Barrelled space

The Lenstra–Lenstra–Lovász (LLL) lattice basis reduction algorithm is a polynomial time lattice reduction algorithm invented by Arjen Lenstra, Hendrik Lenstra and László Lovász in 1982. Given a basis $\mathbf{B}=\{{\mathbf{b}}_1,{\mathbf{b}}_2,\dots ,{\mathbf{b}}_d\}$ with n-dimensional integer coordinates, for a lattice L in Rⁿ with $d\le n$, the LLL algorithm outputs an LLL-reduced (short, nearly orthogonal) lattice basis in time

$O(d^{5}n{\log }^{3}B)$

where B is the largest length of $b_i$ under the Euclidean norm.

The original applications were to give polynomial time algorithms for factorizing polynomials with rational coefficients into irreducible polynomials, for finding simultaneous rational approximations to real numbers, and for solving the integer linear programming problem in fixed dimensions.

LLL reduction

The precise definition of LLL-reduced is as follows: Given a basis

$\mathbf{B}=\{{\mathbf{b}}_1,{\mathbf{b}}_2,\dots ,{\mathbf{b}}_n\},$

define its Gram–Schmidt process orthogonal basis

${\mathbf{B}}^{*}=\{{\mathbf{b}}_1^{*},{\mathbf{b}}_2^{*},\dots ,{\mathbf{b}}_n^{*}\},$

and the Gram-Schmidt coefficients

${\mu }_{i,j}=\frac{⟨{\mathbf{b}}_i,{\mathbf{b}}_j^{*}⟩}{⟨{\mathbf{b}}_j^{*},{\mathbf{b}}_j^{*}⟩}$, for any $1\le j<i\le n$.

Then the basis $B$ is LLL-reduced if there exists a parameter $\delta$ in (0.25,1] such that the following holds:

1. (size-reduced) For $1\le j<i\le n:\left|{\mu }_{i,j}\right|\le 0.5$. By definition, this property guarantees the length reduction of the ordered basis.
2. (Lovász condition) For k = 2,3,..,n $:\delta \Vert {\mathbf{b}}_{k-1}^{*}{\Vert }^2\le \Vert {\mathbf{b}}_k^{*}{\Vert }^2+{\mu }_{k,k-1}^2\Vert {\mathbf{b}}_{k-1}^{*}{\Vert }^2$.

Here, estimating the value of the $\delta$ parameter, we can conclude how well the basis is reduced. Greater values of $\delta$ lead to stronger reductions of the basis. Initially, A. Lenstra, H. Lenstra and L. Lovász demonstrated the LLL-reduction algorithm for $\delta =\frac{3}{4}$. Note that although LLL-reduction is well-defined for $\delta =1$, the polynomial-time complexity is guaranteed only for $\delta$ in (0.25,1).

The LLL algorithm computes LLL-reduced bases. There is no known efficient algorithm to compute a basis in which the basis vectors are as short as possible for lattices of dimensions greater than 4. However, an LLL-reduced basis is nearly as short as possible, in the sense that there are absolute bounds $c_i>1$ such that the first basis vector is no more than $c_1$ times as long as a shortest vector in the lattice, the second basis vector is likewise within $c_2$ of the second successive minimum, and so on.

LLL Algorithm

The following description is based on Template:Harv.

INPUT:

$▹$ a lattice basis ${\mathbf{b}}_1,{\mathbf{b}}_2,\dots ,{\mathbf{b}}_n\in Z^m$,

$▹$ parameter $\delta$ with $\frac{1}{4}<\delta <1$

PROCEDURE:

Perform Gram-Schmidt:

