Correction to
-----[WRONG STATEMENT:]
WLOG up to rotation, flips, and uniform scaling we can consider just the
parallelograms generated by the complex numbers 1 and tau where Im(tau) > 0.
-----
I should not have included flips; a correct statement is:
------
WLOG, up to rotation and uniform scaling we can consider just the parallelograms
generated by the complex numbers 1 and tau where Im(tau) > 0.
-----
(This way, amphicheiral mirror-image parallelograms correspond to distinct tau
in the upper half plane.)
-----Original Message-----
>From: Dan Asimov <dasimov(a)earthlink.net>
>Sent: Jan 13, 2018 8:48 PM
>To: math-fun <math-fun(a)mailman.xmission.com>
>Subject: Further dissection thoughts
>
>Further thoughts: The total measure of the cuts needed for a dissection
>might be an interesting quantity to optimize in typical dissection problems.
>
>Maybe the simplest plane dissection problem is
>
> an arbitrary rectangle of unit area —> unit square
>
>or more generally,
>
> an arbitrary parallelogram of unit area —> unit square.
>
>WLOG up to rotation, flips, and uniform scaling we can consider just the
>parallelograms generated by the complex numbers 1 and tau where Im(tau) > 0.
>
>E.g., for dissecting a parallelogram of unit area into the unit square, what is
>the infimum of the total lengths of the sets of cuts that suffice?
>
>(Simple case: an n^2 x 1/n^2 rectangle. Answer appears to be 2n(n-1).)
>
>Q2: Is the infimum of the total cut length necessarily realized by an actual
>dissection?
>
>Q3: A slightly more interesting problem might be to answer the same questions
>for the two tori C/L and C/L' where
>
> L = Z + iZ
>
>(the "square" torus), and
>
> L' = Z + tau*Z
>
>(a parallelogramic one) where tau is a complex number in the upper half plane (Im(tau) > 0).
>
>are two lattices that are subgroups of the complex numbers C (as additive groups).
>
>Example: C/L' = an n^2 x 1/n^2 rectangular torus.
>Answer appears to be 2n^2.)))
>
>Example: C/L' where tau = -1/2 + i*sqrt(3)/2 is the hexagonal torus, the result of
>identifying opposite sides of a regular hexagon. (But we need to first scale it to have
>area = 1, which I think means we need to use tau' = (4/3)^(1/4) * tau.) The hexagonal torus
>is the only shape of torus besides the square one that has "maximal symmetry": Any small
>enough perturbation of the torus results in one with at most equal symmetry.
>
>I don't know the answer for the hexagonal torus, but it is sure to be interesting.
>
>Final questions:
>
>FQ1: What is the *supremum* over all shapes of torus, of all the infima
>of the sets of total lengths of sets of cuts that suffice to dissect?
>
>This might turn out to be infinite.
>
>FQ2: If it's finite, is it realized by an actual parallelogramic torus?
>
>—Dan
>
>-----
>I thought there might be some other interesting things to optimize here.
>
>Given a sphere, what's the largest exocube* it can be dissected into
>using *any* number of pieces? We include infinitely many, so we should say:
>What is the supremum of the exocubes that can be so made?
>
>[Aside: I was momentarily startled to see my friend's computer's autocorrect
>had changed exocubes to excuses, altering the sense a just a bit.]
>
>Alas, I think the answer is that an arbitrarily large exocube could be built.
>
>(((Hmm, what if you went the other way, from a solid cube to an exosphere?
>Again it looks as if there is no upper bound to the size of the exosphere.)))
>
> * * *
>
>But it might be interesting to minimize (find the infimum of) the *total area*
>of *all cuts*. Given a sphere containing volume = V, clearly the exocube must
>have side >= V^(1/3), so the TAC = total area of all cuts must be >= 6*V^(2/3).
>
>But that can't be the infimum, can it?
>
>—Dan
>—————————————————————————————————————————————————————————————————————————————
>* Definition:
> -----------
>An *exocube* looks like a cube from the outside and consists of the closure of
>a cube after their interiors of at most countably many disjoint closed topological
>3-balls have removed from it.
>
>
>Allan Wechsler wrote:
>-----
>It's obviously possible to dissect a sphere into pieces that could be
>reassembled into something that looks like a cube from the outside. Inside,
>it would be hollow, possibly with a bunch of spare pieces rattling around
>inside.
>
>Feasibility sketch: build thin polyhedral planks from the inner part of the
>sphere, with 45-degree bevels where necessary to let them form the cube's
>edges and vertices. All the extra material could be chopped up into
>manageable chunks and hidden in the interior. If there's too much extra
>material, redesign with thinner planks to make the outer cube bigger.
>
>Intuitively, it feels like we could get away with just a few dozen pieces
>or so: maybe 4 to 6 per face, and then an approximately equal number to
>pack inside ... but I don't know. Can anybody provide an explicit
>construction, or make lower-bound arguments about the number of pieces?
>-----
>-----
