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\begin{center}\framebox{\parbox{\boxlength}{\bf
CS 681: Computational Number Theory and Algebra \hfill Lecture 23
\\
Lecturer: Manindra Agrawal \hfill
Notes by: Ashwini Aroskar
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September 30, 2005.
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\vspace{5mm}

\textbf{Recall} $\vert B \vert \geq (\ln n)^r$ \quad where
$r=\frac{\ln n}{\ln k}$
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\section*{Time Complexity of Dixon's Algorithm}

Expected number of iterations in Dixon's algorithm to get a single
pair $(a,b) \leq (\ln n)^r$ \\
\\
Time complexity of the algorithm =\~{O}$(k^3+k^2(\ln n)^r)$\\
\\
The goal now is to choose $k$ such that the time complexity is
minimized.\\
The time complexity for matrix multiplication, using Gaussian
elimination, is $\bigcirc(k^3)$, but this can be reduced for
sparse matrices.\\
\\
\textbf{Sparse Matrix} has $l$ non-zero entries out $n^2$.\\
\textbf{Theorem:} There is an algorithm to invert such a matrix
with time complexity $\bigcirc(nl)$.\\
\\
The $(t+1)$ x $t$ matrix $(\alpha _{ij})$ has at most $(t+1)\log
n$ non-zero entries.\\
So, the time complexity of finding a non-trivial vector in the
null-space of this matrix $= \bigcirc(t^2\ln n) = \bigcirc(k^2\ln
n)$\\
\\
Improved time complexity of Dixon's Algorithm = \~{O}$(k^2(\ln
n)^r)$\\
\\
$k^2(\ln n)^r=e^{2\ln k +r\ln n\ln\ln n}=e^{2\ln k + \frac{\ln
n\ln\ln n}{\ln k}}$\\
Minimum is achieved when $2\ln k = \frac{\ln n\ln\ln n}{\ln k}$,
that is, when $\ln k = \frac{1}{\surd 2}(\ln n \ln\ln
n)^{\frac{1}{2}}$\\
\\
Time complexity = \~{O}$(e^{2\surd 2 (\ln n\ln\ln
n)^{\frac{1}{2}}})$\\
\\
\section*{Quadratic Sieve}
Let $c=\lfloor \surd n \rfloor$\\
Consider numbers of the form $(c+l)^2-n$.\\
$(c+l)^2-n \approx 2cl +l^2$\\
If $l \leq n^\epsilon$ then $(c+l)^2-n = \bigcirc(n^{\frac{1}{2} +
\epsilon})$.\\
We choose $b$ among these numbers as the fraction of $k$-smooth
numbers among them is higher. Although this is strongly implied by
certain conjectures, there is no proof.








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