I am trying to create a Monte Carlo simulation for a physical situation. The situation is a layer of disks are lying on a plane. The disks have a known mean radius and standard deviation, and a known mean center-to-center distance (measured compared to it's nearest $n$ neighbors) and standard deviation. Both follow a normal distribution.

I would like to create a simulation that, given the mean values and standard deviations produces a list of disk radii and center locations that match the statistics. The disks should not overlap and fit within a given boundary.

  • $\begingroup$ What have you tried? $\endgroup$ Oct 4, 2017 at 16:52
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    $\begingroup$ Take the centers, two at a time, and check that the sum of the associated two radii is less than the separation between the centers. Note: You cannot have true normal distribution under your constraints. $\endgroup$ Oct 4, 2017 at 16:58
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    $\begingroup$ start here. mathematica.stackexchange.com/questions/126486/…. I guess this is not quite a dup due to the matching statistics part, but not really on topic with that either. $\endgroup$
    – george2079
    Oct 4, 2017 at 17:14
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    $\begingroup$ your target packing density is a big factor in this problem. If you are going after a fairly dense packing the problem becomes nearly intractable. At low density you can readily pack with your target size and mean spacing statistics. Capturing the spacing distribution can be done but it is a challenge. $\endgroup$
    – george2079
    Oct 4, 2017 at 17:34
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    $\begingroup$ Could you provide some explicit data to play with? Also you could try incorporating RegionDisjoint. See the 'Neat Example' in the documentation here. $\endgroup$
    – Greg Hurst
    Oct 6, 2017 at 17:42

1 Answer 1


To solve the problem I created some function to help visualize the placement and statistics of the disks. For each functions disks is a list with elements {radius, {xCenter, yCenter}} and boundary is a vector {width, height} describing the bounding area, which is centered at the origin.

areaCoverage calculates the percent of the enclosed area covered by particles.

raiusStats calculates relevant statistics of the disk radii.

visualizeLayer creates a graphic with the disks and bounding area, and gives a list of any disks that intersect each other or the boundary.

areaCoverage[disks_, boundary_] := Module[
   {diskArea, boundaryArea},
   diskArea = Pi*Total[disks[[All, 1]]^2];
   boundaryArea = Times @@ boundary;


radiusStats[disks_] := Module[
   radii = disks[[All, 1]] ;

   N[ {Mean[radii], StandardDeviation[radii] } ]

visualizeLayer[disks_, boundary_] := Module[
   {d, b, pairs, intersecting, bInt},
   b = Rectangle[ -boundary/2, boundary/2 ];
   d = Table[Disk[disk[[2]], disk[[1]] ], {disk, disks}];
   pairs = Subsets[d, {2} ];
   pairs = Table[With[{
      intersect = (RegionMeasure@RegionIntersection@pair != 0)
     {pair, intersect}
   {pair, pairs}
  intersecting = Select[pairs, #[[2]] == True &][[All, 1]];

  bInt = Table[With[{
      intersect = ! (RegionUnion@{b, disk} === b)
     {disk, intersect}
    {disk, d}
  bInt = Select[bInt, #[[2]] == True &][[All, 1]];

 {Show[Graphics[ {FaceForm[], EdgeForm[Black], b} ], d // Graphics, 
   ImageSize -> 500], intersecting, bInt}

I started with a list of three disks, adding new ones one-by-one playing with the parameters and placement to match the statistics I wanted.

While not automated, these functions certainly helped create scenarios more quickly than would otherwise be possible. The visualization of the particles made adding new ones quite easy, and most of the time was spent making small adjustments to match the statistics to given values.

Here is a minimum set of data:

boundary = {200, 200};

    { 10, {0, 0} },
    { 10, { 50, 50 } },
    { 10, {-50, -50} },
    { 20, {-40, -50} },
    { 30, {-80, 0} } 
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    $\begingroup$ I think what people would need to work this problem is an example of the statistics you are after. $\endgroup$
    – george2079
    Oct 6, 2017 at 20:49
  • $\begingroup$ @george2079 that's a good point. I was mentioned in the question that I was interested in the distribution of the particle radii, and the center-to-center distance between the n nearest neighbors. The nearest neighbors calculation ended up being overkill, so I switched to coverage percentage instead. $\endgroup$
    – bicarlsen
    Oct 7, 2017 at 20:37

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