 We can use LLL to find a set of lattice vectors. Now the question is, well, how can we use LLL to solve the corresponding lattice problems, particularly the SVP shortest vector problem? And so remember that our LLL algorithm began with a lattice span by some set of basis vectors, and we can apply LLL to get some set of quasi-orthogonal basis vectors, and a corresponding set of Gram-Schmidt basis vectors. And so the natural question to ask is, after we've done all that work, how good is the LLL basis? And we can enter that by answering how effectively it solves the shortest vector problem, finding the shortest non-zero vector in the lattice. So let's take a look at some preliminaries. Suppose we've actually found the LLL basis. Remember we can rely on the following. First of all, mu ij has to be strictly less than a half. Mu is vi vj star over vj star squared. So again, the star vectors are the Gram-Schmidt basis. The unstarred vectors are the lattice basis vectors. And we know that this dot product over the norms is going to be less than a half. The reason we know this is because we've reduced all of our lattice vectors by the rounded value of mu ijv. So if this is greater than a half, we'd still have a further reduction to perform. The other thing that we can rely on is the Lavage condition, norm vk squared, is going to be greater than or equal to three quarters mu k, k minus one squared, and so on. And again, we know this because whenever this failed to be true, we swapped our vectors around. If we have found our LLL basis, we know this has to be true because otherwise we aren't done, and we know this has to be true again because otherwise we aren't done. And there's some immediate deductions we can make from this. So let's begin with the Gram-Schmidt basis vectors. So because our mu's are going to be less than a half, this tells me that a quarter is going to be greater than the square and a little bit of playing around with the inequalities tells me that this expression here, which shows up in the Lavage inequality, is going to be greater than one half. Now, since the Lavage condition has to be satisfied, I know that vk star squared must be greater than or equal to this vk minus one star squared. But I know that this is greater than a half, so that tells me that this is greater than one half of the preceding vectors. And so in effect, this Lavage condition translates into a size requirement on the relative magnitude of the Gram-Schmidt basis vectors. Well, it's all about the basis. So we can apply this inequality repeatedly, and so I can find then, if I were to apply this repeatedly, vk star greater than a half, greater than a half again, and so on, and in general, I can cascade vk star squared must be greater than or equal to one over two squared vk minus two squared. And in general, that tells me I can compare any basis vectors to the first of the Gram-Schmidt basis vector. And that's important, because remember in the Gram-Schmidt process, the first Gram-Schmidt basis vector is going to be the first vector. Well, again, we can say something else here. So remember, our lattice basis vectors vk are related to the Gram-Schmidt basis vectors vk star by the reduction. So we're going to take vk, we're going to subtract off the portion that is parallel to all of the other Gram-Schmidt basis vectors, and we'll rearrange that a little bit. So that tells me vk is the Gram-Schmidt basis vector, plus the dot product have a whole bunch of vectors. Now vk dot vk, well, it's just the dot product with itself, and over on the left-hand side I vk squared, over on the right-hand side I vk star vk star, there's that, and then I have all the other dot products in here. And the thing to remember is that because the star vectors are the Gram-Schmidt basis vectors, they're going to be orthogonal. So the dot product is going to be zero anytime i and j are different. So all of these dot products cancel or drop out, because they're all zero, with one exception, any vector times itself. And so that'll give me mu ii squared, the i star squared, and I'm going to include the sum of all those vectors. Now I know that, again, in general my mu values are less than a half, so that tells me that if I replace this with something that is greater, mu less than a half, mu squared has to be less than a quarter, I'll replace this with something bigger, I get this inequality. And so what that tells me is something very useful. The lattice basis vectors are going to be smaller than the corresponding Gram-Schmidt basis vectors, plus the sum of all the preceding Gram-Schmidt basis vectors. Well we'd like to be able to characterize that a little bit more elegantly, so what we'd like to do is find a bound on the sum of the first k-1 Gram-Schmidt factors. Again, what we have here is we have the sum of the first k-1 Gram-Schmidt basis vectors. We'd like to find a bound on this quantity, because while this inequality is true and valid, it's a little bit ugly. So let's see what we can do. Well the key to remember here is that any Gram-Schmidt basis vector satisfies this relationship to the preceding Gram-Schmidt basis vector. And again remember we can cascade this so that this is also going to be true to squared for the second preceding vector, and so on for all of the others. And eventually we get down to that very first basis vector and the coefficient there, 1 over 2 to the power k minus 1. Well let's transpose those powers in two, because what I have over here on the right hand side are the vectors, are the values that I want to add together. So if I add these values together over on the right hand side I have the sum of the squares of the magnitudes of all my Gram-Schmidt basis vectors, and over on the left hand side I have the magnitude of the kth Gram-Schmidt basis vector with coefficients to 2 squared, 2 cubed, and so on, this sum here. And now I can add all those together. So I'm going to take a look at this. Well this is a geometric series. So I have a geometric series summation formula that I can apply, and so that's going to give me this expression over here, still the k squared, and this is still the sum of the squares of the magnitudes of all those Gram-Schmidt basis vector. And we'll simplify that. And again I'll join this to our earlier result again. We found that the kth lattice vector, magnitude squared, less than or equal to the magnitude of the kth Gram-Schmidt basis vector squared, plus one-quarter the sum of all of the remaining Gram-Schmidt basis vectors, all the preceding Gram-Schmidt basis vectors. Well this sum itself is smaller than this value. So I'm going to replace this with something bigger, and let's see. Well now these are both the k star squares, so I can add these two together. And well this is going to be two to k minus two minus one-half. And I can carry that around, but that's two to k minus two plus a half is definitely going to be smaller than two to k minus one. So I'm going to replace this entire coefficient here with something that's a little bit larger, two to the power of k minus one. And this gives us a much cleaner result of how our lattice basis vectors relate to our Gram-Schmidt basis vectors.