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https://www.blockchain.com/explorer/addresses/btc/17s2b9ksz5y7abUm92cHwG8jEPCzK3dLnT
120 puzzle solved....
120 puzzle solved....
so, send the private key
who solved ? which tools useing to crack keyhunt or BitCrack
if successful block mine from python || how to coinbase send me my reward
@MagicByt3 your topic https://bitcointalk.org/index.php?topic=5440240.0
@irsada your topic https://bitcointalk.org/index.php?topic=5439273.0
my question is
if successful block mine using btc_mine_python_code ,how to coinbase send my reward ### simple python code coinbase send my reward
explain me particular btc_mine_python_code coinbase send my reward
if you don t understand my question simple,i am asking for i need coinbase send my reward in python code
@irsada your topic https://bitcointalk.org/index.php?topic=5439273.0
my question is
if successful block mine using btc_mine_python_code ,how to coinbase send my reward ### simple python code coinbase send my reward
explain me particular btc_mine_python_code coinbase send my reward
if you don t understand my question simple,i am asking for i need coinbase send my reward in python code
Re: Python based Solo miner for CPU | Learn Basic Bitcoin Mining | Just for fun
pip install error
/home/user/Downloads# pip install hashlib
/home/user/Downloads# pip install binascii
/home/user/Downloads# pip install hashlib
Code:
Looking in indexes: https://pypi.org/simple, https://us-python.pkg.dev/colab-wheels/public/simple/
Collecting hashlib
Using cached hashlib-20081119.zip (42 kB)
error: subprocess-exited-with-error
× python setup.py egg_info did not run successfully.
│ exit code: 1
╰─> See above for output.
note: This error originates from a subprocess, and is likely not a problem with pip.
Preparing metadata (setup.py) ... error
error: metadata-generation-failed
× Encountered error while generating package metadata.
╰─> See above for output.
note: This is an issue with the package mentioned above, not pip.
hint: See above for details.
Collecting hashlib
Using cached hashlib-20081119.zip (42 kB)
error: subprocess-exited-with-error
× python setup.py egg_info did not run successfully.
│ exit code: 1
╰─> See above for output.
note: This error originates from a subprocess, and is likely not a problem with pip.
Preparing metadata (setup.py) ... error
error: metadata-generation-failed
× Encountered error while generating package metadata.
╰─> See above for output.
note: This is an issue with the package mentioned above, not pip.
hint: See above for details.
/home/user/Downloads# pip install binascii
Code:
Looking in indexes: https://pypi.org/simple, https://us-python.pkg.dev/colab-wheels/public/simple/
ERROR: Could not find a version that satisfies the requirement binascii (from versions: none)
ERROR: No matching distribution found for binascii
ERROR: Could not find a version that satisfies the requirement binascii (from versions: none)
ERROR: No matching distribution found for binascii
Re: Python based Solo miner for CPU | Learn Basic Bitcoin Mining | Just for fun
Hello Bitcoiners
I want to share a python based solo bitcoin miner which uses ckpool. You can use other pools as well if you want. It is basically like a lottery which has extremly low chances to win but it can be used as a proof of concept. Maybe it has already been shared somewhere and I do not remember its true origin so cannot give proper reference.
You can make changes as you want. This code can help users understand true basics of mining bitcoins.
Installing Dependencies
You must have python. Try with old versions first.
Install dependencies with
Code
Edit Code
You can edit code. Set your Bitcoin address to receive your mining rewards. Set pool host and port. Save code in a .py file.
One of popular solo pool is solo.ckpool.org which has low fee of like 2%. You get 98% of your mining reward.
Run Code
You can run code by opening command prompt in same folder of file and run command
Missing Dependencies and errors
As decribed above you can use pip install command to install any missing dependency. You can google errors and still if you did not find solution maybe your python version is not compatible. Try changing python version.
Future Work
If you improve this code then share here so others can benefit as well and I will update this code as well.
Honor List for Code Improvers
Tribute
I want to pay tribute to original author of this code. I am not original author.
I want to share a python based solo bitcoin miner which uses ckpool. You can use other pools as well if you want. It is basically like a lottery which has extremly low chances to win but it can be used as a proof of concept. Maybe it has already been shared somewhere and I do not remember its true origin so cannot give proper reference.
You can make changes as you want. This code can help users understand true basics of mining bitcoins.
Installing Dependencies
You must have python. Try with old versions first.
Install dependencies with
Code:
pip install hashlib
pip install binascii
If any dependency is missing you can use pip install DependencyName.pip install binascii
Code
Code:
# -*- coding: utf-8 -*-
import socket
import json
import hashlib
import binascii
from pprint import pprint
import random
address = 'YourBitcoinAddress'
nonce = hex(random.randint(0,2**32-1))[2:].zfill(8)
host = 'solo.ckpool.org'
port = 3333
#host = 'pool.mainnet.bitcoin-global.io'
#port = 9223
def main():
print("address:{} nonce:{}".format(address,nonce))
print("host:{} port:{}".format(host,port))
sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
sock.connect((host,port))
#server connection
sock.sendall(b'{"id": 1, "method": "mining.subscribe", "params": []}\n')
lines = sock.recv(1024).decode().split('\n')
response = json.loads(lines[0])
sub_details,extranonce1,extranonce2_size = response['result']
#authorize workers
sock.sendall(b'{"params": ["'+address.encode()+b'", "password"], "id": 2, "method": "mining.authorize"}\n')
#we read until 'mining.notify' is reached
response = b''
while response.count(b'\n') < 4 and not(b'mining.notify' in response):
response += sock.recv(1024)
#get rid of empty lines
responses = [json.loads(res) for res in response.decode().split('\n') if len(res.strip())>0 and 'mining.notify' in res]
pprint(responses)
job_id,prevhash,coinb1,coinb2,merkle_branch,version,nbits,ntime,clean_jobs \
= responses[0]['params']
target = (nbits[2:]+'00'*(int(nbits[:2],16) - 3)).zfill(64)
print('nbits:{} target:{}\n'.format(nbits,target))
# extranonce2 = '00'*extranonce2_size
extranonce2 = hex(random.randint(0,2**32-1))[2:].zfill(2*extranonce2_size) # create random
coinbase = coinb1 + extranonce1 + extranonce2 + coinb2
coinbase_hash_bin = hashlib.sha256(hashlib.sha256(binascii.unhexlify(coinbase)).digest()).digest()
print('coinbase:\n{}\n\ncoinbase hash:{}\n'.format(coinbase,binascii.hexlify(coinbase_hash_bin)))
merkle_root = coinbase_hash_bin
for h in merkle_branch:
merkle_root = hashlib.sha256(hashlib.sha256(merkle_root + binascii.unhexlify(h)).digest()).digest()
merkle_root = binascii.hexlify(merkle_root).decode()
#little endian
merkle_root = ''.join([merkle_root[i]+merkle_root[i+1] for i in range(0,len(merkle_root),2)][::-1])
print('merkle_root:{}\n'.format(merkle_root))
def noncework():
nonce = hex(random.randint(0,2**32-1))[2:].zfill(8) #hex(int(nonce,16)+1)[2:]
blockheader = version + prevhash + merkle_root + nbits + ntime + nonce +\
'000000800000000000000000000000000000000000000000000000000000000000000000000000000000000080020000'
# print('blockheader:\n{}\n'.format(blockheader))
hash = hashlib.sha256(hashlib.sha256(binascii.unhexlify(blockheader)).digest()).digest()
hash = binascii.hexlify(hash).decode()
# print('hash: {}'.format(hash))
if(hash[:5] == '00000'): print('hash: {}'.format(hash))
if hash < target :
# if(hash[:10] == '0000000000'):
print('success!!')
