Function isomorphisms

Monday, 22 January 2018 14:37:00 UTC

Instance methods are isomorphic to functions.

This article is part of a series of articles about software design isomorphisms.

While I have already, in an earlier article, quoted the following parable about Anton, Qc Na, objects, and closures, it's too good a fit to the current topic to pass up, so please pardon the duplication.

The venerable master Qc Na was walking with his student, Anton. Hoping to prompt the master into a discussion, Anton said "Master, I have heard that objects are a very good thing - is this true?" Qc Na looked pityingly at his student and replied, "Foolish pupil - objects are merely a poor man's closures."

Chastised, Anton took his leave from his master and returned to his cell, intent on studying closures. He carefully read the entire "Lambda: The Ultimate..." series of papers and its cousins, and implemented a small Scheme interpreter with a closure-based object system. He learned much, and looked forward to informing his master of his progress.

On his next walk with Qc Na, Anton attempted to impress his master by saying "Master, I have diligently studied the matter, and now understand that objects are truly a poor man's closures." Qc Na responded by hitting Anton with his stick, saying "When will you learn? Closures are a poor man's object." At that moment, Anton became enlightened.

- Anton van Straaten

The point is that objects and closures are two ways of looking at a thing. In a nutshell, objects are data with behaviour, whereas closures are behaviour with data. I've already shown an elaborate C# example of this, so in this article, you'll get a slightly more formal treatment of the subject.

Isomorphism #

In object-oriented design, you often bundle operations as methods that belong to objects. These are isomorphic to static methods, because a lossless translation exists. We can call such static methods functions, although they aren't guaranteed to be pure.

Diagram showing isomorphism between instance method and function.

In the spirit of Refactoring, we can describe each translation as a refactoring, because that's what it is. I don't think the book contains a specific refactoring that describes how to translate from an instance method to a static method, but we could call it Replace Object with Argument.

Going the other way is, on the other hand, already described, so we can use Refactoring's Move Method.

Replace Object with Argument #

While the concept of refactoring ought to be neutral across paradigms, the original book is about object-oriented programming. In object-oriented programming, objects are the design ideal, so it didn't make much sense to include, in the book, a refactoring that turns an instance method into a static method.

Nevertheless, it's straightforward:

  1. Add an argument to the instance method. Declare the argument as the type of the hosting class.
  2. In the method body, change all calls to this and base to the new argument.
  3. Make the method static.
As an example, imagine that you start with an instance method like this Baz method:

public class Foo
    public Bar Baz(Qux qux, Corge corge)
        // Do stuff, return bar
    // Other members...

You'll first introduce a new Foo foo argument to Baz, change the method body to use foo instead of this, and then make the method static. The result is this:

public class Foo
    public static Bar Baz(Foo foo, Qux qux, Corge corge)
        // Do stuff, return bar
    // Other members...

Once you have a static method, you can always move it to another class, if you'd like. This can, however, cause some problems with accessibility. In C#, for example, you'd no longer be able to access private or protected members from a method outside the class. You can choose to leave the static method reside on the original class, or you can make the member in question available to more clients (make it internal or public).

Move Method #

The book Refactoring doesn't contain a recipe like the above, because the goal of that book is better object-oriented design. It would consider a static method, like the second variation of Baz above, a code smell named Feature Envy. You have this code smell when it looks as if a method is more interested in one of its arguments than in its host object. In that case, the book suggests using the Move Method refactoring.

The book already describes this refactoring, so I'm not going to repeat it here. Also, there's no sense in showing you the code example, because it's literally the same two code examples as above, only in the opposite order. You start with the static method and end with the instance method.

C# developers are most likely already aware of this relationship between static methods and objects, because you can use the this keyword in a static method to make it look like an instance method:

public static Bar Baz(this Foo foo, Qux qux, Corge corge)

The addition of this in front of the Foo foo argument enables the C# compiler to treat the Baz method as though it's an instance method on a Fooobject:

var bar = foo.Baz(qux, corge);

This is only syntactic sugar. The method is still compiled as a static method, and if you, as a client developer, wish to use it as a static method, that's still possible.

Functions #

A static method like public static Bar Baz(Foo foo, Qux qux, Corge corge) looks a lot like a function. If refactored from object-oriented design, that function is likely to be impure, but its shape is function-like.

In C#, for example, you could model it as a variable of a delegate type:

Func<FooQuxCorgeBar> baz = (foo, qux, corge) =>
    // Do stuff, return bar

Here, baz is a function with the same signature as the above static Baz method.

Have you ever noticed something odd about the various Func delegates in C#?

They take the return type as the last type argument, which is contrary to C# syntax, where you have to declare the return type before the method name and argument list. Since C# is the dominant .NET language, that's surprising, and even counter-intuitive.

