Spatial reasoning quiz

Next clue: assume the "P-cube" is isomorphic to the open ball B(r) on the real line R.
Prove this is true by demonstrating that removing a single color from the cube (de-coloring a face) leaves the # of combinations unchanged.

How many combinations are lost if 2 faces are un-colored? Does it make a difference if the 2 are adjacent, or opposite faces?

Now I need to know if I can make a triangle that has interchangeable edges and vertices (I can swap the vertices for edges, but only over the whole group of edges/vertices, there are 3 + 3).

It also has a center which is missing. The center of each edge is like a vertex, and each vertex is like an edge (when I choose which is which). The six elements in the total set of edges and vertices has a center which isn't in the triangle (it's imaginary). This corresponds to a tetrahedron's six edges if I map the six elements in the triangle group to adjacent faces of the tetrahedron, i.e. in pairs.

This composition must correspond to a set of 4 faces of the tetrahedron, as mapped by edge pairs back to the triangle group. Each vertex has [two] outer and [one] inner crossing, to an adjacent vertex, or "through the missing center" to an opposite edge of the triangle respectively. This triangle is like the Fano plane, with the central vertex removed but the edges left intact as crossings through T, so each edge/vertex commutes, and each pair can change identites on the 4-sided tetrahedral figure; the set of edge onnections in T, projects onto the face group F of the tetrahedron.

So is a tetrahedron a "perfect" solid? It's one of the regular Platonic solids that fits inside a cube.
Then, can this idea of a complete, perfect "form" apply to the musical theory notion of a perfect interval? Why is a 4th perfect, and in which scales of the diatonic, with 12 intervals in groups of 8?
Does a quaternionic algebra correspond to "algebraic noise", which we call a composition - harmony, rhythm, thematic structure, ...?

Guys, I think the riddle is: "why is music - on a violin, or a piano or guitar - and general relativity related to the group of rotations on a 'sliced' solid? Why is a bunch of cubicles with cubies in them like geometry and algebra on a stick?

You can transform the geometry of cubes into the geometry of other solids - the megaminx and other puzzles do this - this is algebraic geometry, the domain of "Clifford algebras"; this leads into Von Neumann entropy and Shannon channel limits - because you invent one of these (not one that looks remotely like the symbolic logic these guys use, but that's a matter of translating your one into theirs).

There is so much math in algebraic manifolds and tessellations - lattice theory, tilings in the plane (across a section of the Mobius link). So many connections, and of course the symmetry of space in three dimensions; here time on the cube becomes rotations in the plane (think about how long you could take, to get that corner to another slot)...

😀

Hmm. The original question was about thinking out the difference between the number of rotations for a single slice of a cube, and the number of faces a single rotation is applied to. That is, whether one, two or three slices are "in the set" when a single move is made - the one-slice subgroup can only have 3 positions for the single face which is "spun".

Moving up to one more slice, per-move, you have twice as many positions, 6; with all three slices in the set there are 9. Per-slice there are 2 quarter moves, and a single "full" move. So why does the list posted in wikipedia, say there are 9 positions for full moves, over the full set of slices, they got the number of quarter moves right - there must be 6 unique positions.
What difference is there beween quarter and full turns?
You can't tell which way a full turn was made; it 'compresses' the cube into 2-dimensions (it flattens two sides, in a color-sense, by rotating through a half-plane); the q-turns can only quarter the 'circle', or leave a trail of torsion behind, that is, they "lift a path" around the sides so you can see a direction, from any outside view of the twist.

A half-plane move erases this path-information from the surface... This gives the moves (rotations) a symmetry and an antisymmetry in the space.

To bring this idea of a kind of musical notation in the cube group* like notes on a keyboard, into a little focus - I thought the idea of two groups of 4 keys, that form a known interval in the diatonic scale, would relate to the two groups of 4 cubies, in 8 cubicles or positions in the Pocket Cube.

It's like the 8 notes you include in the scale, exclude notes from the scale. To play the two notes called a 4th (include the root note and the 5th semitone above it), you exclude the 4 notes on 4 keys in between the root and 5th notes/keys. Then the cubies, in the smallest actual geometry they can be in and still have edge-exchanges, are excluded keys in two 4ths of "cubic color"; since the underlying structure is independent of the "sticker" it has on it.

In the same sense a harmonic interval's structure is independent of the maj/min key you play it in. Or the "color" of harmony is the keys, or, the structure is the color.

* All the solids that can be made into solid puzzles, with a way to rotate numbers or colors between faces.

If anyone is still following this "quiz", the connections to graph theory, complexity and spatial reasoning are these:

anyone who resolves a scrambled cube, or learns how to, descends into the graph through which a path exists,
the same thing as resolving a directed cyclic graph into an acyclic graph, you want to keep exactly one set of cycles/directions
to get to the center.
When the cube is scrambled you're in orbit; you don't know how many steps there are back to what we call a solved state.

You have to use a pattern-recognition algorithm, to eliminate possible directions through G. These are the related links I have open in Wikipedia (which represent a tree-walk, through a hypertext-space, the links are one-to-one):
<a href="https://en.wikipedia.org/wiki/Total_order" target="_blank" rel="nofollow noopener noreferrer">Total Order</a>
<a href="https://en.wikipedia.org/wiki/Directed_acyclic_graph" target="_blank" rel="nofollow noopener noreferrer">Directed Acyclic Graph</a>
<a href="https://en.wikipedia.org/wiki/St-connectivity" target="_blank" rel="nofollow noopener noreferrer">St Connectivity</a>


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You have to assume that the deterministic TM is you, trying to read-ahead and not have to backtrack, i.e. be able to record your position and momentum through G.

I'm not sure if I'm going the right way, or if it is a valid model but here goes anyway.

Triangles with equal sides are like fundamental or ideal forms of triangles - they always have an invariant ratio of the area to the sides and angles, and the barycenter coincides with all the other triangle centers.

Mapping or embedding a triangle on the surface of a sphere is straightforward via projection from the plane. Embedding a line on the sphere, means the line is a section of a great circle geodesic, so three lines of equal length "in position" is three geodesics. Call any equilateral T on S[sup]2[/sup] abb', where bb' is the base and a the apex. Map bb' to the equator, and reflect abb' along bb' to give the inverted T = a'b'b.

Extend aa' the imaginary line between the upper and lower apex of the parallelogram (a parallelepiped in S[sup]3[/sup]); make these approach the equator on the opposite side of S[sup]2[/sup]. As a - a' gets closer to zero at the equator, the imaginary line between the ||'grams apices approaches the circumference of S.

If bb' is stretched in the same manner lengthwise, but this is the real base of both triangles, the ||'gram eventually covers the sphere. Four sides vanish as this happens.

The two triangles were initially sections on S. The ||'gram formed by two equilateral T figures, has a diagonal equal to the length of each side. The "highest" and "lowest" points are a and a', respectively. The distance between them is also a geodesic section with a known relation to the area, the sides and angles, in Euclidean space when the Ts are 'flattened'.

On the sphere, the distance is stretched along bb' and aa' when the Ts are projected. The curvature means the geometric relation between the sides and angles changes, lines that were parallel in Euclidean space meet at 'poles' when projected onto S.

So projection is like removing a section of the surface, if the projection is a line section, it removes a geodesic from S, three lines in position remove a triangle = create an imaginary area with imaginary sides congruent with the real side(s) of the remaining surface.

So that's how to find the nullspace, and extend/expand it to cover S, so it has a real 'figure' to contract into = two equilaterals adjacent on S. Contraction means you can store information about geodesic sections in T[sup]2[/sup], so we're all good...