print('hash: {}'.format(hash))
payload = bytes('{"params": ["'+address+'", "'+job_id+'", "'+extranonce2 \
+'", "'+ntime+'", "'+nonce+'"], "id": 1, "method": "mining.submit"}\n', 'utf-8')
sock.sendall(payload)
print(sock.recv(1024))
input("Press Enter to continue...")
# else:
# print('failed mine, hash is greater than target')
for k in range(10000000):
noncework()
print("Finished 10M Search. Regaining Information.")
sock.close()
main()
main()
import socket
import json
import hashlib
import binascii
from pprint import pprint
import random
address = 'YourBitcoinAddress'
nonce = hex(random.randint(0,2**32-1))[2:].zfill(8)
host = 'solo.ckpool.org'
port = 3333
#host = 'pool.mainnet.bitcoin-global.io'
#port = 9223
def main():
print("address:{} nonce:{}".format(address,nonce))
print("host:{} port:{}".format(host,port))
sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
sock.connect((host,port))
#server connection
sock.sendall(b'{"id": 1, "method": "mining.subscribe", "params": []}\n')
lines = sock.recv(1024).decode().split('\n')
response = json.loads(lines[0])
sub_details,extranonce1,extranonce2_size = response['result']
#authorize workers
sock.sendall(b'{"params": ["'+address.encode()+b'", "password"], "id": 2, "method": "mining.authorize"}\n')
#we read until 'mining.notify' is reached
response = b''
while response.count(b'\n') < 4 and not(b'mining.notify' in response):
response += sock.recv(1024)
#get rid of empty lines
responses = [json.loads(res) for res in response.decode().split('\n') if len(res.strip())>0 and 'mining.notify' in res]
pprint(responses)
job_id,prevhash,coinb1,coinb2,merkle_branch,version,nbits,ntime,clean_jobs \
= responses[0]['params']
target = (nbits[2:]+'00'*(int(nbits[:2],16) - 3)).zfill(64)
print('nbits:{} target:{}\n'.format(nbits,target))
# extranonce2 = '00'*extranonce2_size
extranonce2 = hex(random.randint(0,2**32-1))[2:].zfill(2*extranonce2_size) # create random
coinbase = coinb1 + extranonce1 + extranonce2 + coinb2
coinbase_hash_bin = hashlib.sha256(hashlib.sha256(binascii.unhexlify(coinbase)).digest()).digest()
print('coinbase:\n{}\n\ncoinbase hash:{}\n'.format(coinbase,binascii.hexlify(coinbase_hash_bin)))
merkle_root = coinbase_hash_bin
for h in merkle_branch:
merkle_root = hashlib.sha256(hashlib.sha256(merkle_root + binascii.unhexlify(h)).digest()).digest()
merkle_root = binascii.hexlify(merkle_root).decode()
#little endian
merkle_root = ''.join([merkle_root[i]+merkle_root[i+1] for i in range(0,len(merkle_root),2)][::-1])
print('merkle_root:{}\n'.format(merkle_root))
def noncework():
nonce = hex(random.randint(0,2**32-1))[2:].zfill(8) #hex(int(nonce,16)+1)[2:]
blockheader = version + prevhash + merkle_root + nbits + ntime + nonce +\
'000000800000000000000000000000000000000000000000000000000000000000000000000000000000000080020000'
# print('blockheader:\n{}\n'.format(blockheader))
hash = hashlib.sha256(hashlib.sha256(binascii.unhexlify(blockheader)).digest()).digest()
hash = binascii.hexlify(hash).decode()
# print('hash: {}'.format(hash))
if(hash[:5] == '00000'): print('hash: {}'.format(hash))
if hash < target :
# if(hash[:10] == '0000000000'):
print('success!!')
print('hash: {}'.format(hash))
payload = bytes('{"params": ["'+address+'", "'+job_id+'", "'+extranonce2 \
+'", "'+ntime+'", "'+nonce+'"], "id": 1, "method": "mining.submit"}\n', 'utf-8')
sock.sendall(payload)
print(sock.recv(1024))
input("Press Enter to continue...")
# else:
# print('failed mine, hash is greater than target')
for k in range(10000000):
noncework()
print("Finished 10M Search. Regaining Information.")
sock.close()
main()
main()
Edit Code
You can edit code. Set your Bitcoin address to receive your mining rewards. Set pool host and port. Save code in a .py file.
One of popular solo pool is solo.ckpool.org which has low fee of like 2%. You get 98% of your mining reward.
Run Code
You can run code by opening command prompt in same folder of file and run command
Code:
python FileName.py
Missing Dependencies and errors
As decribed above you can use pip install command to install any missing dependency. You can google errors and still if you did not find solution maybe your python version is not compatible. Try changing python version.
Future Work
If you improve this code then share here so others can benefit as well and I will update this code as well.
Honor List for Code Improvers
- 1.
- 2.
- 3.
