9 ms·
The cursed d65536
- klyrs 4y agoLook at the humble d10 for inspiration. The trick is to avoid making a spherical-looking die. A fair d65536 could look like a pair of cones glued together, with 32768 sides per cone. Alternatively, one can make a fair 65536-sided cylinder with rounded ends.
- aidenn0 4y agoI know it's not the point of TFA, but you can just roll a d8 6 times, generating 3 bits each time, for a total of 18 bits, and then discard two of them.
- bombcar 4y agoYep, or use 16 dice that have only 0 and 1 on them. The order would be important, so you'd want to color them or otherwise indicate which was high order and which was low.
- anyfoo 4y agoOr do 16 throws of such a d(2) die, i.e. 16 coin flips, in succession. If you really want to do it in one throw, you don't really need to color them all differently, though. Just read them in a specified way, i.e. always from left to right and top to bottom. (You don't even have to always read them the same way as long as how you read them is completely independent from what their faces show.)
- jandrese 4y agoWhy not just roll 4D16? https://www.amazon.com/Lanema-Polyhedral-Dungeons-Dragons-Sided/dp/B08GL79B12 https://www.amazon.com/Lanema-Polyhedral-Dungeons-Dragons-Si...
- aidenn0 4y ago"... for Dungeons and Dragons" ah yes, I use a D16 all the time when playing D&D /s
- bargle0 4y agoIf the d12 is for the great axe, then the d14 is for the greater axe and the d16 is for the greatest axe.
- jeramey 4y agoI'd be a little worried that I might be tempted to unconsciously order the values in some way which would bias the results. But you could build a box with four separate cells and a clear plexiglass top, shake them all up, and always read left-to-right or right-to-left! Sounds like a fun weekend project!
- timando 4y agoAre those dice fair?
- tialaramex 4y agoYou can buy hexadecimal dice ie D16 with the sides labelled 0 through F, I have some, I use them to generate random numbers when it is important to me that I personally have confidence these are random numbers. With hex dice you can generate 16 bits with four rolls, four bits each time. Obviously you're more likely to already have a D6 somewhere, but then you're even more likely to own a coin, and 16 coin tosses is also an effective way to generate a 16-bit number.
- Agentlien 4y ago> Obviously you're more likely to already have a D6 somewhere, but then you're even more likely to own a coin I'm not sure this is still true. At least not here in Sweden. Everyone I know owns multiple board games which come with at least one D6 and in recent years I've had discussions with friends and family about us paying everything electronically nowadays. There has even been talk about the risk to our society of no one having cash anymore. In fact, some weeks ago I needed a coin to show my kids when they asked about this physical money they'd heard of and I couldn't find one.
- aetherson 4y agoThe article suggests flipping coins in the last sentence, which gets you exactly as many bits as you want.
- dragonwriter 4y agoOr you could just roll a d4, generating 2 bits each time, 8 times, and not worry about throwing out anything.
- sp332 4y agoFrom the broken first attempt: Each face has 3 vertices, shared between 6 faces, so the total number of vertices is 65536, and they become faces of my die. This should have tipped you off. If you have six triangular faces around every vertex, you have a flat Euclidean plane, not a sphere with positive curvature. For another example of this: https://m.youtube.com/watch?v=jfSTwqmrQDc https://m.youtube.com/watch?v=jfSTwqmrQDc
- avmich 4y agoThat's why "don't do math after midnight". But I don't get what's wrong with repeated throws of tetrahedron - 2 bits of result at a time - or even icosahedron, with re-throws if one of 4 "wrong" sides is up - 4 bits at a "good" throw...
- thaumasiotes 4y agoYou can also make repeated throws of an octahedron for 3 bits per throw. 16 is not a multiple of 3, but there's no difficulty with just discarding the 17th and 18th bits. So you'd need to roll 6d8 instead of 8d4.
