3
$\begingroup$

Whenever I try to find the intersection between a two regions, one of which is a derived region using RegionUnion, I get a BooleanRegion with which I can do little.

s1 = Sphere[{0, 0, 0}, 0.1];
s2 = RegionUnion[Sphere[{0, 0, 0}, 0.1], Sphere[{0.05, 0, 0}, 0.1]];
l = Line[{{-0.2, 0, 0}, {0.01, 0, 0}}];

Show[Region[s2], Region[Line[{{-0.2, 0, 0}, {0.01, 0, 0}}]]]

enter image description here

RegionIntersection[s2, l]

BooleanRegion[(#1 || #2) && #3 &, {Sphere[{0, 0, 0}, 0.1], Sphere[{0.05, 0, 0}, 0.1], Line[{{-0.2, 0, 0}, {0.01, 0, 0}}]}]

RegionIntersection[s1, l]

Point[{{-0.1, 0, 0}}]

I am sure this has been answered somewhere else, or that I am missing something basic, but I can not find a reference

$\endgroup$
3
$\begingroup$

I agree that at the present time RegionIntersection can produce simplifications where the equivalent BooleanRegion version does not. Here is a function that converts a BooleanRegion object into an equivalent RegionUnion/RegionIntersection construct so that these additional simplifications have a chance to fire:

simplifyRegion[BooleanRegion[predicate_, regions_]] := With[
    {dnf = BooleanConvert[predicate] /. {Or->RegionUnion, And->RegionIntersection}},

    dnf @@ regions
]

Using BooleanConvert produces a boolean predicate in DNF form, which means that RegionIntersection fires before RegionUnion. Here is the output of simplifyRegion on your example:

RegionIntersection[s2,l]
simplifyRegion @ %

BooleanRegion[(#1 || #2) && #3 &, {Sphere[{0, 0, 0}, 0.1], Sphere[{0.05, 0, 0}, 0.1], Line[{{-0.2, 0, 0}, {0.01, 0, 0}}]}]

BooleanRegion[#1 || #2 &, {Point[{{-0.1, 0, 0}}], Point[{{-0.05, 0, 0}}]}]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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