Tribute
I want to pay tribute to original author of this code. I am not original author.
how to mine testnet
any GPU speed, python mining
Re: Can Quantum Computer's destroy Blockchain and Bitcoins[SHA-256 specifically]
my reply for this topic >>> Re: Can Quantum Computer's destroy Blockchain and Bitcoins[SHA-256 specifically]
two type answer:
1. incase future 2022 < 2022 billion year ago , IBM or Google made Quantum Computer's you will crack 100% sha256
2. my_case
i am already solved crack sha256, #note: sha256 is magic its easy to reverse operation # using python3.9
you need sha256 cracking method its available in .com this site# only educational purpose ,i am leaked online
donate me : 1ASzaPvnM5BM2HE5VF1VCgkwX6yUWvJTvW
two type answer:
1. incase future 2022 < 2022 billion year ago , IBM or Google made Quantum Computer's you will crack 100% sha256
2. my_case
i am already solved crack sha256, #note: sha256 is magic its easy to reverse operation # using python3.9you need sha256 cracking method its available in
donate me : 1ASzaPvnM5BM2HE5VF1VCgkwX6yUWvJTvW
What you think what false is because false pubkey is all 03 not 02 ? 02... all thrue...
small change my code
and 99% predict correct answer, for which one small or bigits easy bro ec way to solved
i know 120 puzzle privatekey
but, i am not swap btc 
and i am unlocked other 1btc puzzle , withdrawl my account
,i am happy note:
my problem already solved ,thanks for all participate btc hacker
@albert0bsd
@all
test.txt
hi guys, how to run my test.txt because, every time i copy past and Run ./keyhunt
small update please,once range complete open my test.txt and run the next range
understand my problom ,sorry i speak little english
so, please tell how to run one by one my custom range>>>my test.txt
@all
test.txt
Code:
./keyhunt -m bsgs -f 120 -r 7357a4501ddfe92f46681b20a0:d576e7357a4501ddff92f46681b20ff
./keyhunt -m bsgs -f 120 -r 3fffffffffffffffffffffffffffffff:3fffffffffffffffffffffffffffffffaeaff
./keyhunt -m bsgs -f 120 -r bfffffffffffffffffffffffffffffff0c0325ad0376782ccfddc6e99c28b0f0:bfffffffffffffffffffffffffffffff0c0325ad0376782cffddc6e99c28b0ff
etc...
etc..
etc..
./keyhunt -m bsgs -f 120 -r 3fffffffffffffffffffffffffffffff:3fffffffffffffffffffffffffffffffaeaff
./keyhunt -m bsgs -f 120 -r bfffffffffffffffffffffffffffffff0c0325ad0376782ccfddc6e99c28b0f0:bfffffffffffffffffffffffffffffff0c0325ad0376782cffddc6e99c28b0ff
etc...
etc..
etc..
hi guys, how to run my test.txt because, every time i copy past and Run ./keyhunt
small update please,once range complete open my test.txt and run the next range
understand my problom ,sorry i speak little english
so, please tell how to run one by one my custom range>>>my test.txt
run this code python2
X is incorrect Y output is correct
('fadd', '0xc8333020c4688a754bf3ad462f1e9f1fac80649a463ae4d4c1afd48d20fccffL', '0xb7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777L')
('fsub', '0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798L', '0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8L')
('fdub', '0xc6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5L', '0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52aL')
('fdiv', '0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798L', '0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8L')
my output X value incorrect print why? what is mistake this code
Y value is correct print
update please
Code:
# -*- coding: utf-8 -*-
# run python2
# Elliptic curve point operations (Python 2.7)
def OnCurve(x,y): # Check if the point is on the curve
A = (y*y)%P
B = (x*x*x)%P
C = False
if A == (B + 7):
C = True
return C
def legendre_symbol(a,p):
ls = pow(a, (p - 1) / 2, p)
return -1 if ls == p - 1 else ls
def modsqrt(a,p): # Square root A modulo P
if legendre_symbol(a, p) != 1:
return 0
elif a == 0:
return 0
elif p == 2:
return p
elif p % 4 == 3:
return pow(a, (p + 1) / 4, p)
s = p - 1
e = 0
while s % 2 == 0:
s /= 2
e += 1
n = 2
while legendre_symbol(n, p) != -1:
n += 1
x = pow(a, (s + 1) / 2, p)
b = pow(a, s, p)
g = pow(n, s, p)
r = e
while True:
t = b
m = 0
for m in xrange(r):
if t == 1:
break
t = pow(t, 2, p)
if m == 0:
return x
gs = pow(g, 2 ** (r - m - 1), p)
g = (gs * gs) % p
x = (x * gs) % p
b = (b * g) % p
r = m
def modinv(a,n): # Extended Euclidean Algorithm in elliptic curves
lm, hm = 1,0
low, high = a%n,n
while low > 1:
ratio = high/low
nm = hm - lm * ratio
new = high - low * ratio
hm = lm
high = low
lm = nm
low = new
return lm % n
def ECadd(xp,yp,xq,yq): # EC point addition
m = ((yq-yp) * modinv(xq-xp,P))%P
xr = (m*m-xp-xq)%P
yr = (m*(xp-xr)-yp)%P
return (xr,yr)
def ECdouble(xp,yp): # EC point doubling
LamNumer = 3*xp*xp+Acurve
LamDenom = 2*yp
Lam = (LamNumer * modinv(LamDenom,P)) % P
xr = (Lam*Lam-2*xp) % P
yr = (Lam*(xp-xr)-yp) % P
return (xr,yr)
def ECmul(xs,ys,Scalar): # EC point multiplication
if Scalar == 0 or Scalar >= N: raise Exception("Invalid Scalar/Private Key")
ScalarBin = str(bin(Scalar))[2:]
Qx,Qy=xs,ys
for i in range (1,len(ScalarBin)):
Qx,Qy = ECdouble(Qx,Qy)
if ScalarBin[i] == "1":
Qx,Qy = ECadd(Qx,Qy,xs,ys)
return (Qx,Qy)
def ECsub(xp,yp,xq,yq): # EC point subtraction
X = (((yp+yq)*modinv(xq-xp,P))**2-xp-xq)%P
A = (xp + X + xq)%P
B = modsqrt(A,P)
B1 = P - B
Y = yq - (xq - X) * B
X = X % P
Y = Y % P
if not OnCurve(X,Y):
Y = yq - (xq - X) * B1
Y = Y % P
return X,Y
def ECdiv(Qx,Qy,Scalar): # EC point division
A = (N-1)/Scalar
Px,Py = ECmul(Qx,Qy,A)
Py = P-Py
return Px,Py
P = 2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 - 1
N = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
Gx = 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
Gy = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
Acurve = 0
ukx = 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 # or your enter publickey x,y
uky = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
#you need add ,sub ,div
fadd = ECadd(ukx,uky,ukx,uky) # 1 + 1 =2
print ('fadd',hex(fadd[0]),hex(fadd[1]))
fsub = ECsub(fadd[0],fadd[1],Gx,Gy) # 2 -1 = 1
print ('fsub',hex(fsub[0]),hex(fsub[1]))
fdub = ECdouble(ukx,uky) # 1 * 2 = 2
print ('fdub',hex(fdub[0]),hex(fdub[1]))
fdiv = ECdiv(fdub[0],fdub[1],2) # 2 / 2 = 1
print ('fdiv',hex(fdiv[0]),hex(fdiv[1]))
output is:# run python2
# Elliptic curve point operations (Python 2.7)
def OnCurve(x,y): # Check if the point is on the curve
A = (y*y)%P
B = (x*x*x)%P
C = False
if A == (B + 7):
C = True
return C
def legendre_symbol(a,p):
ls = pow(a, (p - 1) / 2, p)
return -1 if ls == p - 1 else ls
def modsqrt(a,p): # Square root A modulo P
if legendre_symbol(a, p) != 1:
return 0
elif a == 0:
return 0
elif p == 2:
return p
elif p % 4 == 3:
return pow(a, (p + 1) / 4, p)
s = p - 1
e = 0
while s % 2 == 0:
s /= 2
e += 1
n = 2
while legendre_symbol(n, p) != -1:
n += 1
x = pow(a, (s + 1) / 2, p)
b = pow(a, s, p)
g = pow(n, s, p)
r = e
while True:
t = b
m = 0
for m in xrange(r):
if t == 1:
break
t = pow(t, 2, p)
if m == 0:
return x
gs = pow(g, 2 ** (r - m - 1), p)
g = (gs * gs) % p
x = (x * gs) % p
b = (b * g) % p
r = m
def modinv(a,n): # Extended Euclidean Algorithm in elliptic curves
lm, hm = 1,0
low, high = a%n,n
while low > 1:
ratio = high/low
nm = hm - lm * ratio
new = high - low * ratio
hm = lm
high = low
lm = nm
low = new
return lm % n
def ECadd(xp,yp,xq,yq): # EC point addition
m = ((yq-yp) * modinv(xq-xp,P))%P
xr = (m*m-xp-xq)%P
yr = (m*(xp-xr)-yp)%P
return (xr,yr)
def ECdouble(xp,yp): # EC point doubling
LamNumer = 3*xp*xp+Acurve
LamDenom = 2*yp
Lam = (LamNumer * modinv(LamDenom,P)) % P
xr = (Lam*Lam-2*xp) % P
yr = (Lam*(xp-xr)-yp) % P
return (xr,yr)
def ECmul(xs,ys,Scalar): # EC point multiplication
if Scalar == 0 or Scalar >= N: raise Exception("Invalid Scalar/Private Key")
ScalarBin = str(bin(Scalar))[2:]
Qx,Qy=xs,ys
for i in range (1,len(ScalarBin)):
Qx,Qy = ECdouble(Qx,Qy)
if ScalarBin[i] == "1":
Qx,Qy = ECadd(Qx,Qy,xs,ys)
return (Qx,Qy)
def ECsub(xp,yp,xq,yq): # EC point subtraction
X = (((yp+yq)*modinv(xq-xp,P))**2-xp-xq)%P
A = (xp + X + xq)%P
B = modsqrt(A,P)
B1 = P - B
Y = yq - (xq - X) * B
X = X % P
Y = Y % P
if not OnCurve(X,Y):
Y = yq - (xq - X) * B1
Y = Y % P
return X,Y
def ECdiv(Qx,Qy,Scalar): # EC point division
A = (N-1)/Scalar
Px,Py = ECmul(Qx,Qy,A)
Py = P-Py
return Px,Py
P = 2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 - 1
N = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
Gx = 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
Gy = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
Acurve = 0
ukx = 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 # or your enter publickey x,y
uky = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
#you need add ,sub ,div
fadd = ECadd(ukx,uky,ukx,uky) # 1 + 1 =2
print ('fadd',hex(fadd[0]),hex(fadd[1]))
fsub = ECsub(fadd[0],fadd[1],Gx,Gy) # 2 -1 = 1
print ('fsub',hex(fsub[0]),hex(fsub[1]))
fdub = ECdouble(ukx,uky) # 1 * 2 = 2
print ('fdub',hex(fdub[0]),hex(fdub[1]))
fdiv = ECdiv(fdub[0],fdub[1],2) # 2 / 2 = 1
print ('fdiv',hex(fdiv[0]),hex(fdiv[1]))
X is incorrect Y output is correct
('fadd', '0xc8333020c4688a754bf3ad462f1e9f1fac80649a463ae4d4c1afd48d20fccffL', '0xb7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777L')
('fsub', '0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798L', '0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8L')
('fdub', '0xc6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5L', '0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52aL')
('fdiv', '0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798L', '0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8L')
my output X value incorrect print why? what is mistake this code
Y value is correct print
update please
let me give it a try:
still open to correction
still open to correction
Code:
class point_addition:
def __init__ (self,x,y,a,b):
#y**2 = x***3+a*x+b
#x and y co-ordinate needed for calculation
self.a=a
self.b=b
self.x=x
self.y=y
if -y**2 != x***3+ a*x + b:
return ('({}{}) has negative co-ordinate').format(x,y)
elif y**2 != x***3+ a*x + b:
return ('({}{}) has positive co-ordinate').format(x,y)
p1 = x (0300000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63)
p2 = y (0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63)
print(x,y)
still a baby coder I wish to get correction
def __init__ (self,x,y,a,b):
#y**2 = x***3+a*x+b
#x and y co-ordinate needed for calculation
self.a=a
self.b=b
self.x=x
self.y=y
if -y**2 != x***3+ a*x + b:
return ('({}{}) has negative co-ordinate').format(x,y)
elif y**2 != x***3+ a*x + b:
return ('({}{}) has positive co-ordinate').format(x,y)
p1 = x (0300000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63)
p2 = y (0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63)
print(x,y)
still a baby coder I wish to get correction
how to run
what is input,explain please reply thinkeasy123@protonmail.com
0300000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
now if you start mapping points, you will get
....
025699b93fc6e1bd29e09a328d657a607b4155b61a6b5fcbedd7c12df7c67df8f5
03c62c910e502cb615a27c58512b6cc2c94f5742f76cb3d12ec993400a3695d413
0300000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
02c62c910e502cb615a27c58512b6cc2c94f5742f76cb3d12ec993400a3695d413
035699b93fc6e1bd29e09a328d657a607b4155b61a6b5fcbedd7c12df7c67df8f5
....
like this points will shrink back toward their min or max value with filling of 02 or 03
until
0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
....
0300000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
....
0379be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
and only known way to know which point is on which side is "non" hahaha.
chill and happy brute force while calling your luck.
my idea is random publickey x,y to guess for range like this ... 3ffffffffffffffffff : 7ffffffffffffffffff, 5ffffffffffffffffff:6ffffffffffffffffff, 1fffffffffffff:3ffffffffffffffffffffffffff etc....
i know this publickey privatekey range,any one find this publickey range
px: 1a1fd15fce078234aa292fc024178056bf006433c9b4bd208f59eb4c9efec95b py: a18af1fe46980989d3ff75bf9601121151ef46e2cfab8999408319ce8f3be725
any python code for 200% guess range for random public keys...
any update contribute .. share thinkeasy123@protonmail.com
@NotATether send your email please
this is my maill thinkeasy123@protonmail.com chat me
this is my maill thinkeasy123@protonmail.com chat me
@COBRAS
just my idea please make and modify div(point1,point2),multi(point1point2) python code
https://github.com/ragestack/EC-Point-Operations/blob/master/EC_Math.py
https://bitcointalk.org/index.php?topic=5244940.msg58488513#msg58488513
| How this communicate with private keys from ooin2 and point 3 ? Try make example with 110 and 90 bit pubkeys and privkeys ? |
https://github.com/ragestack/EC-Point-Operations/blob/master/EC_Math.py
https://bitcointalk.org/index.php?topic=5244940.msg58488513#msg58488513
how to prediction 100%
modify please update and send live chat me thinkeasy123@protonmail.comi donate 2022$
any alternative method suggest ...
