# ploeh blog danish software design

## Quasigroups

*A brief introduction to quasigroups for object-oriented programmers.*

This article is part of a larger series about monoids, semigroups, and other group-like algebraic structures. In this article, you'll get acquainted with the concept of a quasigroup. I don't think it plays that big of a role in software design, but it *is* a thing, and I thought that I'd cover it briefly with a well known-example.

During all this talk of monoids and semigroups, you've seen that normal arithmetic operations like addition and multiplication form monoids. Perhaps you've been wondering where subtraction fits in.

Subtraction forms a quasigroup.

What's a quasigroup? It's an invertible binary operation.

### Inversion #

What does it mean for a binary operation to be invertible? It means that for any two elements `a`

and `b`

, there must exist two other elements `x`

and `y`

that turns `a`

into `b`

.

This is true for subtraction, as this FsCheck-based test demonstrates:

[Property(QuietOnSuccess = true)] public void SubtractionIsInvertible(int a, int b) { var x = a - b; var y = a + b; Assert.True(a - x == b); Assert.True(y - a == b); }

This example uses the FsCheck.Xunit glue library for xUnit.net. Notice that although FsCheck is written in F#, you can also use it from C#. This test (as well as all other tests in this article) passes.

For any `a`

and `b`

generated by FsCheck, we can calculate unique `x`

and `y`

that satisfy `a - x == b`

and `y - a == b`

.

Subtraction isn't the only invertible binary operation. In fact, addition is also invertible:

[Property(QuietOnSuccess = true)] public void AdditionIsInvertible(int a, int b) { var x = b - a; var y = b - a; Assert.True(a + x == b); Assert.True(y + a == b); Assert.Equal(x, y); }

Here I added a third assertion that demonstrates that for addition, the inversion is symmetric; `x`

and `y`

are equal.

Not only is integer addition a monoid - it's also a quasigroup. In fact, it's a group. Being associative or having identity doesn't preclude a binary operation from being a quasigroup, but these properties aren't required.

### No identity #

No identity element exists for integer subtraction. For instance, *3 - 0* is *3*, but *0 - 3* is *not 3*. Therefore, subtraction can't be a monoid.

### No associativity #

Likewise, subtraction is not an associative operation. You can easily convince yourself of that by coming up with a counter-example, such as *(3 - 2) - 1*, which is *0*, but different from *3 - (2 - 1)*, which is *2*. Therefore, it can't be a semigroup either.

### Summary #

A quasigroup is an invertible binary operation. Invertibility is the only *required* property of a quasigroup (apart from being a binary operation), but if it has other properties (like associativity), it's still a quasigroup.

I haven't had much utility from thinking about software design in terms of quasigroups, but I wanted to include it in case you were wondering how subtraction fits into all of this.

What if, however, you have a binary operation with *no other* properties?

**Next:** Magmas.

## Semigroups accumulate

*You can accumulate an arbitrary, non-zero number of semigroup values to a single value. An article for object-oriented programmers.*

This article is part of a series about semigroups. In short, a *semigroup* is an associative binary operation.

As you've learned in a previous article, you can accumulate an arbitrary number of monoidal values to a single value. A corresponding property holds for semigroups.

### Monoid accumulation #

When an instance method `Op`

forms a monoid, you can easily write a function that accumulates an arbitrary number of `Foo`

values:

public static Foo Accumulate(IReadOnlyCollection<Foo> foos) { var acc = Identity; foreach (var f in foos) acc = acc.Op(f); return acc; }

Notice how this generally applicable algorithm starts with the `Identity`

value. One implication of this is that when `foos`

is empty, the return value will be `Identity`

. When `Op`

is a semigroup, however, there's no identity, so this doesn't quite work. You need a value to start the accumulation; something you can return if the collection is empty.

### Semigroup accumulation #

From Haskell you can learn that if you have a semigroup, you can accumulate any non-empty collection:

`sconcat :: Semigroup a => NonEmpty a -> a`

You can read this as: for any generic type `a`

, when `a`

forms a `Semigroup`

, the `sconcat`

function takes a non-empty list of `a`

values, and reduces them to a single `a`

value. `NonEmpty a`

is a list with at least one element.

### NotEmptyCollection #

You can also define a `NotEmptyCollection<T>`

class in C#:

public class NotEmptyCollection<T> : IReadOnlyCollection<T> { public NotEmptyCollection(T head, params T[] tail) { if (head == null) throw new ArgumentNullException(nameof(head)); this.Head = head; this.Tail = tail; } public T Head { get; } public IReadOnlyCollection<T> Tail { get; } public int Count { get => this.Tail.Count + 1; } public IEnumerator<T> GetEnumerator() { yield return this.Head; foreach (var item in this.Tail) yield return item; } IEnumerator IEnumerable.GetEnumerator() { return this.GetEnumerator(); } }

Because of the way the constructor is defined, you *must* supply at least one element in order to create an instance. You can provide any number of extra elements via the `tail`

array, but one is minimum.

The `Count`

property returns the number of elements in `Tail`

, plus one, because there's always a `Head`

value.

The `GetEnumerator`

method returns an iterator that always starts with the `Head`

value, and proceeds with all the `Tail`

values, if there are any.

### Finding the maximum of a non-empty collection #

As you've already learned, `Math.Max`

is a semigroup. Although the .NET Base Class Library has built-in methods for this, you can use a generally applicable algorithm to find the maximum value in a non-empty list of integers.

private static int Accumulate(NotEmptyCollection<int> numbers) { var acc = numbers.Head; foreach (var n in numbers.Tail) acc = Math.Max(acc, n); return acc; }

Notice how similar this algorithm is to monoid accumulation! The only difference is that instead of using `Identity`

to get started, you can use `Head`

, and then loop over all elements of `Tail`

.

You can use it like this:

var nec = new NotEmptyCollection<int>(42, 1337, 123); var max = Accumulate(nec);

Here, `max`

is obviously `1337`

.

As usual, however, this is much nicer, and more succinct in Haskell:

Prelude Data.Semigroup Data.List.NonEmpty> getMax $ sconcat $ fmap Max $ 42 :| [1337, 123] 1337

That's hardly the idiomatic way of getting a maximum element in Haskell, but it does show how you can 'click together' concepts in order to achieve a goal.

### Aggregate #

Perhaps the observant reader will, at this point, have recalled to him- or herself that the .NET Base Class Library already includes an `Aggregate`

extension method, with an overload that takes a seed. In general, the simpliest `Aggregate`

method doesn't gracefully handle empty collections, but using the overload that takes a seed is more robust. You can rewrite the above `Accumulate`

method using `Aggregate`

:

private static int Accumulate(NotEmptyCollection<int> numbers) { return numbers.Tail.Aggregate( numbers.Head, (x, y) => Math.Max(x, y)); }

Notice how you can pass `Head`

as the seed, and accumulate the `Tail`

using that starting point. The `Aggregate`

method is more like Haskell's `sconcat`

for semigroups than `mconcat`

for monoids.

### Summary #

A semigroup operation can be used to reduce values to a single value, just like a monoid can. The only difference is that a semigroup operation can't reduce an empty collection, whereas a monoid can.

**Next:** Quasigroups

## Bounding box semigroup

*A semigroup example for object-oriented programmers.*

This article is part of a larger series about monoids, semigroups, and other group-like algebraic structures. In this article, you'll see a non-trivial example of a semigroup that's *not* a monoid. In short, a semigroup is an associative binary operation.

### Shapes #

Imagine that you're developing a library of two-dimensional shapes, and that, for various reasons, each shape should have a *bounding box*. For example, here's a blue circle with a green bounding box:

The code for a circle shape could look like this:

public class Circle : ICanHasBox { public int X { get; } public int Y { get; } public int Radius { get; } public Circle(int x, int y, int radius) { this.X = x; this.Y = y; this.Radius = Math.Abs(radius); } public BoundingBox BoundingBox { get { return new BoundingBox( this.X - this.Radius, this.Y - this.Radius, this.Radius * 2, this.Radius * 2); } } }

In addition to the `Circle`

class, you could have other shapes, such as rectangles, triangles, or even irregular shapes, each of which have a bounding box.

### Bounding box unions #

If you have two shapes, you also have two (green) bounding boxes, but perhaps you'd like to find the (orange) bounding box of the union of both shapes.

That's fairly easy to do:

public BoundingBox Unite(BoundingBox other) { var newX = Math.Min(this.X, other.X); var newY = Math.Min(this.Y, other.Y); var newRightX = Math.Max(this.rightX, other.rightX); var newTopY = Math.Max(this.topY, other.topY); return new BoundingBox( newX, newY, newRightX - newX, newTopY - newY); }

The `Unite`

method is an instance method on the `BoundingBox`

class, so it's a binary operation. It's also associative, because for all `x`

, `y`

, and `z`

, `isAssociative`

is `true`

:

`var isAssociative = x.Unite(y).Unite(z) == x.Unite(y.Unite(z));`

Since the operation is associative, it's at least a semigroup.

### Lack of identity #

Is `Unite`

also a monoid? In order to be a monoid, a binary operation must not only be associative, but also have an identity element. In a previous article, you saw how the union of two convex hulls formed a monoid. A bounding box seems to be conceptually similar to a convex hull, so you'd be excused to think that our previous experience applies here as well.

It doesn't.

There's no *identity bounding box*. The difference between a convex hull and a bounding box is that it's possible to define an empty hull as an empty set of coordinates. A bounding box, on the other hand, always has a coordinate and a size.

public struct BoundingBox { private readonly int rightX; private readonly int topY; public int X { get; } public int Y { get; } public int Width { get; } public int Height { get; } // More members, including Unite... }

An identity element, if one exists, is one where if you `Unite`

it with another `BoundingBox`

object, the return value will be the other object.

Consider, then, a (green) `BoundingBox x`

. Any other `BoundingBox`

inside of `x`

, including `x`

itself, is a candidate for an identity element:

In a real coordinate system, there's infinitely many candidates contained in `x`

. As long as a candidate is wholly contained within `x`

, then the union of the candidate and `x`

will return `x`

.

