Lattice Based Cryptography-Week 2
- 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume Week 2: Way To Orthogonalization Masum Billal Dhaka University May 11, 2014 Masum Billal Week 2: Way To Orthogonalization
- 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume Span & Fundamental Parallelepiped The span of a lattice L is the vector space spanned by its vectors. In writing, span(L(B)) = span(B) = {BY : Y ∈ Rn } The fundamental parallelepiped of a lattice L is defined as P(B) = {B · Y : 0 ≤ yi < 1} Masum Billal Week 2: Way To Orthogonalization
- 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume Is A Set Of Vectors A Basis? Let B = {b1, ..., bn} be a set of vectors. We need to know when this forms a basis of a lattice L. Masum Billal Week 2: Way To Orthogonalization
- 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume Is A Set Of Vectors A Basis? Let B = {b1, ..., bn} be a set of vectors. We need to know when this forms a basis of a lattice L. Theorem Let B = {b1, ..., bn} be a set of linearly independent vectors and L be a lattice. Then B is a basis of L iff P(B) ∩ L = {(0)} where (0) indicates the origin i.e. the vector with all coordinates set to 0. Masum Billal Week 2: Way To Orthogonalization
- 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume Proof Proof. Let’s assume first that, B is a basis of L. By definition, L is the set of their integer combinations. Since, P is the integer combination in range [0, 1), we have P(B) ∩ L = {(0)}. Masum Billal Week 2: Way To Orthogonalization
- 6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume Proof Proof. Let’s assume first that, B is a basis of L. By definition, L is the set of their integer combinations. Since, P is the integer combination in range [0, 1), we have P(B) ∩ L = {(0)}. Now, assume that, P(B) ∩ L = {(0)}. We can write a vector x ∈ L as ∑ bixi for xi ∈ R. And by definition, L is closed under vector addition. So, x′ = ∑ (xi − ⌊xi⌋)bi also belongs to L. From our assumption, x′ = (0). Thus, xi − ⌊xi⌋ = 0 i.e. xi is an integer. Hence, B a basis of L. Masum Billal Week 2: Way To Orthogonalization
- 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume Determinant Definition Let Λ = L(B) be a lattice of rank n. Then the determinant of Λ is the n-dimensional volume of its parallelepiped, and is denoted as det(Λ). If B is a full rank basis, then det(Λ) = | det B|. Otherwise, det(Λ) = √ det(BT B). The determinant of a lattice is inverse proportional to its density: the smaller the determinant, the denser the lattice is. In more precise terms, if one takes a large ball K (in the span of Λ) then the number of lattice points inside K approaches vol(K)/det(Λ) as the size of K goes to infinity. Masum Billal Week 2: Way To Orthogonalization
- 8. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume Gram-Schmidt Orthogonalization Gram-Schmidt orthogonalization takes n linearly independent vectors and creates n orthogonal vectors. Masum Billal Week 2: Way To Orthogonalization
- 9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume Gram-Schmidt Orthogonalization Gram-Schmidt orthogonalization takes n linearly independent vectors and creates n orthogonal vectors. Definition Given n linearly independent vectors b1, b2, ..., bn. Their Gram-Schmidt Orthogonalization is the sequence of vectors ¯b1, ¯b2, ..., ¯bn defined as ¯bi = bi − i−1∑ j=0 µij ¯bj where µij = ⟨bi, ¯bj⟩ ⟨ ¯bj, ¯bj⟩ . In words, ¯bi is the component of bi orthogonal to ¯b1, ..., ¯bi−1. Masum Billal Week 2: Way To Orthogonalization
- 10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume Volume In Terms Of Orthogonal Vector If we consider the orthonormal basis of B given by ¯b1 | ¯b1| , ¯b2 | ¯b2| , . . . , ¯bn | ¯bn| we find that, the volume of the parallelepiped generated by Λ is n∏ i=1 |¯bi|. Here, |a| denotes the absolute value of the vector a. Or equivalently, det(Λ) = det B = det(P(B)) = n∏ i=1 |¯bi| Masum Billal Week 2: Way To Orthogonalization
- 11. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume Orthogonal Vectors Theorem Let B is a basis of rank n, and ¯B be its orthogonalized basis. Then, λ1(B) ≥ min i=1,...,n |¯bi| > 0 This theorem is quite useful for finding a short vector in a lattice. Corollary: Let Λ be a lattice. Then there is a constant ε so that, for any two non-equal points x, y ∈ Λ, |x − y| > ε. Masum Billal Week 2: Way To Orthogonalization