It does, however, nicely align with an ML language like F#. As we'll see in a future article, the above baz function translates to an F# function with the type Foo * Qux * Corge -> Bar, and to Haskell as a function with the type (Foo, Qux, Corge) -> Bar. Notice that the return type comes last, just like in the C# Func.

Closures #

...but, you may say, what about data with behaviour? One of the advantages, after all, of objects is that you can associate a particular collection of data with some behaviour. The above Foo class could have data members, and you may sometimes have a need for passing both data and behaviour around as a single... well... object.

That seems to be much harder with a static Baz method.

Don't worry, write a closure:

var foo = new Foo { /* initialize members here */ };
Func<QuxCorgeBar> baz = (qux, corge) =>
    // Do stuff with foo, qux, and corge; return bar

In this variation, baz closes over foo. Inside the function body, you can use foo like you can use qux and corge. As I've already covered in an earlier article, the C# compiler compiles this to an IL class, making it even more obvious that objects and closures are two sides of the same coin.

Summary #

Object-oriented instance methods are isomorphic to both static methods and function values. The translations that transforms your code from one to the other are refactorings. Since you can move in both directions without loss of information, these refactorings constitute an isomorphism.

This is another important result about the relationship between object-oriented design and functional programming, because this enables us to reduce any method to a canonical form, in the shape of a function. From a language like Haskell, we know a lot about the relationship between category theory and functional programming. With isomorphisms like the present, we can begin to extend that knowledge to object-oriented design.

Next: Argument list isomorphisms.


Matt Heuer #

This Isomorphism applies to non-polymorphic Methods. Polymorphic Functions need to be mapped to a Set is static Methods with the same Signature. Is there a functional Construct for this?

2021-02-20 17:57 UTC

Matt, thank you for writing. What makes you write that this isomorphism applies (only, I take it) to non-polymorphic methods. The view here is on the implementation. In C# (and all other statically typed languages that I know that support functions), functions are polymorphic based on signature.

A consumer that depends on a function with the type Func<Foo, Qux, Corge, Bar> can interact with any function with that type.

2021-02-21 17:19 UTC

Unit isomorphisms

Monday, 15 January 2018 07:33:00 UTC

The C# and Java keyword 'void' is isomorphic to a data type called 'unit'.

This article is part of a series of articles about software design isomorphisms.

Many programming languages, such as C# and Java, distinguish between methods that return something, and methods that don't return anything. In C# and Java, a method must be declared with a return type, but if it doesn't return anything, you can use the special keyword void to indicate that this is the case:

public void Foo(int bar)

This is a C# example, but it would look similar (isomorphic?) in Java.

Two kinds of methods #

In C# and Java, void isn't a type, but a keyword. This means that there are two distinct types of methods:

  • Methods that return something
  • Methods that return nothing
In C#, methods that return something declare their return type before the method name:

public Foo Bar(Baz baz, Qux qux)

On the other hand, methods that return nothing must use the special void keyword:

public void Bar(Baz baz, Qux qux)

If you want to generalise, you can use generics like this:

public T Foo<TT1T2>(T1 x, T2 y)

Such a method could return anything, but, surprisingly, not nothing. In C# and Java, nothing is special. You can't generalise all methods to a common set. Even with generics, you must model methods that return nothing in a different way:

public void Foo<T1T2>(T1 x, T2 y)

In C#, for example, this leads to the distinction between Func and Action. You can't reconcile these two fundamentally distinct types of methods into one.

Visual Basic .NET makes the same distinction, but uses the keywords Sub and Function instead of void.

Sometimes, particularly when writing code with generics, this dichotomy is really annoying. Wouldn't it be nice to be able to generalise all methods?

Unit #

While I don't recall the source, I've read somewhere the suggestion that the keyword void was chosen to go with null, because null and void is an English (legal) idiom. That choice is, however, unfortunate.

In category theory, the term void denotes a type or set with no inhabitants (e.g. an empty set). That sounds like the same concept. The problem, from a programming perspective, is that if you have a (static) type with no inhabitants, you can't create an instance of it. See Bartosz Milewski's article Types and Functions for a great explanation and more details.

Functional programming languages like F# and Haskell instead model nothing by a type called unit (often rendered as empty brackets: ()). This type is a type with exactly one inhabitant, a bit like a Singleton, but with the extra restriction that the inhabitant carries no information. It simply is.

It may sound strange and counter-intuitive that a singleton value represents nothing, but it turns out that this is, indeed, isomorphic to C# or Java's void.