In the code example, however, coordinates are 32-bit integers, so for any bounding box `x`

, there's only a finite number of candidates. Even for the smallest possible bounding box, though, the box itself is an identity candidate.

In order to be an identity element, however, the *same* element must behave as the identity element for *all* bounding boxes. It is, therefore, trivial to find a counter-example:

Just pick any other `BoundingBox y`

outside of `x`

. Every identity candidate must be within `x`

, and therefore the union of the candidate and `y`

cannot be `y`

.

In code, you can demonstrate the lack of identity with an FsCheck-based test like this:

[Property(QuietOnSuccess = true)] public Property UniteHasNoIdentity(PositiveInt w, PositiveInt h) { var genCandidate = from xi in Gen.Choose(1, w.Get) from yi in Gen.Choose(1, h.Get) from wi in Gen.Choose(1, w.Get - xi + 1) from hi in Gen.Choose(1, h.Get - yi + 1) select new BoundingBox(xi, yi, wi, hi); return Prop.ForAll( genCandidate.ToArbitrary(), identityCandidate => { var x = new BoundingBox(1, 1, w.Get, h.Get); // Show that the candidate behaves like identity for x Assert.Equal(x, x.Unite(identityCandidate)); Assert.Equal(x, identityCandidate.Unite(x)); // Counter-example var y = new BoundingBox(0, 0, 1, 1); Assert.NotEqual(y, y.Unite(identityCandidate)); }); }

This example uses the FsCheck.Xunit glue library for xUnit.net. Notice that although FsCheck is written in F#, you can also use it from C#. This test passes.

It follows the above 'proof' by first generating an identity candidate for `x`

. This is any `BoundingBox`

contained within `x`

, including `x`

itself. In order to keep the code as simple as possible, `x`

is always placed at the coordinate *(1, 1)*.

The test proceeds to utilise two Guard Assertions to show that `identityCandidate`

does, indeed, behave like an identity for `x`

.

Finally, the test finds a trivial counter-example in `y`

, and verifies that `y.Unite(identityCandidate)`

is not equal to `y`

. Therefore, `identityCandidate`

is *not* the identity for `y`

.

`Unite`

is a semigroup, but not a monoid, because no identity element exists.

### Summary #

This article demonstrates (via an example) that non-trivial semigroups exist in normal object-oriented programming.

**Next:** Semigroups accumulate.

## Semigroups

*Introduction to semigroups for object-oriented programmers.*

This article is part of a larger series about monoids, semigroups, and other group-like algebraic structures. In this article, you'll learn what a semigroup is, and what distinguishes it from a monoid.

Semigroups form a superset of monoids. They are associative binary operations. While monoids additionally require that an identity element exists, no such requirement exist for semigroups. In other words, all monoids are semigroups, but not all semigroups are monoids.

This article gives you an overview of semigroups, as well as a few small examples. A supplemental article will show a more elaborate example.

### Minimum #

An operation that returns the smallest of two values form a semigroup. In the .NET Base Class Library, such an operation is already defined for many numbers, for example 32-bit integers. Since associativity is a property of a semigroup, it makes sense to demonstrate it with a property-based test, here using FsCheck:

[Property(QuietOnSuccess = true)] public void IntMinimumIsAssociative(int x, int y, int z) { Assert.Equal( Math.Min(Math.Min(x, y), z), Math.Min(x, Math.Min(y, z))); }

This example uses the FsCheck.Xunit glue library for xUnit.net. Notice that although FsCheck is written in F#, you can also use it from C#. This test (as well as all other tests in this article) passes.

For mathematical integers, no identity element exists, so the minimum operation doesn't form a monoid. In practice, however, .NET 32-bit integers *do* have an identity element:

[Property(QuietOnSuccess = true)] public void MimimumIntHasIdentity(int x) { Assert.Equal(x, Math.Min(int.MaxValue, x)); Assert.Equal(x, Math.Min(x, int.MaxValue)); }

Int32.MaxValue is the maximum possible 32-bit integer value, so it effectively behaves as the identity for the 32-bit integer minimum operation. All 32-bit numbers are smaller than, or equal to, `Int32.MaxValue`

. This effectively makes `Math.Min(int, int)`

a monoid, but conceptually, it's not.

This may be clearer if, instead of 32-bit integers, you consider BigInteger, which is an arbitrarily large (or small) integer. The *minimum* operation is still associative:

[Property(QuietOnSuccess = true)] public void BigIntMinimumIsAssociative( BigInteger x, BigInteger y, BigInteger z) { Assert.Equal( BigInteger.Min(BigInteger.Min(x, y), z), BigInteger.Min(x, BigInteger.Min(y, z))); }

No identity element exists, however, because no matter which integer you have, you can always find one that's bigger: no maximum value exists. This makes `BigInteger.Min`

a semigroup, but not a monoid.

### Maximum #

Like *minimum*, the *maximum* operation forms a semigroup, here demonstrated by `BigInteger.Max`

:

[Property(QuietOnSuccess = true)] public void BigIntMaximumIsAssociative( BigInteger x, BigInteger y, BigInteger z) { Assert.Equal( BigInteger.Max(BigInteger.Max(x, y), z), BigInteger.Max(x, BigInteger.Max(y, z))); }

Again, like minimum, no identity element exists because the set of integers is infinite; you can always find a bigger or smaller number.

Minimum and maximum operations aren't limited to primitive numbers. If values can be compared, you can always find the smallest or largest of two values, here demonstrated with `DateTime`

values:

[Property(QuietOnSuccess = true)] public void DateTimeMaximumIsAssociative( DateTime x, DateTime y, DateTime z) { Func<DateTime, DateTime, DateTime> dtMax = (dt1, dt2) => dt1 > dt2 ? dt1 : dt2; Assert.Equal( dtMax(dtMax(x, y), z), dtMax(x, dtMax(y, z))); }

As was the case with 32-bit integers, however, the presence of DateTime.MinValue effectively makes `dtMax`

a monoid, but *conceptually*, no identity element exists, because dates are infinite.

### First #

Another binary operation simply returns the first of two values:

public static T First<T>(T x, T y) { return x; }

This may seem pointless, but `First`

*is* associative:

[Property(QuietOnSuccess = true)] public void FirstIsAssociative(Guid x, Guid y, Guid z) { Assert.Equal( First(First(x, y), z), First(x, First(y, z))); }

On the other hand, there's no identity element, because there's no *left identity*. The *left identity* is an element `e`

such that `First(e, x) == x`

for any `x`

. Clearly, for any generic type `T`

, no such element exists because `First(e, x)`

will only return `x`

when `x`

is equal to `e`

. (There are, however, degenerate types for which an identity exists for `First`

. Can you find an example?)

### Last #

Like `First`

, a binary operation that returns the last (second) of two values also forms a semigroup:

public static T Last<T>(T x, T y) { return y; }

Similar to `First`

, `Last`

is associative:

[Property(QuietOnSuccess = true)] public void LastIsAssociative(String x, String y, String z) { Assert.Equal( Last(Last(x, y), z), Last(x, Last(y, z))); }

As is also the case for `First`

, no identity exists for `Last`

, but here the problem is the lack of a *right identity*. The *right identity* is an element `e`

for which `Last(x, e) == x`

for all `x`

. Clearly, `Last(x, e)`

can only be equal to `x`

if `e`

is equal to `x`

.

### Aggregation #

Perhaps you think that operations like `First`

and `Last`

seem useless in practice, but when you have a semigroup, you can reduce any non-empty sequence to a single value. In C#, you can use the Aggregate LINQ method for this. For example

var a = new[] { 1, 0, 1337, -10, 42 }.Aggregate(Math.Min);

returns `-10`

, while

var a = new[] { 1, 0, 1337, -10, 42 }.Aggregate(Math.Max);

returns `1337`

. Notice that the input sequence is the same in both examples, but the semigroup differs. Likewise, you can use `Aggregate`

with `First`

:

var a = new[] { 1, 0, 1337, -10, 42 }.Aggregate(First);

Here, `a`

is `1`

, while

var a = new[] { 1, 0, 1337, -10, 42 }.Aggregate(Last);

returns `42`

.

LINQ has specialised methods like Min, Last, and so on, but from the perspective of behaviour, `Aggregate`

would have been enough. While there may be performance reasons why some of those specialised methods exist, you can think of all of them as being based on the same abstraction: that of a semigroup.

`Aggregate`

, and many of the specialised methods, throw an exception if the input sequence is empty. This is because there's no identity element in a semigroup. The method doesn't know how to create a value of the type `T`

from an empty list.

If, on the other hand, you have a monoid, you can return the identity element in case of an empty sequence. Haskell explicitly makes this distinction with `sconcat`

and `mconcat`

, but I'm not going to go into that now.

### Summary #

Semigroups are associative binary operations. In the previous article series about monoids you saw plenty of examples, and since all monoids are semigroups, you've already seen more than one semigroup example. In this article, however, you saw four examples of semigroups that are *not* monoids.

All four examples in this article are simple, and may not seem like 'real-world' examples. In the next article, then, you'll get a more realistic example of a semigroup that's not a monoid.

**Next:** Bounding box semigroup.

## Monoids accumulate

*You can accumulate an arbitrary number of monoidal values to a single value. An article for object-oriented programmers.*

This article is part of a series about monoids. In short, a *monoid* is an associative binary operation with a neutral element (also known as *identity*).

Recall that a binary operation is an operation involving two arguments of the same type, and returning a value of that type.

public static Foo Op(Foo x, Foo y)

Notice that such an operation reduces two `Foo`

values to a single `Foo`

value.

### Accumulation #

Since you have an operation that can reduce two values to a single value, you can use that single value as the input for yet another binary operation. This enables you to accumulate, or aggregate, an arbitrary number of values.

Consider the instance variation of the above `Op`

method:

public Foo Op(Foo foo)

This is another representation of the operation, but instead of being a static method, it's an instance method on the `Foo`

class.

When `Op`

is a monoid, you can easily write a function that accumulates an arbitrary number of `Foo`

values:

public static Foo Accumulate(IReadOnlyCollection<Foo> foos) { var acc = Identity; foreach (var f in foos) acc = acc.Op(f); return acc; }

You start with the `Identity`

value, which also becomes the return value if `foos`

is empty. Then you simply loop over each value in `foos`

and use `Op`

with the value accumulated so far (`acc`

) and the current element in the sequence.

Once you're done looping, you return the accumulator.

This implementation isn't always guaranteed to be the most efficient, but you can *always* write accumulation like that. Sometimes, a more efficient algorithm exists, but that doesn't change the overall result that you can always reduce an arbitrary number of values whenever a monoid exists for those values.

### Generalisation #

You can do this with any monoid. In Haskell, this function is called `mconcat`

, and it has this type:

mconcat :: Monoid a => [a] -> a

The way you can read this is that for any monoid `a`

, `mconcat`

is a function that takes a linked list of `a`

values as input, and returns a single `a`

value as output.

This function seems both more general, and more constrained, than the above C# example. It's more general than the C# example because it works on any monoid, instead of just `Foo.Op`

. On the other hand, it seems more limited because it works only on Haskell lists. The C# example, in contrast, can accumulate any `IReadOnlyCollection<Foo>`

. Could you somehow combine those two generalisations?

Nothing stops you from doing that, but it's already in Haskell's `Data.Foldable`

module:

fold :: (Monoid m, Foldable t) => t m -> m

The way to read this is that there's a function called `fold`

, and it accumulates any monoid `m`

contained in any 'foldable' data container `t`

. That a data container is 'foldable' means that there's a way to somehow fold, or aggregate, the element(s) in the container into a value.

Linked lists, arrays, and other types of sequences are foldable, as are Maybe and trees.

In fact, there's little difference between Haskell's `Foldable`

type class and .NET's `IEnumerable<T>`

, but as the names suggest, their foci are different. In Haskell, the focus is being able to fold, accumulate, or aggregate a data structure, whereas on .NET the focus is on being able to enumerate the values inside the data structure. Ultimately, though, both abstractions afford the same capabilities.

In .NET, the focal abstraction is the Iterator pattern, which enables you to enumerate the values in the data container. On top of that abstraction, you can derive other behaviour, such as the ability to Aggregate data.

In Haskell, the focus is on the ability to fold, but from that central abstraction follows the ability to convert the data container into a linked list, which you can then enumerate.

### Summary #

You can accumulate an arbitrary number of monoidal values as long as they're held in a container that enables you to 'fold' it. This includes all sorts of lists and arrays.

This article concludes the article series about monoids. In the next series of articles, you'll learn about a related category of operations.

**Next: ** Semigroups.

## Comments

@ploeh as always I loved your blog post but I don't 100% agree on your comparison of the iterator pattern with Foldable - the iterator pattern allows usually sideeffects and you have Traversable for that - you might also like this: http://www.cs.ox.ac.uk/jeremy.gibbons/publications/iterator.pdf …

(Comment submitted via Twitter)

## Endomorphism monoid

*A method that returns the same type of output as its input forms a monoid. An article for object-oriented programmers.*

This article is part of a series about monoids. In short, a *monoid* is an associative binary operation with a neutral element (also known as *identity*). Methods that return the same type of value as their input form monoids over composition. The formal term for such an operation is an endomorphism.

### Scheduling example #

Imagine that you have to develop some functionality for scheduling events in the future. As a concrete example, I recently wrote about adjusting dates while taking bank holidays into account. For instance, if you want to find the latest bank day *before* a given date, you could call the `AdjustToLatestPrecedingDutchBankDay`

method. If you give it a normal bank day (say, a Thursday), it'll simply return the input date, but if you give it a Sunday, it'll return the preceding Friday. That is, unless that particular Friday is a bank holiday, in which case it'll return the Thursday before - as long as that's not also a bank holiday, and so on.

In that previous article, the `AdjustToLatestPrecedingDutchBankDay`

method is an extension method, but you can also model it as an instance method, like this:

public DateTimeOffset Adjust(DateTimeOffset value) { var candidate = value; while (!(IsDutchBankDay(candidate.DateTime))) candidate = candidate.AddDays(-1); return candidate; }

This method would be part of a class that implements an interface:

public interface IDateTimeOffsetAdjustment { DateTimeOffset Adjust(DateTimeOffset value); }

You can make other implementations of this interface. Here's one that adjusts a date and time to business hours:

public class BusinessHoursAdjustment : IDateTimeOffsetAdjustment { private readonly static TimeSpan startOfBussiness = TimeSpan.FromHours(9); private readonly static TimeSpan endOfBusiness = TimeSpan.FromHours(17); public DateTimeOffset Adjust(DateTimeOffset value) { // Warning: May not handle DST changes appropriately! // It's only example code... if (value.TimeOfDay < startOfBussiness) return value - value.TimeOfDay + startOfBussiness; if (endOfBusiness < value.TimeOfDay) return (value - value.TimeOfDay + startOfBussiness).AddDays(1); return value; } }

To keep the example simple, business hours are hard-coded to 9-17.

You could also adapt conversion to UTC:

public class UtcAdjustment : IDateTimeOffsetAdjustment { public DateTimeOffset Adjust(DateTimeOffset value) { return value.ToUniversalTime(); } }

Or add a month:

public class NextMonthAdjustment : IDateTimeOffsetAdjustment { public DateTimeOffset Adjust(DateTimeOffset value) { return value.AddMonths(1); } }

Notice that the `Adjust`

method returns a value of the same type as its input. So far when discussing monoids, we've been looking at binary operations, but `Adjust`

is a *unary* operation.

An operation that returns the same type as its input is called an *endomorphism*. Those form monoids.

### Composing adjustments #

It's easy to connect two adjustments. Perhaps, for example, you'd like to first use `BusinessHoursAdjustment`

, followed by the bank day adjustment. This will adjust an original input date and time to a date and time that falls on a bank day, within business hours.

You can do this in a general-purpose, reusable way:

public static IDateTimeOffsetAdjustment Append( this IDateTimeOffsetAdjustment x, IDateTimeOffsetAdjustment y) { return new AppendedAdjustment(x, y); } private class AppendedAdjustment : IDateTimeOffsetAdjustment { private readonly IDateTimeOffsetAdjustment x; private readonly IDateTimeOffsetAdjustment y; public AppendedAdjustment( IDateTimeOffsetAdjustment x, IDateTimeOffsetAdjustment y) { this.x = x; this.y = y; } public DateTimeOffset Adjust(DateTimeOffset value) { return y.Adjust(x.Adjust(value)); } }

The `Append`

method takes two `IDateTimeOffsetAdjustment`

values and combines them by wrapping them in a private implementation of `IDateTimeOffsetAdjustment`

. When `AppendedAdjustment.Adjust`

is called, it first calls `Adjust`

on `x`

, and then calls `Adjust`

on `y`

with the return value from the first call.

In order to keep the example simple, I omitted null guards, but apart from that, `Append`

should work with any two implementations of `IDateTimeOffsetAdjustment`

. In other words, it obeys the Liskov Substitution Principle.

### Associativity #

The `Append`

method is a binary operation. It takes two `IDateTimeOffsetAdjustment`

values and returns an `IDateTimeOffsetAdjustment`

. It's also associative, as a test like this demonstrates:

private void AppendIsAssociative( IDateTimeOffsetAdjustment x, IDateTimeOffsetAdjustment y, IDateTimeOffsetAdjustment z) { Assert.Equal( x.Append(y).Append(z), x.Append(y.Append(z))); }

As usual in this article series, such a test doesn't *prove* that `Append`

is associative for all values of `IDateTimeOffsetAdjustment`

, but if you run it as a property-based test, it demonstrates that it's quite likely.

### Identity #

In true monoidal fashion, `IDateTimeOffsetAdjustment`

also has an identity element:

public class IdentityDateTimeOffsetAdjustment : IDateTimeOffsetAdjustment { public DateTimeOffset Adjust(DateTimeOffset value) { return value; } }

This implementation simply returns the input value without modifying it. That's a neutral operation, as a test like this demonstrates:

private void AppendHasIdentity(IDateTimeOffsetAdjustment x) { Assert.Equal( x.Append(new IdentityDateTimeOffsetAdjustment()), x); Assert.Equal( new IdentityDateTimeOffsetAdjustment().Append(x), x); }

These two assertions verify that left and right identity holds.

Since `Append`

is associative and has identity, it's a monoid.

This holds generally for any method (or function) that returns the same type as it takes as input, i.e. `T SomeOperation(T x)`

. This matches the built-in library in Haskell, where `Endo`

is a `Monoid`

.

### Conclusion #

A method that returns a value of the same type as its (singular) input argument is called an endomorphism. You can compose two such unary operations together in order to get a composed operation. You simply take the output of the first method and use it as the input argument for the second method. That composition is a monoid. Endomorphisms form monoids.

**Next:** Maybe monoids.

## Comments

Hi, Mark!

I'm really enjoing your series on category theory. I think you're doing a great job in making such abstract concepts accessible to us programmers. However, I found this particular episode puzzling. You claim that any composition of endomorphisms is a monoid, and you also claim to have tested that your `Append`

method is associative. However, it is not hard to come up with a counter-example:

[Fact] public void CounterExample() { IDateTimeOffsetAdjustment weekend = new WeekendAdjustment(); IDateTimeOffsetAdjustment nextMonth = new NextMonthAdjustment(); var a = new AppendedAdjustment(weekend, nextMonth); var b = new AppendedAdjustment(nextMonth, weekend); var startDate = new DateTimeOffset(2019, 1, 5, 0, 0, 0, TimeSpan.Zero); Assert.Equal(a.Adjust(startDate), b.Adjust(startDate)); }

Here, `WeekendAdjustment`

is just a simplified `DutchBankDayAdjustment`

.

It would also be interesting to see how you can do property based testing on this, i.e. how to automatically generate meaningful instances of `IDateTimeOffsetAdjustment`

. When I try to run your test, I get: "IDateTimeOffsetAdjustment is not handled automatically by FsCheck. Consider using another type or writing and registering a generator for it."

Tor, thank you for writing, and for your kind words. I suppose that the `CounterExample`

test fails when one executes it; you don't explicitly write that, but that's what I expect would happen.