• $b_1^{*}:=b_1,B_1:=⟨b_1^{*},b_1^{*}⟩$
• for $i$ from $2$ to $n$ do
  • $b_i^{*}:=b_i$
  • for $j$ from $1$ to $i-1$ do
    • ${\mu }_{i,j}:=\frac{⟨b_i,b_j^{*}⟩}{B_j}$
    • $b_i^{*}:=b_i^{*}-{\mu }_{i,j}{b}_j^{*}$
  • end for
  • $B_i:=⟨b_i^{*},b_i^{*}⟩$
• end for
• $k:=2$ (k is the stage at which the vectors ${\mathbf{b}}_1,{\mathbf{b}}_2,\dots ,{\mathbf{b}}_{k-1}$ are reduced according to size-reduced property 1.)
• if $\left|{\mu }_{i,j}\right|>\frac{1}{2}$ then execute reduction subroutine RED(k,k-1):
  • for $l$ from $k-1$ to $1$ do
    • $r:=\lfloor 0.5+{\mu }_{k,l}\rfloor$
    • $b_k:=b_k-r{b}_l$
    • for $j$ from $1$ to $l-1$ do
      • ${\mu }_{k,j}:={\mu }_{k,j}-r{\mu }_{l,j}$
    • end for
    • ${\mu }_{k,l}:={\mu }_{k,l}-r$
  • end for
• end if
• Calculate ${\mu }_{i,j}$ for 1 $\le j<i\le n$ and $B_i$ for $i$ from 1 to $n$

• while $k\le n$ do
  • Length reduce $b_k$ and correct ${\mu }_{k,j}$ according to reduction subroutine in step 4, for $j$ from 1 till $k-1$
  • if $B_k<(\frac{3}{4}-{\mu }_{k,k-1}^2)B_{k-1}$ then
    • Exchange $b_k$ and $b_{k-1}$
    • $k$:= max $(2,k-1)$
  • else
    • $k:=k+1$
  • end if
• end while

OUTPUT: LLL reduced basis ${\mathbf{b}}_1,{\mathbf{b}}_2,\dots ,{\mathbf{b}}_n$

Example

The following presents an example due to W. Bosma.

INPUT:

Let a lattice basis ${\mathbf{b}}_1,{\mathbf{b}}_2,{\mathbf{b}}_3\in Z^3$, be given by the columns of

$\left[\begin{array}{ccc}1& -1& 3\\ 1& 0& 5\\ 1& 2& 6\end{array}\right]$

Then according to the LLL algorithm we obtain the following:

1.$b_1^{*}=b_1=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],B_1=⟨b_1^{*},b_1^{*}⟩=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=3$

2.For $i=2$ DO:

2.1.For $j=1$ set ${\mu }_{2,1}=\frac{⟨b_2,b_1^{*}⟩}{B_1}=\frac{\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]}{3}=\frac{1}{3}(<\frac{1}{2})$

and $b_2^{*}=b_2-{\mu }_{2,1}{b}_1^{*}=\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right]-\frac{1}{3}\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}\frac{-4}{3}\\ \frac{-1}{3}\\ \frac{5}{3}\end{array}\right].$

2.2$B_2=⟨b_2^{*},b_2^{*}⟩=\left[\begin{array}{c}\frac{-4}{3}\\ \frac{-1}{3}\\ \frac{5}{3}\end{array}\right]\left[\begin{array}{c}\frac{-4}{3}\\ \frac{-1}{3}\\ \frac{5}{3}\end{array}\right]=\frac{14}{3}.$

3. $\mathbf{k}:=2$

4.Here the step 4 of the LLL algorithm is skipped as size-reduced property holds for ${\mu }_{2,1}$

5.For $i=3$ and for $j=1,2$ calculate ${\mu }_{i,j}$ and $B_i$: ${\mu }_{3,1}=\frac{⟨b_3,b_1^{*}⟩}{B_1}=\frac{\left[\begin{array}{c}3\\ 5\\ 6\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]}{3}=\frac{14}{3}(>\frac{1}{2})$

hence $b_3^{*}=b_3-{\mu }_{3,1}{b}_1^{*}=\left[\begin{array}{c}3\\ 5\\ 6\end{array}\right]-\frac{14}{3}\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}\frac{-5}{3}\\ \frac{1}{3}\\ \frac{4}{3}\end{array}\right]$

and ${\mu }_{3,2}=\frac{⟨b_3,b_2^{*}⟩}{B_2}=\frac{\left[\begin{array}{c}3\\ 5\\ 6\end{array}\right]\left[\begin{array}{c}\frac{-4}{3}\\ \frac{-1}{3}\\ \frac{5}{3}\end{array}\right]}{\frac{14}{3}}=\frac{13}{14}(>\frac{1}{2})$

hence $b_3^{*}=b_3^{*}-{\mu }_{3,2}{b}_2^{*}=\left[\begin{array}{c}\frac{-5}{3}\\ \frac{1}{3}\\ \frac{4}{3}\end{array}\right]-\frac{13}{14}\left[\begin{array}{c}\frac{-4}{3}\\ \frac{-1}{3}\\ \frac{5}{3}\end{array}\right]=\left[\begin{array}{c}\frac{-18}{42}\\ \frac{27}{42}\\ \frac{-9}{42}\end{array}\right]=\left[\begin{array}{c}\frac{-6}{14}\\ \frac{9}{14}\\ \frac{-3}{14}\end{array}\right]$ and

