## Archive for the ‘Programming’ Category

### Discrete robot motion planning in a continuous domain

28 February 2019

A robot has a discrete configuration space if it has a discrete set of possible actions at each step, right?

Wrong! In fact, even in a perfectly idealized world where there is zero error in motion, a robot with only discrete actions can effectively reach a continuum of different configurations. This can make motion planning for such robots very interesting.

Here’s a robot I built when I was about 15 years old:

I don’t have this robot anymore.  I gave it to the local jawas in the California desert before moving to Germany.   But now, a weekend’s worth of programming let me bring it back to life in a virtual world.  Plus, I’ve made it more intelligent than it ever was in its previous life.  Bwahhh-ha-ha!

Let me explain what’s interesting about this robot, and why I coded it into a virtual world.

The original robot could not do much.  It could rotate and move forward or backward only in fixed increments, and I could pre-program a sequence of moves ahead of time, which it would then execute.  I know—not very impressive as robots go these days.

But the limitations of this robot are also what make it interesting.  Since it only had discrete motions, and rather coarse ones at that, getting it to move from one location to another without running into things was an interesting challenge!

## An ideal discrete robot

Let’s consider an idealized version of this robot which can do only two things:

1. Move forward or backward by exactly 1 meter.
2. Rotate in place in increments of exactly 45 degrees.

The robot I built was very much like this, although it moved some shorter distance (1m is just for a convenient number), and of course executed neither of the two possible moves with perfect consistency or accuracy. Still, it was rather good at doing what it was designed to.

In fact, we can idealize this robot just a bit more.  Let’s ignore the need for the robot to rotate and just assume it can move instantly in any of eight directions, each separated from 45 from the last.   Thus, if we point the robot Northward, at each stage it can move in any of eight basic compass directions: N, S, E, W, NW, NE, SE, and SW.

In a wide open space with nothing to run into, if you give our idealized robot a pen (like the Logo turtle) and send it out on a random walk, you’ll see something like this:

There’s something important to observe here: the locations where the robot stopped between moves, denoted by dots, can end up being very close together, even though the robot always moves in steps of one meter.

In fact, the robot can get essentially anywhere. The positions the robot can get to form a dense subset of the plane:  Given enough moves, it can stop as close as you wish to any destination point you pick.  I say it can “stop” close to the destination because I don’t want to count it is reaching its destination if it just rolls over the destination in transit without actually stopping there.  That would be cheating!

For example, suppose the robot is at (0, 0) and you want to program it to get to (0, 0.5).  Just telling the robot to move straight toward (0, 0.5) won’t work: it will drive through the point (0, 0.5), but it won’t stop there; it will keep going all the way to (0, 1).  In fact, you’ll never be able to reach (0, 0.5) exactly, but if you’re satisfied with getting there to within 0.1 meters of the destination, then there’s an easy solution with only 3 steps:

The red x marks the goal at (0, 0.5), and the final resting spot of the robot, marked by a red dot , is within 0.1 of this.  More precisely, it’s easy to see that the robot comes to rest at the point $(0,\sqrt{2}-1)$, or approximately (0, 0.4142), which is within our acceptable error of 0.1 m.

If we’re not satisfied getting within 0.1 m, and we’d rather get within, say, 0.05 m (5cm) of the goal, we can still do it with a bit more thought.  Here’s a way to do it with 20 moves, not counting rotations:

Of course, solutions are not at all unique: Here are two more solutions to the same problem:

However, each of these also consists of 20 moves.  In fact, these are all minimal solutions: every solutions to this problem, at the specified precision of 5cm, has at least 20 moves not counting rotations.

Of course, you could also demand even more precision.  If you want to get from (0, 0) to (0, 0.5) to within a distance of 0.01, i.e. within 1cm, given our step size of 1m, you’ll need at least 119 moves:

Yikes!