This is admirably illustrated by F#, which consistently uses unit instead of void. F# is a multi-paradigmatic language, so you can write classes with methods as well as functions:

member this.Bar (x : int) = ()

This Bar method has the return type unit. When you compile F# code, it becomes Intermediate Language, which you can decompile into C#. If you do that, the above F# code becomes:

public void Bar(int x)

The inverse translation works as well. When you use F#'s interoperability features to interact with objects written in C# or Visual Basic, the F# compiler interprets void methods as if they return unit. For example, calling GC.Collect returns unit in F#, although C# sees it as 'returning' void:

> GC.Collect 0;;
val it : unit = ()

F#'s unit is isomorphic to C#'s void keyword, but apart from that, there's nothing special about it. It's a value like any other, and can be used in generically typed functions, like the built-in id function:

> id 42;;
val it : int = 42

> id "foo";;
val it : string = "foo"

> id ();;
val it : unit = ()

The built-in function id simply returns its input argument. It has the type 'a -> 'a, and as the above F# Interactive session demonstrates, you can call it with unit as well as with int, string, and so on.

Monoid #

Unit, by the way, forms a monoid. This is most evident in Haskell, where this is encoded into the type. In general, a monoid is a binary operation, and not a type, but what could the combination of two () (unit) values be, other than ()?

λ> mempty :: ()

λ> mappend () ()

In fact, the above (rhetorical) question is imprecise, since there aren't two unit values. There's only one, but used twice.

Since only a single unit value exists, any binary operation is automatically associative, because, after all, it can only return unit. Likewise, unit is the identity (mempty) for the operation, because it doesn't change the output. Thus, the monoid laws hold, and unit forms a monoid.

This result is interesting when you start to think about composition, because a monoid can always be used to reduce (aggregate) multiple values to a single value. With this result, and unit isomorphism, we've already explained why Commands are composable.

Summary #

Since unit is a type only inhabited by a single value, people (including me) often use the word unit about both the type and its only value. Normally, the context surrounding such use is enough to dispel any ambiguity.

Unit is isomorphic to C# or Java void. This is an important result, because if we're to study software design and code structure, we don't have to deal with two distinct cases (methods that return nothing, and methods that return something). Instead, we can ignore methods that return nothing, because they can instead be modelled as methods that return unit.

The reason I've titled this article in the plural is that you could view the isomorphism between F# unit and C# void as a different isomorphism than the one between C# and Java void. Add Haskell's () (unit) type and Visual Basic's Sub keyword to the mix, and it should be clear that there are many translations to the category theory concept of Unit.

Isormorphisms between the Unit concept and constructs in selected languages: C#, Java, Visual Basic, F#, Haskell.

Unit isomorphism is an example of an interlingual isomorphism, in the sense that C# void maps to F# unit, and vice versa. In the next example, you'll see an isomorphism that mostly stays within a single language, although a translation between languages is also pertinent.

Next: Function isomorphisms.

Software design isomorphisms

Monday, 08 January 2018 08:35:00 UTC

When programming, you can often express the same concept in multiple ways. If you can losslessly translate between two alternatives, it's an isomorphism. An introduction for object-oriented programmers.

This article series is part of an even larger series of articles about the relationship between design patterns and category theory.

There's a school of functional programming that looks to category theory for inspiration, verification, abstraction, and cross-pollination of ideas. Perhaps you're put off by terms like zygohistomorphic prepromorphism (a joke), but you shouldn't be. There are often good reasons for using abstract naming. In any case, one term from category theory that occasionally makes the rounds is isomorphism.

Equivalence #

Don't let the terminology daunt you. An isomorphism is an easy enough concept to grasp. In essence, two things are isomorphic if you can translate losslessly back and forth between them. It's a formalisation of equivalence.

Many programming languages, like C# and Java, offer a multitude of alternative ways to do things. Just consider this C# example from my Humane Code video:

public bool IsSatisfiedBy(Customer candidate)
    bool retVal;
    if (candidate.TotalPurchases >= 10000)
        retVal = true;
        retVal = false;
    return retVal;

which is equivalent to this:

public bool IsSatisfiedBy(Customer candidate)
    return candidate.TotalPurchases >= 10000;

An outside observer can't tell the difference between these two implementations, because they have exactly the same externally visible behaviour. You can always refactor from one implementation to the other without loss of information. Thus, we could claim that they're isomorphic.

Terminology #

If you're an object-oriented programmer, then you already know the word polymorphism, which sounds similar to isomorphism. Perhaps you've also heard the word xenomorph. It's all Greek. Morph means form or shape, and while poly means many, iso means equal. So isomorphism means 'being of equal shape'.

This is most likely the explanation for the term Isomorphic JavaScript. The people who came up with that term knew (enough) Greek, but apparently not mathematics. In mathematics, and particularly category theory, an isomorphism is a translation with an inverse. That's still not a formal definition, but just my attempt at presenting it without too much jargon.

Category theory uses the word object to describe a member of a category. I'm going to use that terminology here as well, but you should be aware that object doesn't imply object-oriented programming. It just means 'thing', 'item', 'element', 'entity', etcetera.

