Skip to main content

performance tuning - Fast Spherical Linear Interpolation of list of quaternions


An accurate way to interpolate between two quaternions is to use Spherical Linear Interpolation (Slerp) because it preserves the unit length, whereas straightforward linear interpolation does not, as shown by this example:


Clear[slerp];
slerp[q1_, q2_, f_] := Module[{omega},

omega = ArcCos[Dot[Normalize@q1, Normalize@q2]]];
If[PossibleZeroQ[omega],
q1,
(Sin[(1 - f) omega] q1 + Sin[f omega] q2)/Sin[omega]
]
]

q1 = {1,0,0,0};
q2 = {0,1,0,0};
q3 = {0,0,1,0};


(* Linear interpolation using Interpolation *)
qint = Interpolation[{{0,q1},{1,q2}},InterpolationOrder->1];

Plot[{Norm@qint[t], Norm@slerp[{1, 0, 0, 0}, {0, 1, 0, 0}, t]}, {t, 0, 1},
AxesLabel -> {"time", "length"}, PlotRange -> {0.7, 1.01},
PlotStyle -> {Automatic, Dashed},
Epilog -> {Text["Linear Interpolation", {0.5, 0.75}],
Text["Spherical Linear Interpolation", {0.5, 0.98}]}
]


Mathematica graphics


What's nice thing about Interpolation, however, is that it is easy to give it a list of vectors to get an interpolating function valid over the specified range, and it works fast:


Interpolation[{{0,q1},{1,q2},{2,q3}},InterpolationOrder->1]


InterpolatingFunction[{{0,2}},<>]



How can I create something like an InterpolatingFunction that preserves the unit-length property?


This is what I came up with, but it seems kludgy and it is very slow (this code does not check input bounds):



Clear[slerpInterpolation]; 
slerpInterpolation[q_List] := Function[{t},
Module[{times, quaternions, pos, u, dt},
times = q[[All, 1]];
quaternions = q[[All, 2]];
pos = Last@Flatten@Position[t - times, x_ /; x >= 0];
dt = times[[pos + 1]] - times[[pos]];
u = (t - times[[pos]])/dt;
slerp[quaternions[[pos]], quaternions[[pos + 1]], u]
]

]

It is too slow in practice, giving about 10 evaluations per second on my laptop. Compare that with regular interpolation:


qlist = Transpose[{Range[100000] - 1, Normalize /@ RandomReal[{-1, 1}, {100000, 4}]}];
sint = slerpInterpolation[qlist];
lint = Interpolation[qlist, InterpolationOrder -> 1];

AbsoluteTiming[Do[sint[541.236], {10}]][[1]]/10



0.0967707



AbsoluteTiming[Do[lint[541.236], {10000}]][[1]]/10000


7.2522*10^-6




Answer



Here's a somewhat complete implementation of Shoemake's spherical linear interpolation that functions completely analogously to Interpolation[] and InterpolatingFunction[]. As already noted, much of the slowness is due to your use of a sequential search. In any event, if you prefer, you could use the built-in interpolation as suggested in the other answer, but I think using a bisection routine is a bit more instructive.


Anyway...



SphericalLinearInterpolation::inddp = "The point `1` is duplicated.";

SphericalLinearInterpolation[data_] :=
Module[{dtr = Transpose[SortBy[data, Composition[N, First]]], diffs, times},
SphericalInterpolatingFunction[data[[{1, -1}, 1]], dtr] /;
If[MemberQ[diffs = Chop[Differences[times = First[dtr]]], 0],
Message[SphericalLinearInterpolation::inddp,
First[Extract[times, Position[diffs, 0]]]]; False, True]]

SphericalInterpolatingFunction::dmval =

"Input value `1` lies outside the domain of the interpolating function.";

MakeBoxes[SphericalInterpolatingFunction[range_, rest__], _] ^:=
InterpretationBox[RowBox[
{"SphericalInterpolatingFunction", "[", "{", #1, ",", #2, "}", ",", "\"<>\"", "]"}],
SphericalInterpolatingFunction[range, rest]] & @@ Map[ToBoxes, range]

SphericalInterpolatingFunction[stuff__][l_List] :=
SphericalInterpolatingFunction[stuff] /@ l


SphericalInterpolatingFunction[range_, __]["Domain"] := range

SphericalInterpolatingFunction[{r_, s_}, data_][t_?NumericQ] :=
(Message[SphericalInterpolatingFunction::dmval, t]; $Failed) /; ! (r <= t <= s)

slerp = Compile[{{q1, _Real, 1}, {q2, _Real, 1}, {f, _Real}},
Module[{n1 = Norm[q1], n2 = Norm[q2], omega, so},
(* vector angle formula by Velvel Kahan *)
omega = 2 ArcTan[Norm[q1 n2 + n1 q2], Norm[q1 n2 - n1 q2]];
If[Chop[so = Sin[omega]] == 0, q1, Sin[{1 - f, f} omega].{q1, q2}/so]]];


SphericalInterpolatingFunction[range_, data_][t_?NumericQ] := Module[{times, quats, k},
{times, quats} = data;
k = GeometricFunctions`BinarySearch[times, t];
slerp[quats[[k]], quats[[k + 1]], Rescale[t, times[[{k, k + 1}]], {0, 1}]]]



See J. Blow's article on why spherical linear interpolation might not always be the best method for interpolating quaternions.


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

Is there a way to do conditional matrix loop using 'continue'

I have the following: n = 3; m = 5; ww = RandomReal[{0, 0.1}, {n, n}]; uu = RandomReal[{0, 1}, {m, n}]; pp = RandomReal[{0, 1}, {n, n}]; ss = RandomInteger[{0, 5}, {m, n}]; Grid[{{"ww", "uu", "pp", "ss"}, {ww // TableForm, uu // TableForm, pp // TableForm, ss // TableForm}}, Spacings -> {5, 2}, Dividers -> All] where I would like to look at every element of matrix ss and produce a matrix tt , with zeroes at the locations in ss which have zeroes, and in all other positions do the following: tt = (-1/Subscript[ww, m]) Log[(1 - uu)/(Subscript[pp, m - 1])], where Subscript[ww, m] is the value at index of ww matrix and where Subscript[pp, m - 1] is the value at index-1 of pp matrix. So for example if the first value ever read from matrix ss happens to be 2, then value taken from matrix ww would be from the row 2, but from pp would be from row 1. Also how to tell difference between a 0 as a valid value from within the matrix elemen...