# InterplationOrder, ListPlot3D and using AbsoluteOption to find value used. Slow using InterpolationOrder

What lead to this, is when I noticed that ListPlot3D was much slower when adding a specific InterpolationOrder->n vs. not and letting the default take care of it.

To find out why it is so much slower (timing is below), I asked Mathematica to tell me what default InterpolationOrder it used for the current plot. So I used the command

      AbsoluteOptions[p, InterpolationOrder]


To do that, which supposed to return the value used in p. From the excellent book Mathematica Navigator:

But Mathematica gave an error and said that InterpolationOrder is not a known option for this plot. But http://reference.wolfram.com/language/ref/ListPlot3D.html shows it there.

So my question is, how does one find what InterpolationOrder is used for current plot of ListPlot3D?

My sub question is actually (the reason why I wanted to find the above), is why ListPlot3D slows down so much when specifying this option vs. not? Here is the MWE

Clear[x, y, z];
nElem = 10; h = 1/(nElem - 1);
grid = N@Table[{i*h, j*h}, {i, -10, 10, h}, {j, -10, 10, h}];
f[x_, y_] := Sin[x*10 y] Exp[-x y];
force = Map[f[#[[1]], #[[2]]] &, grid, {2}];
p = ListPlot3D[force, InterpolationOrder -> 1, AxesLabel -> {x, y, z}]


 AbsoluteOptions[p, InterpolationOrder]
(*error*)

Timing[ListPlot3D[force, InterpolationOrder -> 1, AxesLabel -> {x, y, z}]]
(*5.647236*)

Timing[ListPlot3D[force, AxesLabel -> {x, y, z}]]
(* 0.624004 *)


And as an extra reward, if you can answer how can one use InterpolationOrder -> 1 and still have fast plot, I will up-vote you 2 times if I can.

Version 10.0 on windows 7.

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Do I understand that your question can be summarized as: Why is using InterpolationOrder -> 1 slower than using the default option InterpolationOrder -> None? Is it really surprising that using interpolation is slower? –  Mr.Wizard Aug 28 '14 at 3:01
@Mr.Wizard Yes, but this is one part of it. The other part is why I get an error when query this option? I did not want to split this to 2 questions. But may be I should have. What do you think? Should they be separate? I can do that. ps. the slow down is really large as can be seen by the numbers. A bit of slow down ok, but 10 times slower? –  Nasser Aug 28 '14 at 3:03

The reason that AbsoluteOptions[p, InterpolationOrder] doesn't work is that p is a Graphics object and InterpolationOrder is an option for (e.g.) ListPlot3D, not Graphics. InterpolationOrder guides the creation of the graphic but it does not remain a mutable part of it. I addressed this topic here:

As to the second question we can easily show that using interpolation plots many more points that if we do not, so it is not surprising that this takes considerably longer:

ListPlot3D[force, AxesLabel -> {x, y, z}][[1, 1]] // Length
ListPlot3D[force, AxesLabel -> {x, y, z}, InterpolationOrder -> 1][[1, 1]] // Length

41473

325306

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I see. So none of options such as PlotStyle used can be queried using AbsoluteOptions. Just tried it on normal plot: p0 = Plot[Sin[x], {x, -1, 1}, PlotStyle -> Red]; AbsoluteOptions[p0, PlotStyle] and this also gave an error. This makes AbsoluteOptions not as useful as I thought before. May be this is what the book means by "special options" in the screen shot I posted. Thanks. –  Nasser Aug 28 '14 at 3:16

It is a curious thing, the default appears to do a linear interpolation ( ie. rendering more points than input ), yet specifying InterpolationOrder->1 dramatically increases the number of interpolation points:

simple example:

 data = Table[ Sin[x] , {x, 0, 10}, {y, 0, 10}] // N;
ListPlot3D[data]


 in = ListPlot[ data[[All, 1]] , PlotStyle -> {PointSize[.02], Red}];
GraphicsRow[Show[{ListPlot[#[[2 ;; 3]] & /@
Select[  Cases[ ListPlot3D[data, InterpolationOrder -> #],
x_List /; Length[x] > 4 , Infinity][[1]] ,
#[[1]] == 1 & ] ], in}] & /@ {None, 1, 2}]


note the number of points for Interpolation->1 is the same as for ->2 (and higher)

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