Edit:
my test code small_big_find.py https://drive.google.com/file/d/1pRFfTxGDC0cx9zJOH4oCwPOqSQMR8yK7/view?usp=sharing
I can unlock all Bitcoins.
i dont know share me thinkeasy123@protonmail.com ...
iknow random private key: how to find y is small range or big range atleast guess
...
iknow random private key: how to find y is small range or big range atleast guess
...
THIS code EASY find for small or bigrange
but,wrong Prediction some y value,100% not work
i need imporve
Code:
import bitcoin
y = pubKey.y
y
# 32670510020758816978083085130507043184471273380659243275938904335757337482424
bitcoin.P//2 # largest Y in small range (half-point)
# 57896044618658097711785492504343953926634992332820282019728792003954417335831
if y <= (bitcoin.P//2): # is Y in the small range?
print('true',hex(y))# True
if y > (bitcoin.P//2): # is Y in the big range?
print('False',hex(y))
y = pubKey.y
y
# 32670510020758816978083085130507043184471273380659243275938904335757337482424
bitcoin.P//2 # largest Y in small range (half-point)
# 57896044618658097711785492504343953926634992332820282019728792003954417335831
if y <= (bitcoin.P//2): # is Y in the small range?
print('true',hex(y))# True
if y > (bitcoin.P//2): # is Y in the big range?
print('False',hex(y))
i test some privatekeys, y range big or small
x y
priv: 0x1
pubKey: 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
true 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
priv: 0x2
pubKey: 0xc6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a
true 0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a
priv: 0x3
pubKey: 0xf9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9 0x388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
true 0x388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
priv: 0x4
pubKey: 0xe493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13 0x51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922
true 0x51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922
priv: 0x5
pubKey: 0x2f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4 0xd8ac222636e5e3d6d4dba9dda6c9c426f788271bab0d6840dca87d3aa6ac62d6
False 0xd8ac222636e5e3d6d4dba9dda6c9c426f788271bab0d6840dca87d3aa6ac62d6
priv: 0x6
pubKey: 0xfff97bd5755eeea420453a14355235d382f6472f8568a18b2f057a1460297556 0xae12777aacfbb620f3be96017f45c560de80f0f6518fe4a03c870c36b075f297
False 0xae12777aacfbb620f3be96017f45c560de80f0f6518fe4a03c870c36b075f297
priv: 0x7
pubKey: 0x5cbdf0646e5db4eaa398f365f2ea7a0e3d419b7e0330e39ce92bddedcac4f9bc 0x6aebca40ba255960a3178d6d861a54dba813d0b813fde7b5a5082628087264da
true 0x6aebca40ba255960a3178d6d861a54dba813d0b813fde7b5a5082628087264da
priv: 0x8
pubKey: 0x2f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01 0x5c4da8a741539949293d082a132d13b4c2e213d6ba5b7617b5da2cb76cbde904
true 0x5c4da8a741539949293d082a132d13b4c2e213d6ba5b7617b5da2cb76cbde904
priv: 0x9
pubKey: 0xacd484e2f0c7f65309ad178a9f559abde09796974c57e714c35f110dfc27ccbe 0xcc338921b0a7d9fd64380971763b61e9add888a4375f8e0f05cc262ac64f9c37
False 0xcc338921b0a7d9fd64380971763b61e9add888a4375f8e0f05cc262ac64f9c37
priv: 0xa
pubKey: 0xa0434d9e47f3c86235477c7b1ae6ae5d3442d49b1943c2b752a68e2a47e247c7 0x893aba425419bc27a3b6c7e693a24c696f794c2ed877a1593cbee53b037368d7
False 0x893aba425419bc27a3b6c7e693a24c696f794c2ed877a1593cbee53b037368d7
priv: 0xb
pubKey: 0x774ae7f858a9411e5ef4246b70c65aac5649980be5c17891bbec17895da008cb 0xd984a032eb6b5e190243dd56d7b7b365372db1e2dff9d6a8301d74c9c953c61b
False 0xd984a032eb6b5e190243dd56d7b7b365372db1e2dff9d6a8301d74c9c953c61b
priv: 0xc
pubKey: 0xd01115d548e7561b15c38f004d734633687cf4419620095bc5b0f47070afe85a 0xa9f34ffdc815e0d7a8b64537e17bd81579238c5dd9a86d526b051b13f4062327
False 0xa9f34ffdc815e0d7a8b64537e17bd81579238c5dd9a86d526b051b13f4062327
priv: 0xd
pubKey: 0xf28773c2d975288bc7d1d205c3748651b075fbc6610e58cddeeddf8f19405aa8 0xab0902e8d880a89758212eb65cdaf473a1a06da521fa91f29b5cb52db03ed81
true 0xab0902e8d880a89758212eb65cdaf473a1a06da521fa91f29b5cb52db03ed81
priv: 0xe
pubKey: 0x499fdf9e895e719cfd64e67f07d38e3226aa7b63678949e6e49b241a60e823e4 0xcac2f6c4b54e855190f044e4a7b3d464464279c27a3f95bcc65f40d403a13f5b
False 0xcac2f6c4b54e855190f044e4a7b3d464464279c27a3f95bcc65f40d403a13f5b
priv: 0xf
pubKey: 0xd7924d4f7d43ea965a465ae3095ff41131e5946f3c85f79e44adbcf8e27e080e 0x581e2872a86c72a683842ec228cc6defea40af2bd896d3a5c504dc9ff6a26b58
true 0x581e2872a86c72a683842ec228cc6defea40af2bd896d3a5c504dc9ff6a26b58
priv: 0x7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1
pubKey: 0x3b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63 0xc0c686408d517dfd67c2367651380d00d126e4229631fd03f8ff35eef1a61e3c
False 0xc0c686408d517dfd67c2367651380d00d126e4229631fd03f8ff35eef1a61e3c
priv: 0x7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2
pubKey: 0xc62c910e502cb615a27c58512b6cc2c94f5742f76cb3d12ec993400a3695d413 0x17f3dadd767275ddd3b23f46723631778bf01dadaebb9a953cf068712457c010
true 0x17f3dadd767275ddd3b23f46723631778bf01dadaebb9a953cf068712457c010
how to prediction 100%
modify please update and send live chat me thinkeasy123@protonmail.comi donate 2022$
any alternative method suggest ...
