it seems appropriate to present the unique nontrivial magic hexagon with its numbers expressed in base 6. final number = 31 (base 6) = 19 (base 10). magic sum = 102 (base 6) = 38 (base 10).
3 25 30
31 11 1 15
24 2 5 10 13
20 4 12 22
14 21 23
first, create a random smooth shape by picking random weights for the first few spherical harmonics and summing them. reject if the object is not convex: determining this seems difficult. can the coefficients be chosen to favor convexity? previously 2D.
then, uniformly randomly sample some points on the surface and construct tangent planes, forming a random convex polyhedron (might not be closed). uniform sampling of surface points on an arbitrary surface seems difficult. maybe a random walk of small steps each of approximately constant geodesic length. how many steps?
or, permit starting with a smooth shape not necessarily convex. reject a tangent plane if its inward half space does not contain the origin.
extend this construction to higher dimensions. what are the hyper-spherical harmonics? curse of dimensionality probably somehow happens.
a 4D polytope (polychoron) is interesting because its cells are random polyhedra whose faces fit together pairwise but not further in 3D. how can a collection of specified irregular polyhedra be manufactured? 3D printing is of course one way. or, mill some faces, but include nubs on one or more faces. then, reorient the piece upside down, holding it in place from below by the nubs. mill the remaining faces. then, somehow remove the nubs.
mill two negative halves then cast then join somehow?
create a net and fold (origami)?
the edges of a random convex 3D polyhedron define a random planar graph. do analogous graphs for convex polytopes in higher dimensions have a property analogous to planarity?
Delaunay triangulation is another way to get a random planar graph, but faces are generally always triangles. but are most of the faces produced by the tangent method described above also triangles?
3D convex hull of random points results in a convex polyhedron, and this generalizes to arbitrary dimensions. I think most faces will again be triangles (simplices). can the points be sampled in a manner to get interesting shapes? for example, after the first few points, avoid new points that are way outside the convex hull so far, drastically changing the polyhedron.
aside: the minimal surface in 3D that contains all given points is not necessarily the convex hull as it is in 2D. the minimal surface might have hyperbolic paraboloid faces.
a random convex polytope generates a random linear or quadratic programming problem.
applying a linear transformation to a convex polytope maintains its convexity. proof: assume false, meaning that after the linear transformation there is a line segment that starts inside, goes outside, then comes back inside (definition of non-convex). apply the inverse linear transformation, which is linear. the line segment remains a line segment. it still goes in out in. the existence of such a line segment in the initial polytope means it was not convex.
consider relaxing the requirement of convexity and only require non-self-intersecting. one can have holes and other weird features, though perhaps some so weird that it might be good to forbid them. what weird features, especially in higher dimensions? polygonal faces can be non-convex with holes. that is probably OK, but faces with intersecting edges (e.g., pentagram) are not OK.
we rearrange the standard triple-angle trigonometric identity for sine to make it more clear the cubic polynomial that you need to solve for the one-third angle trigonometric identity.
4*(sin x)^3 - 3*(sin x) + (sin(3x)) = 0
given sin(3x) in the constant term, solve for (sin x) by your favorite root-finding algorithm, for example, Newton's method.
what is a good starting guess for Newton's method? maybe use the first few terms of the Taylor expansion of sine.
although you could use the cubic formula directly to solve the cubic equation, that requires computing the cube root of a complex number. the standard technique for doing that involves sine and cosine of 1/3 of an angle, so that feels like cheating. previously.
similarly for cosine:
4*(cos x)^3 - 3*(cos x) - (cos(3x)) = 0
slightly more explicitly, these therefore are the one-third angle trigonometric identities:
sin(y/3) = RootOf(4*z^3 - 3*z + sin(y) == 0, z)
cos(y/3) = RootOf(4*z^3 - 3*z - cos(y) == 0, z)
here are the half-angle identities written using the same notation:
sin(y/2) = RootOf(2*z^2 - (1-cos(y)) == 0, z)
= +- sqrt((1-cos(y))/2)
cos(y/2) = RootOf(2*z^2 - (1+cos(y)) == 0, z)
= +- sqrt((1+cos(y))/2)
it is very convenient that the triple angle identity for sine is a polynomial in only sines, so requires only solving a single cubic equation. worse would have been having to solve a simultaneous system of two cubic equations in (sin x)^3, (cos x)^3, lower powers, and cross terms.