Finding a plan to get from one location to another is an example of “motion planning” in robotics, and doing this kind of motion planning is one of the ways I had fun with the original robot when I was a teenager.  I learned to do some of this intuitively just by getting a feel for how to use combinations of moves.  Of course, that kind of intuition never let me do anything so complicated as finding the 119 moves above!

## Motion planning on a hypercube lattice

The virtual reincarnation of my old robot, despite the disadvantage of lacking any corporeal existence, has one big advantage over its former self: it can solve the motion planning problem itself.

To  do this, it uses a classic AI tool—A* search.  This is an algorithm for finding minimal paths through weighted graphs, and is commonly used for route finding in map programs, solving puzzles, and planning complex sequences of actions.

In the robot’s case, although it occupies a 2d world which we identify with a subset of $\mathbb{R}^2$, the A* search is carried out instead on the 4-dimensional lattice $\mathbb{Z}^4$ representing the four directions it can travel in the physical space: the N-S, E-W, NE-SW, and SE-NW directions.

In other words, the robot really thinks of itself as living on a hypercube lattice.  You may be used to seeing pictures of hypercubes in one of their 3d projections, like this:

However, if you project a hypercube down into two dimensions in such a way that all of the edges have the same length in the projection, you can get this:

This picture of a hypercube is something I could draw with my robot!  All of the angles are multiples of 45 degrees, and all of the edges have the same length.  (Plus, I can draw it in one pass: all of the vertices are even, so there exists an Euler circuit.)

Of course, there’s a deep reason for this: the real configuration space of the robot, still ignoring the robot’s orientation, is a projection of a hypercube lattice into $\mathbb{R}^2$.  The vertices of this lattice are the states the robot can occupy at rest, while the edges represent the moves, or allowed transitions between states.

The entire hypercube lattice is like an ordinary cubical lattice:

but one dimension higher, and instead of 8 cubes meeting at each vertex, it has 16 hypercubes meeting at each vertex.  This is rather complicated to draw, and I won’t show a picture of the hypercube lattice.  However, for your amazement, here is a picture I drew of its projection into two dimensions:

There’s nothing wrong with this picture—The lattice’s projection is dense in $\mathbb{R}^2$, as my robot shows, remember! :-)

Explicitly, the basic projection $\mathbb{Z}^4 \to \mathbb{R}^2$ can be written

$(m, n, a, b) \mapsto (x,y)$

where

$\begin{array}{rcl} x &=& m + \frac{1}{2}(a+b)\sqrt{2} \\ y &=& n + \frac{1}{2}(a-b)\sqrt{2}\end{array}$

The eight possible moves—or in other words, the edges in the hypercube lattice—correspond to incrementing or decrementing one of the coordinates $m, n, a, b$.  You can check that these project exactly to the eight allowed robot motions.

I combine this projection into $\mathbb{R}^2$ with a translation and rotation, so that the robot’s current position in the lattice is always (0,0,0,0).

## A* Search implementation

To solve the motion planning problem, the virtual robot uses A* search on the hypercube lattice.  This searches through the tree of paths starting at the robot’s current position (0,0,0,0) in the lattice $\mathbb{Z}^4$ until it finds a minimal length route to some point in the lattice whose projection into $\mathbb{R}^2$ is less than the specified tolerance away from the destination in $\mathbb{R}^2$.  At each stage, the allowed moves are those edges in the hypercube lattice whose projections do not either leave the “room” the robot is contained in or intersect any obstacles.

A* search also relies on a “heuristic distance”—a way of conservatively estimating distance remaining to to the goal.  This is because the tree of paths is traversed in order of total estimated distance for a given route.  The advantage of this over a  breadth first search, which would in principle also find a minimal solution, is that far fewer paths need to be explored.

An effective heuristic distance in the case of the robot, at least for relatively small problems, comes from again using the projection from $\mathbb{Z}^4$ to $\mathbb{R}^2$ and checking the Euclidean distance to the specified goal.

The big point here is that although the robot’s shortest path between two points in a room might not look to us much like a straight line:

it’s still a minimal path in a 4 dimensional lattice, projected down to the plane.

