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I am currently trying to solve and understand a problem about packing the maximum of squares into a circle. What I am trying to do, is to calculate the maximum number of variable sized squares (with fixed side) into a circle of variable diameter. For example the result you get at WolframAlpha typing 'pack squares of side 50mm into a circle of diameter 300mm':

Densest known packing:

enter image description here

Furthermore, I would like to be able to add a variable for horizontal and vertical spacing between each square. I have found this to be exactly what I am trying to understand and build.

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Hi! and welcome to Mma.SE! It's not clear to me what you're asking and a more specific question might better attract answers. Is there some part of Mathematica you don't understand? Or would you like someone build such a program for you? Or do you have some other specific question? (In case you haven't seen the, here are some guidelines for asking questions.) –  Michael E2 Aug 17 '13 at 13:30
this relates to the problem in general…. According to you can fit 29 unit squares inside a circle with 6.825 diameter.… wolframalpha says otherwise –  Jonie Aug 17 '13 at 14:28

1 Answer 1

(* i, j, -n ...0 ...n*)
(* calculate a square center with vertical and horizontal spacings*)
center[i_, j_, s_, spx_, spy_] := {i (s + spx), j (s + spy)}
(* calculate the vartices of a square given the center*)
verts[{cx_, cy_}, s_] := {cx, cy} + # s/2 & /@ Tuples[{-1, 1}, 2]
(* calculate all squares to consider, given a circle center*)
(* note: it's overestimating, but it doesn't matter *)
allSquares[cirCenterX_, cirCenterY_, circR_, s_, spx_, spy_] := 
                         N@center[i, j, s, spx, spy] /. Solve[(#.#) &@({cirCenterX, cirCenterY} - 
                                  center[i, j, s, spx, spy]) < (circR + s + spx + spy)^2, {i, j}, 
(* calculate all verttices to consider, given a circle center*) 
allVerts[cirCenterX_, cirCenterY_, circR_, s_, spx_, spy_] := 
                     N@verts[#, s] & /@ allSquares[cirCenterX, cirCenterY, circR, s, spx, spy]
(* test if a square is inside a given circle *)
testVertInsideCirc[cirCenterX_, cirCenterY_, circR_, vert_] := 
                     Norm[{cirCenterX, cirCenterY} - vert] <= circR

(* set a problem*)
s = 10; (* square side *)
r = 50; (* circle radius *)
spx = 2/10; (*horiz spacing*)
spy = 3/10; (*vert spacing*)
(* calculate all possible involved squares for a given problem*)
allvs = allVerts[0, 0, r, s, spx, spy];
(* Solve*)
nm = NMaximize[{Tr[ Boole /@  And @@@ Map[testVertInsideCirc[ccx, ccy, r, #] &, N@allvs, {2}]],
               {-s/2 <= ccx <= s/2, -s/2 <= ccy <= s/2}}, {ccx, ccy},
               Method -> "DifferentialEvolution"]

{59., {ccx -> -3.33338, ccy -> -0.0896481}}

Equivalent problem solved with your linked reference:

Mathematica graphics


Drawing our solution:

With[{cx = ccx /. nm[[2]], cy = ccy /. nm[[2]]}, 
     Graphics[{EdgeForm[Directive[Thick, Black]], 
              Flatten@Transpose@{And @@@ 
                     Map[testVertInsideCirc[cx, cy, r, #] &, N@allvs, {2}] /. 
                                                      {True -> Blue, False -> Yellow}, 
                                    Rectangle @@@ (allvs[[All, {1, 4}]])}, 
              Red, Thick, Circle[{cx, cy}, r], PointSize[Large], Point[{cx, cy}]}]]

Mathematica graphics

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This is awesome! But I have tried playing with the spacing settings a little. As I go smaller, calculations fail. –  Alex Aug 27 '13 at 18:49
@Alex Could you please try it again? I corrected a minor bug. If you still have problems please post the example so I can look for the problem –  belisarius is forth Sep 7 '13 at 19:14
I seem to get calculation errors, when I use s=1,r=150,spx=spy=1/1000 for example. I am not quite sure, but it seems, that the s variable is causing this? –  Alex Sep 19 '13 at 17:07

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