- tgb 4y agoPedantry: it doesn’t have to be flat - for example a triangulated parabola could also have that configuration. You only get a topological result from just knowing edge and vertex counts. Now if the triangles are identical equilateral then you’re in business. What if the triangles are all congruent but not equilateral? Can that even happen? That’s a fun one, so I won’t spoil it.
- sp332 4y agoOh, do you think it could make a sphere then? If the angle at each triangle corner is a bit less than 60 degrees?
- tgb 4y agoNot a sphere, we know that can’t happen due to the topological constraint you brought up. Instead picture a float plane tiled by equilateral triangles. Now mark each vertex as either low, middle, or high such that every triangle has one of each. Push the low ones below the plane and the high vertices above the plane by some amount x. Now it’s a bumpy plane full of triangles, each one is isosceles and all identical. I think that’s the only way to do this, but maybe there are more. Could we get a hyperbolic plane this way? Normally you squeeze extra triangles around each vertex to do that so I doubt it but maybe.
- woliveirajr 4y ago> XKCD joked about the problem of secure generation of random 32 bit numbers by rolling a 65536-sided die Well, 2^16 = 65536, so it's a 16-bit number. To get a random 32-bit number you'd need to roll more than once....
- recursive 4y agoXKCD got it right. aleph.se got it wrong.
- kazinator 4y ago> This is of course hilariously cursed. It will look almost perfect, but very rarely give numbers outside the expected range. If the D65538 is fair, it is usable: you just discard the two unwanted values when they show up and roll again. Those faces could be labelled as "roll again". If you have a uniform source of random numbers from 1 to N, you can get a uniform distribution from 1 to M < N simply by discarding values obtained above M. Those values are just "rain that fell elsewhere".
- stkdump 4y agoThis process has an infinitely small (but non-zero) chance of never terminating.
- kazinator 4y agoThe die also has a very small chance of landing on an edge between two faces, which also requires a repeated roll.
- davidgay 4y agoBut that probability rapidly drops below the probability that you will drop dead from some random cause while waiting for the roll's result :)
- phinnaeus 4y agoI think the most likely cause of death in that case is being crushed by your d65536
- a1369209993 4y agoI think that would be much less likely than being crushed by your d65538, given the premise.
- Smaug123 4y agoThis is a rather sloppy way of saying "this process may fail to terminate, though the set of nonterminating outcomes has probability 0". No real number is infinitely small but nonzero.
- tlb 4y agoI'd like a physics analysis of how flat/hard a surface you need to roll a D65536, how long it would take to settle, and how good a microscope you'd need to read the top face. Is there, like, 99designs for physics questions? I'd happily pay $99 for the answer, then post it here like I worked it out myself.
- hnuser123456 4y agoa 1ft diameter d65536 could be broken into surfaces with area 4.45 square millimeters, or a little over 2x2mm. With good eyes and enough squinting, a microscope might not be needed. maybe fill it with liquid with a tiny air bubble to find the top face 1in diameter, however, gives 0.175mmx0.175mm per face, or about the width of a hair by the width of a hair.
- thematrixturtle 4y agoI wonder how much bias engraving or even painting the numbers on the die would introduce? The weight of "1" will be noticeably different to "58888".
- eesmith 4y agoUse a 7-segment font and two paints with the same density, with one paint the same as the background color. For each digit, paint in all 7 segments. For each number, draw all 5 digits.
- JoeAltmaier 4y agoI'd look at the bottom face. Because you have a reference (the table) for the selected side, not just guessing which side is toppiest.
- samch 4y agoProbably says something about me that my immediate reaction to the number 65536 wasn’t 2^16, but rather the frequency of the timer crystal you had to swap into a Radio Shack tone dialer to make a “red box”[1] that would let you make free phone calls on US pay phones by simulating the analog coin code sound of a quarter being deposited. Fun fact that the “red box” was the first hackaday article[2] posted way back in 2004. I learned about this back in the days of 2600 magazine. Fun stuff! [1] http://www.techfreakz.org/2600faq.html#redbox http://www.techfreakz.org/2600faq.html#redbox [2] https://hackaday.com/2004/09/05/radioshack-phone-dialer-red-box/ https://hackaday.com/2004/09/05/radioshack-phone-dialer-red-...
- scrame 4y agoI just coded it in basic and the recorded it to a mini-casette lecture-recorder. Unfortunately, by that point the tones just triggered an operator call.
- reaperducer 4y agoThe DTMF chips that generated regular Touch Tones™ were capable of generating all kinds of fun sounds simply by extending the key matrix. You could build a red box, pink box, blue box, and others by modifying a standard tone dialer. The easiest thing to do was to add a fourth column of keys, which supposedly gave you the ability to use the ABCD digits required for military networks. I never tried that, though. For those of you who missed out on that era, a tone dialer was a little palm-sized box that had a small Touch Tone keypad on one side, and a speaker on the other. If you had a rotary phone, after dialing, you could hold it up to the mouthpiece and use many of the fancy features that came with the invention of Touch Tone, like using FŌN cards, or listening to your messages on your answering machine.
- skykooler 4y agoInteresting, most timer crystals I'm familiar with are 32768 hz.
- bryanrasmussen 4y agoI guess it says you're more into hardware than you're into math, when I saw the title I thought 'this must be some really weird Dungeons and Dragons article'
- pacaro 4y agoI'm surprised that nobody has mentioned the d120 which AFAICT is the largest fair dice that you can just go out and buy. Sadly it's a smidge under 7 bits, but you can "reasonably" roll up 32 bits of entropy with a mere 6 rolls (and still have 9 bits left over) https://www.wired.com/2016/05/mathematical-challenge-of-designing-the-worlds-most-complex-120-sided-dice/ https://www.wired.com/2016/05/mathematical-challenge-of-desi...
- krallja 4y agoRight - also, what’s wrong with flipping a fair coin sixteen times? Or a D16 four times?
- bluthcon 4y agoI'm a fan of the bipyramid with 32,768 facets on each side. I keep it next to my d34 and my box of metallic d7s.
- Syzygies 4y agoThere's an assumption here that a polytope with equal area faces will be fair. I'd like to see a proof. My mathematical intuition is that the probability of stopping on a face depends on the extent that nearby faces slow a rotation, which depends on the angle of attack across each face. Without symmetry, this will vary by face.
- skykooler 4y agoThe article explicitly says that equal areas are not required for fairness; and gives this explanation for wanting equal areas: "Dice that are fair by symmetry (and hence under any reasonable throwing dynamics) always have to have an even number of sides and belong to certain symmetry groups (Diaconis & Keller 1989)."
- anderskaseorg 4y agoThe four tetrahedral Goldberg polyhedra (https://en.wikipedia.org/wiki/Goldberg_polyhedron https://en.wikipedia.org/wiki/Goldberg_polyhedron) GP_III(27, 166), GP_III(38, 159), GP_III(41, 157), GP_III(102, 107) each have 4 triangles and 65532 hexagons. Most small regions of them would look roughly like the XKCD.
- thyrsus 4y agoI believe that there are fewer than factorial(65536) particles in our universe[0], so I reject using a die with that many faces to equalize the probabilities. As other comments point out, a mere 65536 faces is on the verge of practicality. [0]https://www.popularmechanics.com/space/a27259/how-many-particles-are-in-the-entire-universe/ https://www.popularmechanics.com/space/a27259/how-many-parti...
- bruce343434 4y agoA better approach would be to calculate the amount of needed nodes that would junction the vertices, and then push them apart (simulated annealing). Though it won't be a regular geometry.
- knome 4y agoit would be easier to avoid edges altogether and regularly space flat circular areas leaving curved space between them. you can have any arbitrary number of faces if you give up on the faces touching.
- chpatrick 4y agoHEALPix (from NASA!) is a great way to uniformly subdivide a sphere. https://healpix.sourceforge.io/ https://healpix.sourceforge.io/ You can't make a d65536 with it but you can make a 74^2 * 12 = d65712.