from cryptography.fernet import Fernet def encrypt_message(public_key, message): # Get the public key in bytes public_key_bytes = bytes.fromhex(public_key) # Generate a symmetric key symmetric_key = Fernet.generate_key() # Encrypt the message using the symmetric key f = Fernet(symmetric_key) encrypted_message = f.encrypt(message.encode()) # Encrypt the symmetric key using the public key rsa_key = RSA.import_key(public_key_bytes) cipher_rsa = PKCS1_OAEP.new(rsa_key) encrypted_key = cipher_rsa.encrypt(symmetric_key) return encrypted_key, encrypted_message

import ecdsa # Generate a private key private_key = ecdsa.SigningKey.generate(curve=ecdsa.SECP256k1) # Get the public key public_key = private_key.get_verifying_key() # Encode the keys to hexadecimal format private_key_hex = private_key.to_string().hex() public_key_hex = public_key.to_string().hex() print(f'Private key: {private_key_hex}') print(f'Public key: {public_key_hex}') from cryptography.fernet import Fernet def encrypt_message(public_key, message): # Get the public key in bytes public_key_bytes = bytes.fromhex(public_key) # Generate a symmetric key symmetric_key = Fernet.generate_key() # Encrypt the message using the symmetric key f = Fernet(symmetric_key) encrypted_message = f.encrypt(message.encode()) # Encrypt the symmetric key using the public key rsa_key = RSA.import_key(public_key_bytes) cipher_rsa = PKCS1_OAEP.new(rsa_key) encrypted_key = cipher_rsa.encrypt(symmetric_key) return encrypted_key, encrypted_message from cryptography.fernet import Fernet def encrypt_message(public_key, message): # Get the public key in bytes public_key_bytes = bytes.fromhex(public_key) # Generate a symmetric key symmetric_key = Fernet.generate_key() # Encrypt the message using the symmetric key f = Fernet(symmetric_key) encrypted_message = f.encrypt(message.encode()) # Encrypt the symmetric key using the public key rsa_key = RSA.import_key(public_key_bytes) cipher_rsa = PKCS1_OAEP.new(rsa_key) encrypted_key = cipher_rsa.encrypt(symmetric_key) return encrypted_key, encrypted_message

Can Government Hack your Bitcoin?

It is theoretically possible for a government or other entity to attempt to gain control of the Bitcoin network through the use of powerful computers, also known as “mining rigs.” This could potentially be done by creating and operating a large number of mining rigs that are able to perform the necessary calculations to validate transactions and create new blocks faster than other miners on the network.

However, the decentralized nature of the Bitcoin network makes it very difficult for any single entity to gain control. In order to successfully take over the network, an attacker would need to control a majority of the mining power, which is currently distributed among many different miners and mining pools around the world. The amount of mining power needed to achieve this level of control is extremely large and would require a significant investment in hardware and electricity.

Furthermore, the Bitcoin network is designed to be resistant to attacks by making it more difficult for miners to create new blocks as the mining power of the network increases. This means that an attacker would need to constantly increase their mining power in order to stay ahead of the rest of the network, which would be very costly and may not be feasible in the long run.

In summary, while it is theoretically possible for a government or other entity to attempt to take over the Bitcoin network through the use of powerful computers, the decentralized nature of the network and the high cost of such an attack make it very unlikely to be successful

import ecdsa # Generate a private key private_key = ecdsa.SigningKey.generate(curve=ecdsa.SECP256k1) # Get the public key public_key = private_key.get_verifying_key() # Encode the keys to hexadecimal format private_key_hex = private_key.to_string().hex() public_key_hex = public_key.to_string().hex() print(f'Private key: {private_key_hex}') print(f'Public key: {public_key_hex}')
SHA256 Hash: e318616353da94fe72e8ec38284bc7f992cb52a40d735e88f31f16467642bc11

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