17 ms·

Explaining Huffman’s Impossible Pyramid

66 points by signa11 5y ago

dsr_ 5y ago
People are brainwashed by their prior experience: they have seen so many puzzles with specific assumptions that they assume them to exist here, as well. The quasi-pyramid is impossible if you assume that there are no hidden nodes and edges. That assumption is not an explicit requirement, so you can posit any number of nodes and edges that are hidden from the specific perspective of the drawing -- it could be as crenellated as the interior of an English muffin.
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petargyurov 5y ago
I didn't really get it at first, in fact, I might still not be getting it, but if you draw a line from A to C, then you can see that the triangle ABC will not be similar to DEF... when it should be? I don't know.
- GuB-42 5y ago
  The triangles don't have to be similar to make it a real pyramid. They only have to be similar of you want the truncation to be made parallel to the base. For it to be a real pyramid, you only need the side edges lines to meet at the apex, which always happen when the top and bottom triangles are similar. Similar triangles are a sufficient, but not necessary condition.
- kd0amg 5y ago
  The issue is about which edges are in the same plane as which other edges. If this is a polyhedron, AD and BE must be coplanar, and BE and CF must be coplanar. Extending the edges into lines, they don't form a 3-way intersection. G gets projected to the same screen point as some point on line AD, but the only way G can be on line AD is if AD and FC are coplanar. If all three of AD, BE, and FC are coplanar, the entire figure is a flat pentagon (and DE, BE, and FE are not edges). If G isn't on line AD, then BE and CF can't be a side of the same truncated pyramid.
Jtsummers 5y ago
https://aitopics.org/download/classics:E29CE08E https://aitopics.org/download/classics:E29CE08E - Source of the image The image being discussed is on page 16 of the PDF (p. 310 by the document's page numbers). EDIT: From my very quick, pre-coffee and woken-up-early-by-the-cats reading: To everyone saying, "Well, if we choose a different set of assumptions it becomes possible." Yes. The discussion in the paper I link above goes into the assumptions used and rationale for why figures would be impossible in the context of his discussion. Basically, if you start with the image and treat it as an accurate (as accurate as it can be, a necessarily lossy process in most cases) 2d representation of a 3d object/scene, is there a valid 3d interpretation? In the case of the pyramid, with the assumption that it is an accurate drawing of a 3d pyramid, it's an impossible 3d pyramid. You'd have to add more information for it to become possible.
- Jtsummers 5y ago
  Too late to edit again, but here is one of the critical assumptions from the original paper: > One assumption we shall make throughout this paper is that all pictures are taken from a 'general position'; that is, that a slight change of the position from which the picture is taken would not change the number of lines in the picture or the configurations in which they come together. In the case of pictures of polyhedra this eliminates the possibility of pictures in which two vertices of the objects in the scene are, by coincidence, represented at the same point in the picture, or two edges in the scene are seen as a single line in the picture, or a vertex is seen exactly in line with an unrelated edge. [p. 298] This is important, since, again, it addresses a lot of the comments here on how to make the image represent a possible object/scene. With this assumption, the "pyramid" is impossible. In the next paragraph (same page) Huffman goes on to address this: > Furthermore, if this assumption leads us to judge as impossible an object or set of objects which we know to exist (and therefore by definition 'possible') we can conclude that the camera was probably not in a general position (or that some other assumption was unjustified). In that case we can either move the camera slightly and retake the picture, or go to an augmented list of local configurations which are possible and reanalyze the picture accordingly.
  - SilasX 5y ago
psanan 5y ago
I personally don’t see the image as “impossible”, in terms of seeing it as a projection of a 3D object. I didn’t interpret ABDE as being flat (A, B, D, and E coplanar), and I didn’t expect G, I, and H to be intersections of lines in 3D.
- jerf 5y ago
  If my "geometric intuition" is working properly, the "problem" is that the figure in the picture wouldn't meet in a point. There would be a line at the top, and it wouldn't be a pyramid. But there's nothing "impossible" about that. The impossibility simply seems to be an assertion of impossibility. It feels like it's a problem similar to spending to much time doing "2 + 5 = _" problems and thinking the equality symbol is directional, in this case, spending too much time looking at figures that do meet at a point and thinking that is obligatory for all figures.
  - thaumasiotes 5y ago
    > If my "geometric intuition" is working properly, the "problem" is that the figure in the picture wouldn't meet in a point. There would be a line at the top, and it wouldn't be a pyramid. But there's nothing "impossible" about that. The proof given in the article seems fine. Assuming the figure has three flat faces, the arrangement of those faces is impossible. A figure such as you describe, with a line on the top, would not be ruled out by the proof, but the depicted figure cannot match that description. For a quick summary-style restatement of the proof: 1. Consider the three sides (as opposed to the top and bottom) of the shape to be flat. Each of them will come to a separate point. Those three points are labeled G, H, and I. 2. We can easily show that the point G lies in the same plane as each side of the shape. We can symmetrically show that this is also true of H and of I. 3. When G, H, and I are the same point, this doesn't restrict the sides in any meaningful way - no matter what the "angles" of three planes are, you can always translate them such that they'll all intersect at an arbitrary point. 4. But when G, H, and I are all different points, there is only a single plane that contains them all. ("Three points determine a plane".) This tells us that the three faces of such a shape would all be coplanar, which obviously can't happen. ----- (5. You are positing that, for example, G and H might coincide while I is a different, second point. But the depicted figure doesn't satisfy that description.)
