Do modules get special handling of tail recursion?

On Thu, Jan 22, 2026 at 5:50 PM Father Horton fatherhorton@gmail.com
wrote:

That's why I put a question mark. I don't know.

On Thu, Jan 22, 2026 at 4:54 PM Adrian Mariano avm4@cornell.edu wrote:

Evidently not, as drawpt is eligible for tail recursion as written.

On 1/22/26 15:15, Father Horton via Discuss wrote:

Haha. The grand wizard around here is Adrian!

I was able to further obfuscate my solution. Gemini figured out the (1 +
sqrt(3)) / 3) part to remove one translate(). Then I figured out how to
remove three more translate()s, having had to /guess/ the sqrt(2):

difference() {
     union() { cube([2, 1, 1]); cube([1, 1, 2]); }
     translate(2 * [sqrt(2), (1 + sqrt(3)) / 3, (1 + sqrt(3)) / 3])
         rotate([0, atan(sqrt(2)), 45]) cube(4, center=true);
}
difference() {
     cube([2, 2, 1]);
     translate([1, 0, 1]) rotate([0, 45, 0]) cube(3);
}

-Curt

On 1/22/26 01:49, Guenther Sohler via Discuss wrote:

The last statement in drawpt is an if() statement, not a recursion.  But I
suspect that modules don't give special handling to tail recursion, even if
more generously interpreted.  I think it's not something that people
normally try to do---thousands deep recursion in a module.  My application
here is rather weird and there's probably a better way to approach the
problem of exploring the group generated by 45 deg rotations.  I just took
a quick/lazy one, figuring that if there was some structure it might pop
out.  It could be rewritten as a function that computes all the points.
That would get tail recursion, and then you could run arbitrarily many
points.  And you could check how close the best match is to [1,1,1].  But
for the picture, looping drawpt() isn't an inferior choice, in terms of
quality of the result.

On Thu, Jan 22, 2026 at 6:42 PM Curt McDowell via Discuss <
discuss@lists.openscad.org> wrote:

just with the points list:

a=[[0,0,0],[1,0,0],[1,0,0],[2,0,0],[2,2,0],[0,2,0]];
b=[[0,0,1],[0,0,1],[1,1,1],[2,1,1],[2,2,1],[0,2,1]];
c=[[0,1,1],[1,1,1],[1,1,1],[1,2,1],[0,2,1]];
d=[[0,1,2],[0,1,2],[1,2,2],[1,2,2],[0,2,2]];
s1=[a,b];
s2=[c,d];
color("blue") for(p=[s1,s2,cpo(s1),cpo(s2)]) for(p1=p) p_line3dc(p1,.02);
for(p=[s1,s2]) swp(p);

//module to draw a polyline in 3d space (loop closed)
module p_line3dc(path,d,rec=0,$fn=20){
for(i=[0:len(path)-1])
let(
i_plus=i<len(path)-1?i+1:0
)
hull(){
translate(path[i])if(rec==0)sphere(d/2); else cube(d,true);
translate(path[i_plus])if(rec==0)sphere(d/2);else cube(d,true);
}
}

function faces(sol)=
//    calculate the faces for the vertices with shape l x m with first and
the last end closed
let(
l=len(sol),
m=len(sol[0]),
n1=[for(i=[0:m-1])i],
n2=[for(i=[0:l-2]) each ([ for(j=[0:m-1])
each
j<m-1?[[(j+1)+im,j+im,j+(i+1)m],[(j+1)+im,j+(i+1)m,(j+1)+(i+1)m]]:
[[0+im,j+im,j+(i+1)m],[0+im,j+(i+1)*m,0+(i+1)*m]]
])],
n3=[for(i=[0:m-1])i+(l-1)*m],
n4=[for(i=[len(n3)-1:-1:0])n3[i]],
n=[n1,each (n2),n4]
)n;

function vertices(sol)=
[each for (p=sol)p];

// module for rendering the polyhedron with ends closed
module swp(sol){
let(
v1=vertices(sol),
f1=faces(sol)
)
polyhedron(v1,f1,convexity=10);
}

//function to change the orientation of points of a prism.
function cpo(prism)=[for(i=[0:len(prism[0])-1])[for(p=prism)p[i]]];

[image: Screenshot 2026-01-23 at 6.39.23 AM.png]

On Wed, 21 Jan 2026 at 19:01, William F. Adams via Discuss <
discuss@lists.openscad.org> wrote:

There /must/ be an if() for tail recursion or it would never stop. Any
(and multiple) recursive call can be optimized from anywhere within a
function as long as the function immediately returns the results of the
call (in this case children) without doing anything with them.

module drawpt(p,max,count=0)
{
    move(p) sphere(r=.01); /Tail elimination pseudo-code:/
    dir=rand_int(1,3,1)[0];
    if (count<max)                                   if count < max
    drawpt(  dir==0 ? zrot(45,p)                         arg0 = (dir
== 0 ? ...)
           : dir==1 ? xrot(45,p)                         arg1 =
max;  arg2 = count + 1
           : /*dir==2*/  yrot(45,p),max,count+1);        goto top
without popping frame
}                                                   pop frame and
return children

I agree it would be very unusual to need it. More important is that
function() does implement tail recursion!

-Curt

On 1/22/26 16:07, Adrian Mariano via Discuss wrote:

Jordan,

When we rotate [1, 0, 0] we could hope to arrive at [1, 1, 1] * √3 / 3
using only rotations of 45 degrees, But I don't believe it's possible.
Every matrix representing a rotation by 45 is in the Abelian matrix
field /M3 ℚ(√2//)/ because /sin(45) = cos(45) = //√2 / 2. S/o
regardless of how many multiplies, a coordinate could never occur that
is a multiple of √3.

Thanks for filing #6555 https://github.com/openscad/openscad/issues/6555!

-Curt

On 1/22/26 10:38, Jordan Brown via Discuss wrote:

On 1/22/2026 2:54 PM, Adrian Mariano wrote:

I forgot the distinction.  I don't know, probably not.

OpenSCAD turns all your input into primitive shapes (circle, cylinder, polyhedron, etc), operations (union, intersection, and difference), and placement/resize (translate, rotate, scale, etc get turned into multmatrix). It doesn't try to be very smart about this and optimize (at this stage, at least); IIRC translate wrapping a translate results in two nested multmatrix operations, when that could obviously be one (if nothing was using the intermediate stage).

But notice I didn't list modules nor loops. Every module instantiation must result in the full tree of primitives, operations, and matrix operations because those are the only things the intermediate form can represent. It can't even do math; every value must be a literal number. It's simply impossible to write a subtree and reuse it. Thus it is impossible to have tail call elimination because you can't even represent calls!

• Cory

On January 22, 2026 4:07:09 PM PST, Adrian Mariano via Discuss discuss@lists.openscad.org wrote: