The Challenge

Create an terminating expression in SKI Combinator Calculus in less than 200 combinators (S, K, I) that reduces to the expression with the most combinators.

There will be no limit on how many parenthesis/applications can be used.


SKI expressions are created using S, K, I and parenthesis. They are reduced like so:

(((Sx)y)z) => ((xz)(yz))
((Kx)y) => x
(Ix) => x

When parenthesis with more than two expressions are in an expression, they are assumed to be nested to the left.

(xyz) = ((xy)z)


The score will be the number of combinators in the output, the goal is for this to be as large as possible.


Here is an example of a reduction in SKI using the rules stated above.


So the expression (((SI)I)(((SI)I)S)) scores 4.

  • 1
    \$\begingroup\$ you should include explanations about things relevant to the challenge in the challenge rather than as an external resource. also, i can't remember if ski combinators can fall into an infinite cycle, but if so, you should probably clarify the rules about that (if not then sorry) \$\endgroup\$
    – hyper-neutrino
    Dec 29, 2021 at 19:04
  • \$\begingroup\$ @hyper-neutrino I clarified the expression needs to terminate, and explaind the reduction. \$\endgroup\$
    – nph
    Dec 29, 2021 at 19:09
  • \$\begingroup\$ @Noodle9 I added an example, hopefully that will clear some things up. If that doesn't, the wikipedia article might help. Tell me if there is anything specific that is confusing in this. \$\endgroup\$
    – nph
    Dec 29, 2021 at 21:56
  • \$\begingroup\$ Much better, retracted close vote. \$\endgroup\$
    – Noodle9
    Dec 30, 2021 at 0:13
  • 2
    \$\begingroup\$ A much better answer than mine could be drawn out of the number n in this answer. \$\endgroup\$
    – Bubbler
    Dec 30, 2021 at 3:37

1 Answer 1


44 combinators, score: \$\approx f_{\omega+1}(3\uparrow\uparrow\uparrow\uparrow3)\$

(S (S (K S) (S S K)) (K I)) arrow_diag (S (S (K S) K) (S (S (K S) K) I)) S K
where arrow_diag = S (S (S I (K (S (S (K S) (S (K (S I)) K)) (K I)))) (S (K (S I)) K)) I

This reduces to an expression of S (S (S (...(S K)...))) which contains n copies of S, where n is the value obtained by repeatedly applying arrow_diag function (a 1-input function that diagonalizes over the Knuth's up arrow notation, explained below) arrow_diag 3 times to the number 3.

Edit: Fixed the arrow_diag function, and changed 2 to 3 because I found arrow_diag 2 was just 4. Though I guess arrow_diag (arrow_diag (arrow_diag (arrow_diag 2))) would still be insanely large.

The first idea is that we will construct a Church numeral for a very large natural number n, which is \f. \x. f (f (f ... f x)) (f applied n times to x), and then force it by applying S and K to it. Then the normal form would be S (S (... S K)) which has exactly n+1 combinators.

Now, let's find a way to build large numbers. A good start point is the deceptively short power function

-- x^y
exp x y = y x
exp = \x. \y. y x

Then we can repeat the partially applied exp to build tetration and beyond (corresponding to \$x\uparrow\uparrow y\$, \$x\uparrow\uparrow\uparrow y\$ and so on):

-- x^^y = x^(x^...(x^x))
tet x y = y (x^) 1
tet = \x. \y. y (\y. y x) 1
-- x^^^y = x^^(x^^...)
quad x y = y (x^^) 1
quad = \x. \y. y (\y. y (\y. y x) 1) 1
-- in general:
arrow<N> = \x. \y. y (arrow<N-1> x) 1

Then we can try converting these into SKI. Since the next arrow notation involves a partially applied previous one, just partially converting gives helpful insights.

exp = \x. \y. y x
= \x. S I (K x)
tet = \x. \y. y (\y. y x) 1
= \x. \y. y (S I (K x)) I
= \x. S (\y. y (S I (K x))) (K I)
= \x. S (S I (K (S I (K x)))) (K I)
quad = \x. \y. y (\y. y (\y. y x) 1) 1
= \x. \y. y (S (S I (K (S I (K x)))) (K I)) I
= ...

Here we can spot a pattern. Here comes the beauty of pure lambda calculus: we can extract arbitrary subexpression into a lambda. Meanwhile, we can change 1 to y to get a higher number and save 1 combinator.

\x. \y. y (someexpr) 1
= \x. S (S I (K someexpr)) (K I)
= \x. (\e. S (S I (K e)) (K I)) someexpr

\x. \y. y (someexpr) y
= \x. S (S I (K someexpr)) I
= \x. (\e. S (S I (K e)) I) someexpr

tet' = \x. 1 (\e. S (S I (K e)) I) (S I (K x))
arrow<n> = \x. n (\e. S (S I (K e)) I) (S I (K x))

Here arrow<n> x y (roughly) calculates \$x \uparrow^{n+1} y\$.

Now we diagonalize over the arrow notation, by using x, the input value, instead of n.

arrow_diag2 = \x. x (\e. S (S I (K e)) I) (S I (K x))
= S (S I (K (S (S (K S) (S (K (S I)) K)) (K I)))) (S (K (S I)) K)

But it is still a binary function that calculates (roughly) \$x \uparrow^{x+1} y\$, so let's make it unary:

arrow_diag = \x. arrow_diag2 x x
= S arrow_diag2 I
= S (S (S I (K (S (S (K S) (S (K (S I)) K)) (K I)))) (S (K (S I)) K)) I

This has just 22 combinators. To get a non-trivial number (don't forget, we're also adding S K at the end to get a deterministic combinator count), we evaluate

arrow_diag 3 arrow_diag 3 S K
= (\x y. x y x y) arrow_diag 3 S K
= (S (S (K S) (S S K)) (K I)) arrow_diag (S (S (K S) K) (S (S (K S) K) I)) S K

which has just 44 combinators but evaluates to an... extremely large number that I don't know how to express correctly.

  • 2
    \$\begingroup\$ It sounds like your number can probably be expressed in terms of the Ackermann function, if that's any help. \$\endgroup\$
    – N. Virgo
    Dec 30, 2021 at 11:14
  • \$\begingroup\$ If arrow_diag takes in two numbers to return a number, how is it that arrow_diag 3 will return a complete number? \$\endgroup\$
    – nph
    Dec 30, 2021 at 13:01
  • 1
    \$\begingroup\$ @nph The current arrow_diag does take one number and evaluates to a number. The one that takes two numbers is the previous version (arrow_diag2 in the current explanation). \$\endgroup\$
    – Bubbler
    Dec 30, 2021 at 13:24
  • \$\begingroup\$ @Bubbler Thanks. \$\endgroup\$
    – nph
    Dec 30, 2021 at 13:45
  • \$\begingroup\$ I think this number is about f_(omega+1)(3^^^^3) in The fast-growing hierarchy \$\endgroup\$
    – nph
    Dec 30, 2021 at 13:50

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