in general, to derive a polynomial for cos(n*x) in powers of only (cos x), expand de Moivre's formula cos(n*x) + i*sin(n*x) = (cos(x) + i*sin(x))^n and collect real terms. the powers of (sin x) in the real terms will only be even. this is very convenient: substitute the Pythagorean identity (sin^2 = 1 - cos^2) to transform the even powers of (sin x) into (cos x). the resulting polynomials in (cos x) (called Chebyshev polynomials) can be solved for (cos x) by your favorite root-finding method.
for sine, there are two possibilities:
if n is even, we still work with the collected real terms as above. the powers of (cos x) in the real terms will be only even. transform the even powers of (cos x) into (sin x) via the Pythagorean identity to get a polynomial in (sin x) with constant term that involves cos(n*x). (for example, note how the half-angle identity for sin(x/2) involves (cos x).)
if n is odd, collect imaginary terms. the powers of (cos x) will all be even, so convert them to (sin x) by the Pythagorean identity, yielding a polynomial in (sin x) with a constant term involving sin(n*x).
the Pythagorean identity and de Moivre's identity are certainly two of the most wonderful things in mathematics.
a cube can be oriented in 24 ways (octahedral group, isometries of a cube). vote two letters off the island, then mark a cube with 24 letters, 4 to a face.
one way to reduce the alphabet down to 24 letters is merge I/J and U/V as done on Roman stone inscriptions. or, the least frequent letters in English text (not word lists) are J Q X Z, so choose some subset and a protocol for what to do if you need them. previously, penalizing collisions. previously, eliminate X and Z if only considering initial letters.
for a square cube face scaled to fit between (0,0) and (1,1) on a coordinate plane, let each letter occupy the area (0.125,0) to (0.5,0.5) (and rotations). a collection of such cubes, kind of like children's letter blocks, can spell out text. one needs to learn to pay attention only to the bottom left quarter of each face.
or, divide each square face by diagonals into isosceles right triangles and fit letters into triangles. should the upright letter go in the top or bottom triangle?
cubes pack nicely. an ordered collection of oriented cubes can store information in a volume. it requires only minimal markings on a cube to establish orientation, for example, just one dot in the corner of just one face. all the markings on d6 dice are of course sufficient. but it seems difficult to read or write data encoded in a packed collection of oriented cubes. layers of rhombic dodecahedra fit in divots of the previous layer so don't accidentally slide around but require a container with divots.
previously, on storing information in cubes. previously vaguely similar, marking letters on a twisty cube, which coincidentally also required first reducing the alphabet to 24 letters.
the following Perl script feeding GP uses Lagrange polynomials to interpolate a polynomial between given points. input is stdin (or a file or files specified on the command line after the script), one point per line, X and Y coordinates separated by whitespace. Pari/GP does the heavy lifting of (automatically) multiplying out and simplifying the polynomial. if inputs are integers or rational numbers, Pari/GP automatically does arbitrary precision arithmetic. (if the inputs are floating point, be careful, as this method is not numerically stable. you may wish to increase Pari/GP's floating point precision.)
perl -nlwae 'push @x,$F[0]; push @y,$F[1]; END{ for $i(0..$#x){ $l=1; for $j(0..$#x){ next if $i==$j; $l.="*((x-($x[$j]))/($x[$i]-($x[$j])))";} $_.="+" if defined$_; $_.="($y[$i])*($l)";} print;}' | gp -q
Because perl is only manipulating strings (no BigInt needed), the inputs may be any expressions that Pari/GP can evaluate. all the extra parentheses in the script support this. (the parentheses were originally needed to support negative inputs.) here we demonstrate the script recovering a general quadratic when given input of algebraic expressions.
