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I am facing this problem in calculating the order of a process which involves ECC point addition: $P+Q$ , scalar multiplication: $aP$, and selecting random points in the group. The group is of prime order say $q$.

What is the time complexity order of these operations? Is it $O(n)$, $O(log(n))$ or something different?

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    $\begingroup$ I notice that there is no question in this post. $\;$ $\endgroup$
    – user991
    Commented Apr 23, 2014 at 12:41
  • $\begingroup$ Are you wondering about order of operations in elliptic curves? $\endgroup$
    – mikeazo
    Commented Apr 23, 2014 at 13:31
  • $\begingroup$ Yes. I want to know the complexity order of these operations like for eg O(n) ,O(log(n)) etc $\endgroup$
    – Prasanth
    Commented Apr 25, 2014 at 3:49
  • $\begingroup$ @Prasanth where $n$ is? The size of the underlying field? $\endgroup$
    – mikeazo
    Commented Apr 25, 2014 at 12:32

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It depends on the multiplication algorithm used. Let $n$ be the bit size of the finite field used, and let $O(n^k)$ be the complexity of the multiplication algorithm used (e.g. $k=2$ for schoolbook multiplication, or $k=1.585$ for Karatsuba multiplication). Then:

  • Point addition: $O(n^k)$ (a constant number of multiplications)
  • Scalar multiplication: $O(n^{k+1})$
  • Selecting random point: depends on the method, but usually the same as scalar multiplication.

As pointed out by poncho, the point addition (and the scalar multiplication) can require a multiplication inverse, whose complexity again depends on the algorithm and may be larger than the complexity of multiplication. For example, binary euclidean takes $O(n^2)$, Fermat's little theorem takes $O(n^{k+1})$.

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    $\begingroup$ Actually, it is considerably more complex than that. For example, if you're adding two points in homologous format, and want the result in homologous format, there's a multiplicative inverse involved, and that's more expensive than all the rest of the operations combined. $\endgroup$
    – poncho
    Commented Apr 25, 2014 at 13:24
  • $\begingroup$ @poncho you're right, I've edited the answer. $\endgroup$
    – Conrado
    Commented Apr 25, 2014 at 13:36
  • $\begingroup$ Can you please elaborate how point addition is polynomial time? $\endgroup$ Commented Dec 4, 2020 at 5:45
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    $\begingroup$ @user134470 point addition is a fixed number of field multiplications, squarings and additions (and inversion, sometimes). Therefore the complexity of addition is the complexity of these field operations (a constant factor won't affect the complexity) $\endgroup$
    – Conrado
    Commented Dec 4, 2020 at 12:06

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