Talk:2626: d65536

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 whole 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 2ⁿ. 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)

Postcript: Ok, so this is my idea for face-placing. Take a D8 (octahedron) and divide each of its 8 originally triangular faces into 8192 smaller faces (alternatively, start with a cube and progressively truncate its corners towards the same end). This is not a divisible by three number (neither can you put one in the centre, the rest are divisble by three and can surround it symmetrically), but you don't need strict rotational symmetry in any way. The opposing side can reflect/copy the non-symmetry as required to create any useful symmetry across the whole of the structure (and make floored-base/upmost-face pairings, amongst other things).

As long as you make the faces equally likely to land on and stay on (could be hyperstellated as a slightly flat irregular 8192agon-based right-pyramid with the pyramid-faces of adjacent sides matching or meshing edges with those of each other, or a complicated mostly-hexagonal mesh, or a triangular one that's a limited fragment of a fine geodesic-like bulged pattern) by some suitable scheme governing area, aspect ratio and inter-face angle of incidence (probably normalising features to touch the unit sphere, for a start) then it should do it fairly and with exactly 65536 faces. I leave the fine-tweaking up to someone else. 172.70.162.5 12:59, 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)

The lowest-value coin of all is the Tiyin from Uzbekistan. Some 3,038 equate to one UK penny (and 2,000 tot up to one US cent) from https://www.bbc.co.uk/news/magazine-21572359. Arachrah (talk) 15:13, 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)

You can reduce bias by taking two not quite fair coins. Flip them together. If both heads, or both tails, then record a 0. If different, record a 1. Arachrah (talk) 15:13, 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)

Because computer binary counting starts with ZERO (and in this case ends with 65535) one has to subtract from 65535. This die would not have a 65536 and it would have a zero. Inquirer (talk) 22:38, 31 May 2022 (UTC)

Notice that the parent comment says “assuming standard dice design” and standard dice start with 1 and end with their d number: a standard d4 has faces 1,2,3,4; a standard d6 has faces 1,2,3,4,5,6; a standard d10 1,2,3,[…],10; and a standard d100 has faces 1 through 100. Which is why rolling 2xd10 does NOT yield the same results as rolling 1xd100, because one cannot roll a 1 OR a 100 with two d10’s, and other numbers are over represented. This is why I have both a d12 and a d100 in my set of dice…John (talk) 11:40, 2 June 2022 (UTC)

D10s are frequently (IME) numbered 0..9, with underlines to distinguish 6 from 9, whether or not that is as a deliberate sop to pairing up for percentiles ("It's the red dice first...", "It isn't!", "It is, it's always the red dice first, that's a 9 percent fail, not at 90 percent success!"). The 0 opposite 9, 1 opposite 8, etc.

If you're rolling 1..10s (depends upon the system being used for... and arguments about the original intentions, possibly, where 10 of something or zero of something turning up/discovered means a lot to what exactly happens next) you treat them as full 10s, and double-zero can (again, hopefully understood in advance) be a "natural 100", rather than a natural-zero, to someone's obvious advantage or detriment.

Similarly, games with D4 may use 0..3 as its outcomes, or D3s (D6/2s) have 0,0,1,1,2,2 as faces. Or if you are somehow stuck with a 1-based die of the appropriate (or divisible) range, and need to play something that requires a 0-based one, you agree that MAXNUM->ZERO or N=N-1.

From my recollections of the truly huge collection of multicoloured, multisized and variously multifaceted dice that was dragged out for wargaming/role-playing sessions, years of accumulations from various sources, the D6s (spotted, not "Hit", "Missfire", <arrows for miss direction>, or otherwise for other uses), D8s, D12s and D20s were habitually 1-upwards, and Ds with 3, 4 and 10 almost all 0-upwards. D6s were the most numerous, but D10s well up there (enough for colour-mismatched pairs for each of a whole battalion of participants, even if the Hit/Miss die had to be passed around at need, and the few big yellow D20s might need to be offered onwards/rolled on someone's behalf on occasion).

In short, the 'standard' is perhaps more firmly that opposite sides add up to double the average (not possible with D4, which is also "point-marked", not one number per side), though I'd like to check a D3 to see if it's 0 opposite 2 (x2) and 1 opposite 1 (which also happens to be a 0..5-dice with the opposition rule, but modded by 3) or perhaps in a basic {1..6}mod3 conversion 1 opposes 0(=6), 2 opposite 2(=5), 0(=3) opposite 1(=4). Maybe next time I go a gaming I'll get the chance. 172.70.86.44 12:48, 2 June 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)

Interesting to consider the 'necessary minimum'. Simplify to a "wheel of fortune" (just one axis of continual rotation) it would depend upon the potential variation of imparted rotation. If (say) 'aiming' at two whole rotations has a (perhaps 'normal') spread of variance that relates to ±½ rotational uncertainty at the 1st and 3rd quartile of probability then the sub-first and above-third 'tails' might wrap around to (roughly) equalise the chances that 2±(whatever fraction) spins lands just about anywhere just about equally. Aiming at four whole rotations (similary ±1 spin at the given quartiles, and the tailing chancs 'filling in' above 5 rotations and below 3) would smooth things out, all else equal, but takes twice as much perceived/attempted effort for not much more 'randomising'.

Similarly, requiring 10 full rolls (maybe honestly aiming for 10, but allowing it to be 7.5 or less if not obviously 'just nudged') seems overkill, in the single dimension.

