Earlier quoted context omitted.
Sure, and I addressed your point there[1], that this still requires a non-standard, new definition of "impossible figure". Furthermore, I don't see how it introduces information that "isn't evident in the image" when several people automatically assumed a 4 sided base (me among them) without trying to be clever. [1] https://news.ycombinator.com/item?id=29879862
"assumed", because it wasn't presented . It wouldn't need to be assumed if it were present. That's pretty much the definition of assuming.
Explaining Huffman’s Impossible Pyramid
61–63 of 63 posts
Re: Explaining Huffman’s Impossible Pyramid
#62Earlier quoted context omitted.
"assumed", because it wasn't presented . It wouldn't need to be assumed if it were present. That's pretty much the definition of assuming.
It's "assumed" only in the sense that the form of any occluded part of an image is "assumed". By that standard, "ACFD is a face/plane" isn't presented either, and, relative to the image alone, is "assumed" in the same sense.
To summarize: If this were a real image of a real object, then it is, by definition, a possible object, but it is not, per the image, a possible pyramid. The camera would have to be shifted (or lighting altered or something changed) in order to discern what it actually was (which could be a pyramid if some distortion were discovered, but could also be any of a number of other possible objects, including, as one person suggested, the corner of a room through a peculiarly shaped viewport).
Re: Explaining Huffman’s Impossible Pyramid
#63In order for the shape to be impossible ADEB, BEFC and ACFD have to be coplanar. If you add a hidden AF or CD edge, making ACFD not coplanar, it becomes possible.
To make things simple, let's assume a simple, orthogonal projection (discard the z). We will use degrees of freedom for that. All vertices have their x, y position fixed, that's their 2D coordinates in the drawing, we don't know z yet because it is discarded by the projection. The problem becomes: for each vertex, find the z coordinate. The constraints are: ADEB, BEFC are coplanar, DBFE are not (making the entire thing a flat shape would be cheating).
So, make the z coordinates for D, B, F, E anything we want, it will make a small tetrahedron, nothing wrong with that. Now, because A has to be in the DEB plane, its z coordinate is fixed, but it can be calculated, no problem. Same thing with C, it is in the BEF plane and its z coordinate is fixed and can be calculated. You can do that in every case. So as long as you are not looking behind the "pyramid", you can always find a 3D shape that matches the projection.
Now if we add the constraint that ACFD are coplanar, that's when you have a problem. All points are fixed, and you have to play with your 4 degrees of freedom that are the z coordinates of D, B, F, E to make ACFD coplanar. Playing with the z, you can do translation, scaling on the z axis and shearing on xz and yz, that's 4 independent transforms, all your degrees of freedom are used up. None of them help making your vertices coplanar, except if you flatten everything, which, as said before, is cheating.
Now, maybe we can also make the impossible possible using fancier projections. Perspective projection has a FOV parameter, it may be an extra degree of freedom we can play with, but that enough maths for today.
EDIT: missed a transform, now, it works