In category theory, a morphism is a mapping or translation of an object to another object. If, for all objects, there's an inverse morphism that leads back to the origin, then it's an isomorphism.

Isomorphism diagram.

In this illustration, the blue arrows going from left to right indicate a single morphism. It's a mapping of objects on the blue left side to objects on the green right side. The green arrows going from right to left is another morphism. In this case, the green right-to-left morphism is an inverse of the blue left-to-right morphism, because by applying both morphisms, you end where you started. It doesn't matter if you start at the blue left side or the green right side.

Another way to view this is to say that a lossless translation exists. When a translation is lossless, it means that you don't lose information by performing the translation. Since all information is still present after a translation, you can go back to the original representation.

Software design isomorphisms #

When programming, you can often solve the same problem in different ways. Sometimes, the alternatives are isomorphic: you can go back and forth between two alternatives without loss of information.

Martin Fowler's book Refactoring contains several examples. For instance, you can apply Extract Method followed by Inline Method and be back where you started.

There are many other isomorphisms in programming. Some are morphisms in the same language, as is the case with the above C# example. This is also the case with the isomorphisms in Refactoring, because a refactoring, by definition, is a change applied to a particular code base, be it C#, Java, Ruby, or Python.

Other programming isomorphisms go between languages, where a concept can be modelled in one way in, say, C++, but in another way in Clojure. The present blog, for instance, contains several examples of translating between C# and F#, and between F# and Haskell.

Being aware of software design isomorphisms can make you a better programmer. It'll enable you to select the best alternative for solving a particular problem. Identifying programming isomorphisms is also important because it'll enable us to formally think about code structure by reducing many alternative representations to a canonical representation. For these reasons, this article presents a catalogue of software design isomorphisms:

In general, I've tried to name each isomorphism after its canonical representation. For instance, by unit isomorphisms, I mean isomorphisms to the unit value. It is, however, not an entirely consistent naming strategy.

Many more software design isomorphisms exist, so if you revisit this article in the future, I may have added more items to this catalogue. In no way should you consider this catalogue exhaustive.

Summary #

An isomorphism is a mapping for which an inverse mapping also exists. It's a way to describe equivalence.

In programming, you often have the choice to implement a particular feature in more than one way. These alternatives may be equivalent, in which case they're isomorphic. That's one of the reasons that many code bases come with a style guide.

Understanding how code is isomorphic to other code enables us to reduce many alternatives to a canonical representation. This makes analysis easier, because we can narrow our analysis to the canonical form, and generalise from there.

Next: Unit isomorphisms.


Funny thing, I had a discussion about refactoring as an isomorphism a bit ago. While I like the idea of using isomorphisms to reason about code, I still stand by the point that refactoring is an isomorphism limited to functionality; i.e. refactoring the code may change its other aspects (readability is the first one to come to mind, performance is the second), so two different representations are no longer totally equal. Or, as another argument, using Inline Method in fact loses some information (the descriptive method name, limited variable scopes), so the translation is not (sorry) lossless.
2018-01-13 08:39 UTC

Sergey, thank you for writing. Good point, you're right that viewed as a general-purpose translation, Inline Method is indeed lossy. When you look at the purpose of refactoring code, the motivation is mostly (if not always) to make the code better in some way. Particularly when the purpose is make the code more readable, a refactoring introduces clarity. Thus, going the opposite way would remove that clarity, so I think it's fair to argue that such a change would be lossy.

It's perfectly reasonable to view a refactoring like Inline Method as a general-purpose algorithm, in which case you're right that it's lossy. I don't dispute that.

My agenda with this article series, however, is different. I'm not trying to make multiple angels dance on a pinhead; it's not my intent to try to redefine the word refactoring. The purpose with this series of articles is to describe how the same behaviour can be implemented in many different ways. The book Refactoring is one source of such equivalent representations.

One quality of morphisms is that there can be several translations between two objects. One such translation could be the general-purpose refactoring that you so rightly point out is lossy. Another translation could be one that 'remembers' the name of the original method.

Take, as an example, the isomorphism described under the heading Roster isomorphism in my article Tuple monoids. When you consider the method ToTriple, you could, indeed, argue that it's lossy, because it 'forgets' that the label associated with the first integer is Girls, that the label associated with the second integer is Boys, and so on. The reverse translation, however, 'remembers' that information, as you can see in the implementation of FromTriple.

This isn't necessarily a 'safe' translation. You could easily write a method like this:

public static Roster FromTriple(Tuple<intintstring[]> triple)
    return new Roster(triple.Item2, triple.Item1, triple.Item3);

This compiles, but it translates triples created by ToTriple the wrong way.