teddyh 5y ago
The implied fact, which makes it an impossible figure, is the assumption that ACFD is a plane. If, on the other hand, there is an edge CD or an edge AF, the figure becomes possible. That’s what my intuition tells me, in any case.
- SilasX 5y ago
  Yes! Thank you! I was wondering what's so impossible about this and I think that's what I was missing -- yes, if you go with the assumption that the back side is a plane, then it's not possible, but that was never my initial assumption. I interpreted it as like a ... 3D trapezoid (?), with a rear support for D and fourth point on its base (though maybe that has its own issue). This is IMHO very different from usual "impossible figure" drawings where there is no interpretation that makes it work.
  - aaroninsf 5y ago
    I am in agreement with y'all. In particular with you observation that this is a different class of "impossible figures." Though I can also understand the counter-argument, that one can also construct objects which from a privileged perspective also "technically" are possible solutions for such figures. I believe the (compelling) counter-counter-argument is that such solutions are AFAIK unique to privileged perspectives, but in this case, there is a whole set of such perspectives. You can rotate the thing through quite a range and still assert you're looking at an impossible pyramid. Adding a single edge to defeat the premise that it's a "pyramid" is I suspect a formalizable distinction which reduces the impossibility (as others have said) to whether you want to hinge all on the word "pyramid."
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mannykannot 5y ago
Here, 'impossible' is defined very loosely, as 'not what it appears to be'. But what does it appear as? Try imagining it, per Necker Cubes, as a concave shape.
JKCalhoun 5y ago
> I read about Huffman’s Pyramid from the consistently excellent blog Futility Closet. Yes, I can also recommend Futility Closet for HN readers. https://www.futilitycloset.com https://www.futilitycloset.com
msandford 5y ago
I understand the argument some people are making as "my intuition tells me it's impossible because $X" and that makes sense. My intuition tells me that there's an extra node not pictured, let's call it Z. There's a triangle ADZ that's out of view and a planar surface DZCF that mirrors DABE. I'm sure my intuition is wrong too, but without pulling out some kind of CAD I can't really see why.
- whatshisface 5y ago
  The front plane is bent. You can bend stuff in real life, but it's implied by the drawing that it's supposed to be a plane.
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drcongo 5y ago
The drawing appears to represent a polyhedron with two triangular faces and three quadrilateral faces. It appears to me to be 1 triangular face and 4 quadrilateral ones. The one they list as ABC should be ABCX where X is a hidden fourth corner on the base. Which would render the shape possible.
- drcongo 5y ago
  Downvoted, but can anyone explain how it's impossible if the base is quadrilateral? [1] [1] https://imgur.com/a/C3tDi50 https://imgur.com/a/C3tDi50
  - Jtsummers 5y ago
    I upvoted your original comment because it shouldn't have been downvoted. But I'd suggest reading the paper I linked elsewhere in this discussion which helps to explain why Huffman called it impossible. The image is "impossible" under the assumptions of that paper. If it were known to be a real object, then it implies that something about the image is wrong (there is hidden information, like the edge you add in your diagram, or the viewport is unusual as described in another comment). Which would mean the camera or lighting needs to be adjust to reveal this hidden information to make the "pyramid" (potentially no longer a pyramid) possible.
    - SilasX 5y ago
      Okay, it's fair to call it impossible if you restrict the domain like that. But, per my other comment, this feels like a non-standard use of "impossible figure", as the conventional meaning is such that, just from the image, there is no reasonable figure that results in that image. That's not the case here. https://news.ycombinator.com/item?id=29879788 https://news.ycombinator.com/item?id=29879788
      - Jtsummers 5y ago
        The context of the creation of this "impossible pyramid" is where the definition for "impossible" comes from here. Under Huffman's assumptions (the image is not deliberately misleading, there is no hidden information, each side lies entirely within a plane, and the image is meant to represent a polyhedron), the pyramid is, indeed, impossible. Relaxing his assumptions (which he discusses) would make it possible, but then, well, you've relaxed the assumptions (there's a hidden edge not seen, the image is meant to mislead, each side does not, in fact, lie entirely within a plane, etc.).
motohagiography 5y ago
