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While I was trying to solve this problem, I noticed a strange result of RegionDistance. The following is a contour plot of the distance from the tilted square. Is it a bug? Should I avoid using Translate to specify a region? It seems Rotate is OK in this case.

I'm using "14.1.0 for Mac OS X ARM (64-bit) (July 16, 2024)".

Clear["Global`*"];
L = 10;
square = Rotate[Translate[Rectangle[-{L,L}/2, {L,L}/2], {3, -3}], 0.3];

dist = RegionDistance[square];

Show[ContourPlot[dist[{x, y}], {x, -L, 2 L}, {y, -2 L, L}, 
  Exclusions -> None], Graphics[{LightYellow, square}]]

square

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2
  • 1
    $\begingroup$ I am using 14.1 Mac OS Sonoma and see the same thing. Interestingly if you increase the rotation the difference gets even bigger up to a maximum at Pi $\endgroup$
    – Dunlop
    Commented Nov 23 at 5:21
  • $\begingroup$ Not a bug since rotate not always get a proper region ( for later version maybe fully support all the graphics object be an region) $\endgroup$
    – cvgmt
    Commented Nov 23 at 5:39

3 Answers 3

Reset to default
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Edit

Finally we found that add the center {0, 0} in the Rotate do the job.

Clear["Global`*"];
L = 10;
square = 
  Rotate[Translate[Rectangle[-{L, L}/2, {L, L}/2], {3, -3}], 
   0.3, {0, 0} (*add here *)];

dist = RegionDistance[square];

Show[ContourPlot[dist[{x, y}], {x, -L, 2 L}, {y, -2 L, L}, 
  Exclusions -> None], Graphics[{LightYellow, square}]]

enter image description here

Edit

  • Compare with the five results, I confirm that upto 14.1 version, it is recommend that not use the Rotate as the last transformation.
Clear["Global`*"];
L = 10;
rectangle = Rectangle[-{L, L}/2, {L, L}/2];
square = Rotate[Translate[rectangle, {3, -3}], 0.3];
square2 = 
  GeometricTransformation[rectangle, 
   RotationTransform[.3]@*TranslationTransform[{3, -3}]];
square3 = 
  GeometricTransformation[rectangle, 
   TranslationTransform[RotationMatrix[.3] . {3, -3}]@*
    RotationTransform[.3]];
square4 = 
  TransformedRegion[rectangle, 
   RotationTransform[.3]@*TranslationTransform[{3, -3}]];
square5 = 
  TransformedRegion[rectangle, 
   RotationTransform[.3] . TranslationTransform[{3, -3}]];
dist = RegionDistance[square];
dist2 = RegionDistance[square2];
dist3 = RegionDistance[square3];
dist4 = RegionDistance[square4];
dist5 = RegionDistance[square5];
{ContourPlot[dist[{x, y}], {x, -L, 2 L}, {y, -2 L, L}, 
  Exclusions -> None, Epilog -> {LightYellow, square}],
 ContourPlot[dist2[{x, y}], {x, -L, 2 L}, {y, -2 L, L}, 
  Exclusions -> None, Epilog -> {LightYellow, square2}],
 ContourPlot[dist3[{x, y}], {x, -L, 2 L}, {y, -2 L, L}, 
  Exclusions -> None, Epilog -> {LightYellow, square3}], 
 ContourPlot[dist4[{x, y}], {x, -L, 2 L}, {y, -2 L, L}, 
  Exclusions -> None, Epilog -> {LightYellow, square4}], 
 ContourPlot[dist5[{x, y}], {x, -L, 2 L}, {y, -2 L, L}, 
  Exclusions -> None, Epilog -> {LightYellow, square5}]}

enter image description here

Original

  • Rotate at first.
Clear[square];
square = 
  Translate[Rotate[Rectangle[-{L, L}/2, {L, L}/2], .3], {3, -3}];
RegionQ[square]