## Real robots

Obviously, my virtual robot is an idealization.  For the real robot it is based on, two kinds of uncertainty come into play: uncertainty in the current configuration, and uncertainty in the moves from one configuration to another.  The planning problem must interact with the “robot localization” problem.

First, there is uncertainty in the robot’s current position and orientation.  For example, even assuming the robot can move with perfect accuracy, optimal planning is very sensitive to errors in angular placement (“attitude”).  If you want to get from (0,0) to (1, 2), and the robot’s placement is very accurate, it will move like a knight:

But rotate its starting position just 2 degrees to the left, and you’ll need 16 moves to get to the same point with an accuracy of ±0.05 m:

Rotate it 2 degrees to the right, and you’ll need 20 moves to get the same accuracy:

Small perturbations in the position of the robot obviously has a similar effect.

More importantly, the moves themselves are not completely accurate.  On the real robot, I had a potentiometer that adjusted the relative speed of the two side wheels to calibrate the forward motion.  Imperfect calibration means not driving in a completely straight line.  Similarly, during a rotation, there was surely also some tiny change in position as well.

When we account for both types of uncertainty, there is a probability distribution for the initial configuration of the robot, including position and orientation.  There is also a probability distribution for the result of each possible rotation or translation move.  Using Bayesian probability, these two distributions are convolved each time we make a move, in order to get the probability distribution for the updated configuration.

The accumulation of errors make accurate motion planning impossible unless we give the robot the ability to “sense” its environment, use sensory data to update its configuration distribution, and adjust the plan accordingly.  But that’s a discussion for another time.

### Fractals and Monads – Part 3

19 February 2019

I promised more examples of using monads to draw fractals.  Let’s do some!

Today I’ll explain a simple method to draw lots of fractal pictures with minimal code, using the idea of Kleisli powers I introduced in the previous part. This will be easy to implement in Haskell, and accessible even if you’re not a Haskell programmer.

To do this, I’ll first introduce another well known way of drawing fractal pictures, using random motion.  If you haven’t seen this before, it’s cool in its own right, but then we’ll change the rules slightly and generate fractals instead using monads.  I’ll again be using the power set and list monads this time, so if you’ve read the first two parts, this should be familiar.

## The Chaos Game

In the “Chaos Game” you pick a fixed set of affine transformations of the plane, usually all contraction mappings, and then watch what happens to a single point in the plane when you apply a randomly chosen sequence from our set of transformations.

For example, in the best-known version of this game we take just three transformations, each a scaling of the plane by a factor of 1/2, but each around a different fixed point. More precisely, fix points A, B, and C, the vertices of a triangle, and define three corresponding transformations $f_A, f_B, f_C$ by

$\displaystyle f_\mathbf{p}(\mathbf{x}) = \frac{\mathbf{p} + \mathbf{x}}{2} \quad \text{for } \mathbf{p} \in \{A, B, C\}$

Here’s how the game typically starts:

I’ve picked a random starting point and a random sequence which begins B, A, C, B, B, B, A, A, B, … .  You can see that at each step, we’re moving in a straight line half way to the next point in the sequence.

At first, it doesn’t sound doesn’t seem like anything very exciting is happening.  But if we keep going, we find something interesting.  To show the pattern more clearly, I’ll stop drawing the lines between steps, and just keep track of the successive points.  I’ll also shrink the dots a little, to avoid them running into each other.  Here’s what a typical random orbit looks like after 3000 iterations:

A picture of the Sierpinski triangle is already emerging.  After 10,000 iterations it looks like this:

And after 50,000 iterations it’s getting quite clear:

There are tons of variations of the “chaos game,” which also go by the less flashy but more descriptive name of “Iterated Function Systems.”  But let’s get to the monads.

## Removing the Randomness with Monads

What does the Chaos Game have to do with monads?  While it’s fun to generate fractal pictures using random motion, we can also trade this random process for a completely  deterministic one in which we iterate a single morphism in the Kleisli category of a monad.

