# Table 1 Algorithm of the structured lattice reduction

Structured Lattice Reduction (SLR)
Input: $H → H R = h R , 1 , ⋯ , h R , 4 , h ̄ R , 1 , ⋯ , h ̄ R , 4$
[1st Stage of the SLR]
1. Go to OLR-block with $H R , α = h R , 1 , h R , 3 , h ̄ R , 1 , h ̄ R , 3$
2. Output: $H ′ R = h ′ R , 1 , h ′ R , 2 , h ′ R , 3 , h ′ R , 4 , h ′ ̄ R , 1 , h ′ ̄ R , 2 , h ′ ̄ R , 3 , h ′ ̄ R , 4$
[2nd Stage of the SLR]
3. Go to OLR-block with $H ′ R , α = h ′ R , 1 , h ′ R , 4 , h ′ ̄ R , 1 , h ′ ̄ R , 4$
4. Output: $H ″ R = h ″ R , 1 , h ″ R , 2 , h ″ R , 3 , h ″ R , 4 , h ″ ̄ R , 1 , h ″ ̄ R , 2 , h ″ ̄ R , 3 , h ″ ̄ R , 4$
Function of OLR-block
Input: Hin, α= [hα, 1, hα, 2, hα, 3, hα, 4]
[Initial Sorting with Hin,α]
1. Hin,αS α = [s1, s2, s3, s4]
s.t. s i = hα, θ(i)and |s i | ≤ |si+1|, for 1 ≤ i < 4
[Conventional LLL-LR with S α ]
2. S α = [s1, s2, s3, s4] → G α = [g1, g2, g3, g4]
Order of columns is changed during LLL
s.t. π(i) → i, for 1 ≤ i ≤ 4, (i.e. g i sπ(i))
[Re-ordering]
3. Re-ordering caused by the LLL
s.t. U α = [u1, u2, u3, u4], (uπ(i)= g i )
4. Re-ordering caused by the initial sorting
s.t. $H ′ in , α = h ′ α , 1 , h ′ α , 2 , h ′ α , 3 , h ′ α , 4 , h ′ α , θ ( i ) = u i$
[Remaining Basis Generation]
5. $H ′ in , α → H ′ in , β = h ′ β , 1 , h ′ β , 2 , h ′ β , 3 , h ′ β , 4$
[Output]
At 1st stage: $H out , 1 = h ′ α , 1 , h ′ β , 1 , h ′ a , 2 , h ′ β , 2 , h ′ α , 3 , h ′ β , 3 , h ′ α , 4 , h ′ β , 4$
At 2nd stage: $H out , 2 = h ′ α , 1 , h ′ β , 1 , h ′ β , 2 , h ′ a , 2 , h ′ α , 3 , h ′ β , 3 , h ′ β , 4 , h ′ α , 4$ 