On the other hand, the pair of translations that I do show in the article is an isomorphism. The point is that an isomorphism exists; not that it's the only possible set of morphisms.

The same argument can be applied to specific pairs of Extract Method and Inline Method. As a general-purpose algorithm, I still agree that Inline Method is lossy. That doesn't preclude that specific pairs of translations exist. For instance, in an article, I discuss how some people refactor Guard Clauses like this:

if (subject == null)
    throw new ArgumentNullException("subject");

to something like this:

Guard.AgainstNull(subject, "subject");

Again, an isomorphism exists: a translation from a Null Guard to Guard.AgainstNull, and another from Guard.AgainstNull to a Null Guard. Those are specific incarnations of Extract Method and Inline Method.

This may not be particularly useful as a refactoring, I admit, but that's also not the agenda of these articles. The programme is to show how particular software behaviour can be expressed in various different ways that are equivalent to each other.

2018-01-14 14:26 UTC

Colour-mixing magma

Tuesday, 02 January 2018 08:36:00 UTC

Mixing RGB colours forms a magma. An 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 an example of a magma, which is a binary operation without additional constraints.

RGB colours #

The opening article about monoids, semigroups, and their friends emphasised Eric Evans' pigment mixing example from Domain-Driven Design. The following article series then promptly proceeded to ignore that example. The reason is that while the example has Closure of Operations, it exhibits precious few other properties. It's neither monoid, semigroup, quasigroup, nor any other named binary operation, apart from being a magma.

Instead of pigments, consider a more primitive, but well-understood colour model: that of RGB colours. In C#, you can model RGB colours using a struct that holds three byte fields. In my final code base, I ended up implementing ==, !=, Equals, and so on, but I'm not going to bore you with all of those details. Here's the RgbColor constructor, so that you can get a sense of the type:

private readonly byte red;
private readonly byte green;
private readonly byte blue;
public RgbColor(byte red, byte green, byte blue)
{ = red; = green; = blue;

As you can see, RgbColor holds three byte fields, one for red, green, and blue. If you want to mix two colours, you can use the MixWith instance method:

public RgbColor MixWith(RgbColor other)
    var newRed = ((int) + (int) / 2m;
    var newGreen = ((int) + (int) / 2m;
    var newBlue = ((int) + (int) / 2m;
    return new RgbColor(

This is a binary operation, because it's an instance method on RgbColor, taking another RgbColor as input, and returning RgbColor. Since it's a binary operation, it's a magma, but could it be another, stricter category of operation?

Lack of associativity #

Could MixWith, for instance, be a semigroup? In order to be a semigroup, the binary operation must be associative, and while it can be demanding to prove that an operation is always associative, it only takes a single counter-example to prove that it's not:

public void MixWithIsNotAssociative()
    // Counter-example
    var x = new RgbColor( 67, 108,  13);
    var y = new RgbColor( 33, 114, 130);
    var z = new RgbColor( 38, 104, 245);

This unit test passes, thereby demonstrating that MixWith is not associative. When you mix x with y, you get #326F48, and when you mix that with z you get #2C6C9E. On the other hand, when you mix y with z you get #246DBC, which, combined with x, gives #346C64. #2C6C9E is not equal to #346C64, so the NotEqual assertion passes.

Because of this counter-example, MixWith isn't associative, and therefore not a semigroup. Since monoid requires associativity as well, we can also rule out that MixWith is a monoid.

Lack of invertibility #

While MixWith isn't a semigroup, could it be a quasigroup? In order to be a quasigroup, a binary operation must be invertible. This means that for any two elements a and b, there must exist two other elements x and y that turns a into b.

This property must hold for all values involved in the binary operation, so again, a single counter-example suffices to demonstrate that MixWith isn't invertible, either:

public void MixWithIsNotInvertible()
    // Counter-example
    var a = new RgbColor( 94,  35, 172);
    var b = new RgbColor(151, 185,   7);
    Assert.False(RgbColor.All.Any(x => a.MixWith(x) == b));
    Assert.False(RgbColor.All.Any(y => y.MixWith(a) == b));

This test also passes. It uses brute force to demonstrate that for all RgbColor values, there's no x and y that satisfy the invertibility property. The test actually takes a while to execute, because All returns all 16,777,216 possible RgbColor values:

private static RgbColor[] all;
private readonly static object syncLock = new object();
public static IReadOnlyCollection<RgbColor> All
        if (all == null)
            lock (syncLock)
                if (all == null)
                    var max = 256 * 256 * 256;
                    all = new RgbColor[max];
                    foreach (var i in Enumerable.Range(0, max))
                        all[i] = (RgbColor)i;
        return all;

For performance reasons, the All property uses lazy initialisation with double-checked locking. It simply counts from 0 to 256 * 256 * 256 (16,777,216) and converts each integer to an RgbColor value using this explicit conversion:

public static explicit operator RgbColor(int i)
    var red = (i & 0xFF0000) / 0x10000;
    var green = (i & 0xFF00) / 0x100;
    var blue = i & 0xFF;
    return new RgbColor((byte)red, (byte)green, (byte)blue);

The bottom line, though, is that the test passes, thereby demonstrating that for the chosen counter-example, no x and y satisfies the invertibility property. Therefore, MixWith isn't a quasigroup.

