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09:25:04 <river> https://www.youtube.com/watch?v=le5uGqHKll8 hangman is like wordle
10:31:28 <river> https://www.youtube.com/watch?v=64pA31_WJa0 vsauce2 the L game, this channel also has a video analyzing sprouts game
11:25:12 <river> https://youtu.be/Y7hm0Xeicus?t=546 oh that's interesting, simon tatham did research on Conways soliders
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13:01:43 <river> I didn't realize that "grist for the mill" was a saying
13:02:00 <river> grist is used in homestuck
13:22:16 <river> platonic solids give us 4, 6, 8, 12, 20
13:22:27 <river> dipyramids give us 8, 12, 16, 20, 24, ...
13:22:50 <river> trapezohedra gives us: 6, 10, 14, 18, 22
13:22:56 <river> Rhombic Triacontahedron: 30
13:23:10 <river> disdyakis triacontahedron: 120
13:23:19 <river> i am not sure if that is every dice, but anyway
13:23:41 <river> you can multi-label a dice to divide its face number
13:24:04 <river> so you can put each label 4 times, to get a 5 sided dice
13:24:21 <river> how do we get a 7 sided dice?
13:24:30 <river> i dont like the prism dice
13:24:56 <river> looks like a trapezohedra /2 gives 7
13:25:15 <river> so the next difficult number is 9
13:25:39 <river> looks like the multi-label system is very effective, and we can get a dice for 4,5,6,7,8,9,10,.. easily
13:28:10 <river> https://en.wikipedia.org/wiki/Isohedral_figure isohedral face transitive shapes
13:34:09 <river> i would propose that we should label dices starting from 0 rather than 1
13:35:01 <river> it makes it easier to reduce the result modulo whatever
13:40:28 <river> something i was wondering though, is can we multi-label multiple dice in such a way that when rolled together and added up we can get a uniform distribution?
13:40:34 <river> and does this provide new numbers
13:40:50 <river> (the dice on their own may not be fair)
13:53:43 <river> https://www.shapeways.com/product/F4S34GX6T/d12-balanced-steam-clock?optionId=171620058&li=marketplace
13:53:54 <river> you can 3d print metal dice
14:03:48 <esolangs> [[Esolang:Introduce yourself]] M https://esolangs.org/w/index.php?diff=93374&oldid=93361 * Kilgharrah * (+355) /* Introductions */
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14:09:29 <esolangs> [[Kak--]] https://esolangs.org/w/index.php?diff=93375&oldid=93356 * ChuckEsoteric08 * (+96)
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14:15:24 <esolangs> [[Kak+]] N https://esolangs.org/w/index.php?oldid=93377 * ChuckEsoteric08 * (+727) Created page with "'''Kak+''' is a Two-demensional version of [[Kak]] Made by [[User:ChuckEsoteric08]] ==Commands== {| class="wikitable" |- ! Command !! Meaning |- | ! || move memory pointer to..."
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18:06:16 <esolangs> [[Mountain]] https://esolangs.org/w/index.php?diff=93397&oldid=93378 * ChuckEsoteric08 * (+67)
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21:48:04 <oerjan> <river> i would propose that we should label dices starting from 0 rather than 1 <-- as i recall that's common for d10s, which you then use in pairs to throw 0-99 values
21:48:49 <oerjan> well one of the pair might be marked 0,10,...,90
21:51:15 <oerjan> heh, i see that also answers your next question
21:52:49 <oerjan> generalized, it gives you a way to multiply dice sizes
21:53:13 <oerjan> so in some sense, prime sizes are hardest
21:54:48 <int-e> The Cursed Remainder Theorem
21:55:13 <oerjan> <river> (the dice on their own may not be fair) <-- not sure if this helps you get more or not
21:56:49 <oerjan> someone's almost certainly investigated this
22:00:51 <int-e> There's a related exercise... where you relabel two D6 such that the distribution of the sum is the same as rolling two standard D6
22:03:37 <oerjan> is there any way other than adding a constant to the labels of one and subtracting it from the other?
22:03:38 <int-e> > let d1 = [[1,2,3,4,5,6],[1,2,3,4,5,6]]; d2 = [[1,2,2,3,3,4],[1,3,4,5,6,8]]; dsum = sort . map sum . sequence in dsum d1 == dsum d2
22:04:26 <int-e> oerjan: it was phrased in a way that rules that out
22:06:06 <int-e> And there's a pen&paper solution based around x^5+x^4+x^3+x^2+x+1 = (x+1)(x^2+x+1)(x^2-x+1); the modified dice correspond to (x+1)(x^2+x+1) and (x+1)(x^2+x+1)(x^2-x+1)^2
22:06:36 <int-e> which is one of the first applications of generating functions that I saw, so it stuck
22:08:48 <int-e> oerjan: I guess the labels were restricted to positive integers
22:09:11 <river> so I can have a coin with sides 1 and 3
22:09:26 <river> plus a 5 side dice with sides 1, 2, 5, 6, 9
22:09:55 <int-e> > let dsum = sort . map sum . sequence in dsum [[1,3],[1,2,5,6,9]]
22:10:17 <river> i didn't want a gap
22:10:35 <fizzie> If you have a set of N coins where the two sides of coin i are labeled 0 and 2^i, are those called bitcoins?
22:10:57 <int-e> damn it, I laughed
22:11:17 <int-e> river: it happens, this was just my way of pointing it out
22:11:46 <river> how do you handle a -
22:11:50 <river> (x^2-x+1) like here?
22:12:09 <int-e> you don't. but the resulting polynomials have no negative coefficients left
22:12:14 <river> what's that corresponding to in terms of dice side
22:12:30 <int-e> they're x^3 + 2*x^2 + 2*x + 1 and x^7 + x^5 + x^4 + x^3 + x^2 + 1
22:12:30 <river> you gotta get rid of them by grouping them with things
22:14:22 <river> i think the best thing to do is write a program which looks for good factorization based dice
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22:59:39 <fizzie> I think if you have two normally labeled d4's (sides 1, 2, 3, 4) that you'd want to be unfair in such a way that their sum is uniformly distributed over (2, 3, 4, 5, 6, 7, 8), you can't. Or at least WA -- after some renaming to get it to even try -- claims to me there's no real solutions to a1*b1 = a1*b2 + a2*b1, a1*b1 = a1*b3 + a2*b2 + a3*b1, ..., a1*b1 = a4*b4, a1+a2+a3+a4 = 1, b1+b2+b3+b4 = 1,
22:59:41 <fizzie> which I think would need to be true.
23:00:19 <fizzie> And in the simpler case of having just two coins labeled 1, 2 and wanting the sum to be uniformly distributed over (1, 2, 3), it's definitely the case that for {a1*b1 = a1*b2 + b1*a2, a1*b1 = a2*b2, a1+a2 = 1, b1+b2 = 1} there's no real solutions too. Because after some simplifications, that just leaves b1 = 1-a1 and a1^2 - a1 + 1/3 = 0.
23:01:46 <fizzie> (Presumably there's a less pedestrian way of handling it too.)