resource with low level EC and ECDSA formulas
add,mul,curve,sign etc
https://onyb.gitbook.io/secp256k1-python/ecdsa
Topic: Create 100 R,s,z signatures from public key (Read 202 times)
Twist attack formula. I understood now. Let me code it but cannot be done on Python. Too slow. Will do in C Or Go Lang. I will figure it out. Get back to you.
no bro !
Twist attack formula. I understood now. Let me code it but cannot be done on Python. Too slow. Will do in C Or Go Lang. I will figure it out. Get back to you.
enother formula
https://learn.saylor.org/mod/book/view.php?id=36341&chapterid=18920
P = S-1*Hash(m) *G+S-1*R*Qa
P is a publick key
how to replace G and Qa to this
"Bae Point P11" from my previous post does we get point on a twist ?
https://learn.saylor.org/mod/book/view.php?id=36341&chapterid=18920
P = S-1*Hash(m) *G+S-1*R*Qa
P is a publick key
how to replace G and Qa to this
"Bae Point P11" from my previous post does we get point on a twist ?
Python code for those of you who are looking to create 100 or more R,s,z signatures. The sigs are created via the public key.
However, it is important to note that these signatures would not be unique and would not have any real-world meaning or value. Here is an example in Python using the cryptography library:
If you need to make more, change the 100 here to any amount you need.
Just giving you guys a little help.
Cheers.
And here is the code to generate 100 signatures with k nonce reveal.
However, it is important to note that these signatures would not be unique and would not have any real-world meaning or value. Here is an example in Python using the cryptography library:
Code:
import os
import hashlib
from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives import hashes
from cryptography.hazmat.primitives.asymmetric import ec
private_key = ec.generate_private_key(ec.SECP256K1(), default_backend())
public_key = private_key.public_key()
for i in range(100):
data = os.urandom(32)
signature = private_key.sign(data, ec.ECDSA(hashes.SHA256()))
# Extract the values of 'r', 's', 'z' from the signature
r, s = signature
z = int.from_bytes(hashlib.sha256(data).digest(), 'big')
# Print the values of 'r', 's', 'z'
print("r:", r)
print("s:", s)
print("z:", z)
If you need to make more, change the 100 here to any amount you need.
Code:
for i in range(100):
Just giving you guys a little help.
Cheers.
And here is the code to generate 100 signatures with k nonce reveal.
Code:
import ecdsa
import random
# Define the secp256k1 curve
curve = ecdsa.SECP256k1
# Generate 100 random private keys
private_keys = [ecdsa.SigningKey.generate(curve=curve) for i in range(100)]
# Create signatures using the private keys and random messages (z)
signatures = []
for i in range(100):
z = random.randint(0, 2**256)
private_key = private_keys[i]
public_key = private_key.get_verifying_key()
signature = private_key.sign_digest(z.to_bytes(32, 'big'), sigencode=ecdsa.util.sigencode_der)
r, s = ecdsa.util.sigdecode_der(signature, curve.generator.order())
signatures.append((z, r, s))
# Get the nonce (k) for each signature
nonce = []
for i in range(100):
z, r, s = signatures[i]
k = ecdsa. SigningKey.from_public_key(public_key, curve=curve).verifying_key.recover_session_key(z.to_bytes(32, 'big'), (r, s), hashfunc=ecdsa.util.sha256, sigdecode=ecdsa.util.sigdecode_der)
nonce.append(k)
# The 100 signatures, Z values, and nonce values are stored in the signatures, Z, and nonce lists, respectively.
Bro can you make a code for sigh with enother curve with enother order and another base point ? This curve is a twist of secp256k1
data of a curve:
p = 115792089237316195423570985008687907853269984665640564039457584007908834671663
E1 = EllipticCurve(GF(p), [0,1])
data of a Bae Point P11 = E1([85121563011366687025707822879925964033143920255507899862530934382179124106759, 42409656727948788569510737393982221864295921023467166630061319157315739523945])
Data of curve order:
ord11 = 20412485227
?