It reminds me of a variation of the bridges of konigsberg problem where in this diagram as presented, it has an even number of vertices and edges, and an implied even number of faces, but the vertices have an odd number of edges, which breaks the need for the number of edges per vertex to be even to do a hamiltonian path. (not a mathematician, can't remember this rule specifically) I'd wonder if there is some topology theorem that generalizes hamiltonian paths into higher dimensions, where the difference between a graph and a solid is whether it has a hamiltonian path. e.g. this shape is "impossible" as represented because of the lack of a hamiltonian path.
whatshisface 5y ago
I can see how it's impossible. If the top and right faces are planar, the front face can't be. It becomes possible again if you cut the front face into two triangles, that being the "hidden edge" the author mentions. The pyramidal explanation sounds like it's using a theorem that I didn't know before, but without proving it. You could say the explanation deserves its own explanation. It's an interesting fact, that the lines have to intersect, nonetheless.
pentagonalhole 5y ago
It's a corner of a room, as seen through a pentagonal hole. The pentagonal hole is the planar pentagon ABCFD. The corner of the room is at E, where 3 planes meet. ED, EF and EB are the visible parts of the edges where two of the planes meet. The only constraint on the shape of the pentagonal hole is to be possible to be positioned such that, from the viewer point of view, the vertices B, F, D to be seen as if they are on the edges of the room corner.
Claudus 5y ago
Funny, when I first looked at the sketch, I saw the front left face as curved (and extending the lines would result in a flat vertical surface at the top that spirals upward).
righttoolforjob 5y ago
The analysis is wrong. There is no reason to believe that AD and BE intersect, etc.
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- cylon13 5y ago
  If AD and BE don't intersect then the quadrilateral ADBE doesn't lie in a plane, and instead you have a curved surface there, which is not what you'd expect from the diagram. The idea is that if you assume all of these surfaces are flat, nothing lines up and the shape can't exist. Certainly you can make something that looks like this, but it won't have perfectly flat surfaces everywhere.
  - righttoolforjob 5y ago
    It states that "The drawing appears to represent a polyhedron with two triangular faces and three quadrilateral faces." There is no reason to believe it actually is a quadrilateral. The initial question just gives you a drawing... It is not a pyramid either, since all the lines would intersect at the apex. So what you have is a pyramid-like drawing and some wrongful analysis of it ¯\_(ツ)_/¯ The only other interpretation is that this is just a rough sketch and actually G=H=I and the analysis again is incorrect, since GHI isn't a triangle, it's the same point, the apex, which thus can exist in both planes.
dragontamer 5y ago
If you draw a "dotted line" from "A to C", it becomes blatantly obvious that the backside must be curved or warped in some degree (where the "dotted line" represents the bottom base). Assuming that the front-faces are flat, the triangle on the top of the figure is too warped to match the readily assumed triangular base. In fact, my brain is bouncing between many different interpretations of the base when I saw this figure (first one is a quadrilateral base, as discussed in the blogpost) I don't consider myself to be that good with geometry, but I did play with CAD for a bit back in college, which probably built my instincts up. I imagine that machinists and mechanical engineers would instinctively see the "impossibility" of flat-faces and a 3-sided bottom base. Us computer engineers really won't work with real-world geometry enough to really get the instincts that those mechanical engineers get though. That's fine. They can taunt us with their superior geometry skills while I'll taunt them with my superior C++ skills! -------- I did require the additional guidelines to see the problem. But an additional set of guidelines (ex: dotted-triangle base of ABC) would really make it more obvious.
- red_trumpet 5y ago
  If the top triangle is not parallel to the base (which you picture by your method), then the two triangles don't match, even in a real picture.
  - dragontamer 5y ago
    The "impossible" part is trying to imagine a "flat" ADFC face. There is no such "ADFC" face that can result in a triangle in the shape of DEF. ------- To imagine what "ADFC" could possibly look like, I suggest trying to visualize the ABC-base at first, and then trying to imagine different "ADFC" faces to match the ABC-triangle with the DEF-triangles. The only way you can get this shape to look like this, is if ADFC is bent or twisted somehow. No "straight/flat" ADFC could possibly connect the bottom with the top.
analog31 5y ago
I don't think it contradicts the discussion, but there might be a rotation of the figure where all the lines appear to intersect. For this reason, while pyramid is demonstrably impossible, the converse is not the case: seeing the lines intersect doesn't necessarily mean it's possible.
fallingfrog 5y ago
It's an object that could exist, except that ABED would not be a plane. It's a twisted plane. I don't know what the name of that shape is but you can take a plane and twist it in a way that every point along a straight line between any two points on the surface is also on the surface, but yet the surface is not an actual plane. Edit: Go to wolfram alpha and graph graph y=(z-zx)/2 from x=-1 to x=3 and z=-1 to z=1 and you'll see an example.
dandanua 5y ago
The "impossibility" proof actually uses 3 assumptions: 1. A, D, E, B lie in one plane 2. B, E, F, C lie in one plane 3. A, D, F, C lie in one plane Any of these assumptions can be wrong. It's easy to perceive that number 3 is not flat, if you think that 1 and 2 are flat.
GuB-42 5y ago
That thing has bothered me for a while and I think I finally found why this is impossible. In 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

