## Scala type level encoding of the SKI calculus

29 01 2010

In one of my posts on type level meta programming in Scala the question of Turing completeness came up already. The question is whether Scala’s type system can be used to force the Scala compiler to carry out any calculation which a Turing machine is capable of. Various of my older posts show how Scala’s type system can be used to encode addition and multiplication on natural numbers and how to encode conditions and bounded loops.

Motivated by the blog post More Scala Typehackery which shows how to encode a version of the Lambda calculus which is limited to abstraction over a single variable in Scala’s type system I set out to further explore the topic.

### The SKI combinator calculus

Looking for a calculus which is relatively small, easily encoded in Scala’s type system and known to be Turing complete I came across the SKI combinator calculus. The SKI combinators are defined as follows:

$Ix \rightarrow x$,
$Kxy \rightarrow x$,
$Sxyz \rightarrow xz(yz)$.

They can be used to encode arbitrary calculations. For example reversal of arguments. Let $R \equiv S(K(SI))K$. Then

$R x y \equiv$
$S(K(SI))K x y \rightarrow$
$K(SI)x(Kx)y \rightarrow$
$SI(Kx)y \rightarrow$
$Iy(Kxy) \rightarrow$
$Iyx \rightarrow yx$.

Self application is used to find fixed points. Let $\beta \equiv S(K\alpha)(SII)$ for some combinator $\alpha$. Then $\beta\beta \rightarrow \alpha(\beta \beta)$. That is, $\beta\beta$ is a fixed point of $\alpha$. This can be used to achieve recursion. Let $R$ be the reversal combinator from above. Further define

$A_0 x \equiv c$ for some combinator $c$ and
$A_n x \equiv x A_{n-1}$.

That is, combinator $A_n$ is the combinator obtained by applying its argument to the combinator $A_{n-1}$. (There is a bit of cheating here: I should actually show that such combinators exist. However since the SKI calculus is Turing complete, I take this for granted.) Now let $\alpha$ be $R$ in $\beta$ from above (That is we have $\beta \equiv S(KR)(SII)$ now). Then

$\beta\beta A_0 \rightarrow c$

and by induction

$\beta\beta A_n \rightarrow \beta\beta A_{n-1} \rightarrow c$.

### Type level SKI in Scala

Encoding the SKI combinator calculus in Scala’s type system seems not too difficult at first. It turns out however that some care has to be taken regarding the order of evaluation. To guarantee that for all terms which have a normal form, that normal form is actually found, a lazy evaluation order has to be employed.

Here is a Scala type level encoding of the SKI calculus:

trait Term {
type ap[x <: Term] <: Term
type eval <: Term
}

// The S combinator
trait S extends Term {
type ap[x <: Term] = S1[x]
type eval = S
}
trait S1[x <: Term] extends Term {
type ap[y <: Term] = S2[x, y]
type eval = S1[x]
}
trait S2[x <: Term, y <: Term] extends Term {
type ap[z <: Term] = S3[x, y, z]
type eval = S2[x, y]
}
trait S3[x <: Term, y <: Term, z <: Term] extends Term {
type ap[v <: Term] = eval#ap[v]
type eval = x#ap[z]#ap[y#ap[z]]#eval
}

// The K combinator
trait K extends Term {
type ap[x <: Term] = K1[x]
type eval = K
}
trait K1[x <: Term] extends Term {
type ap[y <: Term] = K2[x, y]
type eval = K1[x]
}
trait K2[x <: Term, y <: Term] extends Term {
type ap[z <: Term] = eval#ap[z]
type eval = x#eval
}

// The I combinator
trait I extends Term {
type ap[x <: Term] = I1[x]
type eval = I
}
trait I1[x <: Term] extends Term {
type ap[y <: Term] = eval#ap[y]
type eval = x#eval
}


Further lets define some constants to act upon. These are used to test whether the calculus actually works.

trait c extends Term {
type ap[x <: Term] = c
type eval = c
}
trait d extends Term {
type ap[x <: Term] = d
type eval = d
}
trait e extends Term {
type ap[x <: Term] = e
type eval = e
}


Eventually the following definition of Equals lets us check types for equality:

case class Equals[A >: B <:B , B]()

Equals[Int, Int]     // compiles fine
Equals[String, Int] // won't compile


Now lets see whether we can evaluate some combinators.