Lack of identity #

Since MixWith is neither associative nor invertible, it's not really any named algebraic construct, other than a magma. It's neither group, semigroup, quasigroup, monoid, loop, groupoid, etc. Does it have any properties at all, apart from being a binary operation?

It doesn't have identity either, which you can illustrate with another counter-example:

public void MixWithHasNoIdentity()
    var nearBlack = new RgbColor(1, 1, 1);
    var identityCandidates = from e in RgbColor.All
                             where nearBlack.MixWith(e) == nearBlack
                             select e;
    // Verify that there's only a single candidate:
    var identityCandidate = Assert.Single(identityCandidates);
    // Demonstrate that the candidate does behave like identity for
    // nearBlack:
    Assert.Equal(nearBlack, nearBlack.MixWith(identityCandidate));
    Assert.Equal(nearBlack, identityCandidate.MixWith(nearBlack));
    // Counter-example
    var counterExample = new RgbColor(3, 3, 3);

The counter-example starts with a near-black colour. The reason I didn't pick absolute black (new RgbColor(0, 0, 0)) is that, due to rounding when mixing, there are eight candidates for absolute black, but only one for nearBlack. This is demonstrated by the Assert.Single assertion. identityCandidate, by the way, is also new RgbColor(1, 1, 1), and further Guard Assertions demonstrate that identityCandidate behaves like the identity for nearBlack.

You can now pick another colour, such as new RgbColor(3, 3, 3) and demonstrate that identityCandidate does not behave like the identity for the counter-example. Notice that the assertion is Assert.NotEqual.

If an identity exists for a magma, it must behave as the identity for all possible values. That's demonstrably not the case for MixWith, so it doesn't have identity.

Commutativity #

While MixWith is neither associative, invertible, nor has identity, it does have at least one property: it's commutative. This means that the order of the input values doesn't matter. In other words, for any two RgbColor values x and y, this assertion always passes:


Since x.MixWith(y) is equal to y.MixWith(x), MixWith is commutative.

Summary #

The MixWith operation is a commutative magma, but while, for example, we call an associative magma a semigroup, there's no fancy word for a commutative magma.

In this article, you got another, fairly realistic, example of a binary operation. Throughout the overall article series on monoids, semigroup, and other group-like algebraic structures, you've seen many examples, and you've learned how to analyse binary operations for the presence or absence of various properties. The present article concludes the series. You can, however, continue reading the even more overall article series.

Next: Functors, applicatives, and friends


At first, the lack of associativity felt counterintuitive: If I take equals parts of three colors, it shouldn't matter in which order I mix them. Then I realized this function doesn't take equal parts of all three. Basically the first mixture mixes one unit of each of two colors, resulting in two units of mixture. Then the second mixture takes one unit of the first mixture and one unit of the third color. That's why it's not associative!

2022-04-20 07:30 UTC

Mark, thank you for writing. I never gave that question that much attention when I wrote the article, but that makes total sense. Thank you for explaining it.

2022-04-20 07:56 UTC

Rock Paper Scissors magma

Thursday, 28 December 2017 11:22:00 UTC

The Rock Paper Scissors game forms a magma. An 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 an example of a magma, which is a binary operation without additional constraints.

Rock Paper Scissors #

When my first child was born, my wife and I quickly settled on a first name, but we couldn't agree on the surname, since we don't share the same surname. She wanted our daughter to have her surname, and I wanted her to have mine. We couldn't think of any rational arguments for one or the other, so we decided to settle the matter with a game of Rock Paper Scissors. I lost, so my daughter has my wife's surname.

Despite that outcome, I still find that Rock Paper Scissors is a great way to pick between two equally valid alternatives. You could also flip a coin, or throw a die, but most people have their hands handy, so to speak.

In C#, you can model the three shapes of rock, paper, and scissors like this:

public abstract class Rps
    public readonly static Rps Rock = new R();
    public readonly static Rps Paper = new P();
    public readonly static Rps Scissors = new S();
    private Rps() { }
    private class R : Rps { }
    private class P : Rps { }
    private class S : Rps { }
    // More members...

I've seen more than one example where people use an enum to model the three shapes, but I believe that this is wrong, because enums have an order to them, including a maximum and a minimum value (by default, enum is implemented with a 32-bit integer). That's not how Rock Paper Scissors work, so instead, I chose a different model where Rock, Paper, and Scissors are Singletons.

