Support arbitrarily offset tiling

[Luapulu]

Currently only the last tile can be relaxed. What about adding a strategy RelaxEdgeTile, that allows offsetting the tile grid by an arbitrary offset?

[johnnychen94]

Could you clarify what's the expected results with offset? Currently, there are two strategies after #23:

julia> A = rand(8);

julia> TileIterator(axes(A), RelaxLastTile((3, )))
3-element TileIterator{1,Tuple{TiledIteration.CoveredRange{StepRange{Int64,Int64},TiledIteration.LengthAtMost}}}:
 (1:3,)
 (4:6,)
 (7:8,)

julia> TileIterator(axes(A), RelaxStride((3, )))
3-element TileIterator{1,Tuple{TiledIteration.CoveredRange{TiledIteration.RoundedRange{LinRange{Float64}},TiledIteration.FixedLength}}}:
 (1:3,)
 (3:5,)
 (6:8,)

[Luapulu]

Ah yes, the idea is to return

(1:1,) # offset of 1
(2:4,)
(5:7,)
(8:8,)

For my purposes I’d only be interested in relaxing the last tile, but in principle this offset is independent of which if the two other strategies are used.

[johnnychen94]

but in principle, this offset is independent of which if the two other strategies are used.

Yes, the following almost gives the expected result.

julia> A = rand(8);

julia> offset = 1
1

julia> R1 = TileIterator(map(r->r[begin:begin+1-offset],  axes(A)), (3, ))
1-element TileIterator{1,Tuple{TiledIteration.CoveredRange{StepRange{Int64,Int64},TiledIteration.LengthAtMost}}}:
 (1:1,)

julia> R2 = TileIterator(map(r->r[begin+offset:end],  axes(A)), (3, ))
3-element TileIterator{1,Tuple{TiledIteration.CoveredRange{StepRange{Int64,Int64},TiledIteration.LengthAtMost}}}:
 (2:4,)
 (5:7,)
 (8:8,)

But the logic seems a little bit more complicated for higher-dimensional cases; I guess recursion is needed. Also, it's not very clear to me where is the best place to add this functionality while still keep tiles lazily calculated. TileIterator(ax, strategy; offset=ntuple(_->0, length(ax)) might probably work but I'm not sure if this can be made performant.