66 points by signa11 5y ago

dsr_ 5y ago

People are brainwashed by their prior experience: they have seen so many puzzles with specific assumptions that they assume them to exist here, as well. The quasi-pyramid is impossible if you assume that there are no hidden nodes and edges. That assumption is not an explicit requirement, so you can posit any number of nodes and edges that are hidden from the specific perspective of the drawing -- it could be as crenellated as the interior of an English muffin.

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petargyurov 5y ago

I didn't really get it at first, in fact, I might still not be getting it, but if you draw a line from A to C, then you can see that the triangle ABC will not be similar to DEF... when it should be? I don't know.

GuB-42 5y ago
The triangles don't have to be similar to make it a real pyramid. They only have to be similar of you want the truncation to be made parallel to the base. For it to be a real pyramid, you only need the side edges lines to meet at the apex, which always happen when the top and bottom triangles are similar. Similar triangles are a sufficient, but not necessary condition.
kd0amg 5y ago
The issue is about which edges are in the same plane as which other edges. If this is a polyhedron, AD and BE must be coplanar, and BE and CF must be coplanar. Extending the edges into lines, they don't form a 3-way intersection. G gets projected to the same screen point as some point on line AD, but the only way G can be on line AD is if AD and FC are coplanar. If all three of AD, BE, and FC are coplanar, the entire figure is a flat pentagon (and DE, BE, and FE are not edges). If G isn't on line AD, then BE and CF can't be a side of the same truncated pyramid.

Jtsummers 5y ago

https://aitopics.org/download/classics:E29CE08E https://aitopics.org/download/classics:E29CE08E - Source of the image The image being discussed is on page 16 of the PDF (p. 310 by the document's page numbers). EDIT: From my very quick, pre-coffee and woken-up-early-by-the-cats reading: To everyone saying, "Well, if we choose a different set of assumptions it becomes possible." Yes. The discussion in the paper I link above goes into the assumptions used and rationale for why figures would be impossible in the context of his discussion. Basically, if you start with the image and treat it as an accurate (as accurate as it can be, a necessarily lossy process in most cases) 2d representation of a 3d object/scene, is there a valid 3d interpretation? In the case of the pyramid, with the assumption that it is an accurate drawing of a 3d pyramid, it's an impossible 3d pyramid. You'd have to add more information for it to become possible.