The `Append`

operation is, indeed, not commutative. This is, however, not a requirement for monoids, or even for groups. Some monoids, such as addition and multiplication, are also commutative, while others, like list concatenation or, here, the endomorphism monoid, aren't.

When it comes to the FsCheck properties, I admit that I cheated slightly with the code listing in the article. I did this because the properties are a bit more complicated than what I show, and I was concerned that the extra infrastructure surrounding those tests would detract from the overall message.

The associativity property in its entirety looks like this:

[Property(QuietOnSuccess = true)] public Property AppendIsAssociative() { return Prop.ForAll( GenerateAdjustment().Three().ToArbitrary(), t => AppendIsAssociative(t.Item1, t.Item2, t.Item3)); } private void AppendIsAssociative( IDateTimeOffsetAdjustment x, IDateTimeOffsetAdjustment y, IDateTimeOffsetAdjustment z) { Assert.Equal( x.Append(y).Append(z), x.Append(y.Append(z)), new DateTimeOffsetAdjustmentComparer()); }

Where `GenerateAdjustment`

is defined like this:

private static Gen<IDateTimeOffsetAdjustment> GenerateAdjustment() { return Gen.Elements<IDateTimeOffsetAdjustment>( new IdentityDateTimeOffsetAdjustment(), new BusinessHoursAdjustment(), new DutchBankDayAdjustment(), new NextMonthAdjustment(), new UtcAdjustment()); }

Not only did I omit all that extra scaffolding, I also pretended that `IDateTimeOffsetAdjustment`

objects could be compared using their default implementations of `Equals`

, which they meaningfully can't. Notice that the full property listed here also asserts using a custom equality comparer:

private class DateTimeOffsetAdjustmentComparer : IEqualityComparer<IDateTimeOffsetAdjustment> { public bool Equals(IDateTimeOffsetAdjustment x, IDateTimeOffsetAdjustment y) { var rnd = new System.Random(); var dt = new DateTimeOffset(rnd.Next(), TimeSpan.Zero); return x.Adjust(dt) == y.Adjust(dt); } public int GetHashCode(IDateTimeOffsetAdjustment obj) { return 0; } }

The custom comparer's `Equals`

method creates a random number of ticks and uses it to create a `DateTimeOffset`

value. It then calls `Adjust`

on both objects, which are considered equal if they produce the same result.

The test for identity is defined in a similar fashion; it also uses `GenerateAdjustment`

and `DateTimeOffsetAdjustmentComparer`

.

Thank you for such a prompt and thorough reply. You are of course correct, I have been confusing associativity with commutativity. I didn't run into the same mistake during the list concatenation article, though, maybe because list concatenation more obviously is associative and not commutative. In the current article, however, I intuitively felt that the operations needed to be commutative, but your reply clears that up.

It is also helpful to see the extra scaffolding around your property based test. The article itself seemed to imply that instances of `IDateTimeOffsetAdjustment`

could be automatically generated. Your approach to set up such a test will come in handy now that I'm convinced that I should let more of my tests be property based, even in C#!

Tor, I made the same mistake several times when I started researching all of this. I think that there's something intuitive and fundamental about commutativity, so at a superficial glance, it seems odd that we can do without it, while at the same time requiring a less intuitive property like associativity.

When it comes to computation, though, I think that it's exactly the nature of associativity that enables us to decompose big problems into several small problems. The order of how you deal with those small problems doesn't matter, as long as you apply the intermediate results in the correct order.

Dear Mark!

These additional explanations and contextual information in the comment sections deserve a place in the original text. To keep the text uncluttered this site uses clickable popups: . If you click on some of the numbers, you'll see what I mean. You might consider adding this feature to the text.

Best regards, Norbert.

## Function monoids

*Methods are monoids if they return monoids. An article for object-oriented programmers.*

This article is part of a series about monoids. In short, a *monoid* is an associative binary operation with a neutral element (also known as *identity*).

### Functions #

In statically typed C-languages, like C# or Java, methods are typically declared like this:

public Foo Bar(Baz baz, Qux qux)

As you'll see in another article, however, you can refactor any method to a method that takes a single argument as input, and returns a single (possibly complex) value as output. In abstract form, we can write it like this:

public Out1 Op1(In1 arg)

Another way to abstract a method is by using generics:

public T Op1<T1, T>(T1 x)

Another article demonstrates how this is similar to a generic function. In F#, for instance, the type of the function would be written as `'a -> 'b`

, whereas in Haskell it'd be written `a -> b`

. The way to read this is that the function takes a value of the generic type `T1`

/`'a`

/`a`

as input, and returns a value of the generic type `T`

/`'b`

/`b`

as output. For the rest of this article, I'll favour the Haskell syntax, since it has minimal noise.

To be clear, however, although I favour the Haskell syntax because of its succinctness, I don't mean to imply that the functions that I discuss are exclusively pure - think of an F# function `'a -> 'b`

which may or may not be pure.

### Binary combination of functions #

A function `a -> b`

is a monoid if `b`

is a monoid. This means that you can combine two functions with the same type. In an object-oriented context, it means that you can combine two methods with the same signature into one method as long as the return type forms a monoid.

Consider the following (facetious) C# example. You're trying to establish how secure a GUID is. Primes are important in cryptography, so the more primes a GUID contains, the better... right?

private const string primes = "2357bd"; public static int CountPrimes(Guid g) { return g.ToString("N").Count(primes.Contains); }

The `CountPrimes`

method counts the number of prime digits in a given GUID. So far so good, but you also think that hexadecimal notation is more exotic than decimal notation, so surely, the digits A-F are somehow more secure, being more unfamiliar. Thus, you have a method to count those as well:

private const string letters = "abcdef"; public static int CountLetters(Guid g) { return g.ToString("N").Count(letters.Contains); }

Good, but both of these numbers are, clearly, excellent indicators of how secure a GUID is. Which one should you choose? Both of them, of course!

Can you combine `CountPrimes`

and `CountLetters`

? Yes, you can, because, while GUIDs don't form a monoid, the return type `int`

forms a monoid over addition. This enables you to write a `Combine`

method:

public static Func<Guid, int> Combine( Func<Guid, int> f, Func<Guid, int> g) { return x => f(x) + g(x); }

Notice that `Combine`

takes two `Func<Guid, int>`

values and return a new `Func<Guid, int>`

value. It's a binary operation. Here's an example of how to use it:

var calculateSecurity = Combine(CountPrimes, CountLetters); var actual = calculateSecurity(new Guid("10763FF5-E3C8-46D1-A70F-6C1D9EDA8120")); Assert.Equal(21, actual);

Now you have an excellent measure of the security strength of GUIDs! 21 isn't *that* good, though, is it?

Antics aside, `Combine`

is a binary function. Is it a monoid?

### Monoid laws #

In order to be a monoid, `Combine`

must be associative, and it is, as the following FsCheck property demonstrates:

[Property(QuietOnSuccess = true)] public void CombineIsAssociative( Func<Guid, int> f, Func<Guid, int> g, Func<Guid, int> h, Guid guid) { Assert.Equal( Combine(Combine(f, g), h)(guid), Combine(f, Combine(g, h))(guid)); }

In this property-based test, FsCheck generates three functions with the same signature. Since functions don't have structural equality, the easiest way to compare them is to call them and see whether they return the same result. This explains why the assertion invokes both associative combinations with `guid`

. The test passes.

In order to be a monoid, `Combine`

must also have an identity element. It does:

public static Func<Guid, int> FuncIdentity = _ => 0;

This is simply a function that ignores its input and always returns `0`

, which is the identity value for addition. The following test demonstrates that it behaves as expected:

[Property(QuietOnSuccess = true)] public void CombineHasIdentity(Func<Guid, int> f, Guid guid) { Assert.Equal(f(guid), Combine(FuncIdentity, f)(guid)); Assert.Equal(f(guid), Combine(f, FuncIdentity)(guid)); }

As was the case with `CombineIsAssociative`

, in order to assert that the combined functions behave correctly, you have to call them. This, again, explains why the assertion invokes the combined functions with `guid`

. This test passes as well.

`Combine`

is a monoid.

### Generalisation #

While the above C# code is only an example, the general rule is that any function that returns a monoid is itself a monoid. In Haskell, this rule is articulated in the standard library:

instance Monoid b => Monoid (a -> b)

This means that for any monoid `b`

, a function `a -> b`

is also (automatically) a monoid.

### Summary #

A function or method with a return type that forms a monoid is itself a monoid.

**Next: ** Endomorphism monoid.

## Tuple monoids

*Tuples of monoids form monoids. Data objects of monoids also form monoids. An article for object-oriented programmers.*

This article is part of a series about monoids. In short, a *monoid* is an associative binary operation with a neutral element (also known as *identity*). This article starts off with some easy-to-understand, but abstract results. Once these are established, however, you'll see how to use them in a relatable example, so keep reading!

### Tuples #

A tuple is a group of elements. In statically typed programming languages, each element has a type, and the types don't have to be the same. As an example, in C#, you can create a tuple like this:

Tuple<int, string> pair = Tuple.Create(42, "Foo");

This creates a tuple where the first element must be an `int`

and the second element a `string`

. In the example, I've explicitly declared the type instead of using the `var`

keyword, but this is only to make the type clearer (since you don't have an IDE in which to read the code).

The `pair`

tuple is a two-tuple, which means that it must have exactly two elements, of the types given, but you can also create larger tuples:

Tuple<string, bool, int> triple = Tuple.Create("Bar", false, 42);

This is a three-tuple, but conceptually, tuples can have any size.

### Pairs of monoids #

A *pair* (a two-tuple) forms a monoid if both elements form a monoid. Haskell formalises this by stating:

`instance (Monoid a, Monoid b) => Monoid (a, b)`

The way to read this is that for any monoid `a`

and any monoid `b`

, the pair `(a, b)`

is also a monoid.