This design works for the example, although I'm not entirely happy with it. The problem is that Rock, Paper, and Scissors should be a finite set, but by making Rps abstract, another developer could, conceivably, create additional derived classes. A finite sum type would have been better, but this isn't easily done in C#. In a language with algebraic data types, you can make a prettier implementation, like this Haskell example. F# would be another good language option.

Binary operation #

When playing the game of Rock Paper Scissors, each round is a throw that compares two shapes. You can model a throw as a binary operation that returns the winning shape:

public static Rps Throw(Rps x, Rps y)
    if (x == Rock && y == Rock)
        return Rock;
    if (x == Rock && y == Paper)
        return Paper;
    if (x == Rock && y == Scissors)
        return Rock;
    if (x == Paper && y == Paper)
        return Paper;
    if (x == Paper && y == Scissors)
        return Scissors;
    if (x == Paper && y == Rock)
        return Paper;
    if (x == Scissors && y == Scissors)
        return Scissors;
    if (x == Scissors && y == Rock)
        return Rock;
    return Scissors;

To a C# programmer, perhaps the method name Throw is bewildering, because you might expect the method to throw an exception, but I chose to use the domain language of the game.

Because this method takes two Rps values as input and returns an Rps value as output, it's a binary operation. Thus, you already know it's a magma, but could it, also, be another, stricter binary operations, such as a semigroup or quasigroup?

Lack of associativity #

In order to be a semigroup, the binary operation must be associative. You can easily demonstrate that Throw isn't associative by use of a counter-example:

public void ThrowIsNotAssociative()
    // Counter-example
        Rps.Throw(Rps.Throw(Rps.Paper, Rps.Rock), Rps.Scissors),
        Rps.Throw(Rps.Paper, Rps.Throw(Rps.Rock, Rps.Scissors)));

This unit test passes, thereby demonstrating that Throw is not associative. The result of paper versus rock is paper, which, pitted against scissors yields scissors. On the other hand, paper versus the result of rock versus scissors is paper, because rock versus scissors is rock, and rock versus paper is paper.

Since Throw isn't associative, it's not a semigroup (and, by extension, not a monoid). Could it be a quasigroup?

Lack of invertibility #

A binary operation must be invertible in order to be a quasigroup. This means that for any two elements a and b, there must exist two other elements x and y that turns a into b.

This property must hold for all values involved in the binary operation - in this case Rock, Paper, and Scissors. A single counter-example is enough to show that Throw is not invertible:

public void ThrowIsNotInvertible()
    // Counter-example
    var a = Rps.Rock;
    var b = Rps.Scissors;
    Assert.False(Rps.All.Any(x => Rps.Throw(a, x) == b));
    Assert.False(Rps.All.Any(y => Rps.Throw(y, a) == b));

This (passing) unit test utilises an All property on Rps:

public static IReadOnlyCollection<Rps> All
    get { return new[] { Rock, Paper, Scissors }; }

For a counter-example, pick Rock as a and Scissors as b. There's no value in All that satisfies the invertibility property. Therefore, Throw is not a quasigroup, either.

Lack of identity #

Since Throw is neither associative nor invertible, it's not really any named algebraic construct, other than a magma. It's neither group, semigroup, quasigroup, monoid, loop, groupoid, etc. Does it have any properties at all, apart from being a binary operation?

It doesn't have identity either, which you can illustrate with another counter-example:

public void ThrowHasNoIdentity()
    // Find all identity candidates for Rock
    var rockIdentities =
        from e in Rps.All
        where Rps.Throw(e, Rps.Rock) == Rps.Rock
           && Rps.Throw(Rps.Rock, e) == Rps.Rock
        select e;
    // Narrow for Paper
    var paperIdentities =
        from e in rockIdentities
        where Rps.Throw(e, Rps.Paper) == Rps.Paper
           && Rps.Throw(Rps.Paper, e) == Rps.Paper
        select e;
    // None of those candidates are the identity for Scissors
    var scissorIdentities =
        from e in paperIdentities
        where Rps.Throw(e, Rps.Scissors) == Rps.Scissors
           && Rps.Throw(Rps.Scissors, e) == Rps.Scissors
        select e;

First, you use Rps.All to find all the values that behave as an identity element for Rps.Rock. Recall that the identity is an element that doesn't change the input. In other words it's a value that when combined with Rps.Rock in Throw still returns Rps.Rock. There are two values that fulfil that property: Rps.Rock and Rps.Scissors. Those are the two values contained in rockIdentities.

In order to be an identity, the value must behave as a neutral element for all possible values, so next, filter rockIdentities to find those elements that also behave as identities for Rps.Paper. Between Rps.Rock and Rps.Scissors, only Rps.Rock behaves like an identity for Rps.Paper, so paperIdentities is a collection containing only the single value Rps.Rock.

