𝐷𝑎𝑡𝑎𝐷𝑎𝑡𝑎ₗₑₙ 𝑆𝑝𝑎𝑛 𝑆𝑝𝑎𝑛ₗₑₙ 𝒯 𝒯ₗₑₙ 𝒫 𝒫ₗₑₙ 𝒪 𝒪⁺ 𝒪⁻

𝐷𝑎𝑡𝑎𝐷𝑎𝑡𝑎ₙ 𝑆𝑝𝑎𝑛 𝑆𝑝𝑎𝑛ₙ 𝒯 𝒯ₙ 𝒫 𝒫ₙ 𝒪 𝒪⁺ 𝒪⁻ ⁺𝒪ₙ ⁻𝒪ₙ

We accept a data sequence𝐷𝑎𝑡𝑎𝒮 of type Vector{T} and of length𝐷𝑎𝑡𝑎𝒮ₙ (𝐷𝑎𝑡𝑎𝒮[begin:end], length(𝐷𝑎𝑡𝑎𝒮) ==𝐷𝑎𝑡𝑎𝒮ₙ). We are given a window specification that includes its length, the span of any tiling, and more.

Given a data seqeunce of N elements and a window that spans W elements (W <= N), ccompletewindows, rremainingindices = fldmod(N, W) if iszero(remainingelements) the data sequence is covered exactly with ccompletewindows otherwise, the data sequence is nearly fully covered with ccompletewindows, leaving rremaining_indices

N = c_complete_windows * W + r_remaining_indices
0 = c_complete_windows * W + r_remaining_indices - N
c_complete_windows * W = N - r_remaining_indices
c_complete_windows = div((N - r_remaining_indices), W)
W = div((N - r_remaining_indices), c_complete_windows)
r_remaining_indices = N - c_complete_windows * W

The preceeding assumes that the window always advances by 1 index. Use A as the whole number of indices (1 <= A <= N-1-W) that window always advances. With A = N-1-W, there is exactly one advance, from index 1 to index 1+N-1-W = N-W the repositioned window now starts at index N-W and spans W indices, N-W+W == N and the window has nowhere more to traverse.

What value of A allows exactly 2 advances? A1 = N-1-W, if iseven(A1) A2 = div(A1,2) or, if isodd(N-W), A2 = div(N-W-1, 2)