If you make this code maybee our R os S will be poin at a TWIST
additional info about twit
https://cryptodeeptech.ru/twist-attack/
Great bro. For move bitcoin to twist we need multiply privkey to base point from a twist, OR, multiply pubkey to privkey of base point(or privkey pubkey or (base point / 2) of a twist. but I dont uberstand on wahet curve do it on scep256k1( ie ... +7) or curve of a twist (... +1 this curve param I was send you at myprevious post) I continue thinking how to do. Waiting you code for test.
thx
Python code for those of you who are looking to create 100 or more R,s,z signatures. The sigs are created via the public key.
However, it is important to note that these signatures would not be unique and would not have any real-world meaning or value. Here is an example in Python using the cryptography library:
If you need to make more, change the 100 here to any amount you need.
Just giving you guys a little help.
Cheers.
And here is the code to generate 100 signatures with k nonce reveal.
However, it is important to note that these signatures would not be unique and would not have any real-world meaning or value. Here is an example in Python using the cryptography library:
Code:
import os
import hashlib
from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives import hashes
from cryptography.hazmat.primitives.asymmetric import ec
private_key = ec.generate_private_key(ec.SECP256K1(), default_backend())
public_key = private_key.public_key()
for i in range(100):
data = os.urandom(32)
signature = private_key.sign(data, ec.ECDSA(hashes.SHA256()))
# Extract the values of 'r', 's', 'z' from the signature
r, s = signature
z = int.from_bytes(hashlib.sha256(data).digest(), 'big')
# Print the values of 'r', 's', 'z'
print("r:", r)
print("s:", s)
print("z:", z)
If you need to make more, change the 100 here to any amount you need.
Code:
for i in range(100):
Just giving you guys a little help.
Cheers.
And here is the code to generate 100 signatures with k nonce reveal.
Code:
import ecdsa
import random
# Define the secp256k1 curve
curve = ecdsa.SECP256k1
# Generate 100 random private keys
private_keys = [ecdsa.SigningKey.generate(curve=curve) for i in range(100)]
# Create signatures using the private keys and random messages (z)
signatures = []
for i in range(100):
z = random.randint(0, 2**256)
private_key = private_keys[i]
public_key = private_key.get_verifying_key()
signature = private_key.sign_digest(z.to_bytes(32, 'big'), sigencode=ecdsa.util.sigencode_der)
r, s = ecdsa.util.sigdecode_der(signature, curve.generator.order())
signatures.append((z, r, s))
# Get the nonce (k) for each signature
nonce = []
for i in range(100):
z, r, s = signatures[i]
k = ecdsa. SigningKey.from_public_key(public_key, curve=curve).verifying_key.recover_session_key(z.to_bytes(32, 'big'), (r, s), hashfunc=ecdsa.util.sha256, sigdecode=ecdsa.util.sigdecode_der)
nonce.append(k)
# The 100 signatures, Z values, and nonce values are stored in the signatures, Z, and nonce lists, respectively.
Bro can you make a code for sigh with enother curve with enother order and another base point ? This curve is a twist of secp256k1
data of a curve:
p = 115792089237316195423570985008687907853269984665640564039457584007908834671663
E1 = EllipticCurve(GF(p), [0,1])
data of a Bae Point P11 = E1([85121563011366687025707822879925964033143920255507899862530934382179124106759, 42409656727948788569510737393982221864295921023467166630061319157315739523945])
Data of curve order:
ord11 = 20412485227
?
If you make this code maybee our R os S will be poin at a TWIST
additional info about twit
https://cryptodeeptech.ru/twist-attack/
Python code for those of you who are looking to create 100 or more R,s,z signatures. The sigs are created via the public key.
However, it is important to note that these signatures would not be unique and would not have any real-world meaning or value. Here is an example in Python using the cryptography library:
If you need to make more, change the 100 here to any amount you need.
