| … | … | @@ -204,20 +204,43 @@ observed, i.e. "in practice") |
| 204 | 204 | Now, cluster them "by similarity". One of the easiest ways fo doing this |
| 205 | 205 | is by cosine-similarity. Cosine similarity requires vectors; where do |
| 206 | 206 | the vectors come from? They arise entirely naturally, in many ways. For |
| 207 | | example, consider all jigsaws that have the label "A" for the center |
| 208 | | vertex; these will have a large variety of labels "p,q,r..." on the |
| 209 | | connectors. How many are there? Say there are `N(A | p,q,r...)` of |
| 210 | | these. If we hold "A" fixed, while allowing "p,q,r..." to vary, this |
| 211 | | forms a vector! The ket `|p,q,r,...)` is the basis elt of the vector. |
| 212 | | Suppose the central vertex is labelled by "B": then `N(B | p,q,r...)` |
| 207 | example, consider all jigsaws that have the label $A$ for the center |
| 208 | vertex; these will have a large variety of labels $p,q,r...$ on the |
| 209 | connectors. How many are there? Say there are $N(A | p,q,r...)$ of |
| 210 | these. If we hold $A$ fixed, while allowing $p,q,r...$ to vary, this |
| 211 | forms a vector! The ket $|p,q,r,...)$ is the basis elt of the vector. |
| 212 | Suppose the central vertex is labelled by $B$: then $N(B | p,q,r...)$ |
| 213 | 213 | is another vector, and the dot product is easy: |
| 214 | | ``` |
| 214 | $ |
| 215 | 215 | A . B = Sum_{p,q,r..} N(A | p,q,r...) N(B | p,q,r...) |
| 216 | | ``` |
| 216 | $ |
| 217 | 217 | In practice, (i.e. experimentally measured) the dot products are |
| 218 | 218 | gaussian distributed, i.e. the data is distributed uniformly on a very |
| 219 | 219 | high dimensional sphere. |
| 220 | 220 | |
| 221 | ### Clustering |
| 222 | Having a bizzillion points uniformly distributed on a high dimensional |
| 223 | sphere is interesting but useless. The first thing we want to do is |
| 224 | dimensional reduction by clustering. There are surely 101 different |
| 225 | clustering algos. Pick one; pick several. |
| 226 | |
| 227 | Consider K-means. Take the high-dim sphere, sprinkle K random points |
| 228 | on it, and assign vectors to the closest one. Duh. That is, each |
| 229 | vertex label in the set {A, B, ...} is assigned to one of K different |
| 230 | clusters. |
| 231 | |
| 232 | Unfortunately, this K-means clustering is horrble and ugly, because |
| 233 | although it reduced the bras $(A|$ to a small number, it did not reduce |
| 234 | the kets $|p,q,r...)$, of which there are still a gazzilion. Well, |
| 235 | sort-of. Given that each p,q,r,... is a vertex indicator plus a sex |
| 236 | label, K-means clustering of the bras will give $K^m$ kets, where $m$ is |
| 237 | the mean number of connectors on a jigsaw. In total, then, there are |
| 238 | $K^{m+1}$ such collections. |
| 239 | |
| 240 | Whatever. There are nicer clustering algos (which are more complex) that |
| 241 | can maybe do a better job. That's not the point. |
| 242 | |
| 243 | ### Axioms! |
| 221 | 244 | |
| 222 | 245 | |
| 223 | 246 | |