Jtsummers 5y ago
Too late to edit again, but here is one of the critical assumptions from the original paper: > One assumption we shall make throughout this paper is that all pictures are taken from a 'general position'; that is, that a slight change of the position from which the picture is taken would not change the number of lines in the picture or the configurations in which they come together. In the case of pictures of polyhedra this eliminates the possibility of pictures in which two vertices of the objects in the scene are, by coincidence, represented at the same point in the picture, or two edges in the scene are seen as a single line in the picture, or a vertex is seen exactly in line with an unrelated edge. [p. 298] This is important, since, again, it addresses a lot of the comments here on how to make the image represent a possible object/scene. With this assumption, the "pyramid" is impossible. In the next paragraph (same page) Huffman goes on to address this: > Furthermore, if this assumption leads us to judge as impossible an object or set of objects which we know to exist (and therefore by definition 'possible') we can conclude that the camera was probably not in a general position (or that some other assumption was unjustified). In that case we can either move the camera slightly and retake the picture, or go to an augmented list of local configurations which are possible and reanalyze the picture accordingly.
- SilasX 5y ago

psanan 5y ago

I personally don’t see the image as “impossible”, in terms of seeing it as a projection of a 3D object. I didn’t interpret ABDE as being flat (A, B, D, and E coplanar), and I didn’t expect G, I, and H to be intersections of lines in 3D.

jerf 5y ago
If my "geometric intuition" is working properly, the "problem" is that the figure in the picture wouldn't meet in a point. There would be a line at the top, and it wouldn't be a pyramid. But there's nothing "impossible" about that. The impossibility simply seems to be an assertion of impossibility. It feels like it's a problem similar to spending to much time doing "2 + 5 = _" problems and thinking the equality symbol is directional, in this case, spending too much time looking at figures that do meet at a point and thinking that is obligatory for all figures.
- thaumasiotes 5y ago
  > If my "geometric intuition" is working properly, the "problem" is that the figure in the picture wouldn't meet in a point. There would be a line at the top, and it wouldn't be a pyramid. But there's nothing "impossible" about that. The proof given in the article seems fine. Assuming the figure has three flat faces, the arrangement of those faces is impossible. A figure such as you describe, with a line on the top, would not be ruled out by the proof, but the depicted figure cannot match that description. For a quick summary-style restatement of the proof: 1. Consider the three sides (as opposed to the top and bottom) of the shape to be flat. Each of them will come to a separate point. Those three points are labeled G, H, and I. 2. We can easily show that the point G lies in the same plane as each side of the shape. We can symmetrically show that this is also true of H and of I. 3. When G, H, and I are the same point, this doesn't restrict the sides in any meaningful way - no matter what the "angles" of three planes are, you can always translate them such that they'll all intersect at an arbitrary point. 4. But when G, H, and I are all different points, there is only a single plane that contains them all. ("Three points determine a plane".) This tells us that the three faces of such a shape would all be coplanar, which obviously can't happen. ----- (5. You are positing that, for example, G and H might coincide while I is a different, second point. But the depicted figure doesn't satisfy that description.)

teddyh 5y ago

The implied fact, which makes it an impossible figure, is the assumption that ACFD is a plane. If, on the other hand, there is an edge CD or an edge AF, the figure becomes possible. That’s what my intuition tells me, in any case.

SilasX 5y ago
Yes! Thank you! I was wondering what's so impossible about this and I think that's what I was missing -- yes, if you go with the assumption that the back side is a plane, then it's not possible, but that was never my initial assumption. I interpreted it as like a ... 3D trapezoid (?), with a rear support for D and fourth point on its base (though maybe that has its own issue). This is IMHO very different from usual "impossible figure" drawings where there is no interpretation that makes it work.
- aaroninsf 5y ago
  I am in agreement with y'all. In particular with you observation that this is a different class of "impossible figures." Though I can also understand the counter-argument, that one can also construct objects which from a privileged perspective also "technically" are possible solutions for such figures. I believe the (compelling) counter-counter-argument is that such solutions are AFAIK unique to privileged perspectives, but in this case, there is a whole set of such perspectives. You can rotate the thing through quite a range and still assert you're looking at an impossible pyramid. Adding a single edge to defeat the premise that it's a "pyramid" is I suspect a formalizable distinction which reduces the impossibility (as others have said) to whether you want to hinge all on the word "pyramid."