Perhaps this is easiest to understand with a C# example. Consider a tuple of the type `Tuple<int, string>`

. Integers form monoids under both addition and multiplication, and strings are monoids under concatenation. Thus, you can make `Tuple<int, string>`

form a monoid as well. For instance, use the multiplication monoid to define this binary operation:

public static Tuple<int, string> CombinePair( Tuple<int, string> x, Tuple<int, string> y) { return Tuple.Create(x.Item1 * y.Item1, x.Item2 + y.Item2); }

For this particular example, I've chosen multiplication as the binary operation for `int`

, and the string concatenation operator `+`

for `string`

. The point is that since both elements are monoids, you can use their respective binary operations to return a new tuple with the combined values.

This operation is associative, as the following FsCheck property demonstrates:

[Property(QuietOnSuccess = true)] public void CombinePairIsAssociative( Tuple<int, string> x, Tuple<int, string> y, Tuple<int, string> z) { Assert.Equal( CombinePair(CombinePair(x, y), z), CombinePair(x, CombinePair(y, z))); }

This property passes for all the `x`

, `y`

, and `z`

values that FsCheck generates.

The `CombinePair`

operation has identity as well:

public static Tuple<int, string> PairIdentity = Tuple.Create(1, "");

Again, you can use the identity value for each of the elements in the tuple: `1`

for the multiplication monoid, and `""`

for string concatenation.

This value behaves as the identity for `CombinePair`

, at least for all non-null string values:

[Property(QuietOnSuccess = true)] public void CombinePairHasIdentity(Tuple<int, NonNull<string>> seed) { var x = Tuple.Create(seed.Item1, seed.Item2.Get); Assert.Equal(CombinePair(PairIdentity, x), CombinePair(x, PairIdentity)); Assert.Equal(x, CombinePair(x, PairIdentity)); }

Again, this test passes for all `seed`

values generated by FsCheck.

The C# code here is only an example, but I hope it's clear how the result generalises.

### Triples of monoids #

In the above section, you saw how pairs of monoids form a monoid. Not surprisingly, triples of monoids also form monoids. Here's another C# example:

public static Tuple<string, bool, int> CombineTriple( Tuple<string, bool, int> x, Tuple<string, bool, int> y) { return Tuple.Create( x.Item1 + y.Item1, x.Item2 || y.Item2, x.Item3 * y.Item3); }

The `CombineTriple`

method is another binary operation. This time it combines two triples to a single triple. Since both `string`

, `bool`

, and `int`

form monoids, it's possible to combine each element in the two tuples to create a new tuple. There's more than one monoid for integers, and the same goes for Boolean values, but here I've chosen multiplication and Boolean *or*, so the identity is this:

public static Tuple<string, bool, int> TripleIdentity = Tuple.Create("", false, 1);

This triple simply contains the identities for string concatenation, Boolean *or*, and multiplication. The operation is associative, but I'm not going to show this with a property-based test. Both tests for associativity and identity are similar to the above tests; you could consider writing them as an exercise, if you'd like.

This triple example only demonstrates a particular triple, but you can find the generalisation in Haskell:

instance (Monoid a, Monoid b, Monoid c) => Monoid (a, b, c)

This simply states that for monoids `a`

, `b`

, and `c`

, the tuple `(a, b, c)`

is also a monoid.

### Generalisation #

At this point, it can hardly come as a surprise that quadruples and pentuples of monoids are also monoids. From Haskell:

instance (Monoid a, Monoid b, Monoid c, Monoid d) => Monoid (a, b, c, d) instance (Monoid a, Monoid b, Monoid c, Monoid d, Monoid e) => Monoid (a, b, c, d, e)

The Haskell standard library stops at pentuples (five-tuples), because it has to stop somewhere, but I'm sure you can see how this is a general rule.

### Data objects as monoids #

If you're an object-oriented programmer, you probably don't use tuples much in your day-to-day work. I'd even suggest that you shouldn't, because tuples carry too little information to make good domain objects. For example, if you have `Tuple<int, string, string>`

, what do the elements mean? A better design would be to introduce a small Value Object called `Customer`

, with `Id`

, `FirstName`

, and `LastName`

properties.

(In functional programming, you frequently use tuples, because they're useful for 'gluing' generic functions together. A Haskell programmer may instead say that they are useful for composing parametrically polymorphic functions, but the meaning would be the same.)

As on object-oriented developer, then why should you care that tuples of monoids are monoids?

The reason this is interesting in object-oriented programming is that there's a strong relationship between tuples and data objects (Value Objects or Data Transfer Objects). Consider the `Customer`

examples that I sketched out a few paragraphs above. As you'll learn in a future article, you can refactor a tuple to a class, or a class to a tuple.

### Example: Roster #

In Denmark, where I live, learning to swim is a mandatory part of the school curriculum. Teachers take the children to the nearby swimming stadium and teach them to swim. Since this is an activity outside of school, teachers would be prudent to keep a roster of the children. Modelled as a class, it might look like this:

public class Roster { public int Girls { get; } public int Boys { get; } public IReadOnlyCollection<string> Exemptions { get; } public Roster(int girls, int boys, params string[] exemptions) { Girls = girls; Boys = boys; Exemptions = exemptions; } // ... }

Some children may be temporarily exempt from a swimming lesson, perhaps because of a passing medical condition. This changes from lesson to lesson, so the roster keeps track of them separately. Additionally, the boys will need to go in the men's changing rooms, and the girls in the women's changing rooms. This is the reason the roster keeps track of the number of boys and girls separately.

This, however, presents a logistical problem, because there's only one teacher for a class. The children are small, so need the company of an adult.

The way my children's school solved that problem was to combine two groups of children (in Danish, *en klasse*, a class), each with their own teacher - one female, and one male.

To model that, the `Roster`

class should have a `Combine`

method:

public Roster Combine(Roster other) { return new Roster( this.Girls + other.Girls, this.Boys + other.Boys, this.Exemptions.Concat(other.Exemptions).ToArray()); }

Clearly, this is easy to implement. Just add the number of girls together, add the number of boys together, and concatenate the two lists of exemptions.

Here's an example of using the method:

[Fact] public void UsageExample() { var x = new Roster(11, 10, "Susan", "George"); var y = new Roster(12, 9, "Edward"); var roster = x.Combine(y); var expected = new Roster(23, 19, "Susan", "George", "Edward"); Assert.Equal(expected, roster); }

The `Combine`

method is an instance method on the `Roster`

class, taking a second `Roster`

as input, and returning a new `Roster`

value. It's a binary operation. Does it also have identity?

Yes, it does:

public static readonly Roster Identity = new Roster(0, 0);

Notice that the `exemptions`

constructor argument is a `params`

array, so omitting it means passing an empty array as the third argument.

The following properties demonstrate that the `Combine`

operation is both associative and has identity:

[Property(QuietOnSuccess = true)] public void CombineIsAssociative(Roster x, Roster y, Roster z) { Assert.Equal( x.Combine(y).Combine(z), x.Combine(y.Combine(z))); } [Property(QuietOnSuccess = true)] public void CombineHasIdentity(Roster x) { Assert.Equal(x, Roster.Identity.Combine(x)); Assert.Equal(x, x.Combine(Roster.Identity)); }

In other words, `Combine`

is a monoid.

This shouldn't surprise us in the least, since we've already established that tuples of monoids are monoids, and that a data class is more or less 'just' a tuple with named elements. Specifically, the `Roster`

class is a 'tuple' of two addition monoids and the sequence concatenation monoid, so it follows that the `Combine`

method is a monoid as well.

### Roster isomorphism #

In future articles, you'll learn more about isomorphisms between various representations of objects, but in this context, I think it's relevant to show how the `Roster`

example is isomorphic to a tuple. It's trivial, really:

public Tuple<int, int, string[]> ToTriple() { return Tuple.Create(this.Girls, this.Boys, this.Exemptions.ToArray()); } public static Roster FromTriple(Tuple<int, int, string[]> triple) { return new Roster(triple.Item1, triple.Item2, triple.Item3); }

This pair of methods turn a `Roster`

into a triple, and a corresponding triple back into a `Roster`

value. As the following two FsCheck properties demonstrate, these methods form an isomorphism:

[Property(QuietOnSuccess = true)] public void ToTripleRoundTrips(Roster x) { var triple = x.ToTriple(); Assert.Equal(x, Roster.FromTriple(triple)); } [Property(QuietOnSuccess = true)] public void FromTripleRoundTrips(Tuple<int, int, string[]> triple) { var roster = Roster.FromTriple(triple); Assert.Equal(triple, roster.ToTriple()); }

This isn't the only possible isomorphism between triples and `Roster`

objects. You could create another one where the `string[]`

element goes first, instead of last; where boys go before girls; and so on.

### Summary #

Tuples of monoids are also monoids. This holds for tuples of any size, but all of the elements has to be monoids. By isomorphism, this result also applies to data objects.

**Next:** Function monoids.

## Comments

Hi Mark, I have trouble understanding your usage of the term 'monoid' in this post. You apply it to the types string, bool, and int when you say that a tuple of those "monoids" is a monoid as well. But up to this point you made it very clear, that a type is NOT a monoid. A function can be a monoid. So, would it be more correct to say that a tuple of certain functions, which are monoids, is a monoid as well?

Punkislamist, thank you for writing. You're entirely correct that a monoid is an associative binary operation with identity. It's a function, not a type. If this article is unclear, the fault is all mine.

Not surprisingly, this topic is difficult to write about. The text has to be exact in order to avoid confusion, but since I'm only human, I sometimes make mistakes in how I phrase my explanations. While I've tried to favour the phrase that *a type forms a monoid*, I can see that I've slipped up once or twice in this article.

Some types form more than a single monoid. Boolean values, for instance, form exactly four monoids. Other types, like integers, form an infinite set of monoids, but the most commonly used integer monoids are addition and multiplication. Other types, particularly *unit*, only form a single monoid.

Why do I talk about types, then? There's at least two reasons. The first is the practical reason that most statically typed languages naturally come with a notion of types embedded. One could argue, I think, that types are a more fundamental concept than functions, since even functions have types (for instance, in Haskell, we'd characterise a binary operation with the type `a -> a -> a`

).

A more abstract reason is that category theory mostly operates with the concepts of *objects* and *morphisms*. Such objects aren't objects in the sense of object-oriented programming, but rather correspond to types in programming languages. (Actually, a category theory *object* is a more fluffy concept than that, but that's the closest analogy that I'm aware of.)

