In the OpenSCAD documentation for minkowski(), examples are given for 2
child objects, ref: https://en.wikibooks.org/wiki/
OpenSCAD_User_Manual/Transformations#minkowski

But what if you supply 3 or more, for example:

$fn=50;
minkowski()
{
cube([10,10,10]);
cylinder(r=2,h=10);
sphere(r=4);
}

This seems to be accepted, but it is not obvious what happens in this
case. It looks like the same as below?

$fn=50;
minkowski()
{
minkowski() {
cube([10,10,10]);
cylinder(r=2,h=10);
}
sphere(r=4);
}

In other words, when more than 2 child objects are supplied the child
objects are subject to a series of implicit minkowski operations until the
list is exhausted. Is this a correct interpretation?

Carsten Arnholm

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Discuss@lists.openscad.org
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In other words, when more than 2 child objects are supplied the child objects are subject to a series of implicit minkowski operations until the list is exhausted. Is this a correct interpretation?

Nophead is right: Minkowski sums are commutative, but they’re implemented by accumulating binary sums as you describe.

The exception is if all operators are convex. In that case we can exploit the fact that the "convex hulls of a minkowski sum” equals the “minkowski sum of convex hulls”. ..and simply collect vertices and calculate a single convex hull:
https://en.wikipedia.org/wiki/Minkowski_addition