**(Please note that in this blog as well, when I say that consider a vector**$\lambda_i$

**, I actually want you to consider a vector of length**$\lambda_i$

**. And by the way I normally mean lengths in**$l_2$

**norm. Can you please keep this information in your mind as you go along; otherwise things might seem a little, or way too much, strange)**

So, finally we are inching towards our target. Next, we seek to attack the big guns - the LLL algorithm. It has a bearing upon nearly all aspects of lattice based cryptography. We will in particular investigate the connection with SVP and another problem whose name has not been mentioned so far in this blog - CVP or the Closest Vector Problem.

This post will majorly delve upon 3 different topics. First, we would talk about LLL algorithm, the ideas leading to its development, the notion of reduced basis and what is it good for. Next, we would talk a bit about SVP and CVP sketching a little more background about the problems. Finally, we will see what LLL has to say about these problems. These are the 3 major topics discussed here but they are not in any particular order. In fact, there will be a zig-zagging through these 3 major areas throughout this post.

So, lets get started. First I will tell develop a little of the historical context behind the development of the LLL algorithm. In early 1980's, H.W. Lenstra (not HomeWork Lenstra!) was working on a problem which asked him to determine whether or not a given triangle (in $\RR^2$) with rational end points contained a lattice point. He began with a vague idea of saying that "fatter triangles will contain" a lattice point; "skinnier ones might not" and developed quite a wonderful solution.

In his own words "The solution essentially consists of denying that "special" triangles exist. If the triangle K looks a bit weird, why not apply a nonsingular linear transformation τ such that the triangle τ[Κ] looks better?" To be specific, he chooses τ so that the transformed triangle is equilateral. (Later on, he changed it to a right angled isosceles triangle) (I am just copy/pasting notes from his original paper - its extremely readable). He goes on to say "Clearly, $K$ contains an element of $Z^2$ if and only if the new triangle τ[Κ] contains an element of the lattice $L$ = τ($Z^2$) . If $e_1$, $e_2$ are the standard basis vectors of $\RR^2$, then τ$(e_1)$, τ$(e_2)$ form a basis for $L$ in the sense that $L = Z*τ(e_1) + Z*τ(e_2)$."

Lenstra, thus notes "The problem has now been shifted from the triangle to the lattice." You just need to find the basis for this lattice which can be done in quite a few ways in poly time. (Maybe this statement needs explanation. See previous posts for this). Then he invokes the idea of covering radius which basically tells him the minimum radius a circle sitting in the plane with a lattice point at its center must have so that the series of all such circles spans the $2-d$ plane. The crucial thing is - this covering radius can be calculated and we are okay with using any polynomial upper bound on the covering radius.

Thus, after having found one such value, Lenstra just checks if the triangle is big enough to inscribe a circle with radius greater than equal to the above value. In case, the triangle is big enough - good, end of story, the triangle did contain a lattice point. In case, the triangle is not big enough, there are a series of lines parallel to the one basis vector hitting the triangle. There are a polynomial number of such lines. Just check for these lines, one at a time, if someone among them manages to get one of their lattice points inside the triangle.

This idea can be immediately generalised to higher dimensions which is where the rudimentary steps towards LLL first developed from. The problem goes like this - given a convex body (defined by a set of linear inequalities) in higher ($n \geq 3$) dimensions you are asked to determine if it contains a $n$-dimensional lattice point.

Lenstra's idea goes like - transform the convex body into a "cuter convex body". Look for lattice points with the given basis inside the transformed body the following way. Find the covering radius (or some polynomial upper bound on covering radius). Determine whether the transformed body can inscribe a $n-dimensional$ sphere. If it can, some lattice point is there inside the body. If it cannot, the body is hit by span of vectors apart from $b_1$ (which form a hyperplane). These hyperplanes are equi-spaced at gaps of $b_1$. Just see whether some of these hyperplanes contain the desired lattice point.

What led to the discovery of the algorithm was, of course, the transformation step. Lenstra found that in effect, his approach kept on improving the angle between the incident vectors by pushing them closer to being orthogonal. He wrote to Lovasz about his concerns who later came up with the classic reduction scheme and its equally amazing analysis.