In category theory, a particular monoid is an object in the category of monoids. For example, the integer addition monoid is an object in the category of monoids, as is the string concatenation monoid, etcetera.

When you consider a 'raw' programming language type like C#'s `int`

, you're correct that it's not a monoid. It's just a type. The same goes for Haskell's corresponding `Int32`

type. As primitive values, we could say that the type of 32-bit integers is an object in some category (for example, the category of number representations). Such an object is not a monoid.

There exists, however, a morphism (a 'map') from the 32-bit integer object to the addition monoid (which is an object in the category of monoids). In Haskell, this morphism is the data constructor `Sum`

:

Prelude Data.Monoid> :t Sum Sum :: a -> Sum a

What this states is that `Sum`

is a function (i.e. a morphism) that takes an object `a`

and turns it into an object `Sum a`

. We have to be careful here, because `Sum a`

is a Haskell *type*, whereas `Sum`

is the function that 'elevates' an object `a`

to `Sum a`

. The names are similar, but the roles are different. This is a common idiom in Haskell, and have some mnemonic advantages, but may be confusing until you get the hang of it.

We can think of `Sum a`

as equivalent to the category theory object *addition in the category of monoids*. That's also how it works in Haskell: `Sum a`

is a monoid:

Prelude Data.Monoid> Sum 40 <> Sum 2 Sum {getSum = 42}

In Haskell, `<>`

is the polymorphic binary operation; exactly what it does depends on the object (that is: the type) on which it operates. When applied to two values of `Sum a`

, the result of combining 40 and 2 is 42.

To be clear, `Sum`

isn't the only morphism from the category of number representations to the category of monoids. `Product`

is another:

Prelude Data.Monoid> :t Product Product :: a -> Product a Prelude Data.Monoid> Product 6 <> Product 7 Product {getProduct = 42}

Thus, there *is* a relationship between types and monoids, but it's most apparent in programming languages that are geared towards that way of thinking (like Haskell). In C#, it's difficult to translate some of these concepts into code, because C#'s type system isn't up to the task. Instead, when we consider a type like `int`

, I think it's pragmatic to state that the type forms one or more monoids. I've also encountered the phrase that *it gives rise to a monoid*.

While you can represent a monoid with a C# interface, I've so far tried to avoid doing so, as I'm not sure whether or not it's helpful.

Hi Mark, I did not expect to recieve such an exhaustive answer. That is incredible, thank you so much! It did clear up my confusion as well. Since most of these terms and concepts are new to me, even a slight inconsistency can be really confusing. But with your additional explanation I think I got a good understanding of the terms again.

Your explanations of these concepts in general are very well written and make it easy for people unfamiliar with this topic to understand the terms and their significance. Thanks again for writing!

Punkislamist, thank you for those kind words. I'm happy to hear that what I wrote made sense to you; it makes sense to me, but I forgot to point out that I'm hardly an expert in category theory. Writing out the above answer helped clarify some things for me as well; as is common wisdom: you only really understand a topic when you teach it.

## Convex hull monoid

*The union of convex hulls form a monoid. Yet another non-trivial monoid example, this time in F#.*

This article is part of a series about monoids. In short, a *monoid* is an associative binary operation with a neutral element (also known as *identity*).

If you're reading the series as an object-oriented programmer, I apologise for the digression, but this article exclusively contains F# code. The next article will return with more C# examples.

### Convex hull #

In a past article I've described my adventures with finding convex hulls in F#. The convex hulls I've been looking at form the external convex boundary of a set of two-dimensional points. While you can generalise the concept of convex hulls to *n* dimensions, we're going to stick to two-dimensional hulls here.

If you have two convex hulls, you can find the convex hull of both:

Here, the dark green outline is the convex hull of the two lighter-coloured hulls.

Finding the convex hull of two other hulls is a binary operation. Is it a monoid?

In order to examine that, I'm going to make some changes to my existing code base, the most important of which is that I'm going to introduce a `Hull`

type. The intent is that if points are contained within this type, then only the convex hull remains. It'd be better if it was possible to make the
case constructor private, but if one does that, then the `hull`

function can no longer be inlined and generic.

type Hull<'a> = Hull of ('a * 'a) list

With the addition of the `Hull`

type, you can now add a binary operation:

// Hull<'a> -> Hull<'a> -> Hull<'a> let inline (+) (Hull x) (Hull y) = hull (x @ y)

This operation explicitly uses the `+`

operator, so I'm clearly anticipating the turn of events here. Nothing much is going on, though. The function pattern-matches the points out of two `Hull`

values. `x`

and `y`

are two lists of points. The `+`

function concatenates the two lists with the `@`

operator, and finds the convex hull of this new list of points.

### Associativity #

My choice of operator strongly suggests that the `+`

operation is a monoid. If you have three hulls, the order in which you find the hulls doesn't matter. One way to demonstrate that property is with property-based testing. In this article, I'm using Hedgehog.

[<Fact>] let ``Hull addition is associative`` () = Property.check <| property { let! (x, y, z) = Range.linear -10000 10000 |> Gen.int |> Gen.tuple |> Gen.list (Range.linear 0 100) |> Gen.tuple3 (hull x + hull y) + hull z =! hull x + (hull y + hull z) }

This automated test generates three lists of points, `x`

, `y`

, and `z`

. The `hull`

function uses the Graham Scan algorithm to find the hull, and part of that algorithm includes calculating the cross product of three points. For large enough integers, the cross product will overflow, so the property constrains the point coordinates to stay within -10,000 and 10,000. The implication of that is that although `+`

is associative, it's only associative for a subset of all 32-bit integers. I could probably change the internal implementation so that it calculates the cross product using bigint, but I'll leave that as an exercise to you.

For performance reasons, I also arbitrarily decided to constrain the size of each set of points to between 0 and 100 elements. If I change the maximum count to 1,000, it takes my laptop 9 seconds to run the test.

In addition to Hedgehog, this test also uses xUnit.net, and Unquote for assertions. The `=!`

operator is the Unquote way of saying *must equal*. It's an assertion.

This property passes, which demonstrates that the `+`

operator for convex hulls is associative.

### Identity #

Likewise, you can write a property-based test that demonstrates that an identity element exists for the `+`

operator:

[<Fact>] let `` Hull addition has identity`` () = Property.check <| property { let! x = Range.linear -10000 10000 |> Gen.int |> Gen.tuple |> Gen.list (Range.linear 0 100) let hasIdentity = Hull.identity + hull x = hull x + Hull.identity && hull x + Hull.identity = hull x test <@ hasIdentity @> }

This test generates a list of integer pairs (`x`

) and applies the `+`

operator to `x`

and `Hull.identity`

. The test passes for all `x`

that Hedgehog generates.

What's `Hull.identity`

?

It's simply the empty hull:

module Hull = let identity = Hull []

If you have a set of zero 2D points, then the convex hull is empty as well.

The `+`

operator for convex hulls is a monoid for the set of coordinates where the cross product doesn't overflow.

### Summary #

If you consider that the `Hull`

type is nothing but a container for a list, it should come as no surprise that a monoid exists. After all, list concatenation is a monoid, and the `+`

operator shown here is a combination of list concatenation (`@`

) and a Graham Scan.

The point of this article was mostly to demonstrate that monoids exist not only for primitive types, but also for (some) more complex types. The `+`

operator shown here is really a set union operation. What about intersections of convex hulls? Is that a monoid as well? I'll leave that as an exercise.

**Next:** Tuple monoids.

## Comments

Is that true that you could replace hull with any other function, and (+) operator would still be a monoid? Since the operator is based on list concatenation, the "monoidness" is probably derived from there, not from function implementation.

Mikhail, thank you for writing. You can't replace `hull`

with any other function and expect list concatenation to remain a monoid. I'm sorry if my turn of phrase gave that impression. I can see how one could interpret my summary in that way, but it wasn't my intention to imply that this relationship holds in general. It doesn't, and it's not hard to show, because we only need to come up with a single counter-example.

One counter example is a function that always removes the first element in a list - unless the list is empty, in which case it simply returns the empty list. In Haskell, we can define a `newtype`

with this behaviour in mind:

Prelude Data.Monoid Data.List> newtype Drop1 a = Drop1 [a] deriving (Show, Eq)

For my own convenience, I wrote the entire counter-example in GHCi (the Haskell REPL), but imagine that the `Drop1`

data constructor is hidden from clients. The normal way to do that is to not export the data constructor from the module. In GHCi, we can't do that, but just pretend that the `Drop1`

data constructor is unavailable to clients. Instead, we'll have to use this function:

Prelude Data.Monoid Data.List> drop1 = Drop1 . drop 1

The `drop1`

function has the type `[a] -> Drop1 a`

; it takes a list, and returns a `Drop1`

value, which contains the input list, apart from its first element.

We can attempt to make `Drop 1`

a monoid:

Prelude Data.Monoid Data.List> :{ Prelude Data.Monoid Data.List| instance Monoid (Drop1 a) where Prelude Data.Monoid Data.List| mempty = drop1 [] Prelude Data.Monoid Data.List| mappend (Drop1 xs) (Drop1 ys) = drop1 (xs ++ ys) Prelude Data.Monoid Data.List| :}

Hopefully, you can see that the implementation of `mappend`

is similar to the above F# implementation of `+`

for convex hulls. In F#, the list concatenation operator is `@`

, whereas in Haskell, it's `++`

.

This compiles, but it's easy to come up with some counter-examples that demonstrate that the monoid laws don't hold. First, associativity:

Prelude Data.Monoid Data.List> (drop1 [1..3] <> drop1 [4..6]) <> drop1 [7..9] Drop1 [5,6,8,9] Prelude Data.Monoid Data.List> drop1 [1..3] <> (drop1 [4..6] <> drop1 [7..9]) Drop1 [3,6,8,9]

(The `<>`

operator is an infix alias for `mappend`

.)

Clearly, `[5,6,8,9]`

is different from `[3,6,8,9]`

, so the operation isn't associative.

