/jp/ - Otaku Culture

>>17995227
And I'm saying there wouldn't be a paradox if you don't accept the law of excluded middle, that an object created by Yukari is such that either Yukari can lift it or can't lift it.
Let R be the set of objects created by Yukari so that she can't lift it, then the definition of R is that R ∈ R ≡ ~(R ∈ R), which is a contradiction only if ~(R ∈ R) ≡ R ∉ R, i.e. an instance of the law of excluded middle.
People that don't presuppose or even reject LEM are called constructivists.

These are quite interesting, from the symmetry point of view.
>>17488740
There actually aren't any symmetries for this one. The outside star has the same symmetry group D_6 as the hexagons while the inside has that of the square, D_4. The catch is that if the displacement angle of the square in the middle from the principle axis (y axis, say) is not an integer multiple of the internal angle of hexagon minus that of the square (modulo 2pi), then the symmetry group is trivial, with only the identity e in it.
>>17488746
Here we can find 4 axes of symmetry, making the symmetry group D_4. Also one thing that's very interesting is that this is an n = 1 framing of two Hopf links Hopf-linked together, which can be characterized by the trace of twists c_{c_{ij}c_{ji}}c_{c_{ji}c_{ij}}c_{c_{kl}c_{lk}}c_{c_{lk}c_{kl}} in the ribbon category.
>>17488749
The symmetry group here consists of no reflections (due to the braiding pattern) but only rotations by pi/3 This makes the symmetry group the cyclic group of order 6, i.e. Z_6. I don't see any way that we can isotopically make the pattern recover the full D_6 symmetry.
>>17488751
There are only 2 axes of reflection and rotations by pi, which makes the symmetry group D_2.
>>17488757
The symmetry group is the same as that of the triangle, D_3.
>>17488763
This one is D_6.
>>17488767
Similar to Sanae's case the symmetries for this pattern only consists of rotations in D_4 and no reflections due to the braiding pattern, which makes the symmetry group Z_4. However if the braids were mirrored (so that the ones going over are near each other and the ones going under are far apart) then we recover the full D_4 symmetry.
>>17488770
This is just D_4. Indeed compared to Marisa's case the octagon in the middle has a displacement angle of 0 (mod 2pi) compared to the square on the outside, so the symmetry group is D_4 ∩ D_8 = D_4.
>>17488787
Left as an exercise for the reader.

Are these all by the way? If not please make more, they're very nice.

>>16992749
Reading PDFs at 2am.