Why fimplicit3 draw the same function differently

f=@(x,y,z) (x+y+z).^2
fimplicit3(f)

We actually know that this equals to draw the graph of ，but the code turns out to be this:

What happened to the function？

[Dyuman Joshi]

Seems like a bug -

f=@(x,y,z) (x+y+z).^2;
fun=fimplicit3(f)

fun = 
  ImplicitFunctionSurface with properties:

     Function: @(x,y,z)(x+y+z).^2
    EdgeColor: [0 0 0]
    LineStyle: '-'
    FaceColor: 'interp'

  Show all properties

No output if Mesh Density is changed, and the axes limits go from the default [-5 5] to [0 1];

figure
fimplicit3(f,'MeshDensity',50)

Specifying interval gives the same result as above

figure
fimplicit3(f,[-5 5],'MeshDensity',50)

I tested the same in my R2021a and it has the same behaviour as above.

Answer 1

[Bruno Luong]

This is NOT a bug.

fimplicit3 plot the surface f(x,y,z) = 0, do you must help it by having a function that explicitly cross the level f=0, because it approximate by a tensorial tri-linear and find where it corss the level. But your function is defined as (x+y+z).^2 is always >= 0. Thus fimplicit have a hard time to estimate the surface.

If you add a tiny positive error in your square function it does not cross at all the level f=0.

If you want to determine the surface x+y+z = 0, then do it straight manner

h = @(x,y,z) x+y+z;
fimplicit3(h) 

Do not square the function, this is stupid way to chanllenge fimplicit3 numerically.

[Dyuman Joshi]

"Do not square the function, this is stupid numerically."

How about symbolically?

syms f(x,y,z)
h = (x+y+z)^2;
fimplicit3(h)

"because it approximate by a tensorial tri-linear and find where it corss the level."

Do you have a source for this?

[Bruno Luong]

Under the hood is numerical surface level, regradless where as your function is symbolic or not.

No I don't have a source but I program myself such level surface. There is nothing sophisticate behind, one just need to decide the cube in voxel and look where the edges cross the level, just like contour in 2D.