How can I draw a polygon from a set of angles?

Question

In recreational mathematics, polytans are polygons formed by edge-connecting isosceles right triangles. Order-n polytans are those constructed from n such triangles. My question is this:

Given a set of n angles and an N x N grid, how do I draw a graph using a set of angles? I am trying to find if the angles form the perimeter of a polytan (i.e., given a starting point on the plane, does progressively iterating through the angles get me back to the point of origin?). For example, given these three sets: {{270, 315, 315}, {315, 270, 315}, {315, 315, 270}}, the output would show two isosceles right triangles: {270, 315, 315} and {315, 315, 270}, and one non-polygon: {315, 270, 315}, as seen in the image below. In the image, the initial reference line is {{0,0}->{0,1} (on the y-axis), and the point of origin is {0,1}.

I need the constructions to be graphs so that I can check whether two polytans of the same order are identical, but just in different orientations (like the two triangles in the image). Vertex count, vertex degree, edge-connectedness, etc., would also be useful.

Answer 1

I think you can regard each one of your edges/vertices as a complex number, so walk along the path given an angle set will be equivalent to adding the corresponding number set successively. And if the total is $0$, it means the path is closed, thus a polytan.

This is a possible implementation:

Clear[polytanPlot]
polytanPlot[angleSet_] :=
 Module[{stepVecSet},

   (* step set represented as complex numbers: *)

  stepVecSet = 
   If[IntegerQ[#/(π/2)], 1, Sqrt[2]] Exp[# I] I & /@ 
    Mod[Accumulate[angleSet - π], 2 π];

   (* closed path criterion: *)

  If[Chop[N[Total[stepVecSet]]] != 0,
   Style[Row[{Style[angleSet, 13],  " is not a valid polytan angle set!"}], 15, Red],

   (* draw the polytan: *)

   pts = Through[{Re, Im}@Accumulate[stepVecSet]]\[Transpose];
   Graphics[{FaceForm[], EdgeForm[{Thick, Blue}],
     Polygon[pts],
     Darker[Green], AbsolutePointSize[15], Point[pts]},
    Frame -> True, PlotLabel -> Style[angleSet, 13]]
   ]]

polytanPlot /@ FullSimplify[
   {{270, 315, 315}, {315, 270, 315}, {315, 315, 270}} π/180
   ] // Column

Cases[polytanPlot /@ RandomChoice[π/4 Range[7], {1000, 10}],  Graphics[__], ∞]

Edit:

For the test data set asked by OP in the comment:

testSets = Select[Tuples[{0,45,90,135,180,225,270,315}, 5], Total[#] == (5 + 2)*180 &];
Cases[polytanPlot /@ FullSimplify[testSets π/180], Graphics[__], ∞]

Note the interior angles at some vertices are $0$, so adjacent edges there get overlapped. Also some edges intersect each other at some non-vertex point (e.g. in the last graph above), makes it look like triangles with length $1$ hypotenuse, although they are merely self-intersected polytans.

If we want to avoid polytans with overlapped edges or with sub-polytans in it, we can test the closed path criterion on every continuous sub-set of the input angle set. This modified version might decrease the number of self-intersected result too, if not totally prevent.

Clear[polytanPlot2]
polytanPlot2[angleSet_] :=
 Module[{stepVecSet, len, nosubCircuitQ, pts},
  stepVecSet = If[IntegerQ[#/(π/2)], 1, Sqrt[2]] Exp[# I] I & /@
    Mod[Accumulate[angleSet - π], 2 π];
  len = Length[stepVecSet];
  nosubCircuitQ = True;
  Table[
   If[#, nosubCircuitQ = False; Goto["label"]] &[
    Chop[N[Total[stepVecSet[[i ;; j]]]]] == 0],
   {i, len - 1}, {j, i + 1, If[i == 1, len - 1, len]}];
  Label["label"];
  If[nosubCircuitQ && Chop[N[Total[stepVecSet]]] == 0,
   pts = Through[{Re, Im}@Accumulate[stepVecSet]]\[Transpose];
   Graphics[{FaceForm[], EdgeForm[{Thick, Blue}],
     Polygon[pts],
     Darker[Green], AbsolutePointSize[15], Point[pts]},
    Frame -> True, PlotLabel -> Style[angleSet, 13]],
   Style[Row[{Style[angleSet, 13], 
      " is not a valid polytan angle set!"}], 15, Red]
   ]]

Cases[polytanPlot2 /@ FullSimplify[testSets π/180], Graphics[__], ∞]

Self-intersected results are still found:

Cases[polytanPlot /@ RandomChoice[π/4 Range[7], {2000, 15}],  Graphics[__], ∞]

Edit 2:

According to J.M.'s comment, we can use an undocumented function Graphics`Mesh`SimplePolygonQ[] to achieve the purpose of avoiding both overlappings and self-intersections:

Clear[polytanPlot3]
polytanPlot3[angleSet_] :=
 Module[{stepVecSet, pts, polyg},
  stepVecSet = If[IntegerQ[#/(π/2)], 1, Sqrt[2]] Exp[# I] I & /@
    Mod[Accumulate[angleSet - π], 2 π];

  If[Chop[N[Total[stepVecSet]]] != 0,

   Style[Row[{Style[angleSet, 13], " is not a valid polytan angle set!"}], 15, Red],

   pts = Through[{Re, Im}@Accumulate[stepVecSet]]\[Transpose];
   polyg = Polygon[pts];
   If[Graphics`Mesh`SimplePolygonQ[polyg],

    Graphics[{FaceForm[], EdgeForm[{Thick, Blue}], polyg,
      Darker[Green], AbsolutePointSize[15], Point[pts]
         }, Frame -> True, PlotLabel -> Style[angleSet, 13]],

    Style[Row[{Style[angleSet, 13], " is not a simple polytan angle set!"}], 15, Red]
    ]
   ]]

Cases[polytanPlot3 /@ RandomChoice[π/4 Range[7], {2000, 15}],  Graphics[__], ∞]