$B_3=⟨b_3^{*},b_3^{*}⟩=\left[\begin{array}{c}\frac{-6}{14}\\ \frac{9}{14}\\ \frac{-3}{14}\end{array}\right]\left[\begin{array}{c}\frac{-6}{14}\\ \frac{9}{14}\\ \frac{-3}{14}\end{array}\right]=\frac{126}{196}=\frac{9}{14}$

6.While $k\le 3$ DO

6.1 Length reduce $b_3$ and correct ${\mu }_{3,1}$ and ${\mu }_{3,2}$ according to reduction subroutine in step 4:

For $\mid {\mu }_{3,1}\mid >\frac{1}{2}$ EXECUTE reduction subroutine RED(3,1):

i.$r=\lfloor 0.5+{\mu }_{3,l}\rfloor =5$ and $b_3=b_3-5b_1=\left[\begin{array}{c}3\\ 5\\ 6\end{array}\right]-\left[\begin{array}{c}5\\ 5\\ 5\end{array}\right]=\left[\begin{array}{c}-2\\ 0\\ 1\end{array}\right]$

ii.${\mu }_{3,1}={\mu }_{3,l}-r{\mu }_{1,1}=\frac{-1}{3}(<\frac{1}{2})$

iii.Set ${\mu }_{3,1}={\mu }_{3,1}-r=\frac{14}{3}-5=\frac{-1}{3}$

For $\mid {\mu }_{3,2}\mid >\frac{1}{2}$ EXECUTE reduction subroutine RED(3,2):

i.$r=\lfloor 0.5+{\mu }_{3,2}\rfloor =1$ and $b_3=b_3-b_2=\left[\begin{array}{c}3\\ 5\\ 6\end{array}\right]-\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right]=\left[\begin{array}{c}4\\ 5\\ 4\end{array}\right]$

ii.Set ${\mu }_{3,2}={\mu }_{3,2}-r{\mu }_{2,2}=\frac{13}{14}-1=\frac{-1}{14}$

iii.${\mu }_{3,2}={\mu }_{3,2}-1=\frac{-1}{14}(<\frac{1}{2})$

6.2 As $B_3<(\frac{3}{4}-{\mu }_{3,2}^2)B_2$ takes place, then

6.2.1 Exchange $b_3$ and $b_2$

6.2.2 $k$:= 2

Apply a SWAP, continue algorithm with the lattice basis, which is given by columns

$\left[\begin{array}{ccc}1& 4& -1\\ 1& 5& 0\\ 1& 4& 2\end{array}\right]$

Implement the algorithm steps again. 1.$b_1^{*}=b_1=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],B_1=3$

2. ${\mu }_{2,1}=\frac{⟨b_2,b_1^{*}⟩}{B_1}=\frac{\left[\begin{array}{c}4\\ 5\\ 4\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]}{3}=\frac{13}{3}(>\frac{1}{2})$

3.$b_2^{*}=b_2-{\mu }_{2,1}{b}_1^{*}=\left[\begin{array}{c}4\\ 5\\ 4\end{array}\right]-\frac{13}{3}\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}\frac{-1}{3}\\ \frac{2}{3}\\ \frac{-1}{3}\end{array}\right]$.

4.$B_2=⟨b_2^{*},b_2^{*}⟩=\frac{2}{3}$.