In general, an alternative to making random selections between equally likely steps is to take all of the possible steps simultaneously.  That is, we can get rid of the randomness at the cost of replacing the ordinary functions $\mathbb{R} \to \mathbb{R}$ with a multi-valued function.  Monads give us a nice way to handle this: we’ve seen before that multi-valued functions are just morphisms in the Kleisli category of the power set monad, and we can compose them using Kleisli composition.

In our case, the functions used in the chaos game are $f_A, f_B, f_C$, which move points closer to A, B, C, respectively.  We can combine these into a single function

$f \colon \mathbb{R}^2 \to \mathcal{P}\mathbb{R}^2$

given by

$f(\mathbf{p}) = \{ f_A(\mathbf{p}), f_B(\mathbf{p}), f_C(\mathbf{p}) \}$

We then have only to iterate this.

The resulting code gives us pictures similar to the ones for the original chaos game, but much more regular and symmetric, since we’ve avoided making any random choices by making all possible choices.  If we also start out in a symmetric state, picking (0,0) as our starting point, and letting the fixed points A, B, and C form an equilateral triangle centered around this point, then at each generation thereafter we get a picture that has all of the symmetries of a triangle.

Here’s generation 6, with $3^6 = 729$ points:

Here’s generation 8, with $3^8=6,561$ points:

And here’s generation 10, with $3^{10} = 59,049$ points:

Comparing this to the picture from the chaos game with 50,0000 iterations, the two are very similar, though the “monadic” version has sharper edges, and none of the small scale noise of the random version.

## The code

To implement this in Haskell, I need to define what a point in the plane is, and define a few operations on points.  I’ll define a new data type, Point, which really just consists of pairs of real (or more accurately, floating point) numbers, and I’ll define vector addition and scalar multiplication on these in the obvious way:

type Point = (Float,Float)

add :: Point -> Point -> Point
add (x1,y1) (x2,y2) = (x1+x2,y1+y2)

mult :: Float -> Point -> Point
mult r (x,y) = (r*x,r*y)

Sure, I could use a linear algebra library instead; I’m implementing these operations myself mainly because it’s easy, and perhaps more friendly and instructive for those new to Haskell.

Using these operations, we can build a function that scales the plane around a given point. I’ll do this more generally than in the examples above, because it will be useful for other versions of the chaos game which I’ll discuss later.  If $\mathbf{p}\in \mathbb{R}^2$ is a point in the plane, and $r$ is a real number, then the function

$\displaystyle f_{r,\mathbf{p}}\colon\mathbf{q} \mapsto (1-r)\mathbf{p} + r\mathbf{q}$

scales $\mathbb{R}^2$ by a factor $r$ around the point $\mathbf{p}$. Here’s a Haskell implementation of this function:

scaleAround :: Float -> Point -> Point -> Point
scaleAround r p q = ((1-r) mult p) add (r mult q)

This can be read as saying we take two parameters, a real number r and a point p, and this results in a function from points to points, which is really just $f_{r,\mathbf{p}}$.  The  marks around add and mult let us use these as “infix” operators, as opposed to prefix operators like in Polish notation; for example 3 mult (1,0) means the same as mult 3 (1,0), and both give the result (3,0).

Now, we can make a list-valued function that will generate the Sierpinski triangle.  First, let’s fix the vertices of our triangle, which will just be a list of Points:

triVertices = [(sin(2*k*pi/3), cos(2*k*pi/3)) | k <- [0..2]]

This is so much like the mathematical notation for sets that it needs no explanation; just squint until the <- starts to look like $\in$, and remember that we’re dealing with Haskell lists here rather than sets.