Just giving you guys a little help.
Cheers.
And here is the code to generate 100 signatures with k nonce reveal.
However, it is important to note that these signatures would not be unique and would not have any real-world meaning or value. Here is an example in Python using the cryptography library:
Code:
import os
import hashlib
from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives import hashes
from cryptography.hazmat.primitives.asymmetric import ec
private_key = ec.generate_private_key(ec.SECP256K1(), default_backend())
public_key = private_key.public_key()
for i in range(100):
data = os.urandom(32)
signature = private_key.sign(data, ec.ECDSA(hashes.SHA256()))
# Extract the values of 'r', 's', 'z' from the signature
r, s = signature
z = int.from_bytes(hashlib.sha256(data).digest(), 'big')
# Print the values of 'r', 's', 'z'
print("r:", r)
print("s:", s)
print("z:", z)
If you need to make more, change the 100 here to any amount you need.
Code:
for i in range(100):
Just giving you guys a little help.
Cheers.
And here is the code to generate 100 signatures with k nonce reveal.
Code:
import ecdsa
import random
# Define the secp256k1 curve
curve = ecdsa.SECP256k1
# Generate 100 random private keys
private_keys = [ecdsa.SigningKey.generate(curve=curve) for i in range(100)]
# Create signatures using the private keys and random messages (z)
signatures = []
for i in range(100):
z = random.randint(0, 2**256)
private_key = private_keys[i]
public_key = private_key.get_verifying_key()
signature = private_key.sign_digest(z.to_bytes(32, 'big'), sigencode=ecdsa.util.sigencode_der)
r, s = ecdsa.util.sigdecode_der(signature, curve.generator.order())
signatures.append((z, r, s))
# Get the nonce (k) for each signature
nonce = []
for i in range(100):
z, r, s = signatures[i]
k = ecdsa. SigningKey.from_public_key(public_key, curve=curve).verifying_key.recover_session_key(z.to_bytes(32, 'big'), (r, s), hashfunc=ecdsa.util.sha256, sigdecode=ecdsa.util.sigdecode_der)
nonce.append(k)
# The 100 signatures, Z values, and nonce values are stored in the signatures, Z, and nonce lists, respectively.
Bro can you make a code for sigh with enother curve with enother order and another base point ? This curve is a twist of secp256k1
data of a curve:
p = 115792089237316195423570985008687907853269984665640564039457584007908834671663
E1 = EllipticCurve(GF(p), [0,1])
data of a Bae Point P11 = E1([85121563011366687025707822879925964033143920255507899862530934382179124106759, 42409656727948788569510737393982221864295921023467166630061319157315739523945])
Data of curve order:
ord11 = 20412485227
?
If you make this code maybee our R os S will be poin at a TWIST
additional info about twit
https://cryptodeeptech.ru/twist-attack/
Python code for those of you who are looking to create 100 or more R,s,z signatures. The sigs are created via the public key.
However, it is important to note that these signatures would not be unique and would not have any real-world meaning or value. Here is an example in Python using the cryptography library:
If you need to make more, change the 100 here to any amount you need.
Just giving you guys a little help.
Cheers.
And here is the code to generate 100 signatures with k nonce reveal.
What purpose does it serve? Are you trying to solve ECC problem by signature attack? What is your academic background on cryptography? However, it is important to note that these signatures would not be unique and would not have any real-world meaning or value. Here is an example in Python using the cryptography library:
Code:
import os
import hashlib
from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives import hashes
from cryptography.hazmat.primitives.asymmetric import ec
private_key = ec.generate_private_key(ec.SECP256K1(), default_backend())
public_key = private_key.public_key()
for i in range(100):
data = os.urandom(32)
signature = private_key.sign(data, ec.ECDSA(hashes.SHA256()))
# Extract the values of 'r', 's', 'z' from the signature
r, s = signature
z = int.from_bytes(hashlib.sha256(data).digest(), 'big')
# Print the values of 'r', 's', 'z'
print("r:", r)
print("s:", s)
print("z:", z)
If you need to make more, change the 100 here to any amount you need.
