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lecture12

# lecture12 - 1 15.083J/6.859J Integer Optimization Lecture...

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1 15.083J/6.859J Integer Optimization Lecture 12: Lattices III

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2 6.4 The approximate nearest vector problem In this section, we use Algorithm 6.3 to produce a reduced basis of a lattice, and then use this reduced basis to solve approximately the nearest vector problem in L introduced in Eq. (6.2) In particular, let L be a lattice and x Q n not belonging to L . The nearest vector problem is NP -hard. Algorithm 6.4 that we present next finds a b ∈ L such that b x (2 n 1) 1 / 2 min { b x : b ∈ L \ { 0 }} . (6.12) Algorithm 6.4 Approximate nearest vector algorithm Input : A reduced basis b 1 , . . . , b n of a lattice L , the Gram-Schmidt orthogonalization ˜ b 1 , . . . , ˜ b n and x Q n . Output : A vector b ∈ L satisfying Eq. (6.12) and rational multipliers λ i , i = 1 , . . . , n with | λ i | ≤ 1 / 2 , such that b x = n i =1 λ i ˜ b i . Algorithm : 1. Set z n +1 = 0 and x n +1 = x . 2. For i = n, . . . , 1 : (a) Compute σ 1 ,i , . . . σ i,i Q such that x i +1 = i j =1 σ j,i ˜ b j . (b) Set λ i = σ i,i σ i,i , where σ i,i denotes the nearest integer to σ i,i . (c) Set z i = z i +1 + σ i,i b i . (d) Set x i = x i +1 σ i,i b i + λ i ˜ b i . 3. Return b = z 1 and λ 1 , . . . , λ n . Example 6.9 We revisit Example 6.6. We have seen that the vectors b 1 = (1 , 1) , b 2 = (2 , 1) constitute a reduced basis of the lattice L ( b 1 , b 2 ). The cor- responding Gram-Schmidt orthogonalization is ˜ b 1 = (1 , 1) and ˜ b 2 = (1 . 5 , 1 . 5). Let x = (7 . 5 , 3 . 0) . We start the approximate nearest vector algorithm by setting z 3 = 0 and x 3 = x . We then set i = 2 and compute multipliers σ 1 , 2 , σ 2 , 2 such that x = σ 1 , 2 ˜ b 1 + σ 2 , 2 ˜ b 2 . This gives σ 1 , 2 = x ˜ b 1 ˜ b 1 2 = 5 . 25 , and σ 2 , 2 = x ˜ b 2 ˜ b 2 2 = 1 . 5 .
Sec. 6.4 The approximate nearest vector problem 3 This leads to λ 2 = σ 2 , 2 σ 2 , 2 = 1 1 . 5 = 0 . 5 z 2 = z 3 + σ 2 , 2 b 2 = 0 + b 2 = (2 , 1) x 2 = x 3 σ 2 , 2 b 2 + λ 2 ˜ b 2 = x b 2 0 . 5 ˜ b 2 = (4 . 75 , 4 . 75) . We continue for i = 1. We compute σ 1 , 1 such that x 2 = σ 1 , 1 ˜ b 1 , leading to σ 1 , 1 = 4 . 75, λ 1 = σ 1 , 1 σ 1 , 1 = 0 . 25, and z 1 = z 2 + σ 1 , 1 b 1 = (7 , 4) . Note that b = z 1 = (7 , 4) ∈ L and b x = λ 1 ˜ b 1 + λ 2 ˜ b 2 . We next show that Algorithm 6.4 correctly finds b ∈ L satisfying Eq. (6.12). Theorem 6.10 The vector b = z 1 computed by the approximate nearest vector algorithm belongs to L and satisfies b x (2 n 1) 1 / 2 min { b x : b ∈ L \ { 0 }} . Proof. Let b 1 , . . . , b n be the reduced basis of L , which is part of the input of the approximate nearest vector algorithm. Let ˜ b 1 , . . . , ˜ b n denote its Gram- Schmidt orthogonalization. Solving the recursions in Steps 2c and 2d of the approximate nearest vector algorithm, we obtain x 1 = x n +1 n i =1 σ i,i b i + n i =1 λ i ˜ b i z 1 = z n +1 + n i =1 σ i,i b i . Since x 1 = x 2 σ 1 , 1 b 1 + λ 1 ˜ b 1 = σ 1 , 1 ˜ b 1 σ 1 , 1 b 1 + λ 1 ˜ b 1 , and ˜ b 1 = b 1 , it follows x 1 = 0 . Moreover, since x n +1 = x , x = n i =1 σ i,i b i n i =1 λ i ˜ b i .

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