Difference between revisions of "Talk:2626: d65536"

Explain xkcd: It's 'cause you're dumb.
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::I believe the suggested scheme would be to take a dodecahedron or icosohedron (either of the two duals can be used to start with) and then subdivide each face in such a manner that equally-sized (but differently distorted) hexagons – and 12 little regular pentagons of identical area fitting in at the old dodecahedron centre/the old icosahedron vertex – emerge from the required segmentation/vertex-truncation and readjustment the radiality of all new mid-edge vertices (or maybe the newer-edges' centres or the newer-faces' centres) to touch the unit sphere. If done symmetrically, it should be entirely fair.
 
::I believe the suggested scheme would be to take a dodecahedron or icosohedron (either of the two duals can be used to start with) and then subdivide each face in such a manner that equally-sized (but differently distorted) hexagons – and 12 little regular pentagons of identical area fitting in at the old dodecahedron centre/the old icosahedron vertex – emerge from the required segmentation/vertex-truncation and readjustment the radiality of all new mid-edge vertices (or maybe the newer-edges' centres or the newer-faces' centres) to touch the unit sphere. If done symmetrically, it should be entirely fair.
 
::The face-count might be troublesome, though. The twelve necessary pentagonal faces leaves 65524 hexagons, to split evenly between* either 12 or 20 zones, and it should be obvious that neither is possible**, in whole numbers, given the starting point of 2<sup>n</sup> faces...
 
::The face-count might be troublesome, though. The twelve necessary pentagonal faces leaves 65524 hexagons, to split evenly between* either 12 or 20 zones, and it should be obvious that neither is possible**, in whole numbers, given the starting point of 2<sup>n</sup> faces...
::(* - you can, and probably will in this design, have some that cross between two of the top-level polygons, but you can fully 'donate' as many as you then fully ''get'' donated from the next face around, so it might as well be just counted as a group of while tiles on an a set of Escher-like interlocking 'rough' polygons.)
+
:::(* - you can, and probably will in this design, have some that cross between two of the top-level polygons, but you can fully 'donate' as many as you then fully ''get'' donated from the next face around, so it might as well be just counted as a group of while tiles on an a set of Escher-like interlocking 'rough' polygons.)
::(** - If you're using 12 zones, that's 3x4x(however many in the zone + one corner each) and there's no factor of 3 in '''any''' value that is 2usup>n</sup>. Arranging into 20 symmetrical zones (5x4), you will find that 65524 isn't divisible by 5, either...)
+
:::(** - If you're using 12 zones, that's 3x4x(however many in the zone + one corner each) and there's no factor of 3 in '''any''' value that is 2usup>n</sup>. Arranging into 20 symmetrical zones (5x4), you will find that 65524 isn't divisible by 5, either...)
 
::You could probably arrange an N-ahedron with the number of faces being 12+(12a) or 12+(20b), for some higher value (a bit of mental arithmatic suggests 65592 might be that value) and mark all the 'excess' faces (56?) with "Roll Again!". Or perhaps some pithy motivational slogans that also convey roughly the same meaning... :P [[Special:Contributions/172.70.162.5|172.70.162.5]] 11:32, 31 May 2022 (UTC)
 
::You could probably arrange an N-ahedron with the number of faces being 12+(12a) or 12+(20b), for some higher value (a bit of mental arithmatic suggests 65592 might be that value) and mark all the 'excess' faces (56?) with "Roll Again!". Or perhaps some pithy motivational slogans that also convey roughly the same meaning... :P [[Special:Contributions/172.70.162.5|172.70.162.5]] 11:32, 31 May 2022 (UTC)
  

Revision as of 11:34, 31 May 2022


I wonder: can we even make a fair polyhedron with 65536 faces? In Randal's illustration, the faces seem to be irregular hexagons. 172.70.130.105 21:37, 30 May 2022 (UTC)