In fact, in the two dimensional case, this is precisely what the algorithm did. It turns out this two dimensional case was known long before. It had been solved by the prince of Mathematics, Carl Friedrich Gauss. It is from Gauss's algorithm that we begin our tour of this topic. Our target is to "reduce" a lattice basis to another equivalent basis which makes some questions easier to answer.

What reduction scheme to follow depends on the situation you are in i.e., on the problem you are trying to solve. No one reduction scheme works "exactly" for all cases. So, we compromise and stay happy with LLL reduction which is "reasonable" to some extent in that it ensures that you do not loose too much. Gauss's algorithm, which we will outline shortly, is just a special case of LLL reduction scheme.

But first let me point out a few reduction schemes in advance which do not work. As you now know, the successive minima vectors do not determine a Basis for the lattice in general. In two-dimensional case, however, they do. So, basing the definition on the basis of successive minima vectors is out. Similarly basing it on orthogonalized set of vectors is out. Also, we have a bad luck in that a mutually orthogonal system of vectors need not be short enough as some problems (SVP or CVP) require. LLL algorithm however finds a nice way out which we will detail a little more shortly; but first its time we turned to Gauss's algorithm. (In what follows, we assume that we are talking $l_2$ norm; though Gauss's algorithm is easily extendable to other norms too.)

Gauss starts off by defining a condition for "reduced

*ness*" in

*2*dimensions for the given basis vectors $b_1$ and $b_2$ which basically says that the basis is reduced if the diagonals of the parallelepiped determined by it are both larger than its sides. That is, $||b_1||, ||b_2|| \leq ||b_1 + b_2||, ||b_1 - b_2||$.

The definition is rooted in the fact that we know that for $2$ dimensional case any lattice Basis can be reduced to the basis with successive minima as its basis vectors. Now I will make clear the relation between reduced base and successive minima of the lattice.

In the given setting ($2-dimensional$ lattices), a basis is reduced if and only if the basis vectors are of length $\lambda_1$ and $\lambda_2$ in some order. How do we prove this statement? Well, you just need to keep the definition of reduced basis in mind and proceed straightaway for the forward direction. Do you see the forward direction proof yet? It just says that if your basis is $b_1$ and $b_2$ with $||b_1|| = \lambda_1$ and $||b_2|| = \lambda_2$ then this basis is reduced. Its too clear - the above statement looks like the proof itself.

Its backward direction which is some sport. It says that if you have a reduced basis, then the vectors $b_1$ and $b_2$ in it better have lengths $\lambda_1$ and $\lambda_2$. Why is that? First write out any lattice vector $Bx$ in this basis as $x_1*b_1 + x_2*b_2$ where $x_1$ and $x_2$ are integers. Assume without loss of generality that $||b_1|| \leq ||b_2||$. You want to show that any vector $Bx$ is larger than either of $b_1$ and $b_2$.

For this, you will also need the following easy to prove lemma that if $||x|| \leq ||x + y||$, then $||x + y|| \leq ||x + \alpha*y||$ where $\alpha > 1$. The lemma basically says that as you move further out along a line, so that a little motion increases your distance from some point $P$, further motion in that direction will increase your distance more from $P$.

Now, out of my laziness, and partly because I think I have more important things to discuss here in this blog, I would not complete this proof (unless you want me to).

Now we finally come to Gauss's algorithm. We will follow it up with its complexity analysis. Gauss's idea is very reminiscent of Euclidean algorithm for GCD calculation. You are invited to try spotting the similarities. The algorithm can be seen as a greedy algorithm which iteratively "shrinks" the longest vector in the given basis to find an equivalent one. It continues as long as the reduced vector becomes shorter than the other one and then it stops.

Note that in the algorithm below we subtract from $b_1$ a quantity close to $\mu_{2,1}$ times the vector $b_2$. Subtracting $b_2$ would have orthogonalized it, subtracting this much would sort of orthogonalize it which is another property a "dream basis" should have.