As before, our function for generating this fractal combines all of the $f_{1/2,\mathbf{p}}$ where $\mathbf{p}$ ranges over the vertices in the triangle:

f :: Point -> [Point]
f q = [scaleAround 1/2 p q | p <- triVertices]

The first line is the type signature of the function; it just says f takes a point as input, and returns a list of points as output.  This is our morphism in the Kleisli category of the list monad.  The rest of the code just amounts to taking Kleisli powers of this morphism.  Recall that code from last time:

kpow :: Monad m => Int -> (a -> m a) -> a -> m a
kpow 0 _ = return
kpow n f = f <=< kpow (n-1) f

That’s everything we need.  Let’s pick our starting point in $\mathbb{R}$ to be the origin, set the number of generations to 10, and calculate the output data:

startPt = (0,0) :: Point
nGens = 10

outputData = kpow nGens f $startPt The last line here takes the 10th Kleisli power of f, and then applies that to the starting point (0,0). The result is the list of points, and if you plot those points, you get the picture I already posted above: ## More examples Once you know the trick, you can build new versions of the chaos game to generate other fractals. Here’s a fun one. Before, we started with an equilateral triangle and scaled around each of its vertices by a factor of 1/2; Now let’s take a regular hexagon and scale around each of its vertices by a factor of 1/3. To build this using our monadic methods, all we really need to change in the code I’ve written above is which vertices and scaling factor we’re using: hexVertices = [(sin(2*k*pi/6), cos(2*k*pi/6)) | k <- [0..5]] f q = [scaleAround 1/3 p q | p <- hexVertices] The result? It’s a Koch curve! Or rather, it’s lots of copies of the Koch curve at ever smaller scales. This is just the fractal that I’ve written about before, starting with “Fun with the Koch Snowflake” which I wrote when I first discovered it. It’s like a “Sierpinski Koch snowflake” because it can also be built by starting with a Koch snowflake and the repeatedly removing a Koch snowflake from the middle to leave six new Koch snowflakes behind. The reason this slightly modified chaos game gives the Sierpinski Koch snowflake is essentially that, at each stage, the Koch snowflake splits into six new Koch snowflakes, each a third of the size of the original, and distributed in a hexagonal pattern. How does this generalize? There’s a whole family of fractals like this, which I posted about before, and there’s a version of the chaos game for each one. I explained in Fun with the Koch snowflake how this replication rule for triangles: leads to the Sierpinski Koch Snowflake we found above: Similarly, I explained how this replication rule for squares: leads to an analogous fractal with octagonal symmetry: leads to this: fractal with decagonal symmetry. This sequence continues. To get the transformations needed for the chaos game, we just need to calculate how much smaller the polygons get in each successive generation. If you stare at the replication rules above, you can see that pattern I’m using: the edge of the parent polygon is congruent to two edges of the child polygon plus a segment extending from one vertex to the next-to-adjacent vertex of the child polygon. Solving this, the correct scaling factor is $\displaystyle \frac{1}{2 + 2 \sin\left(\frac{(n-2)\pi}{2n}\right)}$. The chaos game—and the tamer “monadic” variant on it—for these fractals just amounts to scaling by this factor around each of the vertices in a regular $2n$-gon. By the way, you might be tempted (I was!) to include some rotations in addition to the scale transformations, since in the polygon-based versions, only half of the polygons have the same angular orientation as their parent. However, these rotations turn out not to matter when we’re using points as our basic geometric objects rather than polygons. Again, there’s little code we need to modify. First, since we need polygons with any number of sides, let’s make a function that takes an integer and gives us that many vertices distributed around the unit circle, with the first vertex always at 12 O’Clock: polyVertices :: Int -> [Point] polyVertices n = [(sin(2*k*pi/nn), cos(2*k*pi/nn)) | k <- [0..(nn-1)]] where nn = fromIntegral n  If you’re new to Haskell, this should all be familiar by now, except perhaps the last line. Haskell will complain if you try dividing a floating point number by an integer since the types don’t match, so the integer n has to be converted to a float nn; this makes k also a float, since the notation [0..x] actually means the list of all of the floating point integers from 0 to x, if x is a floating point number, and then everything in sight is a float, so the compiler is happy. We then have to change our generating morphism to: f :: Int -> Point -> [Point] f n q = [scaleAround r p q | p <- polyVertices (2*n)] where r = 1/(2 + 2*sin((nn-2)*pi/(2*nn))) nn = fromIntegral n and remember that the generating morphism is now not just f but f n. Now we can set the starting point, the number of “sides” of the basic polygon, and the number of generations, as we wish: startPt = (0.2,0.5) :: Point m = 5 -- number of generations n = 7 -- number of "sides" outputData = kpow m (f n)$ startPt

