I will think about some ways and make them public here. I will probably have to drink some wine to be more creative.
The first idea would be:
If someone manages to get a mathematical closed formula for the x-value, he could just paste that thing into wolfram alpha and get k out of it right away.
Meaning:
You have two points (x,y) - one is called "other" and one is called "self"
The point addition of those points (x3,y3) can be calculated as follows.
l = ( ( other.__y - self.__y ) * inverse_mod( other.__x - self.__x, p ) ) % p
x3 = ( l * l - self.__x - other.__x ) % p
y3 = ( l * ( self.__x - x3 ) - self.__y ) % p
After this step, other.x is set to x3 and other.y is set to y3.
We have all we need (initial state):
p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2FL (i guess)
other.x = 0x4641b45737ee8e11ae39899060160507d61a30928b0d3e37b6aede29b4ed807bL
other.y = 0xb61b706b81dbb5512c556dfd16815cced84e2fa12b5c8b6440057355f0df2a12L
self.x = 0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798L
self.y = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8L
All you gotta know is: How often do you have to perform the above "point addition" so that x3=self.x
The inversemod of this value (to the basis of p) is the k.
So does that mean how many times the point addition is run (lets say its count) so that x3 = self.x and do the inverse_mod(count, p) = k?
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