(The following has been taken from the book Algorithmic Cryptanalysis which contains a very good discussion of this algorithm)

**Gauss's Algorithm****Require: Initial lattice basis**($b_1$, $b_2$)

if $b_1 < b_2$ then

**Exchange**$b_1$ and $b_2$

**end if**

repeat

repeat

Find integer $c$ that minimizes $b_1$ $-$ $cb_2$ by:

$c \leftarrow \lfloor \mu_{2,1} \rceil$

Let $b_1 \leftarrow b_1- cb_2$

**Swap**$b_1$ and $b_2$

**until**$||b_1|| \leq ||b_2||$

**Output**($b_1$, $b_2$) as reduced basis

You can find $c$ quickly by binary search. The norm of the longest vector reduces at every step and so the algorithm terminates; the norm cannot go below $\lambda_1$. Further, each iteration of the algorithm clearly takes polynomial time. So, in order to show that the algorithm is a poly time algorithm, all we need to do is just to show that the total number of iterations is polynomial in size of the input. And it can be shown that this must be the case as every iteration decreases the size of the longest vector by $1/2$. I would not go into the details but just mention that it can be proved by first noticing that every iteration shrinks the longest vector by a multiplicative factor.

But in order to develop better intuition for what we want to discuss, it is nice to do the complexity survey the following way. Consider a variation of Gauss's algorithm which we call $t$-Gauss algorithm. (Here $t \geq 1$ is a new parameter to complicate Gauss's algorithm and to simplify our discussion when we talk about LLL).

**t-Gauss's Algorithm****Require: Initial lattice basis**($b_1$, $b_2$)

**Paramter**$t \geq 1$

if $b_1 < b_2$ then

**Exchange**$b_1$ and $b_2$

**end if**

repeat

repeat

Find integer $c$ that minimizes $b_1$ $-$ $cb_2$ by:

$c \leftarrow \lfloor \mu_{2,1} \rceil$

Let $b_1 \leftarrow b_1- cb_2$

**Swap**$b_1$ and $b_2$

**until**$||b_1|| \leq t*||b_2||$ ==> The only step differing from Gauss's algo

**Output**($b_1$, $b_2$) as reduced basis

The algorithm above is cute in the sense that it ensures that the shortest vector shrinks by a factor of t in all iterations except the last one.

Why? Think about it. After the swap step, $b_2$ is your new (shrunk) vector and the previous value of $b_2$ is stored in $b_1$. As long as you remain in the loop now, you must have $t||b_2|| \leq ||b_1||$. That is, $t$ times the new (shrunk) vector is less than the ||b_2|| that you began the round with (or equivalently shortest vector at the start of round, $b_2$, shrunk in size to $b_2$ at the start of next round.

At the risk of overexplaining which might obscure things, I will keep quiet and just state that by a similar reasoning you can see that the longest vector also must decrease in every round by a factor of $t$ (except maybe the first round). Also, you can notice that the longest vector in a round becomes the shortest vector in the next one. That is to say, the longest one shrinks by atleat a factor of $t$.

As a consequence, this algorithm has polynomial number of iterations upper bounded by

**O**$((log(max (b_1, b_2))/t )$. You can derive an expression for number of iterations and by throwing in a bit of analysis, you can show that this algorithm is polynomial for $t=1$ which is the classical Gauss algorithm.

This wraps up a chapter (and if it does not, let me sweep it under the rug)

Next, we have come one more step closer to our target - starting LLL. But I did not tell you the expanded form, did I? It is Lenstra-Lenstra and Lovasz algorithm after its inventors.

So let me grab a coke and come back.

Great so I am back. Coke's gone too.

As I told you earlier, there is no single reduction scheme that works "exactly" for all cases in poly time. We are instead happy with a "reasonable" algorithm which works most of the time. The presentation here follows closely Oded Regev's lecture notes on the subject. We will also discuss a bit from this lecture. So, let me mention the three major characteristics that LLL algorithm shares with Gauss's algorithm. I will sketch some high level points first. It is instructive to keep them in mind as you study the algorithm, but I will repeat them when needed so you do not really need to worry.