Just remember that the final list is a list of $(2n)^m$ points, so don’t blame me if you make these numbers big and run out of memory. :-)

In any case, with code like this, you can generate analogs of the Sierpinski Koch snowflake based on squares:

Pentagons, which start “overlapping”:

Hexagons, where the overlapping becomes perfect:

and more.  Heptagons, where the overlapping gets a bit more severe:

Octagons:

and so on.

But enough of that.  Next time, I’ll move on to fractals of a different sort …

### Fractals and Monads – Part 2

30 January 2019

Last time I discussed the power set monad and how it shows up in certain methods of building fractals.  This time I’ll go a bit further, and also explain some Haskell code I actually use to generate fractal pictures like this:

To do this, I’ll introduce Kleisli categories, the list monad, and a bit about how monads work in Haskell!

Recall we were considering fractals that can be built using a “replication” function

$f\colon X \to \mathcal{P}X$

on some set of “shapes” $X$.  As an example, we had considered a map that takes any triangle in the plane to sets of three new triangles as shown here:

and we noted that “iterating” this map indefinitely gives us all of the triangular “holes” in a Sierpinski triangle:

I also described how iterating this map, if we do everything carefully, involves the structure of the power set monad.

In fact, considering maps like $f$ leads us directly to the “Kleisli category” associated to the power set monad.  I want to explain this here, because in computer science people actually tend to think of monads essentially in terms of their associated Kleisi categories, and that’s how monads are implemented in Haskell.

Every monad has an associated Kleisli category, but since I’ve been working with the power set monad $(\mathcal{P}, \eta, \mu)$ in the category of sets, I’ll start with that special case.

First, in the Kleisli category, which we’ll denote $\mathrm{Set}_\mathcal{P}$, the objects are the same as the original category $\mathrm{Set}$: they’re just sets.  A morphism from the set $X$ to the set $Y$, however, is a function

$f\colon X \to \mathcal{P}Y$

For now, at least, I’ll avoid making up any special notation for morphisms in the Kleisli category, and instead freely regard $f\colon X \to \mathcal{P}Y$ either as a morphism from $X$ to $\mathcal{P}Y$ in the category $\mathrm{Set}$ of sets, or as a morphism from $X$ to $Y$ in the Kleisli category $\mathrm{Set}_\mathcal{P}$; these are really the same thing.  If this is confusing, whenever you’re thinking of $f$ as a morphism in the Kleisli category, just think of the $\text{}\mathcal{P}\text{''}$ as decoration on the arrow , rather than on the codomain object $Y$.

Given two functions $f\colon X \to \mathcal{P} Y$ and $g\colon Y \to \mathcal{P} Z$, the Kleisli composite $g\bullet f$ is defined by

$g\bullet f := \mu_Z\circ \mathcal{P} g\circ f$

Checking the domain and codomain of each of the functions on the right, we can see that this indeed gives a morphism from $X$ to $Z$ in the Kleisli category:

$\qquad \qquad X \stackrel f \longrightarrow \mathcal{P} Y \stackrel {\mathcal{P} g} \longrightarrow \mathcal{P} \mathcal{P} Z \stackrel {\mu_Z} \longrightarrow \mathcal{P} Z$

You can check that $\bullet$ is associative, and that the maps $\eta_X\colon X \to \mathcal{P} X$ are identity morphisms, so we really do get a category. That’s all there is to the Kleisli category.

