2
$\begingroup$

Why does the first one return EmptyRegion[2] while the second one works? I expected the result is Line[{{0, 0}, {1, 1}}] as the second one. 

RegionIntersection[
 Polygon[{{-1, 1}, {1, 1}, {1, 0}, {-1, 0}, {-1, 1}}], 
 Line[{{0, 0}, {1, 1}}]]

enter image description here

RegionIntersection[Polygon[{{-1, 1}, {1, 1}, {1, 0}, {-1, 0}}], 
 Line[{{0, 0}, {1, 1}}]]

enter image description here

Aslo how does EmptyRegion is defined? I read the docs but still not quite clear. It seems like a point is not empty in R but empty in R2.

How would I define a polygon region like above but without the boundary?

Improve this question
$\endgroup$
8
  • $\begingroup$ On V12 I got the output Line[{{0, 0}, {1, 1}}] from the first one; see image $\endgroup$
    – bmf
    Commented Apr 5, 2022 at 6:25
  • $\begingroup$ @bmf that is strange. Mine is "12.2.0 for Microsoft Windows (64-bit) (December 12, 2020)" $\endgroup$
    – hana
    Commented Apr 5, 2022 at 6:33
  • $\begingroup$ thanks for letting me know. Mine is 12.0.0. Maybe others can check on other versions. If nobody checks, I will run a check on 13.0.0 in the morning $\endgroup$
    – bmf
    Commented Apr 5, 2022 at 6:40
  • $\begingroup$ got the same output as OP (EmptyRegion[2] and Line[{{0, 0}, {1, 1}}]]) on v13.0.0.0 $\endgroup$
    – thorimur
    Commented Apr 5, 2022 at 6:48
  • 2
    $\begingroup$ RegionEqual[Polygon[{{-1, 1}, {1, 1}, {1, 0}, {-1, 0}}], Polygon[{{-1, 1}, {1, 1}, {1, 0}, {-1, 0}, {-1, 1}}]] is True indicate that it is a bug. $\endgroup$
    – cvgmt
    Commented Apr 5, 2022 at 9:41

1 Answer 1

Reset to default
2
$\begingroup$ 
ln1 = Line[{{0, 0}, {1, 1}}];
pts0 = {{-1, 1}, {1, 1}, {1, 0}, {-1, 0}, {-1, 1}}
p0 = Polygon@pts0
p1 = Polygon[{{-1, 1}, {1, 1}, {1, 0}, {-1, 0}}]
p2 = CanonicalizePolygon[p0]

These are regions nevertheless:

RegionQ /@ {p0, p1, p2}

{True, True, True}

but a proper polygon produces the desired results for RegionIntersection. The polygon p1 you defined just happened to be a correct polygon with no repeating points.

RegionIntersection[#, ln1] & /@ {p0, p1, p2}

{EmptyRegion[2], Line[{{0, 0}, {1, 1}}], Line[{{0, 0}, {1, 1}}]}

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks, actually the first one had problem so I removed the repeating points. I thought the first one should work as well. $\endgroup$
    – hana
    Commented Apr 5, 2022 at 7:24
  • $\begingroup$ Frankly your question has not been answered. What I have is a workaround only. If something is recognized as a Region as I showed above, then it must play nice with other regions. $\endgroup$
    – Syed
    Commented Apr 5, 2022 at 11:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.