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I have the following points:

aa={{238.5, 394.5}, {195.5, 441.5}, {219.5, 397.5}, {216.5, 
 398.5}, {246.5, 397.5}, {265.5, 476.5}, {275.5, 450.5}, {273.5, 
 435.5}, {274.5, 461.5}, {212.5, 447.5}, {221.5, 457.5}}

Now, I want to draw a closed curve from those line. So I used the following code:

Graphics[{{Blue, Point[aa]}, JoinedCurve[Line[aa], CurveClosed -> True]}]

But, it gives me curve like this:

enter image description here

I want the curve like this way:

enter image description here

Please let me know, how to do it. And, I want it to do automatically, not point by point. Please let me know. Thanks.

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3 Answers 3

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Note that the above method doesn't always work. Here is an alternative:

aa = RandomReal[{-1, 1}, {20, 2}];
Graphics[Line[aa[[Last[FindShortestTour[aa]]]]]]

enter image description here

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    $\begingroup$ @Dr.belisarius Ah, didn't mean to sound competitive - just meant as an alternative! I am always learning from your answers anyway +1 :) $\endgroup$
    – martin
    Commented Mar 7, 2016 at 22:36
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    $\begingroup$ Don't worry, I was joking $\endgroup$
    – Dr. belisarius
    Commented Mar 7, 2016 at 22:39
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    $\begingroup$ it looks like a cute cat😃 $\endgroup$
    – partida
    Commented May 9, 2016 at 13:48
  • $\begingroup$ This also guarantees the curve is closed which is a plus $\endgroup$
    – VivaLaPanda
    Commented May 9, 2018 at 17:48
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curve = FindCurvePath[aa]
ListLinePlot[aa[[curve[[1]]]], AspectRatio -> Automatic]

Mathematica graphics

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4
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Why not directly use ListCurvePathPlot?

ListCurvePathPlot@aa

enter image description here

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  • $\begingroup$ It appears that ListCurvePathPlot doesn't guarantee that the curve is closed. $\endgroup$
    – VivaLaPanda
    Commented May 8, 2018 at 4:03
  • $\begingroup$ @AdrianSmith Yeah, actually it's mentioned in the Details and Options of the document: ListCurvePathPlot can generate disconnected curves in certain cases. But it at least works for OP's case. $\endgroup$
    – xzczd ♦
    Commented May 8, 2018 at 5:26

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