enter image description here

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5
  • $\begingroup$ Thank you. I will Rotate first for such cases. @Syed's answer suggests that TransformedRegion works as I expected (i.e. translate first, and then rotate). I wonder if Rotate is designed for Graphic Primitives but not for Regions --- as you commented.... $\endgroup$
    – A. Kato
    Commented Nov 23 at 6:28
  • $\begingroup$ Very nice! Could you explain why just adding the center in the Rotate solves this problem? $\endgroup$
    – A. Kato
    Commented Nov 23 at 13:05
  • 1
    $\begingroup$ @A.Kato I still do not know why, but I found that Clear[a, b, t, square, squareX, dist, distX]; {a, b} = {4, 5}; t = .5; square = Rotate[Translate[rectangle, {a, b}], t, {a, b}]; squareX = Translate[Rotate[rectangle, t], {a, b}]; dist = RegionDistance[square]; distX = RegionDistance[squareX]; {ContourPlot[dist[{x, y}], {x, -L, 2 L}, {y, -2 L, L}, Exclusions -> None, Epilog -> {LightYellow, square}], ContourPlot[distX[{x, y}], {x, -L, 2 L}, {y, -2 L, L}, Exclusions -> None, Epilog -> {LightYellow, squareX}]} are the same. $\endgroup$
    – cvgmt
    Commented Nov 23 at 13:48
  • $\begingroup$ Thank you for testing various codes. I ran your code (in your comment) with rectangle = Rectangle[{0, 0}, {10, 1}]. Then the result was different. It seems that RegionDistance ignores/forgets the fact that Rotate[g,t] will rotate graphics around the center of g, not the origin {0,0} of the coordinate system. $\endgroup$
    – A. Kato
    Commented Nov 24 at 7:31
  • 1
    $\begingroup$ @A.Kato Thanks, you are right, we still need to add {0,0} in Rotate. L = 10; {a, b} = {4, 5}; t = .5; rectangle = Rectangle[{0, 0}, {10, 1}]; square = Rotate[Translate[rectangle, {a, b}], t, {a, b}]; squareX = Translate[Rotate[rectangle, t, {0, 0}], {a, b}]; dist = RegionDistance[square]; distX = RegionDistance[squareX]; {ContourPlot[ dist[{x, y}], {x, -L, 2 L}, {y, -2 L, L}, Exclusions -> None, Epilog -> {LightYellow, square}], ContourPlot[distX[{x, y}], {x, -L, 2 L}, {y, -2 L, L}, Exclusions -> None, Epilog -> {LightYellow, squareX}]} $\endgroup$
    – cvgmt
    Commented Nov 24 at 8:07
3
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I am using v12.2.0.

Clear["Global`*"];
L = 10;
square = TransformedRegion[Rectangle[-{L, L}/2, {L, L}/2], 
   RotationTransform[0.3] . TranslationTransform[{3, -3}]];
dist = RegionDistance[square];
Show[ContourPlot[dist[{x, y}], {x, -L, 2 L}, {y, -2 L, L}, 
  Exclusions -> None], Graphics[{LightYellow, square}]]

RegionDistance to transformed region

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4
  • $\begingroup$ Thank you very much. I learned that TransformdRegion is the appropriate tool here. Just to make sure, TransformdRegion[reg, A.B] applies first B, and then A. Am I right? $\endgroup$
    – A. Kato
    Commented Nov 23 at 6:18
  • $\begingroup$ I can't say for sure. I deleted my previous comment as translation is not linear. $\endgroup$
    – Syed
    Commented Nov 23 at 7:21
  • 1
    $\begingroup$ @A.Kato RotationTransform[0.3] . TranslationTransform[{3, -3}] == AffineTransform[{RotationMatrix[.3], RotationMatrix[.3] . {3, -3}}] , it means that RotationTransform[0.3] . TranslationTransform[{3, -3}] is also a AffineTransform $\endgroup$
    – cvgmt
    Commented Nov 23 at 12:11
  • $\begingroup$ @cvgmt, Thanks for the explanation and effort. $\endgroup$
    – Syed
    Commented Nov 25 at 6:57
2
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An attention should be payed on what the output of Rotate is. Rotate acts immediately on the expression no matter what the expression is.

So I recommend to use GeometricTransformation with RotationMatrix or RotationTransform for a serious manipulation of geometric objects.

Rotate[Translate[Rectangle[-{L, L}/2, {L, L}/2], {3, -3}], 0.3]
Translate[Rotate[Rectangle[-{L, L}/2, {L, L}/2], 0.3], {3, -3}]

enter image description here

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