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mannykannot 5y ago

Here, 'impossible' is defined very loosely, as 'not what it appears to be'. But what does it appear as? Try imagining it, per Necker Cubes, as a concave shape.

JKCalhoun 5y ago

> I read about Huffman’s Pyramid from the consistently excellent blog Futility Closet. Yes, I can also recommend Futility Closet for HN readers. https://www.futilitycloset.com https://www.futilitycloset.com

msandford 5y ago

I understand the argument some people are making as "my intuition tells me it's impossible because $X" and that makes sense. My intuition tells me that there's an extra node not pictured, let's call it Z. There's a triangle ADZ that's out of view and a planar surface DZCF that mirrors DABE. I'm sure my intuition is wrong too, but without pulling out some kind of CAD I can't really see why.

whatshisface 5y ago
The front plane is bent. You can bend stuff in real life, but it's implied by the drawing that it's supposed to be a plane.

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drcongo 5y ago

The drawing appears to represent a polyhedron with two triangular faces and three quadrilateral faces. It appears to me to be 1 triangular face and 4 quadrilateral ones. The one they list as ABC should be ABCX where X is a hidden fourth corner on the base. Which would render the shape possible.

drcongo 5y ago
Downvoted, but can anyone explain how it's impossible if the base is quadrilateral? [1] [1] https://imgur.com/a/C3tDi50 https://imgur.com/a/C3tDi50
- Jtsummers 5y ago
  I upvoted your original comment because it shouldn't have been downvoted. But I'd suggest reading the paper I linked elsewhere in this discussion which helps to explain why Huffman called it impossible. The image is "impossible" under the assumptions of that paper. If it were known to be a real object, then it implies that something about the image is wrong (there is hidden information, like the edge you add in your diagram, or the viewport is unusual as described in another comment). Which would mean the camera or lighting needs to be adjust to reveal this hidden information to make the "pyramid" (potentially no longer a pyramid) possible.
  - SilasX 5y ago
    Okay, it's fair to call it impossible if you restrict the domain like that. But, per my other comment, this feels like a non-standard use of "impossible figure", as the conventional meaning is such that, just from the image, there is no reasonable figure that results in that image. That's not the case here. https://news.ycombinator.com/item?id=29879788 https://news.ycombinator.com/item?id=29879788
    - Jtsummers 5y ago
      The context of the creation of this "impossible pyramid" is where the definition for "impossible" comes from here. Under Huffman's assumptions (the image is not deliberately misleading, there is no hidden information, each side lies entirely within a plane, and the image is meant to represent a polyhedron), the pyramid is, indeed, impossible. Relaxing his assumptions (which he discusses) would make it possible, but then, well, you've relaxed the assumptions (there's a hidden edge not seen, the image is meant to mislead, each side does not, in fact, lie entirely within a plane, etc.).

motohagiography 5y ago

It reminds me of a variation of the bridges of konigsberg problem where in this diagram as presented, it has an even number of vertices and edges, and an implied even number of faces, but the vertices have an odd number of edges, which breaks the need for the number of edges per vertex to be even to do a hamiltonian path. (not a mathematician, can't remember this rule specifically) I'd wonder if there is some topology theorem that generalizes hamiltonian paths into higher dimensions, where the difference between a graph and a solid is whether it has a hamiltonian path. e.g. this shape is "impossible" as represented because of the lack of a hamiltonian path.

whatshisface 5y ago

I can see how it's impossible. If the top and right faces are planar, the front face can't be. It becomes possible again if you cut the front face into two triangles, that being the "hidden edge" the author mentions. The pyramidal explanation sounds like it's using a theorem that I didn't know before, but without proving it. You could say the explanation deserves its own explanation. It's an interesting fact, that the lines have to intersect, nonetheless.

pentagonalhole 5y ago

It's a corner of a room, as seen through a pentagonal hole. The pentagonal hole is the planar pentagon ABCFD. The corner of the room is at E, where 3 planes meet. ED, EF and EB are the visible parts of the edges where two of the planes meet. The only constraint on the shape of the pentagonal hole is to be possible to be positioned such that, from the viewer point of view, the vertices B, F, D to be seen as if they are on the edges of the room corner.

