Table 1 Algorithm of the structured lattice reduction

[1st Stage of the SLR]

   1. Go to OLR-block with $H_{R,\alpha }=\left[h_{R,1},h_{R,3},{\bar{h}}_{R,1},{\bar{h}}_{R,3}\right]$

   2. Output: ${H^{\prime }}_R=\left[{h^{\prime }}_{R,1},{h^{\prime }}_{R,2},{h^{\prime }}_{R,3},{h^{\prime }}_{R,4},{\overline{h^{\prime }}}_{R,1},{\overline{h^{\prime }}}_{R,2},{\overline{h^{\prime }}}_{R,3},{\overline{h^{\prime }}}_{R,4}\right]$

[2nd Stage of the SLR]

   3. Go to OLR-block with ${H^{\prime }}_{R,\alpha }=\left[{h^{\prime }}_{R,1},{h^{\prime }}_{R,4},{\overline{h^{\prime }}}_{R,1},{\overline{h^{\prime }}}_{R,4}\right]$

   4. Output: ${H^{″}}_R=\left[{h^{″}}_{R,1},{h^{″}}_{R,2},{h^{″}}_{R,3},{h^{″}}_{R,4},{\overline{h^{″}}}_{R,1},{\overline{h^{″}}}_{R,2},{\overline{h^{″}}}_{R,3},{\overline{h^{″}}}_{R,4}\right]$

Function of OLR-block

Input: H_{in, α}= [h_{α, 1}, h_{α, 2}, h_{α, 3}, h_{α, 4}]

[Initial Sorting with H_in,α]

1. H_in,α→ S _α = [s₁, s₂, s₃, s₄]

s.t. s _i = h_{α, θ(i)}and |s _i | ≤ |s_i+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 ↔ s_π(i))

[Re-ordering]

3. Re-ordering caused by the LLL

s.t. U _α = [u₁, u₂, u₃, u₄], (u_π(i)= g _i )

   4. Re-ordering caused by the initial sorting

   s.t. ${H^{\prime }}_{\mathsf{\text{in}},\alpha }=\left[{h^{\prime }}_{\alpha ,1},{h^{\prime }}_{\alpha ,2},{h^{\prime }}_{\alpha ,3},{h^{\prime }}_{\alpha ,4}\right],\left({h^{\prime }}_{\alpha ,\theta \left(i\right)}=u_i\right)$

[Remaining Basis Generation]

   5. ${H^{\prime }}_{\mathsf{\text{in}},\alpha }\to {H^{\prime }}_{\mathsf{\text{in}},\beta }=\left[{h^{\prime }}_{\beta ,1},{h^{\prime }}_{\beta ,2},{h^{\prime }}_{\beta ,3},{h^{\prime }}_{\beta ,4}\right]$

[Output]

At 1st stage: $H_{\mathsf{\text{out}},1}=\left[{h^{\prime }}_{\alpha ,1},{h^{\prime }}_{\beta ,1},{h^{\prime }}_{a,2},{h^{\prime }}_{\beta ,2},{h^{\prime }}_{\alpha ,3},{h^{\prime }}_{\beta ,3},{h^{\prime }}_{\alpha ,4},{h^{\prime }}_{\beta ,4}\right]$

At 2nd stage: $H_{\mathsf{\text{out}},2}=\left[{h^{\prime }}_{\alpha ,1},{h^{\prime }}_{\beta ,1},{h^{\prime }}_{\beta ,2},{h^{\prime }}_{a,2},{h^{\prime }}_{\alpha ,3},{h^{\prime }}_{\beta ,3},{h^{\prime }}_{\beta ,4},{h^{\prime }}_{\alpha ,4}\right]$