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Bard117
Bard117
4:57 PM
Would one of the mods be able to close my question on mars mission trouble? It's a poorly worded question that I'd like to potentially restructure. I tried awarding the bounty to delete the question but it is unable to be deleted now.
 
user61230
@Bard117 Done!
 
Bard117
Bard117
Thank you
 
 
2 hours later…
Mike Earnest
Mike Earnest
6:52 PM
@xnor So you're asking for which numbers k it's possible to design two loaded k-sided dice, labeled 1,2,...,k, so that their sum is uniformly distributed over 2,3,...,2k.
 
xnor
xnor
yes, exactly
oh, not uniformly distributed but distributed as the sum of two equal-weighted dice with sides 1 through k
but we want a non-trivially different design
 
Mike Earnest
Mike Earnest
ohhhh gotcha, that's interesting, I hadn't thought about it
 
xnor
xnor
so if we shift the dice down from 0 to k-1
they have generating function (x^k-1)/(x-1)
also ignoring scaling
so the two-dice sum g.f. is the square of that
the question then is whether we can split it into two factors whose coefficients are non-negative real numbers
we can't do that for prime k because (x^k-1)/(x-1) is irreducible
we can do it for k=6 as the problem shows, and i think for twice an odd prime in general
 
Mike Earnest
Mike Earnest
7:20 PM
When k is a perfect square, I think it is doable
let k = h^2
we can factor 1 + x + ... + x^{h^2 - 1}
sorry, I mean we can factor (1 + x + ... + x^{h^2 - 1} )^2 as
(1 + x + ... + x^{h-1})^2 times
(1 + x^h + x^{2h} + ... + x^{(h-1)h})^2
both of these factor has positive coef's, and represent h^2 sided dice
for example, when k = 9, the labels on each dice are
0,1,1,2,2,2,3,3,4 and
0,3,3,6,6,6,9,9,12
@xnor but that's all I've figured out
actually, I think a similar construction works as long as k isnt square free
 

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