5.For $\mid {\mu }_{2,1}\mid >\frac{1}{2}$ EXECUTE reduction subroutine RED(2,1):

i.$r=\lfloor 0.5+{\mu }_{2,l}\rfloor =4$ and $b_2=b_2-4b_1=\left[\begin{array}{c}4\\ 5\\ 4\end{array}\right]-\left[\begin{array}{c}4\\ 4\\ 4\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$

ii.Set ${\mu }_{2,1}={\mu }_{2,1}-4{\mu }_{1,1}=\frac{13}{3}-4=\frac{1}{3}(<\frac{1}{2})$

6. As $B_2<(\frac{3}{4}-{\mu }_{2,1}^2)B_1$ takes place, then

7. Exchange $b_2$ and $b_1$

OUTPUT: LLL reduced basis

$\left[\begin{array}{ccc}0& 1& -1\\ 1& 0& 0\\ 0& 1& 2\end{array}\right]$

Applications

The LLL algorithm has found numerous other applications in MIMO detection algorithms and cryptanalysis of public-key encryption schemes: knapsack cryptosystems, RSA with particular settings, NTRUEncrypt, and so forth. The algorithm can be used to find integer solutions to many problems.

In particular, the LLL algorithm forms a core of one of the integer relation algorithms. For example, if it is believed that r=1.618034 is a (slightly rounded) root to a quadratic equation with integer coefficients, one may apply the LLL reduction to the lattice in $R^4$ spanned by $[1,0,0,10000r^2],[0,1,0,10000r],$ and $[0,0,1,10000]$. The first vector in the reduced basis will be an integer linear combination of these three, thus necessarily of the form $[a,b,c,10000(a{r}^2+br+c)]$; but such a vector is "short" only if a, b, c are small and $a{r}^2+br+c$ is even smaller. Thus the first three entries of this short vector are likely to be the coefficients of the integral quadratic polynomial which has r as a root. In this example the LLL algorithm finds the shortest vector to be [1, -1, -1, 0.00025] and indeed $x^2-x-1$ has a root equal to 1.6180339887…(The Golden Ratio)

Implementations

LLL is implemented in

• Arageli as the function lll_reduction_int
• fpLLL as a stand-alone implementation
• GAP as the function LLLReducedBasis
• LiDIA as the function/method lll in the LT package
• Macaulay2 as the function LLL in the package LLLBases
• Magma as the functions LLL and LLLGram (taking a gram matrix)
• Maple as the function IntegerRelations[LLL]
• Mathematica as the function LatticeReduce
• Number Theory Library (NTL) as the function LLL
• PARI/GP as the function qflll
• Sage as the method LLL driven by fpLLL and NTL

See also

• Coppersmith method

Notes

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References

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  A vendor's stamp duty has been launched on industrial property for the primary time, at rates ranging from 5 per cent to 15 per cent. The Authorities might be trying to reassure the market that they aren't in opposition to foreigners and PRs investing in Singapore's property market. They imposed these measures because of extenuating components available in the market." The sale of new dual-key EC models will even be restricted to multi-generational households only. The models have two separate entrances, permitting grandparents, for example, to dwell separately. The vendor's stamp obligation takes effect right this moment and applies to industrial property and plots which might be offered inside three years of the date of buy. JLL named Best Performing Property Brand for second year running
  
  The data offered is for normal info purposes only and isn't supposed to be personalised investment or monetary advice. Motley Fool Singapore contributor Stanley Lim would not personal shares in any corporations talked about. Singapore private home costs increased by 1.eight% within the fourth quarter of 2012, up from 0.6% within the earlier quarter. Resale prices of government-built HDB residences which are usually bought by Singaporeans, elevated by 2.5%, quarter on quarter, the quickest acquire in five quarters. And industrial property, prices are actually double the levels of three years ago. No withholding tax in the event you sell your property. All your local information regarding vital HDB policies, condominium launches, land growth, commercial property and more
  
  There are various methods to go about discovering the precise property. Some local newspapers (together with the Straits Instances ) have categorised property sections and many local property brokers have websites. Now there are some specifics to consider when buying a 'new launch' rental. Intended use of the unit Every sale begins with 10 p.c low cost for finish of season sale; changes to 20 % discount storewide; follows by additional reduction of fiftyand ends with last discount of 70 % or extra. Typically there is even a warehouse sale or transferring out sale with huge mark-down of costs for stock clearance. Deborah Regulation from Expat Realtor shares her property market update, plus prime rental residences and houses at the moment available to lease Esparina EC @ Sengkang

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