$ echo -e 'x1 a*x1^2+b*x1+c\nx2 a*x2^2+b*x2+c\nx3 a*x3^2+b*x3+c' | perl lagrangepolynomial.pl | gp -q
a*x^2 + b*x + c
this was inspired by a "guess the next word" puzzle, so input expressions such as 27*(27*(27*(27*(8)+5)+12)+12)+15 (the word "hello" encoded in big-endian base 27) also work. it is always possible to guess a next value (or word) of a sequence by interpolating a polynomial to the previous values. (the answer might not be what the puzzle poser is looking for. unclear what to do with a negative predicted next value when values are encoded words.) it is also possible to justify any next value by adding it to the interpolation.
runtimes with exact arithmetic:
( echo "allocatemem(10^9)" ; for i in `seq 1 100` ; do echo $i $RANDOM ; done | perl lagrangepolynomial.pl ) | nice time gp -q > /dev/null
100 points: 1.4 seconds
150 points: 7 s
200 points: 22 s
250 points: 56 s
300 points: 120 s
400 points: 400 s
450 points: 655 s
500 points: 1016 s
can runtime be improved?
fly anywhere, unimpeded by the rods. you can see quite far because the rods are thin. slide a cursor that rides on the rods to the exit.
2D mazes are usually overhead view, providing lots of visibility; thin rods with unlimited camera movement approximates that for 3D.
it might be difficult to tell which of a pair of visually intersecting rods is closer. add artificial visual cues.
consider a horizontal row of dots equally spaced 1 unit apart. put another such row above it, also with horizontal spacings of 1 unit. let the vertical spacing between rows be A = sqrt(15)/4 ~= 0.968 .
let the upper row slide horizontally at a constant velocity. at some snapshot in time, consider a dot in the upper row, and the dot in the lower row that it is closest to. if the upper dot is directly above the lower dot, that is, their relative horizontal offset is zero, then the distance between them is A. if their relative horizontal offset is 0.25, then the distance between them is 1 by Pythagoras. if their relative horizontal offset is 0.5, then the distance between them is sqrt(19)/4 ~= 1.090 . this is the maximum possible separation: if the offset is greater, then it gets closer to the next dot in the row. thus, half the time (0 to 0.25) the vertical distance is less than the horizontal distance between dots (namely 1), and the other half of the time (0.25 to 0.5), the vertical distance is greater.
motivation is dots in motion but staying well separated. rows can be stacked.
previously: dots arranged in rings instead of rows.
if we want the average distance to the nearest dot in the other row to equal 1, then I don't think there is a closed form solution for the vertical space h between rows. Mathematica:
N[Solve[ 2*Integrate[ Sqrt[h^2+x^2], {x, 0, 1/2}, Assumptions -> Element[h, Reals] && h > 0]==1, {h}, Reals], 50]
yields h ~= 0.95813624081219179188949156285292561053238539746725
(Inverse Symbolic Calculator finds nothing.)
if the rows of dots are not moving, then both the square lattice and equilateral triangular lattice achieve equal separation vertically and horizontally.
solutions to the Thomson problem up to 12 points (excluding 11) have nice symmetry. (though I feel that the solution for 1 point also lacks pretty symmetry.)
2 . digon . vertex types: 1
3 . equilateral triangle . vertex types: 1
4 . regular tetrahedron . vertex types: 1
5 . triangular dipyramid . vertex types: 2
6 . regular octahedron . vertex types: 1
7 . pentagonal dipyramid . vertex types: 2
8 . square antiprism . vertex types: 1
9 . triaugmented triangular prism . vertex types: 2
10 . gyroelongated square dipyramid . vertex types: 2
12 . regular icosahedron . vertex types: 1
"vertex types" refers to vertex transitivity. we do not care about face or edge types of the convex hull polyhedron because the Thomson problem is about points (vertices).
https://www.mathpages.com/home/kmath005/kmath005.htm describes some (locally) minimum energy configurations.
maybe we care about the "balanced" entries in this Wikipedia table in which magnitude(sum(r[i])) = 0 (precisely), where r[i] are the positions of the charges as vectors. the table has no citation, but Laszlo Hars has replicated energy results for small solutions.
assuming we trust the Wikipedia table, the balanced sequence continues 2 3 4 5 6 7 8 9 10 12 14 15 16 17 18 20 22 23 24 27 28 29 30 32 which does not have an OEIS number.
there is also the sequence 2 3 4 6 8 12 24 in which the solution has only one vertex type. 24 is snub cube. there might be some more entries between 12 and 24, but probably not. the sequence probably ends at 24.
not all icosahedral geodesic spheres appear in the minimum energy table. GP(2,0) = 42, GP(3,0) = 92, GP(4,0) = 162 are the first few missing. clearly GP(n,0) seems to not be preferred. which geodesic spheres are more likely to be minimum energy solutions? this is likely similar to how cube is not minimum energy for 8 points, instead square antiprism. do icosahedral geodesic spheres continue to appear as minimum energy configurations, or is there a final geodesic sphere which is a solution?
