We have so far seen a couple of schemes for identifying clusters in sets of data which are subsets whose members are in some sense similar to each other. Specifically, we have looked at the k means and the shared near neighbours algorithms which implicitly define clusters by the closeness of each datum to the average of each cluster and by their closeness to each other respectively.
Note that both of these algorithms use a heuristic, or rule of thumb, to assign data to clusters, but there's another way to construct clusterings; define a heuristic to measure how close to each other a pair of clusters are and then, starting with each datum in a cluster of its own, progressively merge the closest pairs until we end up with a single cluster containing all of the data. This means that we'll end up with a sequence of clusterings and so before we can look at such algorithms we'll need a structure to represent them.

Ho there Sir R-----! Will you join me for a cold tankard of ale to refresh yourself on this warm spring evening?

And, might I hope, for a little sport?

I should not have doubted it for a moment sir!

This fine weather reminds me of the time I spent as the Empress's trade envoy to the market city of Argos, famed almost as much for the remarkable, if somewhat fragile, mechanical contraptions made by its artificers and the most reasonably priced jewellery sold by its goldsmiths as for its fashion for tiny writing implements.

Last time we saw how we could use the list of the nearest neighbours of a datum in a set of clustered data to calculate its strengths of association with their clusters. Specifically, we used the k nearest neighbours algorithm to define those strengths as the proportions of its k nearest neighbours that were members of each cluster or with a generalisation of it that assigned weights to the neighbours according to their positions in the list.
This time we shall take a look at a clustering algorithm that uses nearest neighbours to identify clusters, contrasting it with the k means clustering algorithm that we covered about four years ago.

Recall that the Baron and Sir R-----'s most recent wager first had the Baron place three coins upon the table, choosing either heads or tails for each in turn, after which Sir R----- would follow suit. They then set to tossing coins until a run of three matched the Baron's or Sir R-----'s coins from left to right, with the Baron having three coins from Sir R----- if his made a match and Sir R----- having two from the Baron if his did.

When the Baron described the manner of play to me I immediately pointed out to him that it was Penney-Ante, which I recognised because one of my fellow students had recently employed it to enjoy a night at the tavern entirely at the expense of the rest of us! He was able to do so because it's an example of an intransitive wager in which the second player can always contrive to make a choice that will best the first player's.

A little under four years ago we saw how we could use the k means algorithm to divide sets of data into distinct subsets, known as clusters, whose members are in some sense similar to each other. The interesting thing about clustering is that even though we find it easy to spot clusters, at least in two dimensions, it's incredibly difficult to give them a firm mathematical definition and so clustering algorithms invariably define them implicitly as the subsets identified by this algorithm.
The k means algorithm, for example, does so by first picking k different elements of the data as cluster representatives and then places every element in the cluster whose representative is nearest to it. The cluster representatives are then replaced by the means of the elements assign to it and the process is repeated iteratively until the clusters stop changing.
Now I'd like to introduce some more clustering algorithms but there are a few things that we'll need first.

My fellow students and I have lately been wondering whether we might be able to employ Professor B------'s Experimental Clockwork Mathematical Apparatus to fashion an ethereal orrery, making a model of the heavenly bodies with equations rather than brass.
In particular we have been curious as to whether we might construct such a model using nought but Sir N-----'s law of universal gravitation, which posits that those bodies are attracted to one another with a force that is proportional to the product of their masses divided by the square of the distance between them, and laws of motion, which posit that a body will remain at rest or move with constant velocity if no force acts upon it, that if a force acts upon it then it will be accelerated at a rate proportional to that force divided by its mass in the direction of that force and that it in return exerts a force of equal strength in the opposite direction.

A little over a year ago we added a bunch of basic computer science algorithms to the ak library, taking inspiration from the C++ standard library. Now, JavaScript isn't just short of algorithms in its standard library, it's also lacking all but a few data structures. In fact, from the programmer's perspective, it only has hash maps which use a function to map strings to non-negative integers that are then used as indices into an array of sets of key-value pairs. Technically speaking, even arrays are hash maps of the string representation of their indices, although in practice the interpreter will use more efficient data structures whenever it can.
As clever as its implementation might be, it's unlikely that it will always figure out exactly what we're trying to do and pick the most appropriate data structure and so it will occasionally be worth explicitly implementing a data structure so that we can be certain of its performance.
The first such data structure that we're going to need is a min-heap which will allow us to efficiently add elements in any order and remove them in ascending order, according to some comparison function.

Sir R-----, my good friend! Come shake the snow from your boots and join me by the hearth for a draught of warming spirits!

And will you also join me in a wager whilst you let the fire chase the chill from your bones?

Fine fellow! Stout fellow!

I have in mind a game that reminds me of my raid upon the vault of Heaven, which I mounted in order to make amends to the Empress for my failure to snatch the Amulet of Yendor from the inner circle of Hell.

We have recently seen how we can define dependencies between random variables with Archimedean copulas which calculate the probability that they each fall below given values by applying a generator function φ to the results of their cumulative distribution functions, or CDFs, for those values, and applying its inverse to their sum.
Like all copulas they are effectively the CDFs of vector valued random variables whose elements are uniformly distributed when considered independently. Whilst those Archimedean CDFs were relatively trivial to implement, we found that their probability density functions, or PDFs, were somewhat more difficult and that the random variables themselves required some not at all obvious mathematical manipulation to get right.
Having done all the hard work implementing the ak.archimedeanCopula, ak.archimedeanCopulaDensity and ak.archimedeanCopulaRnd functions we shall now use them to implement some specific families of Archimedean copulas.

When last they met, the Baron challenged Sir R----- to evade capture whilst moving rooks across and down a chessboard. Beginning with a single rook upon the first file and last rank, the Baron should have advanced it to the second file and thence downwards in rank in response to which Sir R----- should have progressed a rook from beneath the board by as many squares and if by doing so had taken the Baron's would have won the game. If not, Sir R----- could then have chosen either rook, barring one that sits upon the first rank, to move to the next file in the same manner with the Baron responding likewise. With the game continuing in this fashion and ending if either of them were to take a rook moved by the other or if every file had been played upon, the Baron should have had a coin from Sir R----- if he took a piece and Sir R----- one of the Baron's otherwise.