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In Mathematica we have ListSurfacePlot3D for reconstructing surface from list of points in 3D. But I can not find something similar if I want to reconstruct a curve instead of a surface. For example if I have this list of points:

{{0.`,0.`,-1.`},{0.06120871905481365`,0.2397127693021015`,-0.9689124217106447`},{0.22984884706593015`,0.42073549240394825`,-0.8775825618903728`},{0.4646313991661485`,0.49874749330202717`,-0.7316888688738209`},{0.7080734182735712`,0.4546487134128408`,-0.5403023058681397`},{0.9005718077734668`,0.2992360720519783`,-0.3153223623952687`},{0.9949962483002227`,0.0705600040299336`,-0.0707372016677029`},{0.9682283436453982`,-0.17539161384480992`,0.17824605564949209`},{0.826821810431806`,-0.37840124765396416`,0.4161468365471424`},{0.6053978997153898`,-0.48876505883254856`,0.6281736227227391`},{0.35816890726838696`,-0.4794621373315693`,0.8011436155469337`},{0.14566511285437003`,-0.35277016278519596`,0.9243023786324636`},{0.01991485667481699`,-0.13970774909946293`,0.9899924966004454`},{0.011706187135988248`,0.10755999404390776`,0.9941296760805463`},{0.12304887282834767`,0.32849329935939453`,0.9364566872907963`},{0.32668234108248706`,0.4689999883873694`,0.8205593573395608`},{0.5727500169043067`,0.4946791233116909`,0.6536436208636119`},{0.8010059513424118`,0.39924355631174513`,0.4460874899137928`},{0.9555651309423384`,0.20605924262087827`,0.2107957994307797`},{0.9985860780981893`,-0.03757556023090465`,-0.03760215288797655`},{0.9195357645382262`,-0.2720105554446849`,-0.28366218546322625`},{0.7377684639979962`,-0.4398478799858351`,-0.5120854772418407`},{0.49778715100597465`,-0.49999510327535174`,-0.70866977429126`},{0.258347620623497`,-0.4377260873442142`,-0.8611924171615208`},{0.07807302063375395`,-0.26828645900021747`,-0.960170286650366`},{0.`,0.`,-1.`}}

Notice that the curve should be a closed loop as the first and last points are the same. I want to find a smooth curve that passes exactly through these points. Something like this:

enter image description here

Any suggestion how to do it?

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    $\begingroup$ Interpolate with option PeriodicInterpolation->True. $\endgroup$
    – Henrik Schumacher
    Jul 12, 2020 at 12:54

1 Answer 1

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Following Henrik's suggestion, how about

 ff = Interpolation[#, PeriodicInterpolation -> True] & /@ Transpose[dat];
{ListPointPlot3D[dat, PlotStyle -> Orange],
ParametricPlot3D[#[t] & /@ ff//Evaluate, {t, 1, Length[dat]}]}//
  Show[#,BoxRatios -> {1, 1, 1}, Axes -> None] &

enter image description here

Note that following the OP's request, you can encapsulate this into a function:

ListCurvePlot3D[dat_] := 
 Module[{t, ff = 
    Interpolation[#, PeriodicInterpolation -> True] & /@ 
     Transpose[dat]},
  {ListPointPlot3D[dat, PlotStyle -> Orange], 
    ParametricPlot3D[#[t] & /@ ff // Evaluate, {t, 1, Length[dat]}]} //
    Show]
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  • $\begingroup$ OP mentions I want to find a smooth curve, but this has corners. Perhaps a BSplineFunction would be more appropriate? $\endgroup$
    – flinty
    Jul 12, 2020 at 12:56
  • $\begingroup$ There is no interpolation. The curve is not smooth. $\endgroup$
    – azerbajdzan
    Jul 12, 2020 at 12:57
  • $\begingroup$ What is the purpose of PeriodicInterpolation -> True when it works exactly same without it? $\endgroup$
    – azerbajdzan
    Jul 12, 2020 at 13:15
  • $\begingroup$ @azerbajdzan i/ I tend to listen to what Henrik says because he is very smart. ii/ I guess you requested the path to be periodic: the documentation states that if this option is used you can evaluate t at values higher than Length[dat] and remain on the curve. I assume that it means internally it is using periodic functions to do the interpolation. For instance {t, -5, Length[dat] + 5} $\endgroup$
    – chris
    Jul 12, 2020 at 13:21
  • $\begingroup$ I see... Now I wonder why they did not make one command like they did for surface. If we have ListSurfacePlot3D we could also have ListCurvePlot3D. Thanks. $\endgroup$
    – azerbajdzan
    Jul 12, 2020 at 13:28

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