Hackle's blog
between the abstractions we want and the abstractions we get.


Sorry to disappoint - Fin here is not French and does not mean I am done bloggine (yet) :)

Fin is a type in Idris that represents Numbers strictly less than some bound. The name comes from "finite sets". according to the source code here (you might not want to go there yet if you are new to Fin or Idris, just like me).

Let's look at its definition.

data Fin : (n : Nat) -> Type where
    FZ : Fin (S k)
    FS : Fin k -> Fin (S k)

You'll find Fin has two constructors, and FS is defined recursively in terms of FZ. FZ is kind of special, its argument must be in the form of S k, which means it can never be 0 as 0 cannot be represented as such. This is Idris' way of saying the argument to FZ must be greater than 0.

Clever, but nothing fancy, except...

The first shock

The acute observers will soon find, although this type has a parameter in (n : Nat), neither of its constructors takes any parameter. In another word, we cannot construct a value for this type! (Edit: actually we can, by specifying the value of the implicit parameter, such as FZ {k=0})

Try this in the Idris REPL:

Idris> :module Data.Fin
*Data/Fin> FZ
(input):Can't infer argument k to FZ
*Data/Fin> FS
(input):Can't infer argument k to FS
*fin> FS FZ
(input):Can't infer argument k to S, Can't infer argument k to FZ

Because there no way we can construct it, we cannot give the type parameter (yet). What is going on here? How this is useful at all?

An example

Let's try to understand it with a typical example.

import Data.Fin
import Data.Vect

indexFin : {n: Nat} -> Fin n -> Vect n a -> a
indexFin FZ (x :: xs) = x
indexFin (FS s) (x :: xs) = indexFin s xs

Given a Vect of length n, the indexFin will not take any invalid index that's equal or greater than n, which saves us the blush of index-out-of-range errors. It is called as such:

*fin> indexFin 3 [1,2,3,4]
4 : Integer
*fin> indexFin 0 [1,2,3,4]
1 : Integer
*fin> indexFin 3 [1,2,3]
(input):1:10:When checking argument prf to function Data.Fin.fromInteger:
        When using 3 as a literal for a Fin 3
                3 is not strictly less than 3

Note how Idris converts numerics to Fin values, which begs the question, what are 3, 0, 3 in Fin?

A sliver of light

Still with the REPL, let's try this:

*fin> the (Fin 4) 3
FS (FS FZ) : Fin 4

Thanks to the the keyword, we can specify that 3 is a value of type Fin 4!

Let's try break it

Notice above there is an error message When using 3 as a literal for a Fin 3, 3 is not strictly less than 3, which of course tells us that 3 cannot be Fin 3, as confirmed with this:

When using 3 as a literal for a Fin 3
        3 is not strictly less than 3

Well, if 3 has to be less than a number, sure we are not short of options?

*fin> the (Fin 4) 3
FS (FS (FS FZ)) : Fin 4
*fin> the (Fin 5) 3
FS (FS (FS FZ)) : Fin 5
*fin> the (Fin 6) 3
FS (FS (FS FZ)) : Fin 6

OMG... what is going on here? so 3 can be Fin 4, Fin 5 or Fin 6???

Well that's right! The trick is with the type parameter to Fin which we never used to construct any Fin value - it's left for interpretation.

Take the above example, try to figure out the type of FZ in each example?

take the (Fin 4) 3 as an example. The REPL interprets it as FS (FS (FS FZ)) : Fin 4, if we reverse-engineer a bit, FZ would have to be Fin 1, so FS FZ is of type Fin 2 and so on.

Now your turn to try Fin 5 and Fin 6 - and you'll find FZ is interpreted as Fin 2 and Fin 3 respectively.

Getting closer...

Finite set

Next let's count how many values can we construct for a certain Fin type.

From what limited I have learnt from Idris, is to always try the base case first. so

*Data\Fin> the (Fin 0) FZ
(input):1:1-14:When checking argument value to function Prelude.Basics.the:
        Type mismatch between
                Fin (S k) (Type of FZ)
                Fin 0 (Expected type)

                Type mismatch between
                        S k

Well, if FZ cannot work, then FS FZ won't either. It seems Fin 0 can have no values. Which makes good sense in the context of our function indexFin - an empty Vect has no value, thus cannot return anything for any index.

(Hold on - don't you want to try indexFin with an empty Vect?)

Let's try Fin 1:

*Data\Fin> the (Fin 1) FZ
FZ : Fin 1
*Data\Fin> the (Fin 1) (FS FZ)
(input):1:14-18:When checking an application of constructor Data.Fin.FS:
        Type mismatch between
                Fin (S k) (Type of FZ)
                Fin 0 (Expected type)

                Type mismatch between
                        S k

So Fin 1 can have 1 value only, which is FZ, and Fin 2, 2 values, FZ and FS FZ. Fin 3, 3. It's all making sense now isnt' it?


Fin is a clever type that by leaving its type parameter to interpretation, limit the number of values that a certain type can have. There is a good name to call these allowed values: inhabitants.

Well I didn't think it would take me so long to explain Fin! But rest assured, it took me much longer to grok the whole thing and reach my WOW moment.

Further reading

See Fin for an alternative version of the indexFin function.