GA Review edit

This review is transcluded from Talk:Midsphere/GA1. The edit link for this section can be used to add comments to the review.

Reviewer: Pi.1415926535 (talk · contribs) 05:12, 1 May 2023 (UTC)Reply

GA review (see here for what the criteria are, and here for what they are not)

  1. It is reasonably well written.
    a. (prose, spelling, and grammar):  
    b. (MoS for lead, layout, word choice, fiction, and lists):  
  2. It is factually accurate and verifiable.
    a. (reference section):  
    b. (citations to reliable sources):  
    c. (OR):  
    d. (copyvio and plagiarism):  
    Earwig just finds a couple mirrors
  3. It is broad in its coverage.
    a. (major aspects):  
    b. (focused):  
  4. It follows the neutral point of view policy.
    Fair representation without bias:  
  5. It is stable.
    No edit wars, etc.:  
  6. It is illustrated by images and other media, where possible and appropriate.
    a. (images are tagged and non-free content have non-free use rationales):  
    b. (appropriate use with suitable captions):  
  7. Overall:
    Pass/fail:  
    An impressively accessible math article. Just a couple minor suggestions.

(Criteria marked   are unassessed)

Comments edit

  • Add alt text for images
    • I am very sympathetic to the goal of accessibility, but achieving this through alt text is not a mechanical requirement in general, nor in the GA criteria. MOS:ALT does ask that the combination of captions plus alt text adequately describe the image. I have heard users of screen readers complain that alt text that adds no useful information to the caption can be more annoying clutter than useful. I think that adding, for instance, "A polyhedron and its midsphere" to an image for which this is the first text in the caption would be merely redundant, and that describing irrelevant details of the image would not be helpful for readers of any kind. So could you please be more specific about what information you think could be conveyed through alt text that is not already conveyed by the caption, and that should be so conveyed? —David Eppstein (talk) 05:06, 2 May 2023 (UTC)Reply
    I do understand your point here. I think a description of the geometry itself would be useful, especially for less technical readers. Perhaps something like this? An irregular polyhedron with several triangular and quadrilateral faces visible. A blue sphere of approximately the same size is tangent to each edge of the polyhedron. The portions of the sphere outside the polyhedron form circular caps on each face. Several red circles on the face of the sphere connect points where it is tangent to the edges. Pi.1415926535 (talk) 05:52, 2 May 2023 (UTC)Reply
  • The cube/octahedron dual image would be better placed under the Properties heading.
  • It would be great to give the values for an example Crelle's tetrahedron - I think that would make the generation easier to grasp.
  • Absolutely not required to meet the GA criteria, but images illustrating a Crelle's tetrahedron and the relation between an example polyhedral graph and its canonical polyhedron would be excellent to add at some point.
    • I added an image of four tangent spheres, and a (newly uploaded) image of the planar circle packing generated by stereographic projection of horizon spheres. The four-sphere Crelle tetrahedron, especially, could be better, but I'm not currently set up for easy generation of new 3d images and I didn't find anything better already on commons. —David Eppstein (talk) 06:21, 2 May 2023 (UTC)Reply

Source check edit

  • 2: passed
  • 5: passed
  • 10: offline source, so
  • 11: passed
  • 17: passed
  • 18: passing on good faith - we're at the limits of my mathematical understanding here.

@Pi.1415926535: I think I've responded to everything; please take another look. —David Eppstein (talk) 06:46, 4 May 2023 (UTC)Reply

@David Eppstein: Just wanted to see your thoughts on my suggested type of alt text. Pi.1415926535 (talk) 18:15, 4 May 2023 (UTC)Reply
Mostly harmless, but a little inaccurate. I added different alt text to the images. I'm not convinced that it is going to add any useful information for people using screen-readers, though, rather than just cluttering the text with unhelpful descriptions. —David Eppstein (talk) 20:50, 4 May 2023 (UTC)Reply
Thanks for taking a look. Passing now, great work once again. Pi.1415926535 (talk) 21:39, 4 May 2023 (UTC)Reply

Where does the name "canonical polyhedron" come from? edit

This article cites Ziegler (1995), but Ziegler doesn't use this name. What Ziegler says explicitly is:

The primal-dual version can in fact be used to prove Steinitz’ theorem in a very strong form:
  • Every 3-connected planar graph is the graph of a 3-polytope   whose edges touch the unit sphere,
  • There is a "canonical" representation of this form for every polytope.
Theorem 4.13. (see Schramm [483])
For every planar 3-connected graph, there is a representation as the graph of a 3-polytope whose edges are all tangent to the unit sphere   and such that   is the barycenter of the contact points.
This representation is unique up to rotations and reflections of the polytope in   In particular, in this representation every combinatorial symmetry of the graph is realized by a symmetry of the polytope.

jacobolus (t) 23:21, 5 February 2024 (UTC)Reply

One source seems to be G. Hart (1997) "Calculating Canonical Polyhedra", Mathematica in Education and Research 6(3): 5–10. –jacobolus (t) 00:30, 6 February 2024 (UTC)Reply
The earliest form of the MathWorld page on the same topic that I can find [1] credits Hart. But Ziegler's phrasing is very close to this already. Schramm's 1992 "how to cage an egg", another early reference for this, does use the word "canonical" twice, but not for exactly this meaning. —David Eppstein (talk) 00:32, 6 February 2024 (UTC)Reply
It seems like the name "Koebe polyhedron" is sometimes given to any polyhedron which has a "midsphere". Is this name widespread enough to be worth mentioning? –jacobolus (t) 00:38, 6 February 2024 (UTC)Reply

The confusing canonical polyhedron's definition edit

The article says about the definition of canonical polyhedron in the following:

Any polyhedron with a midsphere, scaled so that the midsphere is the unit sphere, can be transformed in this way into a polyhedron for which the centroid of the points of tangency is at the center of the sphere. The result of this transformation is an equivalent form of the given polyhedron, called the canonical polyhedron, with the property that all combinatorially equivalent polyhedra will produce the same canonical polyhedra as each other, up to congruence

The words seems a bit confusing and WP:TECHNICAL as I read, an example is "centroid of the points of tangency". Can you provide some explanation both in roughly speaking and fully meaning here, @David Eppstein? Also, MathWorld says that a polyhedra is canonical provided that the edges are touching the sphere as well, but I remember that MathWorld is not good source, so I leave it to you in this case. I was trying to understand the definition. Dedhert.Jr (talk) 12:46, 17 March 2024 (UTC)Reply

MathWorld says "and the center of gravity of their contact points is the center of that sphere". This is the same as our "centroid of the points of tangency is at the center of the sphere". Scale and translate the polyhedron so that its midsphere is a unit sphere centered at the origin. Take all the points where the edges touch the sphere. Average their coordinates. The resulting average is itself a point, somewhere in space. If it's the origin, you're done: you have a canonical polyhedron. —David Eppstein (talk) 16:59, 17 March 2024 (UTC)Reply
"Average their coordinates"? I don't get it in this case. Also, should you provide an example of canonical polyhedron in this case, an example is gyrate rhombicosidodecahedron? The infobox says it is canonical, but no source has ever mentioned this fact nowadays. Dedhert.Jr (talk) 01:36, 18 March 2024 (UTC)Reply
You know. The average of (1,1,1), (-1,2,-7), and (3,-4,2) is ( (1,1,1) + (-1,2,-7) + (3,-4,2) )/3 = (1-1+3/3,1+2-4/3,1-7+2/3). Add up the vectors of Cartesian coordinates, coordinate by coordinate, and then divide by how many you added. Just like you compute any other kind of average.
In the case of the gyrate rhombicosidodecahedron, the rhombicosidodecahedron has enough symmetry to force the average of the points of tangency to be the center of the sphere, as long as you have a midsphere at all. And then spinning the cupolas doesn't change the existence of a midsphere or the position of the average of the tangency points. For the rhombicosidodecahedron itself, it's not completely obvious that it has a midsphere because it's not edge-transitive, but there's a simple argument that can be used to show it: because it is vertex-transitive, there is a circumsphere, through all vertices. And then the faces that are squares have their edge midpoints equally far from the circumcenter, by the symmetry of the square. But each edge of the rhombicosidodecahedron belongs to a square, so all edge midpoints are equally far from the circumcenter, and the circumcenter is also a midcenter.
I doubt you find these arguments in any published source, though, so it is probably original research. —David Eppstein (talk) 06:45, 18 March 2024 (UTC)Reply
Ooooooohhhh... My bad. I was expecting that the word "average" has alternative meaning other than statistical definition. Dedhert.Jr (talk) 11:03, 18 March 2024 (UTC)Reply