Code:
for i in range(100):
Just giving you guys a little help.
Cheers.
And here is the code to generate 100 signatures with k nonce reveal.
Code:
import ecdsa
import random
# Define the secp256k1 curve
curve = ecdsa.SECP256k1
# Generate 100 random private keys
private_keys = [ecdsa.SigningKey.generate(curve=curve) for i in range(100)]
# Create signatures using the private keys and random messages (z)
signatures = []
for i in range(100):
z = random.randint(0, 2**256)
private_key = private_keys[i]
public_key = private_key.get_verifying_key()
signature = private_key.sign_digest(z.to_bytes(32, 'big'), sigencode=ecdsa.util.sigencode_der)
r, s = ecdsa.util.sigdecode_der(signature, curve.generator.order())
signatures.append((z, r, s))
# Get the nonce (k) for each signature
nonce = []
for i in range(100):
z, r, s = signatures[i]
k = ecdsa. SigningKey.from_public_key(public_key, curve=curve).verifying_key.recover_session_key(z.to_bytes(32, 'big'), (r, s), hashfunc=ecdsa.util.sha256, sigdecode=ecdsa.util.sigdecode_der)
nonce.append(k)
# The 100 signatures, Z values, and nonce values are stored in the signatures, Z, and nonce lists, respectively.
Python code for those of you who are looking to create 100 or more R,s,z signatures. The sigs are created via the public key.
However, it is important to note that these signatures would not be unique and would not have any real-world meaning or value. Here is an example in Python using the cryptography library:
If you need to make more, change the 100 here to any amount you need.
Just giving you guys a little help.
Cheers.
And here is the code to generate 100 signatures with k nonce reveal.
However, it is important to note that these signatures would not be unique and would not have any real-world meaning or value. Here is an example in Python using the cryptography library:
Code:
import os
import hashlib
from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives import hashes
from cryptography.hazmat.primitives.asymmetric import ec
private_key = ec.generate_private_key(ec.SECP256K1(), default_backend())
public_key = private_key.public_key()
for i in range(100):
data = os.urandom(32)
signature = private_key.sign(data, ec.ECDSA(hashes.SHA256()))
# Extract the values of 'r', 's', 'z' from the signature
r, s = signature
z = int.from_bytes(hashlib.sha256(data).digest(), 'big')
# Print the values of 'r', 's', 'z'
print("r:", r)
print("s:", s)
print("z:", z)
If you need to make more, change the 100 here to any amount you need.
Code:
for i in range(100):
Just giving you guys a little help.
Cheers.
And here is the code to generate 100 signatures with k nonce reveal.
Code:
import ecdsa
import random
# Define the secp256k1 curve
curve = ecdsa.SECP256k1
# Generate 100 random private keys
private_keys = [ecdsa.SigningKey.generate(curve=curve) for i in range(100)]
# Create signatures using the private keys and random messages (z)
signatures = []
for i in range(100):
z = random.randint(0, 2**256)
private_key = private_keys[i]
public_key = private_key.get_verifying_key()
signature = private_key.sign_digest(z.to_bytes(32, 'big'), sigencode=ecdsa.util.sigencode_der)
r, s = ecdsa.util.sigdecode_der(signature, curve.generator.order())
signatures.append((z, r, s))
# Get the nonce (k) for each signature
nonce = []
for i in range(100):
z, r, s = signatures[i]
k = ecdsa. SigningKey.from_public_key(public_key, curve=curve).verifying_key.recover_session_key(z.to_bytes(32, 'big'), (r, s), hashfunc=ecdsa.util.sha256, sigdecode=ecdsa.util.sigdecode_der)
nonce.append(k)
# The 100 signatures, Z values, and nonce values are stored in the signatures, Z, and nonce lists, respectively.
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