使用 Hessian 曲线实现 ECDSA

如何解决使用 Hessian 曲线实现 ECDSA

我一直在尝试使用 Python 的 Hessian 曲线来实现 ECDSA（椭圆曲线数字签名算法），但是签名没有得到验证。 ECDSA 可以在这里得到最好的解释，https://en.wikipedia.org/wiki/Elliptic_Curve_Digital_Signature_Algorithm。

我知道代码太长，但我已经尝试了几个月。非常感谢您的帮助。

代码运行如下：

def inv(a,m):
    m0 = m 
    y = 0
    x = 1
  
    if (m == 1) : 
        return 0
  
    while (a > 1) : 
  
        # q is quotient 
        q = a // m 
  
        t = m 
  
        # m is remainder now,process 
        # same as Euclid's algo 
        m = a % m 
        a = t 
        t = y 
  
        # Update x and y 
        y = x - q * y 
        x = t 
  
  
    # Make x positive 
    if (x < 0) : 
        x = x + m0 
  
    return x 

def checkPoint(point,P,D):
    
    if ( point[0]**3 + point[1]**3 +  point[2]**3) % P == ( D * point[0] * point[1] * point[2] ) % P :
        return True
    
    return False


def doubleAndAdd( point,i,P):

    addend = point
    result = None
    
    for b in bits(i) :
        if b:
            result = point_add(result,addend,P)
        addend = point_double(addend,P)
    return result

def findOrder(point,Pinfinity,P):
    
    for i in range(1104601):
        Gprime = doubleAndAdd(point,P)
        if Gprime == Pinfinity:
            print(i," ",Gprime)
            return i

def point_add( result,P) : 

    if result is None or addend is None: 
        return result or addend
     
   
    X3 = addend[0] * addend[2] * result[1]**2 - result[0] * result[2] * addend[1]**2
    Y3 = addend[1] * addend[2] * result[0]**2 - result[1] * result[2] * addend[0]**2
    Z3 = addend[0] * addend[1] * result[2]**2 - result[0] * result[1] * addend[2]**2

    return ( X3 % P,Y3 % P,Z3 % P)

def bits(n):

    while n:
        yield n & 1
        n >>= 1

def point_double(addend,P): 
    
    if addend is None:
        return None 
    
    X3 = addend[1] * ( addend[2]**3 - addend[0]**3 )
    Y3 = addend[0] * ( addend[1]**3 - addend[2]**3 )
    Z3 = addend[2] * ( addend[0]**3 - addend[1]**3 )  

    return ( X3 % P,Z3 % P)

if __name__ == '__main__':
    print("The equation of the curve is x^3 + y^3+z^3  = Dxyz")
    print("The value of D enter")
    D=int(input())
    P = 1051
    print("One of the coordinate of the curve is (x,y,z). This point is e1")
    print("Please choose another prime number")
    Q = int(input())
    
    Pinfinity = (1,1050,0)
    print("Enter the points")
    x=int(input())
    y=int(input())
    z=int(input())
    point = (x,z)
    print("Is the point on the curve?",checkPoint(point,D))
    order = findOrder(point,P)
    print(order)
    print("Choose a random integer d")
    d = int(input())
    print("We calculate e2 by multiplying e1 with d")
    e2 = doubleAndAdd(point,d,P)
    print(e2)
    print("Is the point on the curve?",checkPoint(e2,D))
    print("The public key of Alice is:"+str(D)+","+str(P)+","+str(Q)+","+str(point)+","+str(e2))
    print("Choose a random number between 1 and "+str(Q-1))
    R = int(input())
    point2 = doubleAndAdd(point,R,P)
    print("Point for signature:"+str(point2))
    S1 = point2[0] % Q
    print("First Signature:"+str(S1))
    print("Enter message M:")
    M = int(input())
    print("Calculating the second signature: S2")
    print("Calculating the inverse of the random number r with respect to Q")
    R_inverse = inv(R,Q)
    print("The inverse of R with respect to Q:"+str(R_inverse))
    Signature2 =  ((M + (d*S1)%Q)* R_inverse)% Q
    print("Second Signature:"+str(Signature2))
    print("Inverse of Signature2:"+str(inv(Signature2,Q)))
    A = (M* inv(Signature2,Q))%Q
    print("The value of A:"+str(A))
    B = (inv(Signature2,Q)*S1)%Q
    print("The value of B:"+str(B))
    temp1 = doubleAndAdd(point,A,P)
    temp2 = doubleAndAdd(e2,B,P)
    T = point_add(temp1,temp2,P)
    print("Is the point on the curve? temp1",checkPoint(temp1,D))
    print("Is the point on the curve?temp2",checkPoint(temp2,D))
    print("Is the point on the curve?T",checkPoint(T,D))
    print(T)
    print(T[0]%Q)
    if T[0]%Q == S1:
        print("Signature Verified")
    else:
        print("Signature Not Verified")

示例输入如下：

D 的值：6

另一个质数：31

点数：4,2,6

选择一个随机整数 d: 9

在 1 到 30:5 之间选择一个随机数

输入消息 M:6