Simplify 2D polyhedral Regions like Interval

Question

I don't think there's a built in "RegionSimplify" for polyhedral Regions and derived akin to how Interval automatically simplifies its argument expression:

Interval[{0, 1}, {1, 2}]

and

Interval[{0, 1}]~IntervalUnion~Interval[{1, 2}]

yield

Interval[{0, 2}]

In contrast, RegionUnion remains unevaluated, even in 2D:

Line[{{0, 0}, {1, 0}}]~ RegionUnion ~ Line[{{1, 0}, {2, 0}}]

(* RegionUnion[Line[{{0, 0}, {1, 0}}], Line[{{1, 0}, {2, 0}}]] *) 

From

Rectangle[{0, 0}, {1, 1}] ~ RegionUnion ~ 
  Rectangle[{1, 0}, {2, 1}] // RegionBoundary // DiscretizeRegion

As expected, the Rectangles' common segment is not part of the output.

How to expose or compute the ideal normal form for 2D (at least) polyhedral Regions like these basic examples:

Overlapping lines:

Line[{{0, 0}, {2, 0}}]~ RegionUnion ~ Line[{{1, 0}, {3, 0}}] (* // RegionSimplify *)

(*  Line[{{0,0},{3,0}}] *)

Union of rectangles:

 Rectangle[{0, 0}, {1, 1}] ~ RegionUnion ~ 
      Rectangle[{1, 0}, {2, 1}] (* // RegionSimplify *)

(* Rectangle[{0, 0}, {2, 1}] *)

Boundary:

 Rectangle[{0, 0}, {1, 1}] ~ RegionUnion ~ 
      Rectangle[{1, 0}, {2, 1}] // RegionBoundary  (* // RegionSimplify *)

(* Line[{{0,0},{2,0},{2,1},{0,1},{0,0}}]  *)

The simplification should work for all 2D polyhedral regions including cones (which btw the current language does not support correctly):

ConicHullRegion[{{0, 0}}, {{1, 0}, {1, 1}}] ~RegionUnion ~ 
 ConicHullRegion[{{0, 0}}, {{0, 1}, {1, 1}}] (* // RegionSimplify *)

(* ConicHullRegion[{{0, 0}}, {{1, 0}, {0, 1}}] *)

For convex regions and convex-preserving operations (eg intersection) should suffice to consider the extreme set (vertices and rays), but in general regions aren't convex (and union doens't preserve it).

(Note: RegionIntersection of the same Rectangles hangs the kernel w/ msg: Last::nolast: "{} has a length of zero and no last element")

Answer 1

eps = 0.001;
repl = (z_Symbol == val_) :> 
   (val - eps/2 <= z <= val + eps/2);

r1 = Line[{{0, 0}, {1, 0}}];
r2 = Line[{{1, 0}, {2, 0}}];

r3 = ImplicitRegion[RegionMember[
     RegionUnion[r1, r2], {x, y}] //
    Simplify, {x, y}] /. repl

ImplicitRegion[(x | y) \[Element] Reals && -0.0005 <= y <= 0.0005 && 
  0 <= x <= 2, {x, y}]

r3 // DiscretizeRegion

r4 = Line[{{0, 0}, {2, 0}}];
r5 = Line[{{1, 0}, {3, 0}}];

r6 = ImplicitRegion[RegionMember[
     RegionUnion[r4, r5], {x, y}] //
    Simplify, {x, y}] /. repl

ImplicitRegion[(x | y) \[Element] Reals && -0.0005 <= y <= 0.0005 && 
  0 <= x <= 3, {x, y}]

r6 // DiscretizeRegion

r7 = Rectangle[{0, 0}, {1, 1}];
r8 = Rectangle[{1, 0}, {2, 1}];
r9 = RegionUnion[r7, r8];

RegionMember[r9, {x, y}] // Simplify

(x | y) \[Element] Reals && 0 <= x <= 2 && 0 <= y <= 1

r9 // DiscretizeRegion

r10 = Rectangle[{0, 0}, {1, 1}];
r11 = Rectangle[{1, 0}, {2, 1}];
r12 = RegionUnion[r10, r11];