This is better than my question, which was simply if you could tile a sphere with these. 172.70.211.36 23:01, 30 May 2022 (UTC)
Definitely possible, just create two identical right pyramids with a 32768-gon base and glue the bases together. Clam (talk) 23:53, 30 May 2022 (UTC)
Would this design be fair? Consider a set of 256 lines of latitude overlapping another set, with the second set's polar axis at the equator of the first. Cut flat quadrangles between the intersection points of the lines of latitude. Doesn't use hexagons like the comic does though. 172.70.110.121 09:41, 31 May 2022 (UTC)
Fairness is a given for pyramids (if that's what you're asking). As long as there's enough 'rolling energy' to get either of the pyramids 'facing up', any N-agon base to the pyramids should have enough indeterminate impetous to then finally roll around a bit to end up with any of those exposed faces on top ::(Interesting to note that for odd-numbered N-agonal bases, like that in a D10, you need to offset the bases and instead of sticking to the triangular faces base-to-base you now have kite-shapes that interlock in a serration that is no longer strictly planar along the axis's perpendiculars.)
That might need a selection of the pyramidal slope. A very wide pair of bases with very little tip-'elevation' (to fit tightly within an oblate spheroid) should transition very well between same-pyramid faces, like a bulgy button, but one with highly acute tip-angle (prolate, likewise) might find the dominant behaviour to be tip-to-tip tipping, more like a toggle-fastener. OTOH, for odd-numbered end-agons it would probably ratchett to subsequent sides as it tips back and forth so long as it has enough energy to it.
If you're asking about lines of latitude intersecting, consider that near the poles of either latitudinal reference the division of the other reference-system is going to be spliced more irregularly and thus give varying degrees of stability to rest upon.
(Also, do you have a latitudinal line that crosses both pairs of poles, or are you deliberately moving them by half a phase (1/512th of the relevent circumference) so that you at least don't have them entirely coincident.)
I believe the suggested scheme would be to take a dodecahedron or icosohedron (either of the two duals can be used to start with) and then subdivide each face in such a manner that equally-sized (but differently distorted) hexagons – and 12 little regular pentagons of identical area fitting in at the old dodecahedron centre/the old icosahedron vertex – emerge from the required segmentation/vertex-truncation and readjustment the radiality of all new mid-edge vertices (or maybe the newer-edges' centres or the newer-faces' centres) to touch the unit sphere. If done symmetrically, it should be entirely fair.
The face-count might be troublesome, though. The twelve necessary pentagonal faces leaves 65524 hexagons, to split evenly between* either 12 or 20 zones, and it should be obvious that neither is possible**, in whole numbers, given the starting point of 2n faces...
(* - you can, and probably will in this design, have some that cross between two of the top-level polygons, but you can fully 'donate' as many as you then fully get donated from the next face around, so it might as well be just counted as a group of while tiles on an a set of Escher-like interlocking 'rough' polygons.)
(** - If you're using 12 zones, that's 3x4x(however many in the zone + one corner each) and there's no factor of 3 in any value that is 2usup>n</sup>. Arranging into 20 symmetrical zones (5x4), you will find that 65524 isn't divisible by 5, either...)
You could probably arrange an N-ahedron with the number of faces being 12+(12a) or 12+(20b), for some higher value (a bit of mental arithmatic suggests 65592 might be that value) and mark all the 'excess' faces (56?) with "Roll Again!". Or perhaps some pithy motivational slogans that also convey roughly the same meaning... :P 172.70.162.5 11:32, 31 May 2022 (UTC)


I don't know why it's so big? Seems like it should have a diameter of approx. 1 meter. 172.70.130.105 21:37, 30 May 2022 (UTC)

Cueball is 50 pixels high. The ball is 340 px high. Assuming Cueball is an average-height male (1.7m), and is standing the same distance from the viewer as the center of the ball, roughly how large is each face of the polygon? Area of a sphere is 4.pi.r.r, r=0.85, so 9.08 m^2 or 9080000 mm^2, divide by number of faces, get 277 mm^2, so we get 1.6cm to a side. If I did that right, then you're right: those are fairly large faces. --172.69.70.39 05:58, 31 May 2022 (UTC)
I ran the calculations for the Trivia section. I used 12pt font which gave each number an area of 1/6 square inch (about 1 square cm) 162.158.106.237 06:57, 31 May 2022 (UTC)

Should the title and picture file use "d" or the comic's difficult to type "ᴅ"? While False (talk) 21:55, 30 May 2022 (UTC)

Since xkcd uses small caps as lowercase letters, the "ᴅ" should just be considered xkcd-font for "d", and as such need not be used on the title, which is not using the xkcd font.
Ah! While False (talk) 06:15, 31 May 2022 (UTC)

If you really did want to generate a 16 bit integer with physical dice, it would be much simpler to roll a hex die four times. Clayot (talk) 23:30, 30 May 2022 (UTC)

Rolling a binary die 16 times would also work. You can get binary dice for 1¢ each. 108.162.245.69 01:31, 31 May 2022 (UTC)
Those 1¢ "dices" are not exactly guaranteed to be random. -- Hkmaly (talk) 06:12, 31 May 2022 (UTC)
They seem as random as other dice? Am I wrong? 172.70.230.63 09:33, 31 May 2022 (UTC)

I think the hardest part (or maybe second-hardest part) is figuring out which facet is the one on top. 162.158.78.109 00:46, 31 May 2022 (UTC)

Roll it on a glass table, check from below which face it's landed on instead. Wait until it has settled safely, though, or it might land on your face! 172.70.90.227 04:58, 31 May 2022 (UTC)
Good plan. Assuming standard dice design, subtract the value from 65537 to get the value of the uppermost face. --172.69.70.39 05:58, 31 May 2022 (UTC)

What material should it be to be light enough to easily roll it but cheap enough that doing the 1,5 meters doest cost a fortune ? Sorry if the question is not clear. 141.101.69.30 05:50, 31 May 2022 (UTC)

I recommend making it hollow. You could probably do something like this for $3000 if you made it out of 1/8th inch acrylic plate. 162.158.106.237 07:02, 31 May 2022 (UTC)
At first I thought aluminum for sturdiness, but really you could make this out of cardboard for dirt cheap, lasercutting precise shapes, but you'd have to design its structural frame to keep it intact, exchanges design effort for price. 172.70.230.63 09:32, 31 May 2022 (UTC)

I disagree with this dice being really random. Like, sure, if thrown correctly, but that's going to be quite hard. -- Hkmaly (talk) 06:12, 31 May 2022 (UTC)

True. For a rolled die to be random, it needs to roll far enough so that the initial orientation no longer governs the outcome. Say, ten times the circumference, or about 150 meters? -- Dtgriscom (talk) 10:28, 31 May 2022 (UTC)

Should it be related to https://xkcd.com/221/ ? 162.158.183.246 08:07, 31 May 2022 (UTC)

I'm going to wait, I think - I don't think there's room in my attic for this as well as all the Betamax kit, my drawers full of MiniDiscs and my Zune collection. No, I'll sit tight - I'm hearing encouraging things about the introduction of the Magic 65536-Ball... Yorkshire Pudding (talk) 09:41, 31 May 2022 (UTC)