create a deck of cards representing the 48 elements of the symmetry group of a cube (isometries of a cube) (automorphism group of a cube). 48 is a good number of cards for a deck, not too difficult to shuffle.
there are two suits, corresponding (roughly) to mirror images.
pairs of cards can be added to yield another: the group action. how can addition be facilitated to make it easier for humans? perhaps print the appropriate line of the addition table on the face of each card. how can be group elements be compactly identified?
describe a group element by a decomposition into rotations of 90 degrees along axes (and zero or one reflection), like Rubik's cube algorithms. but this is not compact.
an illustration of axis and angle is compact.
augment the deck with d6 dice: two of different colors, mirror images of each other. how easy is to acquire a d6 die that is the mirror image of the conventional d6 die?
find the card corresponding to a given rotation of a die relative to a canonical unrotated state. maybe this is the mechanic of games.
or, consider a 60-card deck of just the orientation-preserving symmetries of the regular dodecahedron or icosahedron; 60 is small enough to comfortably shuffle. previously. you would only need one d12 (or d20) die. maybe two, if you want to keep track of before and after a rotation.
do d12 and d20 dice have standardized numbering?
how can one describe a rotation of (say) a dodecahedron? starting from a canonical orientation, move the face labeled X to the top in a canonical orientation, then orient the X top face to one of 5 possible orientations. equivalently, top face and "front" face adjacent to top.
what rectangle aspect ratios permit finite tessellations of regular hexagons to be neatly inscribed within them?
assuming one edge of hexagons parallel to one edge of rectangle, this should be easy. in other words, close-packed clusters of regular hexagons circumscribed by a rectangle, two (opposite) sides of the rectangle coincident with edges of outermost hexagons, and the two other (opposite) sides of the rectangle coincident with outermost hexagon vertices.
the point at coordinates (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24) in 24-dimensional space is exactly 70 units from the origin, because the sum of the first 24 squares is 4900. (this amazing relation is somehow related to the 24-dimensional Leech lattice, a different lattice than our hypercube lattice of all points with integer coordinates.) the point is one of 905754973269671819419476064818886581168 lattice points 70 units from the origin. that large number divided by 24 factorial is approximately 1459839320625890.4 : there are many more lattice points on a hypersphere of radius 70 than just the permutations of the coordinates of our special point.
our special point is one of 370123270431636382177726611489013547080401 lattice points inside or on the boundary of a 24-dimensional ball of radius 70. this volume is approximately 408.6 times larger than the above surface "area" of lattice points. in other words, this outermost layer of lattice points does not dominate the total lattice points. this matches lower dimensional intuition, which is rare when considering volumes in high dimensions.
future work: actual volume, surface area, and ratio of a 24-dimensional sphere of radius 70. this should be easy.
the above results are derived from the b-file of OEIS A000156. I confirmed the list up to entry 1000 (with 48 hours of computing). I do not know how its larger entries (up to entry 10000) were calculated; there must be some algorithmic tricks beyond what I implemented.
I discovered A000156 by calculating its first few terms (future post kiqanzug), then searching OEIS for the continuation. this a standard way to use the database.
consider a brick in N dimensions composed of unit hypercubes. for example, let N=3. each unit cube is labeled 0 or 1, subject to the following monotonicity constraints:
is constraint 2 necessary, or does it logically follow from 1 ?
how many valid labelings of the cubes in the brick are there?
I suspect the number is much less than total number of possible unconstrained labelings. what is a compact way of describing a coloring, more compact than giving the label of every unit cube?
inspired by trying to generalize Conway's Game of Life cellular automata to 3D. let neighbors through a corner of a cell, edge, and face each contribute a different amount of influence to the "number of neighbors" a cell has. we want more generality than a weighted sum, but we want to keep rules to be monotonic: for example, if a threshold has been crossed for "too many neighbors", then it remains crossed for "more" neighbors. (previously, 2D.) in 3D, there can be 0 to 8 corner neighbors, 0 to 12 edge, and 0 to 6 face. these together which define a 9 x 13 x 7 brick. label each of its 819 unit subcubes "OK" or "too crowded". without the monotonicity constraints, there are 2^819 possible labelings, far too large to evaluate all of them by brute force.
