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26
NOTES.md
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26
NOTES.md
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@ -0,0 +1,26 @@
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# NOTES
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New approach for the 5-cells:
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Pick a tetrahedron of an inscribed 600-cell with vertices A, B, C, D
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This gives pairs of vertices:
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AB
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AC
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AD
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BC
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BD
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CD
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Each of these gives rise to seven pairs of 5-cells which are on neighboring vertices
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of the 5 600-cells.
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Try enumerating these and inspecting them to find one or more coherent sets of four
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5-cells which lie on one tetrahedron from each of the 600-cells.
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(I expect there to be more than one, like how there are two ways to partition the
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120-cell vertices into 600-cells)
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@ -105,12 +105,10 @@ export const LAYERS120 = {
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};
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export const CELL120_CELL5 = {
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"1": [ 258, 1, 510, 304, 431 ],
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"2": [ 185, 93, 222, 295, 372 ],
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"1": [ 1, 93, 37, 157 ],
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}
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// Schoute's partition via https://arxiv.org/abs/1010.4353
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export const PARTITION600 = {
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1
explore_120
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1
explore_120
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@ -199,7 +199,6 @@ function try_120_5_cells_fails(cell120, cells, l) {
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console.log("overlap");
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console.log(c);
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return true;
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}
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}
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return false;
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@ -269,27 +268,24 @@ function node_dist(cell120, aid, bid) {
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return dist(a, b);
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}
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function print_row(v1, v2, p, v5) {
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console.log(`${v1.id},${v2.id},${p},${v5[1]},${v5[2]},${v5[3]},${v5[4]},${v5[5]}`);
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}
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// for a pair of vertices which are on the same inscribed 600 cell,
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// this returns all 7 pairs of 5-cells which contain v1 and v2 and
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// which are also evenly spaced (ie every pair of vertices on the
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// same 600-cell is one edge apart)
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function follow_600(cell120, all5) {
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const v = cell120.nodes[0];
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console.log("Start vertex:");
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console.log(v);
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const v5s = all5.filter((c5) => c5[v.label] === v.id);
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console.log(`Vertex ${v.id} belongs to these 5-cells:`);
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console.log(v5s);
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const n600s = neighbours600(cell120, v.id);
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const n600id = n600s[0];
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const n600 = cell120node(cell120, n600id);
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console.log("One 600-cell neighbour:");
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console.log(n600);
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const DIST600 = round_dist(node_dist(cell120, v.id, n600id));
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const nv5s = all5.filter((c5) => c5[v.label] === n600id);
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console.log(`Vertex ${n600id} belongs to these 5-cells:`);
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console.log(nv5s);
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console.log("Distances for each pair of 5-cells from the two sets:");
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for( const v5a of v5s ) {
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for( const v5b of nv5s ) {
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function find_adjoining_5cells(cell120, all5, v1, v2) {
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const DIST600 = round_dist(node_dist(cell120, v1.id, v2.id));
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const v15s = all5.filter((c5) => c5[v1.label] === v1.id);
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const v25s = all5.filter((c5) => c5[v2.label] === v2.id);
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let p = 0;
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const c5pairs = [];
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for( const v5a of v15s ) {
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for( const v5b of v25s ) {
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let match = true;
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const d = {};
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for( const label in v5a ) {
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@ -299,19 +295,76 @@ function follow_600(cell120, all5) {
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}
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}
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if( match ) {
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console.log("--- pair ---");
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console.log(v5a);
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console.log(v5b);
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c5pairs.push([ v5a, v5b ]);
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}
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}
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}
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return c5pairs;
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}
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function tetras(cell120, all5, v) {
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// given a vertex v, find all of the 600-cell tetras it's on
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const n600s = neighbours600(cell120, v.id);
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// need to find all sets of three neighbours which are neighbours: there
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// should be 20 of these because they're faces of an icosahedron
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console.log(v.id);
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const tetras = new Set;
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for( const v2id of n600s ) {
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// find mutual neighbours of the first two
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const n2600s = neighbours600(cell120, v2id);
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const mutuals = n2600s.filter((nid) => {
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return nid != v2id && nid != v.id && n600s.includes(nid)
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});
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for( const nm of mutuals ) {
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const nnms = neighbours600(cell120, nm);
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const mutuals2 = nnms.filter((nid) => {
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return nid != nm && nid != v2id && nid != v.id && mutuals.includes(nid)
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});
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for( const m2 of mutuals2 ) {
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const t = [ v.id, v2id, nm, m2 ];
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t.sort((a, b) => a - b);
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const tstr = t.join(',');
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tetras.add(tstr);
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}
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}
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}
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console.log(tetras);
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}
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/* // find mutuals which are mutuals of each other
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const m0 = mutuals[0];
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const mn = neighbours600(cell120, m0);
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const mm = mn.filter((nid) => nid != m0 && mutuals.includes(nid));
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return [ v1.id, v2id, m0, mm[0] ];
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}*/
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function str5cell(c5) {
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return ["1","2","3","4","5"].map((l) => String(c5[l]).padStart(3, '0')).join('-');
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}
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function tetra_sets(cell120, all5, tetra) {
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// given a tetrahedron on a 600-cell, find the sets of adjacent 5-cells on
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// all of the pairs
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// this is ass-backwards. Need to find tetras on the other 4 vertices of a 5-cell
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const vs = tetra.map((tid) => cell120node(cell120, tid));
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const pairs = [[0,1], [0,2], [0, 3], [1, 2], [1, 3], [2, 3]];
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for( const p of pairs ) {
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const v1 = vs[p[0]];
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const v2 = vs[p[1]];
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const c5pairs = find_adjoining_5cells(cell120, all5, v1, v2);
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console.log(v1.id, v2.id);
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console.log(c5pairs.map((p) => str5cell(p[0]) + " " + str5cell(p[1])));
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}
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}
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const cell120 = POLYTOPES.cell120_inscribed();
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const all5 = gather_5cells(cell120);
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follow_600(cell120, all5);
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const v1 = cell120.nodes[0];
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tetras(cell120, all5, v1);
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tetras(cell120, all5, cell120.nodes[1]);
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@ -473,14 +473,14 @@ export const cell120_inscribed_cell5 = () => {
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const links = [];
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for( const p of [ 1, 2 ]) {
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/* for( const p of [ 1, 2 ]) {
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const nodes600 = nodes.filter((n) => n.label === p);
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const links600 = auto_detect_edges(nodes600, 12);
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links600.map((l) => l.label = p);
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links.push(...links600);
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}
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for( const c5 in CELLINDEX.CELL120_CELL5 ) {
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*/ for( const c5 in CELLINDEX.CELL120_CELL5 ) {
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const nodes5 = nodes.filter((n) => CELLINDEX.CELL120_CELL5[c5].includes(n.id));
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const links5 = auto_detect_edges(nodes5, 4);
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links5.map((l) => l.label = 0);
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