While calculating the public key, $d * G = (X_a, Y_a)$ and similarly for $(X_b, Y_b)$, where $d$ is the private key. If calculating $d * G$ means adding $G$ to itself $d$ times, why can't the same be done to guess the private key '$d$' by an intruder – adding $G$ to itself until he gets the public key and hence find the private key? How is the calculation of public key efficient and guessing it computationally infeasible?
In Diffie Hellman key exchange using Elliptic Curve cryptography, how are the public keys calculated
cryptographyelliptic-curvesgroup-theorymodular arithmeticpolynomials
Best Answer
Current day elliptic curve cryptosystems use keys 200+ bits long. In this context it means that your group has size $2^{200} \approx 10^{60},$ so on average you will need to compute $G,2G,\ldots,dG$ until $d \geq 5\times 10^{59}$ until you hit the point $(X_a,Y_a)$ and discover the discrete logarithm $d.$