1) Define a LLL reduced basis

2) Present an algorithm to find that basis

3) Analyse the running time.

First allow me to mention that now we are no longer talking full rank lattices. Lattices will, in general, have fewer (or equal) vectors as compared to the dimension.

Next, let me define the notion of reducedness we use. (Just reiterating. Keep in your mind that its the Basis which is being reduced; saying you are reducing a lattice does not make sense in our context)

We say that a basis $(b_1, b_2, b_3 ..... b_n) \in \RR^{m*n}$ is $\delta$-LLL reduced (for a parameter $\delta$ about which we will talk later) if the following holds true

1) $|\mu_{i,j}| \leq 1/2$ for all $i > j$ where $\mu_{i,j}$ are the Gram-Schmidt coefficients

2) for any consecutive vectors $b_i$, $b_{i+1}$ the following holds

$\delta$ times Projection of $b_i$ on span of $(b_i*, b_{i+1}*....b_n*)$ is

*less than or equal to*Projection of $b_{i+1}$ on the same span.

You might be having some sense of deja vu, or more precisely you might be feeling that the 2nd requirement for a Basis to be $\delta$-LLL reduced can be written more clearly. (After all, isn't projection of $b_i$ on span of $(b_i*, b_{i+1}*....b_n*)$ something nice? Yes, it is. It is your Gram-Schmidt orthogonalized vector $b_i*$. Think about it. You took away from $b_i$ the components it would leave along $b_1, b_2....b_{i-1}$ by projecting it on the above span. This gives $b_i*$). More formally, we put it this way. We define a family of projection functions, $\pi_i(x)$, for each index $i$ which measures the projection of vector $x$ onto span $(b_i*, b_{i+1}*....b_n*)$.

It is given as $\pi_i(x)$ = $\sum_{j=1}^{j=n}($<$x, b_j^*$>$/$<$b_j^*, b_j^*$>$)$$b_j^*$

We then note $\pi_i(b_i) = b_i*$. Thus, the 2nd requirement for LLL reduction says that $\delta*b_i^* \leq b_{i+1}^*$ for all consecutive vectors $b_i$ and $b_{i+1}$.

Also, I would like you to notice these characteristics in Gauss's algorithm wherein you have

1) $\mu_{2,1} \leq 1/2$

2) $||b_1||| \leq ||b_2||$

You are invited to verify this.

Okay. I guess you would want a cleaner way to understand the first requirement for a basis to be LLL reduced. Do not fear, thats what we do next. Any guesses on how will I try making the content of this requirement clear? How do you think I will do that?

(Relax, its not actually a totally unthinkable way. Its just an alternative convenient representation which helps bind the ideas in your mind; I did too much about ado about nothing [:)] )

By representing the vectors in the "clean basis". We represent the basis vectors in the Gram-Schmidt Orthogonal system to achieve this. If you would recall, in this basis the original Basis vectors are written this way

$\begin{pmatrix}||b_1^*||&\mu_{2,1}||b_1^*||&...&\mu_{n,1}||b_1^*||\\0&||b_2^*||&...&\mu_{n,2}||b_2^*||\\...&...&...&...\\...&...&...&...\\...&...&...&...\\0&0&...&||b_n^*||\end{pmatrix}$

Quoting from Regev's lecture "Here, column $i$ shows co-ordinates of $b_i$ in orthonormal basis. The first condition in the definition of LLL-reduced basis guarantees that the absolute value of any off diagonal element is at most half the value written in the diagonal element of the same row. This can be written as"

$\begin{pmatrix}||b_1^*||&\leq1/2||b_1^*||&...&\leq1/2||b_1^*||\\0&||b_2^*||&...&\leq1/2||b_2^*||\\...&...&...&...\\...&...&...&...\\...&...&...&...\\0&0&...&||b_n^*||\end{pmatrix}$

For illustrating the $2nd$ condition consider the $2*2$ submatrix formed with the upper left entry at index $(i,i)$.

