• theblueredditrefugee [she/her, fae/faer]@hexbear.netOP
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      3 months ago

      Actually it doesn’t, it merely projects all genders onto the male/female plane. I don’t think this is incorrect for this purpose, for instance if we consider a xenogender entirely orthogonal to the two standard basis vectors, projection onto the male/female plane yields coordinates of (0,0). It would be reductive to say that this is equivalent to other genders that map to this point, however, it still correctly yields the result of a tie. This is the natural outcome of neither player betting on a xenogender

  • Lichtblitz@discuss.tchncs.de
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    3 months ago

    “greater than" and “less than” are missing the “or equal to”. A difference of exactly 1 should be sufficient for an absolute victory. I’m also missing a definition of what negative numbers on an axis mean, since they are currently implied to reach absolute victories and are not excluded by the definition. I’ll show myself out 😅

    • We simply define the two dimensional space as an empirical observation, where the median man is located at position (0,1). So there potentially exist men at position (0.1,1.1) but the epicenter of masculinity is at this coordinate - similarly so for female. In the construction of this simplified geometric space we have no guarantees except the median property of the two most popular genders. Since nonbinary genders frequently defy notions of categorization there is no reason to believe that the projection of an extremely high dimensional gender into this space will yield numbers with any meaningful properties. All we can say for certain is that a Manhattan distance greater than or equal to one is sufficient for an absolute win.

    • In this case you’re just fucked lol.

      If the kid has a relatively clear “average” gender, (e.g. spends 50% of the time as simply female, 25% of the time simply xenogender, 25% simply demimale) and this average doesn’t change over time, you could compute the average gender and then use the aforementioned algorithm to solve. BUT if the kid tends to move genders over years or even decades you’re never gonna have a clear winner.

      If the kid comes out as genderfluid your best bet is to just call it a draw and both players get their money back

            • Huh, I’m not familiar with that terminology. In my sphere of familiarity, “vector” mostly just means “list of numbers augmented with a dot product and matrix multiplication”. And the dimensionality of the space is just the number of values in the list, like (0,0,0,0) is a 4d vector.

              My mental model of gender is basically a collection of points in an extremely high dimensional space, and we’re just saying “find the average man, find the average woman, and use these two vectors as basis vectors for a low dimensional space which we project the entire space into for that purpose of resolving the bet”

  • morte [she/her]@hexbear.net
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    3 months ago

    What if we embed genders in hyperdimensional vectors and train a neural network to classify someone’s “unique” gender

    • Hmm, interesting idea, but what if the kid incorporates a xenogender axis orthogonal to all previously known xenogender axes? My algorithm would project away this axis and it would have no effect on the computed output, but a neural network would have randomly determined weights connected to the input corresponding to that axis, resulting in unexpected results