Claudus 5y ago

Funny, when I first looked at the sketch, I saw the front left face as curved (and extending the lines would result in a flat vertical surface at the top that spirals upward).

righttoolforjob 5y ago

The analysis is wrong. There is no reason to believe that AD and BE intersect, etc.

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cylon13 5y ago
If AD and BE don't intersect then the quadrilateral ADBE doesn't lie in a plane, and instead you have a curved surface there, which is not what you'd expect from the diagram. The idea is that if you assume all of these surfaces are flat, nothing lines up and the shape can't exist. Certainly you can make something that looks like this, but it won't have perfectly flat surfaces everywhere.
- righttoolforjob 5y ago
  It states that "The drawing appears to represent a polyhedron with two triangular faces and three quadrilateral faces." There is no reason to believe it actually is a quadrilateral. The initial question just gives you a drawing... It is not a pyramid either, since all the lines would intersect at the apex. So what you have is a pyramid-like drawing and some wrongful analysis of it ¯\_(ツ)_/¯ The only other interpretation is that this is just a rough sketch and actually G=H=I and the analysis again is incorrect, since GHI isn't a triangle, it's the same point, the apex, which thus can exist in both planes.

dragontamer 5y ago

If you draw a "dotted line" from "A to C", it becomes blatantly obvious that the backside must be curved or warped in some degree (where the "dotted line" represents the bottom base). Assuming that the front-faces are flat, the triangle on the top of the figure is too warped to match the readily assumed triangular base. In fact, my brain is bouncing between many different interpretations of the base when I saw this figure (first one is a quadrilateral base, as discussed in the blogpost) I don't consider myself to be that good with geometry, but I did play with CAD for a bit back in college, which probably built my instincts up. I imagine that machinists and mechanical engineers would instinctively see the "impossibility" of flat-faces and a 3-sided bottom base. Us computer engineers really won't work with real-world geometry enough to really get the instincts that those mechanical engineers get though. That's fine. They can taunt us with their superior geometry skills while I'll taunt them with my superior C++ skills! -------- I did require the additional guidelines to see the problem. But an additional set of guidelines (ex: dotted-triangle base of ABC) would really make it more obvious.

red_trumpet 5y ago
If the top triangle is not parallel to the base (which you picture by your method), then the two triangles don't match, even in a real picture.
- dragontamer 5y ago
  The "impossible" part is trying to imagine a "flat" ADFC face. There is no such "ADFC" face that can result in a triangle in the shape of DEF. ------- To imagine what "ADFC" could possibly look like, I suggest trying to visualize the ABC-base at first, and then trying to imagine different "ADFC" faces to match the ABC-triangle with the DEF-triangles. The only way you can get this shape to look like this, is if ADFC is bent or twisted somehow. No "straight/flat" ADFC could possibly connect the bottom with the top.

analog31 5y ago

I don't think it contradicts the discussion, but there might be a rotation of the figure where all the lines appear to intersect. For this reason, while pyramid is demonstrably impossible, the converse is not the case: seeing the lines intersect doesn't necessarily mean it's possible.

fallingfrog 5y ago

It's an object that could exist, except that ABED would not be a plane. It's a twisted plane. I don't know what the name of that shape is but you can take a plane and twist it in a way that every point along a straight line between any two points on the surface is also on the surface, but yet the surface is not an actual plane. Edit: Go to wolfram alpha and graph graph y=(z-zx)/2 from x=-1 to x=3 and z=-1 to z=1 and you'll see an example.

dandanua 5y ago

The "impossibility" proof actually uses 3 assumptions: 1. A, D, E, B lie in one plane 2. B, E, F, C lie in one plane 3. A, D, F, C lie in one plane Any of these assumptions can be wrong. It's easy to perceive that number 3 is not flat, if you think that 1 and 2 are flat.

GuB-42 5y ago

That thing has bothered me for a while and I think I finally found why this is impossible. In 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