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# 格基规约¶

## Lenstra – Lenstra – Lovasz¶

### basic introduction¶

The LLL algorithm is to find a set of bases on the lattice, which satisfies the following effects. Moreover, the following properties of the base generated by this method are very useful. ### Simple application¶

Here I will give a second example from LLL paper. Given n real numbers $\alpha_i,...,\alpha_n$, find the rational linear approximation of the n numbers, ie find n numbers $m_i$, so that $\sum\limits_{i=1}^{n }m_i\alpha_i$ is equal to 0 as much as possible. We can construct a matrix like this, where $a_i$ is a rational approximation of $\alpha_i$.

A = \left[ \begin{matrix} 1 & 0 & 0 & \cdots & 0 & ca_1 \\ 0 & 1 & 0 & \cdots & 0 & c a_2 \\ 0 & 0 & 1 & \cdots & 0 & c a_3 \\\vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 &0 & \cdots & 1 & c a_n \\ \end{matrix} \right]

The matrix is n*(n+1), we can find the determinant corresponding to this lattice according to the method of finding the determinant.

$Det (L) = sqrt {AA}$ ^ T

We further consider such a matrix

A = \left[ \begin{matrix} 1 & 0 & 0 & \cdots & 0 & a_1 \\ 0 & 1 & 0 & \cdots & 0 & a_2 \\ 0 & 0 & 1 & \cdots & 0 & a_3 \\\vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 &0 & \cdots & 1 & a_n \\ \end{matrix} \right]

Then

AA^T = \left[ \begin{matrix} 1+a_1^2 & a_1a_2 & a_1a_3 & \cdots & a_1a_n \\ a_2a_1 & 1+a_2^2 & a_2a_3 & \cdots & a_2a_n \\ a_3a_1 & a_3a_2 & 1+a_3^2 & \cdots & a_3a_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_na_1 & a_na_2 &a_na_3 & \cdots & 1+a_n^2 \\ \end{matrix} \right]

Further, let's try it from low-dimensional to high-dimensional (strictly prove that you can consider adding a row and a column, the upper left corner is 1), and the determinant of the lattice is

$\sqrt{1+\sum\limits_{i=1}^n\alpha_i^2}$

Can refer to the following proof of the postgraduate Yuge Then after the LLL algorithm, we can get

$||b_1|| \leq 2^{\frac{n-1}{4}} (1+\sum\limits_{i=1}^n\alpha_i^2)^{\frac{1}{2(n+1)}}$

In general, the latter item tends to 1 when it is opened n times, because $a_i$ is a constant and is generally not related to n, so

$||b_1|| \leq 2^{\frac{n-1}{4}}*k$

k is relatively small. In addition, $b_1$ is a linear combination of the original vectors, then

$b_1[n]=\sum\limits_{i=1}^{n}m_ic*a_i=c\sum\limits_{i=1}^{n}m_i*a_i$

Obviously if c is large enough, then the subsequent summation must be small enough to satisfy the above constraints.

## Reference¶

• Survey: Lattice Reduction Attacks on RSA