Except, of course that you also need enough distance (on top of whatever factor you consider practical as a variation-wrapping value, which might not be the ½-in-2 I give) to also roll sideways. If for some reason you really don't want to roll 65536 or 1 (or is it 65535 and 0?), which may be on polar-opposite faces, you might make sure that they are directly to the left and right before you propel the die forwards a little, not caring which distribution of numbers is on/near the rolling-equator (2 is acceptible to you, and 65533, etc; other very low/high values conceivably placed on that thin band of "wheel-like chance" but you're just avoiding the very largest and smallest, or specifically just the one of them) but knowing that it's more unlikely to easily present the exact face(s) you dislike than it might be in a truly 'fair' roll.

Perhaps the best thing is to have a rolling track to send the thing down that puts it the required "two or so rotations" forward to then either hit a wall or climb slightly up a slope (at a roughly 45 degree angle) that then sends it back roughly sideways to the original vector for a similar distance with a perpendicular or even composite moment of rolling rotation, to bring 'initially axial' numbers fully into play... And that dog-leg would require a sligthly shorter length from launch-position to where the thoroughly mixed-up final stopping point should be, whilst significantly foiling the master-manipulators who actually try to arrange an initial setup that favours better final results (rather than just nudge it, uncaring, for a result not as totally random but certainly not more predominently of desired-for ranges than otherwise). 141.101.99.8 12:28, 31 May 2022 (UTC)

At what point does the structural material the die is composed of, combined with its mass, create a smoothing effect that will destroy the fairness of the die. I mean a small plastic die is no problem. A 2-ton acrylic die would start grinding off the edges of some faces with every roll, would it not? 172.69.69.122 13:35, 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)

The number of sides on the die inside the ball is not what determines the name of the ball. It's the exterior housing which is colored in the manner of an Eight Ball. The classic design uses a d20, and is still called an Eight Ball, not a Twenty Ball. 172.70.130.195 18:00, 31 May 2022 (UTC)

Full disclosure: I don't have any of those things in my attic. And I'm not entirely sure, but I don't think Randall thinks rolling a d65536 is genuinely the hardest part of generating random 16-bit numbers. And Grape Nuts contain neither grapes nor nuts. Yorkshire Pudding (talk) 23:37, 31 May 2022 (UTC)

I'm suprised the hidden message points to 2624. I would've thought it would point to 2626 to refer to itself. Maybe things didn't get published as intended? Or maybe Randall really just wanted to point people to the Voyager comic? Linux2647 (talk) 18:13, 31 May 2022 (UTC)

I'm no ASCII expert, but from the description provided I'm pretty sure the comic URL would require the number representing "26" to show up twice. A die with, say, two 13,359 faces would obviously not be fair. If only Randall had published this as #2625 or #2627! (Or maybe he planned to publish it last week and had to shuffle his schedule after finalizing this comic?) GreatWyrmGold (talk) 18:24, 31 May 2022 (UTC)

Probably the latter, seeing as it doesn't actually line up so that any of them are actually "26". The numbers are xk-cd-.c-om-/2-62-4/, so the 26 and 24 aren't lined up like that. 172.70.126.87 19:38, 31 May 2022 (UTC)

Perl code to decode the ASCII:

perl -E 'for (30827, 25444, 11875, 28525, 12082, 13874, 13359) { print chr($_ >> 8), chr($_ & 0xff) }; print "\n"'

Remember that "die" is singular and "dice" is plural Mushrooms (talk) 09:30, 1 June 2022 (UTC)

Might be worth offering some easier alternatives..

Needless to say, 16 ordered coins (eg ordered by the date of their minting) provide a much easier alternative to a d65536.

Other (easy to find) d p^2 - such as 8 ordered d4, or 5 ordered d8 and a coin (or bit-shave a dice.) It's true some rules need to be applied (the highest number of the die is treated as a 0, and the order of the dice is strictly followed).

For example using a rainbow spectrum ordering of 6d8, I role: 7, 1, 2, 8, 3, 5. Each dice represents 3 bits - 111-001-010-000-011-101. We shave off the last two bits (because we want 16 bits, not 18, for a d65536) 1110 0101 0000 0111. Hex = E507, which is decimal 58,631.

While it takes a few seconds for a human to convert the number it is quite trivial to write a program to convert an image capture.

There are also specialist dice - such as d16 'hexidice' which provide 4 bits per ordered die, and far less human calculation. There are even d256 hex dice made, but they suffer the same problem that a d65536 would have.

20040302 (talk)

how are you trivially converting image captures? i still use my keyboard. what's the update? 162.158.79.120 18:26, 1 June 2022 (UTC)

I found an octahedral Goldberg with 65538 sides, 6 squares and 65532 irregular hexagons, notated GP₄(128,0). Can be constructed by chamfering a cube 7 times (Conway cccccccC). I don't think anything can be closer. SDSpivey (talk) 04:37, 2 June 2022 (UTC)

I have had two designs in mind:

• Start with octahedron, subdivide each face triangle into 16,384 triangular faces (subdivide all original edges into 128ths, effectively, while "halving into quarter-sized triangles), pairing the triangles up into half that many 'diamonds' then inflating to touch the bounding sphere, keeping the diamonds planar and possibly equal area, but will also distort them (some into kites, others may be utterly irregular) so it'll be a (literal?) balancing act to make all tippings between faces equally possible.
• Start with a tetrahedron, subdivide likewise, stick with the triangles, have the same issue of inflating the faces towards the unit sphere while balancing the resulting distortions.

I tried to work out something that seems to exactly match the image (or as exactly as I perceive it to be) with mildly distorted hexagons - albeit with the necessary pentagonal substitutions to carry the curvature - formed of (a similarly subdivided set of) sub-triangles drawn out on the original surfaces of either icosahedron or a snub-stellated extension onwards the dodecahedron, but I couldn't get the numbers to add up. Which is not to say that there isn't a way to do it. I was going to spend more time on this issue later, so if I get past the issues, then... Watch this space? 172.70.162.77 13:17, 2 June 2022 (UTC)