Returning to the sort of function we had been considering for generating fractals:

$f\colon X \to \mathcal{P}X$

these are clearly just endomorphisms in the Kleisli category.  This also makes it clear what we mean by “iterating” such a function: it’s an endomorphism, so we can just compose it with itself as many times as we want.  In particular, analogously to how we define “powers” of ordinary functions by repeated composition, we can define the nth Kleisli power of $f$ recursively by

$\displaystyle f^{{\bullet}n} = \left\{\begin{array}{cc}\eta_X & n=0\\ f \bullet f^{{\bullet}(n-1)}& n\ge 1 \end{array}\right.$

That is, the 0th power of $f$ is the identity, while the nth power of is the Kleisli composite of $f$ with its (n-1)st power.

For example, consider the set $T$ of all equilateral triangles in the plane and consider the map $f\colon T \to \mathcal{P}T$ defined so that $f(t)$ has six elements, each 1/3 the size of, and linked together around the boundary of the original triangle $t$, as shown here:

Then, starting with an arbitrary triangle $t \in T$ the images under successive Kleisli powers of the generating endomorphism:

$f^{\bullet 0}(t), f^{\bullet 1}(t), f^{\bullet 2}(t), f^{\bullet 3}(t), f^{\bullet 4}(t), \ldots$

look like this:

Let’s do some Haskell—If you don’t know Haskell, don’t worry; I’ll explain what everything means. If you do know Haskell, this will be super easy, perhaps even obvious, though I may write things in ways you won’t expect, since I’m transitioning over from the categorical definition of monad, rather than diving in with how monads are actually defined in Haskell.

The code we really need is almost embarrassingly short, mainly because monads are extremely well suited to the task at hand.  What we need most is just to be able to compute Kleisli powers!

Here’s how the formula for the Kleisli power of an endomorphism becomes a function in Haskell:

import Control.Monad

kpow :: Monad m => Int -> (a -> m a) -> a -> m a
kpow 0 _ = return
kpow n f = f <=< kpow (n-1) f


Let’s go through this line by line.

The first line just loads some tools for working with monads; there’s not much I need to say about that now.

The next line is the first one of real content; it’s the “type signature” of the function “kpow”.  The first bit, Monad m, just says that m is any monad; Haskell knows about several monads, and you can also define your own.   Then, the rest of that line:

Int -> (a -> m a) -> a -> m a

can be interpreted as declaring that we have a function that takes one integer (Int), the “power,” and one element of $\hom(a, ma)$, (i.e. something of type a -> m a), and gives us another element of $\hom(a, ma)$.  So if I were to write this line in standard mathematical notation, I would write:

$\mathrm{kpow}:\mathbb{N} \times \hom(a, ma) \to \hom(a, ma)$

where $m$ is any monad, and $a$ is any type.  We take an integer and a Kleisli endomorphism for the monad m, and return another endomorphism. Notice there’s a slight discrepancy here, in that an Int is any integer, whereas I used the natural numbers $\mathbb{N}$.  But the integer really is intended to be non-negative.  If you fed a negative integer into kpow, the recursion would never terminate; so, my code doesn’t forbid it, but you shouldn’t do it.

The last two lines of the code give the actual formula for the Kleisli power. If you just look at my earlier formula:

$\displaystyle f^{{\bullet}n} = \left\{\begin{array}{cc}\eta_X & n=0\\ f \bullet f^{{\bullet}(n-1)}& n\ge 1 \end{array}\right.$

and compare this to these two lines:

kpow 0 _ = return
kpow n f = f <=< kpow (n-1) f


you can probably figure out all of the Haskell syntax just by matching up symbols.

The first line says “whenever the first argument of kpow is 0, the result is “return.” This may be confusing if you’re used to how the keyword “return” is used in many other programming languages. Sorry; it’s not my fault! In Haskell “return” means “$\eta_x$ for any object $x$“, where the objects in this case are always data types.

The second line says “the nth Kleisli power of f is the composite of f with the (n-1)st power of f”, just as in the other formula. You probably guessed that the Kleisli composite $f \bullet g$ is written f <=< g in Haskell.