Equivalently, identity fails as well:

Prelude Data.Monoid Data.List> mempty <> drop1 [1..3] Drop1 [3] Prelude Data.Monoid Data.List> drop1 [1..3] Drop1 [2,3]

Again, `[3]`

is different from `[2,3]`

, so `mempty`

isn't a proper identity element.

It was easy to come up with this counter-example. I haven't attempted to come up with more, but I'd be surprised if I accidentally happened to pick the only counter-example there is. Rather, I conjecture that there are infinitely many counter-examples that each proves that there's no general rule about 'wrapped' lists operations being monoids.

## Money monoid

*Kent Beck's money TDD example has some interesting properties.*

*monoid* is an associative binary operation with a neutral element (also known as *identity*).

In the first half of Test-Driven Development By Example Kent Beck explores how to develop a simple and flexible Money API using test-driven development. Towards the end, he arrives at a design that warrants further investigation.

### Kent Beck's API #

The following treatment of Kent Beck's code is based on Yawar Amin's C# reproduction of Kent Beck's original Java code, further forked and manipulated by me.

The goal of Kent Beck's exercise is to develop an object-oriented API able to handle money of multiple currencies, and for example be able to express operations such as *5 USD + 10 CHF*. Towards the end of the example, he arrives at an interface that, translated to C#, looks like this:

public interface IExpression { Money Reduce(Bank bank, string to); IExpression Plus(IExpression addend); IExpression Times(int multiplier); }

The `Reduce`

method reduces an `IExpression`

object to a single currency (`to`

), represented as a `Money`

object. This is useful if you have an `IExpression`

object that contains several currencies.

The `Plus`

method adds another `IExpression`

object to the current object, and returns a new `IExpression`

. This could be money in a single currency, but could also represent money held in more than one currency.

The `Times`

method multiplies an `IExpression`

with a multiplier. You'll notice that, throughout this example code base, both multiplier and amounts are modelled as integers. I think that Kent Beck did this as a simplification, but a more realistic example should use `decimal`

values.

The metaphor is that you can model money as one or more *expressions*. A simple expression would be *5 USD*, but you could also have *5 USD + 10 CHF* or *5 USD + 10 CHF + 10 USD*. While you can reduce some expressions, such as *5 CHF + 7 CHF*, you can't reduce an expression like *5 USD + 10 CHF* unless you have an exchange rate. Instead of attempting to reduce monetary values, this particular design builds an expression tree until you decide to evaluate it. (Sounds familiar?)

Kent Beck implements `IExpression`

twice:

`Money`

models an amount in a single currency. It contains an`Amount`

and a`Currency`

read-only property. It's the quintessential Value Object.`Sum`

models the sum of two other`IExpression`

objects. It contains two other`IExpression`

objects, called`Augend`

and`Addend`

.

*5 USD + 10 CHF*, you can write:

IExpression sum = new Sum(Money.Dollar(5), Money.Franc(10));

where `Money.Dollar`

and `Money.Franc`

are two static factory methods that return `Money`

values.

### Associativity #

Did you notice that `Plus`

is a binary operation? Could it be a monoid as well?

In order to be a monoid, it must obey the *monoid laws*, the first of which is that the operation must be associative. This means that for three `IExpression`

objects, `x`

, `y`

, and `z`

, `x.Plus(y).Plus(z)`

must be equal to `x.Plus(y.Plus(z))`

. How should you interpret equality here? The return value from `Plus`

is another `IExpression`

value, and interfaces don't have custom equality behaviour. Either, it's up to the individual implementations (`Money`

and `Sum`

) to override and implement equality, or you can use test-specific equality.

The xUnit.net assertion library supports test-specific equality via custom comparers (for more details, see my Advanced Unit Testing Pluralsight course). The original Money API does, however, already include a way to compare expressions!

The `Reduce`

method can reduce any `IExpression`

to a single `Money`

object (that is, to a single currency), and since `Money`

is a Value Object, it has structural equality. You can use this to compare the values of `IExpression`

objects. All you need is an exchange rate.

In the book, Kent Beck uses a 2:1 exchange rate between CHF and USD. As I'm writing this, the exchange rate is 0.96 Swiss Franc to a Dollar, but since the example code consistently models money as integers, that rounds to a 1:1 exchange rate. This is, however, a degenerate case, so instead, I'm going to stick to the book's original 2:1 exchange rate.

You can now add an Adapter between `Reduce`

and xUnit.net in the form of an `IEqualityComparer<IExpression>`

:

public class ExpressionEqualityComparer : IEqualityComparer<IExpression> { private readonly Bank bank; public ExpressionEqualityComparer() { bank = new Bank(); bank.AddRate("CHF", "USD", 2); } public bool Equals(IExpression x, IExpression y) { var xm = bank.Reduce(x, "USD"); var ym = bank.Reduce(y, "USD"); return object.Equals(xm, ym); } public int GetHashCode(IExpression obj) { return bank.Reduce(obj, "USD").GetHashCode(); } }

You'll notice that this custom equality comparer uses a `Bank`

object with a 2:1 exchange rate. `Bank`

is another object from the *Test-Driven Development* example. It doesn't implement any interface itself, but it does appear as an argument in the `Reduce`

method.

In order to make your test code more readable, you can add a static helper class:

public static class Compare { public static ExpressionEqualityComparer UsingBank = new ExpressionEqualityComparer(); }

This enables you to write an assertion for associativity like this:

Assert.Equal( x.Plus(y).Plus(z), x.Plus(y.Plus(z)), Compare.UsingBank);

In my fork of Yawar Amin's code base, I added this assertion to an FsCheck-based automated test, and it holds for all the `Sum`

and `Money`

objects that FsCheck generates.

In its present incarnation, `IExpression.Plus`

is associative, but it's worth noting that this isn't guaranteed to last. An interface like `IExpression`

is an extensibility point, so someone could easily add a third implementation that would violate associativity. We can tentatively conclude that `Plus`

is currently associative, but that the situation is delicate.

### Identity #

If you accept that `IExpression.Plus`

is associative, it's a monoid candidate. If an identity element exists, then it's a monoid.

Kent Beck never adds an identity element in his book, but you can add one yourself:

public static class Plus { public readonly static IExpression Identity = new PlusIdentity(); private class PlusIdentity : IExpression { public IExpression Plus(IExpression addend) { return addend; } public Money Reduce(Bank bank, string to) { return new Money(0, to); } public IExpression Times(int multiplier) { return this; } } }

There's only a single identity element, so it makes sense to make it a Singleton. The private `PlusIdentity`

class is a new `IExpression`

implementation that deliberately doesn't do anything.

In `Plus`

, it simply returns the input expression. This is the same behaviour as zero has for integer addition. When adding numbers together, zero is the identity element, and the same is the case here. This is more explicitly visible in the `Reduce`

method, where the identity expression simply reduces to zero in the requested currency. Finally, if you multiply the identity element, you still get the identity element. Here, interestingly, `PlusIdentity`

behaves similar to the identity element for multiplication (*1*).

You can now write the following assertions for any `IExpression x`

:

Assert.Equal(x, x.Plus(Plus.Identity), Compare.UsingBank); Assert.Equal(x, Plus.Identity.Plus(x), Compare.UsingBank);

Running this as a property-based test, it holds for all `x`

generated by FsCheck. The same caution that applies to associativity also applies here: `IExpression`

is an extensibility point, so you can't be sure that `Plus.Identity`

will be the identity element for all `IExpression`

implementations someone could create, but for the three implementations that now exist, the monoid laws hold.

`IExpression.Plus`

is a monoid.

### Multiplication #

In basic arithmetic, the multiplication operator is called *times*. When you write *3 * 5*, it literally means that you have 3 five times (or do you have 5 three times?). In other words:

3 * 5 = 3 + 3 + 3 + 3 + 3

Does a similar relationship exist for `IExpression`

?

Perhaps, we can take a hint from Haskell, where monoids and semigroups are explicit parts of the core library. You're going to learn about semigroups later, but for now, it's interesting to observe that the `Semigroup`

typeclass defines a function called `stimes`

, which has the type `Integral b => b -> a -> a`

. Basically, what this means that for any integer type (16-bit integer, 32-bit integer, etc.) `stimes`

takes an integer and a value `a`

and 'multiplies' the value. Here, `a`

is a type for which a binary operation exists.

In C# syntax, `stimes`

would look like this as an instance method on a `Foo`

class:

public Foo Times(int multiplier)

I named the method `Times`

instead of `STimes`

, since I strongly suspect that the *s* in Haskell's `stimes`

stands for `Semigroup`

.

Notice how this is the same type of signature as `IExpression.Times`

.

If it's possible to define a universal implementation of such a function in Haskell, could you do the same in C#? In `Money`

, you can implement `Times`

based on `Plus`

:

public IExpression Times(int multiplier) { return Enumerable .Repeat((IExpression)this, multiplier) .Aggregate((x, y) => x.Plus(y)); }

The static `Repeat`

LINQ method returns `this`

as many times as requested by `multiplier`

. The return value is an `IEnumerable<IExpression>`

, but according to the `IExpression`

interface, `Times`

must return a single `IExpression`

value. You can use the `Aggregate`

LINQ method to repeatedly combine two `IExpression`

values (`x`

and `y`

) to one, using the `Plus`

method.

This implementation is hardly as efficient as the previous, individual implementation, but the point here isn't about efficiency, but about a common, reusable abstraction. The exact same implementation can be used to implement `Sum.Times`

:

public IExpression Times(int multiplier) { return Enumerable .Repeat((IExpression)this, multiplier) .Aggregate((x, y) => x.Plus(y)); }

This is literally the same code as for `Money.Times`

. You can also copy and paste this code to `PlusIdentity.Times`

, but I'm not going to repeat it here, because it's the same code as above.

This means that you can remove the `Times`

method from `IExpression`

:

public interface IExpression { Money Reduce(Bank bank, string to); IExpression Plus(IExpression addend); }

Instead, you can implement it as an extension method:

public static class Expression { public static IExpression Times(this IExpression exp, int multiplier) { return Enumerable .Repeat(exp, multiplier) .Aggregate((x, y) => x.Plus(y)); } }

This works because any `IExpression`

object has a `Plus`

method.

