Added notes/planning for 120-cell partition
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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 ],
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[ 29, 101 ],
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[ 29, 105 ],
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[ 29, 153 ],
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[ 29, 165 ],
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[ 45, 101 ],
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[ 45, 105 ],
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[ 45, 169 ],
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[ 45, 181 ],
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[ 101, 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 ],
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[ 33, 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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* test this labelling manually (ie colour one set of 60-angle vertices)
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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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