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Update README.MD
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Bill Buchanan
@@ -370,129 +370,45 @@ Now modify the code to implement the SECP192R1 and also for the SECP521R1 curve.
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The code to implement Curve 25519 for key exchange (X25519) is:
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```python
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P = 2 ** 255 - 19
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A24 = 121665
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def bytes_to_int(bytes):
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from cryptography.hazmat.primitives import hashes
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from cryptography.hazmat.primitives.asymmetric.x25519 import X25519PrivateKey
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from cryptography.hazmat.primitives.kdf.hkdf import HKDF
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from cryptography.hazmat.primitives import serialization
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from cryptography.hazmat.backends import default_backend
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import binascii
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import sys
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result = 0
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for b in bytes:
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result = result * 256 + int(b)
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return result
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Bob_private_key = X25519PrivateKey.generate()
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Alice_private_key = X25519PrivateKey.generate()
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size=32 # 256 bit key
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Bob_shared_key = Bob_private_key.exchange(Alice_private_key.public_key())
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Bob_derived_key = HKDF(algorithm=hashes.SHA256(),length=size,salt=None,info=b'',backend=default_backend()).derive(Bob_shared_key)
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Alice_shared_key = Alice_private_key.exchange(Bob_private_key.public_key())
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Alice_derived_key = HKDF(algorithm=hashes.SHA256(),length=size,salt=None,info=b'',backend=default_backend()).derive(Alice_shared_key)
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print ("Name of curve: Curve 25519")
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vals = binascii.b2a_hex(Bob_private_key.private_bytes(serialization.Encoding.Raw,serialization.PrivateFormat.Raw,serialization.NoEncryption()))
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print (f"\nBob private key value: {vals}")
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vals=Bob_private_key.public_key()
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enc_point=binascii.b2a_hex(vals.public_bytes(encoding=serialization.Encoding.PEM,format=serialization.PublicFormat.SubjectPublicKeyInfo)).decode()
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print("Bob's public key: ",enc_point)
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vals = binascii.b2a_hex(Alice_private_key.private_bytes(serialization.Encoding.Raw,serialization.PrivateFormat.Raw,serialization.NoEncryption()))
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print (f"\nAlice private key value: {vals}")
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vals=Alice_private_key.public_key()
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enc_point=binascii.b2a_hex(vals.public_bytes(encoding=serialization.Encoding.PEM,format=serialization.PublicFormat.SubjectPublicKeyInfo)).decode()
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print("Alice's public key: ",enc_point)
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def int_to_bytes(value, length):
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result = []
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for i in range(0, length):
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result.append(value >> (i * 8) & 0xff)
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return result
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def cswap(swap, x_2, x_3):
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dummy = swap * ((x_2 - x_3) % P)
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x_2 = x_2 - dummy
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x_2 %= P
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x_3 = x_3 + dummy
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x_3 %= P
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return (x_2, x_3)
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# Based on https://tools.ietf.org/html/rfc7748
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def X25519(k, u):
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x_1 = u
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x_2 = 1
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z_2 = 0
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x_3 = u
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z_3 = 1
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swap = 0
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for t in reversed(range(255)):
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k_t = (k >> t) & 1
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swap ^= k_t
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x_2, x_3 = cswap(swap, x_2, x_3)
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z_2, z_3 = cswap(swap, z_2, z_3)
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swap = k_t
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A = x_2 + z_2
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A %= P
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AA = A * A
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AA %= P
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B = x_2 - z_2
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B %= P
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BB = B * B
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BB %= P
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E = AA - BB
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E %= P
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C = x_3 + z_3
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C %= P
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D = x_3 - z_3
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D %= P
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DA = D * A
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DA %= P
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CB = C * B
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CB %= P
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x_3 = ((DA + CB) % P)**2
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x_3 %= P
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z_3 = x_1 * (((DA - CB) % P)**2) % P
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z_3 %= P
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x_2 = AA * BB
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x_2 %= P
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z_2 = E * ((AA + (A24 * E) % P) % P)
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z_2 %= P
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x_2, x_3 = cswap(swap, x_2, x_3)
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z_2, z_3 = cswap(swap, z_2, z_3)
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return (x_2 * pow(z_2, P - 2, P)) % P
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def decodeScalar25519(k):
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k_list = [(b) for b in k]
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k_list[0] &= 248
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k_list[31] &= 127
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k_list[31] |= 64
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return decodeLittleEndian(k_list)
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def decodeLittleEndian(b):
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return sum([b[i] << 8*i for i in range( 32 )])
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def unpack2(s):
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if len(s) != 32:
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raise ValueError('Invalid Curve25519 scalar (len=%d)' % len(s))
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t = sum((ord(s[i]) ) << (8 * i) for i in range(31))
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t += (((ord(s[31]) ) & 0x7f) << 248)
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return t
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def pack(n):
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return ''.join([chr((n >> (8 * i)) & 255) for i in range(32)])
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def clamp(n):
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n &= ~7
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n &= ~(128 << 8 * 31)
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n |= 64 << 8 * 31
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return n
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# Return nP
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def multscalar(n, p):
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n = clamp(decodeScalar25519(n))
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p = unpack2(p)
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return pack(X25519(n, p))
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# Start at x=9. Find point n times x-point
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def base_point_mult(n):
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n = clamp(decodeScalar25519(n))
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return pack(X25519(n, 9))
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print ("\nBob's derived key: ",binascii.b2a_hex(Bob_derived_key).decode())
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print("Alice's derived key: ",binascii.b2a_hex(Alice_derived_key).decode())
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```
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A sample of this is [here](https://repl.it/@billbuchanan/simpleecdh).
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@@ -500,20 +416,6 @@ A sample of this is [here](https://repl.it/@billbuchanan/simpleecdh).
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Do Bob and Alice end up with the same key?
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How large are the random numbers that Bob and Alice generate?
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Do you think that this program will be secure? How might Eve discover the shared secret?
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Estimate the time it would take her to discover the key if she can try one billion keys per second:
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How would you modify that program so that it was more secure?
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### D.2
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We used Curve 25519 in D.1. Can you modify the code so that it uses secp256k1? The code for secp256k1 is defined in the secp256k1.py file [here](https://asecuritysite.com/encryption/python_secp256k1ecdh2).
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