RegionMember[r12, {x, y}] // Simplify

(x | y) \[Element] Reals && 0 <= x <= 2 && 0 <= y <= 1

r12 // DiscretizeRegion

n = 5;
rectangles = 
  Rectangle @@@ 
   Transpose[{RandomInteger[{0, 10}, {n, 2}], 
     RandomInteger[{11, 20}, {n, 2}]}];

Graphics[{Opacity[.25], rectangles}]

r13 = RegionUnion @@ rectangles;

RegionMember[r13, {x, y}]

(x | y) \[Element] 
  Reals && ((2 <= x <= 13 && 3 <= y <= 14) || (2 <= x <= 14 && 
     4 <= y <= 11) || (3 <= x <= 16 && 
     8 <= y <= 19) || (4 <= x <= 13 && 
     9 <= y <= 13) || (5 <= x <= 17 && 8 <= y <= 13))

r13 // DiscretizeRegion

EDIT: Example of disconnected regions

r14 = Rectangle[{0, 0}, {1, 1}];
r15 = Rectangle[{1.1, 1.1}, {1.6, 1.7}];
r16 = RegionUnion[r14, r15];

RegionMember[r16, {x, y}] // Simplify

(x | y) \[Element] 
  Reals && ((0 <= x <= 1 && 0 <= y <= 1) || (1.1 <= x <= 1.6 && 
     1.1 <= y <= 1.7))

r16 // DiscretizeRegion

SECOND EDIT: Additional example--regions with holes.

r1 = Rectangle[{0, 0}, {1, 1}];
r2 = Rectangle[{.25, .25}, {.5, .5}];
r3 = RegionDifference[r1, r2];
r4 = Disk[{1.5, 1.25}, .75];
r5 = Rectangle[{1.25, 1.25}, {1.75, 1.35}];
r6 = RegionDifference[r4, r5];
r7 = RegionUnion[r3, r6];

RegionMember[r7, {x, y}]

(x | y) [Element] Reals && (((-1.5 + x)^2 + (-1.25 + y)^2 <= 0.5625 && ! (1.25 <= x <= 1.75 && 1.25 <= y <= 1.35)) || (0 <= x <= 1 && 0 <= y <= 1 && ! (0.25 <= x <= 0.5 && 0.25 <= y <= 0.5)))

DiscretizeRegion[r7]

Answer 2

(version 10) I made function lastUnion , firstUnion and RectangleRegionSimplify instead of RegionSimplify. I think you might want to make RectangleRegionSimplify.

lastUnion[coo_] := 
 Flatten[Replace[SplitBy [ coo, Last], {f_, ___, l_} -> {f, l}, 1], 1]
firstUnion[coo_] := 
 Flatten[Replace[SplitBy [ coo, First], {f_, ___, l_} -> {f, l}, 1], 1]
RectangleRegionSimplify[p__] := Module[{pts, rst},
  pts = MeshPrimitives[
     BoundaryDiscretizeRegion[
      DiscretizeRegion[RegionUnion@p], MaxCellMeasure -> \[Infinity]],
      2][[1, 1]];
  rst = Append[pts, First[pts]];
  rst = N[Rationalize[Chop[rst, 10^-6]], 5];
  rst // firstUnion // lastUnion // Rationalize // N // Line
]

And I made some sample Rectangle's as Region. Rectangle is not always Region, so I randomly selected those as Region.

n = 5;
data = Transpose[{RandomInteger[{0, 10}, {n, 2}], 
    RandomInteger[{11, 20}, {n, 2}]}];
fig1 = {ColorData["Atoms", "ColorList"][[1 ;; n]], 
    Rectangle @@@ data} // Transpose;
Graphics[{EdgeForm[Gray], Opacity[0.3], fig1}]

I think you can check for your code with following like this.

fig2 = RectangleRegionSimplify[RegionUnion @@ Rectangle @@@ data]

Line[{{4., 8.}, {5., 8.}, {5., 5.}, {7., 5.}, {7., 1.}, {12., 1.}, {12., 3.}, {16., 3.}, {16., 5.}, {19., 5.}, {19., 14.}, {15., 14.}, {15., 19.}, {4., 19.}, {4., 8.}}]

Graphics[{%, PointSize[Large], Red, Point @@ %}]

Caution

RegionQ[Rectangle[{0, 0}, {1, 3}]]

True

RegionQ[Rectangle[{2, 0}, {1, 3}]]

False