2D: 5 x 5 brick (0 ... 4 edge neighbors and corner neighbors), so 2^25 = 32 million unconstrained labelings. counting those which satisfy the monotonicity constraints can be done by brute force. finding those which yield "good" cellular automata is harder.
future work: 3 labels for unit cubes: "too isolated", "OK", "too crowded". we need to come up with a new definition for monotonicity among the 3 ordered labels, probably not difficult. a second brick describing births from dead cells. perhaps constrain the conditions suitable for birth to be a subset of conditions suitable for life.
consider the axis-aligned square between (0,0) and (4,4). we will draw a binary tree inside this box.
the root node is centered on the top edge, at (2,4). draw an edge from the root to one child at (1,2) and another edge from the root to the other child at (3,2).
next, create two copies of the box so far shrunk linearly by a factor of 2. place one between (0,0) and (2,2), and the other between (2,0) and (4,2). the children of the large box become the roots of the small boxes.
repeat recursively, producing an infinite binary tree in finite space. (this is an example of an iterated function system (IFS).)
inspired by the period-doubling logistic fractal, which requires a lot of computation to produce nice images. this is easier.
perhaps scale things so that the scale factor is not 2 but the Feigenbaum constant.
instead of straight lines, consider semicircles. it will look like arches built on arches, a design seen in Roman aqueducts. or, not semicircles but catenaries.
a square (or rectangle) fundamental polygon is good for 2D games.
do weird things happen at the corners of the square fundamental polygon for an objects larger than a point (e.g., a sprite) moving around the space?
sphere: Y
torus: N
real projective plane: Y
Klein bottle: N
cylinder: N
Moebius strip: N
hemisphere: Y
square: N
things are not weird if the neighborhood around corners behaves like flat (Euclidean) space. investigate by tracing a small circle around each corner. in flat space, we expect a 360-degree circle. in the answers above, Y = (weird things happen at corners; there is global or total curvature; small circles are less than 360 degrees); N = (corners behave like flat space; circles are 360 degrees).
if you have a fundamental polygon with more edges than a square, e.g., hexagon, you can probably get small circles larger than 360 degrees, some sort of hyperbolic space. you might be able to get some corners with positive curvature that cancel out other corners with negative curvature for zero global curvature. but it would still be awkward for games, because local curvature is what we care about for rendering sprites.
even though weird things happen at corners of sphere, projective plane, and hemisphere when drawn as a square fundamental polygon, they are all manifolds (locally Euclidean), so there are other ways of depicting them so that small circles remain approximately small circles throughout, smearing the curvature or weirdness throughout the surface.
or, just put an obstacle over the weird points.
"Convex Regular Faced" is usually applied to polytopes in 4D and above, but it is a useful term for 3D as well, subsuming the Platonic solids, Archimedean solids, the pyramids, prisms, and antiprisms that are uniform polyhedra, and the Johnson solids.
consider a finite subset of CRF polyhedra whose faces have 3, 4, 5, 6, 8, or 10 sides. these are all the faces seen in the finite categories (not the pyramids, prisms, or antiprisms). (it is a little surprising that the regular dodecagon is seen only in the infinite categories.)
consider a further subset: eliminate polyhedra which can be constructed by pasting together two or more other polyhedra in this set (pasting only matching faces). inspired by the partition of Johnson solids between those which can be constructed by "cut and paste" and those which are "elementary". we seek the elementary CRF polyhedra.
for example, the regular octahedron is eliminated because it can be constructed from two square pyramids glued base to base, and square pyramids are CRF.
is there a unique minimal set?
these elementary polyhedra (replicated as many times as needed) might be a fun set of building blocks. how should faces physically attach? given a not-necessarily convex arrangement of them glued by matching faces, determine whether the solid intersects itself.
both people lying on their side, next to each other. 4 possibilities: big spoon, little spoon, face to face, back to back. we do not count mirror images (who is on which side), even though maybe we should because which hand is free (the dextrous one or not) might matter for touching activities in cuddle.
bodies stacked vertically, each person facing up or down: 8 possibilities.
the above do not consider variations of where the legs go (future post tiohqbtv).
one person on their front or back, other person on their side facing in: 4 possibilities.
one person on their back, knees up and legs resting on top of other person's legs, other person on their side facing in, legs underneath the other person's knees: 2 possibilities. we do consider legs (distinguishing from the above 4) because they make a big difference on how the position feels -- here, the person on their back feels very much on top. the person on their side gets good access with their free arm to the other person's genitals.
below are some more, similar in spirit to side-by-side back-to-back counted above. possibly less exciting, but mere physical contact can be entertaining.
one person on their front or back, other person on their side facing away: 4 possibilities.
one person on their front or back, other person on their front or back, side by side: 4 possibilities.