As I've already admitted, this is likely to be less efficient than specialised implementations of `Times`

. In Haskell, this is addressed by making `stimes`

part of the typeclass, so that implementers can implement a more efficient algorithm than the default implementation. In C#, the same effect could be achieved by refactoring `IExpression`

to an abstract base class, with `Times`

as a public virtual (overridable) method.

### Haskell sanity check #

Since Haskell has a more formal definition of a monoid, you may want to try to port Kent Beck's API to Haskell, as a proof of concept. In its final modification, my C# fork has three implementations of `IExpression`

:

`Money`

`Sum`

`PlusIdentity`

data Expression = Money { amount :: Int, currency :: String } | Sum { augend :: Expression, addend :: Expression } | MoneyIdentity deriving (Show)

You can formally make this a `Monoid`

:

instance Monoid Expression where mempty = MoneyIdentity mappend MoneyIdentity y = y mappend x MoneyIdentity = x mappend x y = Sum x y

The C# `Plus`

method is here implemented by the `mappend`

function. The only remaining member of `IExpression`

is `Reduce`

, which you can implement like this:

import Data.Map.Strict (Map, (!)) reduce :: Ord a => Map (String, a) Int -> a -> Expression -> Int reduce bank to (Money amt cur) = amt `div` rate where rate = bank ! (cur, to) reduce bank to (Sum x y) = reduce bank to x + reduce bank to y reduce _ _ MoneyIdentity = 0

Haskell's typeclass mechanism takes care of the rest, so that, for example, you can reproduce one of Kent Beck's original tests like this:

λ> let bank = fromList [(("CHF","USD"),2), (("USD", "USD"),1)] λ> let sum = stimesMonoid 2 $ MoneyPort.Sum (Money 5 "USD") (Money 10 "CHF") λ> reduce bank "USD" sum 20

Just like `stimes`

works for any `Semigroup`

, `stimesMonoid`

is defined for any `Monoid`

, and therefore you can also use it with `Expression`

.

With the historical 2:1 exchange rate, 5 Dollars + 10 Swiss Franc, times 2, is equivalent to 20 Dollars.

### Summary #

In chapter 17 of his book, Kent Beck describes that he'd been TDD'ing a Money API many times before trying out the expression-based API he ultimately used in the book. In other words, he had much experience, both with this particular problem, and with programming in general. Clearly this is a highly skilled programmer at work.

I find it interesting that he seems to intuitively arrive at a design involving a monoid and an interpreter. If he did this on purpose, he doesn't say so in the book, so I rather speculate that he arrived at the design simply because he recognised its superiority. This is the reason that I find it interesting to identify this, *an existing example*, as a monoid, because it indicates that there's something supremely comprehensible about monoid-based APIs. It's conceptually 'just like addition'.

In this article, we returned to a decade-old code example in order to identify it as a monoid. In the next article, I'm going to revisit an example code base of mine from 2015.

**Next: ** Convex hull monoid.

## Comments

You'll notice that, throughout this example code base, both multiplier and amounts are modelled as integers. I think that Kent Beck did this as a simplification, but a more realistic example should use decimal values.

Actually, in a lot of financial systems money is stored in cents, and therefore as integers, because it avoids rounding errors.

Great articles btw! :)

Hrvoje, thank you for writing. Yes, it's a good point that you could model the values as cents and rappen, but I think I recall that Kent Beck's text distinctly discusses *dollars* and *francs*. I am, however, currently travelling, without access to the book, so I can't check.

The scenario, as simplistic as it may be, involves currency exchange, and exchange rates tend to involve much smaller fractions. As an example, right now, one currency exchange web site reports that 1 CHF is 1.01950 USD. Clearly, representing the U.S. currency with cents would incur a loss of precision, because that would imply an exchange rate of 102 cents to 100 rappen. I'm sure arbitrage opportunities would be legion if you ever wrote code like that.

If I remember number theory correctly, you can always scale any rational number to an integer. I.e. in this case, you could scale 1.01950 to 101,950. There's little reason to do that, because you have the `decimal`

struct for that purpose:

"The Decimal value type is appropriate for financial calculations that require large numbers of significant integral and fractional digits and no round-off errors."All of this, however, is just idle speculation on my point. I admit that I've never had to implement complex financial calculations, so there may be some edge cases of which I'm not aware. For all the run-of-the-mill eCommerce and payment solutions I've implemented over the years,

`decimal`

has always been more than adequate.
Although exchange rates are typically represented as decimal fractions, it does not follow that amounts of money should be, even if the amounts were determined by calculations involving that exchange rate.

The oversimplified representation of foreign exchange (FX) in Kent Beck's money examples has always struck me as a particularly weak aspect (and not simply because they are integers; that's the least of the problems). You could argue that the very poor modelling of FX is tolerable because that aspect of the problem domain is not the focus in his example. But I think it's problematic because it can lead you to the wrong conclusion about the design of the central parts of the model. Your conclusion that it might be a good idea not to represent a money amount as an integer is an example - I believe it's the wrong conclusion, and that you've been led to it by the completely wrong-headed way his example represents FX.

The nature of foreign exchange is that it is a transaction with a third party. Some entity (perhaps a bank, or the FX trading desk within an company that may or may not be a financial institution (large multinational firms sometimes have their own FX desks) or maybe a friend who has some of the kind of currency you need in her purse) agrees to give you a specific amount of one currency if you give them a specific amount of some other currency, and there is usually an accompanying agreement on the timescale in which the actual monies are to be transferred. (There will sometimes be more than two currencies involved, either because you're doing something complex, or just because you agree to pay a commission fee in some currency that is different from either the 'to' or 'from' currency.) The amounts of actual money that changes hands will invariably be some integer multiple of the smallest available denomination of the currencies in question.

There may well be a published exchange rate. It might even form part of some contract, although such an advertised rate is very often not binding because markets can move fast, and the exchange rate posted when you started negotiation could change at any moment, and might not be available by the time you attempt to reach an agreement. In cases where a published exchange rate has some reliable meaning, it will necessarily come with a time limit (and unless this time limit is pretty short, the time window itself may come at a price - if someone has agreed to sell you currency for a specific price within some time window, what you have there is in effect either a future or an option, depending on whether you are allowed to decide not to complete the transaction).

One very common case where a 'current' exchange rate does in fact apply is when using a credit or debit card abroad. In this case, somewhere in the terms and conditions that you agreed to at some point in the past, it will say that the bank gets to apply the current rate for some definition of current. (The bank will generally have freedom to define what it means by 'current', which is one of the reasons you tend not to get a very good deal on such transactions.) And there will be rules (often generally accepted conventions, instead of being explicitly set out in the contract) about how the rate is applied. It will necessarily involve some amount of rounding. When you bought something on your credit card in a foreign currency, it will have been for a precise amount in that currency - merchants don't get to charge you Pi dollars for something. And when the bank debits your account, they will also do so by a precise amount - if you've ever used a card in this way you'll know that you didn't end up with some fractional number of cents or pennies or whatever in your account afterwards. So the exchange rate you got in practice will very rarely be exactly the advertised one (unless it's such a large transaction that the amounts involved have more decimal places than the 'current' exchange rate, or, by sheer coincidence, the numbers worked out in such a way that you happened to get the exact exchange rate advertised.).

So although you will often see published exchange rates with multiple decimal places, the actual exchange rate depends entirely on the agreement you strike with whoever it is that is going to give you money in the currency you want in exchange for money in the currency you have. The actual exchanges that result from such agreements do not involve fractional amounts.

Where does this leave Kent's example? Fundamentally, 'reducing' a multi-currency expression to a single-currency result will need to create at least one FX transaction (possibly several). So you'll need some sort of mechanism for agreeing the terms of those transactions with the other party or parties. And realistically you'd want to do something to minimize transaction costs (e.g., if you perform multiple USD to GBP conversions, you'll want to handle that with a single FX transaction), so you'll need some sort of logic for managing that too. It's certainly not going to be as simple as looking up the bank's rate.

Ian, thank you for writing. Much of what you write about foreign exchange matches the little I know. What interested me about Kent Beck's example was that his intuition about good programming lead him to a monoidal design.

It seems to me that your criticism mostly targets how the exchange itself is implemented, i.e. the `Reduce`

method, or rather, its `bank`

argument. In its current form, the `Bank`

implementation is indisputably naive.

Would a more sophisticated `Bank`

implementation address some of the problems? What if, instead of calling it `Bank`

, we called it `Exchange`

?

Already in its current form, the `Bank`

implementation is nothing but a dictionary of exchange rates, defined by a `from`

and a `to`

currency. It follow that the *USD/CHF* entry isn't the same as the *CHF/USD* entry. They don't have to be each others' inverses. Doesn't this, already, enable arbitrage?

Another change that we could add to a hypothetical more sophisticated `Exchange`

class would be to subtract a fee from the returned value. Would that address one of the other concerns?

Furthermore, we could add a time limit to each dictionary of exchange rates.

It's not my intent to claim that such a model would be sufficient to implement an international bank's foreign exchange business, but that's not the scenario that Kent Beck had in mind. The introduction to *Test-Driven Development By Example* explicitly explains that the scenario is a bond portfolio management system. Doesn't the overall API he outlines sufficiently address that?

## Comments

Thanks

Yacoub, thank you for writing. The operation used here isn't the

intersection, but rather theunionof two bounding boxes; that's the reason I called the method`Unite`

.Hello Mark. I am aware of this, but maybe I did not explain my self correctly.

What I am trying to say is that when coming up with a counter-example, we should choose a BoundingBox y wholly outside of x (not just partially outside of x).

If we choose a BoundingBox y partially outside of x, then the intersection between x and y (the BoundingBox z = the area shared between x and y) is a valid identity element.

Yacoub, I think you're right; sorry about that!

Perhaps I should have written

Just pick any otherWould that have been correct?`BoundingBox y`

partially or wholly outside of the candidate.That would be correct. I am not sure though if this is the best way to explain it.

y being wholly ourside of x seems better to me.

Yacoub, I've corrected the text in the article. Thank you for the feedback!