Re: List of all Bitcoin addresses ever used - NOW BACK online!
@LoyceV
i am download some tsv file those site
https://gz.blockchair.com/bitcoin/inputs/
http://addresses.loyce.club/
how to extract .tsv file in terminal explain and best tool viewer suggest
thanks,
read this topic any update https://bitcointalk.org/index.php?topic=5379443.0
solve this i pay for 2022$
i am download some tsv file those site
https://gz.blockchair.com/bitcoin/inputs/
http://addresses.loyce.club/
how to extract .tsv file in terminal explain and best tool viewer suggest
thanks,
read this topic any update https://bitcointalk.org/index.php?topic=5379443.0
solve this i pay for 2022$
i speak little bit englsih
@NotATether help me this is your post 10/10/21 i need tips and python code: https://bitcointalk.org/index.php?topic=5364939.msg58145174#msg58145174
Implement a check before you do the modulus (I assume you made your own Point class) looking something like this:
This will make your point addition much slower but it's also the only way to detect whether the addition went outside the field.
my problem for any random public key add,sub.div,multple output is unknow 3rd publickey (x,y)ok, next 3rd publickey (x,y) in get y value find even odd value is small or big range how to find any math algebra code
small range means 3rd publickey y odd or even value in this range n/2 fist half#0 to 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0
big range means 3rd publickey y odd or even value is outof range n/2 2nd half #7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0 to fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
what is relation math behind [ random x,y add,sub,div,mul ] y get find small range or big range
Privatekey : 1
first x: c994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb
second x: 79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 private base x
third x: bcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb
even y: 483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8 small range y private key: 1
odd y: b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777 big range y private key: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364140
Privatekey : 2
first x: c360a6d0b34ce6df4135ee7d59f87b33d2fad8cce43837ef3e995b6ed89250e1
second x: 769ad99b0ac59bb38e84d114104707f3d08d98e78ed88b6915ba9ad5cafd0898
third x: c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 private base x
even y: 1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a small range y private key: 2
odd y: e51e970159c23cc65c3a7be6b99315110809cd9acd992f1edc9bce55af301705 big range y private key: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd036413f
Privatekey: 6
first x: 19cab650e04db19581801eb9e6c50b54f6a51b9223f6040c894f936926e302c3
second x: fff97bd5755eeea420453a14355235d382f6472f8568a18b2f057a1460297556 private base x
third x: e63bcdd9aa535fc65e3aa731e3e8bed786649d3e56a15a6847aaf28078f38045
even y: 51ed8885530449df0c4169fe80ba3a9f217f0f09ae701b5fc378f3c84f8a0998 big range y private key: 6
odd y: ae12777aacfbb620f3be96017f45c560de80f0f6518fe4a03c870c36b075f297 small range y private key: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd036413b
iknow random private key: how to find y is small range or big range atleast guess
yellow color is private key is base x value , then you find which one small range y bigrange y
test1:
first x: 1a1fd15fce078234aa292fc024178056bf006433c9b4bd208f59eb4c9efec95b private base x
second x: fc5eddb62317c03b0c0b501b3cbc7849f8babc4ac5515d4d25f4430a0aeb71c0
third x: e98150ea0ee0bd9049cb80249f2c075f4844df8170f9e5924ab1d1a75615bd43
even y: 5e750e01b967f6762c008a4069feedeeae10b91d30547666bf7ce63070c4150a
odd y: a18af1fe46980989d3ff75bf9601121151ef46e2cfab8999408319ce8f3be725
test2:
fist x: e6d4662474e3fae342e8cf460ff685fa5dcf99de42e6629320acae4e8bdb98d8
second x: 46ae596ce6268177c099b0c935759812fd4717661368960f44267c4a2f850be7
third x: d27d406ea4f583a4fc7d7ff0ba93e1f2a4e94ebba9b1075d9b2cd565449f539f private base x
even y: 2f090a097441f7d57a0af087fd8a6243b227d8a838e4751b290933588ac3b568
odd y: d0f6f5f68bbe082a85f50f7802759dbc4dd82757c71b8ae4d6f6cca6753c46c7
guys do you not understand my problom reply me i am exlain more
use this script modify script: y from x https://bitcointalk.org/index.php?topic=5358400.msg57874787#msg57874787
Edit:
any one solve this problom i well donate for you 2022$ == 0.043 current btc price #
@NotATether help me this is your post 10/10/21 i need tips and python code: https://bitcointalk.org/index.php?topic=5364939.msg58145174#msg58145174
1. Is it possible to know when the point addition process has gone outside of the field?
Implement a check before you do the modulus (I assume you made your own Point class) looking something like this:
Code:
x = # x point calculation
y = # y point calculation
if x > p_order:
print("X out of range: {}".format(x))
if y > p_order:
print("Y out of range: {}".format(y))
x %= p
y %= p
y = # y point calculation
if x > p_order:
print("X out of range: {}".format(x))
if y > p_order:
print("Y out of range: {}".format(y))
x %= p
y %= p
This will make your point addition much slower but it's also the only way to detect whether the addition went outside the field.
