245 lines
5.7 KiB
Markdown
245 lines
5.7 KiB
Markdown
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Steps forward -
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1. algorithm which, given a face, finds the two dodecahedra it belongs to
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2. using this, generate a list of all 120 dodecahedra:
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1.
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For a face: there are five edges, and ten other faces sharing an edge.
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These edges are in two sets: one for each dodecahedron. The sets are defined
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by them sharing vertices which aren't in the first face.
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Go around a set of five, by pairs: for each pair, find the other neighbour -
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this gives the next five faces.
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There's only one face left, which is defined by the shared other vertices of
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the last five.
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2. have tried manual labelling and it will drive me crazy before I finish
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3. Automated approach based on what I've got so far:
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- should be possible to colour a single dodecahedron from a single face and
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one other vertex (to pick a chirality)
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- write a function to do this - the compound-of-four-tetrahedra map can be
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more or less hard coded: follow the pattern 1-2-3-4-5, 3-4-5-1-2, etc out
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from the inner ring, and map the original face's permutation
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From this:
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- colour the first dodecahedron, picking a chirality
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- the next vertices from each of this dodecahedron's vertices have colours
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which come from the first dodeca
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- these can be used to colour the next layer of dodecahedra
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- and so on
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Alternatively:
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- do it by the discrete Hopf fibration, one fibre at a time
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/// old shit below that didn't work VVVV
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Chords: 1.74806 - the 120-cell has 7200 chords of this length
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Looking for a way to partition the 600 vertices of the 120 cell into five
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disjoint 600-cells, each of which has 120 vertices.
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(there are 10 such 600-cells so two ways to do the partition I guess)
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a 600-cell has 720 edges! optimistically this means that each chord in the
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collection of 7200 belongs to one and only one of the 600-cells.
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the way forward:
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I need to take the 7200 chords (pairs of nodes) and divide them into sets
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which are connected to one another - with any luck, each of these will be
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one of the 10 600-cells
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Then need to sort these 10 sets of 120 vertices into the two sets of 5
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collate chords by node
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Each 120-cell vertex has 24 of the chord3s from it - as a 600-cell has 12
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edges to each vertex, this suggests that each 120-vertex belongs to two
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600-cells with a disjoint set of vertices
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Next algorithm - gather each 600-cell
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use the chords as the basis for this.
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n1 -> 24 chords -> add these 24 neighbours
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bad luck - traversing chord3s from the first vertex reaches all 600 vertices-
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which isn't suprising as the two 5 disjoint sets overlap. Sigh.
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Use the angles between the chords? seems a bit complex
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Get the angles from the 600-cell model. Use these to separate out the sets of
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24 chords from a point on the 120-cell.
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Notes from dinner:
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- all of the 60-degree angles are chords joining the vertices of the tetrahedra
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- there should be two sets of these
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for eg - this works for the chords from 1!
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[ 25, 41 ],
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[ 25, 97 ],
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[ 25, 109 ],
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[ 25, 157 ],
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[ 25, 161 ],
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[ 41, 97 ],
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[ 41, 109 ],
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[ 41, 173 ],
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[ 41, 177 ],
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[ 97, 113 ],
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[ 97, 161 ],
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[ 97, 177 ],
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[ 37, 53 ],
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[ 37, 93 ],
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[ 37, 113 ],
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[ 37, 157 ],
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[ 37, 161 ],
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[ 53, 93 ],
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[ 53, 113 ],
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[ 53, 173 ],
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[ 53, 177 ],
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[ 173, 177 ]
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[ 93, 109 ],
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[ 93, 157 ],
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[ 93, 173 ],
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[ 109, 157 ],
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[ 109, 173 ],
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[ 113, 161 ],
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[ 113, 177 ],
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[ 157, 161 ],
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[ 29, 45 ], 5
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[ 29, 101 ], 101
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[ 29, 105 ], 105
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[ 29, 153 ], 153
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[ 29, 165 ], 165
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[ 45, 101 ],
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[ 45, 105 ],
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[ 45, 169 ], 169
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[ 45, 181 ], 181
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[ 101, 117 ], 117
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[ 101, 165 ],
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[ 101, 181 ],
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[ 105, 153 ],
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[ 105, 169 ],
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[ 33, 49 ], 33 49
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[ 33, 89 ], 89
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[ 33, 117 ],
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[ 33, 153 ],
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[ 33, 165 ],
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[ 49, 89 ],
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[ 49, 117 ],
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[ 49, 169 ],
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[ 49, 181 ],
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[ 169, 181 ],
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[ 89, 105 ],
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[ 89, 153 ],
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[ 89, 169 ],
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[ 117, 165 ],
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[ 117, 181 ],
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[ 153, 165 ],
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So each of these is one of the two icosahedral pyramids from node 1.
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Doing this manually for the rest of the partition is possible, but could it
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be automated based on angles?
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Plan for Sunday:
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* use the existing label_subgraph to make a function which partitions the
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60-angle chords into two groups (like I did manually above)
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// this is done and seems to work
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* test this labelling manually (ie colour one set of 60-angle vertices)
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// done this with the manual labels and it looks good
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* make another labeling routine which can fill out the rest of the 600-cell
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from the starting dodecahedron, by only following chords which are at 60
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to the entering chord
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Then the big algorithm does the following:
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- start from node 1, find 60-angles, pick one partition at random, label that 600-cell
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- find the next unlabelled node
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- find 60-angles, partition them, pick a partition with no unlabelled cells and label that 600-cell
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- repeat the previous step for the remaining three 600-cells
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Alternative, more manual option: just write the second labelling routine and
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do the rest by hand
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[ 25, 41 ],
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[ 25, 97 ],
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[ 25, 109 ],
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[ 25, 157 ],
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[ 25, 161 ],
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[ 41, 97 ],
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[ 41, 109 ],
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[ 41, 173 ],
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[ 41, 177 ],
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[ 97, 113 ],
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[ 97, 161 ],
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[ 97, 177 ],
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[ 37, 53 ],
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[ 37, 93 ],
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[ 37, 113 ],
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[ 37, 157 ],
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[ 37, 161 ],
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[ 53, 93 ],
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[ 53, 113 ],
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[ 53, 173 ],
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[ 53, 177 ],
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[ 173, 177 ]
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[ 93, 109 ],
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[ 93, 157 ],
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[ 93, 173 ],
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[ 109, 157 ],
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[ 109, 173 ],
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[ 113, 161 ],
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[ 113, 177 ],
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[ 157, 161 ],
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25 41 97 109
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157 161 173 177
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113 37 53 93
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Another idea - look at colouring the vertices of a dodecahedron according to
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the four compound tetrahedra and see if this can be repeated automatically to
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neighbouring cells
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