$\begin{pmatrix}||b_i^*||&\mu_{i+1,i}||b_i^*||\\0&||b_{i+1}^*||\end{pmatrix}$

This property just ensures that the $2nd$ column in the above submatrix is almost as long as the first one.

Now I request you to revisit the conditions for a Basis to be LLL-reduced. Keep those in mind for the following discussion which shows how closely LLL is related with SVP (Shortest Vector Problem). It can be used to give an approximate answer to this problem in higher dimensions. The approximation factor is exponential but nonetheless, it is a tremendous achievement. To make the discussion concerete, we would set $\delta = 3/4$.

Assume we have a LLL reduced basis which, by definition, satisfies the conditions I asked you to keep track of.

We know that :

$\delta*||\pi_i(b_i)||^2 \leq ||\pi_i(b_{i+1})||^2$

$\Rightarrow \delta||b_i||^2 \leq ||b_{i+1}^* + \mu_{i+1,i}b_i^*||^2$

$= ||b_{i+1}^*||^2 + ||\mu_{i+1,i}b_i^*||^2$

$= ||b_{i+1}^*||^2 + \mu_{i+1,i}^2||b_i^*||^2$

$ \leq ||b_{i+1}^*||^2 + 1/4*||b_i^*||^2$

Rearranging terms, we get

$(\delta - 1/4)*||b_i^*||^2 \leq ||b_{i+1}^*||^2$

which can be rewritten as

$||b_i^*||^2 \leq \alpha ||b_{i+1}^*||^2$ where $\alpha = \delta - 1/4$

For $\delta = 3/4$, we get $\alpha = 2$. And thus, we have $b_i$ is atmost twice as long as $b_{i+1}$.

And finally we obtain $\lambda_1 \geq min(b_i) \geq 2^n||b_1||$.

Neat! isn't it?

Now, I give you the algorithm (from Regev's notes)

**INPUT:**Lattice basis $b_1,b_2...,b_n \in Z^n$

**OUTPUT:**$\delta$-LLL-reduced basis for L(B)

*Start*

**: compute**$b_1^*, b_2^*,..,b_n^*$

**Reduction Step:***for*$i = 2$ to $n$

*do*

*for*$j = i-1$ to $1$

*do*

*$b_i \leftarrow b_i - c_{i,j}bj$ where $c_{i,j} =$ <$b_i, b_j^*$>$/$<$b_j^*, b_j^*$>*

**Swap Step:***if*$\exists i$ s.t. $\delta||b_i||^2 > ||\mu_{i+1,i}b_i^* + b_{i+1}^*||^2$

*then*

$b_i \leftrightarrow b_{i+1}$

*goto*

*start*

**Output**$b_1, b_2,...,b_n$

Thus, you can see that LLL algorithm is trying out an "extension" (or more precisely relaxation) of Gauss's algorithm in which we want condition 1 satisfied for all $\mu_{i,j}$'s and for condition 2, apart from the factor $\delta$, we ensure that the final reduced base is "cute" in the sense that its Gram-Schmidt Orthogonalization ensures that $b_i$ is not much bigger than its predecessor (its atmost $1/\delta$ times longer) and when this chain winds up all the way to $b_1^*$, it says that $b_1 = b_1^*$ is not too long.

**, you can visualize LLL algorithm as an effort at combining Gram-Schmidt Orthogonalization with Gauss reduction for full rank lattice of dimension $2$. Quoting from Algorithmic Cryptanalysis, "it [LLL] considers the lattice generated by the orthogonal projection of two consecutive vectors $b_i$ and $b_{i+1}$ on the vector subspace generated by the previous vectors $(b_1, b_2,...,b_{i-1})$ and applies one iteration of t-Gauss reduction to this projected lattice. In order to also modify the high dimension lattice, all swap and translation operations are pulled back from the projected space and applied to the vectors $b_i$ and $b_{i+1}$ themselves. In addition, these vectors may also be changed by adding to them some vector from the sublattice with basis $(b_1, b_2,...,b_{i-1})$. The goal of this operation is to make sure that $b_i$ is not too far from the corresponding projected vector $b_i^*$".**

*At a high level*You have been with me this far. Please have a little more patience, once you get the ideas right; you will still be the same; but you will like the subject more. You may need a break; but before that reflect upon the fact that we have completed the first two items in the list

1) Define a LLL reduced basis

2) Present an algorithm to find that basis

3) Analyse the running time.