my problem for any random public key add,sub.div,multple output is unknow 3rd publickey (x,y)ok, next 3rd publickey (x,y) in get y value find even odd value is small or big range how to find any math algebra code
small range means 3rd publickey y odd or even value in this range n/2 fist half#0 to 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0
big range means 3rd publickey y odd or even value is outof range n/2 2nd half #7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0 to fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
what is relation math behind [ random x,y add,sub,div,mul ] y get find small range or big range
Privatekey : 1
first x: c994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb
second x: 79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 private base x
third x: bcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb
even y: 483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8 small range y private key: 1
odd y: b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777 big range y private key: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364140
Privatekey : 2
first x: c360a6d0b34ce6df4135ee7d59f87b33d2fad8cce43837ef3e995b6ed89250e1
second x: 769ad99b0ac59bb38e84d114104707f3d08d98e78ed88b6915ba9ad5cafd0898
third x: c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 private base x
even y: 1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a small range y private key: 2
odd y: e51e970159c23cc65c3a7be6b99315110809cd9acd992f1edc9bce55af301705 big range y private key: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd036413f
Privatekey: 6
first x: 19cab650e04db19581801eb9e6c50b54f6a51b9223f6040c894f936926e302c3
second x: fff97bd5755eeea420453a14355235d382f6472f8568a18b2f057a1460297556 private base x
third x: e63bcdd9aa535fc65e3aa731e3e8bed786649d3e56a15a6847aaf28078f38045
even y: 51ed8885530449df0c4169fe80ba3a9f217f0f09ae701b5fc378f3c84f8a0998 big range y private key: 6
odd y: ae12777aacfbb620f3be96017f45c560de80f0f6518fe4a03c870c36b075f297 small range y private key: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd036413b
iknow random private key: how to find y is small range or big range atleast guess
yellow color is private key is base x value , then you find which one small range y bigrange y
test1:
first x: 1a1fd15fce078234aa292fc024178056bf006433c9b4bd208f59eb4c9efec95b private base x
second x: fc5eddb62317c03b0c0b501b3cbc7849f8babc4ac5515d4d25f4430a0aeb71c0
third x: e98150ea0ee0bd9049cb80249f2c075f4844df8170f9e5924ab1d1a75615bd43
even y: 5e750e01b967f6762c008a4069feedeeae10b91d30547666bf7ce63070c4150a
odd y: a18af1fe46980989d3ff75bf9601121151ef46e2cfab8999408319ce8f3be725
test2:
fist x: e6d4662474e3fae342e8cf460ff685fa5dcf99de42e6629320acae4e8bdb98d8
second x: 46ae596ce6268177c099b0c935759812fd4717661368960f44267c4a2f850be7
third x: d27d406ea4f583a4fc7d7ff0ba93e1f2a4e94ebba9b1075d9b2cd565449f539f private base x
even y: 2f090a097441f7d57a0af087fd8a6243b227d8a838e4751b290933588ac3b568
odd y: d0f6f5f68bbe082a85f50f7802759dbc4dd82757c71b8ae4d6f6cca6753c46c7
guys do you not understand my problom reply me i am exlain more
use this script modify script: y from x https://bitcointalk.org/index.php?topic=5358400.msg57874787#msg57874787
Edit:
any one solve this problom i well donate for you 2022$ == 0.043 current btc price #
i speak little bit englsih
@NotATether help me this is your post 10/10/21 i need tips and python code: https://bitcointalk.org/index.php?topic=5364939.msg58145174#msg58145174
Implement a check before you do the modulus (I assume you made your own Point class) looking something like this:
This will make your point addition much slower but it's also the only way to detect whether the addition went outside the field.
my problem for any random public key add,sub.div,multple output is unknow 3rd publickey (x,y)ok, next 3rd publickey (x,y) in get y value find even odd value is samll or big range how to find any math algebra code
small range means 3rd publickey y odd or even value in this range n/2 fist half#0 to 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0
big range means 3rd publickey y odd or even value is outof range n/2 2nd half #7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0 to fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
what is relation math behind [ random x,y add,subdiv,mul ] y get find small range or big range
Privatekey : 1
first x: c994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb
second x: 79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 private base x
third x: bcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb
even y: 483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8 small range y private key: 1
odd y: b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777 big range y private key: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364140
Privatekey : 2
first x: c360a6d0b34ce6df4135ee7d59f87b33d2fad8cce43837ef3e995b6ed89250e1
second x: 769ad99b0ac59bb38e84d114104707f3d08d98e78ed88b6915ba9ad5cafd0898
third x: c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 private base x
even y: 1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a small range y private key: 2
odd y: e51e970159c23cc65c3a7be6b99315110809cd9acd992f1edc9bce55af301705 big range y private key: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd036413f
Privatekey: 6
first x: 19cab650e04db19581801eb9e6c50b54f6a51b9223f6040c894f936926e302c3
second x: fff97bd5755eeea420453a14355235d382f6472f8568a18b2f057a1460297556 private base x
third x: e63bcdd9aa535fc65e3aa731e3e8bed786649d3e56a15a6847aaf28078f38045
even y: 51ed8885530449df0c4169fe80ba3a9f217f0f09ae701b5fc378f3c84f8a0998 big range y private key: 6
odd y: ae12777aacfbb620f3be96017f45c560de80f0f6518fe4a03c870c36b075f297 small range y private key: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd036413b
iknow random private key: how to find y is small range or big range atleast guess
yellow color is private key is base x value , then you find which one small range y bigrange y
test1:
first x: 1a1fd15fce078234aa292fc024178056bf006433c9b4bd208f59eb4c9efec95b private base x
second x: fc5eddb62317c03b0c0b501b3cbc7849f8babc4ac5515d4d25f4430a0aeb71c0
third x: e98150ea0ee0bd9049cb80249f2c075f4844df8170f9e5924ab1d1a75615bd43
even y: 5e750e01b967f6762c008a4069feedeeae10b91d30547666bf7ce63070c4150a
odd y: a18af1fe46980989d3ff75bf9601121151ef46e2cfab8999408319ce8f3be725
test2:
fist x: e6d4662474e3fae342e8cf460ff685fa5dcf99de42e6629320acae4e8bdb98d8
second x: 46ae596ce6268177c099b0c935759812fd4717661368960f44267c4a2f850be7
third x: d27d406ea4f583a4fc7d7ff0ba93e1f2a4e94ebba9b1075d9b2cd565449f539f private base x
even y: 2f090a097441f7d57a0af087fd8a6243b227d8a838e4751b290933588ac3b568
odd y: d0f6f5f68bbe082a85f50f7802759dbc4dd82757c71b8ae4d6f6cca6753c46c7
do you not understand my problom reply me i am exlain more
@NotATether help me this is your post 10/10/21 i need tips and python code: https://bitcointalk.org/index.php?topic=5364939.msg58145174#msg58145174
1. Is it possible to know when the point addition process has gone outside of the field?
Implement a check before you do the modulus (I assume you made your own Point class) looking something like this:
Code:
x = # x point calculation
y = # y point calculation
if x > p_order:
print("X out of range: {}".format(x))
if y > p_order:
print("Y out of range: {}".format(y))
x %= p
y %= p
y = # y point calculation
if x > p_order:
print("X out of range: {}".format(x))
if y > p_order:
print("Y out of range: {}".format(y))
x %= p
y %= p
This will make your point addition much slower but it's also the only way to detect whether the addition went outside the field.
