Complexity of Lattice Problems: A Cryptographic Perspective by Daniele Micciancio

By Daniele Micciancio

Lattices are geometric gadgets that may be pictorially defined because the set of intersection issues of an enormous, standard n-dimensional grid. De­ spite their obvious simplicity, lattices cover a wealthy combinatorial struc­ ture, which has attracted the eye of significant mathematicians during the last centuries. no longer strangely, lattices have came across quite a few ap­ plications in arithmetic and laptop technology, starting from quantity idea and Diophantine approximation, to combinatorial optimization and cryptography. The research of lattices, particularly from a computational standpoint, was once marked by means of significant breakthroughs: the improvement of the LLL lattice aid set of rules via Lenstra, Lenstra and Lovasz within the early 80's, and Ajtai's discovery of a connection among the worst-case and average-case hardness of convinced lattice difficulties within the past due 90's. The LLL set of rules, regardless of the fairly terrible caliber of the answer it supplies within the worst case, allowed to plot polynomial time suggestions to many classical difficulties in desktop technological know-how. those contain, fixing integer courses in a hard and fast variety of variables, factoring polynomials over the rationals, breaking knapsack dependent cryptosystems, and discovering suggestions to many different Diophantine and cryptanalysis problems.

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3) by 8 and get IIx + yll < IIx + aY11 as Since 8 > 0, we can claimed in the lemma. 0 We can now establish a relation between reduced bases and the successive minima of the lattice. 2. 2 Let [a, bJ be a lattice basis, and let Al and A2 be the successive minima of the lattice. Then, [a, bJ is reduced if and only if a and b have norm A1 and A2. , IIall = Al and IIbll = A2. By definition of Al we know that Iia - bll and lIa + bll are at least as large as lIali. Moreover, since [a, bJ is a basis, b is linearly independent from a, and therefore each of a - b and a + b is linearly independent from a.

Proof: Consider the subsequence (ai-I, ai, ai+r). In order to keep notation simple, we rename this sequence as (a, b, e). L 2: 1 and f = ±1. Lb. L = 1. This case is not possible because lie - bll = lIall < IIbll would contradict the assumption that [b, e] is well ordered. 31 Approximation algorithms • Case € = -1,J,L = 2. Also not possible because lie - bll = 11- a+ bll would contradict either lIa - bll < IIbil or IIbil < lib - ell· • Case € = -1, J,L > 2. In this case, lIell = II - a+ J,Lbil which, by ~ lIall < IIbll, is strictly bigger J,Lllbll - lIall than • Case € = 1, J,L ~ 2.

C(B). We want to determine if equality holds. The comparison can be easily performed using oracle A as follows. c(B))2 ~ r + 1. Then, we call oracle A on input 47 Closest Vector Problem (B', t, vr+T). If the oracle returns NO, then dist(t, £(B')) > vr+T ~ dist(t, £(B)). ;r < dist(t, £(B)) ::; dist(t, £(B')) ::; vr+T, and therefore dist(t, £(B)) = dist(t, £(B')) because both dist(t, £(B))2 and dist(t, £(B'))2 are integers. Now notice that if Xl is even for some closest vector Bx, then dist(t,£(B)) = dist(t,£(B')), while if Xl is odd for all closest vectors Bx, then dist(t, £(B)) < dist(t, £(B')).

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