我也看得云里雾里，
但是ECC和RSA并列为非对称加密双雄，
还是很有必要了解一下的。
RSA是用质数分解，ECC是用离散的椭圆方程解，安全度更高。
而且，这个ECC的加法乘法规则，和普通都不一样，
其解是属于一个什么阿贝尔群(一听就知道高级啦)。
百度的文章，下面这个比较详细。
https://www.sohu.com/a/216057858_465483
from hashlib import sha256
def sha256d(string):
if not isinstance(string, bytes):
string = string.encode()
return sha256(sha256(string).digest()).hexdigest()
def inv_mod(b, p):
if b < 0 or p <= b:
b = b % p
c , d = b, p
uc, vc, ud, vd, temp = 1, 0, 0, 1, 0
while c != 0:
temp = c
q, c, d = d // c, d % c, temp
uc, vc, ud, vd = ud - q * uc, vd - q * vc, uc, vc
assert d == 1
if ud > 0:
return ud
else:
return ud + p
def leftmost_bit(x):
assert x > 0
result = 1
while result <= x:
result = 2 * result
return result // 2
print(inv_mod(2, 23))
print(3*inv_mod(1, 23) % 23)
def show_points(p, a, b):
return [(x, y) for x in range(p) for y in range(p) if (y*y - (x*x*x + a*x + b)) % p == 0]
print(show_points(p=29, a=4, b=20))
def double(x, y, p, a, b):
l = ((3 * x * x + a) * inv_mod(2 * y, p)) % p
x3 = (l * l - 2 * x) % p
y3 = (l * (x - x3) - y) % p
return x3, y3
print(double(1, 4, p=5, a=2, b=3))
def add(x1, y1, x2, y2, p, a, b):
if x1 == x2 and y1 == y2:
return double(x1, y1, p, a, b)
l = ((y2 - y1) * inv_mod(x2 - x1, p)) % p
x3 = (l * l - x1 -x2) % p
y3 = (l * (x1 - x3) - y1) % p
return x3, y3
print(add(1, 4, 3, 1, p=5, a=2, b=3))
def get_bits(n):
bits = []
while n != 0:
bits.append(n & 1)
n >> 1
return bits
class CurveFp(object):
def __init__(self, p, a, b):
""" y^2 = x^3 + a*x + b (mod p)."""
self.p = p
self.a = a
self.b = b
def contains_point(self, x, y):
return (y * y - (x * x * x + self.a * x + self.b)) % self.p == 0
def show_all_points(self):
return [(x, y) for x in range(self.p) for y in range(self.p) if
(y * y - (x * x * x + self.a * x + self.b)) % self.p == 0]
def __repr__(self):
return "Curve(p={0:d}, a={1:d}, b={2:d})".format(self.p, self.a, self.b)
class Point(object):
def __init__(self, curve, x, y, order=None):
self.curve = curve
self.x = x
self.y = y
self.order = order
# self.curve is allowed to be None only for INFINITY:
if self.curve:
assert self.curve.contains_point(x, y)
if order:
assert self * order == INFINITY
def __eq__(self, other):
"""Is this point equals to another"""
if self.curve == other.curve \
and self.x == other.x \
and self.y == other.y:
return True
else:
return False
def __add__(self, other):
"""Add one point to another point."""
if other == INFINITY:
return self
if self == INFINITY:
return other
assert self.curve == other.curve
if self.x == other.x:
if (self.y + other.y) % self.curve.p == 0:
return INFINITY
else:
return self.double()
p = self.curve.p
l = ((other.y - self.y) * \
inv_mod(other.x - self.x, p)) % p
x3 = (l * l - self.x - other.x) % p
y3 = (l * (self.x - x3) - self.y) % p
return Point(self.curve, x3, y3)
def __mul__(self, other):
e = other
if self.order:
e = e % self.order
if e == 0:
return INFINITY
if self == INFINITY:
return INFINITY
e3 = 3 * e
negative_self = Point(self.curve, self.x, -self.y, self.order)
i = leftmost_bit(e3) // 2
result = self
while i > 1:
result = result.double()
if (e3 & i) != 0 and (e & i) == 0:
result = result + self
if (e3 & i) == 0 and (e & i) != 0:
result = result + negative_self
i = i // 2
return result
def __rmul__(self, other):
"""Multiply a point by an integer."""
return self * other
def __repr__(self):
if self == INFINITY:
return "infinity"
return "({0},{1})".format(self.x, self.y)
def double(self):
"""the double point."""
if self == INFINITY:
return INFINITY
p = self.curve.p
a = self.curve.a
l = ((3 * self.x * self.x + a) * \
inv_mod(2 * self.y, p)) % p
x3 = (l * l - 2 * self.x) % p
y3 = (l * (self.x - x3) - self.y) % p
return Point(self.curve, x3, y3)
def invert(self):
return Point(self.curve, self.x, -self.y % self.curve.p)
INFINITY = Point(None, None, None)
p, a, b = 29, 4, 20
curve = CurveFp(p, a, b)
p0 = Point(curve, 3, 1)
print(p0*2)
print(p0*20)输出：
12 3 [(0, 7), (0, 22), (1, 5), (1, 24), (2, 6), (2, 23), (3, 1), (3, 28), (4, 10), (4, 19), (5, 7), (5, 22), (6, 12), (6, 17), (8, 10), (8, 19), (10, 4), (10, 25), (13, 6), (13, 23), (14, 6), (14, 23), (15, 2), (15, 27), (16, 2), (16, 27), (17, 10), (17, 19), (19, 13), (19, 16), (20, 3), (20, 26), (24, 7), (24, 22), (27, 2), (27, 27)] (3, 1) (2, 0) (24,7) (15,27)