The last one remains. And the end of the story is absolutely rewarding.

Had your break? That was fast. So, we are now down to the last point wherein we analyze the run time of the algorithm we developed. But first, its not even clear that the algorithm terminates. And it it does, does it indeed compute a $\delta$-LLL reduced basis correctly?

So lets dive right in. Say the algorithm terminates. How do we determine whether the output be correct? Thats easy, just check if the output basis satisfies the two conditions for a basis to be $\delta$-LLL reduced.

Condition $2$ is trivially satisfied. The algorithm was so scared to death by this condition, that it wont stop unless this condition is satisfied. (See the Swap step). Condition $1$ is also satisfied. This is because of the way algorithm operates at every $b_i$. For each $b_i$, the algorithm makes knowledge of the Gram Schmidt representation of the Basis vectors to ensure that all $\mu_{i,j}$ for $j < i$ are less than $1/2$ in the reduction step. This is achieved by subtracting suitable amounts of $b_j$'s for $j < i$ from $b_i$. Note that the order (

*from i-1 to 1*)is crucial.

So, if we do not have anything to swap, we are home. If we have - ah those bad days. Now we would have to recompute Gram-Schmidt Orthogonalized system for the new basis. It would involve re-doing the Reduction Step and the Swap Step. Bad days.

But wait you say. Do I not need to recompute the Gram-Schmidt Orthogonalized system after every iteration of the reduction step? The answer is no. After all, such operations, $b_i \leftrightarrow b_i + ab_j$ never disturb the orthogonalized system. It stays the same.

Does the algorithm stop? You know it does. (Otherwise you would not be studying it.) We prove a stronger statement here by demonstrating not just that it terminates, but by showing the time that it takes to terminate.

Now, the last piece. Run time.

We have to show that the run time is polynomial in input size. To this end, we will first bound the number of iterations. Next, we will bound the running time of a single iteration.

For every iteration performed by the algorithm, one swap step is used. That is to say, that the number of iterations is same as the same as the number of swap operations. How do we bound this number?

In fact, we are going to see the idea that Lovasz used in his analysis, but it is instructive to see the more general model (or that dreadful word, paradigm) his approach for analysing this algorithm falls into. Below, we are going to use a sort of "amortized analysis" for bounding the number of iterations this algorithm goes through. We are reverse-engineering the process. We know the algorithm terminates. We want to find out a bound for this. In algorithmic studies, we often find it convenient to associate with every intermediate state of the algorithm a number which indicates the distance from the target. Without loss of generality, we assume that to begin with the target is at zero and it "stays" at that value (this might depend on the application - corresponding to a stupid move in some cases, the target might move further away. In our case however, we will choose a target and will craft intermediate steps carefully enough so that target stays at the same value.) Demonstrating that this target always decreases gives you an answer for the fact that the algorithm does terminate. Demonstrating that every "careful algorithmic step" decreases target by a particular amount gives a bound on the run time of the algorithm.

So, what do you think we choose as target in this application? What about your intermediate checkpoints? Indeed, your intermediate checkpoint is everytime you orthogonalize the system or every new iteration, but what do you associate an integer value with? And again with what do you associate your target value?

This is where Lovasz steals the ball from most of the mortals. See if you like his analysis. Since, Lovasz knows that the algorithm swaps adjacent vectors when they are out of order, he knows this step must be improving upon the "distance" to the target. So, we should associate a number with some quantity which "captures this information". Hmm, Enough fuss...what should target be anyway? Ah! dont be impatient. Think. Once you well move past a vector $b_{i+1}$, the effect of subsequent vectors on $b_{i}$ vanishes. Something tells us that we should associate an integer with the current Basis. (It is the obvious choice!). Further, its nice to associate it with the determinant of the basis as in its orhtogonalized form, determinant will be a piece of cake. But it has to capture what happens at a swap step. So, it will be like a product of determinants.

