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Author SHA1 Message Date
Mike Lynch
506bf1cdfe Added a function which finds all of the 600-tetras from a vertex of the 120-cell 2025-12-30 16:04:41 +11:00
Mike Lynch
f5afdff3bb More exploration of the 5-cell inscriptions 2025-12-30 08:43:34 +11:00
Mike Lynch
10de708c19 Added NOTES.md 2025-12-30 08:43:04 +11:00
5 changed files with 109 additions and 31 deletions

26
NOTES.md Normal file
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@ -0,0 +1,26 @@
# NOTES
New approach for the 5-cells:
Pick a tetrahedron of an inscribed 600-cell with vertices A, B, C, D
This gives pairs of vertices:
AB
AC
AD
BC
BD
CD
Each of these gives rise to seven pairs of 5-cells which are on neighboring vertices
of the 5 600-cells.
Try enumerating these and inspecting them to find one or more coherent sets of four
5-cells which lie on one tetrahedron from each of the 600-cells.
(I expect there to be more than one, like how there are two ways to partition the
120-cell vertices into 600-cells)

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@ -105,12 +105,10 @@ export const LAYERS120 = {
};
export const CELL120_CELL5 = {
"1": [ 258, 1, 510, 304, 431 ],
"2": [ 185, 93, 222, 295, 372 ],
"1": [ 1, 93, 37, 157 ],
}
// Schoute's partition via https://arxiv.org/abs/1010.4353
export const PARTITION600 = {

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@ -161,9 +161,9 @@ export function gather_5cells(cell120) {
}
function audit_5cells(cells) {
// this verifies that for each label (a 600-cell set), each of its
// this verifies that for each label (a 600-cell set), each of its
// vertices is in exactly 7 5-cells. It checks out.
['1','2','3','4','5'].map((l) => {
const sets = {};
for( const cell of cells ) {
@ -199,7 +199,6 @@ function try_120_5_cells_fails(cell120, cells, l) {
console.log("overlap");
console.log(c);
return true;
}
}
return false;
@ -269,27 +268,24 @@ function node_dist(cell120, aid, bid) {
return dist(a, b);
}
function print_row(v1, v2, p, v5) {
console.log(`${v1.id},${v2.id},${p},${v5[1]},${v5[2]},${v5[3]},${v5[4]},${v5[5]}`);
}
// for a pair of vertices which are on the same inscribed 600 cell,
// this returns all 7 pairs of 5-cells which contain v1 and v2 and
// which are also evenly spaced (ie every pair of vertices on the
// same 600-cell is one edge apart)
function follow_600(cell120, all5) {
const v = cell120.nodes[0];
console.log("Start vertex:");
console.log(v);
const v5s = all5.filter((c5) => c5[v.label] === v.id);
console.log(`Vertex ${v.id} belongs to these 5-cells:`);
console.log(v5s);
const n600s = neighbours600(cell120, v.id);
const n600id = n600s[0];
const n600 = cell120node(cell120, n600id);
console.log("One 600-cell neighbour:");
console.log(n600);
const DIST600 = round_dist(node_dist(cell120, v.id, n600id));
const nv5s = all5.filter((c5) => c5[v.label] === n600id);
console.log(`Vertex ${n600id} belongs to these 5-cells:`);
console.log(nv5s);
console.log("Distances for each pair of 5-cells from the two sets:");
for( const v5a of v5s ) {
for( const v5b of nv5s ) {
function find_adjoining_5cells(cell120, all5, v1, v2) {
const DIST600 = round_dist(node_dist(cell120, v1.id, v2.id));
const v15s = all5.filter((c5) => c5[v1.label] === v1.id);
const v25s = all5.filter((c5) => c5[v2.label] === v2.id);
let p = 0;
const c5pairs = [];
for( const v5a of v15s ) {
for( const v5b of v25s ) {
let match = true;
const d = {};
for( const label in v5a ) {
@ -299,19 +295,76 @@ function follow_600(cell120, all5) {
}
}
if( match ) {
console.log("--- pair ---");
console.log(v5a);
console.log(v5b);
c5pairs.push([ v5a, v5b ]);
}
}
}
return c5pairs;
}
function tetras(cell120, all5, v) {
// given a vertex v, find all of the 600-cell tetras it's on
const n600s = neighbours600(cell120, v.id);
// need to find all sets of three neighbours which are neighbours: there
// should be 20 of these because they're faces of an icosahedron
console.log(v.id);
const tetras = new Set;
for( const v2id of n600s ) {
// find mutual neighbours of the first two
const n2600s = neighbours600(cell120, v2id);
const mutuals = n2600s.filter((nid) => {
return nid != v2id && nid != v.id && n600s.includes(nid)
});
for( const nm of mutuals ) {
const nnms = neighbours600(cell120, nm);
const mutuals2 = nnms.filter((nid) => {
return nid != nm && nid != v2id && nid != v.id && mutuals.includes(nid)
});
for( const m2 of mutuals2 ) {
const t = [ v.id, v2id, nm, m2 ];
t.sort((a, b) => a - b);
const tstr = t.join(',');
tetras.add(tstr);
}
}
}
console.log(tetras);
}
/* // find mutuals which are mutuals of each other
const m0 = mutuals[0];
const mn = neighbours600(cell120, m0);
const mm = mn.filter((nid) => nid != m0 && mutuals.includes(nid));
return [ v1.id, v2id, m0, mm[0] ];
}*/
function str5cell(c5) {
return ["1","2","3","4","5"].map((l) => String(c5[l]).padStart(3, '0')).join('-');
}
function tetra_sets(cell120, all5, tetra) {
// given a tetrahedron on a 600-cell, find the sets of adjacent 5-cells on
// all of the pairs
// this is ass-backwards. Need to find tetras on the other 4 vertices of a 5-cell
const vs = tetra.map((tid) => cell120node(cell120, tid));
const pairs = [[0,1], [0,2], [0, 3], [1, 2], [1, 3], [2, 3]];
for( const p of pairs ) {
const v1 = vs[p[0]];
const v2 = vs[p[1]];
const c5pairs = find_adjoining_5cells(cell120, all5, v1, v2);
console.log(v1.id, v2.id);
console.log(c5pairs.map((p) => str5cell(p[0]) + " " + str5cell(p[1])));
}
}
const cell120 = POLYTOPES.cell120_inscribed();
const all5 = gather_5cells(cell120);
follow_600(cell120, all5);
const v1 = cell120.nodes[0];
tetras(cell120, all5, v1);
tetras(cell120, all5, cell120.nodes[1]);

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@ -473,14 +473,14 @@ export const cell120_inscribed_cell5 = () => {
const links = [];
for( const p of [ 1, 2 ]) {
/* for( const p of [ 1, 2 ]) {
const nodes600 = nodes.filter((n) => n.label === p);
const links600 = auto_detect_edges(nodes600, 12);
links600.map((l) => l.label = p);
links.push(...links600);
}
for( const c5 in CELLINDEX.CELL120_CELL5 ) {
*/ for( const c5 in CELLINDEX.CELL120_CELL5 ) {
const nodes5 = nodes.filter((n) => CELLINDEX.CELL120_CELL5[c5].includes(n.id));
const links5 = auto_detect_edges(nodes5, 4);
links5.map((l) => l.label = 0);