From 4ba8b0617c14b78174176753fcb7388bb6c0e430 Mon Sep 17 00:00:00 2001
From: Mike Lynch <mike@mikelynch.org>
Date: Sun, 6 Aug 2023 08:41:44 +1000
Subject: [PATCH] Added notes/planning for 120-cell partition

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 docs/notes_120_cell.md | 139 +++++++++++++++++++++++++++++++++++++++++
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+
+
+Chords: 1.74806 - the 120-cell has 7200 chords of this length
+
+Looking for a way to partition the 600 vertices of the 120 cell into five
+disjoint 600-cells, each of which has 120 vertices.
+
+(there are 10 such 600-cells so two ways to do the partition I guess)
+
+a 600-cell has 720 edges! optimistically this means that each chord in the
+collection of 7200 belongs to one and only one of the 600-cells.
+
+
+the way forward:
+
+I need to take the 7200 chords (pairs of nodes) and divide them into sets
+which are connected to one another - with any luck, each of these will be
+one of the 10 600-cells
+
+Then need to sort these 10 sets of 120 vertices into the two sets of 5
+
+
+collate chords by node
+
+Each 120-cell vertex has 24 of the chord3s from it - as a 600-cell has 12
+edges to each vertex, this suggests that each 120-vertex belongs to two 
+600-cells with a disjoint set of vertices
+
+Next algorithm - gather each 600-cell
+
+use the chords as the basis for this.
+
+n1 -> 24 chords -> add these 24 neighbours
+
+bad luck - traversing chord3s from the first vertex reaches all 600 vertices-
+which isn't suprising as the two 5 disjoint sets overlap. Sigh.
+
+Use the angles between the chords? seems a bit complex
+
+Get the angles from the 600-cell model. Use these to separate out the sets of
+24 chords from a point on the 120-cell.
+
+Notes from dinner:
+
+- all of the 60-degree angles are chords joining the vertices of the tetrahedra
+  - there should be two sets of these
+
+for eg - this works for the chords from 1!
+
+    [ 25, 41 ],
+    [ 25, 97 ],
+    [ 25, 109 ],
+    [ 25, 157 ],
+    [ 25, 161 ],  
+    [ 41, 97 ],
+    [ 41, 109 ],
+    [ 41, 173 ],
+    [ 41, 177 ],
+ 	[ 97, 113 ],
+ 	[ 97, 161 ],
+    [ 97, 177 ],
+    [ 37, 53 ],
+    [ 37, 93 ],
+    [ 37, 113 ],
+    [ 37, 157 ],
+    [ 37, 161 ],
+    [ 53, 93 ],
+    [ 53, 113 ],
+    [ 53, 173 ],
+    [ 53, 177 ],
+	[ 173, 177 ]
+    [ 93, 109 ],
+	[ 93, 157 ],
+	[ 93, 173 ],   
+ 	[ 109, 157 ],
+ 	[ 109, 173 ],
+	[ 113, 161 ],
+	[ 113, 177 ],
+    [ 157, 161 ],  
+
+    [ 29, 45 ], 
+    [ 29, 101 ],
+    [ 29, 105 ],
+    [ 29, 153 ],
+    [ 29, 165 ],
+    [ 45, 101 ],
+    [ 45, 105 ],
+    [ 45, 169 ],
+    [ 45, 181 ],
+	[ 101, 117 ],
+	[ 101, 165 ],
+	[ 101, 181 ],
+    [ 105, 153 ],
+    [ 105, 169 ],
+    [ 33, 49 ],
+    [ 33, 89 ],
+    [ 33, 117 ],
+    [ 33, 153 ],
+    [ 33, 165 ],
+    [ 49, 89 ],
+    [ 49, 117 ],
+    [ 49, 169 ],
+    [ 49, 181 ],
+	[ 169, 181 ],
+    [ 89, 105 ],
+    [ 89, 153 ],
+    [ 89, 169 ],
+    [ 117, 165 ],
+    [ 117, 181 ],
+	[ 153, 165 ], 
+    
+
+So each of these is one of the two icosahedral pyramids from node 1.
+
+Doing this manually for the rest of the partition is possible, but could it
+be automated based on angles?
+
+
+
+Plan for Sunday:
+
+* use the existing label_subgraph to make a function which partitions the
+  60-angle chords into two groups (like I did manually above)
+
+* test this labelling manually (ie colour one set of 60-angle vertices)
+
+* make another labeling routine which can fill out the rest of the 600-cell
+  from the starting dodecahedron, by only following chords which are at 60
+  to the entering chord
+
+Then the big algorithm does the following:
+
+- start from node 1, find 60-angles, pick one partition at random, label that 600-cell
+
+- find the next unlabelled node
+
+- find 60-angles, partition them, pick a partition with no unlabelled cells and label that 600-cell
+
+- repeat the previous step for the remaining three 600-cells
\ No newline at end of file