Here’s a dictionary relating the notation I’ve been using to Haskell notation for the same thing:

I didn’t need to use $\mu$, a.k.a. join explicitly in the definition of Kleisli powers, since it’s already wrapped into the definition of Kleisli composition.

To do some examples I’ll use the list monad, which is a built-in monad in Haskell. This is very much like the power set monad, since “lists” are like sets in many ways, so I can introduce this pretty quickly. We really just need a very basic idea of how the list monad works.

A “list” of elements of “type” $a$ can be written written simply as

$[x_1, x_2, \ldots x_n] \quad \text{where } x_i\in a$

for a finite list, or $[x_1, x_2, \ldots]$ for an infinite list.

Making lists of elements of a given type clearly gives a functor analogous to the power set functor: for any function $f\colon a \to b$ we get a function that just acts element-wise on lists:

$([x_1,x_2,\ldots, x_n]) \mapsto [f(x_1),f(x_2), \ldots f(x_n)]$

mapping a list of elements of type $a$ into a list of elements of type $b$.

Based on what we’ve done with sets, you can probably guess how the two natural transformations in the list monad work, too.

First, any single item can be regarded as a list with only one item on it, so

$\eta_a(x) = [x]$

In Haskell, for example, you could take an element of type Int, say the integer 5, and turn it into a list of integers by using return 5. As long as Haskell has a context to know it is working in the list monad rather than some other monad, this will output [5]. Voilà!

Second, the natural transformation $\mu$ turns lists of lists into lists.  If you’ve got a list of to-do lists, you can always turn it into a single list by first writing down the items on the first list, in order, then continuing with the items on the second list, and so on.

In Haskell, If you enter join [[1,2,3],[4,5]], you’ll get [1,2,3,4,5] as output. That’s $\mu$.

There’s one noteworthy subtlety, which is that infinite lists are also allowed. So, if you have a list of lists:

[Fun stuff to do, Work to do]

and the first item happens to be an infinite list, then if you try joining these lists up into one in the way I described, you’ll never get to any items on the second list. This is one way $\mu$ in the list monad differs from $\mu$ in the power set monad: In the former, as soon as one of the lists is infinite, everything after that gets thrown away. This last part is not obvious; you certainly could devise ways to join a list of infinite lists together and include all of the items, but that’s not how things work in this monad.

You can easily work out how Kleisli composition works from this, so we’ve now got everything we need.

## Cantor set in the List Monad.

Let’s do an example: the Cantor set. I’ll represent closed intervals on the real line by ordered pairs in Haskell. So, when I write (0.3, 1.5), I mean the set $\{x\in \mathbb{R} : 0.3 \le x \le 1.5\}$. I know, round brackets for closed intervals is weirdly non-standard, but in Haskell, square brackets are reserved for lists, and it’s better to keep things straight.

To generate the Cantor set, all we need is the Kleisli endomorphism in the list monad that describes the general step. This can be defined by a one-line Haskell function:
 f (p,q) = [(p, p + (q-p)/3), (q-(q-p)/3, q)] 
If we apply this to the unit interval (0, 1), here’s what we get:
 [(0.0,0.3333333333333333),(0.6666666666666667,1.0)] 
So this works: it’s mapping the closed interval $[0,1]$ to a list of two closed intervals: $[0, 1/3]$ and $[2/3, 1]$.

Now we can compute any step in the construction of the Cantor set by computing Kleisli powers. Let’s just do one. Once you’ve loaded the function kpow, enter this in Haskell:
 kpow 3 f (0,1) 
and—kapow!—you’ll get this output:

[(0.0,3.7037037037037035e-2),
(7.407407407407407e-2,0.1111111111111111),
(0.2222222222222222,0.25925925925925924),
(0.2962962962962963,0.3333333333333333),
(0.6666666666666667,0.7037037037037037),
(0.7407407407407408,0.7777777777777778),
(0.888888888888889,0.9259259259259259),
(0.962962962962963,1.0)]
`

Ugly isn’t it? But convert to fractions and you’ll see this is doing just the right thing.

I’ll do more examples in the next post.