 Okay, so let's take a look at another example of LLL, this time for a 3-vector lattice. And so I'll have 3 vectors that span our lattice, a 15, 23, 11, 46, and 32, 1, 1. So let's go ahead and take a look at that. So again, it doesn't really matter what order I put the vectors in. So I'll go ahead and use the stated order. And I have below my basis vectors for the orthogonal basis produced by the Gram-Schmidt process. So I'll start off with my first vector as my first working vector. And I want to reduce V2, so that means I need to compute the Gram-Schmidt basis. And in practice, if this is my working vector, Vk is my working vector, I only need the Gram-Schmidt basis as far as the vector Vk star. And so I don't need to calculate the rest of them. So I'm working with V2. I do need to calculate V2 star, but I don't need to worry about this third vector because I'm not going to use it. Now I'll use my first basis vector to try and reduce the second basis vector. And I get a computation, 31, negative 8, negative 8, and that's my new V2 vector. And I'll check on the Lavage condition. So my magnitude squared is the first basis vector. My magnitude squared is the second basis vector. My dot product, again, that's my second basis vector. My first Gram-Schmidt basis over the square of the magnitude of the first Gram-Schmidt basis vector, 0.211. And finally, my 3 quarters minus mu2, 1 squared, 0.701. And I'll check the Lavage condition. Is this squared greater than or equal to 3 quarters minus mu squared times the first magnitude squared? And it turns out that that is, in fact, going to be the case. So I like these two vectors as they are, and I can move on to trying to find the third, trying to reduce that third basis vector. So again, I'll recompute my Gram-Schmidt basis. I can still use my initial vector. My second basis vector, I do need to find my second Gram-Schmidt basis vector and my third Gram-Schmidt basis vector. Now, again, because my V2 has changed, I do actually need to recompute, or at least verify, that my second basis vector hasn't changed. So I'll compute that second basis vector. And because V3 is my working basis vector, then I do have to compute my third Gram-Schmidt basis vector. So again, I'll subtract off those components. And so V3 is going to be this amount here. And so now I want to use V1 to reduce V3. So I'll find that dot product. So again, this is V3 dotted with the first Gram-Schmidt basis vector over the square of the magnitude of that first Gram-Schmidt basis vector. So when I do that, I'm going to get, as my first reduction, 17 negative 2210. Now, I want to reduce the V2 to reduce the new 17 negative 2210. So again, the strange thing here is that my dot product is actually going to be using the second Gram-Schmidt basis vector, but I'm going to be reducing it according to the second lattice vector. So that's my dot product V3 with the basis vector, Gram-Schmidt basis vector, over the square of the magnitude, the dot product of the basis vector with itself. But again, I'm going to be reducing that by the second actual lattice vector. So again, these three vectors here are the basis, but this vector is going to be the actual lattice basis. And that works out to be negative 14, negative 14, 2, and that's my third basis vector. So now I have the original two basis vectors, my new third basis vector, and again, my Gram-Schmidt basis vectors. So I'll check the Lavage condition. So again, that's just comparing these two vectors. So the magnitude of the basis vector squared, the magnitude of the basis vector squared, the dot product, and again, note that V3 has changed, so my dot product is going to change, and 3 quarters minus the square of mu, again, 0.71. And so checking our Lavage condition, we find that the magnitude squared is not greater than 3 quarters minus mu, 3, 2, squared, V2 squared. So that says I need to swap the vectors V2 and V3. I'll move this one, make this the working vector as our second vector, and then this vector gets moved to the end, and we'll ignore it until we fix things. So I'll make that swap, that resets our Gram-Schmidt basis, because we've now changed the vectors, and now I can continue. I'm going to try, this is my working vector. I'm going to try and reduce this using the preceding lattice vectors. So I'll apply my Gram-Schmidt reduction process. So again, I'll take V1 as my first Gram-Schmidt basis vector. I'll do the reduction to find my second Gram-Schmidt basis vector. I don't actually need to find the third Gram-Schmidt basis vector, because this is my working vector, and I don't really need anything past that point. So I'll use V1 to reduce V2. There's my new reduced vector, and I'll check the Lavage condition. Again, I have my V1 squared, I have my V2 squared, and I'm going to compute those two values, 0.36. I have my new value of 32 mu squared, and again, because my second working vector squared is not greater than or equal to this quantity, then that tells me I need to swap my two vectors, V1 and V2. So I'll make that swap, and again, I'm going to start again. So I'm going to find the Gram-Schmidt basis vectors again. I'll reduce V2 using this. That gives me a new vector, and I'll check on the Lavage condition one more time, and this does hold in this particular case, and so that says I can keep both of these as my working vector, and I'm going to now reduce my third vector. So I'll find the Gram-Schmidt basis. I'll reduce V3 using V1, first of all. So again, here's the dot product lattice vector with the Gram-Schmidt basis vector over the square of the magnitude, but then reduced by the lattice basis vector. And again, because of the way the Gram-Schmidt process works, V1 is the same as V1 star. But that's not generally going to be true for any of the others. So that reduction gives me 32, 1, 1 as my new vector V3 so far, but now I want to reduce it by the second lattice basis vector, 13, 5, 7. So again, I'm going to take 32, 1, 1. I'm going to subtract the dot product, and here's where it makes a difference. That's the dot product of our current V3 with the Gram-Schmidt basis V2 over the Gram-Schmidt basis V2 magnitude squared. That's the dot product of itself. But I'm going to do the reduction by a multiple of the second lattice basis vector. So again, here, this second vector that I'm using is going to be very different from the vector I'm actually subtracting. And that gives me the result 6, negative 9, 15 as our final version of V3. And so now I check the Lavash condition. So second lattice vector, magnitude squared. Third Gram-Schmidt basis vector, magnitude squared. Dot product, 3 quarters minus dot product squared. And I check the Lavash condition, and I see the V3 and V2 do satisfy the Lavash condition. So I like this vector. I'll keep it here. I like this vector. I'll keep it there. I like this vector. I've kept it there. And I've gone through all three of my basis vectors. And so they will form our reasonably orthogonal set for our lattice spanned by this not-so-reasonably orthogonal set.