P1, P2, P3, P4 are defined by 3pts for each as all planes 3 points define a plane.
P1 (A, B, H)
P2 (B, C, D)
P3 (D, E, F)
P4 (F, G, H)
The question is : does exist a point P which is on P1, P2, P3 and P4 ?
I think, we have two possibilities to determine it :
1/ Mathematical with a system of equations (each plane is determined by a equation) and its resolution.
2/ MOI (or CAD drawing)
if we do planes P1, P2, P3 and P4 with points A,B,C,D,E,F,G,H with their 3 points, we can do intersections of these 4 planes. If we have only a point... we win and have P !
With your file, the result is a small thetahedron... P does not exist.
You have the original structure Yes!
You fill the hole Yes!
You have only planes Yes!
You have not 4 planes but 8 so better twice! :)
And you have only quads Yes!
And there is an infinity of this solution! Excellent!
Burmann's Triangles wins ! ;) Bermuda Triangle ? :)
Now the best will be a little script or Elephant for automatise the down levels
or maybe a tricky use of existing functions (Loft, Sweep, NetWork...) for have directly plane surfaces from false surfaces! (dream :)
I have seen during this little exercise than NetWork has numerous parameters! (Uniform with number of Points...)
By hands it's 20 seconds for the last 4 top quads ;)
For this : here by hands its nothing for 12 quads but when you have 200-300 it's another story :)
It's true Quads "plane" built by external perimeter!
All blue faces are planar!
Constant angle by row...red faces are not planes! Angle not constant on the red original surface!
Taking the same length for the segment than the red structure...but as the angle is constant by rows... at the end some is missing at the right side! :)
Right side faces are a little deformed but now all yellow volume is quasi wraped!