Without any more ado, we notice that the integer can be associated with the following product $\Pi_{i=1}^ndet(\Lambda_k)^2$ where $\Lambda_k$ is the lattice spanned by the first $k$ basis vectors. Since swapping ensures that "out of order" vectors obey $2nd$ condition of $\delta$-LLL reduced basis, we also get the reduction of determinant value as a free bonus!

We are decreasing the integer associated with the basis at every iteration! That too by a multiplicative factor. Thats an achievement. We can see that the algorithm has got something like $ln(d_0)*C$ iterations for some constant $C$ and some $d_0$ which we define shortly.

To see that what I said above is indeed happening notice the following ration after the $ith$ step. Assume that at the beginning of this iteration, $d_i$ is the value associated with the basis and after the swap, it becomes $d_i$'.

$d_i$' is the determinant of the lattice $\Lambda_i$' where $\Lambda_i$' $= L[b_1, b_2,...., b_{i-1},b_{i+1}]$. Consider the ratio $d_i/d_i$'. The determinants for $j < i$ are not modified and cancel out. The determinants for $j > i$ are same in absolute value. (Swap 2 columns). Thus, the only thing that stand uncancelled is, when $j = i$.

Thus, $d_i$

**'**$/d_i = det(\Lambda_i$'$)/det(\Lambda_i)$

$ = det([b_1,....,b_{i-1}, b_{i+1}])^2/det([b_1,...,b_i])^2$

$ = ((\Pi_{j=1}^{j=i-1})^2*||\mu_{i+1,i}b_i^* + b_{i+1}||^2)/\Pi_{j=1}^{j=i}||b_j||^2$

$ = ||\pi_i(b_{i+1})||^2/||\pi_i{b_i}||^2$

$ \leq \delta$ by definition of $\delta$-LLL reduced basis

Thus, we notice, that after every swap, we inch closer to the target. The factor of inching closer is $1/\delta$. And thus the claim the number of iterations is $ln(d_0)/ln(1/\delta)$ is vindicated.

Now, we also have to bound the run time of a single iteration. How do you think we should go about it? Well, since we intend to find an upperbound on the number of steps, lets take a pessimistic approach. The number of arithmetic operations performed in each iteration is clearly polynomial. Thus, in order to prove a polynomial upper bound, we just need to show that the size of the numbers involved in the entire computation is also bounded by a polynomial in the size of the input. How worse can that get? (The pessimistic question)

By the way, I hope you have heard of the new (or quantum) difference a pessimist and optimist. The classical version goes this way. The pessimist "Glass is half empty". The Optimist "Glass is half full".The Quantum version. The Pessimist "It can't get any worse this". The Optimist "It can!"

So back to the problem. How worse can it, in fact, get? The LLL algorithm uses rational numbers. So we are under a two-sided attack. We have to bound both precision and the magnitude of the involved quantities.

(I will write the details in this space soon enough. For now, you can just assume it and rest).

Now we have seen the SVP connection. Whats the connection with CVP? As you might have guessed, this LLL algorithm can be used for an approximate solution to CVP as well. Here is how. Given a Basis $B$ and a target vector $t$ you need to find a lattice vector, $Bx$, that is closest to $t$. To this end, you just add $t$ to the basis and find the and run the reduction step of the LLL algorithm on this basis.

Quoting from Goldwasser-Micciano's book "Now you find a vector $x \in L(B)$ such that $t - x$ can be written as $t^* + \sum_{i=1}^nc_ib_i^*$ where $t^*$ is component of $t$ orthogonal to span$(b_1, b_2,...,b_n)$ and $|c_i| \leq 1/2$ $\forall$ $i = 1,2,..,n$."

This incidentally is the high level sketch of the approx-CVP algorithm as it can be shown that the distance of $t$ from $x$ is within an exponential approximation of the optimal. More on this in m future posts.

There is a lot more to come on this post. So, please wait. I will do that whenever I get time. (Yeah, you got it. I publish the post then edit and re-edit till it gets to the final version!)

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