Is Rps.Rock an identity for Rps.Scissors, then? It's not. scissorIdentities is empty. There's no element in Rps.All that acts an identity for all values in Rps.All. Therefore, by brute force, the test ThrowHasNoIdentity demonstrates (as it says on the tin) that throw has no identity.

Commutativity #

While Throw is neither associative, invertible, nor has identity, it does have at least one property: it's commutative. This means that the order of the input values doesn't matter. In other words, for any two Rps values x and y, this assertion always passes:

    Rps.Throw(x, y),
    Rps.Throw(y, x));

Since Rps.Throw(x, y) is equal to Rps.Throw(y, x), Throw is commutative.

Summary #

The Rock Paper Scissors Throw operation is a commutative magma, but while, for example, we call an associative magma a semigroup, there's no fancy word for a commutative magma.

Next: Colour-mixing magma.


Wednesday, 27 December 2017 08:32:00 UTC

A binary operation with no constraints on its behaviour is called a magma. An introduction for object-oriented programmers.

In the overall article series about group-like algebraic structures, you've so far seen examples of monoids, semigroups, and quasigroups. Common to all of these structures is that they are binary operations governed by at least one law. The laws are different for the different categories, but there are rules.

What if you have a binary operation that follows none of those rules?

Monoids are a subset of semigroups, which are subsets of magmas. Quasigroups are also a subset of magmas, but can overlap semigroups and monoids.

All binary operations are magmas. If they have additional properties, we may call them quasigroups, or monoids, or some such, depending on the specific properties, but they're still all magmas. This is the most inclusive category.

You've already seen examples of monoids and semigroups, but what about magma examples? In a sense, you've already seen those as well, because all the examples you've seen so far have also been magma examples. After all, since all monoids are magmas, all the monoid examples you've seen have also been magma examples.

Still, it's not that hard to come up with some programming examples of magmas that aren't semi- or quasigroups. In the next articles, you'll see some examples.

Particularly the second example is fairly realistic, which demonstrates that as programmers, we can benefit from having vocabulary that enables us to describe any binary operation that doesn't obey any particular laws. In fact, establishing a vocabulary has been my primary motivation for writing this article series.

Next: Rock Paper Scissors magma


Monday, 18 December 2017 13:31:00 UTC

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 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.


Onur Sahin #

Is division over real numbers also a quasigroup? I think so, for any number a and b we have x and y such that:
a / x = b
y / a = b
I think division over integers is not a quasigroup since for 5 and 10 there is no x such that 5 / x = 10 since 0.5 is not an integer.

2022-03-27 13:45 UTC

Onur, thank you for writing. In general, when pondering pure mathematics rather than their programming aspects, I'm no authority. You'd be better off researching somewhere else. The Wikipedia article on quasigroups isn't too bad for that purpose, I find.

According to that article, division of non-zero rational or real numbers is an invertible binary operation.

As far as I can tell (I'm not a mathematician) you're correct that integer division doesn't form a quasigroup. When, as you suggestion, you pick a = 5 and b = 10, there's no integer x for which 5 / x = 10.

2022-03-29 7:00 UTC

Semigroups accumulate

Monday, 11 December 2017 08:28:00 UTC

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]

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(
        (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

Monday, 04 December 2017 08:40:00 UTC

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:

Circle with 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
            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.

Union of two shapes with bounding boxes.

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(
        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:

Bounding box with candidates 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:

Bounding box identity element 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(
        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 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.


Thank you for writing about category theory. I just have a small note. "Just pick any other BoundingBox y partially or wholly outside of x." I think that one should pick a BoundingBox y wholly outside of x. Other wise the intersection between x and y would return x or y when we pass it to x.Unite or y.Unite respectively.
2017-12-08 16:04 UTC

Yacoub, thank you for writing. The operation used here isn't the intersection, but rather the union of two bounding boxes; that's the reason I called the method Unite.

2017-12-09 12:55 UTC

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.

2017-12-09 13:05 UTC

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

Perhaps I should have written Just pick any other BoundingBox y partially or wholly outside of the candidate. Would that have been correct?

2017-12-09 13:57 UTC

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.

2017-12-09 14:15 UTC

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

2017-12-09 15:21 UTC


Monday, 27 November 2017 12:39:00 UTC

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.

Monoids are a subset of semigroups.

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)
        Math.Min(Math.Min(x, y), z),
        Math.Min(x, Math.Min(y, z)));

This example uses the FsCheck.Xunit glue library for 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)
        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)
        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<DateTimeDateTimeDateTime> dtMax =
        (dt1, dt2) => dt1 > dt2 ? dt1 : dt2;
        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)
        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)
        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.

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