  // Ic -> c
Equals[I#ap[c]#eval, c]

// Kcd -> c
Equals[K#ap[c]#ap[d]#eval, c]

// KKcde -> d
Equals[K#ap[K]#ap[c]#ap[d]#ap[e]#eval, d]

// SIIIc -> Ic
Equals[S#ap[I]#ap[I]#ap[I]#ap[c]#eval, c]

// SKKc -> Ic
Equals[S#ap[K]#ap[K]#ap[c]#eval, c]

// SIIKc -> KKc
Equals[S#ap[I]#ap[I]#ap[K]#ap[c]#eval, K#ap[K]#ap[c]#eval]

// SIKKc -> K(KK)c
Equals[S#ap[I]#ap[K]#ap[K]#ap[c]#eval, K#ap[K#ap[K]]#ap[c]#eval]

// SIKIc -> KIc
Equals[S#ap[I]#ap[K]#ap[I]#ap[c]#eval, K#ap[I]#ap[c]#eval]

// SKIc -> Ic
Equals[S#ap[K]#ap[I]#ap[c]#eval, c]

// R = S(K(SI))K  (reverse)
type R = S#ap[K#ap[S#ap[I]]]#ap[K]
Equals[R#ap[c]#ap[d]#eval, d#ap[c]#eval]


Finally lets check whether we can do recursion using the fixed point operator from above. First lets define $\beta$.

  // b(a) = S(Ka)(SII)
type b[a <: Term] = S#ap[K#ap[a]]#ap[S#ap[I]#ap[I]]


Further lets define some of the $A_n$s from above.

trait A0 extends Term {
type ap[x <: Term] = c
type eval = A0
}
trait A1 extends Term {
type ap[x <: Term] = x#ap[A0]#eval
type eval = A1
}
trait A2 extends Term {
type ap[x <: Term] = x#ap[A1]#eval
type eval = A2
}


Now we can do iteration on the type level using a fixed point combinator:

  // Single iteration
type NN1 = b[R]#ap[b[R]]#ap[A0]
Equals[NN1#eval, c]

// Double iteration
type NN2 = b[R]#ap[b[R]]#ap[A1]
Equals[NN2#eval, c]

// Triple iteration
type NN3 = b[R]#ap[b[R]]#ap[A2]
Equals[NN3#eval, c]


Finally lets check whether we can do ‘unbounded’ iteration.

trait An extends Term {
type ap[x <: Term] = x#ap[An]#eval
type eval = An
}
// Infinite iteration: Smashes scalac's stack
type NNn = b[R]#ap[b[R]]#ap[An]
Equals[NNn#eval, c]


Well, we can ;-)

$scalac SKI.scala Exception in thread "main" java.lang.StackOverflowError at scala.tools.nsc.symtab.Types$SubstMap.apply(Types.scala:3165)
at scala.tools.nsc.symtab.Types$SubstMap.apply(Types.scala:3136) at scala.tools.nsc.symtab.Types$TypeMap.mapOver(Types.scala:2735)


### 15 responses

29 01 2010

Well done! Very cool.

29 01 2010

Magnificent.

30 01 2010

Wow. That’s very impressive! Did you have to use the -Yrecursion switch?

30 01 2010

No, that’s without -Yrecursion.

30 01 2010

Words fail me.

1 02 2010

AFAICS some things have changed with 2.8 regarding -Yrecursion. In many situation where I needed -Yrecursion before, the code compiles fine without it now.

31 01 2010

[…] This post was mentioned on Twitter by Daniel Spiewak, Daniel Spiewak, Daniel Spiewak, Planet Scala, Bubbl Scala Feed and others. Bubbl Scala Feed said: Scala type level encoding of the SKI calculus « Michid’s Weblog http://ff.im/-f9cFI […]

1 02 2010

Cool! Didn’t know that was possible, actually :-)
Looks like syntax for anonymous type functions and/or partial type application could make this a little more pleasant… bumped them a little higher on my TODO list ;-)

2 02 2010

Great, some syntax for partial type application would definitely help!

18 06 2010

[…] Level Programming: Equality 18 06 2010 Apocalisp has a great series on Type Level Programming with Scala. At some point the question came up whether it is possible to […]

5 12 2010

[…] is cool you can prove the Scala type system is turing complete. I’m sure Monads are cool and useful thing, but I haven’t grasped them, and I know many […]

12 06 2011

[…] all types for which there is a corresponding type level acceptor function are expressible. Since type level calculations in Scala are Turing complete, it should be possible to find an acceptor for any recursive function. This means that – in […]

21 01 2013

[…] seen this before, and it absolutely blew my mind. So I asked Jorge for a link to the proof. The link he sent me is a really beautiful blog post. It doesn't just prove that Scala type inference is Turing […]

28 02 2013

[…] “Scala type level encoding of the SKI calculus” at Michid’s Weblog; and […]

26 06 2013

[…] too). In other words, subtyping is undecidable for Scala (in fact, Scala’s type system is Turing-complete, i.e. you can make Scala’s type checker perform any computation, including an infinite […]