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96 lines
2.1 KiB
Python
96 lines
2.1 KiB
Python
# Don't trust me with cryptography.
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"""
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Implementation of https://gist.github.com/RubenSomsen/be7a4760dd4596d06963d67baf140406
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Bob (Mint):
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A = a*G
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return A
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Alice (Client):
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Y = hash_to_curve(secret_message)
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r = random blinding factor
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B'= Y + r*G
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return B'
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Bob:
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C' = a*B'
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(= a*Y + a*r*G)
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return C'
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Alice:
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C = C' - r*A
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(= C' - a*r*G)
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(= a*Y)
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return C, secret_message
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Bob:
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Y = hash_to_curve(secret_message)
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C == a*Y
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If true, C must have originated from Bob
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"""
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import hashlib
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from secp256k1 import PrivateKey, PublicKey
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def hash_to_curve(message: bytes):
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"""Generates a point from the message hash and checks if the point lies on the curve.
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If it does not, it tries computing a new point from the hash."""
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point = None
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msg_to_hash = message
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while point is None:
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try:
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_hash = hashlib.sha256(msg_to_hash).digest()
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point = PublicKey(b"\x02" + _hash, raw=True)
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except:
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msg_to_hash = _hash
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return point
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def step1_alice(secret_msg: str, blinding_factor: bytes = None):
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Y = hash_to_curve(secret_msg.encode("utf-8"))
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if blinding_factor:
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r = PrivateKey(privkey=blinding_factor, raw=True)
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else:
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r = PrivateKey()
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B_ = Y + r.pubkey
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return B_, r
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def step2_bob(B_, a):
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C_ = B_.mult(a)
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return C_
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def step3_alice(C_, r, A):
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C = C_ - A.mult(r)
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return C
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def verify(a, C, secret_msg):
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Y = hash_to_curve(secret_msg.encode("utf-8"))
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return C == Y.mult(a)
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### Below is a test of a simple positive and negative case
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# # Alice's keys
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# a = PrivateKey()
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# A = a.pubkey
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# secret_msg = "test"
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# B_, r = step1_alice(secret_msg)
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# C_ = step2_bob(B_, a)
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# C = step3_alice(C_, r, A)
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# print("C:{}, secret_msg:{}".format(C, secret_msg))
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# assert verify(a, C, secret_msg)
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# assert verify(a, C + C, secret_msg) == False # adding C twice shouldn't pass
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# assert verify(a, A, secret_msg) == False # A shouldn't pass
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# # Test operations
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# b = PrivateKey()
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# B = b.pubkey
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# assert -A -A + A == -A # neg
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# assert B.mult(a) == A.mult(b) # a*B = A*b
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