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So according to the wikipedia page https://en.wikipedia.org/wiki/General_number_field_sieve the algorithm has complexity $$\exp \left( \left(\sqrt[\leftroot{1}\uproot{0}3]{\frac{64}{9}} + o(1) \right) (\ln n)^{\frac{1}{3}}(\ln \ln n)^{\frac{2}{3}} \right)$$ but I am not too familiar with this notation.

I understand the big-O notation for complexity if somebody could explain how this is similar and what it means for the running time it would be appreciated!!

Also I know that this is subexponential in the size of $n$ running time, but I am unsure how to show this just from this complexity.

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  • $\begingroup$ yeah $\exp(x)=e^x$ its just when $x$ is a large function it gets clustered so I decided to leave it in this form $\endgroup$ – Matt Mar 26 at 0:02
  • $\begingroup$ Err, no. I thought that was the problem. You say you understand the notation, yet that's the notation for the sub-ex type. It's just a bit slower that exponential. It's probably me, but what's the question again? $\endgroup$ – Paul Uszak Mar 26 at 0:11
  • $\begingroup$ Oh sorry for the confusion! Well I understand the usual notation I've seen in the past for complexity, such as $O(n)$ and $O(n^2)$ etc. but I am unsure how this related to that notation exactly. For example, is there an $x$ such that this expression is the same as $O(x)$, and if not what does it mean by "complexity", does it mean plugging in a value of $n$ gives us the approximate time taken to run and thats all? $\endgroup$ – Matt Mar 26 at 0:17
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Since the numeric constant in the exponent is approximately $1.923 $ the expression for the average number of iterations, say time complexity $T$ given by $$T=\exp \left( \left(\sqrt[\leftroot{1}\uproot{0}3]{\frac{64}{9}} + o(1) \right) (\ln n)^{\frac{1}{3}}(\ln \ln n)^{\frac{2}{3}} \right)$$ is proportional to (for $n$ large enough so that $o(1)$ term is negligible) $$T\propto \exp \left(1.923 \times (\ln n)^{\frac{1}{3}}(\ln \ln n)^{\frac{2}{3}} \right)$$ which is upper bounded as (ignoring a multiplicative constant in front) $$T\leq \exp \left(2 (\ln n)^{\frac{1}{3}}(\ln \ln n)^{\frac{2}{3}} \right)=T_0^2.$$

Clearly $T$ is polynomial/exponential if and only if $T_0$ is.

Now consider the more general expression $$f(\alpha,n)=\exp \left( (\ln n)^{\alpha}(\ln \ln n)^{1-\alpha} \right),\quad \alpha \in [0,1],$$ and note that when $\alpha=1,$ $f(1,n)=\exp(\ln n)=n,$ hence it is exponential in the bitsize $\log_2 n$ of the input $n.$

Also note that when $\alpha=0,$ $f(0,n)=\exp(\ln\ln n)=\ln n,$ hence it is polynomial in the bitsize $\log_2 n$ of the input $n.$

However we have $T_0=f(1/3,n)$ with $\alpha=1/3$ above so the complexity we have is a linear combination in the exponent of polynomial and exponential complexities. This is referred to as subexponential but superpolynomial complexity.

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This is the growth curve of the expectation value of the number of steps in the algorithm to factor $n$, or asymptotic expected running time. The $o(1)$ term means a function that converges to zero as $n \to \infty$. Here a ‘step’ is the number of iterations of the loops in the algorithm; each ‘step’ corresponds to an unspecified number of (say) bit operations, and we are not considering storage or communication costs. For a more detailed area*time cost analysis, see, e.g., the batch NFS paper.

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