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\begin{center}\framebox{\parbox{\boxlength}{\bf
CS 681: Computational Number Theory and Algebra \hfill Lecture 26
\\\\
Analysis of Discrete Log Problem using Index Calculus and Number Field Sieve
\\\\
Lecturer: Manindra Agrawal \hfill Scribe: Sudeepa Roy
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October 10, 2005
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\section{Introduction }
In this lecture we will first do the analysis of the algorithm to solve discrete log problem using index calculus stated in the last lecture. Next we will briefly discuss \textit{Number Field Sieve}.
\section{Analysis of Discrete Log Problem}
Here we use the same notations as in the previous lecture. To recapitulate, the problem was to compute $m$ from the equation $e = g^m$ given $g$ and $e$, and the idea was to find $r$ and $s$, such that 
\begin{enumerate}
	\item $g^re^s = 1$
	\item $\gcd(s, order(g)) = 1$
\end{enumerate}

After the step 4 of the algorithm, we have $k$ triples $(r_i, s_i, u_i)$ with $u_i = \prod\limits_{j=1}^t {p_j}^{\alpha_{ij}}$, where $p_1, \cdots, p_t$ are all the primes $\leq k$.\\\\
We know $\vec{\beta}$ is such that $\sum\limits_{i=1}^k \beta_i\alpha_{ij} = 0~ (\mod p - 1~)$\\\\
$r = \sum\limits_{i = 1}^k \beta_ir_i$ and $s = \sum\limits_{i = 1}^k \beta_is_i$\\\\
Then\\\\
$g^re^s$\\\\
$= g^{\sum\limits_i \beta_ir_i}e^{\sum\limits_i\beta_is_i}$\\
$= \prod\limits_{i=1}^k(g^{r_i}e^{s_i})^{\beta_i}$ \\
$= \prod\limits_{i=1}^k u_i^{\beta_i}$\\
$= \prod\limits_{i=1}^k \prod\limits_{j=1}^t{p_j}^{\alpha_{ij}\beta_i}$\\
$= \prod\limits_{j=1}^t \prod\limits_{i=1}^k{p_j}^{\alpha_{ij}\beta_i}$\\
$= \prod\limits_{j=1}^t {p_j}^{\sum\limits_{i=1}^k\alpha_{ij}\beta_i}$\\
$= 1~ (\mod p~)$\\\\
We also need $\gcd(s, p-1)=1$. $\\Pr[\gcd(s, p-1) = 1] = \frac{\phi(p-1)}{p-1} \geq \frac{1}{2}$.\\\\
Time complexity of the algorithm is \\
$O(e^{2(\ln p)^{\frac{1}{2}}(\ln\ln p)^{\frac{1}{2}})}$.
 
\section{Number Field Sieve}
Number Field Sieve is a very fast method to factor integer $n$ and improvement over quadratic sieve reducing the exponent. The time complexity is $O(e^{c(\ln n)^{\frac{1}{3}}(\ln\ln n)^{\frac{2}{3}})}$ where $c \approx 1.93$.
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