся в банкротстве:
20 штук двигателей Limbach L550E - 50 л.с.
без транспорта 0,5 BTC / штука
с транспортом в европейские страны, Турцию, Китай: дополнительно 0,5 BTC / штука
мы продаем не менее 3 штук за раз.
Привет всем.
Это гостевой аукцион.
На сегодняшний день у нас есть для продажи со складов торговой компании, находящейся в банкротстве:
20 штук двигателей Limbach L550E - 50 л.с.
без транспорта 0,5 BTC / штука
с транспортом в европейские страны, Турцию, Китай: дополнительно 0,5 BTC / штука
мы продаем не менее 3 штук за раз.
Это гостевой аукцион.
На сегодняшний день у нас есть для продажи со складов торговой компании, находящейся в банкротстве:
20 штук двигателей Limbach L550E - 50 л.с.
без транспорта 0,5 BTC / штука
с транспортом в европейские страны, Турцию, Китай: дополнительно 0,5 BTC / штука
мы продаем не менее 3 штук за раз.
Привет всем.
Это гостевой аукцион.
На сегодняшний день у нас есть для продажи со складов торговой компании, находящейся в банкротстве:
20 штук двигателей Limbach L550E - 50 л.с.
без транспорта 0,5 BTC / штука
с транспортом в европейские страны, Турцию, Китай: дополнительно 0,5 BTC / штука
мы продаем не менее 3 штук за раз.
Это гостевой аукцион.
На сегодняшний день у нас есть для продажи со складов торговой компании, находящейся в банкротстве:
20 штук двигателей Limbach L550E - 50 л.с.
без транспорта 0,5 BTC / штука
с транспортом в европейские страны, Турцию, Китай: дополнительно 0,5 BTC / штука
мы продаем не менее 3 штук за раз.
Привет всем.
Это гостевой аукцион.
На сегодняшний день у нас есть для продажи со складов торговой компании, находящейся в банкротстве:
20 штук двигателей Limbach L550E - 50 л.с.
без транспорта 0,5 BTC / штука
с транспортом в европейские страны, Турцию, Китай: дополнительно 0,5 BTC / штука
мы продаем не менее 3 штук за раз.
Re: Proposal to Address Dormant Bitcoin:Recycling Lost Coins into the Mining Process
2. Some say quantum computers could crack private keys of UTXOs created in Satoshi's time. By the time such computers exist, nobody will be using the old algorithms (the community will be forced to transition to something else). But would it be fair that the UTXOs from 2009 go to the quantum computer developer? Wouldn't it be better to change the consensus regarding UTXOs made with old algorithms when transitioning to new ones?
answer:
First of all, these mythical quantum computers. They are not and will not be useful for such tasks. All those fancy scientific studies are tainted with elaborate theories. Notice that so far, they can only confirm algorithms that have already been devised. The states of the so-called qubits as 0, 1, or unknown, don't really matter. If it were otherwise, why would people keep creating new supercomputers when they could have a super quantum computer for half the cost? Please stop scaring people with these quantum computers and what they cannot do. If you had experience with such computers (and I do), you would know that it's a fairy tale, like something from moss and ferns.
answer:
First of all, these mythical quantum computers. They are not and will not be useful for such tasks. All those fancy scientific studies are tainted with elaborate theories. Notice that so far, they can only confirm algorithms that have already been devised. The states of the so-called qubits as 0, 1, or unknown, don't really matter. If it were otherwise, why would people keep creating new supercomputers when they could have a super quantum computer for half the cost? Please stop scaring people with these quantum computers and what they cannot do. If you had experience with such computers (and I do), you would know that it's a fairy tale, like something from moss and ferns.
No there is no point in secp256k1 where x==y
but there is one where x==y+1
x=103219894018170979103981239500535823206309202530631329673674059809050911020508
y=103219894018170979103981239500535823206309202530631329673674059809050911020507
(x**3+7)%P==(y**2)%P => True
but there is one where x==y+1
x=103219894018170979103981239500535823206309202530631329673674059809050911020508
y=103219894018170979103981239500535823206309202530631329673674059809050911020507
(x**3+7)%P==(y**2)%P => True
So. if we are talking abouy the curve and we have Fp and N :
field Fp is defined by :
Fp = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE FFFFFC2F
order:
Fn = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE BAAEDCE6 AF48A03B BFD25E8C D0364141
your point x,y where
x==103219894018170979103981239500535823206309202530631329673674059809050911020508
y==103219894018170979103981239500535823206309202530631329673674059809050911020507
it is real point as half_mod Fp/Fn , I got problem to explain my english is not so good to technical explain.