26 total. perhaps they can be mapped to letters of the alphabet.
not considering non-parallel bodies or bodies not aligned head to toe, for example, 69.
minimum spanning tree, Delaunay triangulation, traveling salesman path, traveling salesman circuit, Euclidean Steiner tree.
on a plane, sphere, flat torus, flat Klein bottle.
(it feels strange to call it the Euclidean Steiner Tree problem when on a non-Euclidean manifold such as a sphere. however, we do need the modifier to specify the problem is about points in space and not the Steiner Tree problem on graphs.)
3D: Delaunay is no longer a triangulation but a graph, the dual of Voronoi as it always is.
in 3D space, 3D flat torus, 3-sphere.
isotropy in chess piece movement is elegant, but because armies start on opposite sides and pawns can't move backwards, forward and backward movement are inherently different in chess. shogi has this reflected in piece types, for example, the gold general which can advance diagonally but not retreat diagonally. we enumerate possible fairy chess pieces which are left-right symmetric but not necessarily forward-backward symmetric. previously on anisotropy.
compounds of deconstructed wazir, ferz, rook, bishop, dabbaba, strong dabbabarider, knight, and alfil. product of two C-shaped regions around a piece.
inner C: 3^5 = 243
outer C (jumping pieces): 3^3 * 2^6 = 1728
dabbabarider, the only jumping rider we choose to include, grew the outer C from 2^9 = 512.
one can also think of it as symmetries inducing equivalence classes among directions.
there are now 15 variations of knight (not including immobile stone) because of its 4 possible amounts of forward or backward movement. this includes crab but not pinwheel knight.
total is 3^8 * 2^6 = 419904, a huge increase from isotropic 72. this includes piece types which can only move backward and piece types confined to their starting rank or file. perhaps cull these, because they seem practically (nearly) useless unless the game has drops like shogi.
also consider culling multiply colorbound.
raise the total to the Nth power to investigate N-piece endgames. there is a question of what pawns may promote to (future post mpyfvglu). finding interesting positions and mechanics will be art.
degrees missing from being flat at vertex:
tetrahedron 180
octahedron 120
cube 90
icosahedron 60
dodecahedron 36
(angle defect)*(number of vertices) = 720 degree = 4*pi radian, a theorem of Descartes.
by this metric, dodecahedron is the most flat, the least confusing (at corners) for a map of a sphere. (does that mean it has a lot of distortion inside each face?) create a tool to display the earth on a dodecahedral net. drag any point to anywhere on the net and reproject. also need rotation. move pentagons to choose among many possible nets. what countries fit neatly inside adjacent pentagons? I don't have a good feel of what is the area and extent of 1/12 or 2/12 of a sphere (1 or 2 pentagons).
move portions of pentagons? the dissection of a pentagon by a pentagram (star) might be useful, as well as the dissection from the center into pie pieces.
vertex-transitive polyhedra have the same angle defect at every vertex. the Archimedean solids are another (the other?) family of vertex-transitive polyhedra. by the theorem of Descartes, the most flat therefore is the polyhedron with the most vertices. excluding the prisms and antiprisms, the truncated icosidodecahedron has the most vertices with 120. decagon, hexagon, and square meet at each of its vertices; angle defect is 6 degrees. how much map distortion is in its 12 relatively large decagons? (30 squares and 20 hexagons are its other faces. 62 faces total.)
previously, pillars of inaccessibility: the tile diagonally opposite you is blank, perhaps annotated with curved arrows indicating edges connect. moving your immersed character locally induces a different net. the faces of a net could discretely change adjacency, or walking around a vertex could cause the net to continuously reproject itself, keeping the inaccessible gap diagonally across from you. not sure if the latter works.
future work: solid angular defect of 4D regular polytopes.
