SVP is no harder than CVP

We write A ≤T B to denote that A is Turing-reducible to B; A is Turing-reducible to a problem B if given an oracle for B, one can compute answers to A. In this case we will show taht if we can compute correct answers for CV Pγ , we can use this to compute the γ-shortest vector of some fixed lattice. Note that the decision formulation of CV P is NP-hard while (the natural decision version of) SV P is in NP; thus we have a generic reduction of SV P to CV P (through, say, 3-SAT). However this reduction is inefficient in that it requires solutions to large instances of CV P in order to compute the corresponding SV P instance. In particular, this reduction gives no sense of the relationship between approximating CV P and approximating SV P . When we prove Threorem 3, we give a reduction that is dimension-preserving, approximation-preserving and deterministic. The known converse to Theorem 3 has none of these three (desirable) properties. First we give a sense of the reduction: The shortest vector of a lattice B is the closest non-zero vector to the origin. As CV P (B, 0) = 0, the trivial reduction won’t work. Instead, we will generate sublattices of B by “knocking out” certain vectors and ask for the closest vector in this sublattice to the “hole” in the lattice we’ve created. Let B be a lattice basis; consider sublattices spanned by bases B, . . . , B where the ith basis B is {b1, . . . , bi−1, 2bi, bi+1, . . . , bn}. Claim 4. bi 6∈ L(B)