my problem for any random public key add,sub.div,multple output is unknow 3rd publickey (x,y)ok, next 3rd publickey (x,y) in get y value find even odd value is samll or big range how to find any math algebra code
small range means 3rd publickey y odd or even value in this range n/2 fist half#0 to 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0
big range means 3rd publickey y odd or even value is outof range n/2 2nd half #7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0 to fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
what is relation math behind [ random x,y add,subdiv,mul ] y get find small range or big range
Privatekey : 1
first x: c994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb
second x: 79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 private base x
third x: bcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb
even y: 483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8 small range y private key: 1
odd y: b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777 big range y private key: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364140
Privatekey : 2
first x: c360a6d0b34ce6df4135ee7d59f87b33d2fad8cce43837ef3e995b6ed89250e1
second x: 769ad99b0ac59bb38e84d114104707f3d08d98e78ed88b6915ba9ad5cafd0898
third x: c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 private base x
even y: 1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a small range y private key: 2
odd y: e51e970159c23cc65c3a7be6b99315110809cd9acd992f1edc9bce55af301705 big range y private key: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd036413f
Privatekey: 6
first x: 19cab650e04db19581801eb9e6c50b54f6a51b9223f6040c894f936926e302c3
second x: fff97bd5755eeea420453a14355235d382f6472f8568a18b2f057a1460297556 private base x
third x: e63bcdd9aa535fc65e3aa731e3e8bed786649d3e56a15a6847aaf28078f38045
even y: 51ed8885530449df0c4169fe80ba3a9f217f0f09ae701b5fc378f3c84f8a0998 big range y private key: 6
odd y: ae12777aacfbb620f3be96017f45c560de80f0f6518fe4a03c870c36b075f297 small range y private key: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd036413b
iknow random private key: how to find y is small range or big range atleast guess
yellow color is private key is base x value , then you find which one small range y bigrange y
test1:
first x: 1a1fd15fce078234aa292fc024178056bf006433c9b4bd208f59eb4c9efec95b private base x
second x: fc5eddb62317c03b0c0b501b3cbc7849f8babc4ac5515d4d25f4430a0aeb71c0
third x: e98150ea0ee0bd9049cb80249f2c075f4844df8170f9e5924ab1d1a75615bd43
even y: 5e750e01b967f6762c008a4069feedeeae10b91d30547666bf7ce63070c4150a
odd y: a18af1fe46980989d3ff75bf9601121151ef46e2cfab8999408319ce8f3be725
test2:
fist x: e6d4662474e3fae342e8cf460ff685fa5dcf99de42e6629320acae4e8bdb98d8
second x: 46ae596ce6268177c099b0c935759812fd4717661368960f44267c4a2f850be7
third x: d27d406ea4f583a4fc7d7ff0ba93e1f2a4e94ebba9b1075d9b2cd565449f539f private base x
even y: 2f090a097441f7d57a0af087fd8a6243b227d8a838e4751b290933588ac3b568
odd y: d0f6f5f68bbe082a85f50f7802759dbc4dd82757c71b8ae4d6f6cca6753c46c7
do you not understand my problom reply me i am exlain more
this is normal math sqrt√(x)^2,
test value: 4^2 = 16, next root return 4
-8^2= 64,next root return 8
how to bitcoin Publickey x and y use to sqrt example: √(x,y)^2
how to calculate sqrt for bitcoin publickey teach me please
example1:x,y
px: e493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13 py: 51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922 #privatekey 4
px: 421f5fc9a21065445c96fdb91c0c1e2f2431741c72713b4b99ddcb316f31e9fc py: 2b90f16d11dabdb616f6db7e225d1e14743034b37b223115db20717ad1cd6781 #privatekey 4^2 = 16
ans
px: e493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13 py: 51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922 #privatekey √16 = 4
example2:
px: 2f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01 py: a3b25758beac66b6d6c2f7d5ecd2ec4b3d1dec2945a489e84a25d3479342132b # -8
px: ed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88 py: e57a6f571288ccffdcda5e8a7a1f87bf97bd17be084895d0fce17ad5e335286e # -8^ = 64
ans
px: 2f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01 py: 5c4da8a741539949293d082a132d13b4c2e213d6ba5b7617b5da2cb76cbde904 # √64 = 8
how does work explain. do u understad my problem ,i speak little english
test value: 4^2 = 16, next root return 4
-8^2= 64,next root return 8
how to bitcoin Publickey x and y use to sqrt example: √(x,y)^2
how to calculate sqrt for bitcoin publickey teach me please
example1:x,y
px: e493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13 py: 51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922 #privatekey 4
px: 421f5fc9a21065445c96fdb91c0c1e2f2431741c72713b4b99ddcb316f31e9fc py: 2b90f16d11dabdb616f6db7e225d1e14743034b37b223115db20717ad1cd6781 #privatekey 4^2 = 16
ans
px: e493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13 py: 51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922 #privatekey √16 = 4
example2:
px: 2f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01 py: a3b25758beac66b6d6c2f7d5ecd2ec4b3d1dec2945a489e84a25d3479342132b # -8
px: ed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88 py: e57a6f571288ccffdcda5e8a7a1f87bf97bd17be084895d0fce17ad5e335286e # -8^ = 64
ans
px: 2f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01 py: 5c4da8a741539949293d082a132d13b4c2e213d6ba5b7617b5da2cb76cbde904 # √64 = 8
how does work explain. do u understad my problem ,i speak little english
this is normal math sqrt√(x)^2,
test value: 4^2 = 16, next root return 4
-8^2= 64,next root return 8
how to bitcoin Publickey x and y use to sqrt example: √(x,y)^2
how to calculate sqrt for bitcoin publickey teach me please
example1:x,y
px: e493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13 py: 51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922 #privatekey 4
px: 421f5fc9a21065445c96fdb91c0c1e2f2431741c72713b4b99ddcb316f31e9fc py: 2b90f16d11dabdb616f6db7e225d1e14743034b37b223115db20717ad1cd6781 #privatekey 4^2 = 16
ans
px: e493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13 py: 51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922 #privatekey √16 = 4
example2:
px: 2f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01 py: a3b25758beac66b6d6c2f7d5ecd2ec4b3d1dec2945a489e84a25d3479342132b # -8
px: ed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88 py: e57a6f571288ccffdcda5e8a7a1f87bf97bd17be084895d0fce17ad5e335286e # -8^ = 64
ans
px: 2f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01 py: 5c4da8a741539949293d082a132d13b4c2e213d6ba5b7617b5da2cb76cbde904 # √64 = 8
how does work explain. do u understad my problem ,i speak little english
test value: 4^2 = 16, next root return 4
-8^2= 64,next root return 8
how to bitcoin Publickey x and y use to sqrt example: √(x,y)^2
how to calculate sqrt for bitcoin publickey teach me please
example1:x,y
px: e493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13 py: 51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922 #privatekey 4
px: 421f5fc9a21065445c96fdb91c0c1e2f2431741c72713b4b99ddcb316f31e9fc py: 2b90f16d11dabdb616f6db7e225d1e14743034b37b223115db20717ad1cd6781 #privatekey 4^2 = 16
ans
px: e493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13 py: 51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922 #privatekey √16 = 4
example2:
px: 2f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01 py: a3b25758beac66b6d6c2f7d5ecd2ec4b3d1dec2945a489e84a25d3479342132b # -8
px: ed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88 py: e57a6f571288ccffdcda5e8a7a1f87bf97bd17be084895d0fce17ad5e335286e # -8^ = 64
ans
px: 2f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01 py: 5c4da8a741539949293d082a132d13b4c2e213d6ba5b7617b5da2cb76cbde904 # √64 = 8
how does work explain. do u understad my problem ,i speak little english